Hrbacek K., Jech T. Introduction to Set Theory 1999

0}, where R is the set of all real numbers. View elements of P as polar coordinates of points in the plane, and definp a relation on P by: (r, 1) "' (r', 1') if and only if r = r' and r - 1.1 is an integer multiple of 211'. Show that ,. . ., is an equivalence relation on P. Show that each equivalence class contains a unique pair (r. -y) with 0 ~ 1 < 2rr. The set of all such pairs is therefore a set of representatiws for""·

33

5. ORDERINGS

5.

Orderings

Orderings are another frequently encountered type of relation. 5.1 Definition A binary relation R in A is antisymm.etric if for all a. b E A. aRb and bRa imply a = b. 5.2 Definition A binary relation R in A which is reflexive, antisymmetric. and transitive is called a (partial) ordering of A. The pair (A, R) is called an ordered set.

aRb can be read as "a is less than or equal to b" or "b is greater than or equal to a" (in the ordering R). So, every element of A is less than or equal to itself. If a is less than or equal to b, and, at the same time, b is less than or equal to a, then a = b. Finally, if a is less than or equal to b and b is less than or equal to c, a has to be less than or equal to c. 5.3 Example (a) :::; is an ordering on the set of all (natural, rational, real) numbers. (b) Define the relation ~A in A as follows: x ~A y if and only if x ~ y and x, y E A. Then ~A is an ordering of the set A. (c) Define the relation 2A in A as follows: x 2A y if and only if x 2 y and x, y E A. Then 2A is also an ordering of the set A. (d) The relation I defined by: n I m if and only if n divides m. is an ordering of the set of all positive integers. (e) The relation IdA is an ordering of A.

The symbols :::; or ~ are often used to denote orderings. A different description of orderings is sometimes convenient. For example. instead of the relation :::; between numbers, we might prefer to USf' the relation < (strictly less). Similarly, we might use CA (proper subset) instead of ~A- Any ordering can be described in either one of these two mutually interchangeable ways. 5.4 Definition A relation S in A is asymmetric if aSb implies that bSa does not hold (for any a, bE A). That is, aSb and bSa can never both be true. 5.5 Definition A relation S in A is a strict ordering if it is asymmetric and transitive.

We now establish relationships between orderings and strict orderings. 5.6 Theorem (a) Let R be an ordering of A; then the relation S defined in A by

aSb is a strict ordering of A.

if and only if aRb and a ofi b

34

CHAPTER 2. RELATIONS, FUNCTIONS, AND ORDERINGS

{b) LetS be a strict ordering of A; then the relation R defined in A by aRb

if and only if aSb or a = b

is an ordering of A. We say that the strict ordering S corresponds to the ordering R and vice versa. Proof. (a) Let us show that Sis asymmetric: Assume that both aSb and bSa hold for some a, b E A. Then also aRb and bRa, so a = b (because R is antisymmetric). That contradicts the definition of aSb. We leave the verification of transitivity of S to the reader. (b) Let us show that R is antisymmetric: Assume that aRb and bRa. Becaww aSb and bSa cannot hold simultaneously (S is asymmetric). we conclude that a = b. Reflexivity and transitivity of R are verified similarly. 0

5. 7 Definition Let a, b E A, and let S be an ordering of A. We say that a and b are comparable in the ordering S if a S b or b S a. We say that a and b are incomparable if they are not comparable (i.e., if neither a S b nor b S a holds). Both definitions can be stated equivalently in terms of the corresponding stric:t ordering <; for example, a and b are incomparable in < if a =I= b and neither a < b nor b < a holds. 5.8 (a) (b) (c) (d)

Example Any two real numbers are comparable in the ordering :<:;. 2 and 3 are incomparable in the ordering /. Any two distinct a,b E A are incomparable in IdA. If the set A has at least two elements, then there are incomparable elements in the ordered set (P(A), ~P(A))·

5.9 Definition An ordering S {or <) of A is called linear or total if any two elements of A are comparable. The pair (A,:<:;) is then called a linearly orderFd set. So the ordering S of positive integers is total, while / is not. ~ A, where A is ordered by :<:;. B is a chain in A if any two elements of B are comparable.

5.10 Definition Let B

For example, the set of all powers of 2 (i.e., {2°, 2 1 ,2 2 , 23 , ... }) is a chain in the set of all positive integers ordered by /. A problem that arises quite often is to find a least or greatest element among certain elements of an ordered set. Closer scrutiny reveals that there are several different notions of "least" and "greatest."

5. ORDERINGS

35

5.11 Definition Let S be an ordering of A, and let B <;;:A. (a) bE B is the least element of Bin the orderingS if b S x for every x E B. (b) bE B is a minimal element of B in the ordering S if there exists no x E B such that x S b and x =I b. (a') Similarly, bE B is the greatest element of B in the ordering S if. for every X E B, X b. (b') b E B is a maximal element of B in the ordering S if there exists no x E B such that b S x and x f; b.

s

5.12 Example Let N be the set of positive integers ordered by divisibility relation I· Then 1 is the least element of N, but N has no greatest element. Let B be the set of all positive integers greater (in magnitude) than 1, B = {2, 3, 4, ... }. Then B does not have a least element in I (e.g., 2 is not the least element because 2 I 3 fails), but it has (infinitely) many minimal elements: numbers 2, 3, 5, etc. (exactly all prime numbers) are minimal. B has neither greatest nor maximal elements. We list some of the properties of least and minimal elements in Theorem 5.13. The proof is left as an exercise.

5.13 Theorem Let A be ordered by s, and let B <;;: A. (a) B has at most one least element. (b) The least element of B (if it exists) is also minimal. (c) If B is a chain, then every minimal element of B is also least. The theorem remains true if the words "least" and "minimal" are replaced by "greatest" and "maximal", respectively.

5.14 Definition Let S be an ordering of A, and let B <;;: A. (a) a E A is a lower bound of Bin the ordered set (A, S) if a S x for all x E B. (b) a E A is called an infimum of B in (A, S) [or the greatest lower bound of B in (A, S)] if it is the greatest element of the set of all lower bounds of B in (A, s). Similarly, (a') a E A is an upper bound of B in the ordered set (A, S) if x S a for all X E B. (b') a E A is called a supremum of B in (A, s) [or the least upper bound of B in (A, s)] if it is the least element of the set of all upper bounds of B in (A,s).

Note that the difference between the least element of B and a lower bound of B is that the second notion does not require b E B. A set can have many lower bounds. But the set of all lower bounds of B can have at most one greatest element, soB can have at most one infimum. We now summarize some properties of suprema and infima.

36

CHAPTER 2. RELATIONS, FUNCTIONS, AND ORDERINGS

5.15 Theorem Let (A,:::;) be an ordered set and let B ~A. (a) B has at most one infimum. (b) If b is the least element of B, then b is the infimum of B. (c) If b is the infimum of B and b E B, then b is the least element of B. {d) bE A is an infimum of B in (A, S) if and only if

{i) b :::; x for all x E B. {ii) If b':::; x for all x E B, then b' S b. The theorem remains true if the words "least" and "infimum" are replau•d by the words "greatest" and "supremum" and "S" is replaced by ":;:,·· in (i) and (ii).

Proof. (a) We proved this in the remark preceding the theorem. (b) The least element b of B is certainly a lower bound of B. If b' is any lowPr bound of B, b' :::; b because b E B. So b is the greatest element of the sPt of all lower bounds of B. (c) This is obvious. (d) This is only a reformulation of the definition of infimum. D We use notations inf(B) and sup( B) for the infimum of Band the supremum of B, if they exist. If B is linearly ordered, we also use min( B) and max( B) to denote the minimal {least) and the maximal (greatest) elements of B, if they exist.

5.16 Example Let ::; be the usual ordering of the set of real numbers; let B 1 = {x I 0 < x < 1}, B2 = {x I 0 ::; x < 1}, B 3 = {x I x > 0}, and B 4 ={xI x < 0}. Then B1 has no least element and no greatest element, but any b ::; 0 is a lower bound of B 1 , so 0 is the greatest lower bound of B 1 ; i.e .. 0 = inf(B). Similarly, any b :;:, 1 is an upper bound of B 1 , so 1 = sup(B 1 ). The set B 2 has a least element; so 0 == min(B2) == inf(B 2 ); it does not have a greatest element. Nevertheless, sup(B2) = 1. The set B3 has neither a greatest Plement nor a supremum (actually B 3 has no upper bound in <::: ): of cours<'. inf(B 3 ) = 0. Similarly, B 4 has no lower bounds, hence no infimum. 5.17 Definition An isomorphism between two ordered sets ( P. <) and (Q. -<) is a one-to-one function h with domain P and range Q such that for all p 1 . P:l E P PI < P2

if and only if

h(pt) -< h(p2).

If an isomorphism exists between (P, <)and (Q, -<),then (P. <) and (Q, -<)are isomorphic. We study isomorphisms in a more general setting in Chapter 3. At this point, let us make the following observation.

5. ORDERINGS

37

5.18 Lemma Let (P, <) and (Q, -<) be linearly ordered sets, and let h be a oneto-one function with domain P and range Q such that h(p1) -< h(p 2 ) whenever PI < P2· Then h is an isomorphism between (P. <) and (Q, -<). Proof. We have to verify that if Pl,P2 E Pare such that h{pJ) -< h(p2). then PI < P2· But if PI is not less than p2, then, because < is a linear ordering of P, either PI= P2 or P2
Exercises

5.1 (a) Let R be an ordering of A, S be the corresponding strict ordering of A, and R* be the ordering corresponding to S. Show that R* = R. (b) Let 8 be a strict ordering of A, R be the corresponding ordering, and 8* be the strict ordering corresponding to R. Then 8* = S. 5.2 State the definitions of incomparable elements. maximal, minimal, greatest, and least elements and suprema and infima in terms of strict orderings. 5.3 Let R t)e an ordering of A. Prove that R- 1 is also an ordering of A, and forB~ A, (a) a is the least element of B in R- 1 if and only if a is the greatest element of B in R; (b) similarly for (minimal and maximal) and (supremum and infimum). 5.4 Let R be an ordering of A and let B ~ A. Show that R n B 2 is an ordering of B. 5.5 Give examples of a finite ordered set (A. :5) and a subset B of A so that (a) B has no greatest element. (b) B has no least element. (c) B has no greatest element, but B has a supremum. (d) B has no supremum. 5. 6 (a) Let (A, <) be a strictly ordered set and b f/: A. Define a relation -< in B =AU {b} as follows:

x-< y

if and only if

(x, yEA and x < y) or (x

E

A andy= b).

Show that -
if and only if

x, y E A 1 and x < 1 y

or

x, y E A2 and x <2 y

or

x E A 1 andy E A2.

CHAPTER 2. RELATIONS, FUNCTIONS. AND ORDERINGS

38

Show that -< is a strict ordering of Band -< nAi =<1. -< nA~ =< 2 . (Intuitively, -< puts every element of A1 before every element of A 2 and coincides with the original orderings of A 1 and A:d 5.7 Let R be a reflexive and transitive relation in A (R is called a preorde.ring of A). Define E in A by

aEb

if and only if aRb and bRa.

Show that E is an equivalence relation on .4. Define the relation R/ E in A/E by [a]e R/ E [b]e if and only if aRb. Show that the definition does not depend on the choice of representatives for [a]e and [b]e. Prove that R/ E is an ordering of A/ E. 5.8 Let A = P(X), X # 0. Prove: (a) Any 5 ~A has a supremum in the ordering ~A; supS= U 5. (b) Any S ~A has an infimum in ~A; inf S = n5 if 5 # 0; inf0 =X. 5.9 Let Fn(X, Y) be the set of all functions mapping a subset of X into Y [i.e., Fn(X, Y) = Uzc;x yz]. Define a relation 5 in Fn(X. Y) by

f 5 g if and only if f

~

g.

(a) Prove that 5 is an ordering of Fn(X. Y). (b) Let F ~ Fn(X, Y). Show that sup F exists if and only if F is a compatible system of functions; then sup F = U F. 5.10 Let A# 0; let Pt(A) be the set of all partitions of A. Define a relation ~in Pt(A) by 51

5.11

5.12 5.13 5.14

~

52 if and only if for every C E 51 there is D E 52 such that C

~

D.

(We say that the partition 51 is a refinement of the partition 52 if 51 ~ S2 holds.) (a) Show that ~ is an ordering. {b) Let 5 1 ,52 E Pt(A). Show that {51 ,52 } has an infimum. [Hint: Define 5 = {C n DIcE 51 and DE 52}.] How is the equivalence relation Es related to the equivalences Es, and Es 2 ? (c) LetT~ Pt(A). Show that infT exists. (d) LetT~ Pt(A). Show that supT exists. [Hint: LetT' be the set of all partitions 5 with the property that every partition from T is a refinement of 5. Show that supT' = infT.] Show that if (P, <) and (Q, -<) are isomorphic strictly ordered sets and < is a linear ordering, then -< is a linear ordering. The identity function on Pis an isomorphism between (P. <)and (P. <). If his isomorphism between (P, <)and (Q, -<),then h- 1 is an isomorphism between (Q. -<) and ( P, <). If f is an isomorphism between ( P1 ,
Chapter 3

Nat ural Numbers 1.

Introduction to Natural Numbers

In order to develop mathematics within the framework of the axiomatic set theory, it is necessary to define natural numbers. We all know natural numbers intuitively: 0, 1, 2, 3, ... , 17, ... , 324, etc., and we can easily give examples of sets having zero, one, two, or three elements: 0 has 0 elements. {0} or, in general, {a} for any a, has one element. {0,{0}}, or {{{0}},{{{0}}}}, or, in general, {a,b} where a f; b, has two elements, etc. The purpose of the investigations in this section is to supplement this intuitive understanding by a rigorous definition. To define number 0, we choose a representative of all sets having no elements. But this is easy, since there is only one such set. We define 0 = 0. Let us proceed to sets having one element (singletons): {0}, {{0}}, { {0, {0}} }; in general, {x}. How should we choose a representative? Since we already defined one particular object, namely 0, a natural choice is {0}. So we define

1 = {0}

= {0}.

Next we consider sets with two elements: {0, {0}}, { {0}, {0, {0}} }, { {0}, { {0}} }, etc. By now, we have defined 0 and 1, and 0 :/; l. We single out a particular two-element set, the set whose elements are the previously defined numbers 0 and 1: 2 = {0, 1} = {0,{0}}. It should begin to be obvious how the process continues: 3

= {0, 1, 2} = {0, {0}, {0, {0}}}

4

= {0,1,2,3} = {0,{0},{0,{0}},{0,{0},{0,{0}}}}

5 = {0, 1,2,3,4} etc. 39

40

CHAPTER 3. NATURAL NUMBERS

The idea is simply to define a natural number n as the set of all smaller natural numbers: {0, 1, ... , n - 1 }. In this way, n is a particular set of n elements. This idea still has a fundamental deficiency. We have defined 0, 1, 2, 3, 4, and 5 and could easily define 17 and-not so easily-324. But no list of such definitions tells us what a natural number is in generaL We need a statement of the form: A set n is a natural number if .... We cannot just say that a set n is a natural number if its elements are all the smaller natural numbers, because such a "definition" would involve the very concept being defined. Let us observe the construction of the first few numbers again. We defined 2 = { 0, 1}. To get 3, we had to adjoin a third element to 2, namely, 2 itself: 3 = 2U {2} == {0, 1} U {2}. Similarly, 4=3u{3} = {0,1,2}u{3}, 5 = 4U {4}, etc. Given a natural number n, we get the "next" number by adjoining one mon"' element ton, namely, n itself. The procedure works even for 1 and 2: 1 = OU{O}. 2 = 1 U {1}, but, of course. not for 0, the least natural number. These considerations suggest the following.

1.1 Definition The successor of a set xis the set S(x) =xU {x}. Intuitively, the successor S(n) of a natural number n is the "one bigger"" number n+ 1. We use the more suggestive notation n+ 1 for S(n) in what follows. We later define addition of natural numbers (using the notion of successor) in such a way that n + 1 indeed equals the sum of n and 1. Until then, it is just a notation, and no properties of addition are assumed or implied by it. We can now summarize the intuitive understanding of natural numbers a.<; follows: (a) 0 is a natural number. {b) If n is a natural number, then its successor n + 1 is also a natural numlwr. (c) All natural numbers are obtained by application of (a) and (b). i.e .. by starting with 0 and repeatedly applying the successor operation: 0. 0 + 1 = 1. 1 + 1 = 2, 2 + 1 = 3. 3 + 1 = 4, 4 + 1 = 5, ... etc.

1.2 Definition A set I is called inductive if (a) 0 E I. (b) If n E I, then ( n + 1) E I. An inductive set contains 0 and, with each element, also its successor. According to (c), an inductive set should contain all natural numbers. The precise meaning of (c) is that the set of natural numbers is an inductive set which contains no other elements but natural numbers, i.e .. it. is the smallest inductivP set. This leads to the following definition.

41

1. INTRODUCTION TO NATURAL NUMBERS 1.3 Definition The set of all natural numbers is the set

N = { x I x E I for every inductive set I}. The elements of N are called natural numbers. Thus a set x is a natural number if and only if it belongs to every inductive set. We have to justify the existence of N on the basis of the Axiom of Comprehension {see the remarks following Example 3.12 in Chapter 1). but that is easy. Let A be any particular inductive set; then clearly

N = {x E A I x E I for every inductive set I}. The only remaining question is whether there are any inductive sets at all. The intuitive answer is, of course, yes: the set of natural numbers is a prime example. But a careful look at the axioms we have adopted so far indicates that existence of infinite sets (such as N) cannot be proved from them. Roughly, the reason is that these axioms have a general form: "For every set X, there exists a set Y such that ... ," where, if the set X is finite, the set Y is also finite. Since the only set whose existence we postulated outright is 0, which is finite, all the other sets whose existence is required by the axioms are also finite. (See Section 2 of Chapter 4 for a more rigorous elaboration of these remarks.) The point is that we need another axiom. The Axiom of Infinity.

An inductive set exists.

Some mathematicians object to the Axiom of Infinity on the grounds that a collection of objects produced by an infinite process (such as N) should not be treated as a completed entity. However, most people with some mathematical training have no difficulty visualizing the collection of natural numbers in that way. Infinite sets are basic tools of modern mathematics and the essence of set theory. No contradiction resulting from their use has ever been discovered in spite of the enormous body of research founded on them. Therefore, we treat the Axiom of Infinity on a par with our other axioms. We now have the set of natural numbers N at our disposal. Before proceeding further, let us check that the set N is indeed inductive.

1.4 Lemma N is inductive. If I is any inductive set, then N

~

I.

Proof. 0 E N because 0 E I for any inductive I. If n E N, then n E I for any inductive I, so (n + 1) E I for any inductive I, and consequently (n + 1) E N. This shows that N is inductive. The second part of the Lemma follows immediately from the definition of N. 0 The next step is to define the ordering of natural numbers by size. As it is our guiding idea to define each natural number as a set of smaller natural numbers, we are led to the following.

42

CHAPTER 3. NATURAL NUMBERS

1.5 Definition The relation< on N is defined by: m < n if and only if mE n. It is of course necessary to prove that < is indeed a linear ordering and that the ordered set (N, <)actually has the properties that we expect natural numbers to have. The needed theory is developed in the rest of this chapter. Exercises

1.1 x

2.

~

S(x) and there is no z such that x C z C S(x).

Properties of Natural Numbers

In the preceding section we defined the set N of natural numbers to be the least set such that (a) 0 E N and (b) if n E N, then (n + 1) E N. We also defined m < n to mean m E n. It is the goal of this section to show that thP above-mentioned concepts really behave in the familiar ways. We begin with a fundamental tool for study of natural numbers. the wellknown principle of proof by mathematical induction. Let P(x) be a property (possibly with parameters). Assume that (a) P(O) holds. (b) For all n EN, P(n) implies P(n + 1). Then P holds for all natural numbers n. The Induction Principle.

Proof. This is an immediate consequence of our definition of N. Tlw assumptions (a) and (b) simply say that the set A = { n E N \ P ( n)} is 0 inductive. N ~ A follows. The following lemma establishes two simple properties of natural numbers and gives an example of a simple proof by induction.

2.1 Lemma (i) 0 S n for all n E N. (ii) For all k, n E N, k < n + 1 if and only if k < n or k = n.

(i) We let P(x) to be the property "0 S x" and proceed to establish Proof. the assumptions of the Induction Principle. (a) P(O) holds. P(O) is the statement "0 S 0," which is certainly true {0 = 0). (b) P(n) implies P(n + 1). Let us assume that P(n) holds, i.e., 0 :::; n. This means, by definition of<, that 0 =nor 0 En. In either case, 0 E nu{n} = n + 1, so 0 < (n + 1) and P(n + 1) holds. Having proved (a) and (b) we use the Induction Principle to conclude that P(n) holds for all n EN, i.e., 0:::; n for all n EN. (ii) This part does not require induction. It suffices to observe that k E n U {n} if and only if k E n or k = n. 0

2. PROPERTIES OF NATURAL NUMBERS

43

The proof of the next theorem provides several other, somewhat more complicated examples of inductive proofs.

2.2 Theorem (N, <) is a linearly ordered set.

Proof. (i) The relation < is tronsitive on N. We have to prove that for all k, m, n E N, k < m and m < n imply k < n. We proceed by induction on n; i.e., we use as P(x) the property "for all k, mE N, if k < m and m < x, then k < x." (a) P(O) holds. P(O) asserts: for all k, m E N, if k < m and m < 0, then k < 0. By Lemma 2.1{i), there is nomE N such that m < 0, so P(O) is trivially true. (b) Assume P(n), i.e., assume that for all k, mE N, if k < m and m < n, then k < n. We have to prove P(n + 1), i.e., we have to show that k < m and m < (n + 1) imply k < (n + 1). But if k < m and m < (n + 1), then by Lemma 2.1(ii) m
44

CHAPTER 3. NATURAL NUMBERS

a parameter; that is, we are going to apply the Induction Principle to tlw property P(x): "If n < x, then n + 1 ~ x"]. If m = 0, the statement ·'if n < 0, then n+ 1 ~ 0" is true (since its assumption must be false). AssumP P(m), i.e., if n < m, then (n + 1) ~ m. To prove P(m + 1), assume that n < m + 1; then n < m or n = m. If n < m, n + 1 ~ m by the inductive assumption, and son+ 1 < m + 1. If n == m then of course n + 1 == m + 1. In either case P(m + 1) is proved. We conclude that P(m) holds for all mEN, as needed. Finally, we finish the proof of (iii) by observing that thP assumptions (a) and (b) of the Induction Principle have now been established. 0 The reader should study the preceding proof carefully for its varied applications of the Induction Principle. In particular, the proof of part (iii) is an example of "double induction": In order to prove a statement depending on two variables, m and n, we proceed by induction on one of them (n); thP proof of the induction assumption {b) then in itself requires induction on the othPr variable m (for fixed n). See Exercise 2.13. Before proceeding further, we state and prove another version of the Induction Principle that is often more convenient.

The Induction Principle, Second Version. Let P(x) be a property (pos:;iblv with parameters). Assume that. f(H all n EN, (*)

If P(k) holds for all k < n, then P(n).

Then P holds for all natural numbers n. In other words, in order to prove P(n) for all n E N, it sufficps to prow P(n) (for all n EN) under the assumption that it holds for all smaller natural numbers. Proof. Assume that (*) is true. Consider the property Q(n): P(k) holds for all k < n. Clearly Q(O) is true (there are no k < 0). If Q(n) holds. tlwn Q(n + 1) holds: If Q(n) holds, then P{k) holds for all k < n, and consequently also fork== n [by(*)]. Lemma 2.1(ii) enables us to conclude that P(k) holds for all k < n + 1, and therefore Q(n + 1) holds. By the Induction Principle. Q(n) is true for all n EN; since fork EN there is some n > k (e.g .. n == k + 1 ). we have P(k) true for all k EN, as desired. []

The ordering of natural numbers by size has an additional important property that distinguishes it from, say, ordering of integers or rational numbers by size. 2.3 Definition A linear ordering -< of a set A is a well-orderin_q if Pwry nonempty subset of A has a least element. The ordered set (A.-<) is th<'n called a well-ordered set.

2. PROPERTIES OF NATURAL NUMBERS

45

Well-ordered sets form a backbone of set theory and we study them extensively in Chapter 6. At this point, we prove only: 2.4 Theorem (N, <) is a well-ordered set. Proof. Let X be a nonempty subset of N; we have to show that X ha.<> a leaEt element. So let us aEsume that X does not have a least element and let us consider N - X. The crucial step is to observe that if k E N - X for all k < n, then n E N - X: otherwise, n would be the lf:'ast element of X. By tlw second version of the Induction Principle we conclude that n E N- X holds for all natural numbers n [P(x) is the property "x E N - X"] and therefor~:' that X = 0, contradicting our initial aEsumption. 0

We conclude this section with another property of the ordering <. 2.5 Theorem If a nonempty set of natural numbers has an upper bound in thr ordering <, then it has a greatest element. Let A~ N, A# 0 be given; let B = {k E N!k is an upper bound Proof. of A}; we aEsume that B # 0. By Theorem 2.4, B haE a }eaEt element n, so n = sup{A). The proof is completed by showing that n E A. [See Theorem 5.15(c) in Chapter 2. with "supremum" in place of "infimum," etc.] Trivial induction proves that either n = 0 or n = k + l for some k E N (see Exercise 2.4). Assume that n 1. A: we then have n > m for all m E A. Since A # 0, it means that n I= 0. TlH•refor<'. n = k + 1 for some k E N, which gives k :2 m for all m E A [Lemma 2.l(ii) again!]. Thus k is an upper bound of A and k < n, a contradiction. 0 Exercises

2.1 Let n EN. Prove that there is no kEN such that n < k < n + 1. 2.2 Use Exercise 2.1 to prove for all m, n E N: if m < n, then m + 1 S:: n. Conclude that m < n implies m + 1 < n + 1 and that therefore the successor S(n) = n + 1 defines a one-to-one function on N. 2.3 Prove that there is a one-to-one mapping of N onto a proper subset of N. [Hint: Use Exercise 2.2.] 2.4 For every n E N, n # 0, there is a unique k E N such that n = k + 1. 2.5 For every n E N, n f= 0, 1, there is a unique k E N such that n = {k+l)+l. 2.6 Prove that each natural number is the set of all smaller natural numbf:'rs. i.e., n ={mEN I m < n}.

[Hint: Use induction to prove that all elements of a natural number are natural numbers.] 2.7 For all m,n EN m
if and only if m c n.

CHAPTER 3. NATURAL NUMBERS

46

f : N --+ N such that for all n E N, f(n) > f(n + 1). {There is no infinite decreasing sequence of natural numbers.) If X c:;; N, then (X,< n X 2 ) is well-ordered. In Exercise 5.6 of Chapter 2, let A= N, b = N. Prove that-< as defined there is a well-ordering of B = N U {N}. Notice that x-< y if and only if x E y holds for all x, y E B. Let P(x) be a property. Assume that kEN and (a) P(k) holds. (b) For all n ~ k, if P(n) then P(n + 1). Then P(n) holds for all n ~ k. (Finite Induction Principle) Let P{x) be a property. Assume that k E N and (a) P(O). (b) For all n < k, P(n) implies P(n + 1). Then P(n) holds for all n $ k. {Double Induction) Let P(x.y) be a property. Assume

2.8 Prove that there is no function

2.9 2.10

2.11

2.12

2.13

(**) If P(k, l) holds for all k, l E N such that k < m or (k l < n), then P{m, n) holds.

m and

Conclude that P(m, n) holds for all m, n EN.

3.

The Recursion Theorem

Our next task is to show how to define addition, multiplication, and other familiar operations of arithmetic. To facilitate this, we develop an important general method for defining functions on N. We begin with some new terminology. A sequence is a function whose domain is either a natural number or N. A sequence whose domain is some natural number n E N is called a finite sequence of length n and is denoted (a,li
or

(a,li=O,l, ... ,n-1)

or

(ao,at,····an-1)·

In particular, ()( = 0) is the unique sequence of length 0, the empty sequence. Seq( A) = UnEN An denotes the set of all finite sequences of elements of A (Prove that it exists!) If the domain of a sequence is N, we call it an infinitt~ sequence and denote it (a;liEN)

or

(a,li=0,1,2 ... )

or

(a;)~o·

So infinite sequences of elements of A are just members of AN. The notation simply specifies a function with an appropriate domain, whose value at i is a,. We also use the notation {a, I i E N}, {a,}~ 0 , etc., for the range of the sequence (a, I i EN). Similarly. {a, /t < n} or {ao,aJ, ... ,an-d df:'notes thf' range of (a, I i < n). Let us now consider two examples of infinite sequences.

47

3. THE RECURSION THEOREM (a) The sequences: N ....... N is defined by

so == l; Sn+l

== n 2

for all n E N.

(b) The sequence f : N ....... N is defined by fo == 1; fn+l == fn

X

(n + 1)

for all n EN.

The two definitions, in spite of superficial similarity, exhibit a crucial difference. The definition of s gives explicit instructions how to compute sx for any x E N. More precisely, it enables us to formulate a property P such that Sx

== y if and only if P(x, y);

namely, let P be "either x == 0 and y == 1 or, for some n E N, x = n + 1 andy= n 2 ." The existence and uniqueness of a sequences satisfying (a) then immediately follows from our axioms:

s = {(x,y) EN x N

I P(x,y)}.

In contraEt, the instructions supplied by the definition of f tell us only how to compute fx provided that the value off for some smaller number (namely, x - 1) waE already computed. It is not immediately obvious how to formulate a property P, not involving the function f being defined, such that fx

=y

if and only if P(x, y).

We might view the definition (b) as giving conditions the sequence f ought to satisfy: "f is a function on N to N which satisfies the 'initial condition': fo = 1, and the 'recursive condition': for all n EN, fn+l = fn x (n + 1)." Such definitions are widely used in mathematics; the reader might wish to draw a parallel, e.g., with the implicit definitions of functions in calculus. However, a definition of this kind is justified only if it is possible to show that there exists some function satisfying the required conditions, and that then• do not exist two or more such functions. In calculus, this is provided for by the Implicit Function Theorem. We now state and prove an analogous result for our situation. For any set A, any a E A, and any function g A x N -. A, there exists a unique infinite sequence f : N ....... A such that {a) fo =a; (b) fn+l == g(fn, n) for all n EN.

The Recursion Theorem

In Example {b), we had A= N, a= 1, and g(u,v) = u x (v + 1). The set a is the "initial value" of f. The role of g is to provide instructions for computing fn+l assuming fn haE been already computed.

48

CHAPTER 3. NATURAL NUMBERS

The proof of the Recursion Theorem consists of devising an explicit definition of f. Consider again Example (b); then fn is the n-factorial; and au explicit definition of f can be readily written down: fo == 1

and

fm == 1 x 2 x · · · x {m- 1) x m if m

f=

0 and mE N.

The problem is in making "· · ·" precise. It can be resolved by stating tlwt f,, is the result of a computation

1X1 !1xl]x2 [1

X

1

[1

X

1 X 2 X···

X

2]

X

3

X

(m- l)j

X

m.

A computation is a finite sequence starting with the "initial value'' of f and repeatedly applying g. In the example above, them-step computation t is a finitP sequence that is of length m + 1 where to= 1 and tk+l = t1.: x (k + l) = y(t~.:. k·) for all k < m, k :::=: 0. The rigorous explicit definition off thf:'n is: fm = tm

where t is an m-step computation {baEed on a= 1 and g).

The problem of the existence and uniqueness off is reduced to the problem of showing that there is precisely one rn-step computation for each m E N. We now proceed with the formal proof of the Recursion Theon•m. As this theorem and its generalizations are among the most important methods of s<'t theory, the reader should study the preceding intuitive example and tlw pruuf itself carefully.

Proof The existence of f. A function t: (rn + 1) ->A is called an m-stPp computation based on a and g if t 0 = a, and, for all k such that 0 -:; k· < 111. t/.:+ 1 = g(t~.:,k). Notice that t C:: N x A. Let

F = {t Let

f

=

E

P(N x A)

It

is an m-step eornputation for some natural number m}.

UF.

3.1 Claim f is a function. It suffices to show that the system of functions F is compatible-sef~ Theorem 3.12 in Chapter 2. So lett, u E F, domt = n EN, dom 11 =mE N. Assutm•. e.g., n -:; m; then n C:: m, and it suffices to show that t~.: == UJ.: for all k < tL This can be done by induction (in the form stated in Exercise 2.12). Snrdy. t 0 = a = u 0 . Next let k be such that k + 1 < n, and assume f~.: = tt~,:. Thren t~,:+ 1 = g(tk.k) = g(u~.:,k) = Uk+l· Thus tk = uk for all k < n.

49

3. THE RECURSION THEOREM

3.2 Claim domf = N; ranf

~A.

We know immediately that dom f ~ N and ran f ~ A. To show that dom f = N, it suffices to prove that for each n E N there is an n-step computation t. We use the Induction Principle. Clearly, t = { {0, a)} is a 0-step computation. Assume that t is an n-step computation. Then the following function t+ on (n + 1) + 1 is an (n +I)-step computation: tt = tk

if k ::::; n:

t;:-+l = g(tn, n).

We conclude that each n EN is in the domain of some computation t E F, soN<;; UtEFdomt = domf.

3.3 Claim f satisfies conditions (a) and (b). Clearly, fo =a since to= a for all t E F. To show that fn+l = g(!n, n) for any n E N, let t be an (n + 1)-step computation; then tk = fk for all k E dom t. so fn+l = tn+l = g(tn, n) = g(fn, n). The existence of a function f with properties rf:'quired by the Recursion Theorem follows from Claims 3.1, 3.2, and 3.3. The uniqueness of f. Let h: N -> A be such that (a') ho = a; (b') hn+l = g(hn, n) for all n EN. We show that fn = hn for all n E N, again using induction. Certainly fo = a= ho. If fn = hn, then fn+l = g(fn,n) = g(hn,n) = hn+l; therefore, f = h, as claimed. 0 As a typical example of the use of the Recursion Theorem, we prove that the propertif:'s of the ordering of natural numbers by size established in thf:' prl:'vious section uniquely characterize the ordered set (N. <). Theorem Let (A,-<) be a nonempty linearly ordered set with the properties: For every p E A, there is q E A such that q ~ p. Every nonempty subset of A has a -<-least dement. Every nonempty subset of A that has an upper bound has a -<-greatest element. Then (A,-<) is isomorphic to (N, <).

3.4 (a) (b) (c)

Proof. We construct the isomorphism f using the Recursion TheorPm. Let a be the least element of A and let g(x, n) be the least element of A grf:'ater than x {for any n). Then a E A and g is a function on A x N into A: notice that g(x. n) is defined for any x E A because of assumptions (a) and (b) in Theorem 3.4, and does not depend on n. The Recursion Theorem guarantees the existence of a function f : N ....... A such that

CHAPTER 3. NATURAL NUMBERS

50

{i) fo = a= the leaEt element of A. {ii) fn+l = g(fn,n) =the leaEt element of A greater than f,. It is obvious that fn -< fn+I for each n E N. By induction, we get J, -< fm whenever n < m (see Exercise 3.1). Consequently, f is a one-to-one function. It remains to show that the range of f is A. If not, A - ran f :f: 0; let p be the leaEt element of A - ran f. The set B = {q E A I q -< p} haE an upper bound p, and is nonempty (otherwise. p would be the least element of A, but then p = f 0 ). Let q be the greatest element of B [it exists by assumption (c) in Theorem 3.4]. Since q-< p, we haw q = fm for some m E N. However, it is now eaEily seen that p is the least element of A greater than q. Therefore, p = fm+ 1 by the recursive condition {ii). Consequently, p E ran f, a contradiction. 0 In some recursive definitions, the value of fn+ 1 depends not only 011 f,. but also on /k for other k < n. A typical example is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, .... Here fo = 1, h = 1, and fn+I = fn + fn-I for n > 0. The following theorem formalizes this apparently more general recursive construction. 3.5 Theorem For any set S and any function g : Seq(S) unique sequence f : N -> S such that

fn

= g(f f n)

=

g((fo, ... .fn-t))

Notice that, in particular, fo Fibonacci sequence, we let

g(t) =

1 { tn-1

=

g(f f 0)

=

->

S there exists a

for all n EN. g( ())

= g(0).

To obtain tlw

if t is a finite sequence of length 0 or 1:

+ tn-2

if t is a finite sequence of length n > 1.

Proof. The idea is to use the Recursion Theorem to define the sequence (Fn In EN)= (f f nInE N). So let us define Fo = (); Fn+I = Fn U {(n,g{Fn))}

for all n EN.

The existence of the sequence (Fn I n E N) follows from the Recursion Theorem with A = Seq(S), a= (), and G : A x N -> A defined by

G(t, n) =

t u { (n, g(t )) } if t is a sequence of length { () otherwise.

n;

It is eaEy to prove by induction that each Fn belongs to sn and that F, ~ Fn+l for all n EN. Therefore (set:' Exercise 3.1), {Fn In EN} is a compatible

3. THE RECURSION THEOREM

51

system of functions. Let f = UnEN Fn; then clearly f: N-> Sand f f n = Fn for all n EN; we conclude that fn = Fn+l(n) = g(Fn) = g(f f n), as needed. 0 Further examples on the use of this version of the Recursion Theorem can be found in Chapter 4. At present, we state yet another, "parametric." vf:'rsion that allows us to use recursion to define functions of two variables. 3.6 Theorem Let a : P

-> A and g : P x A x N -> A be functions. There exists a unique function f : P x N -> A such that {a) f(p, 0) = a(p) for all p E P; {b) f(p, n + 1) = g(p,J(p, n), n) for all n E N and p E P.

The reader may prefer the notation fp,O in place of f(p, 0), etc. Proof. This is just a "parametric" version of the proof of the Recursion Theorem. Define an m-step computation to be a function t: P x {m + 1)-> A such that, for all p E P,

t(p, 0)

= a(p)

and

t(p, k + 1) = g(p, t(p, k), k)

for all k such that 0 :::; k < m. Then follow the steps in the proof of the Recursion Theorem, always carrying p along. Alternatively, one can deduce the parametric version directly from the Recursion Theorem (see Exercise 3.4). 0

We now have all the machinery needed to define addition of natural numbers. as well as other arithmetic operations. We do so in the next Section. Exercises

3.1 Let f be an infinite sequence of elements of A, where A is ordered by -<. Assume that fn -< fn+l for all n E N. Prove that n < m implies fn -< fm for all n, m E N. [Hint: Use induction on m in the form of Exercise 2.11, with k = n+ 1.] 3.2 Let (A,-<) be a linearly ordered set and p,q E A. We say that q is a successor of p if p -< q and there is no r E A such that p -< r -< q. Note that each p E A can have at most one successor. Assume that (A,-<) is nonempty and has the following properties: {a) Every p E A has a successor. {b) Every nonempty subset of A has a -<-least element. (c) If p E A is not the -<-least element of A, then p is a successor of some q EA. Prove that (A,-<) is isomorphic to (N, <). Show that the conclusion need not hold if one of the conditions (a)-{ c) is omitted. 3.3 Give a direct proof of Theorem 3.5 in a way analogous to the proof of the Recursion Theorem.

CHAPTER 3. NATURAL NUMBERS

52

3.4 Derive the "parametric" version of the Recursion Theorem (Theorem 3.6) from the Recursion Theorem. [Hint: Define F: N __.. AP by rt>rursion:

Fo =a

E

AP;

Fn+I = G(Fn, n)

g(p. :r(p). n) for where G : AP x N -+ AP is defined by G(x, n)(p) x E AP, n EN. Then set f(p,n) = Fn(p).j 3 .vr. Prove the following version of the Recursion Theorem: Let g be a function on a subset of A x N into A. a E A. Then there is a unique sequence f of elements of A such that (a) fo =a; (b) fn+l = g(fn,n) for all n EN such that (n + 1) E domf: (c) f is either an iufinitP sPquence or f is a finitP s~·quPncc• of lPngt h k + 1 and g(fk, k) is undefined. [Hint: Let A= Au {a} wlwre a rf. A. Define g: A X N -+A as follows:

_(

)

g x,n =

{g(x, n)

a

if defined; otherwise.

Use the Recursion Theorem to get the corresponding infinitf:' sequence]. If 71 = U. for some I E N. consider 7 f I for tlw least such I.J 3.6 Provf:': If X<:;; N, then thew is a one-to-orw (finite or infinit<') seqUPIHP f such that ranf =X. [Hillt: Use Exercise 3.5.J

4.

Arithmetic of Natural Numbers

As an application of the Recursion Theorem we now show how to define addition of natural numbers, and we use Induction to prove basic properties of addition. In similar fashion, one can introdun~ other arithmPtic opPrations (seP Exercis<'s). 4.1 Theorem There is a unique function + : N x N (a) +(m.O) = m for all mEN: (b) +(m,n + 1) = +(m,n) + 1 fm· all m,n EN.

Proof.

N. ct(p)

=p

-+

N such that

In the parametric version of the Recursion TlworPm h•t A = I' '· '·-' for all p E N and g(p . .r. n) = :!' + 1 for all J!, :r. n E N.

Notice that letting n = 0 in (b) lead:> to +(m,O+ 1) = +(m.O) + 1: but +(m,O) = m by (a) and 0 + 1 = S(O) = 1 by the definition of the numlH'r 1 Thus we have+ (m, 1) = m + 1 = S(m) and, as we claimed parlier. the successor of m E N is really the sum of m and 1, justifying our notation for it. Of coms••. we write m + n instead of +(m, n) in the 1;equel. The defining properties of addition can now bP rest.ated a.s

53

4. ARITHMETIC OF NATURAL NUMBERS

+0 =

4.2

m

m.

4.3

m + (n + l) = (m + n) + 1.

Properties of recursively defined functions are typically est.abli:;lwd by induction. As an example, we prove the commutative law of addition. 4.4 Theorem Addition is commutative; i.e., for all m,n E N,

(4.5)

m

+n

= n

+ m.

Proof. Let us say that n commutes if {4.5) holds for all mE N. \\'e prow that every n E N commutes, by induction on n. To show that 0 commutes, it suffices to show that 0 + m = rn for all m [because then 0 + m = m + 0 by 4.2]. Clearly, 0 + 0 = 0, and if 0 + m = m. then 0 + (m + 1) = {0 + m) + 1 = m + 1, so the claim follows by induction (on m). Let. us now assume that n commutes, and let. us show that. n + 1 commutes. We prove, by induction on m, that (4.6)

m + (n + 1) = (n + 1) + m

for all TilE N.

If m = 0, then (4.6) holds, as we have already shown. Thus let us assunw that {4.6) holds form, and let us prove that

(4.7)

(m + 1) + (n + 1) = (n + 1) + (m

+ 1).

We derive (4. 7) as follows: (m + 1) + (n +I)= ((m + 1) + n) + 1

[by 4.3]

= (n + (m + 1)) + 1

[since n commutes]

={(n+m)+1)+1

[by 4.3]

=

+ n) + 1) + 1 (m + (n + 1)) + 1 ((m

[since n commutes]

=((n+1)+m)+1

[by 4.3] [since ( 4.6) holds for m]

=(n+1)+(m+1)

[by 4.3].

=

0 Detailed development of the arithmetic of natural numbers is beyond the scope of this book. Readers who are so inclined may continue in it by working the exercises at the end of this section, and thus convince themselws that axiomatic set theory has the power to establish rigorously all of the usual laws of arithmetic. From there one can proceed to define such notions as divisibility and prime numbers and prove fundamental results of elementary number t.lwor~·. such as the existence and uniqueness of a decomposition of each natural numlwr into a product of prime numbers.

54

CHAPTER 3. NATURAL NUMBERS

Having established the natural numbers and their arithmetic. one can proceed to define rigorously, first, the set Z of integers, then the set Q of rational numbers, and derive their corresponding arithmetic laws. These constructions are well known from algebra, and most readers probably encountered tlwm there. For completeness, we describe them in some detail in Section l of Chapter 10, which can be studied at this point. In any case, we view the arithmPtic of integers and rationals as rigorously established from now on. We conclude this section with a remark about axiomatic arithmetic of natural numbers. The theory of arithmetic of natural numbers can be developed axiomatically. The accepted system of axioms is due to Giuseppe Peano. Tlw undefined notions of Peano arithmetic are the constant 0, the unary operation S, and the binary operations + and ·. The axioms of Peano arithmetic are thP following: {Pl) If S(n) = S(m), then n = m. {P2) S(n) -I 0. {P3) n + 0 = n. (P4) n + S(m) = S(n + m). {P5) n · 0 = 0. (P6) n · S(m) = (n · m) + n. (P7) If n "I 0, then n = S(k) for some k. {P8) The Induction Schema. Let A be an arithmetical property (i.e., a property expressible in terms of+. ·. S, 0). If 0 has the property A and if A(k) implies A(S(k)) for every k, then every number has the property A. It is not difficult to verify that natural numbers and arithmetic operations, as we defined them, satisfy all of the Peano axioms (Exercist' 4.8). Most of the needed facts have been proved already. Exercises

4.1 Prove the associative law of addition: (k

+ m) + n

= k

+ (m + n) for all k, m. n EN.

4.2 If m, n, kEN, then m < n if and only if m + k < n + k. 4.3 If m, n E N then m :S n if and only if there exists k E N such that n = m + k. This k is unique. so we can denote it n- m. the differencr of n and m. This is how subtraction of natural numbers is defined. 4.4 There is a unique function · (multiplication) from N x N toN such that

m · 0 = 0 for all m E N; m · (n + 1) = m · n + m for all m, n E N. 4.5 Prove that multiplication is commutative, associative, and distributive over addition. 4.6 If m, n, k E N and k > 0, then m < n if and only if m · k < n · k. 4.7 Define exponentiation of natural numbers as follows: m0 = 1

for all mE N (in particular, 0°

mn+l = mn · m

=

1);

for all rn. n E N (in particular, on

=

0 for n > 0).

5. OPERATIONS AND STRUCTURES

55

Prove the usual laws of exponents. 4.8 Verify the axioms of Peano arithmetic. The needed results can be found in the text and exercises. 4.9 For each finite sequence (ki I 0 :S i < n) of natural numbers define 1 l:(ki I 0 :S i < n) (more usually denoted Eo::;i
l::o =0; L(ko) L(ko, ... , kn)

5.

=

ko;

= L(ko, ...

, kn-1) + kn

for n 2: 1.

Operations and Structures

The functions +, ·, etc., are usually referred to as operations. The aim of this section is to define the general concept of operation and to study its properties. Each of the aforementioned operations assigns to a pair of objects (numbers. sets) a third object of the same kind (their sum, difference, union, etc.). The order may make a difference, e.g., 7- 2 and 2- 7 are two different objects. We therefore submit the following definition. 5.1 Definition A binary operation on S is a function mapping a subset of S 2 into S. Nonletter symbols such as +, x, *• D., etc., are often used to denote operations. The value (result) of the operation* at (x, y) is then denoted x * y rather than *(x,y). There are also operations, such as square root or derivative, which are applied to one object rather than to a pair of objects. We introduce the following definitions. 5.2 Definition A unary operation on S is a function mapping a subset of S into S. A ternary operation on Sis a function on a subset of S 3 into S.

5.3 Definition Let f be a binary operation on Sand A~ S. A is closed under the operation f if for every x, y E A such that f(x, y) is defined, f(x, y) E A. One can give similar definitions in the case of unary or ternary operations.

5.4 Example (a) Let + be the operation of addition on the set of all real numbers. Then + is defined for all real numbers a and b. The set of all real numbers, as well as the set of all rational numbers and the set of all integers, are closed under +. The set of even natural numbers is closed under +, but the set of odd natural numbers is not.

CHAPTER 3. NATURAL NUMBERS

56

(b) Let 7 be the operation of division on the set of all real numbers. + is not defined for (a, b) where b = 0. The set of all rational numbers is closPd under 7, but the set of all integers is not. (c) Let S be a set; define binary operations Us and ns on 8 as follows; (i) If X, y E Sand XU yES. then X Us y =XU y. (ii) If x, y E S and x n y E S, then x ns y = :r n y. If we take S = P(A) for some A, Us and ns are defined for every pair (x, y) E 8 2 . When developing a particular branch of mathematics, we are usually interested in sets of objects together with some relations between them and operations on them. For example, in number theory or analysis, we study the set of all (natural or real) numbers together with operations of addition and multiplication, relation of less than, etc. In geometry, we study the set of all points and lines with relations of incidence and between and operation of intersection. de. We introduce the notion of structure to describe this situation abstractly. In general, a structure consists of a set A and of several relations and opNations on A. For example, we consider structures with two binary relations and two operations, say a unary operation and a binary operation. Let A be a set. let R 1 and R2 be binary relations in A, and let f be a unary operation and g a binary operation on A. We make use of the five-tuple (A. R 1 • R 2 • f. g) to rknot<" the structure thus defined.

5.5 Example (a) Every ordered set is a structure with one binary relation. (b) (A, UA, nA, ~A) is a structure with two binary operations and one binarv relation. (c) Let R be the set of all real numbers. (R, +. -. x. 7) is a structur<' with four binary operations. We now proceed to give a general definition of a structure. In Chapter 2. we introduced unary (1-ary), binary (2-ary), and ternary (3-ary) relations. The reason we did not talk about n-ary relations for arbitrary n is simplv that we could not then handle arbitrary natural numbers. This obstacle is now remowd. and the present section gives precise definitions of n-tuples, n-ary relations and operations, and n-fold cartesian products, for all natural numbers n. We start with the definition of an ordered n-tuple. Recall that orderPd pair (a 0 , a!) has been defined in Section 1 of Chapter 2 as a set that uniquely determines its two coordinates, a 0 and a 1 : i.e., (a 0 , a I)= (b 0 , bJ)

if and only if

ao

= bo

and a 1

= b1 .

We called a0 the first coordinate of (a 0 ,a1), and a 1 its second coordinatP. In analogy. ann-tuple (ao, a1, .... a,_ 1 ) should be a set that uniqm•ly detPrmirws its n coordinates, a 0 , a 1 , . . . • an-I; that is. we want (*) (ao ....

,an-d= (bo, .... bn-d if and only if a,= b, for all t =

0 ..... n -- l.

57

5. OPERATIONS AND STRUCTURES

But we already introduced a notion that satisfies (*): it is the sequence of length n, (a 0 , a 1 , ... , an-J). The statement that (ao, ... , an-t)

= (bo, ...

, bn- I) if and only if a,

= b,

for all i

= 0, ...

,n - 1

is just a reformulation of equality of functions, as in Lemma 3.2 in Chapter 2. We therefore define n-tuples as sequences of length n. For each i, 0 :S i < n, a; is called the (i + 1)st coordinate of (ao,aJ, ... ,an- 1). So ao is thP first coordinate, a 1 is the second coordinate, ... , an-I is the n-th coordinate. [Really, a more consistent terminology would be zeroth, first, ... , (n - 1)st coordinate. but the custom is to talk about the first and second coordinates of points in thP plane, etc.] We notice that the only 0-tuple is the empty sequence () = 0. having no coordinates. 1-tuples are sequences of the form (a 0 ) [i.e., sets of the form { (0, u 0 )} ]: it usually causes no confusion if one does not distinguish between a 1-tuple (a 0 ) and an element ao. If (A, I 0 :S i < n) is a finite sequence of sets, then the n-fold cartesian product f1o
rr

A, I for some ao, ... ,an-I. a= {ao, ... 'an-r)and P(ao, .... a,_J) }.

o::•
CHAPTER 3. NATURAL NUMBERS

58

cartesian product, Ao X At and no<•<2 A,, two definitions of binary relations and operations, etc. However, there is a canonical one-to-one correspondence between ordered pairs and 2-tuples that preserves first and second coordinates. Define 8((ao,at)) = (ao,at); then 8 is a one-to-one mapping on A 0 x A 1 onto no<•<2 A; and X is a first (second, respectively) coordinate of (ao,al) if and only if x is a first (second, respectively) coordinate of (a 0 , a 1 ). For almost all practical purposes, it makes no difference which definition one uses. Therefore. we do not distinguish between ordered pairs and 2-tuples from now on. Similar remarks hold about the relationships between triples and 3-tuples, etc. In fact. it is possible to define n-tuples for any natural number n along the lines of Chapter 2. The idea is to proceed recursively and let (ao) {ao , a 1 ,

•. . ,

= ao;

an) = ( ( ao, at, . . . , an- J), an)

for all n E N, n 2 1.

However, making this precise involves certain technical problems. We do not yet have a Recursion Theorem sufficiently general to cover this type of definition (but see Exercise 4.2 in Chapter 6). As we have no need for an alternative definition of n-tuples, we refer the interested reader to Exercise 5.17. We ust' notations (a 0 , ... , an-!) and (ao, ... , an-d interchangeably from now on. We continue this section by generalizing the notion of a structure. First. a type T is an ordered pair {(ro, .... 1'm-I), Uo, ... , f n-I)) of finite sequences of natural numbers, where r, > 0 for all i -s; m- 1, i 2 0. A structure of type T is a triple 21 = (A, (Ro, ... , Rm-1), (Fo, ... , Fn-1)) where R.; is an ri-ary relation on A for each i 'S: m - 1 and Fj is an frary operation on A for each j "S: n - 1; in addition, we require F1 'I f/J if f 1 = 0 Note that if f 1 = 0, F1 is a 0-ary operation on A. Following our earlier remarks, Fi is a constant, i.e., just a distinguished element of A. For example, lJ1 = (N. ( <), {0, +, ·)) is a structure of type ( (2), (0, 2, 2) ); it carries one binary relation. one constant, and two binary operations. !R = (R, (), (0, 1. +. -. x.-;-)) is a structure of type ((),(0,0,2,2,2,2)), etc. We often present the structure as a (1 + m + n)-tuple, for example (N, <, 0, +, ·), when it is understood which symbols represent relations and which operations. We call A the universe of the structure 21. We now introduce a notion that is of crucial importance in the theory of structures. (In the special case of ordered sets, compare with Definition 5.17 iu Chapter 2.) 5.6 Definition An isomorphism between structures 21 and 21' =(A'. (RG ..... R:r,_ 1 ), (F0, ... , F~_t)) of the same type T is a one-to-one mapping h on A onto A' such that (a) Ri(ao, ... ,ar,-difandonlyifR;{h{ao), ... ,h(ar,-d)holdsforallao ..... ar, _ 1 E A and i :S m - 1. (b) h(FJ(ao, ... ,af,-d) = F;{f~(ao), ... ,h{at,-d)holdsforallao .... . af,-1 E A and all j -s; n- 1 provided that either side is defined.

5. OPERATIONS AND STRUCTURES

59

For example, let (A,R 1,R2,f,g) and (A',R;,R2,J',g') be structures with two binary relations and one unary and one binary operation. Then h is an isomorphism of (A,R 1 ,R2,f,g) and (A',R~,R2,f',g') if all of the following requirements hold: (a) h is a one-to-one function on A onto A'(b) For all a,b E A, aR1b if and only if h(a)R'1 h(b). (c) For all a, bE A, aR2b if and only if h(a)R2h(b). (d) For all a E A, f(a) is defined if and only if f'(h(a)) is defined and h(f(a)) = f'(h(a)). (e) For all a, bE A, g(a, b) is defined if and only if g'(h(a), h(b)) is defined and h(g(a, b))= g'(h(a), h(b)). The structures are called isomorphic if there is an isomorphism between them. 5. 7 Example Let A be the set of all real numbers, -s:A be the usual ordering of real numbers, and + be the operation of addition on A. Let A' be the set of all positive real numbers, -s:A' be the usual ordering of positive real numbers, and x be the operation of multiplication on A'. We show that the structures (A, -s:A, +) and (A', SA'• x) are isomorphic. To that purpose, let h be the function

h(x) =ex

for x E A.

We prove that his an isomorphism of (A, SA,+) and (A', SA', x). We have to prove: (a) his a one-to-one function on A onto A'. But dearly his a function, dam h = A, and ran h = A'. If x 1 "I x 2 , then ex' "I ex 2 , so h is one-to-one. (In this example, we assume knowledge of some facts from elementary calculus.) (b) Let x 1 ,x2 E A; then x1 SA x2 if and only if h(xt) SA' h(x2). But the function ex is increasing, so x1 S x2 if and only if ex' -s: ex 2 is indeed true. (c) Let x 1 , x 2 E A; then x 1 +x2 is defined if and only if h(xi) x h(x2) is defined and h(x 1 + x2) = h(xJ) x h(x2). First, notice that both + on A and x on A' are defined for all ordered pairs. Now, h(x 1 +x 2) = ex' +x 2 = ex' x er 2 = h(xt) x h(x2). D The significance of establishing an isomorphism between two structures lies in the fact that the isomorphic structures have exactly the same properties as far as the relations and operations on the structures are concerned. Therefore. if the properties of these relations and operations are our only interest, it does not make any difference which of the isomorphic structures we study. 5.8 Example Let (A, R) and (B, S) be isomorphic (R and S are binary relations). R is an ordering of A if and only if S is an ordering of B. A has a least element in R if and only if B has a least element in S.

Proof Let h be an isomorphism of (A, R) and (B, S). Assume that R is an ordering of A. We prove that Sis an ordering of B. Hence, let b1, b2, b3 E B and b1Sb 2, b2Sb3. Since his onto B, there exist a1,a2,a3 E A such that b1 = h(at),

60

CHAPTER 3. NATURAL NUMBERS

b2 = h(a2), and b2 = h(a3). Because a 1 Ra2 holds if and only if h(at)Sh(a 2 ) holds, i.e., if and only if b1Sb2 holds, we conclude that a 1 Ra 2 and similarly a2Ra3. But R is transitive in A, so a1Ra3. But then h(at)Sh(a 3 ), i.e., b1 Sb: 1 . We leave the proof of reflexivity and antisymmetry to the reader. Similarly. one proves that if S is an ordering of B, then R is an ordering of A. Now let .4 have a least element. We claim that B has a least element. Let a be the lPast element of A, i.e, aRx holds for all x E A. Then h(a) is the least element of B. as the following consideration shows. If y E B. then y = h(J.:) for sonw :r E A (h is onto B). But, for this x, aRx holds. Correspondingly, h(a)Sh(x) holds: and we see that h(a)Sy holds for all y E B. Thus h(a) is the least elPment of B. D

An isomorphism between a structure 21 and itself is called an automorphism of 21. The identity mapping on the universe of 21 is trivially an automorphism of 21. The reader can easily verify that the structure (N, <) has no other automorphisms. On the other hand, the structure (Z, <), where Z is the set of all integers, has nontrivial automorphisrns. In fact, they arP precisely tht• functions J,., hE Z, where J,.(x) = x +h. Some further properties of automorphisms are listed in Exercise 5.12. Consider a fixed structure 21 = (A.(Ro, ... ,Rm-d-(Fo .... . Fn_t)). A sPt B ~ A is called closed if the result of applying any operation to elements of B is again in B, i.e., if for all j :S n -1 and all ao, . .. , a f, _ 1 E B, FJ (a 0 , . . . . a/, -1) E B provided that it is defined. In particular, all constants of 21 belong to B. Lr•t C ~ A; the closure of C, to be denoted C, is the least closed set containing all elements of C:

C

=

n{B <;; A I C

~ B and B is closed}.

Notice that A is a closed set containing C, so the system whose iut.erse<"tion defines Cis nonempty. It is trivial to check that C is dosed; by dPfinition. tiH'H. C is indeed the least closed set containing C.

5.9 Example (a) Every set B <;;;;A is closed if 21 has no operations. (b) Let R be the set of all real numbers and let C = {0}. The set of all natural numbers N is the closure of C in the structure (R, f) where f is the successor function defined by f(x) = x + 1 for all real numbers x. (c) Let C = {0, 1 }; the set of all integers Z is the closure of C in the struc:tm<' (R. +.-)or in (R. +. -, x). (d) The set of all rationals Q is the closure of f/J in (R. 0, 1, +,-, x, + ). The notion of closure is important in algebra, logic, and other areas of rnathpmatics. The next theorem shows how to construct the closure of a set '"from below."

5. OPERATIONS AND STRUCTURES

61

5.10 Theorem Let 21 = {A,(Ro, ...• Rm-I),(Fo, ... ,Fn- 1 )) be a structure and let C ~ A. If the sequence (C; I i E N) is defined recursively by

Co=C; Ci+l

then

= C; U Fo[C{"] U · · · U Fn-dC["-'],

C = U~o C;.

Of course, the notation AoU· · ·UAn-1 is a shorthand for Uo<•
belongs to F3 [C{j ~ C1+1 ~ 6, so 6 is closed. It remains to prove the opposite inclusion 6 ~ C. Clearly, C 0 = C ~ C. If C; ~ C, then FJ [Cf'] ~ C for each j S: n- 1 because C is closed, and therefore also Ct+ 1 ~C. We conclude using the Induction PrinciplP that C, ~ C for all i E N and, finally, 6 = U:::o C; ~ C, as required. 0 We close this section with a theorem that is often used to prove that all elements of a closure have some property P. It can be viewed as a generalization of the Induction Principle, which is its special case for the structure (N, S) (S is the successor operation) and C = {0}. 5.11 Theorem Let P(x) be a property. Assume that (a) P(a) holds for all a E C. {b) For each j S n- 1, ifP(a 0 ), .•. , P(aJ,-d hold and F1 (a 0 , . . . • aJ,-d defined, then P(FJ(ao, ... ,at,-d) holds. Then P(x) holds for all x E C.

IS

Proof. The assumptions (a) and (b) postulate that the set B = {:r E A I P(x)} is closed and C ~B. 0 Exercises

5.1 Which of the following sets are closed under operations of addition. subtraction, multiplication, and division of real numbers? (a) The set of all positive integers. (b) The set of all integers.

62

CHAPTER 3. NATURAL NUMBERS

(c) The set of all rational numbers. (d) The set of all negative rational numbers. (e) The empty set. 5.2 Let * be a binary operation on A. (a) * is called commutative if, for all a, bE A, whenever a* b is defined. then b *a is also defined and a* b = b *a. (b) * is called associative if, for all a, b, c E A, (a* b)* c = a* (b *c) holds whenever the expression on one side of the equality sign is defined (the other expression then must also be defined). Which of the operations mentioned in this section are commutative or associative? 5.3 Let * and /:::, be operations in A. We say that * is distributive over 6 if a* (b 6 c) = (a* b) 6 (a* c) for all a, b, c E A for which the expression on one (and then also on the other) side is defined. For example, multiplication of real numbers is distributive over addition. but addition is not distributive over multiplication. 5.4 Let A -I 0, B = P(A). Show that (B, Ua. na) and (B, na, Ua) arP isomorphic structures. [Hint: Set h(x) = B- x; notice that Ua in the first structure corresponds to na in the second structure and vice versa.) 5.5 Refer to Example 5.7 for notation. (a) There is a real number a E A such that a+ a= a (namely. a = 0). Prove from this that there is a' E A' such that a' x a' = a'. Find this a'. (b) For every a E A there is bE A such that a+ b = 0. Show that for every a' E A' there is b' E A' such that a' x b' = 1. Find this b'. 5.6 Let z+ and be, respectively, the sets of all positive and negatiw integers. Show that (Z+. <.+)is isomorphic to (Z-. >.+)(where< is the usual ordering of integers). 5.7 Let R be a set all elements of which are n-tuples. Prove that R is
z-

63

5. OPERATIONS AND STRUCTURES

5.12

5.13

5.14

5.15

(ao, ... ,am-1)*(bo, ... ,bn-1) ==(co, ... ,Cm+n-1) wherec, =a, for 0::; i < m and c; == b;-m form::; i ::; m + n- 1; conv((ao, ... , am-d)== (bo, ... , bm-1) where b, = am-t-1 for 0::; i:Sm-1. Prove that for those a, b, c E B where the appropriate operations are defined length(a *b)= length(a) + Iength(b); length(tail(a)) = length(a)- 1; a =head( a)* tail( a); head(a *b)= head(a); tail(a *b) = tail(a) * b; a * (b * c) == (a * b) * c; conv(conv(a)) ==a; conv(a) == conv(tail(a)) *head( a). Let 21, IE, and
FR((a1, a2), (bt. b2)) ==(at. b2) if a2 = b1 (and is undefined otherwise). Show that the closure of R in (A 2 , FR) is a transitive relation. Show that if R is reflexive and symmetric, then its closure is an equivalence relation. 5.16 Let B = Seq(A) where A= {a,b,c, ... ,x,y,z} is the set of all letters in the English alphabet. Refer to Exercise 5.11. Identify sequences of length 1 with elements of A, so that A <::;; B. Now let F be a binary operation defined as follows: if x, y E B andy E A (i.e., it has length 1), then F(x, Y) = (F is undefined otherwise).

y*x *y

Let C be the closure of A in the structure (B, F). Prove that c = conv(c)

64

CHAPTER 3. NATURAL NUMBERS

for all c E C. {The elements of C are called palindromes.) [Hint: UsP Theorem 5.11.] 5.17 Define n-tuples so that (a) {ao) = ao. (b) (ao,at,··· ,an)= ((ao .... ,an-d, an) for all n 2 1. [Hint: Say that a is an n-tuple if there exist finite sequences f and (ao, ... , an-i) of length n such that {i) fo = ao. (ii) f,+l =(f.. a,+!) for all i < n -1, i 2 0. (iii) a= fn-1· Show that for each n-tuple a there is a unique pair of finite sequences f. (a 0 , ... , an-!) of length n such that (i), (ii), and (iii) hold; we then writP a= (a 0 , ... ,an-d· Show that for every finite sequence (ao ..... an-t) of length n there is a unique n-tuple a = ( ao, ... , a,._ 1). Then prow (a) and {b).]

Chapter 4

Finite, Countable, and Uncountable Sets 1.

Cardinality of Sets

From the point of view of pure set theory, the most basic question about a set is: How many elements does it have? It is a fundamental observation that we can define the statement "sets A and B have the same number of elements" without knowing anything about numbers. To see how that is done, consider the problem of determining whether the set of all patrons of some theater performance has the same number of elements as the set of all seats. To find the answer, the ushers need not count the patrons or the seats. It is enough if they check that each patron sits in one, and only one, seat, and each seat is occupied by one, and only one, theatergoer. 1.1 Definition Sets A and B are equipotent (have the same cardinality) if there is a one-to-one function f with domain A and range B. We dl'note this by IAI = IBI. 1.2 Example

(a) {0,{0}} and {{{0}},{{{0}}}} are equipotent; let !(0) = {{0}}. f({0}) =

{{{0}}}. {b) {0} and {0, {0}} are not equipotent. (c) The set of all positive real numbers is equipotent with the set of all negative real numbers; set f(x) = -x for all positive real numbers x. 1.3 Theorem

(a) A is equipotent to A. (b) If A is equipotent to B, then B is equipotent to A. (c) If A is equipotent to B and B is equipotent to C, then A is equipotent to C. 65

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

66

Proof (a) IdA is a one-to-one mapping of A onto A. (b) If f is a one-to-one mapping of A onto B, f- 1 is a one- to-one mapping of B onto A. (c) Iff is a one-to-one mapping of A onto Band g is a one-to-one mapping of B onto C, then go f is a one-to-one mapping of A onto C.

c Similarly, the following definition is very intuitive. 1.4 Definition The cardinality of A is less than or equal to the cardinality of

B {notation: IAI -::; IBI) if there is a one-to-one mapping of A into B. Notice that IAI -::; IBI means that IAI = ICI for some subset C of B. Wt> also write lA! < IBI to mean that IAI -::; IBI and not IAI = IBI, i.e., that there is a one-to-one mapping of A onto a subset of B, but there is no one-to-one mapping of A onto B. Notice that this is not the same thing as saying that there exists a one-to-one mapping of A onto a proper subset of B; for example. there exists a one-to-one mapping of the set N onto its proper subset (ExercisP 2.3 in Chapter 3), while of course INI = INI. Theorem 1.3 shows that the property IAI = IBI behaves like an equivalence relation: it is reflexive, symmetric, and transitive. We show next that the property IAI ~ IBI behaves like an ordering on the "equivalence classes" under equipotence. 1.5 Lemma

(a) If IAI -::; IBI and IAI == ICI, then ICI -::; IBI. (b) If IAI ~ IBI and IBI == ICI, then IAI S ICI. (c) IAI -::; IAI. (d) If IAI-::; IBI and IBI-::; ICI, then IAI ~ ICI.

Proof

Exercise 1.1.

0

We see that -::; is reflexive and transitive. It remains to establish antisymmetry. Unlike the other properties, this is a major theorem. 1.6 Cantor-Bernstein Theorem

If lXI -::; IYI and IYI S lXI, then lXI =

IYI·

Proof If IX I -::; IYI, then there is a one-to-one function f that maps X into Y; if IYI S lXI, then there is a one-to-one function g that maps Y into X To show that lXI = IYI we have to exhibit a one-to-one function which maps X onto Y. Let us apply first f and then g; the function g o f maps X into X and is one-to-one. Clearly, g[f[X]] ~ g[Y] ~X; moreover, since f and g are one-toone, we have lXI = lg[f[X]]I and IYI = lg[Y]I. Thus the theorem follows from the next Lemma 1.7. (Let A= X, B = g[Y], A 1 = g[f[X]].) [J

1. CARDINALITY OF SETS

67

1.7 Lemma If A1 <;; B <;;A and JA1l = JAJ, then

IBI

= JAJ.

The proof can be followed with the help of this diagram:

Proof. Let f be a one-to-one mapping of A onto A 1 . By recursion, we define two sequences of sets:

and (In the diagram, then An's are the squares and the Bn 's are the disks.) Let Ao=A, Bo=B, and for each n,

(*)

An+!= f[An],

Bn+! = f[Bn]·

Since Ao 2 Bo 2 A1, it follows from (*),by induction, that for each n, An :::J An+ 1· We let, for each n, and

00

C =

UCn,

D = A -C.

n=D

(Cis the shaded part of the diagram.) By(*), we have f[Cn]

= Cn+!,

00

f[C]

=

UCn. n=!

Now we are ready to define a one-to-one mapping g of A onto B:

g(x) = {f(x) X

~f x E C; !f XED.

so

CHAPTER 4. FINITE, COUNTABLE. AND UNCOUNTABLE SETS

68

Both g [ C and g [ D are one-to-one functions, and their ranges are disjoint.. Thus g is a one-to-one function and maps A onto f[CJ U D = B. 0 We now see that ~ has all the attributes of an ordering. A natural question to ask is whether it is linear, i.e., whether (e) JAI ~ JBJ or JBJ ~ JAI holds for all A and B. It is known that the proof of {e) requires the Axiom of Choice. We return to this subject in Chapter 8; in the meantime, we refrain from using (e) in any proofs. By now, we have established the basic properties of cardinalities, without actually defining what cardinalities are. In principle, it is possible to continut· the study of the properties JAI = JBJ and JAI ~ JBI in this vein. without Pwr defining JAI; one can view JAI = IBI simply as shorthand for the property '·A is Pquipotent to B." etc. However, it is both conceptually and notationally useful to define JAI, "the number of elements of the set A." as an actual object of sd theory, i.e., a set. We therefore make the following assumption.

1.8 Assumption There are sets called cardinal numbers (or cardinals) with the property that for every set X there is a unique cardinal lXI (the cardinal number· of X, the cardinality of X) and sets X and Y are t:>quipotent if and only if lXI is equal to IYJ. In effect, we are assuming existence of a unique "representative·· for !'ach class of mutually equipotent sets. The Assumption is harmless in the sensp that we only use it for convenience, and could formulate and prove all our theorems without it. The Assumption can actually be proved with the help of the Axiom of Choice, and we do that in Chapter 8. Moreover, for certain classes of sets. e!lfdinal numbers can be defined, and the Assumption proved, even without tlw Axiom of Choice. The most important case is that of finite sets. We study finite sets and their cardinal numbers in detail in the next section. Exercises

1.1 Prove Lemma 1.5. 1.2 Prove: (a) If JAJ < JBJ and JBI ~ (b) If IAI ~ JBI and JBI < 1.3 If A~ B, then JAI ~ jBJ. 1.4 Prove: (a) JA x Bl = jB x AI.

1.5 1.6

JCJ, JCJ,

then then

JAJ < JCI. JAI < JCJ.

(b) j(A X B) X Cj = jA X (B X C)j. (c) JAI ~lAx Bl if B I 0. Show that lSI~ IP(S)I. [Hint: lSI= I{ {a} I a E S}j.] Show that JAI ~ JA 5 j for any A and any S I 0. [Hint:

Consider constant functions.] 1.7 If S ~ T, then JA 5 j ~ IATJ; in particular, JAn I ~ JAm I if n ~ m. [Hint: Consider functions that have a fixed constant value on T- S.]

2. FINITE SETS

69

1.8 JTJ ~ JSTJ if JSJ ~ 2. [Hint: Pick u, v E S, u ~ v, and, for each t E T, consider ft : T ~ S such that ft(t) = u, j 1 (x) = v otherwise.] 1.9 If JAJ ~ JBJ and if A is nonempty then there exists a mapping f of B onto A. It is somewhat peculiar that the proof of a fundamental, general result. like the Cantor-Bernstein Theorem, should require the use of such specific sets as natural numbers. In fact, it does not. The following sequence of exercises leads to an alternative proof, and in the process, acquaints the reader with the fixed-point property of monotone functions, an important result in itself. Let F be a function on P(A) into P(A). A set X ~ A is called a fixed point of F if F(X) = X. The function F is called monotone if X ~ Y ~ A implies F(X) ~ F(Y). 1.10 Let F: P(A) --> P(A) be monotone. Then F has a fixed point. [Hint: LetT= {X~ A I F(X) ~X}. Notice that T ~ 0. Let X= nT and prove that X E T, F(X) E T. Therefore, F(X) c X is impossible.] 1.11 Use Exercise 1.10 to give an alternative proof of the Cantor-Bernstein Theorem. [Hint: Prove Lemma 1.7 as follows: let F: P(A) -+ P(A) lw defined by F(X) = (A- B) U f[X]. Show that F is monotone. Let C be a fixed point ofF, i.e., C = (A- B) U f[C], and let D = A - C. Define gas in the original proof and show that it is one-to-one and onto B.] 1.12 Prove that X in Exercise 1.10 is the least fixed point ofF, i.e., if F(X) = X for some X~ A, then X~ X. The remaining exercises show that the two proofs are not that different, after all. A function F : P(A) --> P(A) is continuous if F(U;EN X,) = U,EN F(X,) holds for any nondecreasing sequence of subsets of A. ((X, J i E N) is nondecreasing if X,~ X 1 holds whenever i ~ j.) 1.13 Prove that F used in Exercise 1.11 is continuous. [Hint: Set> Exercise 3.12 in Chapter 2.] 1.14 Prove that if X is the least fixed point of a monotone continuous function F : P(A) --> P(A), then X = UiEN X, where we define recursively

Xo

=

0, X,+l

=

F(X;).

The reader should compare the proof of Lemma 1. 7 with the construction of the least fixed point for F(X) = (A- B) U f[X] in Exercise 1.14.

2.

Finite Sets

Finite sets can be defined as those sets whose size is a natural number.

2.1 Definition A set S is finite if it is equipotent to some natural number n E N. We then define JSJ =nand say that S has n elements. A set is infinite if it is not finite.

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

70

By our definition, cardinal numbers of finite sets are the natural numbers. Obviously, natural numbers are themselves finite sets, and )nl = n for all n E N. However, we have to verify that the cardinal number of a finite set is unique. This follows from Lemma 2.2 2.2 Lemma Ifn EN, then there is no one-to-one mapping ofn onto a proper subset XC n.

Proof By induction on n. For n = 0, the assertion is trivially trut>. Assuming that it is true for n, let us prove it for n + 1. If the assertion is falst> for n + 1, then there is a one-to-one mapping f of n + 1 onto some X c n + 1. There are two possible cases: Either n E X or n ¢ X. If n ¢ X. then X <;: 11. and f r n maps n onto a proper subset X - {f (n)} of n, a contradiction. If n E X, then n = f (k) for some k ~ n. We consider the function g on n defined as follows: for all i i= k, i < n; if i = k < n. The function g is one-to-one and maps n onto X - { n}, a proper subset of n. a contradiction. D 2.3 (a) (b) (c)

Corollary If n i= m, then there is no one-to-one mapping of n onto m. If IS) = n and )SI = m, then n = m. N is infinite.

Proof (a) Ifni= m, then either n C m or m C n, by Exercise 2.7 in Chapter 3. so there is no one- to-one mapping of n onto m. (b) Immediate from (a). (c) By Exercise 2.3 in Chapter 3, the successor function is a one-to-one mapping of N onto its proper subset N- {0}. [J

Another noteworthy observation is that if m, n E N and m < n (in tliP usual ordering of natural numbers by size defined in Chapter 3), then m C n. so m = )m) < lnl = n, where < is the ordering of cardinal numbers defined in the preceding section. Hence there is no need to distinguish between the two orderings, and we denote both by <. The rest of the section studies properties of finite sets and their cardinals in more detail. 2.4 Theorem If X is a finite set and Y <;: X. then Y is finite.

IYI

~

lXI.

Moreover.

2. FINITE SETS

71

Proof We may assume that X= {xo, ... ,Xn-1 }, where (xo, ... ,Xn-1) is a one-to-one sequence, and that Y is not empty. To show that Y is finite, we construct a one-to-one finite sequence whose range is Y. We use the Recursion Theorem in the version from Exercise 3.5 in Chapter 3. Let ko =the least k such that

Xk

k,+ 1 =the least k such that k

E Y;

> k,, k < n, and

Xk

E Y

(if such k exists).

We leave it to the reader to verify that this definition fits the pattern of Exercise 3.5 in Chapter 3. [Hint: A= n = {0, 1, ... , n- 1}; a= min{k E n I Xk E Y}; g(t, i) = min{k E n I k > t and Xk E Y} if such k exists; undefined otherwise.] This defines a sequence (ko, ... ,km- 1). When we let y, = Xk, for all i < m, then Y = {y; I i < m}. The reader should verify that m ::; n (by induction, k, 2: i whenever defined, so especially m- 1::; km-1 :S: n- 1). D

2.5 Theorem If X is a finite set and f is a function, then Moreover, lf[X]I :S: lXI.

f[X] is finitt'.

Proof Let X = { xo, . .. , Xn- d. Again, we use recursion to construct a finite one-to-one sequence whose range is /[X]. Actually, here Wt> use the version with f(n + 1) = g(f r n). We ask the reader to supply the details; the construction goes as follows: k0

= 0,

k;+ 1 =the least k

> k; such that k
(if it exists), and y, = f(xk.). Then ![X] = {yo, ... , Ym- d for some m :S: n. D As a consequence, if (a; I i < n) is any finite sequence (with or without repetition), then the set {a; I i < n} is finite. All the constructions made possible by the Axioms of Comprehension when applied to finite sets yield finite sets. We now show that if X is finite, then P(X) is finite, and if X is a finite collection of finite sets, then U X is finite. Hence addition of the Axiom of Infinity is necessary in order to obtain infinite sets.

2.6 Lemma If X and Y are finite, then XU Y is finite. Moreover, lXI + IYI, and if X andY are disjoint, then IX U Yl = lXI + JYI.

IX U Yl ::;

If X = {Xo, ... , Xn-d and if Y = {yo, ... , Ym-d. where (.To ..... (yo, ... , Ym-1) are one-to-one finite sequences, consider the finitt> sequence z = (xo, ... ,Xn-1, yo, ... • Ym- 1) of length n+m. (Precisely, definE' z = (z, I 0 ::; i < n + m) by Proof

Xn- 1) and

z, = x,

for 0 :S: i

< n,

Z;=Yi-n

forn:S:i
72

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

Clearly, z maps n + m onto XU Y. so XU Y is finite and IX U Yl :S n + m by Theorem 2.5. If X andY are disjoint, z is one-to-one and IX U Yl = n + m. 0

2. 7 Theorem If S is finite and if every X

E

S is finite, then

US

is finite.

Proof. We proceed by induction on the number of elements of S. The statement is true if lSI = 0. Thus assume that it is true for all S with lSI = n. and letS = {Xo, ... ,Xn-l.Xn} be a set. with n + 1 elements, each X, E S being a finite set. By the induction hypothesis, U~-01 X, is finite. and we have

which is, therefore, finite by Lemma 2.6.

0

2.8 Theorem If X is finite, then P(X) is finite. Proof. By induction on lXI. If lXI = 0, i.e., X = 0, then P(X) = {0} is finite. Assume that P(X) is finite whenever lXI = n, and let Y be a set with n + 1 elements: Y ={yo, ... ·Yn}· Let X= {yo, ... ,y,_!}. We not.P that P(Y) = P(X) U U, where U = {u I u <;;: Y and .1/n E u}. WP furtlwr note that lUI== IP(X)I because there is a one-to-one mapping of U onto P(X): f(u) = u- {Yn} for all u E U. Hence P(Y) is a union of two finite sets and. consequently, finite. C The final theorem of the section shows that infinite sets really haw more elements than finite sets. 2.9 Theorem If X is infinite, then lXI > n for all n EN.

Proof It suffices to show that lXI ~ n for all n E N. This can be done by induction. Certainly 0 < lXI. Assume that lXI ~ n: then there is a one-to-one function f : n -> X. Since X is infinite, there exists x E (X - ran f). Dehw· g = f U { (n, x)}; g is a one-to-one function on n + 1 into X. We concludP that lXI ~ n + 1. c; To conclude this section, we briefly discuss another approach to finiteness. The following definition of finite sets does not use natural numbers: A set X is finite if and only if there exists a relation -< such that (a) -< is a linear ordering of X. (b) Every nonempty subset of X has a least and a greatest element in -<. Note that this notion of finiteness agrees with the one we defined using finitf' sequences: If X = { xo, ... , Xn- 1}, then xo -< · · · -< Xn-l describes a linear ordering of X with the foregoing properties. On the other hand. if (X.-<) satisfies (a) and (b), we construct, by recursion, a sequence (f0 . ft .... ). as

2. FINITE SETS

73

in Theorem 3.4 in Chapter 3. As in that theorem, the sequence exhausts all elements of X, but the construction must come to a halt after a finite number of steps. Otherwise, the infinite set {!0 , !J, /2, ... } has no greatest element in

(X,-<). We mention another definition of finiteness that does not involve natural numbers. We say that X is finite if every nonempty family of subsets of X has a ~-maximal element; i.e., if 0 f. U ~ P(X). then there exists z E U such that for no y E U, z C y. Exercise 2.6 shows equivalence of this definition with Definition 2.1. Finally, let us consider yet another possible approach to finiteness. It follows from Lemma 2.2 that if X is a finite set, then there is no one-to-one mapping of X onto its proper subset. On the other hand, infinite sets. such as the :-;et of all natural numbers N, have one-to-one mappings onto a proper subset [e.g., f(n) = n + 1]. One might attempt to define finite sets as those sets which are not equipotent to any of their proper subsets. However, it is impossiblt> to prove equivalence of this definition with Definition 2.1 without using the Axiom of Choice (see Exercise 1.9 in Chapter 8).

Exercises

2.1 If S = { Xo, .. . , Xn- 1 } and the elements of S are mutually disjoint, then I U Sl = l:~:o1 IX;/. 2.2 If X andY are finite, then X x Y is finite, and IX x Yl = lXI · IY[. 2.3 If X is finite, then )P(X)I = 2IXI. 2.4 If X and y are finite, then xY has IXIIYI elements. 2.5 If lXI = n 2: k = IY), then the number of one-to-one functions f: Y-+ X is n · (n- 1) · · · · · (n- k + 1). 2.6 X is finite if and only if every nonempty system of subsets of X has a ~-maximal element. [Hint: If X is finite, lXI = n for some n. If U ~ P(X), let m be the greatest number in {/YI I Y E U}. If Y E U and )Y) = m, then Y is maximal. On the other hand, if X is infinite. lt't U = {Y ~ X I Y is finite}.] 2. 7 Use Lemma 2.6 and Exercises 2.2 and 2.4 to give easy proofs of commutativity and associativity for addition and multiplication of natural numbers, distributivity of multiplication over addition, and the usual arithmetic properties of exponentiation. [Hint: To prove, e.g., the commutativity of multiplication, pick X andY such that lXI = m, fYI = n. By Exercise 2.2, m·n =IX x Yl, n·m = IY x X/. But X x Y andY x X are equipotent.] 2.8 If A, B are finite and X ~ A x B, then lXI = LaEA ka where kn = [X n ({a} x B)f.

74

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

3.

Countable Sets

The Axiom of Infinity provides us with an example of an infinite set - the set N of all natural numbers. In this section, we investigate the cardinality of N; that is, we are interested in sets that are equipotent to the set N. 3.1 Definition A set Sis countable if lSI = INI. A setS is at most countable if lSI~ INI. Thus a set S is countable if there is a one-to-one mapping of N onto S. that. is, if S is the range of an infinite one-to-one sequence.

3.2 Theorem An infinite subset of a countable set is countable. Proof. Let A be a countable set, and let B <;:;; A be infinite. There is an infinite one-to-one sequence (an)~=D• whose range is A. We let bo = ako• where ko is the least k such that ak E B. Having constructed bn, we let bn+ 1 = ak .. ~, . where kn+l is the least k such that ak E B and ak I bt for every i ~ n. Such k exists since B is infinite. The existence of the sequence (bn)~=O follows easily from the Recursion Theorem stated as Theorem 3.5 in Chapter 3. It is easily seen that B = {bn I n E N} and that (bn)~=D is one-to-one. Thus B is ~~~-

0

If a set S is at most countable then it is equipotent to a subset of a countablt:> set; by Theorem 3.2, this is either finite or countable. Thus we have Corollary 3.3. 3.3 Corollary A set is at most countable if and only if it is either finite or countable. The range of an infinite one-to-one sequence is countable. If (an)~=O is an infinite sequence which is not one-to-one, then the set {an }~=D may be finite (e.g., this happens if it is a constant sequence). However, if the range is infinite. then it is countable.

3.4 Theorem The range of an infinite sequence (an)~=O is at most countabif'. i.e., either finite or countable. (In other words, the image of a countable set under any mapping is at most countable.) Proof. By recursion, we construct a sequence (bn) (with either finite or infinite domain) which is one-to-one and has the same range as (an)~=D· We let bo = ao, and, having constructed bn, we let bn+l = ak,.+,, where kn+l is the least k such that ak I bt for all i ~ n. (If no such k exists, then we consider the finite sequence (bi I i ~ n).) The sequence (bi) thus constructed is one-to-one and its range is {an }~=O· 0 One should realize that not all properties of size carry over from finite sets to the infinite case. For instance, a countable set S can be decomposed into two

3. COUNTABLE SETS

75

disjoint parts, A and B, such that IAI = IBI = lSI; that is inconceivable if S is finite (unless S = 0). Namely, consider the set E = {2k I k E N} of all even numbers, and the set 0 = {2k + 1 I k E N} of all odd numbers. Both E and 0 are infinite, hence countable; thus we have INI = lEI = 101 while N = E U 0 and En 0 = 0. We can do even better. Let Pn denote the nth prime number (i.e., p0 = 2, PI = 3, etc.). Let So= {2k IkE N}, sl = {3k IkE N}, ... ' Sn = {p~ IkE N}, .... The sets Sn (n E N) are mutually disjoint countable subsets of N. Thus we haveN 2 u:=O Sn, where ISnl = INI and the Sn's are mutually disjoint. The following two theorems show that simple operations applied to countable sets yield countable sets.

3.5 Theorem The union of two countable sets is a countable set. Let A = {an I n E N} and B = {bn I n E N} be countable. We Proof. construct a sequence (cn)~=O as follows: for all k

c2k = ak and C2k+l = bk

E

N.

Then A U B = {Cn I n E N} and since it is infinite, it is countable.

0

3.6 Corollary The union of a finite system of countable sets is countable. Proof

Corollary 3.6 can be proved by induction (on the size of the system). D

One might be tempted to conclude that the union of a countable system of countable sets in countable. However, this can be proved only if one uses the Axiom of Choice (see Theorem 1.7 in Chapter 8). Without the Axiom of Choice, one cannot even prove the following "evident" theorem: If S = {An I n E N} and IAn I = 2 for each n, then u:=o An is countable! The difficulty is in choosing for each n E N a unique sequence enumerating An. If such a choice can be made, the result holds, as is shown by Theorem 3.9. But first we need another important result.

3. 7 Theorem If A and B are countable, then A x B is countable. It suffices to show that IN x Nl = INI, i.e., to construct either Proof. a one-to-one mapping of N x N onto N or a one-to-one sequence with range N x N. [See Exercise 3.1(b).] (a) Consider the function f(k,n)

= 2k · (2n + 1)- 1.

We leave it to the reader to verify that ~fisN.

f

is one-to-one and that the range

D

76

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

(b) Here is another proof: Construct a sequence of elements of N x N in tht:> manner prescribed by the diagram.

(0,0/0,~0,2/0,3)

(1,0)~1(1

(1,2)

...

// /

(2,0)

(2, 1)

(2,2)

(0, 0 ) - (0.1)

(0,2)

(3, 0)

We have yet another proof:

l l

(1, 0 ) - (1, 1)

(1, 2)

l

(2, 0 ) - (2, 1 ) - (2, 2)

0

3.8 Corollary The cartesian pmduct of a finite number of countable sets zs countable. Consequently, Nm is countable, for every m > 0. Proof.

Corollary 3.8 can be proved by induction.

0

3.9 Theorem Let (An I n E N) be a countable system of at most countab!P sets, and let (an In EN) be a system of enumerations of An; i.e., for each n E N, an= (an(k) IkE N) is an infinite sequence, and An= {an(k) IkE N}. Then u~=O An is at most countable.

X

Proof. Define f : N N --+ u~=O An by f(n, k) = an(k). f maps N onto u~=O An, so the latter is at most countable by Theorems 3.4 and 3.7. As a corollary of this result we can now prove

XN 0

77

3. COUNTABLE SETS

3.10 Theorem lj A is countable, then the set Seq( A) of all finite sequences of elements of A is countable. Proof.

It is enough to prove the theorem for A = N.

As Seq(N) =

U::"=o Nn, the theorem follows from Theorem 3.9, if we can produce a sequence (an / n 2 1) of enumerations of Nn. We do that by recursion. Let g be a one-to-one mapping of N onto N x N. Define recursively a 1 (i) = (i) for all i E N; an+ I (i) = (bo, ... , bn-1, i2) where g(i) = (i 1, i2) and (bo, ... , bn-1) = an(ii), for all i E N.

The idea is to let an+ 1 ( i) be the ( n + 1)-tuple resulting from the concatenation of the (i 1 )th n-tuple (in the previously constructed enumeration of n-tuples, an) with i 2 . An easy proof by induction shows that an is onto Nn, for all n 2 1, and therefore U;:"= 1 Nn is countable. Since N° = { () }, U::"=o Nn is also countable. D 3.11 Corollary The set of all finite subsets of a countable set is countable. Proof. The function F defined by F((ao .... . an- 1 )) = {ao .... . a,_ I} maps the countable set Seq(A) onto the set of all finite subsets of A. The conclusion follows from Theorem 3.4. 0

Other useful results about countable sets are the following. 3.12 Theorem The set of all integers Z and the set of all rational numbers Q are countable. Proof.

Z is countable because it is the union of two countable sets: Z= {0,1,2,3, ... }u{-1,-2,-3, ... }.

Q is countable because the function f: Z x (Z- {0})

---+

Q defined by f(p. q) =

p/q maps a countable set onto Q.

0

3.13 Theorem An equivalence relation on a countable set has at most countably many equivalence classes. Proof. Let E be an equivalence relation on a countable set A. The function F defined by F(a) = [a]E maps the countable set A onto the set A/E. By Theorem 3.4, A/E is at most countable. 0

3.14 Theorem Let 2l be a structure with the universe A, and let C most countable. Then C. the closure of C, is also at most countable.

1:00;

A be at

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

78

Proof. Theorem 5.10 in Chapter 3 shows that C = U::o C., where Co= C and C>+l = C, U Fo[C{"J U · · · U Fn-dC("-']. It therefore suffices to produce a system of enumerations of (Ci I i E N). Let (c(k) I k E N) be an enumeration of C, and let g be a mapping of N onto the countable set ( n + 1) x N x N fo x · · · x N f,.-'. We define a system of enumerations (a, I i E N) recursively as follows: ao(k) = c(k)

for all kEN;

_ {F,(a,(rg), ... , a,(rt'·-l )) a, (q )

a,+l (k) -

if 0 ~ p ~ n- 1; if p = n,

where g(k) = (p, q, (rg, ... , r{/- 1 ), ... , (r~-l, ... , r!·~-/ -I)). The definition of ai+l is designed so as to make it transparent that if ai enumerates C., a,+ 1 enumerates Ci+ 1 (with many repetitions). By induction, a, enumerates C, for each i E N, as required. Ci We conclude this section with the definition of the cardinal number of countable sets. 3.15 Definition IAI = N for all countable sets A. We use the symbol No (aleph-naught) to denote the cardinal number of countable sets (i.e., the set of natural numbers, when it is used as a cardinal number). Here is a reformulation of some of the results of this section in terms of the new notation. 3.16

(a) No> n for all n EN; if No 2 r;. for some cardinal number "'· then "' = N0 or K = n for some n E N (this is Corollary 3.3). (b) If IAI = No, IBI = No, then lA U Bl =No, lA x Bl = No (Theorems 3.5 and 3.7). (c) If !AI= No, then I Seq(A)I =No (Theorem 3.10). Exercises

3.1 Let lAd= IA2I. IB1I = IB2I· Prove: (a) If A, n A2 = 0, B, n B2 = 0, then lA, u A2l = IB, u B2l· (b) lA, X A2l = IBI X B21· (c) I Seq( A I) I= ISeq(A2)I. 3.2 The union of a finite set and a countable set is countable. 3.3 If A f. 0 is finite and B is countable, then A x B is countable. 3.4 If A f. 0 is finite, then Seq(A) is countable. 3.5 Let A be countable. The set [A]n = {S ~A I lSI= n} is countable for all n E N, n f. 0.

4. LINEAR ORDERINGS

79

3.6 A sequence (sn)~=O of natural numbers is eventually constant if there is no E N, s E N such that Sn = s for all n 2 n 0 . Show that the set of eventually constant sequences of natural numbers is countable. 3. 7 A sequence (sn)~=O of natural numbers is (eventually) periodic if there are no,p EN, p 2 1, such that for all n 2 no, Sn+p = Sn. Show that the set of all periodic sequences of natural numbers is countable. 3.8 A sequence (sn)~=O of natural numbers is called an arithmetic progression if there is d E N such that Sn+ 1 == Sn + d for all n E N. Prove that the set of all arithmetic progressions is countable. 3.9 For every s = (so, ... , Sn-I) E Seq(N - {0} ), let f(s) = p 0" · · · p~·._-,, where p, is the ith prime number. Show that f is one-to-one and use this fact to give another proof of ISeq(N)I =No. 3.10 Let (S, <) be a linearly ordered set and let (An I n E N) be an infinite sequence of finite subsets of S. Then U::"=o An is at most countable. [Hint: For every n E N consider the unique enumeration (an(k) I k < IAnl) of An in the increasing order.] 3.11 Any partition of an at most countable set has a set of representatiws.

4.

Linear Orderings

In the preceding section we proved countability of various familiar sets, such as the set of all integers Z and the set of all rational numbers Q. The important point we want to make is that we cannot distinguish among the sets N, Z, and Q solely on the basis of their cardinality; nevertheless, the three sets "look" quite different (visualize them as subsets of the real line!). In order to capture this difference, we have to consider the way they are ordered. Then it becomes apparent that the ordering of N by size is quite different from the usual ordering of Z (for example, N has a least element and Z does not), and both are quite different from the usual ordering of Q (for example, between any two distinct rational numbers lie infinitely many rationals, while between any two distinct integers lie only finitely many integers). Linear orderings are an important tool for deeper study of properties of sets; therefore, we devote this section to the theory of linearly ordered sets, using countable sets for illustration. 4.1 Definition Linearly ordered sets (A,<) and (B, -<) are similar (have the same order type) if they are isomorphic, i.e., if there is a one-to-one mapping f on A onto B such that for all a 1 , a 2 E A, a 1 < a 2 holds if and only if f(ai) -< f( a 2 ) holds. (See Definition 5.17 in Chapter 2 and the subsequent Lemma.) Similar ordered sets "look alike"; their orderings have the same properties [see Example 5.8 in Chapter 3]. It follows that (N, <) and (Z, <) are not similar; likewise, (Z, <)and (Q, <)are not similar, and neither are (N, <) and (Q, <). (Here, < is the usual ordering of numbers.) It is trivial to show that the property of similarity behaves like an equivalence relation:

80

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

(a) (A,<) is similar to (A,<). (b) If (A,<) is similar to (B, -<),then (B, -<)is similar to (A,<). (c) If (A 1, t can be linearly ordered in only one way. up to isomorphism. 4.2 Lemma Every linear· ordering on a finite set is a well-o1·dering. Proof. We prove that every nonempty finite subset B of a linearly orden~d set (A,<) has a least element by induction on the number of elements of B. If B has 1 element, the claim is clearly true. Assume that it is true for all n-element sets and let B have n + 1 elements. Then B = { b} U B' where B' has n elements and b ¢: B'. By the inductive assumption, B' has a least element b'. If b' < /J. then b' is the least element of B; otherwise, b is the least element of B. In eithPr case, B has a least element. 0

4.3 Theorem lf(At, en (A 1 - {a 1}, < 1 n(A 1 - {at} f) and (A 2 - {a 2}, <2 n(A2- {a 2 }f). Defiup f: A,--. A 2 by f(a!) = a2; f(a) = g(a)

It is easy to check that

for a E A,- {at}.

f is an isomorphism between (A,,
<2).

c

We conclude that for finite sets, order types correspond to cardinal numbers. As the examples at the beginning of this section show, linear orderings of infinite sets are much more interesting. We next look at. some ways of producing linear orderings that will be useful later.

4. LINEAR ORDERINGS

81

4.4 Lemma If (A,<) is a linear ordering, then (A. <- 1 ) is also a linear ordering. Proof.

Left to the reader (see Exercise 5.3 in Chapter 2).

0

For example, the inverse of the ordering (N, <) is the ordering (N. <-I): . · · 4 <- 1 3 <- 1 2 <-I 1 <-I 0; notice that it is similar to the ordering of negative integers by size:

... -4 < -3 < -2 < -1 and that it is not a well-ordering. 4.5 Lemma Let (A 1,
=

0.

< b if and only if a, b E A 1 and a < 1 b or a, b E A2 and a <2 b or a E A1, bE A2

is a linear ordering. Proof

This is Exercise 5.6 in Chapter 2, so again we leave it to the reader. 0

The set A is ordered by putting all elements of A 1 before all elements of A 2 . We say that the linearly ordered set (A,<) is the sum of the linearly ordered sets (A1,
if and only if a1 <1 b1 or (a1 = b1 and a2 <2 b2)

is a linear ordering. Proof. Transitivity: If (a1.a2) < (b1,b2) and (b1,b 2 ) < (c 1,c2). we have either a1 <1 b1 or (a1 = b1 and a2 <2 b2). In the first case a 1 <1 b1 and b1 S:1 CJ gives a 1 < 1 c 1 . In the second case, either b1 <1 c 1 and a1 <1 CJ again, or b1 = c 1 and b2 < 2 c2 , so that a 1 = c 1 aud a2 <2 c2. In either case we conclude that (a 1 .a 2) < (cJ,c 2 ). Asymmetry: This follows immediately from asymmetry of< 1 and < 2 . Linearity: Given (a I, a 2) and (bt, b2), one of the following cases has to occur:

82

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

(a) a1 <1 b1 [so (a1,a2) < (bJ, b2)]. (b) b1 <1 a1 [so (b1,b2) < (a1,a2)]. (c) at= bt and a2 <2 b2 [so (a1, a2) < (b1, b2)]. (d) at= bt and b2 <2 a2 [so (b 1 ,b2) < (at,a 2 )]. (e) at= bt and a2 = b2 [so (at.a2) = (b 1 ,b2)]. In each case (at, a2) and (bt, b2) are comparable in <.

0

We call < the lexicographic ordering (lexicographic product) of At x A 2 ; the terminology comes from the observation that if At = A2 = {a, b, . .. , z} is the set of all letters and
if and only if diff(J,g) = {i E I If,

f.

gi}

:10

and

/i., <, g;" 0

where i 0 is the least element of diff(f, g) (in the usual ordering < of natural numbers) is a linear ordering of IT.eJ A, (it is called its lexicographic ordering). Proof. Transitivity: Assume that f -< g and g -< h and let io [J 0 , respectively] be the least element of diff(f,g) [diff(g, h), respectively]. If io < j 0 , we have/;,, < 9io and g,11 = h;,., so f,, < h," and io is the least element of diff(f, h). We conclude that f - j 0 are similar. Asymmetry: f -< g and g -< f is impossible because it would mean that f,o < g, 0 and 9io < /i., for io =the least element of diff(f,g) = diff(g, f). Linearity: If diff (f, g) = 0, we have f = g. Otherwise, if io is the least element of diff(J, g), either J,., < g; 0 or f,o > g," holds and, consequently, either f -< g or f >-g. 0

In particular, if (A.,<;) = (A,<) for all i E I = N, -< is the lexicographic ordering of the set AN of all infinite sequences of elements of A. One can also choose to compare second coordinates before comparing th!' first coordinates and so define the antilexicographic ordenng -< of A 1 x A2 :

The proof that -< is a linear ordering is entirely analogous to the lexicographic case. The two orderings are generally quite different; compare, for example. the lexicographic and antilexicographic products of At = N = {0, 1, 2, ... } an(l A2 = {0,1} (both ordered by size):

83

4. LINEAR ORDERINGS ordered lexicographically: (0,1) (1,1) (2,1) (3,1)

1~1~1~1

(0,0)

(1, 0)

(2,0)

(3,0)

ordered antilexicographically: (0, 1)- (1, 1)- (2, 1)- (3, 1)

(0, 0)- (1, 0)- (2, 0)- (3, 0) The first ordering is similar to (N, <) and the second is not [it is the sum of two copies of (N, <)]. The previous results show that there is a rich variety of types of linear orderings on countable sets. It is thus rather surprising to learn that there is a universal linear ordering of countable sets, i.e., such that every countable linearly ordered set is similar to one of its subsets. The rest of this section is devoted to the proof of this important result. 4.8 Definition An ordered set (X,<) is dense if it has at least two elements and iffor all a, b E X, a < b implies that there exists x E X such that a < x < b. Let us call the least and the greatest elements of a linearly ordered set (if they exist) the endpoints of the set. The most important example of a countable dense linearly ordered set is the set Q of all rational numbers, ordered by size. The ordering is dense because, if r, s are rational numbers and r < s, then x = (r + s)/2 is also a rational number, and r < x < s. Moreover, (Q, <) has no endpoints (if r E Q then r + 1, r - 1 E Q and r - 1 < r < r + 1). Other examples of countable dense linearly ordered sets are in Exercises 4.6 and 4.7. However, we prove that all countable linearly ordered sets without endpoints have the same order type.

4.9 Theorem Let (P, --:) and (Q, <) be countable dense linearly ordered sets without endpoints. Then (P, --:) and (Q, <) are similar. Proof Let (Pn I n E N) be a one-to-one sequence such that P = {Pn I n E N}, and let (qn I n E N) be a one-to-one sequence such that Q = {Qn I n E N}. A function h on a subset of P into Q is called a partial isomorphism from P to Q if p --: p' if and only if h(p) < h(p') holds for all p, p' E dom h. We need the following claim: If h is a partial isomorphism from P to Q such that dom h is finite, and if p E P and q E Q, then there is a partial isomorphism hp,q 2 h such that p E dom hp,q• and q E ran hp,q·

84

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

Proof of the claim. Let h = { (p;,, q;,), ... , (P;., q;.)} where p,, -< p, 2 -< · · · -< p,., and thus also q;, < q, 2 < · · · < Q;•. If p ~ dom h, we haw either p -< p;, or p,, -< p -< p,., +, for some 1 ~ e ~ k, or p,. -< p. Take the least natural number n such that Qn is in the same relationship to q,,, .... q,, asp is to p,,, ... , p,.; more precisely, Qn is such that: if p -< pq, then Qn < q,,; if p,, -< p-< p,·+•, then q,, < Qn < q;, +'; and if p,. -< p, then q;. < Qn. The possibility of doing this is guaranteed by the fact that (Q, <) is a dense linear ordering without endpoints. It is clear that h' = h U { (p, Qn)} is a partial isomorphism. If q E ran h', then we are done. If q ¢ ran h', then by the same argument as before (with the roles of P and Q reversed), there is Pm E P such that h' u {(Pm, q)} is a partial isomorphism, and we take the least such m, and let hp,q = h' U {(pm,q)}. The claim is now proved. 0 We next construct a sequence of compatible partial isomorphisms by rermsion:

ho = 0, hn+l = (hn)p.,,q,.

where (hn)p ... q,, is the extension of hn (as provided by the claim) such that. Pn E dom(hn)p .. ,q,., qn E ran(hn)p .. ,q,.· Let h = UnEN hn· It is trivial to verify that h: P---> Q is an isomorphism between (P, -<) and (Q, <).

0

4.10 Theorem Every countable linearly ordered set can be mapped isomorphically into any countable dense linearly ordered set without endpoints. Proof. Theorem 4.10 requires a one-sided version of the previous proof. Let (P, --<) be a countable linearly ordered set and let (Q, <) be a countable dense linearly ordered set without endpoints. For every partial isomorphism h from the ordered set ( P, -<) into Q and for every p E P, we define a partial isomorphism hp ~ h such that p E dom hp· Then we use recursion again. 0

Exercises 4.1 Assume that (A~o <1) is similar to (B 1, -
4. LINEAR ORDERINGS

85

4.4 If (A; I i E N) is an infinite sequence of linearly ordered sets of natural numbers and IA,I ? 2 for all i E N, then the lexicographic ordering of niEN A, is not a well-ordering. 4.5 Let ((A;,<;) I i E I) be an indexed system of mutually disjoint linearly ordered sets, I <:;; N. The relation -< on U,E 1 A, defined by: a -< b if and only if either a, bE A; and a <; b for some i E I or a E A,, b E A 1 and i < j (in the usual ordering of natural numbers) is a linear ordering. If all <, are well-orderings, so is -<. 4.6 Let (Z, <)be the set of all integers with the usual linear ordering. Let-< be the lexicographic ordering of zN as defined in Theorem 4.7. Finally, let FS ~ zN be the set of all eventually constant elements of zN: i.e., (a; I i E N) E FS if and only if there exist n 0 E N. a E Z such that a, = a for all i ? no (compare with Exercise 3.6). Prove that FS is countable and (FS,-< nFS 2 ) is a dense linearly ordered set without. endpoints. 4.7 Let -< be the lexicographic ordering of NN (where N is assumed to be ordered in the usual way) and let P <:;; NN be the set of all eventually periodic, but not eventually constant, sequences of natural numbers (seP Exercises 3.6 and 3.7 for definitions of these concepts). Show that (P.-< nP 2 ) is a countable dense linearly ordered set without endpoints. 4.8 Let (A,<) be linearly ordered. Define-< on Seq( A) by: (ao ..... a"'_ 1 )-< (b 0 , ... , bn-1) if and only if there is k < n such that a, = b, for all i < k and either ak < bk or ak is undefined (i.e., k = m < n). Prove that -< is a linear ordering. If (A,<) is well-ordered, (Seq(A). -<) is also \\"Pllordered. (In this ordering, if a finite sequence extends a shorter sequPnrP. the shorter one comes before the longer one.) 4.9 Let (A,<) be linearly ordered. Define -
86

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

(a) the sum of two copies of (N, <); (b) the sum of (N, <)and (N, <- 1); (c) the lexicographic product of (N, <) with (N, <).

5.

Complete Linear Orderings

We have seen in the preceding section that the usual ordering < of the set Q of all rational numbers is universal among countable linear orderings (Theorem 4.10). However, when arithmetic operations on Q are considered, it becomes apparent that something is missing. For example, there is no rational number x such that x 2 = 2 (Exercise 5.1). Another example of this phenomenon appears when one considers decimal representations of rational numbers. Every rational number has a decimal expansion that is either finite (e.g., 1/4 = 0.25) or infinite but periodic from some place on (e.g., 1/6 = 0.1666 · · ·) (see Section 10.1). Although it is possible to write down decimal expansions O.a 1a2a3 · · · when· (ai)~ 1 is an arbitrary sequence of integers between 0 and 9, unless the sequence is finite or eventually periodic, there is no rational number x such that x = O.a 1a 2a 3 · · ·. (As a specific example, consider 0.1010010001· · · .) It is dear from these considerations that the ordered set (Q, <) has gaps. The notion of a gap can be formulated in terms involving only the linear ordering.

5.1 Definition Let (P, <) be a linearly ordered set. A gap is a pair (A. B) of sets (a) (b) (c)

such that A and Bare nonempty disjoint subsets of PandA U B = P. If a E A and bE B, then a < b. A does not have a greatest element and B does not have a least element.

For example, let B = {x E Q I x > 0 and x 2 > 2} and A = Q - B = { x E Q 1 x ~ 0 or (x > 0 and x 2 < 2)}. It is not difficult to check that (A, B) is a gap in Q (Exercise 5.2). Similarly, an infinite decimal expansion which is not eventually periodic gives rise to a gap (Exercise 5.3). At this point, we ask the reader to recall the notions of upper and lower bound, and supremum and infimum, that were introduced in Chapter 2. We call a nonempty subset of a linearly ordered set P bounded if it has both lower and upper bounds. A set is bounded from above (from below) if it has an upper (lower) bound. Let (A, B) be a gap in a linearly ordered set. The set A is bounded from above because any b E B is its upper bound. We claim that A does not haw a supremum. For if c were a supremum of A then either c would be the greatest element of A or the least element of B, as one can easily verify. On the other hand, let S be a nonempty set, bounded from above. Let A = {x

Ix

~

s for some s E S},

B = {xI x > s for every s E S}.

1. WELL-ORDERED SETS

105

the initial segment of W given by a. (Note that if a is the least element of W, then W[a] is empty.) By Lemma 1.2, each initial segment of a well-ordered set is of the form W[a] for some a E W. Of course, W[a] is also well-ordered by < (more precisely, by < nW[aj2); we usually do not mention the well-ordering relation explicitly when it is understood from the context.

1.3 Theorem Ij(W1,
1.4 Lemma If (W, <) is a well-ordered set and iff: W ---. W is an increasing function, then f(x) ~ x for all x E W. If the set X = {x E W I f(x) < x} is nonempty, it has a least Proof. element a. But then f(a)
Corollary No well-ordered set is isomorphic to an initial segment of itself. Each well-ordered set has only one automorphism, the identity. If W 1 and W2 are isomorphic well-ordered sets, then the isomorphism between WI and w2 is unique.

Proof. (a) Assume that f is an isomorphism between W and W[a] for some a E W. Then f(a) E W[a] and therefore f(a) < a, contrary to the lemma, as f is an increasing function. (b) Let f be an automorphism of W. Both f and f- 1 are increasing functions and so for all x E W, f(x) ~ x and J- 1(x) ~ x, therefore x ~ f(.r). It follows that f(x) = x for all x E W. (c) Let f and g be isomorphisms between Wt and W2. Then f o g- 1 is an automorphism of W2 and hence is the identity map. It follows that f =g. D Let WI and w2 be well-ordered sets. It is a 1.6 Proof of Theorem 1.3 consequence of Lemma 1.4 that the three cases (a), (b), and (c) are mutually

88

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

is dense in C) there is p E P such that c --< p -< d. It is readily seen that sup• Sc -<* p -<*sup* Sd and hence h(c) -<* h(d). From this we conclude that h is an isomorphism using Lemma 5.18 from Chapter 2. Finally, if :r E P. then x =sup Sx = sup• Sx and so h(x) = :r. 0 To prove the existence of a completion, we introduce the notion of a Dedekind cut.

5.4 Definition A cut is a pair (A, B) of sets such that (a) A and B are disjoint nonempty subsets of P and Au B = P. (b) If a E A and b E B, then a < b. We recall that a cut is a gap if, in addition, A does not have a greatest element and B does not have a least element. Notice that since P is dense, it is not possible that both A has a greatest element and B has a least element. Thus the remaining possibilities are that either B has a least element and A does not have a greatest element, or A has a greatest element and B does not have a least element. In either case, the supremum of A exists: In the first case. the supremum is the least element of B, and in the other case. the supremum is the greatest element of A. Hence, we consider only the first case and disregard the cuts where A has the greatest element. 5.5 Definition A cut (A, B) is a Dedekind cut if A does not have a greatest element. We have two types of Dedekind cuts (A, B): (a) Those where B = {x E PIx~ p} for some p E P; we denote (A. B) = [p]. (b) Gaps. Now we consider the set C of all Dedekind cuts (A, B) in (P, <)and order C a.s follows: (A, B) =:(((A', B') if and only if A~ A'. We leave it to the reader to verify that (C, =:(() is a linearly ordered set. If p, q E P are such that p < q, then we have [p] -< [q]. Thus the linearly ordered set (P', --<), where P' = {[p] I p E P}, is isomorphic to (P, <). We intend to show that (C, -<)is a completion of (P', --<). Since (P, <) and (P', -<) are isomorphic, it follows that (P, <) has a completion. It suffices to prove (c') P' is dense in (C, --<), (d') C does not have endpoints, and, of course, (e) ( C, -<) is complete. To show that P' is dense inC, let c, dEC be such that c-< d; i.e., c =(A, B), d =(A', B'), and A C A'. Let pEP be such that pEA' and p ¢A. Moreover, we can assume that pis not the least element of B. Then (A, B) -< [p] -<(A', B') and hence P' is dense in C. [This also shows that (C, -<) is a densely ordered set.]

5. COMPLETE LINEAR ORDERINGS

89

Similarly, if (A, B) E C, then there is p E B that is not the least element of B, and we have (A, B) -< [p]. Hence C does not have a greatest element. For analogous reasons, it does not have a least element. To show that Cis complete, letS be a nonempty subset of C, bounded from above. Therefore, there is (Ao, Bo) E C such that A<:;; A0 whenever (A, B) E S. To find the supremum of S, let As= U{A I (A, B) E S}

and

Bs =P-As= n{B I (A, B) E S}.

It is easy to verify that (As, Bs) is a cut. (Note that Bs is nonempty because Bo <:;; Bs.) In fact, (As, Bs) is a Dedekind cut: As does not have a greatest element since none of the A's does. Since As 2 A for each (A,B) E S, (As,Bs) is an upper bound of S; let us show that (As, Bs) is the least upper bound of S. If (A, B) is any upper bound of S, then A <:;; A for all (A, B) E S, and, so, As = U{A I (A, B) E S} <:;; it hence (As,Bs) ~(A, B). Thus (As,Bs) is the supremum of S. 0 Theorem 5.3 is now proved. In particular, the ordered set (Q, <)of rationals has a unique completion (up to isomorphism); this is the ordered set of real numbers. As the ordering of real numbers coincides with < on Q, it is customary to use the symbol < (rather than -<) for it as well. 5.6 Definition The completion of (Q, <) is denoted (R, <); the elements of R are the real numbers. We get immediately the following characterization of (R, <).

5. 7 Theorem (R, <) is the unique (up to isomorphism) complete linearly ordered set without endpoints that has a countable subset dense in it. Proof Let (C, -<) be a complete linearly ordered set without endpoints, and let P be a countable subset of C dense in C. Then (P, -<) is isomorphic to (Q, <)and by the uniqueness of completion (Theorem 5.3), (C, -<)is isomorphic to the completion of (Q, <), i.e., to (R, <). 0

It remains to define algebraic operations on R and show that they satisfy the usual laws of algebra and that on rational numbers they agree with their previous definitions. As this topic is of more interest to algebra and real analysis than set theory, we do not pursue it here. The reader interested in learning how the arithmetic of real numbers can be rigorously established can at this point read Section 10.2. Exercises

5.1 Prove that there is no x E Q for which x 2 = 2. [Hint: Write x = pfq where p, q E Z are relatively prime, and use p 2 = 2q 2 to show that 2 has to divide both p and q.]

90

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS 5.2 Show that (A, B), where A== {x E Q I x ~ 0 or (x > 0 and x 2 < 2)}. B == {x E Q I x > 0 and x 1 > 2}, is a gap in (Q, <). [Hint: To prow that A does not have a greatest element. given x > 0. :c 2 < 2. find a rationalE> 0 such that (1: + c;) 2 < 2. It suffices to takt· f < .r: such that x 2 + 3xc; < 2.] 5.3 Let O.aJ a2a3 · · · be an infinitP. but not periodic, decimal expansion. Ld A == {x E Q I x ::; O.a 1a1 · · · ak for somp k E N- {0} }, B = {:r E Q : x ~ O.a1a2 · · · ak for all k E N- {0} }. Show that (A. B) is a gap iu (Q, <). 5.4 Show that a dense linearly ordered set (P, <) is complete if and only if every nonempty S s;; P bounded from below has an infimum. 5.5 Let D be dense in (P, <). and let E be dense in (D, <). Show that E i» dense in (P, <). 5.6 Let F be the set of all rational numbers that have a decimal expansion with only a finite number of nonzero digits. Show that F iR densP in Q. 5. 7 Let D (the dyadic rationals) be the set of all numbers m/2n wlwr(' 111 is an integer and n is a natural number. Show that D is dense in Q. 5.8 Prove that the set R- Q of all irrational numbers is dense in R. [Hi11t Given a < b, let x = (a+ /1) /2 if it is irrational. :r == (a+ &) / J2 otherwise. u~e Exercise 5.1.]

6.

Uncountable Sets

All infinite sets whose cardinalities we have determined up to this point turnC'd out to be countable. Naturally, a question arises whether perhaps all infinite sets are countable. If it were so, this book might end with the preceding section. It was a great discovery of Georg Cantor that uncountable sets. in fact. Pxist This discovery provided an impetm; for the development of set theory and becamp a source of its depth and richness.

6.1 Theorem The set R of all real numbers is uncountable.

(R, <) is a dense linear ordering without endpoints. If R werP Proof countable, (R, <) would be isomorphic to (Q, <) by Theorem 4.9. But this is 0 not possible because (R, <) is complete and (Q, <)is not. The proof above relies on the theory of linear orderings developed in Section 4. Cantor's original proof used his famous "diagonalization argument'' Vve giw it below.

Cantor's Proof of Theorem 6.1. Assume that R is countable, i.e .. R is thP range of some infinite sequence (rn ~~ 1 . Let a~n) .a in) a~nl a~nl · · · lw tlw dPcinml expansion of r n. (We assume that a decimal expansion does not contain only t ht> digit 9 from some place on, so each real number has a unique decimal expansion. See Section 10.1.) Let bn = 1 if a~,n) == 0, bn == 0 otherwise: and let r \w tht>

6. UNCOUNTABLE SETS

91

real number whose decimal expansion is O.b1b2b3 · · · . We have bn r f. Tn, for all n = 1, 2, 3, ... , a contradiction.

f. a~,n), hence 0

The combinatorial heart of the diagonal argument (quite similar to Russell's Paradox, which is of later origin) becomes even clearer in the next theorem.

6.2 Theorem The set of all sets of natural numbers is uncountable; in fact, IP(N)I > INJ. Proof. The function f: N--.. P(N) defined by f(n) = {n} is one-to-one, so IN! ~ !P(N)j. We prove that for every sequence (Sn In EN) of subsets of N there is some S ~ N such that S -f. Sn for all n E N. This shows that there is no mapping of N onto P(N), and hence JNI < IP(N)J. We define the set S ~ N as follows: S = { n E N I n ¢ Sn}. The mrmbPr n is used to distinguish S from Sn: If n E Sn, then n ¢ S, and if n ¢ S,, then n E S. In either case, S "f. Sn, as required. 0

Detailed study of uncountable sets is the subject of Chapter 5 (and subsequent chapters). Here we only prove that the set 2N = { 0, 1} N of all infinite sequences of O's and 1 's is also uncountable, and, in fact, has the same cardinality as P(N) and R.

6.3 Theorem IP(N)J = I2NI Proof. For each S {0, 1 }, as follows:

~

= IRJ.

N define the characteristic function of S,

xs (

n)

= {0

xs : N

-->

E

if n S; 1 if n ¢ S.

It is easy to check that the correspondence between sets and their characteristic functions is a one-to-one mapping of P(N) onto {0, 1}N. To complete the proof, we show that IRI ~ IP(N)I and also I2NI ~ IRI and use the Cantor-Bernstein Theorem. (a) We have constructed real numbers as cuts in the set Q of all rational numbers. The function that assigns to each real number r = (A. B) the set A ~ Q is a one-to-one mapping of R into P(Q). Therefore JRI ~ IP(Q)J. As IQI = JNJ, we have IP(Q)I = IP(N)l (Exercise G.3). Hence IRI ~ IP(N)J. (b) To prove J2N I ~ JRI we use the decimal representation of real numbers. ThP function that assigns to each infinite sequence (an)~=o of O's and 1 's tlw unique real number whose decimal expansion is O.a 0 a 1a2 · · · is a one-to-one mapping of 2N into R. Therefore we have J2NI ~ IRJ.

0 We introduced N0 as a notation for the cardinal of N. Due to Theorem 6.3, the cardinal number of R is usually denoted 2N". The set R of all real

92

CHAPTER 4. FINITE, COUNTABLE, AND UNCOUNTABLE SETS

numbers is also referred to as "the continuum"; for this reason, 2N" is called the cardinality of the continuum. In this notation, Theorem 6.2 says that No < 2N". Exercises

6.1 Use the diagonal argument to show that NN is uncountable. [Hint: Consider (an I n E N) where an = (ank I k E N). Define d E NN by dn =ann+ 1.] 6.2 Show that INN I = 2N". [Hint: 2N <:::: NN <:::: P(N x N).] 6.3 Show that IAI = IBI implies IP(A)I = IP(B)I.

Chapter 5

Cardinal Numbers 1.

Cardinal Arithmetic

Cardinal numbers have been introduced in Chapter 4. The present chapter is devoted to the study of their general properties, with the particular emphasis on the cardinality of the continuum, 2No. In this section we set out to define arithmetic operations (addition, multiplication, and exponentiation) on cardinal numbers and investigate the properties of these operations. To define the sum K. + ,\ of two cardinals, we use the analogy with finite sets: If a set A has a elements, a set B has b elements, and if A and B are disjoint, then A U B has a + b elements. 1.1 Definition

K.

+ ,\ = lA UBl where IAI

=

K.,

IBI

= A, and

An B

0. K. + ,\ does =

In order to make this definition legitimate, we have to show that not depend on the choice of the sets A and B. This is the content of the following lemma. 1.2 Lemma If A, B, A', B' are such that then IAUBI = IA'uB'I·

IAI = IA'I, IBI = IB'I,

and

An B

=

0= A'nB',

Proof. Let f and g be, respectively, a one-to-one mapping of A onto A' and of B onto B'. Then f U g is a one-to-one mapping of A U B onto A' U B'. 0

Not only does addition of cardinals coincide with the ordinary addition of numbers in case of finite cardinals, but many of the usual laws of addition remain valid. For example, addition of cardinal numbers is commutative and associative:

(a) "" + ,\ = ,\ + K.. (b) ,.. + (,\ + J..L) = (K. + ,\) + 11· These laws follow directly from the definition. Similarly, the following inequalities are easily established: 93

94

CHAPTER 5. CARDINAL NUMBERS

(c)_.._::;~~:+..\.

(d) If 11:1 ::; 11:2 and ..\1 ::; ..\2, then 11:1 + ..\1 :S 11:2 + ..\2. However, not all laws of addition of numbers hold also for addition of cardinals. In particular, strict inequalities in formulas are rare in case of infinite cardinals and, as is discussed later (Konig's Theorem), those that hold are quite difficult to establish. As an example, take the simple fact that if n # 0, then n + n > n. If 11: is infinite, then this is no longer true: We have seen that N0 + N0 = No !see 3.16(b) in Chapter 4], and the Axiom of Choic:e implies that l'i. + K = l'i. for t>ver:J infinite 11:. The multiplication of cardinals is again motivated by properti('s of multiplication of numbers. If A and B are sets of a and b elements. respectively, then the product A x B has a · b elements.

1.3 Definition

11: ·A==

lAx Bl, where IAI ==

11:

and IBI = >..

The legitimacy of this definition follows from Lemma 1.4.

1.4 Lemma If A, B, A', B' are such that IAJ lA' X B'J. Proof. Let f : A A' x B' as follows:

-+

A', g : B

-+

= IA'I,

JBI

= IB'J,

then JAxBJ =

B' be mappings. We define h : A x B

-+

h(a, b)= (J(a),g(b)).

Clearly, iff and g are one-to-one and onto, so is h.

0

Again, multiplication has some expected properties; in particular, it is commutative and associative. Moreover, the distributive law holds. (e) t~:·A==A·t~:. (f) l'i. · (>. · J.L) = (~~:·A) · Jl· (g) ~~: · (A+ J.L) == ~~:·A+~~:· Jl· The last property is a consequenc(' of the equality A

X

(B u C) = (A

X

B) u (A

X

C),

that holds for any sets A, B, and C. We also have: (h) ~~: ::; ~~: · A if A > 0. (i) If 11: 1 ::; 11:2 and A1 ::; A2, then 11:1 • A1 :S 11:2 · ..\2. To make the analogy between multiplication of cardinals and multiplication of numbers even more complete, let us prove (j) II: + II: = 2 . 11:.

=

Proof If IAI 11:, then 2 · '"' is the cardinal of {0, 1} x A. We note that {0, 1} x A= ({0} x A) U ({1} x A), that I{O} x AI= J{l} x AI== 11:. and that the two summands are disjoint. Hence 2 · -"' = 11: + K. 0

(k)

As a consequence of (j), we haw 11: + 11: <::; 11: · 11:, whenever 11: 2:: 2.

1. CARDINAL ARITHMETIC

95

As in the case of addition, multiplication of infinite cardinals has some properties different from those valid for finite numbers. For example, ~ 0 · ~ 0 = No [see 3.16(b) in Chapter 4]. (And the Axiom of Choice implies that"'· K ="'for all infinite cardinals.) To define exponentiation of cardinal numbers, we note that if A and B are finite sets, with a and b elements respectively, then ab is the number of all functions from B to A. 1.5 Definition "'>. = [A 8

[,

where !AI ="'and IBI = >..

The definition of"'>. does not depend on the choice of A and B. 1.6 Lemma 1/IAI =

IA'I

and IBI = [B'[, then [A 8 ! =!A'

B'

[.

Proof Let f : A --> A' and g : B --> B' be one-to-one and onto. Let B' B' 8 F: A - t A' be defined as follows: If k E A 8 , let F(k) = h, when" hE A' is such that h(g(b)) = f(k(b)) for all bE B, i.e., h = f o k o g- 1 . g

B----B' h = F(k)

kl

A --,----+ A'

f

Then F is one-to-one and maps A 8 onto A'

B'

.

0

It is easily seen from the definition of exponentiation that (!) "' ::; "'>. if A > 0. (m) A ::; "'>. if"'> 1. (n) If "-I ::; "-2 and A1 ::; A2, then K-~ 1 ::; 2 . We also have (0) "' . "' = "'2. To prove (o), it suffices to have a one-to-one correspondence between Ax A. t.Jw set of all pairs (a, b) with a,b E A, and the set of all functions from {0.1} into A. This correspondence has been established in Section 5 of Chapter 3. The next theorem gives further properties of exponentiation.

K-;

1.7 (a) (b) (c)

Theorem "'>.+).1 = "'>.. "-P.· ("-A)J.l = "->. J.l.

("' · >.) ~' =

"'~' · A~' .

Proof. Let"'= [K[, A= ILl, J..t = [M[. To show (a). assume that Land M are disjoint. We construct a one-to-one mapping F of KL x K 111 onto KLu!lt

96

CHAPTER 5. CARDINAL NUMBERS

If (!,g) E KL x KM, we let F(f,g) == f U g. We note that f U g is a function. in fact a member of KLuM, and every hE KLuM is equal to F(!,g) for some (!,g) E KL x KM (namely, f == h r L, g = h r M). It is easily seen that F is one-to-one. To prove (b), we look for a one-to-one map F of K L x M onto ( K l')'\t A typical element of KLxM is a function f: L x M ~ K. We let F assign to f the function g: M ~ KL defined as follows: for all mE M, g(m) = hE KL where h(l) = f(l,m) (for alll E L}. We leave it to the reader to verify that F is one-to-one and onto. For a proof of (c) we need a one-to-one mapping F of KM x LM onto (K x L)M. For each (ft,h) E KM x £M, let F(f 1 ,h) = g: M ~ (K x L) where g(m) = (!1 (m), h(m)), for all mE M. It is routine to check that F is one-to-one and onto. 0

We now have a collection of useful general properties of cardinals, but the only specific cardinal numbers encountered so far are natural numbers, ~ 0 , and 2N". We show next that, in fact, there exist many other cardinalities. The fundamental result in this direction is the general Cantor's Theorem (see Theort->m 6.2 in Chapter 4 for a special case).

1.8 Cantor's Theorem

/X/ < /P(X)/, for every set X.

Proof The proof is a straightforward generalization of the proof of Theorem 6.2 in Chapter 4. Its heart is an abstract form of the diagonalization argument. First, the function f: X ~ P(X) defined by f(x) = {x} is clearly one-toone, and so /X/ ::S /P(X)J. It remains to be proven that there is no mapping of X onto P(X). So let f be a mapping of X into P(X). Consider the set S = {x E X / x 1. f(x)}. We claim that Sis not in the range off SupposP that S = f(z) for some z EX. By definition of S, z E S if and only if z ¢ f(z): so we have z E S if and only if z 1. S, a contradiction. This shows that f is not onto P(X), and the proof of JXJ < [P(X)J is complete. C The first half of Theorem 6.3 of Chapter 4 also generalizes to arbitrary sets.

1.9 Theorem /P(X)/

Proof Chapter 4.

= 21XI,

for every set X.

Replace N by X in the proof of JP(N)J = 2INI in Theorem 6.3 in

0

Cantor's Theorem can now be restated in terms of cardinal numbers as follows: "' < 2" for every cardinal number "-· We conclude this section with an observation that for any set of cardinal numbers, there exists a cardinal number greater than all of them.

1. CARDINAL ARITHMETIC

1.10 Corollary For any system of sets S there is a set Y such that holds for all X E S.

Proof.

I US\2: lXI

97

IYI > lXI

Let Y = P(U S). By Cantor's Theorem, IYI > I U Sl, and clearly for all XES (because X~ US if XES). D

Exercises 1.1 Prove properties (a)-( n) of cardinal arithmetic stated in the text of this section. 1.2 Show that ",0 = 1 and K: 1 = ,.. for all K:. 1.3 Show that 1" = 1 for all ,.. and 0" = 0 for all K: > 0. 1.4 Prove that K:"" ~ 2"."". 1.5 If IAI ~ IB\ and if A "I 0, then there is a mapping of B onto A. We later show, with the help of the Axiom of Choice, that the converse is also true: If there is a mapping of B onto A, then IAI ~ IBI. 1.6 If there is a mapping of B onto A, then 2IAI ~ 21 8 1. [Hint: Given g mapping B onto A, let f(X) = g- 1 [X], for all X~ A.] 1.7 Use Cantor's Theorem to show that the "set of all sets" does not exist. 1.8 Let X be a set and let f be a one-to-one mapping of X into itself such that f[X] C X. Then X is infinite. Call a set X Dedekind infinite if there is a one-to-one mapping of X onto its proper subset. A Dedekind finite set is a set that is not Dedekind infinite. The remaining exercises investigate properties of Dedekind finite and Dedekind infinite sets. 1.9 Every countable set is Dedekind infinite. 1.10 if X contains a countable subset, then X is Dedekind infinite. 1.11 If X is Dedekind infinite, then it contains a countable subset. [Hint: Let x E X - f[X]; define Xo = x, X1 = f(xo), ... , Xn+l = f(xn), .... The set {xn I n E N} is countable.]

Thus Dedekind infinite sets are exactly those that have a countable subset. Later, using the Axiom of Choice, we show that every infinite set has a countable subset; thus Dedekind infinite =infinite. 1.12 If A and B are Dedekind finite, then AU B is Dedekind finite. [Hint:

Use Exercise 1.11.] 1.13 If A and B are Dedekind finite, then Ax B is Dedekind finite. [Hint: Use Exercise 1.11.] 1.14 If A is infinite, then P(P(A)) is Dedekind infinite. [Hint: For each n E N, let Sn = {XC A \IX\= n}. The set {Sn \ n E N} is a countable subset of P(P(A)).]

CHAPTER 5. CARDINAL NUMBERS

98

2.

The Cardinality of the Continuum

We the the (a)

are by now well acquainted with the properties of the cardinal number N0 • cardinality of countable sets. We summarize them here for reference. using concepts of cardinal arithmetic introducPd in the preceding section. "'< No if and only if"' EN. (b) n +No =No+ No= No (n EN). (c) n ·No = No· No = No (n E N. n > 0). (d) N0 = No (n E N, n > 0). In the present section we study the second most important infinite cardin recall that 2N" is indeed the cardinality of th!' set R of all real numbers. 2.1 Theorem

Proof.

IRI

= 2N".

This is the second half of Theorem 6.3 in Chapter 4.

[]

The next theorem summarizes arithmetic properties of the cardinality oft h(' continuum. 2.2 (a) (b) {c)

Theorem n + 2N" =No+ 2N" = 2N" + 2N" = 2N" (n EN). n · 2N" =No· 2N" = 2N" · 2N" = 2N" {n EN, n > 0). (2N"t = (2N"t" = nN" = N~" = 2N" {n EN, n > 0}.

Proof. (a) This follows from the obvious sequence of inequalities

by the Cantor-Bernstein Theorem. (b) Similarly, we have

(c) We have both and []

It is interesting to notice that Theorem 2.2, although an easy c:orollarv of laws of cardinal arithmetic and the Cantor-Bernstein Theorem, has some rathE'r unexpected consequences. For example, 2N" · 2N" = 2N" means that IR x Rl = IRI; however, the set R x R of all pairs of real numbers is in a one-to-one correspondence with the set of all points in the plane (via a cartesian coordinate system). Thus we see that there exists a one-to-one mapping of a straight line

2. THE CARDINALITY OF THE CONTINUUM

99

R onto a planeR x R (and similarly, onto a three-dimensional spaceR x R x R, etc.). These results (due to Cantor) astonished his contemporaries; they seem rather counterintuitive, and the reader may find it helpful to actually construct such a mapping (see Exercise 2.7). The next theorem shows that several important sets have the cardinality of the continuum. 2.3 (a) (b) (c) (d)

Theorem The set of all The set of all The set of all The set of all

points in the n-dimensional space R" has cardinality 2H". complex numbers has cardinality 2H". infinite sequences of natural numbers has cardinality 2H". infinite sequences of real numbers has cardinality 2H".

Proof. (a) IR"I = (2H")n by definition of cardinal exponentiation; (2H")n = 2H" by Theorem 2.2(c). (b) Complex numbers are represented by pairs of reals (see Exercise 2.6 in Chapter 10), so the cardinality of the set of all complex numbers is [RxR[ = (2H" )2 = 2Ho. (c) The set of all infinite sequences of natural numbers is NN and INN I = t{~" = 2Ho. (d) IRNI = (2H")H" = 2H". 0

The next theorem helps to establish further results of this sort. 2.4 Theorem If A is a countable subset of Band fBI

= 2N",

then IB-AI

= 2N".

(Here we remark that using the Axiom of Choice, we are able to show in general that if IAI < IB[, then IB- AI= IBI.) Proof

We can assume without loss of generality that B = R x R. R

/1 L/: --11-------....-..--t:::--R p Xo X

Let P = domA:

P = {x

E R I (x,

y) E A for some y}.

Since IAI = N0 , we have IPI ::S No. Thus there is xo E R such that xo t/. P. Consequently, the set X = {xo} x R is disjoint from A, so X <;;; (R x R) - A. Clearly, !XI = IRI = 2H", and we have !(R x R) -AI 2: 2H" 0

100

2.5 (a) (b) (c)

CHAPTER 5. CARDINAL NUMBERS

Theorem The set o/ all irrational numbers has cardinality 2N". The set of all infinite sets of natural numbers has cardinality 2N". The set of all one-to-one mappings of N onto N has cardinality 2N".

Proof (a) The set of all rationals Q is countable, hence the set R- Q of all irrational numbers has cardinality 2N" by Theorem 2.4. (b) The set of all subsets of N, P(N), has cardinality 2N". and the set of all finite subsets of N is countable (see Corollary 3.11 in Chapter 4), hence the set of all infinite subsets of N has the cardinality of the continuum. (c) Let P be the set of all one-to-one mappings of N onto N; as P ~ N N. clearly IPI ::; 2N". Let E and 0, respectively, be the sets of all even and odd natural numbers. If X ~ E is infinite, define a mapping f x : N _, N as follows: Jx(2k} =the kth element of X (kEN): Jx(2k

+ 1)

=the kth element of N- X (kEN).

Notice that N - X 2 0 is infinite, so f x is a one-to-one mapping of N onto N. Moreover, it is easy to show that X 1 I- X 2 implies / x, I- f x 2 . W!:' thus have a one-to-one correspondence between infinite subsets of E and certain elements of P. Since there are 2N" infinite subsets of E by Theorem 2.5(b), we get IPI 2: 2N" as needed. 0

Yet other similar results are provided by the next theorem. The definitions and basic properties of open sets and continuous functions can be found in Section 3, Chapter 10.

2.6 Theorem (a) The set of all continuous functions on R toR has cardmality 2N". (b) The set of all open sets of reals has cardinality 2N". Proof (a) We use the fact (proved in Theorem 3.11, Chapter 10) that every continuous function on R is determined by its values on a dense set, in particular by its values at rational arguments: If f and g are two continuous functions on R, and if /(q) = g(q) for every rational number q, then f = y. Tlms let C be the set of all continuous real-valued functions on R. Let F bE' a mapping of C into RQ defined by F(f) = / r Q. By the fact. above. F is one-to-one, so ICI :S IRQ!= (2N")N" = 2N". On the other hand. clearly ICI ::::=: 2N" (consider the constant functions). (b) Every open set is a union of a system of open intervals with rational endpoints (see Lemma 3.14 in Chapter 10). There are N0 open intervals with rational endpoints (each such interval is determined by an ordered pair of

2. THE CARDINALITY OF THE CONTINUUM

101

rationals), and hence 21-to such systems. This shows that there are at most 21-t" open sets. On the other hand, if a, bE R, a I- b, then (a, oo) I- (b, oo ), so there are at least 21-t" open sets.

0 The results of this section demonstrate the importance of the cardinal number 21-tn. It should not be surprising that the problem of determining the magnitude of 21-to is of fundamental significance. We know that 21-to is greater than N0 , but how much greater? Cantor conjectured that 21-t" is the next cardinal number after N0 . This is the famous Continuum Hypothesis. The Continuum Hypothesis

that

K-

There is no uncountable cardinal number "' such

< 21-to.

In words, the Continuum Hypothesis asserts that every set of real numbers is either finite or countable, or else it is equipotent to the set of all real numbers. There are no cardinalities in between. In 1900, David Hilbert included the Continuum Problem in his famous list of open problems in mathematics (as Problem 1 ). It is still not fully resolved today. In 1939, Kurt Gi:idel showed that the Continuum Hypothesis is consistent with the axioms of set theory. That is, using the axioms of Zermelo-Fraenkel set theory (including the Axiom of Choice), one cannot refute the Continuum Hypothesis. In 1963, Paul Cohen proved that the Continuum Hypothesis is independent of the axioms. This means that one cannot prove the Continuum Hypothesis from the axioms. Wf' discuss these questions in more detail in Chapter 15. We conclude this section with an example of a set that has cardinality greater than the continuum.

2. 7 Lemma The set of all real-valued functions on real numbers has cardinality 22"" > 21-tn.

Exercises

2.1 Prove that the set of all finite sets of reals has cardinality 2 No. We remark here that the set of all countable sets of reals also has cardinality 21-t", but the proof of this requires the Axiom of Choice. 2.2 A real number x is algebraic if it is a solution of some equation

(*) where ao, ... , an are integers. If x is not algebraic, it is called transcendental. Show that the set of all algebraic numbers is countable and hence the set of all transcendental numbers has cardinality 21-to. 2.3 If a linearly ordered set P has a countable dense subset, then !PI :::; 21-t". 2.4 The set of all closed subsets of reals has cardinality 21-tn.

102

CHAPTER 5. CARDINAL NUMBERS

2.5 Show that, for n > 0, n · 22 "" = No · 22"" (22"" )n = (22"" )N" == (22"" )2" 11 = 22"".

= 2N" . 22"" = 22""

. 22 "" =

2.6 The cardinality of the set of all discontinuous functions is 22 "". [Hint: Using Exercise 2.5, show that [RR- C[ = 22 "" whenever JCI::; 2N".) 2.7 Construct a one-to-one mapping of R x R onto R. [Hint: If a, bE [0. 11 have decimal expansions O.a1a2a3 · · · and O.b 1 b'2b3 · · ·. map the ordPn'd pair (a, b) onto O.a1b1a2b2a3b3 · · · E [0, 1]. Make adjustments to avoid sequences where the digit 9 appears from some place onward.)

Chapter 6

Ordinal Numbers 1.

Well-Ordered Sets

When we introduced natural numbers, we were motivated by the need to formalize the process of "counting": the natural numbers start with 0 and are generated by successively increasing the number by one unit: 0, 1, 2, 3 .... , and so on. We defined the operation of successor by S(.r) =xU {x} and introduced natural numbers as elements of the smallest set containing 0 and closed under S. It is desirable to be able to continue the process of counting beyond natural numbers. The idea is that we can imagine an infinite number w that conws "after" all natural numbers and then continue the counting process into the transfinite: w, w + 1, (w + 1) + 1, and so on. In the present chapter we formalize the process of transfinite counting, and introduce ordinal numbers as a generalization of natural numbers. As is usual in any meaningful generalization, the resulting concept shares many features of natural numbers. Most important, the theorems on induction and recursion are generalized to theorems on transfinite induction and transfinite recursion. As a starting point, we use the fact that each natural number is identified with the set of all smaller natural numbers: n = {m E N I m < 11}. (See Exercise 2.6 in Chapter 3.) By analogy, we let w, the least transfinitE' numbl'r. to be the set N of all natural numbers: w = N = {0, l. 2, ... }. It is easy to continue the process after this "limit" step is made: The operation of successor can be used to produce numbers following w in the same way we used it to produce numbers following 0:

S(w) =wU{w} = {0,1,2, ... ,w}, S(S(w)) = S(w) U {S(w)} = {0, 1,2, ... ,w, S(w)},

etc.

We use the suggestive notation

S(w)=w+1,

S(S(w))=(w+l)+l=w+2, 103

etc.

104

CHAPTER 6. ORDINAL NUMBERS

In this fashion, we can generate greater and greater "numbers": w, w + 1. w + 2, ... , w + n, ... , for all n EN. A number following all w + n can again

be conceived of as a set of all smaller numbers: w · 2 = w + w = {0,1,2, ... ,w,w + 1,w + 2, ... }.

The reader may wish to introduce still greater numbers; e.g., w ·2+ 1 = w

+ w + 1 = {0,1, 2, ... ,w,w + 1,w +

2, ... ,w + w},

w·3=w+w+w = {0, 1,2, ... ,w,w + 1,w + 2, ... ,w + w,w + w + 1, ... }, w·w= {0,1,2, ... ,w,w+1, ... ,w·2,w·2+1, ... ,w·3, ... ,w·4 .... }.

The sets we generate behave very much like natural numbers, in this respect: they are linearly ordered byE, and every nonempty subset has a least element. We called linear orderings with this property well-orderings (see Definition 2.3 in Chapter 3). As the concept of well-ordering is, along with cardinality, one of the most important ideas in abstract set theory, let us recall the definition:

1.1 Definition A set W is well-ordered by the relation < if (a) (W, <) is a linearly ordered set. (b) Every nonempty subset of W has a least element. The sets generated above are all examples of sets well-ordered by the relation E. Later in this chapter we show that all well-orderings can be represented by such sets and we introduce ordinal numbers to serve as order types of wellordered sets. More examples of well-ordered sets can be found in the exercises at the end of this section. The fundamental property of well-orderings is that they can be compared by their "lengths." The precise meaning of this is given in Theorem 1.3. Let (L, <) be a linearly ordered set. A setS~ Lis called an initial segment of L if S is a proper subset of L (i.e., S "I= L) and if for every a E S, all x < a are also elements of S. (For instance, both the set of all negative reals and the set of all non positive reals are initial segments of the set of all real numbers.) 1.2 Lemma If (W, <) is a well-ordered set and if S is an initial segment of ( W, <), then there exists a E W such that S = { x E W I x < a} . Proof. Let X = W- S be the complement of S. AsS is a proper subset of W, X is nonempty, and so it has a least element in the well-ordering <. Let a be the least element of X. If x < a, then x cannot belong to X, as a is its least member, sox belongs to S. If x :2: a, then x cannot be inS because otherwise a would also be in S as S is an initial segment. Thus S = { x E W I x < a}. [] If a is an element of a well-ordered set (W, <), we call the set W[a] = {x E

WI x


1. WELL-ORDEREDSETS

105

the initial segment of W given by a. (Note that if a is the least element of W, then W[a] is empty.) By Lemma 1.2, each initial segment of a well-ordered set is of the form W[a] for some a E W. Of course, W[a] is also well-ordered by < (more precisely, by < nW[a] 2 ); we usually do not mention the well-ordering relation explicitly when it is understood from the context.

1.3 Theorem lf(WI, <1) and (W2, <2) are well-ordered sets, then exactly one of the following holds: (a) either W 1 and W2 are isomorphic, or (b) W 1 is isomorphic to an initial segment of W 2, or (c) W 2 is isomorphic to an initial segment of W 1 • In each case, the isomorphism is unique. This theorem provides the method of comparison for well-orderings mentioned above: we say that wl has smaller order type than w2 if WI is isomorphic to W2 [a] for some a E W 2 . Before we prove Theorem 1.3, we prove the following lemma and state some of its corollaries. A function f on a linearly ordered set (L, <)into Lis increasing if x 1 < x 2 implies f(xi) < f(x 2). Note that an increasing function is one-to-one, and is an isomorphism of (L, <)and (ran/,<).

1.4 Lemma If (W, <) is a well-ordered set and iff : W ---. W is an increasing function, then f(x) ~ x for all x E W. If the set X = { x E W I f ( x) < x} is nonempty, it has a least Proof. element a. But then f(a)
1.5 (a) (b) (c)

Corollary No well-ordered set is isomorphic to an initial segment of itself. Each well-ordered set has only one automorphism, the identity. If W 1 and W2 are isomorphic well-ordered sets, then the isomorphism between WI and w2 is unique.

Proof. (a) Assume that f is an isomorphism between W and W[a] for some a E W. Then f(a) E W[a] and therefore f(a) < a, contrary to the lemma, as f is an increasing function. (b) Let f be an automorphism of W. Both f and f- 1 are increasing functions and so for all x E W, f(x) ~ x and f- 1 (x) ~ x, therefore x ~ f(.r). It follows that f(x) = x for all x E W. (c) Let f and g be isomorphisms between W 1 and W2. Then f o g- 1 is an automorphism of W2 and hence is the identity map. It follows that f =g. 0

Let W1 and W2 be well-ordered sets. It is a 1.6 Proof of Theorem 1.3 consequence of Lemma 1.4 that the three cases (a), (b), and (c) are mutually

106

CHAPTER 6. ORDINAL NUMBERS

exclusive: For example, if W1 were isomorphic to W2[a2] for some a 2 E W2 and at the same time Wz were isomorphic to Wi[a 1 ] for some a 1 E W 1 , then the composition of the two isomorphisms would be an isomorphism of a well-ordered set onto its own initial segment. Also, the uniqueness of the isomorphism in each case follows from Corollary 1.5; thus, we only have to show that one of the three case~ (a). (b). and (c) always holds. We define a set of pairs f t;;; W 1 x W 2 and show that either f or j- 1 is an isomorphism attesting to (a), (b), or (c). Let

f

=

{(x,y) E

w, X w2

I Wi[x] is isomorphic to W2[y]}.

First, it follows from Corollary 1.5 (a) that f is a one-to-one function: If W 1 [x] i~ isomorphic both to W2[y] and to W2[y'], then y = y' because otherwise W2 [y] would be an initial segment of W 2 [y'] (or vice versa) while they are isomorphic. and that is impossible. Hence (x, y) E f and (x, y') E f imply y = y'. A similar argument shows that (x, y) E f and (x', y) E f imply x = x'. Second, x < x' implies f(x) < f(x'): If his the isomorphism between W 1 [.T'] and W 2 [f(x')], then the restriction h f W 1 [x] is an isomorphism between lV1 f:r] and Wz[h(x)], so f(x) = h(x) and f(x) < f(x'). Hence f is an isomorphism between its domain, a subset of W 1 , and ib range, a subset of W2 . If the domain off is W 1 and the range off is W2 • then W 1 is isomorphic to W2 . We show now that if the domain off is not all of M' 1 then it is its initial segment, and the range off is all of W2 . (This is mough to complete the proof as the remaining case is obtained by interchanging the role of W 1 and W2.) So assume that dom f -1- W,. We note that the set S = dom f is an initial segment of W 1 : If x E S and z < x, let h be the isomorphism between W 1 [x] and W2 [J(x)]; then h f WI[z] is an isomorphism between Wi[z] and W2 [h(.::)j. so z E 5. To show that the set T = ran f = W2 , we assume otherwise and, by a similar argument as above, show that Tis an initial segment -of W2 . But then domf = Wi[a] for some a E W,, and ranf = W 2 [b] for some bE Wz. In other words, f is an isomorphism between Wi[a] and Wz[b]. This means. by the definition of J, that (a,b) E /,so a E domf = Wi[a], that is, a< a. a contradiction. C Exercises

1.1 Give an example of a linearly ordered set (L, <) and an initial ~egment S of L which is not of the form {xI x
2. ORDINAL NUMBERS

107

1.7 Let (W, <) be a well-ordered set, and let a f/: W. Extend < to vF' = W U {a} by making a greater than all x E W. Then W has smaller order type than W'. 1.8 The sets W = N x {0, 1} and W' = {0, 1} x N. ordered lexicographically. are nonisomorphic well-ordered sets. (See the remark following Theon'm 4.7 in Chapter 4.)

2.

Ordinal Numbers

In Chapter 3 we introduced the natural numbers to represent both the cardinality and the order type of finite sets and used them to prove theorems on induction and recursion. We now generalize this definition by introducing ordinal numbers. Ordinal numbers continue the procedure of generating larger numbers into transfinite. As was the ca.se with natural numbers. ordinal numbers are defined in such a way that each is well-ordered by the E relation. Moreover. the collection of all ordinal numbers (which as we shall see is not a set) is itself wellordered byE, and contains the natural numbers as an initial segment. And most significantly, ordinal numbers are representatives for all well-ordered sets: every well-ordered set is isomorphic to an ordinal number. Thus ordinal numbers can be viewed as order types of well-ordered sets. 2.1 Definition A set Tis transitive if every element ofT is a subset ofT. In other words, a transitive set has the property that u E v E T implies u E T. For more on transitive sets we refer the reader to the exercises in this section, and to Chapter 14. 2.2 Definition A set a is an ordinal number if (a) a is transitive. (b) a is well-ordered by Eo:· It is a standard practice to use lowercase Greek letters to denote ordinal numbers. Also, the term ordinal is often used for ordinal number. For every natural number m, if k E l E m (i.e., k < l < m), then k E m. Hence every natural number is a transitive set. Also, every natural numbPr is well-ordered by the E relation (because every n E N is a subset of N and N is well-ordered by E). Thus 2.3 Theorem Every natural number is an ordinal. The set N of all natural numbers is easily seen to be transitive, and is also well-ordered by E. Thus N is an ordinal number. 2.4 Definition w = N. We have just given the set N a new name w.

108

CHAPTER 6. ORDINAL NUMBERS

2.5 Lemma If a is an ordinal number, then S(a) is also an ordinal number. S(a) = aU{a} is a transitive set. Moreover, au{a} is well-ordered Proof. by E, o being its greatest element, and a C a U {a} being the initial segment given by a. So S(o) is an ordinal number. 0 We denote the successor of a by a + 1: a+l=S(a)=aU{a}. An ordinal number a is called a successor ordinal if a = (3 + 1 for some (3. Otherwise, it is called a limit ordinal. For all ordinals o and (3, we define a < (3 if and only if o E {3, thus extending the definition of ordering of natural numbers from Chapter 3. The next theorem shows that < does indeed have all the properties of a linear ordering, in fact, of a well-ordering. Theorem Let a, (3, and r be ordinal numbers. If a< (3 and {3 < ,, then a< r· a < {3 and (3 < a cannot both hold. Either a < {3 or a= {3 or (3
2.6 (a) (b) (c) (d)

The proof uses the following lemmas. 2.7 Lemma If a is an ordinal number, then a

f/: a.

None of the axioms of set theory we have considered excludes the existence of sets X such that X E X. However, the sets which arise in mathematical practice do not have this peculiar property. If a Eo then the linearly ordered set (a, E 0 Proof. such that x E x, contrary to asymmetry of Eo:·

)

has an element x = n 0

2.8 Lemma Every element of an ordinal number is an ordinal number. Proof. Let o be an ordinal and let x E a. First we prove that x is transitive. Let u and v be such that u E v E x; we wish to show that u E :r. Since a is transitive and x E a, we have v E a and therefore, also u E a. Thu~ u, v, and x are all elements of a and u E v E x. Since Eo linearly orders a, we conclude that u E x. Second, we prove that Ex is a well-ordering of x. But by transitivity of a we have x ~ a and therefore, the relation Ex is a restriction of the relation E". Since Eo: is a well-ordering, so is Ex· []

2. ORDINAL NUMBERS

109

2.9 Lemma If a and {3 are ordinal numbers such that a C (3, then a E {3.

Proof. Let a C (3. Then {3- a is a nonempty subset of {3, and hence has a least element 1 in the ordering Ea. Notice that 1 ~ a: If not, then any E 1 - a would be an element of {3- a smaller than 1 (by transitivity of (3). The proof is complete if we show that a ~ r (and so a = 1 E (3). Let o E a; we show that o E I· If not, then either 1 E o or 1 = o (both o and 1 belong to (3, which is linearly ordered by E). But this implies that 1 E a, since a is transitive. That contradicts the choice of 1 E {3- a. 0

o

Proof of Theorem 2.6 (a) If a E (3 and (3 E ,, then a E 1 because 1 is transitive. (b) Assume that a E {3 and {3 E a. By transitivity, a E a, contradicting Lemma 2.7. (c) If a and (3 are ordinals, an (3 is also an ordinal (check properties (a) and (b) from the definition) and an {3 ~ a, an (3 ~ {3. If an {3 = a, then a ~ {3 and so a E {3 or a = {3 by Lemma 2.9. Similarly, an (3 = (3 implies (3 E a or {3 =a. The only remaining case, anf] C a and anf] c {3, cannot occur, because then an {3 E a and an (3 E (3, leading to an (3 E an (3 and a contradiction with Lemma 2.7. (d) Let A be a nonempty set of ordinals. Take a E A, and consider the set an A. If an A = 0, a is the least element of A. If an A i= 0, an A ~ a has a least element {3 in the ordering E 0 . Then (3 is the least element of A in the ordering <. (e) Let X be a set of ordinal numbers. Since all elements of X are transitive sets, U X is also a transitive set (see Exercise 2.5). It immediately follows from part (d) of this theorem that E well-orders U X; consequently, U X is an ordinal number. Now let a = S(U X); a is an ordinal number and a f/. X. [Otherwise, we get a ~ U X and, by Lemma 2.9, either a = U X or a E UX. In both cases, a E S(UX) =a, contradicting Lemma 2.7.]

0 The ordinal number U X used in the proof of (e) is called the sup rem urn of X and is denoted sup X. This is justified by observing that U X is the least ordinal greater than or equal to all elements of X: (a) If a E X, then a ~ U X, so a $ U X. (b) If a :S 1 for all a E X, then a ~ r for all a E X and so U X ~ I· i.e.,

ux :S /·

If the set X has a greatest element {3 in the ordering <, then sup X = {3. Otherwise, supX > 1 for all1 E X (and it is the least such ordinal). We see that every set of ordinals has a supremum (in the ordering <). The last theorem of this section restates the fact that ordinals are a generalization of the natural numbers:

llO

CHAPTER 6. ORDINAL NUMBERS

2.10 Theorem The natuml numbers are exactly the finite ordinal number·s.

Proof. We already know (from Theorem 2.3) that every natural number is an ordinal, and of course, every natural number is a finite set. So we only have to prove that all ordinals that are not natural numbers are infinite set~. If a is an ordinal and a f/: N, then by Theorem 2.6(b) it must be the case that n ~ w (because a f. w), so a 2 w because n is transitive. Son lms an inhnitt> subset and hence is infinite. 0 Every ordinal is a well-ordered set, under the well-ordering E. If ct and .J are distinct ordinals, then they are not isomorphic as well-ordered sets because one is an initial segment of the other. We also prove that any well-ordered set is isomorphic to an ordinal number. This, however. requires an introduction of another axiom, and is done in the next section. One final comment. Lemma 2.8 establishes that each ordinal number n has the property that a=

{.8 I {J is an ordinal and {J
If we view a as a set of ordinals, then if a is a successor, say {J + 1, then it has a greatest element, namely {3. If a is a limit ordinal. then it does not haw a greatest element, and a= sup{/3 I {3
2.1 A set X is transitive if and only if X <;;; P(X). 2.2 A set X is transitive if and only if U X ~ X. 2.3 Are the following sets transitive? (a) {0,{0},{{0}}}, (b) {0,{0},{{0}},{0,{0}}}, (c) {0,{{0}}}. 2.4 Which of the following statements are true? (a) If X andY are transitive, then XU Y is transitive. (b) If X and Yare transitive, then X n Y is transitive. (c) If X E Y andY is transitive, then X is transitive. (d) If X<;;; Y andY is transitive, then X is transitive. (e) If Y is transitive and S ~ P(Y), then Y US is transitive. 2.5 If every X E S is transitive, then US is transitive. 2.6 An ordinal a is a natural number if and only if every nonempty suhset of a has a greatest element. 2.7 If a set of ordinals X does not have a greatest element, then sup X is a limit ordinaL 2.8 If X is a nonempty set of ordinals, then X is an ordinal. ~1owover. X is the least element of X.

n

n

3. THE AXIOM OF REPLACEMENT

3.

111

The Axiom of Replacement

As we indicated at the end of Section 2, well-ordered sets can be represented by ordinal numbers. The following theorem states precisely what we mean by "representation." 3.1 Theorem Every well-ordered set is isomorphic to a unique ordinal number. We give a proof of this theorem below. Although the reader may find the proof entirely acceptable, it nevertheless has a deficiency: it uses an assumption which, however plausible, does not follow from the axioms that we haw introduced so far. For that reason it is necessary to introduce an additional axiom. Proof. Let (W, <) be a well-ordered set. Let A be the set of all those a E W for which W[a] is isomorphic to some ordinal number. As no two distinct ordinals can be isomorphic (one is an initial segment of the other). this ordinal number is uniquely determined, and we denote it by a a. Now suppose that there exists a set S such that S = {O:'a I a E A}. The set S is well-ordered by E as it is a set of ordinals. It is also transitive. because if r E aa E S, let 'P be the isomorphism between W[a] and aa and let c = I.{J- 1 (r); it is easy to see that 'P f cis an isomorphism between W[c] and r and so 1 E S. Therefore, Sis an ordinal number, S =a. A similar argument shows that a E A, b < a imply b E A: let 'P be the isomorphism of W[a] and aa. Then 'P f W[b] is an isomorphism of W[b] and an initial segment I of aa. By Lemma 1.2, there exists f] < aa such that I = {r E O:'a I r < ,8} = f]; i.e., f] = O:'b. This shows that b E A and ab < 0' 0 . We conclude that either A = W or A = W[c] for some c E W (Lemma 1.2 again). We now define a function f: A-+ S =a by f(a) = aa. From the definition of S and the fact that b < a implies ab < aa it is obvious that j is an isomorphism of (A,<) and a. If A= W[c], we would thus have c E A. a contradiction. Therefore A= W, and f is an isomorphism of (W, <) and the ordinal a. This would complete the proof of Theorem 3.1, if we were justified to makP the assumption that the set S exists. The elements of S are certainly well specified, but there is no reason why the existence of such a set should formally follow from the axioms that were considered so far. To further illustrate the problem involved, considPr two other PX
(0, {0}, { {0} }, { { {0}} }, ... ). we might define

ao

= 0,

an+l ={an}

for all n EN,

CHAPTER 6. ORDINAL NUMBERS

112

following the general pattern of recursive definitions. The difficulty here is that to apply the Recursion Theorem we need a set A, given in advance, such that g: N x A--+ A defined by g(n,x) = {x} can be used to compute the (n + 1)st term of the sequence from its nth term. But it is not obvious how to prove from our axioms that any set A such that

0 E A,

{0} E A,

{{0}} E A,

{{{0}}} E A, ...

exists. It seems as if the definition of A itself required recursion. Let us consider another example. In Chapter 3, we have postulated existence of w; from it, the sets w + 1 = wu {w}, w+ 2 = (w + l)U {w + 1 }, etc., can easily be obtained by repeated use of operations union and unordered pair. In the introductory remarks in Section 1, we "defined" w + w as the union of w and the set of all w + n for all n E w, and passed over the question of existence of this set. Although intuitively it is hardly any more questionable than the existence of w, the existence of w + w cannot be proved from the axioms we accepted so far. We know that, to each nEw, there corresponds a unique set w + n; as yet. we do not have any axiom that would allow us to collect all these w + n into one set. The next axiom schema removes this shortcoming. The Axiom Schema of Replacement Let P(x, y) be a property such that for every x there is a unique y for which P(x, y) holds. For every set A, there is a set B such that, for every x E A, there is y E B for which P(x, y) holds.

We hope that the following comments provide additional motivation for the schema. 3.2 Let F be the operation defined by the property P; that is, let F(x) denote the unique y for which P(x, y). (See Section 2 of Chapter 1.) The corresponding Axiom of Replacement can then be stated as follows: For every set A there is a set B such that for all x E A, F(x) E B. Of course, B may also contain elements not of the form F(x) for any x E A: however, an application of the Axiom Schema of Comprehension shows that

{y E B

Iy

I P(x,y) holds for some x E A} {y I P(x,y) holds for some x E A}

= F(x) for some x E A}= {y E B =

exists. We call this set the image of A by F and denote it { F( x) simply F[A].

Ix

E A} or

3.3 Further intuitive justification for the Axiom Schema of Replacement can be given by comparing it with the Axiom Schema of Comprehension. The latter allows us to go through elements of a given set A, check for each x E A whether or not it has the property P(x), and collect those x which do into a set. In

3. THE AXIOM OF REPLACEMENT

113

an entirely analogous way, the Axiom Schema of Replacement allows us to go through elements of A, take for each x E A the corresponding unique y having the property P(x, y), and collect all such y into a set. It is intuitively obvious that the set F[A] is "no larger than" the set A. In contrast, all known examples of "paradoxical sets" are "large," on the order of the "set of all sets." 3.4 Let F be again the operation defined by P, as in (3.2). The Axiom of Replacement then implies that the operation F on elements of a given set A can be represented, "replaced," by a function, i.e., a set of ordered pairs. Precisely: For every set A, there is a function f(x) = F(x) for all x EA.

f such that dom f

= A and

We simply let f = { (x, y) E Ax B I P(x, y)}, where B is the set provided by the Axiom of Replacement. We use notation F f A for this uniquely determined function f. Notice that ran(F fA) = F[AJ. We can now complete the proof of Theorem 3.1: We have concluded earlier that in order to prove the theorem, we only have to guarantee the existence of the setS= {aa I a E W}, where for each a E W, aa is the unique ordinal number isomorphic to W[aj. Let P(x, y) be the property: Either x E Wand y is the unique ordinal isomorphic to W[x], or x

f/: W and

y =

0.

Applying the Axiom of Replacement [with P(x, y) as above] we conclude that (for A = W) there exists a set B such that for all a E W there is a E B for which P(a, a) holds. Then we let S ={a E B I P(a,a) holds for some a E W} where F is the operation defined by P.

= F[WJ 0

Using Theorem 3.1, we have: 3.5 Definition If W is a well-ordered set, then the order type of W is the unique ordinal number isomorphic toW. And what about the examples mentioned above? It turns out that we need a more general Recursion Theorem. Compare the following theorem with the Recursion Theorem proved in Chapter 3: 3.6 The Recursion Theorem Let G be an operation. For any set a there is a unique infinite sequence (an I n E N) such that (a) ao =a. (b) an+ I = G(an, n) for all n EN.

CHAPTER 6. ORDINAL NUMBERS

114

With this theorem, the existence of the sequence (0, {0}, { {0} }.... ) and of follows (see Exercise 3.2). We prove this Recursion Theorem, as well
w

+w

Exercises 3.1 Let P(x, y) be a property such that for every x there is at most one y for which P(x, y) holds. Then for every set A there is a set B such that. for all x E A, if P(x, y) holds for some y, then P(x. y) holds tor sonw y E B. 3.2 Use Theorem 3.6 to prove the existence of (a) The set {0, {0}, {{0}}, {{{0} }}, ... }. (b) The set {N, P(N), P(P(N)), ... }. (c) The set w + w = w U { w. w + 1, (w + 1) + 1.... }. 3.3 Use Theorem 3.6 to define Vo = 0; Vn+l = P(Vn) Vw =

(nEw);

U Vn. nEw

3.4 (a) Every x E Vw is finite. (b) Vw is transitive. (c) Vw is an inductive set. The elements of Vw are called hereditarily finite sets. 3.5 (a) If x E Vw andy E Vw, then {x, y} E Vw. (b) If X E Vw, then UX E Vw and P(X) E Vw. (c) If A E Vw and f is a function on A such that f(x) E Vw for each x E A, then f[X) E Vw. (d) If X is a finite subset of Vw. then X E Vw.

4.

Transfinite Induction and Recursion

The Induction Principle and the Recursion Theorem are the main tools for proving theorems about natural numbers and for constructing functions with domain N. We used them both extensively in the previous chapters. In this section, we show how these results generalize to ordinal numbers. Let P(x) be a property (possibly 4.1 The Transfinite Induction Principle with parameters). Assume that, for all ordinal numbers o.: (4.2)

IJP(/3) holds for all /3 < o., then P(o.).

Then P(o.) holds for all ordinals o..

Suppose that some ordinal number 1 fails to have property P, and Proof. let S be the set of all ordinal numbers /3 :S 1 that do not have property P. The

4. TRANSFINITE INDUCTION AND RECURSION set S has a least element o. Since every {3 (4.2) that P(o) holds. a contradiction.

<

Cl'

115

has property P. it follows

b~·

'

It is sometimes convenient to use the Transfinite Induction Principle in a form which resembles more closely the usual formulation of the Induction Principle for N. We do it by treating successor and limit ordinals separately. 4.3 The Second Version of the Transfinite Induction Principle Lr.t P(x) be a property. Assume that (a) P(O) holds. (b) P(a) implies P(a + 1) for all ordinals a. (c) For all limit ordinals a =J 0, ifP(/3) holds for all /3
Proof. It suffices to show that the assumptions (a). (b), and (c) imply (4.2). So let a be an ordinal such that P(/3) for all 13 < o. If a = 0, then P( o) holds by (a). If a is a successor, i.e., if there is 13 < ct such that o = ;3 + 1. WP know that P(/3) holds, so P(o) holds by (b). If a =J 0 is limit, Wf' haw P(a) by (c). D We proceed to generalize the Recursion Theorem. Functions whose domain is an ordinal a are called transfinite sequences of length a. 4.4 Theorem Let n be an ordinal number, A a set, and S = Uad! A" the set of all transfinite sequences of elements of A of length less than 0. Let g : S --+ A be a function. Then there exists a unique function f : n --+ A such that

f(a) = g(f

r a:)

for all a

< n.

The reader might try to prove this theorem in a way entirely analogous to the proof of the Recursion Theorem in Chapter 3. We do not go into the details since this theorem follows from the subsequent more general TransfinitP Recursion Theorem. If 1J is an ordinal and f is a transfinite sequence of length !J, we use the notation f = (ao I a < !J). Theorem 4.4 states that if g is a function on the set of all transfinite sequences of elements of A of length less than with values in A. then there is a transfinitP < o) ). sequence (ao a < n) such that for all a < n, aa = g( (a~

n

1

1 (

4.5 The Transfinite Recursion Theorem Let G be an operation: then the property P stated in (4.6} defines an operation F such that F(o) = G(F r o) for all ordinals a.

Proof. We call t a computation of length a based on G if t is a function. domt =a+ 1 and for all {3::::; o, t(/3) = G(t r {3).

CHAPTER 6. ORDINAL NUMBERS

116

Let P(x, y) be the property (4.6)

{

x is an ordinal number and y

= t(x)

for some computation t of length x

based on G, or x is not an ordinal number and y == 0.

We prove first that P defines an operation. We have to show that for each x there is a unique y such that P(x, y). This is obvious if x is not an ordinal. To prove it for ordinals, it suffices to show. by transfinite induction: For every ordinal o there is a unique computation of length o. We make the inductive assumption that for all /3 < o there is a unique computation of length /3, and endeavor to prove the existence and uniqueness of a computation of length o. Existence: According to the Axiom Schema of Replacement applied to the property "y is a computation of length x" and the set o, there is a set T == {t I t is a computation of length

/3 for some f3 < o}.

Moreover, the inductive assumption implies that for every /3 < o there is a unique t E T such that the length oft is /3. T is a system of functions; let I == UT. Finally, let r = I U { ( o, G(t))}. WP prove that r is a computation of length o. 4. 7 Claim r is a function and dom r

= o + 1.

We see immediately that domt == UteTdomt == U/Jea(/3 + 1) = o; consequently, dom r = dom l U { o} = o + 1. Next, since o ¢ dom t, it is enough to prove that l is a function. This follows from the fact that T is a compatible system of functions. Indeed, let t 1 and t 2 E T be arbitrary, and let dom t 1 == {3 1, dom t 2 == !32 . Assume that, e.g., /3 1 $ /32; then /3 1 ~ /J2, and it suffices to show that t 1(-y) = t 2('y) for all 'Y < {3 1. We do that by transfinite induction. So assume that 'Y < /J1 and t1(8) = t2(8) for all 8 <"f. Then t1 f 1 = t2 f /, and we haw t 1('y) = G(t1 f 'Y) = G(t2 f 'Y) = t2(1). We conclude that t1(1) = t2(-y) for all 'Y < {3 1. This completes the proof of Claim 4. 7. 4.8 Claim r(/3)

= G(r f /3)

for all /3 $ o.

This is clear if /3 = o, as r(o) = G(t) = G(r f o). If {3 < o, pick t E T such that {3 E domt. We have r(/3) = t({3) = G(t f /3) = G(r f {3) because tis a computation, and t ~ r. Claims 4.7 and 4.8 together prove that r is a computation of length o. Uniqueness: Let (]" be another computation of length o; we prove r = (]". As + 1 = dom (]", it suffices to prove by transfinite induction that r('y) = (J"('y) for all 'Y $ o.

r and (]" are functions and dom r = o

4. TRANSFINITE INDUCTION AND RECURSION Assum~ that r(8) = tJ(8) for all tJ('Y). The assertion follows.

o< "(.

Then r('Y)

= G(r

117

f -y)

= G(tJ

f -y) =

This concludes the proof that the property P defines an operation F. Notice that for any computation t, F f dom t = t. This is because for any /3 E dom t. t 6 = t f (.6 + 1) is obviously a computation of length ,6, and so, by the definition ofF, F(/3) = t 6 (j3) = t(f3). To prove that F(a) = G(F fa) for all a, let t be the unique computation of length a; we have F(a) = t(a) = G(t fa) = G(F fa). 0 We need again a parametric version of the Transfinite Recursion Theorem. If F(z,x) is an operation in two variables, we write Fz(x) in place of F(z,x). Notice that, for any fixed z, F z is an operation in one variable. If F is defined by Q(z, x, y), the notations F z[A] and F z fA thus have meaning:

Fz[A] = {y I Q(z,x,y) for some x E A}; Fz fA= {(x,y) I Q(z,x,y) for some x E A}. We can now state a parametric version of Theorem 4.5.

4.9 The Transfinite Recursion Theorem, Parametric Version Let G be an operation. The property Q stated in (4.10} defines an operation F such that F(z,a) = G(z,Fz fa) for all ordinals a and all sets z. Proof Call t a computation of length a based on G and z if t is a function, dom t =a+ 1, and, for all {3::; a, t({3) = G(z, t f ,6). Let Q(z, x, y) be the property

(4.10) xis an ordinal number and y {

= t(x)

based on G and z, or x is not an ordinal number and y

for some computation t of length x

= 0.

Then carry z as a parameter through the rest of the proof in an obvious way. 0 It is sometimes necessary to distinguish between successor ordinals and limit ordinals in our constructions. It is convenient to reformulate the Transfinite Recursion Theorem with this distinction in mind.

4.11 Theorem Let G 1 , G2, and G 3 be operations, and let G be the operation defined in (4 .12} below. Then the property P stated in (4. 6) (based on G) defines an operation F such that

F(a

F(O)

= Gl(0),

+ 1)

= G2(F(a))

F(a) = G3(F fa)

for all a, for all limit a =J 0.

CHAPTER 6. ORDINAL NUMBERS

118

Proof

Define an operation G by

(4.12) G(x) = y if and only if either

(a) (b) (c) (d)

x = 0 andy= GI(0), or xis a function, domx = o + 1 for an ordinal o andy= G 2 (x(o)). or xis a function, domx = n for a limit ordinal o of. 0 andy= G 3 (.r). or xis none of the above andy= 0.

Let P be the property stated in (4.6) in the proof of the Transfinite Recursion Theorem (based on G). The operation F defined by P then satisfies F(o) = G(F f o) for all o. Using our definition of G, it is easy to verify that F has the required properties. 0 A parametric version of Theorem 4.11 is straightforward and we leave it to the reader. We conclude this section with the proofs of Theorems 3.6 and 4.4.

Proof of Theorem 3.6. Let G be an operation. We want to find. for ewrv set a, a sequence (an I n E w) such that ao = a and an+! = G(an. n) for all nE N. By the parametric version of the Transfinite Recursion Theorem 4.11. therP is an operation F such that F(O) =a and F(n+ 1) = G(F(n), n) for all n E N. Now we apply the Axiom of Replacement: There exists a sequence (an In E w) that is equal to F f w. and the Theorem follows. C Proof of Theorem 4.4.

Define an operation G by G(t) = {g(t)

0

if t E S.; otherwise.

The Transfinite Recursion Theorem provides an operation F such that F(o) = G(F f o) holds for all ordinals o. Let f = F f 0. -·' Exercises

4.1 Prove a more general Transfinite Recursion Theorem (Double Recursion Theorem): Let G be an operation in two variables. Then there is an operation F such that F({3, o) = G(F f ({3 x o)) for all ordinals 3 and o. [Hint: Computations are functions on ({3 + 1) x (o + 1).] 4.2 Using the Recursion Theorem 4.9 show that there is a binary operation F such that (a) F(x, 1) = 0 for all x. (b) F(x, n + 1) = 0 if and only if there exist y and z such that x = (y. z) and F(y, n) = 0. We say that x is an n-tuple (where n E w, n > 0) if F(:c, n) = 0. Prow that this definition of n-tuples coincides with the one given in Exercise 5.17 in Chapter 3.

5. ORDINAL ARITHMETIC

5.

119

Ordinal Arithmetic

We now use the Transfinite Recursion Theorem of the preceding section to define addition, multiplication, and exponentiation of ordinal numbers. These definitions are straightforward generalizations of the corresponding definitions for natural numbers. 5.1 Definition- Addition of Ordinal Numbers (a) {3 + 0 = {3. {b) {3 + (a + 1) = ({3 + a) + 1 for all a. {c) {3 +a = sup{{J + 'Y I 'Y < a} for all limit a -1- 0.

For all ordinals {3,

If we let a = 0 in (b), we have the equality {3 + 1 = {3 + 1; the left-hand side denotes the sum of ordinal numbers {3 and 1 while the right-hand side is the successor of {3. To see how Definition 5.1 conforms with the formal version of the Transfinite Recursion Theorem, let us consider operations G 1 , G 2 , and G 3 where G 1 (z,x) = z, G 2 (z,x) = x + 1, and G 3 (z,x) = sup(ranx) if xis a function (and G 3 (z, x) = 0 otherwise). The parametric form of Theorem 4.11 then provides an operation F such that for all z (5.2)

{

F(z,O) = G 1 (z,O) = z. F(z,a + 1) = G2(z,Fz(a)) =_F(z,a) + 1 for all a. F(z,a) = G3(z,Fz fa)= sup(ran(Fz fa))= sup({F(z,-y) for limit a -1- 0.

11· <

n})

If {3 and a are ordinals, then we write {3 +a instead of F({3, a) and see that the conditions (5.2) are exactly the clauses (a), (b), ~md (c) from Definition 5.1. In the subsequent applications of the Transfinite Recursion TheorPm. WP use the abbreviated form as in Definition 5.1, without explicitly formulating the defining operations G 1, G2, and G3 to conform with Theorem 4.11. One consequence of (5.1) is that for all {3, ({3+1)+1={3+2,

({3 etc. Also, we have (if a= {3 w

+ 2) + 1 = {3 + 3,

= w):

+ w = sup{w + n In< w},

and similarly, (w

+ w) + w =

sup{(w

+ w) + n In< w }.

In contrast to these examples, let us consider the sum m + w for m < w. We have m + w = sup{m + n In< w} = w, because, if m is a natural number,

120

CHAPTER 6. ORDINAL NUMBERS

m + n is also a natural number. We see that m + w =I w + m; the addition of ordinals is not commutative. One should also notice that, while 1 =I 2, we have 1 + w = 2 + w. Thus cancellations on the right in equations and inequalities are not allowed. However, addition of ordinal numbers is associative and allows left cancellations. (Lemma 5.4.) In Chapter 4 we defined the sum of linearly ordered sets. We now prove that the ordinal sum defined in 5.1 agrees with the earlier more general definition.

5.3 Theorem Let (W 1 ,
5.4 Lemma (a) If a 1, a: 2, and {3 are ordinals, then a1 < a2 if and only if {3 + Ot < {3 + a 2 . (b) For all ordinals a 1, 02, and {3, {3 + 01 = {3 + 02 if and only if a1 = a2. {c) (a+ {3) + 1 =a+ (/3 + 1) for all ordinals a, {3, and f. Proof.

(a) We first use transfinite induction on a2 to show that a 1 < 02 implies f3 + a 1 < {3 + a 2. Let us then assume that a 2 is an ordinal greater than a 1 and that a 1 < implies /3 + a 1 < !3 + 8 for all < a2. If 02 is a successor ordinal, then a 2 = o+ 1 where o ~ a 1. By the inductive assumption in casP o> a 1, and trivially in case o=OJ. we obtain {3 +a1 ~ !3+ 8 < ({3+8) + 1 = {3 + (8 + 1) = {3 + a 2. If a2 is a limit ordinal, then a 1 + 1 < a2 and we haw {3 + a 1 < (/3 + ai) + 1 = /3 + (a1 + 1) ~ sup{/3 + 8 I 8 < a2} = !3 + a-2 To prove the converse we assume that {3 + a 1 < {3 + n2. If a2 < a 1, thP implication already proved would show !3 + a2 < {3 + !lJ. Since a2 = a1 is also impossible (it implies /3 + a2 = {3 + aJ), we conclude from linearity of < that a 1 < a2. (b) This follows immediately from (a): If a 1 =I a2, then either a1 < a 2 or az < a 1 and, consequently, either !3 + a 1 < {3 + a2 or {3 + a2 < !3 + OJ. If a: 1 = 0:2, then {3 + a 1 = {3 + a2 holds trivially.

o

o

121

5. ORDINAL ARITHMETIC

(c) We proceed by transfinite induction on 'Y· If 1 = 0, then (a + {3) + 0 = a+ {3 = a+ ({3 + 0). Let us assume that the equality holds for "(, and let us prove it for 'Y + 1: (a+

/3) + ('Y + 1) =[(a+ /3) + -y] + 1 =[a+ ({3 + -y)J + 1 =a+ [({3 + -y) + 1] =a+ [f3 + ('Y + 1)]

(we have used the inductive assumption in the second step, and the second clause from the definition of addition in the other steps). Finally, let 'Y be a limit ordinal, 'Y =/= 0. Then (a+ {3) + 'Y = sup{ (a+ {3) + 8 I fJ < -y} =sup{ a+ ({3 + fJ) I < -y}. We observe that sup{{J + fJ I < -y} = {3 + 'Y (the third clause in the definition of addition) and that {3 + 'Y is a limit ordinal (if ~ < {3 + 'Y then ~ ::; {3 + fJ for some 8 < 'Y and so ~ + 1 $ ({3 + o) + 1 = {3 + (o + 1) < {3 + 1 because 'Y is limit). It remains to notice that sup{ a + ({3 + o) I 8 < 'Y} = sup{ a + ~ I ~ < {3 + 'Y} (because {3 + 'Y = sup{{3 + I 8 < 'Y} ), and so we have (a+ {3) + 'Y = sup{ a + ~ I ~ < {3 + 'Y} = a + ({3 + 'Y), again by the third clause in Definition 5.1.

o

o

o

0 The following lemma shows that it is possible to define subtraction of ordinal numbers:

5.5 Lemma !fa::; {3 then there is a unique ordinal

number~

such that

a+~=

{3.

Proof. As a is an initial segment of the well-ordered set {3 (or a = {3), Theorem 5.3 implies that {3 = a+~ where~ is the order type of the set {3- a = {vI a::; 11 < /3}. By Lemma 5.4(b), the ordinal~ is unique. 0

Next, we give a definition of ordinal multiplication:

5.6 Definition- Multiplication of Ordinal Numbers

For all ordinals

{3,

(a) {3 · 0 = 0. {b) {3 · (a + 1) = {3 · a + {3 for all a. (c) {3 ·a= sup{{3 · 'Y I 'Y
CHAPTER 6. ORDINAL NUMBERS

122

For more properties of ordinal products, see Exercises 5.1, 5.2 and 5.7. Ordinal multiplication as defined in 5.6 also agrees with the general definition of products of linearly ordered sets as defined in Chapter 4:

/3 be ordinal numbers. Both the lexicographic and thP antilexicographic orderings of the product a x (3 aTe well-orderings. The md('r type of the antilexicographic ordering of a x (3 is a · (3, while the lexicographic ordering of a x (3 has order type (3 · a.

5.8 Theorem Let a: and

Let --( denote the antilexicographic ordering of a x (3. We define an Proof. isomorphism between (a x (3, -<() and a· (3 as follows: for~ < a and Tf < (3. !Pt f (~, Tf) = a · 17 + E. The range of f is the set {a · Tf + E I Tf < (3 and E < n} = (\ · .1 and f is an isomorphism (we leaVP. the details ~-proved by induction --- to t.liP 0 reader; see also Exercises 5.1, 5.2, 5.7, and 5.8).

5.9 Definition- Exponentiation of Ordinal Numbers (a) (3o = 1. (b) f3<>+l = (3a · /3 for all a. (c) (3a = sup{(3"~

I"(< a}

For all (3,

for all limit a=/= 0.

5.10 Example (a) (31 = (3, (32 = /3. {3, /33 = /32. {3 = /3. {3. /3, etc. (b) 13w = sup{/3n In E w}; in particular, 1"' = 1, 2w = w, 3w = w, ... , nw = w for any nEw, ww = sup{wn In E w} > w.

It should be pointed out that ordinal arithmetic differs substantially from arithmetic of cardinals. Thus, tor instance, 2w = w and ww are countablP ordinals, while 2No = N~" is uncountable. One can use arithmetic operations to generate larger and larger ordinals: 0, 1, 2, 3, ... 'w, w + 1, w + 2, ... 'w. 2, w. 2 + 1, ... 'w. 3, .... w. 4 .. w · w = w 2 , w 2 + 1, ... , w 2 · 2, .... w 3 , ... , w 4 , ... , ww, ww + 1..... ..v"" · :2. ... , ww · w = w...; +I , ... , w ~ , ... , w ~ , ... , w ~ , ... , w ~~ . }. The process can easily be continued. We define t: = sup{ w. w"'. w"'~. w""One can then form r:: + 1, t: + w, t:w, t:'. t:'', etc. Exercises

5.1 Prove the associative law (a· /3) ·"(=a· (/3 · -y). 5.2 Prove the distributive law a· (/3 + -y) =a· /3 +a·"(. 5.3 Simplify: (a) (w + I)+ w. (b)w+w 2 . (c) (w

+ 1)

· w2.

5. ORDINAL ARITHMETIC

123

5.4 For every ordinal a, there is a unique limit ordinal /3 and a unique natural number n such that a= {3 + n. [Hint: {3 = sup{r :Sa II is limit}.) 5.5 Let a :S {3. The equation ~+a = {3 may have 0, 1, or infinitely many solutions. 5.6 Find the least a > w such that~+ a =a for all~ < a. 5.7 (a) If OJ, n2, and {3 are ordinals and {3 # 0, then OJ < n 2 if and only if {3 · OJ < {3 · 02. (b) For all ordinals OJ, n 2, and /3 # 0, /3 ·OJ = /3 · n 2 if and only if OJ

= et2.

5.8 Let a, {3, and 1 be ordinals, and let a < /3. Then (a) a+ 1 :S {3 + 1, (b) a · I :S /3 · I and :S cannot be replaced by < in either inequality. 5.9 Show that the following rules do not hold for all ordinals a, {3, and 1: (a) lfn+1=/3+1,thenn=/3. (b) If 1 > 0 and a· 1 = {3 ·1, then a = {3. (c) ({3 + 1) · a = {3 · a + 1 · a. 5.10 An ordinal a is a limit ordinal if and only if a = w · {3 for some {3. 5.11 Find a set A of rational numbers such that (A, :SQ) is isomorphic to (a, :S) where (a) n=w+1, (b) a= w · 2, (c) a= w · 3, (d) a= ww, (e) a = c:. [Hint: {n -1/m I m,n EN- {0}} is isomorphic to w 2, etc.) 5.12 Show that (w. 2) 2 fo w2 . 2 2. 5.13 (a) a 13 +..,. = ai3 ·a"~'. (b) (a 13 )'1 = a 1h . 5.14 (a) If a :S /3 then a"~'$ j]'l'. (b) If a > 1 and if {3 < 1, then af3 < a..,.. 5.15 Find the least~ such that (a) w + ~ = ~. (b) W·~=~,~-~0. (c)

w{

= ~.

[Hint for part (a): Let ~o = 0, ~n+l = w + ~n. ~ = sup{~n In E w }.] 5.16 (Characterization of Ordinal Exponentiation) Let a and {3 be ordinals. For f: /3-+ a, let s(f) = {~ < /31 !(~) # 0}. Let S(/3, a) = {!If: {3--+ a and s(J) is finite}. Define -< on S(/3, a) as follows: f -< g if and only if there is ~o < {3 such that f(~o) < g(~o) and f(~) = g(() for all ( > ( 0 . Show that (S({3,a), -<)is isomorphic to (a 13 , <). 5.3 Simplify: (a) (w + 1) + w. (b) w + w 2 . (c) (w + 1) · w 2 .

CHAPTER 6. ORDINAL NUMBERS

124

6.

The Normal Form

Using exponentiation, one can represent ordinal numbers in a way similar to decimal expansion of integers. Ordinal numbers can be expressed uniquely in normal form, to be made precise in the theorem below. We apply the normal form to prove an interesting result about so-called Goodstein sequences of integers. First we observe that the ordinal functions o:+t3, o:·t3, and o:f3 are continuous in the second variable: If 1 is a limit ordinal and t3 = supv<-r t3v, then (6.1)

o:+t3=sup(o:+tJ.,),

o:·t3=sup(o:·t3.,),

v<-y

v<-y

o:{3=supo:f3". v<-y

This follows directly from Definitions 5.1, 5.6, and 5.9. As a consequence, we have: 6.2 Lemma

(a} If 0 < o: S 1 then there is a greatest ordinal t3 such that o: · t3 S f. (b} If 1 < o: S 1 then there is a greatest ordinal t3 such that o:f3 S 1Proof Since o: · (t + 1) 2: 1 + 1 > 1, there exists a 6 such that o: · 6 > -y. Similarly, because o:-r+l ?. 1 + 1 >I· there is a 6 with o:6 > "(. The least 6 such that o: · 6 > 1 (or that o:0 > 1) must be a successor ordinal because of (6.1), say 0 6 = t3 + 1. Then t3 is greatest such that o: · t3 S 1 (respectively, o:f3 S 1). The following lemma is the analogue of division of integers: 6.3 Lemma If"' is an arbitrary ordinal and ifo: f. 0, then there exists a unique ordinal t3 and a unique p < o: such that 1 = o: · t3 + p.

Proof Let t3 be the greatest ordinal such that o: · t3 ~ 1 (if o: > 1 then t3 = 0), and let p be the unique p (by Lemma 5.5) such that o: · t3 + p =f. The ordinal p is less than o:, because otherwise we would have o: · (t3 + 1) = o: · t3 + o: ~ o: · t3 + p = 1, contrary to the maximality of !3. To prove uniqueness, let "' = o: · tJ1 +PI = o: · t3z + pz with PI, pz < o:. Assume that 13 1 < tJz. Then t3t + 1 S t3z and we have o:·tJ1 +(o:+pz) = o:-(131 + 1) + P2 'S. o: · t3z + pz = o: · t3I + PI• and by Lemma 5.4(a), PI 2: o: + pz?. o:, a contradiction. Thus tJ1 = tJz, and PI = pz follows by Lemma 5.5. C The normal form is analogous to decimal expansion of integers, with the base for exponentiation being the ordinal w: 6.4 Theorem Every ordinal o: > 0 can be expressed uniquely as

where

!31 > t3z > · · · > t3n, and k1 > 0, kz > 0,

... , kn > 0 are finite.

6. THE NORMAL FORM We remark that it is possible to have o: ==

125

w",

see Exercise 6.1.

Proof. We first prove the existence of the normal form, by induction on o:. The ordinal o: = 1 can be expressed as 1 = w 0 · 1. Now let o: > 0 be arbitrary. By Lemma 6.2(b) there exists a greatest {3 such that wi3 :S o: (if o: < w then {3 == 0). Then by Lemma 6.3 there exist unique J and p such that p < wi3 and o: = wi3 · J + p. As wi3 :S o:, we have 6 > 0 and p < o:. We claim that J is finite. If J were infinite, then o: ~ wi3 · J ~ wi3 · w = w!3+ 1 , contradicting the maximality of {3. Thus let {3 1 = {3 and k 1 = J. If p = 0 then o: = wi3 1 • k 1 is in normal form. If p > 0 then by the induction hypothesis, 13 p = w • · k2 + · · · + w 13" · kn for some /32 > · · · > f3n and finite k2, ... , kn > 0. As p < w!3 1 , we have < wi3 1 and so /31 > /32. It follows that o: = wi3' ·k 1+wi32 ·k2 + · · ·+wi3 .. · kn is expressed in normal form. To prove uniqueness, we first observe that if {3 < {, then w!3 · k < w"~ for every finite k: this is because wi3 · k < wi3 · w = wi3+l :S w"~. From this it easily follows that if o: = wi3 1 • k 1 + · · · + wi3" · kn is in normal form and 1 > {3 1, then o:
We use the normal form to prove an interesting result on Goodstein sequences. Let us first recall that for every natural number a ~ 2, every natural number m can be written in base a, i.e., as a sum of powers of a:

with b1 > · · · > bn and 0 < ki 0 is a sequence mo, m 1 , m 2 , ... of natural numbers defined as follows: First, let m 0 = m, and write mo in base 2:

To obtain m 1 , increase the base by 1 (from 2 to 3) and then subtract 1:

126

CHAPTER 6. ORDINAL NUMBERS

In general, to obtain mk+I from mk (as long as mk =f. 0), write mk in base k + 2, increase the base by 1 (to k + 3) and subtract 1. For example, the weak Goodstein sequence starting at m = 21 is as follows: mo

= 21

m1

= 3

4

m2 = 4

4

m 3 =5

4

+5 · 3 +2 =

642

Tn4

= 6

4

+ 6. 3 + 1 =

1315

m5

= 7 + 7 · 3 = 2422 4 + 8. 2 + 7 = 4119

= 2

4

+ 32 + 42

+ 22 + 1 = 90 -

1 = 44

+4·3+3

= 271

4

Tn6

= 8

ffi7

= 9

ms =

4

+ 9 . 2 + 6 = 6585 10 + 10 · 2 + 5 = 10 025 4

etc. Even though weak Goodstein sequences increase rapidly at first, we have 6.5 Theorem For each m > 0, the weak Goodstein sequence starting at m eventually terminates with mn = 0 for some n. Proof. We use the normal form for ordinals. Let m > 0 and m 0 , m 1 • m 2 , ... be the weak Goodstein sequence starting at m. Its a th term is written in ba.~P a+ 2:

(6.6)

Consider the ordinal obtained by replacing base a + 2 by w in ( 6. 6). It is easily seen that o 0 > o 1 > > · · · > Oa > · · · is a decreasing sequence of ordinals, necessarily finite. Therefore, there exists some n such that o, = 0. But clearly rna ::; Oa for evPr\· a = 0, 1, 2, ... , n. Hence mn = 0.

n2

Remark For the weak Goodstein sequence m 0 , m 1 , m 2 ,.

starting at m = 21 displayed above, the corresponding ordinal numbers are w 4 + w 2 + 1. w~ + w 2 . w 4 + w · 3 + 3, w 4 + w · 3 + 2, w 4 + w · 3 + 1, w 4 + w · 3, w 4 + w · 2 + 7 .... We shall now outline an even stronger result. A number m is written in pun base a 2 2 if it is first written in base a, then so are the exponents and t lw exponents of exponents, etc. For instance, the number 324 written in pure bas~· 3 is 33+ 2 + 33+ 1 . The Goodstein sequence startiug at m > 0 is a sequence m 0 , m1, m2 .... obtained as follows: Let mo = m and write m 0 in pure ba.<;e 2. To define m 1 . replace each 2 by 3, and then subtract 1. In generaL to get mk+ 1 . write rn~,- iu

6. THE NORMAL FORM

127

pure base k + 2, replace each k + 2 by k + 3, and subtract 1. For example, the Goodstein sequence starting at m = 21 is as follows: mo = 21 = 2 m1

33

= 3

m2 = 4

44

50

m3

= 5

m4

= 6 6"

22

+ 22 + 1

+ 3 3 "' 7.6 X 10 12 44 + 44 - 1 = 4 + 4 3 · 3 +

42 · 3 + 4 · 3 + 3"' 1.3 x 10 154

+5 3 · 3 +5 2 · 3 + 5 · 3 + 2"' 1.9 x 10 2184 + 6 3 · 3 + 6 2 · 3 + 6 · 3 + 1 "'2.6 x 10

36305

etc. Goodstein sequences initially grow even more rapidly than weak Goodstein sequences. But still:

6. 7 Theorem For each m > 0, the Goodstein sequence starting at m eventually terminates with mn = 0 for some n. Again, we define a (finite) sequence of ordinals no > o 1 > · · · > ma is written in pure base a + 2, we get Cla by replacing each a+ 2 by w. For instance, in the example above, the ordinals are ww~ +ww +1, ww~ +ww, www +w 3 ·3+w 2 ·3+w·3+3, www +w 3 ·3+w 2 ·3+w·3+2. www + w 3 · 3 + w 2 • 3 + w · 3 + 1, etc. The ordinals Oa are in normal form. and again, it can be shown that they form a (finite) decreasing sequence. Therefore On = 0 for some n, and since ma :S O'a for all a, we have mn = 0. D Proof

O'a

> · · · as follows: When

Exercises

6.1 Show that w' =E. 6.2 Find the first few terms of the Goodstein sequence starting at m

=

28.

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Chapter 7

Alephs 1.

Initial Ordinals

In Chapter 5 we started the investigation of cardinalities of infinite sets. Although we proved several results involving the concept of jXj, the cardinality of a set X, we have not defined lXI itself, except in the case when X is finite or countable. In the present chapter we consider the question of finding "representatives" of cardinalities. Natural numbers play this role satisfactorily for finite sets. We generalized the concept of natural number and showed that the resulting ordinal numbers have many properties of natural numbers, in particular, inductive proofs and recursive constructions on them are possible. However, ordinal numbers do not represent cardinalities, instead, they represent types of wellorderings. Since any infinite set can be well-ordered in many different ways (if at all) (see Exercise 1.1), there are many ordinal numbers of the same cardinality; w, w + 1, w + 2, ... , w + w, ... , w · w, w · w + 1, ... are all countable ordinal numbers; that is, lwl = jw + lj = jw + wl = · · · = ~o- An explanation of the good behavior of ordinal numbers of finite cardinalities ~ that is, natural numbers ~ is provided by Theorem 4.3 in Chapter 4: All linear orderings of a finite set are isomorphic, and they are well-orderings. So for any finite set X, there is precisely one ordinal number n such that jnj = \Xj. We have called this n the cardinal number of X. In spite of these difficulties, it is now rather easy to get representatives for cardinalities of infinite (well-orderable) sets; we simply take the least ordinal number of any given cardinality as the representative of that cardinality.

1.1 Definition An ordinal number a is called an initial ordinal if it is not equipotent to any j3 < a. 1.2 Example Every natural number is an initial ordinal. w is an initial ordinal, because w is not equipotent to any natural number. w + 1 is not initial, because jwj = jw + lj. Similarly, none of w + 2, w + 3, w + w, w · w, ww, ... , is initial.

129

130

CHAPTER 7. ALEPHS

1.3 Theorem Each well-orderable set X is equipotent to a unique initial ord1nal number. Proof. By Theorem 3.1 in Chapter 6, X is equipotent to some ordinal a. Let a 0 be the least ordinal equipotent to X. Then a 0 is an initial ordinal. because lnol = 1!31 for some {3
1.4 Definition If X is a well-order able set, then the cardinal number of X. denoted IX I. is the unique initial ordinal equipotent to X. In particular. IX I = JJ for any countable set X and lXI = n for any finite set of n elements, in agreement with our previous definitions. According to Theorem 1.3, the cardinal numbers of well-orderable sets are precisely the initial ordinal numbers. A natural question is whether there are any other initial ordinals besides the natural numbers and w. The next theorem shows that there are arbitrarily large initial ordinals. Actually. we prove something more general. Let A be any set; A may not be well-orderable itself. but it certainly has some well-orderable subsets; for example, all finite subsets of A are well-orderable.

1.5 Definition For any A. let h(A) be the least ordinal number which is not equipotent to any subset of A. h(A) is called the Hartogs number of A. By definition, h(A) is the least ordinal a such that

lol 1:. IAI.

1.6 Lemma For any A, h(A) is an initial ordinal number. Proof. Assume that \{31 = lh(A)I for some {3 < h(A). Then {3 is equipotent to a subset of A, and {3 is equipotent to h(A). We conclude that h(A) is equipotent to a subset of A, i.e., h(A) < h(A), a contradiction. 0 So far, we have evaded the main difficulty: How do we know that the Hartogs number of A exists? If all infinite ordinals were countable, h(w·) would consist of all ordinals!

1.7 Lemma The Hartogs number of A exists for all A. Proof. By Theorem 3.1 in Chapter 6, for every well-ordered set (W. R) where W <; A, there is a unique ordinal a such that (o:. <) is isomorphi<.: to (W, R). The Axiom Schema of Replacement implies that there exists a set H such that, for every well-ordering R E P(A x A), the ordinal o: isomorphic to it is in H. We claim that H contains all ordinals equipotent to a subset of A. Indeed, if f is a one-to-one function mapping o: into A, we set W = ran f and R = {(f({3),J(t)) I {3 < 1 < o:}. R o;;; Ax A is then a well-ordering isomorphir

1. INITIAL ORDINALS

131

too: (by the isomorphism f). These considerations show that h(A) = {a: E HI o: is an ordinal equipotent to a subset of A}, and justify the existence of h(A) by the Axiom Schema of Comprehension. 0 The preceding developments allow us to define a "scale" of larger and larger initial ordinal numbers by transfinite recursion.

1.8 Definition wo =w;

Wa+l = h(wa) for all o:; Wa = sup{ w13 I {3 < o:} if o: is a limit ordinal, o

-1 0.

The remark following Definition 1.5 shows that lwa+ 1l > lwa I, for each o:. and so lwal < lwi31 whenever o: < /3.

1.9 Theorem (a) W 0 is an infinite initial ordinal number for each o:. (b) If is an infinite initial ordinal number, then =

n

n

Wo

for some

0:.

Proof. (a) The proof is by induction on o. The only nontrivial case is when o: is a limit ordinal. Suppose that lwal = bl for some 1 < W0 ; then there is {3 < o: such that 1 S w13 (by the definition of supremum). But this implies lwo:l = ill : : ; lwpl S lwal and yields a contradiction. (b) First, an easy induction show that o: ::::; Wa for all o:. Therefore, for every infinite initial ordinal fl, there is an ordinal 0 such that fl < Wa (for example, o: = fl + 1). Thus it suffices to prove the following claim: For every infinite initial ordinal fl < Wa, there is some I< D such that fl = w,. The proof proceeds by induction on o. The claim is trivially true for o = 0. If 0: = {3 + 1, fl < W0 = h(w13) implieS that lfll S lw,al, SO either fl = W(3 and we can let 1 = /3, or fl < w13 and existence of 1 < {3 < a follows from the inductive assumption. If o: is a limit ordinal, n < Wa = sup{ w 6 I {3 < a} implies that n < WiJ for some j3 < D. The inductive assumption again guarantees the existence of some 1 < 13 such that fl = w..,. 0

The conclusion of this section is that every well-orderable set is equipotent to a unique initial ordinal number and that infinite initial ordinal numbers form a transfinite sequence Wo: with o: ranging over all ordinal numbers. Infinite initial ordinals are, by definition, the cardinalities of infinite well-orderable sets. It is customary to call these cardinal numbers alephs; thus we define

for each o:.

CHAPTER 7. ALEPHS

132

The cardinal number of a well-orderable set is thus either a natural number or an aleph. In particular, IN!== No in agreement with our previous usage. The reader should also note that the ordering of cardinal numbers by size defined in Chapter 4 agrees with the ordering of natural numbers and alephs as ordinals by < (i.e., E): If lXI = Na and !YI = N13, then !XI < !YI holds if and only if Na < N13 (i.e., Wa E w13) and a similar equivalence holds if one or both of lXI and IYI are natural numbers. In Chapter 5 we defined addition, multiplication, and exponentiation of cardinal numbers; these operations agree with the corresponding ordinal addition. multiplication, and exponentiation, as defined in Chapter 6, as long as the ordinals involved are natural numbers but they may differ for infinite ordinals. For example, wo + wo f. wo if + stands for the ordinal addition, but w0 + wo = wo if + stands for the cardinal addition. The addition of cardinal numbers is commutative, but the addition of ordinal numbers is not. To avoid confusion, we employ the convention of using thew-symbolism when the ordinal operations are involved, and the aleph-symbolism for the cardinal operations. Thus w0 + w0 and 2wo indicate ordinal addition and exponentiation (wo + wo == sup{w + n In< wo} > wo, 2wo = sup{2n I n < wo} = wo), while N0 + N0 and 2N" are cardinal operations (No+ N0 = N0 , 2N" is uncountable). Exercises

1.1 If X is an infinite well-orderable set, then X has nonisomorphic wellorderings. 1.2 If o: and /3 are at most countable ordinals then o: + /3, o: · {3, and o:i3 are at most countable. [Hint: Use the representation of ordinal operations from Theorems 5.3 and 5.8 and Exercise 5.16 in Chapter 6. Another possibility is a proof by transfinite induction.] 1.3 For any set A, there is a mapping of P(A x A) onto h(A). [Hint: Define f(R) == the ordinal isomorphic to R, if R <;;; A x A is a well-ordering of its field; f(R) = 0 otherwise.] 1.4 !AI < IAI + h(A) for all A. 1.5 lh(A)I < IP(P(A x A))! for all A. [Hint: Prove that !P(h(A))I 'S IP(P(A x A))l by assigning to each X E P(h(A)) the set of all wellorderings R <;;; A x A for which the ordinal isomorphic to R belongs to

X.] 1.6 Let h*(A) be the least ordinal o: such that there exists no function with domain A and range o:. Prove: (a) If o: ~ h*(A), then there is no function with domain A and range n. (b) h*(A) is an initial ordinal. (c) h(A) $ h*(A). (d) If A is we!l-orderable, then h(A) = h*(A). (e) h* (A) exists for all A. [Hint for part (e): Show that o: E h*(A) if and only if o: = 0 or o: = the ordinal isomorphic to R, where R is a well-ordering of some partition of A into equivalence classes.]

2. ADDITION AND MULTIPLICATION OF ALEPHS

2.

133

Addition and Multiplication of Alephs

Let us recall the definitions of cardinal addition and multiplication: Let K. and >. be cardinal numbers. We have defined K. + >. as the cardinality of the set XU Y, where JXJ = K., JYJ = >., and X and Y are disjoint:

JXJ + JYI = IX u Yl

if

X n Y = 0.

As we have shown, this definition does not depend on the choice of X and Y. The product K. · >. has been defined as the cardinality of the cartesian product X x Y, where X and Y are any two sets of respective cardinalities K. and >.:

lXI· IYI = IX

X

Yl;

and again, this definition is independent of the choice of X and Y. We verified that addition and multiplication of cardinal numbers satisfy various arithmetic laws such as commutativity, associativity, and distributivity:

+ ,\ = >. + K., >. = >. . K., K. + (,\ + J.L) = (K. + ,\) + J.L, K.. (>. . J.L) = (K.. >.) . J.L, K.. (>. + J.L) = K.. >. + K.. J.L. K.

K. .

Also, if K. and>. are finite cardinals (i.e., natural numbers), then the operations K. + >. and K. · >. coincide with the ordinary arithmetic operations. The arithmetic of infinite numbers differs substantially from the arithmetic of finite numbers and in fact, the rules for addition and multiplication of alephs are very simple. For instance: No

+n =No

for every natural number n. (If we add n elements to a countable set, the result is a countable set.) We even have No+ No= No,

since, for example, we can view the set of all natural numbers as the union of two disjoint countable sets: the set of even numbers and the set of odd numbers. We also recall that No· No= No.

(The set of all pairs of natural numbers is countable.) We now prove a general theorem that determines completely the result of addition and multiplication of alephs.

134 2.1 Theorem N0

CHAPTER 7. ALEPHS ·

N0 =No, for every a.

Before proving the theorem, let us look at its consequences for addition and multiplication of cardinal numbers. 2.2 Corollary For every a and /3 such that a 'S /3, we have

Also, for every positive natural number n.

If a $ /3, then on the one hand, we have N13 = 1 · No 'S N0 · N,J. Proof. and on the other hand, Theorem 2.1 gives No ·No 'S Na ·Na = Na. Thus by th<' Cantor-Bernstein Theorem N0 • N13 = No. The equality n · N0 = N0 is proved similarly. Ci

2.3 Corollary For every a and /3 such that a'S /3, we have

Also, for all natural numbers n. Proof. If a 'S /3, then No 'S Na + Na 'S No + Na = 2 · Na = Na. and tlw :::-J assertion follows. The second part is proved similarly. Proof of Theorem 2.1. We prove the theorem by transfinite induction. For every a, we construct a certain well-ordering -< of the set w(i x w". and show. using the induction hypothesis No · N13 'S No for all /3 < a. that the ordPrtype of the well-ordered set (w 0 x w,, -<) is at most W 0 . Then it follows that N0 · Na 'S N.. , and since N0 ·NQ :::0: N0 , we have NQ · N0 = Na. \Ve construct the well-ordering -< of w 0 x w 0 uniformly for all Wa: that is. we define a property -< of pairs of ordinals and show that -< well-orders W 0 x w" for every Wo· We let (a 1,a2 )-< (iJJ,/32) if and only if: either max{a1,a2} < max{/3J,/3z}, or max{o:,,a2} = max{OJ,i32} and < 01, or max{a1,a2} = max{O!.Oz}, a1 = 61 and az < /3z. We now show that -< is a well-ordering (of any set of pairs of ordinals). First, we show that -< is transitive. Let a1, Oz, {3,, iJ2, -y,, 1"2 be such that (a 1 ,a2)-< (/31,/32) and (0 1,/}z)-< (-y,,-y2)· It follows from the definition of-< that max{a 1,a2} 'S max{i3 1.p2} 'S max{-y1,1"2}; hence max{a1,a2} $ max{: 1,-y2}. If max{a1,a2} < max{rJ,/'2} then (a,,az) -< bi./'2). Thus a..'isurne that max{a1,a2} = max{tl 1 .62} = max{f',,-yz}. Then we haw n, $

o,

2. ADDITION AND MULTIPLICATION OF ALEPHS

135

{3 1 ::; -y 1, and so a1 ::; 1'!· If a1 < 1't. then (a,, az) -< (1' 1, )'2 ); otherwise, we have a, = {3, = 1"1· In this last case, max{a,,az} = max{f3I,f3z} = maxhJ,)'z}, and 01 = {3 1 = -y 1, so, necessarily, az < f3z < -yz, and it follows again that (0 I, az) -< ('/'I , 1"2) · Next we verify that for any Oi,Oz,f3I,f3z, either (a1,a2) -< ({3,,{3z) or ({3 1 ,{3z)-< (a,,az) or (a,,az) = (f3t,f3z) (and that these three cases an• mutually exclusive). This follows directly from the definition: Given (a 1, o 2 ) and ({31, {32 ), we first compare the ordinals max{ a 1, a 2 } and max{{3 1, {32 }, then the ordinals a 1 and {31, and last the ordinals a 2 and {3z. We now show that -< is a well-ordering. Let X be a nonempty set of pairs of ordinals; we find the -<-least element of X. Let 8 be the least maximum of the pairs in X; i.e., let 8 be the least element of the set {max{a,{3} I (n,{J) EX}. Further, let Y = {(a,{3) EX I max{a,{3} = 8}. The set Y is a nonempty subset of X, and for every (a, {3) E Y we have max{a,{3} = 8; moreover, 8 < max{a',{3'} for any (a',{3') E X- Y, and hence (a.{3)-< (a',{3') whenever (o,{3) E Y and (a',{3') EX- Y. ThereforP, the least element of Y, if it exists, is also the least element of X. Now let n 0 lw the least ordinal in the set {a I (a, {3) E Y for some {3}, and let

Z= {(a,{3) E Y Ia =no}. The set Z is a nonempty subset of Y, and we have (o, /1) . . :: (o', (J') whenever (a, {3) E Z and (a', {3') E Y- Z. Finally, let {30 be the least ordinal in the set {{3 I (o 0 ,{3) E Z}. Clearly, (a 0 , {30 ) is the least element of Z, and it follows that (a 0 , {30 ) is the least element of X. Having shown that -< is a well-ordering of Wa x Wo: for every a, we use this well-ordering to prove, by transfinite induction on a, that \wo x wal ::; N,, that is, N, ·Na :'S Na. We have already proved that No · No = N0 , and so our assertion is trw• f(>r a = 0. So let a > 0, and let us assume that Na · Na ::; No for all {3 < a. We prow that \wa x r...•,,\ ::; No. If suffices to show that the order-type of the well-ordPwd set (wa x w,, -<) is at most w 0 . If the order-type of W 0 x Wo were greater than Wo, then there would exist (a1, oz) E W 0 x Wa such that the cardinality of the set is at least Na. Thus it suffices to prove that for any (a,, oz) E W 0 x Wa. we have \XI< N". Let {3 = max{a,,az} + 1. Then {3 E Wa, and, for every (( 1,(2) EX. we have max{(,,(z} ::; max{a,,az} < {3, so 6 E {3 and 6 E 0. In other words. X~ (3 X 3. Let -y
136

CHAPTER 7. ALEPHS

Now it follows that Jw 0 X w0 J :S N0 • Thus we have proved, by transfinit~ induction on a, that N0 · N0 :S N0 for all a. Since N0 :S N,. · N0 , we have N0 · N0 = N0 , and the proof of Theorem 2.1 is complete. 0 Exercises

2.1 Give a direct proof of N0 + N0 = N0 by expressing Wo as a disjoint union of two sets of cardinality N0 . 2.2 Give a direct proof of n · N,. = N0 by constructing a one-to-one mapping of W 0 onto n x w 0 (where n is a positive natural number). 2.3 Show that (a) N~ = N0 , for all positive natural numbers n. (b) I[No]nl = N0 , where [No]" is the set of all n-element subsets of Nn. for all n > 0. (c) I[No]
Chapter 8

The Axiom of Choice 1.

The Axiom of Choice and its Equivalents

In the preceding chapter, we left unanswered a fundamental question: Which sets can be well-ordered? Interestingly, the question was posed in this form only at a later stage in the development of set theory. Cantor considered it quite obvious that every set can be well-ordered. Here is a fairly intuitive "proof" of this "fact." In order to well-order a set A, it suffices to construct a one-to-one mapping of some ordinal A onto A. We proceed by transfinite recursion. Let a be any set not in A. Define

f(O)

= {saome element of A

f(l)

= {saorne element of A- {f(O)}

if A "I 0, otherwise, if A- {f(O)} -:10, otherwise,

etc. Generally,

f(o:)

= {saome element of A- ran(! f o:)

if A -ran(! f o:) -:1 0, otherwise.

Intuitively, f lists the elements of A, one by one, as long as they are available: when A is exhausted, f has value a. We first notice that A does get exhausted at some stage A < h(A), the Hartogs number of A. The reason is that, for o: < (3, if f((3) "I a, then J((3) E A - ran(! f (3), f(o:) E ran(! f (3), and thus f(o:) # !((3). If f(o:) # a were to hold for all o: < h(A), f would be a one-to-one mapping of h(A) into A, contradicting the definition of h(A) as the least ordinal which cannot be mapped into A by a one-to-one function. Let >. be the least o: < h(A) such that f(o:) =a. The considerations of the previous paragraph immediately show that f f A is one-to-one. The "proof"

137

CHAPTER 8. THE AXIOM OF CHOICE

138

is complete if we show that ran(! r A) = A. Clearly ran(! r >.) <;;; A: if ran(! f A) c A, A -ran(! f >.) =!= 0 and f(>.) =!= a, contradicting our definition of>.. Our use of quotation marks indicates that something is wrong with this argument, but it may not be obvious precisely what it is. However, if one tries to justify this transfinite recursion by the Recursion Theorem, say in the form stated in Theorem 4.5 in Chapter 6, one discovers a need for a function G surh that f can be defined by f(a) = G(J fa). Such a function G should haw thP following properties:

G(J fa) E A -ran(! fa) G(f r a)= a

if A- ran(! fa) =/= 0, otherwise.

If A were well-orderable, some such G could easily be defined; e.g.: G(x) ={the -<-le~t element of A- ranx if xis a function and A- ranx =/= 0. a otherwise, where -< is some well-ordering of A. But in the absence of well-orderings on .4. no property which could be used to define such a function G is obvious. More specifically, let S be a system of sets. A function g defined on S i~ ealled a choice function for S if g(X) E X for all nonempty X E S. If we now assume that there is a choice function g for P (A), we are ab!P to fill the gap in the previous proof by defining G(x)

= {g(A- ranx) a

if xis~ function and A- ranx otherwise.

f= 0,

We proved the difficult half of the following theorem, essentially due to Ernst Zermelo:

1.1 Theorem A set A can be well-ordered if and only if the set P(A) of all subsets of A has a choice junction. Proof. The proof of the second half is easy. If -< well-orders A, we a choice function g on P(A) by

g(x)

= {~he least element of x

in the well-ordering -<

defiru~

if X=/= 0, if :r = 0.

0 The problem of well-ordering the set A is now reduced to an equivalent question, that of finding a choice function for P(A). We first notice Theorem 1.2.

1. THE AXIOM OF CHOICE AND ITS EQUIVALENTS

139

1.2 Theorem Every finite system of sets has a choice Junction. Proof Proceed by induction. Let us assume that every system with n elements has a choice function, and let lSI = n + 1. Fix X E S; th~ set S- {X} has n elements, and, consequently, a choice function 9X· If X = 0. g = gx U {(X,0)} is a choice function for S. If X 1=- 0, gx = gx U {(X.x)} is a choice function for S (for any x E X). D The reader might be well advised to analyze the reasons why this proof cannot be generalized to show that every countable system of sets has a choice function. Also, while it is easy to find a choice function for P(N) or P(Q) (why?), no such function for P(R) suggests itself. Although choice functions for infinite systems of sets of real numbers have been tacitly used by analysts at least since the end of the nineteenth century. it took some time to realize that the assumption of their existence is not entirely trivial. The following axiom was formulated by Zermelo in 1904.

Axiom of Choice

There exists a choice function for every system of sets.

Sixty years later, in 1963, Paul Cohen showed that the Axiom of Choice cannot be proved from the axioms of Zermelo-Fraenkel set theory (more on this in Chapter 15). The Axiom of Choice is thus a new principle of set formation; it differs from the other set forming principles in that it is not effective. That is, the Axiom of Choice asserts that certain sets (the choice functions) exist without describing those sets as collections of objects having a particular property. Because of this, and because of some of its counterintuitive consequences (see Section 2), some mathematicians raised objections to its use. We next examine several equivalent formulations of the Axiom of Choic~ and some of the consequences it has in set theory and in mathematics. Following that, at the end of Section 2, we resume discussion of its justification. To keep track of the use of the Axiom of Choice in this chapter, we denote the theorems whose proofs depend on it, and exercises in which it has to be used, by an asterisk. In later chapters, we use the Axiom of Choice without explicitly pointing it out each time.

1.3 Theorem The following statements are equivalent: (a) (The Axiom of Choice) There exists a choice function for every system of sets. (b) Every partition has a set of representatives. (c) If (X, I i E I) is an indexed system of nonempty sets, then there is a function f such that !( i) E X, for all i E I. We remind the reader that a partition of a set A is a system of mutually disjoint nonempty sets whose union equals A. We call X t;;;' A a set of represen· tatives for a partition S of A if, for every C E S, X n C has a unique element. (See Section 4 of Chapter 2 for these definitions.)

140

CHAPTER 8. THE AXIOM OF CHOICE

The statement (c) in Theorem 1.3 can be equivalently formulated as follows (compare with Exercise 5.10 in Chapter 3): (d) If X; -j. 0 for all i E /,then DiE/ X; -j. 0.

Proof. (a) implies (b). Let f be a choice function for the partition S: then X = ran f is a set of representatives for S. Notice that for any C E S. f(C) E XnC, but j(D) f/. XnC forD -j. C [because J(D) E D and DnC = 0]. So X n C ={!(C)} for any C E S. {b) implies (c). Let C; = {i} x X;. Since i -j. i' implies C, n c,. = 0. S = {C, I i E I} is a partition. Iff is a set of representatives for S, f is a set of ordered pairs, and for each i E I, there is a unique x such that ( i, x) E f n C,. But this means that f is a function on I and f(i) EX; for all i E /. (c) implies {a). Let S be a system of sets; set I = S- {0}, Xc = C for all C E /. Then (Xc I C E /) is an indexed system of nonempty sets. If f E DeE/ Xc, f is a choice function for S if 0 f/. S. If 0 E S, then f U { (0, 0)} is a choice function for S. D Theorem 1.3 gives several equivalent formulations of the Axiom of Choice. Many other statements are known to be equivalent to the Axiom, and we use some in the applications. The most frequently used equivalent is the WellOrdering Theorem, which states that every set can be well-ordered (its equivalence with the Axiom follows from Theorem 1.1). Another frequently used version (Zorn's Lemma) is given in Theorem 1.13. But first we present some consequences of the Axiom of Choice.

1.4 Theorem* Every infinite set has a countable subset. Let A be an infinite set. A can be well-ordered, or equivalently. Proof. arranged in a transfinite one-to-one sequence (ao 1 a < n) whose length n is an infinite ordinal. The range C = { aa I a < w} of the initial segment (a 0 I a < w) of this sequence is a countable subset of A. 0

1.5 Theorem* For every infinite set S there exists a unique aleph N, such that lSI= No. Proof. As S can be well-ordered, it is equipotent to some infinite ordinal, and hence to a unique initial ordinal number w 0 . 0 Thus in the set theory with the Axiom of Choice, we can define for any set X its cardinal number lXI as the initial ordinal equipotent to X. Sets X andY are equipotent if and only if lXI is the same ordinal as IYI (i.e., lXI = IYI). Also, the ordering < of cardinal numbers by size agrees with the ordering of ordinals byE: lXI < IYI if and only if /XI E JYI. These considerations rigorously justify Assumption 1. 7 in Chapter 4: There are sets called cardinals with the property that for every set X there is a unique cardinal lXI, and sets X andY are equipotent if and only if lXI is equal to /YI.

1. THE AXIOM OF CHOICE AND ITS EQUIVALENTS

141

As E is a linear ordering (actually a well-ordering) on any set of ordinal numbers, we have the following theorem.

1.6 Theorem* For any sets A and B either

lA/ ::; /BI or /B/ ::; IAI.

1.1 Theorem* The union of a countable collection of countable sets is countable.

(Compare with Theorem 3.9 in Chapter 4.) Let S be a countable Proof. set whose every element is countable, and let A = U S. We show that A is countable. As S is countable, there is a one-to-one sequence (An / n E N) such that S = {An / n E N}. For each n E N, the set An is countable, so there exists a sequence whose range is An. By the Axiom of Choice, we can choose one such sequence for each n. [That is: For each n, let Sn be the set of all sequences whose range is An. Let F be a choice function on {Sn In EN}, and let Sn = F(Sn) for each n.] Having chosen one Sn = (an(k) IkE N) for each n, we obtain a mapping f of N x N onto A by letting f(n, k) = an(k). Since N x N is countable and A is its image under f, A is also countable. 0

1.8 Corollary* The set of all real numbers is not the union of countably many countable sets. Proof.

0

The set R is uncountable.

1.9 Corollary* The ordinalw1 is not the supremum of a countable set of countable ordinals. Proof.

If

{On

In

E

N} is a set of countable ordinals, then its supremum

a= sup{ an

In EN}=

U

On

nEN

is a countable set, and hence a< w1.

Proof.

This follows from Theorem 1.5 and the fact that 2N" > N0 .

0

0

As a result, the Continuum Hypothesis can be reformulated as a conjecturf' that the least uncountable cardinal number.

CHAPTER 8. THE AXIOM OF CHOICE

142

1.11 Theorem* Iff is a function and A is a set, then

IJ[AJI

~

lA I.

Proof. For each b E f[A], let Xb = f- 1 ({b} ). Note that Xb cj 0 aud xb, n xb, = 0 if b, "1- h Take g E f1bEf[A] Xb; then g : f[A] - t A and b, lc h2 implies g(b,) E xb,' g(bl) E xh. so g(b!) f= g(b2}. WP conclude that y is a one-to-one mapping of f[A] into A, and consequently lf[A]I ~!A!. !J

1.12 Theorem* If

lSI~ No and, for all A E

S. IAI ~ Nn. thrn I U Sl ~ ~ .•

This is a generalization of Theorem 1. 7. We assume that S lc 0 aud I v < N"} and for each v < N,.. choosP a transfinite sequence A.,= (av{K:) I K: < N,.) such that A., = {a.,(K:) I K: < N" }. (Cf. Exercise 1.9 in Chapter 4.) Wt• define a mapping f on N" x N0 onto US by f(v, K:) = a.,(K:). By Theorem 1.11. Proof


0

(the last step is Theorem 2.1 in Chapter 7).

WP next derive Zorn's Lemma. another important version of tht• Axiom of Choice.

1.13 Theorem The following statements are equivalent: (a) (The Axiom of Choice) Then~ e:rists a choice function for evn~J systrm of sets. (b) (The Well-Ordering Principle) Every set can be well-ordered. (e) (Zorn's Lemma) If every chain in a partially ordered set ha-5 an upprr· hmtnd. then the partiallJJ ordered set has a maximal element. WP remind the reader that a C'hain is a linearly ordered subset of a partially ordered set (see Section 5 of Chapter 2 for this definition. as well as t lH' ddinitions of ·'ordered set," '·upper bound." "maximal element.'' and otlwr l'Olll't'pts related to orderings). Proof. Equivalence of (a) and (b) follows immediately from Theor('m 1 1: therefore, it is enough to show that (a) implies (c) and (c) implies (a). (a) implies (c). Let (A,~) be a (partially) ordered set in which every chain has an upper bound. Our strategy i:,; to search for a maximal element of (A.~ I by constructing a ~-increasing transfinite sequence of elements of A. W£> fix some b ¢ A and a cl10iet' function g for P(A), and definf> (an I n < h(A)) by tmnsfinite recursion. Given (a~ I ( < n), we consider two <"<1St's. If b # a~ for all ( < o and An = {a E A I a~ -< a holds for all ( < n} cJ 0. Wt' it't a,. = g( A,.); otherwise we let a 0 = b. \Vp leave to the reader tlw easy task of justifying this definition b\' TlH•on•JJl 4.4 in Chapter 7. We note that a" = b for some a < h(A); otherwisP. (a~ ~ < h(A)) would be a one-to-one mapping of h(A) into A. Let,\ be the least o for 1

1. THE AXIOM OF CHOICE AND ITS EQUIVALENTS

143

which a 0 = b. Then the set C = {a~ I ( < ,\} is a chain in (A,~) and so it has an upper bound c E A. If c ...: a for some a E A. we have a E A" i" 0 and aJ.. = g(AJ..) -1- b, a contradiction. Soc is a maximal element of A. (It is easy to see that, in fact, ,\ = {3 + 1 and c = ap.) (c) implies (a). It suffices if we show that every system of nonempty sets S has some choice function. Let F be the system of all functions f for which domf <;;; S and J(X) E X holds for any X E S. The set F is ordered by inclusion <;;;. Moreover, if Fo is a linearly ordered subset of (F, <;;;) (i.e., either f <;;; g or g <;;; f holds for any j, g E Fo), fo = U Fo is a function. (See Theorem 3.12 in Chapter 2.) It is easy to check that fo E F and fo is an upper bound on F0 in (F, <;;;). The assumptions of Zorn's Lemma being satisfied, we conclude that (F.<;;;) has a maximal element f. The proof is complete if we _show that dom 7 = S. If not, select some X E S- domf and x E X. Clearly. J = 7 U {(X, .r.)} E F and contradicting the maximality of f. 0

J::) ],

We conclude this section with a theorem needed in ChaptPr 11.

1.14 Theorem* lf(A, ~)is a linear ordering such that for all x E A, then IAI :S N,..

I{Y E A I y

~

x}l < N,

Proof. Exactly as in the proof of Zorn's Lemma, we construct an increasin~?; sequence (a€ I~ < ,\) of elements of A for which AJ.. = 0, i.e., there is no a E A such that a~ -< a holds for all ( E A. As (A.~) is linearly ordered. this means that, for every a E A, a ~ a~ for some ( E A. (We say that the sequence (a~ I ( < ,\) is cofinal in (A,~).) We have A= Ue
1.1 Prove: If a set A can be linearly ordered, then every system of finite subsets of A has a choice function. (It does not follow from the ZerrneloFraenkel axioms that every set can be linearly ordered.) 1.2 If A can be well-ordered, then P(A) can be linearly ordered. [Hi11t: Let < be a well-ordering of A; for X, Y <;;; A define X -< Y if and only if thP <-least element of X !::,. Y belongs to X.] 1.3* Let (A, :S) be an ordered set in which every chain has an upper bound. Then for every a E A, there is a $-maximal element x of A such that a::; x. 1.4 Prove that Zorn's Lemma is equivalent to the statement: For all (A. $), the set of all chains of (A, :S) has an ~-maximal element. 1.5 Prove that Zorn's Lemma is equivalent to the statement: If A is
CHAPTER 8. THE AXIOM OF CHOICE

144

1.6 A system of sets A has finite character if X E A if and only if every finite subset of X belongs to A. Prove that Zorn's Lemma is equivalent to the following ('Iukey's Lemma): Every system of sets of finite character has an <;;-maximal element. [Hint: Use Exercise 1.5.] 1. 7* Let E be a binary relation on a set A. Show that there exists a function f : A --+ A such that for all x E A, (x, f(x)) E E if and only if there is some y E A such that (x, y) E E. 1.8* Prove that every uncountable set has a subset of cardinality N1 . 1.9* Every infinite set is equipotent to some of its proper subsets. Equivalently, Dedekind finite sets are precisely the finite sets. 1.10* Let (A,<) be a linearly ordered set. A sequence (an I n E w) of elements of A is decreasing if an+ I < an for all nEw. Prove that (A,<) is a wellordering if and only if there exists no infinite decreasing sequence in A. 1.11* Prove the following distributive laws (see Exercise 3.13 in Chapter 2).

n(u

At .• )

=

u (n

tET sES

/EST tET

tET sES

/EST tET

At./(t))·

1.12* Prove that for every ordering~ on A, there is a linear ordering ::; on A such that a ~ b implies a ::; b for all a, bE A (i.e., every partial ordering can be extended to a linear ordering). 1.13* (Principle of Dependent Choices) If R is a binary relation on M f 1/J such that for each x E M there is y E M for which xRy, then there is a sequence (xn I n E w) such that XnRxn+l holds for all n E w. 1.14 Assuming only the Principle of Dependent Choices, prove that every countable system of sets has a choice function (the Axiom of Countable Choice). 1.15 If every set is equipotent to an ordinal number, then the Axiom of Choice holds. 1.16 If for any sets A and B either !AI ::; !B! or !BI ::; !AI, then thE' Axiom of Choice holds. [Hint: Compare A and B = h(A).] 1.17* If B is an infinite set and A is a subset of B such that !AI < !Bi. then

IB- AI=

2.

IB!.

The Use of the Axiom of Choice in Mathematics

In this section we present several examples of the use of the Axiom of Choice in mathematics. We have chosen examples which illustrate the variety of roles the Axiom of Choice plays in mathematics, but which do not require extensive background outside of set theory. Other applications of the Axiom of Choice can be found in the exercises and in most textbooks of general topology, abstract

2. THE USE OF THE AXIOM OF CHOICE IN MATHEMATICS

145

algebra, and functional analysis. The examples are followed by some discussion of importance and soundness of the Axiom of Choice.

2.1 Example. Closure Points. According to the usual definition, a sequence of real numbers (xn I n E N) converges to a E R if for every positive real number c there exists n., E N such that lxn - al < c holds for all natural numbers n 2: n.,. (In this example, Jxl denotes the absolute value of x, and not the cardinality of x.) Let A be a set of real numbers. Advanced calculus textbooks commonly characterize closure points of A in either (or both) of the following ways: (a) a E R is a closure point of A if and only if there exists a sequence (xn I n E N) with values in A, which converges to a. (b) a E R is a closure point of A if and only if for every positive real number c there exists x E A such that lx- al 0, there is n., E N such that lxn - al < e: for all n 2: n.,. In particular, lxn, - al < c and Xn, E A. (b) implies (a). The usual proof proceeds as follows: Let Xn = {x E A I lx- al < 1/n }. By (b), Xn # 0 for all n E N. Let (xn In E N) be a sequence such that Xn E Xn for all n E N. Then each Xn E A and (xn I n E N) converges to a. D The question usually passed over is, What reasons do we have to assume that any such sequence (xn I n E N) exists? Notice that we do not give any property P(x, y) such that P(n, y) holds if and only if y = Xn (for all n E N). Such a property can be exhibited in special cases (e.g., if A is open - seE> Exercise 2.1); however, it has been shown that the equivalence of (a) and (b) for all A <::; R cannot be proved from the axioms of Zermelo-Fraenkel set theory alone. Of course, if we do assume the Axiom of Choice, the fact that Xn =/; 0 for all n E N immediately implies that llnEN Xn =/; 0.

2.2 Example. Continuity of a Function. The standard definition of continuity of a real-valued function of a real variable is as follows: (a) f : R -+ R is continuous at a E R if and only if for every c > 0 there is 8 > 0 such that IJ(x) - f(a)J < c for all x such that lx- al < 8. Continuity is also characterized by the following property: (b) f : R -+ R is continuous at a E R if and only if for every sequence (xn I n E N) which converges to a, the sequence {! (xn) n E N) converges to f(a). It is easy to see that (a) implies (b): If (xn I n E N) converges to a and if c > 0 is given, then first we find o > 0 as in (a), and because (xn I n E N) converges, there exists n 0 such that lxn - al < t5 whenever n 2: n 0 . Clearly, lf(xn)- f(a)l < c for all such n. If we assume the Axiom of Choice, then (b) also implies (a), and hence (a) and (b) are two equivalent definitions of continuity. Suppose that (a) fails. then there exists c > 0 such that for each 8 > 0 there exists an x such that lx- al < t5 J

CHAPTER 8. THE AXIOM OF CHOICE

146

but lf(x)- f(a)i 2 e. In particular, for each k = 1, 2, 3, ... , we can choose some Xk such that !xk- a!< 1/k and if(xk)- f(a)l 2 c. The sequence (xk IkE N) converges to a, but the sequence (f(:rk) I k E N) does not convergP to /(a). so (b) fails as well. 0 As in Example 2.1, it can be shown that the equivalence of (a) and (b) cannot be proved from the axioms of Zermelo-Fraenkel set theory alone.

2.3 Example. Basis of a Vector Space. We assume that the reader is familiar with the notion of a vectm space over a field (e.g., over the tield of rPal numbers). Definitions and basic algebraic properties of vector spaces can b<· found in any text on linear algebra. A set A of vectors is linearly mdependent if no finite linear combination et1 v 1 + · · ·+On 'Vn of elements v1, ... , Vn of A with nonzero coefficients et1 . . . . . n,. from the field is equal to the zero vector. A basis of a vector space Vis a maximal (in the ordering by inclusion) linearly independent subset of V. One of the fundamental facts about vector spaces is 2.4 Theorem* Every vector space has a basis.

Proof. The theorem is a straightforward application of Zorn's Lemma: If C is a C-chain of independent subsets of the given vector space. then the union of C is also an independent set. Consequently, a maximal independent set exists. [For more details, see the special case in the next example.) Ll This theorem cannot be proved in Zermelo-Fraenkel set theory alone, without using the Axiom of Choice.

2.5 Example. Hamel Basis. Consider the set of all real numbers as a vector space over the field of rational numbers. By Theorem 2.4 this vector space has a basis, called a Hamel basis for R. In other words, a set X c;;; R is a Hamel basis for R if every x E R can be expressed in a unique way as X =

r1 ·

X1

+ · · · + rn

· Xn

for some mutually distinct x 1 , ... , X 11 E X and some nonzero rational nunJbers r 1 , ... , rn. Below we give a detailed argument that a set X with the last mentioned property exists. A set of real numbers X is called dependent if there are mutually distinct x 1 , ... , x,. E X and r 1 , ... , rn E Q such that r1 · X1 + · · · + rn · Xn

= 0

and at least one of the coefficients r1, ... , rn is not zero. A set which is not dependent is called independent. Let A be the system of all independent st':'ts of real numbers. We use Zorn's Lemma to show that A has a maximal element

2. THE USE OF THE AXIOM OF CHOICE IN MATHEMATICS

147

in the ordering by c;;;; the argument is then completed by showing that any c;;;-maximal independent set is a Hamel basis. To verify the assumptions of Zorn's Lemma. consider Ao c;;; A linearly ordered by c;;;. Let Xo == UAo; Xo is an upper bound of Ao in (A. c;;;) if Xo E .4. i.e., if X 0 is independent. But this is true: Suppose there were x,, ... , Xn E X 0 and r,, ... , r n E Q, not all zero. such that r, · .r1 + · · · + r n · Xn = 0. Then there would be X1, ... ,Xn E Ao such that X1 EX,, ... . Xn E Xn. Since Ao is linearly ordered by c;;;, the finite subset {X 1, ... , Xn} of Ao would have a c;;;-greatest element, say X,. But then x1, ... ,Xn E X,, so X, would not be independent. By Zorn's Lemma, we conclude that (A, c;;;) has a maximal element X. It remains to be shown that X is a Hamel basis. Suppose that x E R cannot be expressed as r 1 · x1 + · · · + rn · Xn for any r1, ... ,rn E Q and XI, ... ,Xn E X. Then x ¢ X (otherwise x = 1 · x), so X U { x} :J X, and X U { x} is dependent (remember that X is a maximal independent set). Thus there are xl,··· ,Xn EX U {x} and s1, . ... sn E Q, not all zero. such that s 1 · x 1 + · · · + Sn · Xn = 0. Since X is independent. x E {:r 1 .... .Xn }: say. x = .r,, and the corresponding coefficient s, f. 0. But then

= ( - ::) · X l

+ .. · + (- S~~ I )

· X;- I

+ (- S~: I )

·

x,+ 1

+ .. · + (- :~) · .Tn,

where x 1 , ... , x,_ 1 , Xi+!, ... , Xn EX, and the coefficients are rational numbers. This contradicts the assumption on x. Suppose now that some x E R can be expressed in two ways: .T = r 1 · X1 + · · · + Tn · Xn = s1 · Yl + ·· · + Sk · Yk, where XI,·· . . Xn·YI· . .. ·Yk E X. r 1 , • • . , r n, s 1 , . . . , s k E Q - {0}. Then (2.6)

r1 · X1

+ · · · + Tn

·

Xn - S! ·

Yl - · · · -

Sk · Yk =

0.

If {xi,· .. ,xn} f. {y,, ... ,yk} (say, x, ¢ {yl,··· ,yk}), (2.6) can be written as a combination of distinct elements from X with at least one nonzero coefficient (namely, r 1 ), contradicting independence of X. We can conclude that n = k and x 1 = y,,, ... , Xn = y,,. for some one-to-one mapping (i1, ... , in) between indices 1, 2, ... , n. We can thus write (2.6) in the form (r 1 -s,, )·x 1 +· · ·+(rn -s,, )·:r,. = 0. Since x 1 , ..• , Xn are mutually distinct elements of X. we conclude that TJ - S 11 = 0, .... T n - S;,. = 0, i.e., that Tl = S; 1 , . . . , r n = S 1 , . These arguments show that each x E R has a unique expression in the desired form and so X is a Hamel basis. The existence of a Hamel basis cannot be proved in ZermelO-Fraenkel theory alone; it is necessary to use the Axiom of Choice.

sPt

2.7 Example. Additive Functions. A function f : R ~ R is called additive if f(x + y) = f(x) + f(y) for all x, y E R. An example of an additive function if provided by fa where fa(x) = a· x for all x E R. and a E R is fixed.

CHAPTER 8. THE AXIOM OF CHOICE

148

Easy computations indicate that any additive function looks much like fa for some a E R. More precisely, let f be additive, and let us set f(1) =a. We then have

f(2) = f(l)

+ f(l) =a· 2,

f(3) = f(2) + f(l) =a· 3,

and, by induction, f(b) = a·b for all bEN- {0}. Since f{O) + f(O) = f(O+O) = f(O), we get f(O) = 0. Next, f( -b) -r f(b) = f(O) = 0, so f( -b) = - f(b) = a· (-b) for bE N. To compute f(l/n), notice that

a= f(1) = f(1/n) + · · · + f(1/n) = n · f(1/n); n times

consequently, f(l/n) = a·lfn. Continuing along these lines, we can easily prow that f(x) = a· x for all rational numbers x. It is now natural to conjecture that f(x) = a· x holds for all real numbers x; in other words, that every additive function is of the form fa for some a E R. It turns out that this conjectur(' cannot be disproved in Zermelo-Fraenkel set theory, but that it is false if WP assume the Axiom of Choice. We prove the following theorem.

2.8 Theorem* There exists an additive function for all a E R.

Proof. f(x)

f: R __. R such that f f. fa

Let X be a Hamel basis for R. Choose a fixed

=

{r, 0

if X=~~ · Xt otherw1se.

+ · · · + r;

· x,

+ · · · + Tn

xE

· Xn

X. Define

and x, =

x,

It is easy to check that f is additive. Notice also that 0 ¢ X and X is infinite (actually, lXI = 2l't"). We have f(x) = 1 (because x = 1 ·xis the basis representation of x), while f(x) = 0 for any x E X, x f. x (because x does not occur in the basis representation of = 1 · of X.). Iff = fa were to hold for some a E R, we would have f(x) = 1 = a · x, showing a f. 0, and, on the other hand, f(x) = 0 = a· x, showing a = 0. 0

x

x

2.9 Example. Hahn-Banach Theorem. A function f defined on a vector space V over the field R of real numbers and with values in R is called a linmr functional on V if f(o: · u + /3 · v) = o: · f(u) + /3 · f(v) holds for all u, v E V and o:, {3 E R. A function p defined on V and with values in R is called a sublinear functional on V if p(u + v)::; p(u) + p(v) for all u,v E V and

p(o: · u)

= o: · p(u)

for all u E V and o: 2 0.

2. THE USE OF THE AXIOM OF CHOICE IN MATHEMATICS

149

The following theorem, due to Hans Hahn and Stefan Banach, is one of the cornerstones of functional analysis. 2.10 Theorem* Let p be a sublinear functional on the vector space V. and let fo be a linear functional defined on a subspace Vo of V such that fo(v) ::; p(v) for all v E Vo. Then ther-e is a linear functional f defined on V such that f 2 fo and f(v) ::; p(v) for all v E V.

Proof. Let F be the set of all linear functionals g defined on some subspace W of V and such that

fo

~

g and

g(v)::; p(v) for all v

E

W.

We obtain the desired linear functional f as a maximal element of (F, ~). To verify the assumptions of Zorn's Lemma, consider a nonempty F0 ~ F linearly ordered by~. If go= UFo, go is an ~-upper bound on Fo provided go E F. Clearly, go is a function with values in Rand go 2 fo. Since the union of a set of subspaces of v linearly ordered by~ is a subspace of V, dom go = ugEFo dom g is a subspace of V. To show that g0 is linear consider u, v E dom go and n, (3 E R. Then there are g, g' E Fo such that u E dom g and v E dom g'. Since F0 is linearly ordered by ~. we have either g ~ g' or g' ~ g. In the first case. u,v,n·u+fJ·v E domg' andgo(n·u+(3·v) = g'(n·u+fJ·v) = n·g'(u)+P·g'(v) = n · g0 (u) + (3 · g0 (v); the second case is analogous. Finally, g0 (u) = g(u) ::; p(u) for any u E dom go and g E Fo such that u E dom g. This completes the check of go E F. By Zorn's Lemma, (F, ~) has a maximal element f. It remains to be shown that dom f = V. We prove that dom f C V implies that f is not maximal. Fix u E V- dom f; let W be the subspace of V spanned by dom f and u. Since every w E W can be uniquely expressed as w = x + n · u for some x E dom f and n E R, the function fc defined by

fc(w)

= f(x) + n

·c

is a linear functional on W and fc :::> f. The proof is complete if we show that c E R can be chosen so that (2.11)

fc(x + n · u)

= f(x)

+ n · c::; p(x + n · u)

for all x E dom f and n E R. The properties off immediately guarantee (2.11) for n = 0. So we have to choose c so as to satisfy two requirements: (a) For all n > 0 and x E dom f, f(x) + n · c::; p(x + n · u). (b) For all n > 0 andy E domf, f(y} + ( -n) · c::; p(y + ( -n) · u). Equivalently,

f(y)- p(y- n · u)::; n · c::; p(x + n · u)- f(x)

CHAPTER 8. THE AXIOM OF CHOICE

150 and then (2.12)

should hold for all x, y E umn f and n

J(v)

+ f(t)

= f(v

+ t)::;

> 0. But, for all v, t

p(v

E dom

f.

+ t)::; p(v- u) + p(t + u)

and thus f(v)- p(v- u)::; p(t

+ u)-

f(t).

If A= sup{f(v) -p(v-u) I u E domf} and B = inf{p(t +u)- f(t) It E domf}. we have A ::; B. By choosing c such that A ::; c::; B, we can make the id<•ntit_,. (2.12) hold.

2.13 Example. The Measure Problem. An important problem in analvsis is to extend the notion of length of an interval to more complicatPd spts of rPal numbers. Ideally, one would like to have a function J1 defined on P( R). with values in [0. 2>0) U { oo}. and having the following properties: 0) Jl([a, b]) = b - a for any a. b E R. a < b. i) J1(0) =

o. Jl(R)

=

oo.

ii) If {An}~=O is a collection of mutually disjoint subsets of R. then

(This property is called countable additivity or a-additivity of Jl.) iii) If a E R, A ~ R. and A+ a = {x +a (translation invariance of Jl).

Ix

E A}, then Jl(A +a) = Jl(.rl)

Several additional properties of Jl follow immediately from 0)·-iii) (ExPrcisP 2.3): iv) If

An B

=

0 then 11(A U B)= Jl(A) + Jl(B) (finite additivity).

v) If A~ B then Jl(A)::; {t(B) (monotonicity). However, the Axiom of Choke implies that no function J1 with tlw aho\'<'mentioned properties exists:

2. THE USE OF THE AXIOM OF CHOICE IN MATHEMATICS

151

2.14 Theorem* There is no function p. : P(R) ___. [O.oo) U {oo} 1.1.1ith thr properties 0)-v).

Proof.

We define an equivalence relation

x ::::: y

~

on R by:

if and only if x - y is a rational number,

and use the Axiom of Choice to obtain a set of representatives X for easy to se~· that (2.15)

R =

u{x + r

~-

It is

Iris rational};

moreover, if q and T are two distinct rationals, then X + q and X+ r are disjoint. We note that p.(X) > 0: if p.(X) = 0, then p.(X + q) = 0 for every q E Q, and

p.(R)

=L

p.(X

+ q)

= 0,

qEQ

a contradiction. By countable additivity, there is a dosed interval [a. b] such that p.(X n [a. b]) > 0. Let Y = X n [a, b]. Then

U

(2.16)

(Y+q)~[a,b+1]

qEQn[O,!J

and the left-hand side is the union of infinitely many mutunlly disjoint sets Y + q, each of measure p.(Y + q) = p.(Y) > 0. Thus the left-hand sid(' of (2.1G) D has measure oo, contrary to the fact that p.([a, b + 1]) = b + 1 - a. The theorem we just established demonstrates that some of the requirements on p. have to be relaxed. For the purposes of mathematical analysis it is most fruitful to give up the condition that J-L is defined for all subsets of R, and require only that the domain of p. is closed under suitable !iet operations. 2.17 Definition LetS be a nonempty set. A collection 6 ~ P(S) is a(]"algebra of subsets of S if (a) 0 E 6 and S E 6. (b) If X E 6 then S- X E 6. (c) If Xn E 6 for all n, then U~o Xn E 6 and Xn E 6.

n:=O

2.18 Definition A (]"-additive measure on a (]"-algebra 6 of subsets of S is a function p. : 6 __. [0, oo) U { oo} such that

i) p.(0)

= 0, p.(S) > 0.

ii) If {Xn}~=O is a collection of mutually disjoint sets from 6. then

152

CHAPTER 8. THE AXIOM OF CHOICE

The elements of 6 are called J-L-measurable sets. The reader can find some simple examples of a-algebras and a-additive measures in the exercises. In particular, P(S) is the largest a-algebra of subsets of S; we refer to a measure defined on P(S) as a measure on S. The theorem we proved in this example can now be reformulated as follows.

2.19 Corollary Let p. be any a-additive measure on a a-algebra 6 of subsets of R such that (0} [a, b] E 6 and p.([a, b]) = b- a, for all a, bE R, a < b. (iii) If A E 6 then A+ a E 6 and p.(A +a)= J-L(A), for all a E R. Then there exist sets of real numbers which are not p.-measurable.

0

In real analysis, one constructs a particular a-algebra 9R of Lebesgue measurable sets, and a a-additive measure p. on !m, the Lebesgue measure, satisfying properties (0) and (iii) of the Corollary. So existence of Lebesgue nonmeasurable sets is a consequence of the Axiom of Choice. Robert Solovay showed that the Axiom of Choice is necessary to prove this result. Other ways of weakening the properties 0)-iv) of p. have been considered. and led to very interesting questions in set theory. For example. it is possibl<> to give up the requirement iii) (translation-invariance) and ask simply whether there exist any a-additive measures p. on R such that p.([a, b]) = b - a for all a, b E R, a < b. This question has deep connections with the theory of large cardinals, and we return to it in Chapter 13. Another interesting possibility is to give up ii) (countable additivity) and require only iv) (finite additivity). Nontrivial finitely additive measures do exist (assuming the Axiom of Choice) and we study them further in Chapter 11. Let us now resume our discussion of various aspects of the Axiom of Choice. First of all, there are many fundamental and intuitively very acceptable results concerning countable sets and topological and measure-theoretic properties of the real line, whose proofs depend on the Axiom of Choice. We have seen two such results in Examples 2.1 and 2.2. It is hard to image how one could stud_.,· even advanced calculus without being able to prove them, yet it is known that they cannot be proved in Zermelo-Fraenkel set theory. This surely constitutPs some justification for the Axiom of Choice. However, closer investigation of thE> proofs in Examples 2.1 and 2.2 reveals that only a very limited form of th(' Axiom is needed; indeed, all of the results can still be proved if one assumes only the Axiom of Countable Choice. Axiom of Countable Choice system of sets.

There exists a choice function for every countablt>

It might well be that the Axiom of Countable Choice is intuitively justified. but the full Axiom of Choice is not. Such a feeling might be strengthened by realizing that the full Axiom of Choice has some counterintuitiw c:ousC'quPncPs.

2. THE USE OF THE AXIOM OF CHOICE IN MATHEMATICS

153

such as the existence of nonlinear additive functions from Example 2. 7, or the existence of Lebesgue nonmeasurable sets from Example 2.13. Incidentally, none of these consequences follows from the Axiom of Countable Choice. In our opinion, it is applications such as Example 2.9 which mostly account for the universal acceptance of the Axiom of Choice. The Hahn-Banach Theorem, Tichonov's Theorem (A topological product of any system of compact topological spaces is compact.) and Maximal Ideal Theorem (Every ideal in a ring can be extended to a maximal ideal.) are just a few examples of theorems of sweeping generality whose proofs require the Axiom of Choice in almost full strength; some of them are even equivalent to it. Even though it is true that we do not need such general results for applications to objects of more immediate mathematical concern, such as real and complex numbers and functions, the irreplaceable role of the Axiom of Choice is to simplify general topological and algebraic considerations which otherwise would be bogged down in irrelevant set-theoretic detail. For this pragmatic reason, we expect that the Axiom of Choice will always keep its place in set theory. Exercises

2.1 Without using the Axiom of Choice. prow that the two definitions of closure points are equivalent if A is an open set. [Hint: Xn is open, so Xn n Q =/; 0, and Q can be well-ordered.] 2.2 Prove that every continuous additive function f is equal to fa for some

a E R. 2.3 Assume that J.L has properties 0)-ii). Prove properties iv) and v). Also prove: vi) J.L(A u B) vii)

= J.L(A) + J.L(B)

- J.L(A n B).

J.L(U;::o An) ::; I:;::o J.L(An)·

2.4* Let 6 = {X ~ S I lXI ::; No or IS- Xi ::; No}. Prove that 6 is a a-algebra. 2.5 Let
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Chapter 9

Arithmetic of Cardinal Numbers 1.

Infinite Sums and Products of Cardinal Numbers

In Chapter 5 we introduced arithmetic operations on cardinal numbers. It is reasonable to generalize these operations and define sums and products of infinitely many cardinal numbers. For instance, it is natural to expect that 1 + 1 + ·· · = ....__,

t{o

No times

or, more generally,

K+K+···=K·A.

'-...--' >.times

The sum of two cardinal numbers K1 and K2 was defined .as the cardinality of A1 U A2, where A1 and A2 are disjoint sets such that /AJ/ = KJ and /A2/ = ~2· Thus we generalize the notion of sum as follows. 1.1 Definition Let (A; / i E /) be a system of mutually disjoint sets. and lot /A,/ = K, for all i E /. We define the sum of (K, I i E /) by

:EK; = iE 1

juA.j. tEl

The definition of I:,El K; uses particular sets A, (i E /). In the finite case. when I= {1, 2} and KJ + K2 = IA1 U A2l, we have shown that the choice of A1 and A 2 is irrelevant. We have proved that if A~, A~ is another pair of disjoint sets such that lA~ I= Kt, /A~I = K2, then /A! U A2/ =/At U A2/. 155

156

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

In general, one needs the Axiom of Choice in order to prove the corresponding lemma for infinite sums. Without the Axiom of Choice, we cannot exclude t.lw following possibility: There may exist two systems (An In E N), (A~ In E N) of mutually disjoint sets such that each An and each A~ has two elements, but U::o An is not equipotent to U;:o A~! For this reason, and because many subsequent considerations depend heavily on the Axiom of Choice, we use the Axiom of Choice from now on without explicitly saying so each time.

1.2 Lemma If (A, I i E I) and (A; I i E I) are systems of mutually disjoint sets such that IA,I == lA: I for all i E I, then I UEI A.l =I Ue 1 A; I. Proof. For each i E I, choose a one-to-one mapping J, of A, onto A:. Then f = u,E[ f, is a one-to-one mapping of uiE/ A, onto uiE/ A;. 0

This lemma makes the definition of I:iel K; legitimate. Since infinite unions of sets satisfy the associative law (Exercise 3.10 in Chapter 2), it follows that thf' infinite sums of cardinals are also associative (see Exercise 1.1). The operation 2.:: has other reasonable properties, like: If K, $ A, for all i E I, then l.::;e 1 K, $ I:.ei A, (Exercise 1.2). However, if K, < A; for all i E I, it does not necessarily follow that 2.::; K, < 2.::; A; (Exercise 1.3). If the summands are all equal, then the following holds, as in the finite case: If K, = K for all t E A, then

LK·=~=K·A. lE..\

..\time"'

(Verify! See Exercise 1.4.) It is not very difficult to evaluate infinite sums. For example. consider

L

n

= 1 + 2 + 3 + · · · + n + ·· ·

(n E N).

nEN

It is easy to see that this sum is equal to t{o. In fact, this follows from a general theorem.

1.3 Theorem Let A be an infinite cardinal, let numbers, and let K = sup{ Ka I o < A}. Then

L

Ka =A.

t>

Ka

=A. sup{Ka

(o < A) be nonzero cardinal

I 0'
o<>.

Proof. On the one hand, K0 $ K for each a < A, and so 2.::(\<>. ~"a $ I:a<>. K = K ·A. On the other hand, we notice that A = 2.::(\<>. 1 $ I:o<>. K(\. We also have K $ 'L:c.<>. K(\: the sum I:a<-\ K0 is an upper bound of the K(\'s and K is the least upper bound. Now since both"' and A are $ I:o<>. K0 , it follows that K · A, which is the greater of the two, is also $ 2.::(\< >. Ka. The conclusion of Theorem 1.3 is now a consequence of the Cantor-Bernstein Theon~m. ,-

1. INFINITE SUMS AND PRODUCTS OF CARDINAL NUMBERS 1.4 Corollary If

K;

157

(i E I) are cardinal numbers, and if \I\ ~ sup{ K, \ i E I},

then

L

K;

==sup K,.

iE/

tE I

(In particular, the assumption is satisfied if all the

K;

's are mutually distinct.) 0

The product of two cardinals K 1 and K 2 has been defined as the cardinality of the cartesian product A 1 x A2, where A 1 and A 2 are arbitrary sets such that /Ad = K! and /A 2 / = K2. This is generalized as follows.

1.5 Definition Let (A;

Ii

E

I) be a family of sets such that

/A;\

=

K,

for all

i E I. We define the product of (K; / i E I) by

IIK;=!IIA;,. iEI

iEI

We use the same symbol for the product of cardinals (the left-hand side) as for the cartesian product of the indexed family (K; / i E I). It is always clear from the context which meaning the symbol rr has. Again, the definition of sets A;.

rr.EI K,

does not actually depend on the particular

1.6 Lemma If (A; I i E I) and (A~ i E I, then I rriEI A;/= I niEI A~/.

Ii

E I) are such that

/A,\

=

\A;\

for all

Proof. For each i E I, choose a one-to-one mapping f; of A, onto A;. Let be the function on rriE I A; defined as follows: If X = (x; I i E I} E rr,E I A,. let f(x) = (f,(x,) I i E I). Then f is a one-to-one mapping of A, onto [liE/A~. 0

f

nei

The infinite products have many properties of finite products of natural numbers. For instance, if at least one K; is 0, then rriEI K, = 0. The products also satisfy the associative law (Exercise 1. 7); another simple property is that if K; ~ A, for all i E I, then [J,EI K; ~ [liEf..\; (Exercise 1.8). If all the factors K1 are equal to K, then we have, as in the finite case,

(Verify! See Exercise 1.10.) The following rules, involving exponentiation, also generalize from the finite to the infinite case (see Exercises 1.11 and 1.12):

iE I

rr(KA;) iE/

iE/

= KE•EI A,.

CHAPTER 9. ARJTHMETIC OF CARDINAL NUMBERS

158

Infinite products are more difficult to evaluate than infinitl" sums. In sonw special Cast'S. for instance when evaluating the product On<.>. h~,. of illl iucreasiug sequence (r;,,. I o < ,\) of cardinals. some simple rules can bP provPd. We cousidf'r only the following very special casP: 00

II n == 1 · 2 · 3 · · · · · n · · · ·

(n E N).

n=l

First, we note that 00

00

rt=!

n=1

IT n S IT ~o

=

~~"

= 2N".

Conversely, we have

n=2

t=l

n=l

and so we conclude that

1 · 2 · 3 · · · · · n · . · · = 2N". We now prove an important theorem. which can be used to derivP various iut>qualities in cardinal arithmetic. 1. 7 Konig's Theorem < ,\, for· all i E I. then

l'i. 1

If

t>,

and

>., (i

E I) are car·dirw/ numbpr·s. and 1f

L ,

tEl

tE I

First, let us show that 2:,.;:/ f\,1 s n,EJ ,\,, Let (A, I i E I) aud I) be such that )A, I = /'\,' and IB, I = >., for all i E I and tlw .4, 's a!'(' mutually disjoint. We may further fiSsumt> that A, C B, for all i E !. \Vt> hurl a one-to-one mapping f of UtE/ A, into n,E/ B,. \Ve choose d, E B; - A; for each i E I, and define a function f as follows: For each x E UtE/ A;, let ix be the unique i E I such that .r E A,. Ld f(x) =(a, I i E I), where

Proof.

(B, I i

E

if i = i:r.

X

a, =

{

d,

ifi/oi:r.

If .1· /o y, let f(x) =a and f(y) =band let us show that a of b. If i, == ly = 1. then a, = x while b, = y. If ix of iy = i. then a, = d, ~ A while b, = y E A. In either case. f(x) /o f(y) and hence f is one-to-one. Now let us show that 2:t "'• < 0, At· Let B, (i E I) be such that )B,I = ,\,. for all i E /. If the product 0, >., were equal to the sum 2:, f\, 1 • we could find mutually disjoint subsets X, of the cartesian product. 0,E 1 B, such that !X,I = "• ffH all i and

ux.

•E I

=

IIB,. tEl

1. INFINITE SUMS AND PRODUCTS OF CARDINAL NUMBERS

159

B; b;

A,

I

Figure 1 We show that this is impossible. For each i E I, let

A, = {a, I a EX;}

(1.8)

(s~·e Fi~?,ure

1).

For every i E I, we have A; C B., since IA,I :S jX,j == t>, < >., = IB,j. Hence then, exists b, E B, such that b, ¢ A;. Let b = ( b, I i E I). Now Wf' can Pasilv show that b is not a member of any X, (i E I): For any i E I. b, tf. A,. and so by (1.8), b tf. X,. Hence uiE/ X; is not the whole set n.El B;, a contradiction. 0 We use Konig's Theorem in Section 3; at present, Jet us just mention that the theorem (and its proof) are generalizations of Cantor's Theorem which stat<.'s that 2" > K for all t>. If we express K as the infinite sum "- = 1

+ 1 + ·· ·

("-times)

and 2" as the infinite product 2" = 2 · 2 · 2 · · · ·

(K times).

we can apply Konig's Theorem (since 1 < 2) and obtain

"'= 2:1
2".

tE~<

Exercises

1.1 If J, (i E I) are mutually disjoint sets and J =

are cardinals. then

L(L tEl jEJ,

1.2

Kj)

=

L

U,E/ .!,. and if t> 7 (.i E .J)

t>)

.JEJ

(associativity of 2:.::). s >., for all ·i E I, then LiE/ "-i $LiE/>.,.

If,._,

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

160

1.3 Find some cardinals Kn, An (n E N) such that ~"n < I::=O Kn = 2.:::,0 An. 1.4 Prove that K + K + · · · (A times) == A· K. 1.5 Prove the distributive law: A.

1 u.El

for all n, but

(L Kt) = L(A. K,). tE/

tEl

1.6

An

A, I ~ I:tEliA,J.

1.7 If Jt (i E /) are mutually disjoint sets and J = UiEl J., and if Kj (J E J)

are cardinals. then

II KJ) =II II( tEl

Kj

jEJ

JEJ;

rn.

(associativity of Kt ~ A; for all i E /, then

1.8 If

II tEl 1.9 Find some cardinals

n:=o = n:=O Prove that Kn

Kn, An

K;

~II A,. iEl

(n E N) such that

Kn

<

An

for all n, but

An.

K • K · · · · (A times) = K>._ 1.10 1.11 Prove the formula (0,Ef K,)" = niEl(Kt}. [Hint: Generalize the proof of the special case (K~")>. = (K>.)~", given in Theorem 1.7 of ChapterS.]

1.12 Prove the formula

II (K"') = K,L•E

I..\,.

tEl

[Hint: Generalize the proof of the special case K>. · K~" = ">-+~> given iu Theorem 1.7(a) of Chapter 5.] 1.13 Prove that if 1 < K, ~A, for all i E /,then I:.El Kt:::; n.E[ A,. [Answer: 2N' .] 1.14 Evaluate the cardinality of Oo
2.

Regular and Singular Cardinals

Let (o.., I v < {)) be a transfinite sequence of ordinal numbers of length {). We say that the sequence is increasing if o.., < o~" whenever v < J1. < 1'J. If d is a limit ordinal number and if (ov I v < f)) is an increasing sequence of ordinals. we define o = limo,.. = sup{o.., I v
and call o the limit of the increasing sequence. 2.1 Definition An infinite cardinal K is called smgular if there exists an increasing transfinite sequence (o.., I v < 1'J) of ordinals o.., < K whose length {) i:-;

161

2. REGULAR AND SINGULAR CARDINALS

a limit ordinal less than K, and singular is called regular. A subset X

~ K

=

K

limv ..... ~ av. An infinite cardinal that is not

is bounded if sup X <

K,

and unbounded if sup X

= K.

2.2 Theorem Let

K be a regular cardinal. (a) If X~ K is such that lXI < K then X is bounded. Hence every unbounded subset of K has cardinality K. (b) If .X< K and f: A-+ K, then /[A] is bounded.

Proof. (a) This is clear if X has a greatest element. Thus assume that the order type of X is a limit ordinal, and let {a.., I 11 < 19} be an increasing enumeration of X. As 1191 = lXI < K, we have 19 < K, and because K is a regular cardinal. it follows that sup X = lim.., .... ~ a.., < K. (b) As lf[A]j $ A < K, this follows from (a). D

An example of a singular cardinal is the cardinal t(,; we have

where w < t{w and t{n < t{w for each n. Similarly, the cardinals t{w+w• Nw·w, Nw, are singular:

On the other hand, 1{ 0 is a regular cardinal. The following lemma gives a different characterization of singular cardinals.

2.3 Lemma An infinite cardinal less than

K

smaller cardinals:

K

K

is singular if and only if it is the sum of

= LiE I K;,

where III <

K

and

K,

<

K

for all

i E I.

If K is singular, then there exists an increasing transfinite sequence Proof. such that K = lim.., .... ~ a..,, where ~ < K and av < K for all 11 < {). Since every ordinal is the set of all smaller ordinals, we can reformulate this as follows:

If we let A.., = av - u~
sets of cardinality

Kv

= jA..,I

=

lav-

I II < f)) u{
is a sequenc!" of fl"wer than :::: lo..,l < K, and since the

162

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

Av's are mutually disjoint, this shows that K = l : v d Kv as required. Thus tlw condition in Lemma 2.3 is necessary. To show that the condition is sufficient.. let us assume that t> = l::o< >. ~; 0 . where >. is a cardinal less than K, and for all o < >., Ka are cardinals srnallPr than K. By Theorem 1.3, K = .X· supa<>- K0 , and since >. < K. we necessaril:; have t> = supa<>- Ka. Thus the range of the transfinite sequence (tin I o < >.) has supremum K, and since Ka < t> for all o < >., we can find (by transfinitP recursion) a subsequence which is increasing and has limit K. Clearly. the length of the subsequence is a limit ordinal rJ :::; >.. and it follows that t> is singular.

An infinite cardinal t{o is called a successor· cardinal if its index u is 11 successor ordinal, i.e., if t{" = 1{;3+1 for some /3. If"- = Nti then we call N,J+ 1 the successor of K and denote it t>+. If o is a limit ordinal, then t{" is call<•d a limit cardinal. If o > 0 is a limit ordinal, then 1{ 0 is the limit of the sequf'ncP (t{u I /3 < o:). 2.4 Theorem Every successor car·dinal Na+t is a regular cardinal. Proof. cardinals:

Otherwise, t{n+ 1 would be the sum of a smaller number of smaller Nn+1 =

LK,, •El

where Ill < No+ 1, and all i E /, and we have Nn+l =

t>,

L •EI

< No+ 1 for all i K; :::;

E /.

Then

III :::;

N" and ~;, :::; Nn

f(H

L ~{"=No· Ill :::; t{,. · t{, = N". zE/

This is a contradiction, and hence t{a+l is regular.

0

By Theorem 2.4, every singular cardinal is a limit cardinal. Let us look now at limit cardinals. 2.5 Lemma There are arbitrarily large singular· cardinals. Proof.

Let t{<> be an arbitrary cardinal. Consider the sequence

Then and hence Na+w is a singular cardinal greater than N".

0

All uncountable limit cardinals we have seen so far were singular. A question naturally arises whether there are any uncountable regular limit cardinals. Suppose that N" is such a cardinal. Since C} is a limit ordinal. Wf' have t{" = lim

J~a

I{ a,

2.

REGULAR AND SINGULAR CARDINALS

163

i.e., 1{ 0 is the limit of an increasing sequence of length a. Since N0 is regular, we necessarily have a 2: N0 , which together with (I :::; t{o gives

(*) Already this property suggests that Na has to be very large. Although the condition (*) seems to be strong, it is not as strong as it looks. 2.6 Lemma There are arbitrarily large singular cardinals Ncr such that Nn = Proof. sequence:

Let

t{"'

be an arbitrary cardinal. Let us consider the following

for all n E N, and let o = limn~w D'n· It is clear that the sequence (Nn .. N) has limit 1{ 0 . But then we have t{a =

Since

1{ 0

(1.

lim

n--+w

t{o,.

= n--+w lim On+!

! 11

E

= o.

is the limit of a sequence of smaller cardinals of length w, it is singular. 0

An uncountable cardinal number 1{ 0 that is both a limit cardinal and regular is called inaccessible (it is often called weakly inaccessiblro to distinguish this kind from cardinals that are defined by a stronger property - see Section 3). It is impossible to prove that inaccessible cardinals exist using only the axioms of Zermelo-Fraenkel set theory with Choice. 2. 7 Definition If o is a limit ordinal, then the cofinality of a, cf(o), is the least ordinal number {) such that o is the limit of an increasing sequence of ordinals of length 1'J. (Note that cf(a) is a limit ordinal and cf(a) ::; a.) Thus t{" is singular if ) < W 0 and is regular if cf(w 0 ) = w 0 . Let o be a limit ordinal which is not a cardinal number. If we let K = I(II. there exists a one-to-one mapping of"' onto a, or, in other words. a one-to-one sequence (o.., I v < K) of length "' such that {a.., I v < "'} = o. Now wt> can find (by transfinite recursion) a subsequence which is increasing and has limit o. Since the length of the subsequence is at most K, and since "' = lal is less than a (because o is not a cardinal), we conclude that d(a) < a. Thus we have proved the following: cf(w 0

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

164

2.8 Lemma If a limit ordinal a is not a cardinal, then cf(a) < a.

0

As a corollary, we have, for all limit ordinals a,

cf(o) =a if and only if a is a regular cardinal. 2.9 Lemma For every limit ordinal a, cf(cf(a)) = cf(a).

Proof Let 19 = cf(a). Clearly, 19 is a limit ordinal, and cf(19) ::; fJ. We have to show that cf(19) is not smaller than 19. If 1 = cf(fJ) < 19, then there exists an increasing sequence of ordinals (v~ I ( < r) such that lim€~-r v€ = 1?. Since 1'J = cf(a), there exists an increasing sequence of ordinals (a.., I v < tJ) such that limv-t9 a.., = a. Then the sequence (a..,~ I ( < 1) has length 1 and lim~--, a..,~ = a. But 1 < fJ, and we reached a contradiction, since 1'J is supposed to be the least length of an increasing sequence with limit a. 0 2.10 Corollary For every limit ordinal a, cf(a) is a regular cardinal.

0

Exercises

2.1 cf(Nw) = cf(Nw+w) = W. 2.2 cf(Nw 1 ) = Wt, cf(Nw 2 ) = W2· 2.3 Let a be the cardinal number defined in the proof of Lemma 2.6. Show that cf(a) = w. 2.4 Show that cf(a) is the least 1 such that a is the union of 1 sets of cardinality less than jaj. 2.5 Let N0 be a limit cardinal, a > 0. Show that there is an increasing sequence of alephs of length cf(Nct) with limit Net. 2.6 Let "- be a limit cardinal, and let ). < "- be a regular infinite cardinal. Show that there is an increasing sequence (a.., I v < cf(K-)) of cardinals such that limv~cf(~<) a..,="- and cf(a..,) =A for all v.

3.

Exponentiation of Cardinals

While addition and multiplication of cardinals are simple (due to the fact that N0 + Nt3 = Na · N(3 = the greater of the two), the evaluation of cardinal exponentiation is rather complicated. Here, we do not give a complete set of rules (in fact, in a sense, the general problem of evaluation of,_>. is still open), but prove only the basic properties of the operation ,_>.. It turns out that there is a difference between regular and singular cardinals. First, we investigate the operation 2N ... By Cantor's Theorem, 2N .. > N<>; in other words, (3.1)

165

3. EXPONENTIATION OF CARDINALS

Let us recall that Cantor's Continuum Hypothesis is the conjecture that 2N" = N1 . A generalization of this conjecture is the Generalized Continuum Hypothesis: 2N., = Na-+l for all a. As we show, the Generalized Continuum Hypothesis greatly simplifies the cardinal exponentiation; in fact, the operation ,_>. can then be evaluated by very simple rules. The Generalized Continuum Hypothesis can be neither proved nor refuted from the axioms of set theory. (See the discussion of this subject in Chapter 15.) Without assuming the Generalized Continuum Hypothesis, there is not much one can prove about 2N.. except (3.1) and the trivial property: (3.2) The following fact is a consequence of Konig's Theorem. 3.3 Lemma For every a, (3.4) Thus 2N" cannot beN"', since cf(2Nw) =No, but the lemma does not prevent 2N" from being No.~,. Similarly, 2N' cannot be either No.~, or N"' or Nw+w• etc. Proof Let 19 = cf(2N .. ); 19 is a cardinal. Thus 2N .. is the limit of an increasing sequence of length 19, and it follows (see the proof of Lemma 2.3 for details) that

where each "-u is a cardinal smaller than 2N ... By Konig's Theorem (where we let ..\" = 2N .. for all v < 19), we have

and hence 2N., < (2N" )19 • Now if 19 were less than or equal to N0 , we would get 2N" < (2N" )19 ~ (2N" )N" a contradiction.

= 2N"

N., == 2N",

0

The inequalities (3.1), (3.2), and (3.4) are the only properties that can be proved for the operation 2N .. if the cardinal Na- is regular. If N0 is singular, then various additional rules restraining the behavior of 2N., are known. We prove one such theorem here (Theorem 3.5); in Chapter 11 we prove Silver's Theorem (Theorem 4.1): If N0 is a singular cardinal of cofinality cf(N,) ~ N1 , and if 2N( = NHI for all~< a, then 2Na = Na+l·

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

166

3.5 Theorem Let N, be a singular cardinal. Let us assume that the value oj 2N< is the same for· all~< n, say 2N< = Ne. Then 2N .. =No. Note that it is implicit in the theorem that No is greater than N,. instance, if we know that 2N" = Nw+G for all n < :.v, then 2N~ = Nw+5·

For

Since N" is singular. there exists, by Lemma 2.3, a collection (n,, Ill = N'Y is a cardinal less than N", and N" = L,E/ K,. By the assumption, we haVf' 2"· = N11 tor all i E /, and also 2N, = Nri· so

Pmof.

i E I) of cardinals such that n,, < No for all i E /. and

2N .. = 2L.~,

"·

=II 2"· =II No = N~' = (2N, )N, IE l

= 2N, = N;J.

IE l

\VP now approach the problPIH of evaluating N~··. whPn> N,. and N.; an· arbitrary infinite cardinals. First. we make the following obsPrva.t.ion.

3.6 Lemma If n -::::; (3, then N~·' = 2N".

Proof.

Clearly, 2N,, <::: N~''. Sinee N, <::: 2N .. , we also have N~" <::: (2N .. )N;, = 2N .. N., = 2N,,

because Na = max{N,, Nil}· \\'hen trying to evaluate N~" lor n > (3. we find the following useful.

3. 7 Lemma Let n 2: {3 and let S lw the set of all subsf'ts X lXI = N,J. Then lSI= N~".

c;: ""'"

sllt:h that

Pmof. We first show that N~" -:: :; IS/. Let S' be tht• set of all subsets WtJ x Wo such that IX/= Nu. Since No· N, =No, we have /S'I = IS/. \'ow Pvery function f : wo ---> w" is a member of the set S' and hence c;: S'.

X c;:

w::·•

Therefore. N~' 'SIS/. Conversely. if X E S, then thPre exists a function f on W(J such that X is the range of f. \Vp pick one f for each XES and let f = F(X). Clearly·. if X /c }' and f = F(X) andy= F(Y). wP havP X =ran f and Y = m11 y. and so f 1 .'1 Thus F is a one-to-one mapping of S into and t.hereforP l·':ii S N~'

w;:··.

\Ve are now in a position to (,valuate N~" for regular cardinals N.,. uudPr t lw assumption of the Generalized Continuum Hypothesis. 3.8 Theorem Let us assume thr Generalized Continuum Hypothesis. If N., ts

a

n~gular

car·dinal, then if /3 < n, if {3 2 Q.

167

3. EXPONENTIATION OF CARDINALS

Proof. If (3 :2: a, then N~P = 2N,, = Ne+I by Lemma 3.6. So let 13 < o and letS = {X <;::;; Wo I lXI = Ne}. By Lemma 3.7. lSI = N~". By Theorem 2.2(a), every X E Sis a bounded subset of w". Thus. let B = U6
L

2161 .

6
However, for every cardinal N, for every 8 < w,, and we get IBI S:

L

< N,, we have 2N~

6 2\ \ :S

6<..J.,

L

= N,.,+ 1

S: No and so 21 61 S: No

N, = N" · Nn = N".

O
0 We proV<' a similar (but a little more complicated) formula for singular No. but first we need a generalization of Lemma 3.3.

3.9 Lemma For every cardinalr.: > 1 and every n. c:f(r.:N .. ) > N". Exactly like the proof of Lemma 3.3, except that 2N .. is replaced 0 3.10 Theorem Let us assume the Generalized Continuum Hypothesis. If N"

is a singular cardinal, then

ifNe < cf(Nn). if cf(N,) S: Nil S: N". ifNe :2: No. Proof. If (3 :2: a, then N~'' = 2N,, = Ne+I· If Ne < cf(Nn), then every subset X t:;;; w" such that lXI = Np is a bounded subset, and we get N~" = N" by exactly the same argument as in the case of regular N0 . Thus let us assume that cf(N 0 ) :5 Ne ::; N,... On the one hand, we have

On the other hand, cf(N~") have cf(N~")

f-

> Np

by Lemma 3.9. and since No

cf(Nu}, and therefore N~"

f-

N0

.

2 cf(N

0 ).

we

Thus necessarily N~·' =No+ I· 0

If we do not assume the Generalized Continuum Hypothesis. thP situation becomes much more complicated. We only prove the following theorem.

CHAPTER 9. ARITHMETIC OF CARDINAL NUMBERS

168

3.11 Hausdorff's Formula

For every a and every (J,

_,N.,

_,N 11

_,

"n+l = "o · ~'o+l·

Proof. If (3 2: a+ 1, then N:~ 1 = 2N 1', N~" = 2N", and No+I :=;No:=; 2N·'; hence the formula holds. Thus let us assume that (3 :=; a. Since N~" :=; N~'~ 1

and Na+l :S N~~~· it suffices to show that N:: 1 .::; N~'' · Na+l· Each function f : we -> Wn+l is bounded; i.e., there is 1 < Wo+l such that /([,) < -r for all (
Now every 1 <

IU

')'
has cardinality hi :S N0 1-vlI N, , Thus L..,')'
Wa+l

,

and we have (by Exercise 1.6)

"'w"l < "' I

-

0

This theorem enables us to evaluate some simple cases of N~'' (see Exereise 3.5). An infinite cardinal Nn is a strong limit cardinal if 2N,, < N" for all (3 < o. Clearly, a strong limit cardinal is a limit cardinal, since if N, = N,.+ 1 , then 2N~ 2: Na. Not every limit cardinal is necessarily a strong limit cardinal: If 2N" is greater than Nw, then Nw is a counterexample. However, if we assume th{' Generalized Continuum Hypothesis, then every limit cardinal is a strong limit cardinal. 3.12 Theorem If N, is a strong limit cardinal and if cardinals such that li < N0 and .\ < N0 , then li>. < N0 • Proof.

li

and .\ are infimtf'

c:

An uncountable cardinal number K. is strongly inaccessible if it is regular and a strong limit cardinal. (Thus every strongly inaccessible cardinal is weakly inaccessible, and, if we assume the Generalized Continuum Hypothesis, evNy weakly inaccessible cardinal is strongly inaccessible.) The reason why such cardinal numbers are called inaccessible is that they cannot be obtained by t h(' usual set-theoretic operations from smaller cardinals:

3. EXPONENTIATION OF CARDINALS

169

3.13 Theorem Let K. be a strongly inaccessible cardinal. (a) If X has cardinality< K., then P(X) has cardinality < K.. (b) If each X E S has cardinality < li and lSI < K., then US has cardinality

<

li.

(c) If lXI <

li

and

f:

X

-+ K.,

then sup f[X) <

li.

Proof. (a) li is a strong limit cardinal. (b) Let.\= JSI and J1. = sup{JXJJ XES}. Then (by Theorem 2.2(a)) J1. < because K. is regular, and I U SJ ::::; A · J1. < K.. (c) By Theorem 2.2(b).

K.

0 Exercises 3.1 If 2N,, 2: N0 , then N~'' = 2N". 3.2 Verify this generalization of Exercise 3.1: If there is 1 < a such that N~'' 2: N0 , say N~'' = No, then N~" = N0 . 3.3 Let a be a limit ordinal and let Np < cf(N 0 ). Show that if N;" :=; Nn for all (
•
nEA,

i
nEA,

t
The other direction is easy.] 3. 7 Prove that N~ 1 = N~" · 2N 1 .

[Hint: N~• = (l:n
=

fln
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Chapter 10

Sets of Real Numbers 1.

Integers and Rational Numbers

In Chapter 3 we defined natural numbers and their ordering and indicated how arithmetic operations on natural numbers can be defined. The next logical st.Pp in the development of foundations for mathematics is to define integers and rational numbers. The guiding idea in both cases is to make an arithmetic operation that is only partially defined on natural numbers (subtraction in the case of integers, division in the case of rationals) into a total operation, and belongs more properly in the realm of algebra than set theory. WP thus limit ourselves to outlining the main ideas, and leave out almost all proofs. ThosP can be found in most textbooks on abstract algebra. Better still, the readPr may work out some or all of them as exercises. In Exercise 4.3 of Chapter 3 we defined subtraction for those pairs (n. m) of natural numbers where n 2': m. In this case, n- m is the unique natural number k for which n = m + k. If n < m, no such natural number k Pxists. and n - m is undefined. If n- m is to be defined, it has to b<, a ''npw'' objPct; for tlw tinw being, we represent it simply by the ordered pair (n, m). However. intuitively familiar properties of integer arithmetic suggest that different ordered pairs may have to represent the same integer: for example. (2. 5) and (o. 9) both represent -3 [2- 5 = 6- 9 = -3]. In general, (n1, mi) and (n 2 , m 2 ) represent the same integer if and only if n 1 - m 1 = n 2 - m 2 . At this point. this makes sense only intuitively, but it can be rewritten in the form

which involves only addition of natural numbers (which has been previously defined). These remarks motivate the following definitions and results. Let Z' = N x N. Define a relation~ on Z' by (u, b)~ (c. d) if and only if a+ d = b +c. The relation~ is an equivalence relation on Z'. (This has to be checked, of course.) Let Z = Z' / ~ be the set of all equivalence dasses of Z' modulo ~. We call Z the set of all integers; its elenwnts are intP_qi'T·s.

171

172

CHAPTER 10. SETS OF REAL NUMBERS One immediate consequence of the definition is of set-thPoretic intNPst.

1.1 Theorem The set of all integers Z is countable. Proof Theorem 3.13 in Chapter 4. Theorem 3.12 in Chapter 4 gives another proof. 0

Next, define a relation < on Z as follows: [(a, b)] < [(c, d)] if and only if a+ d < b +c. [Recall that intuitively (a, b) represents a-band (c, d) represents c - d, so a - b < c- d should mean a + d < b +c.] One can prove that< is well defined (i.e., truth or falsity of [(a, b)] < [(c, d)] does not depend on the choice of representatives (a, b) and (c, d) but only on their respective equivalence classes) and that it is a linear ordering. Finally, we observe that for each integer [(a, b)] either a ~ b, in which cmw (a, b) ::::: (a - b, 0) (here - stands for subtraction of natural numbers, which is defined in this case), or a < b, in which case (a. b) ::::: (0. b- a). It follows that each integer contains a unique pair of the form (n,O), n E Nor (O.n). n E N- {0}. So [(n, 0)] are the positive integers and [(0, n)J are the negative ones. The mapping F : N -> Z defined by F(n) = [(n,O)] is one-to-one and order-preserving [i.e., m < n implies that F(m) < F(n)]. We identify each integer of the form [(n, 0)] with the corresponding natural number n, and denotP each integer of the form [(0, n)] by -n. So, e.g., -3 = [(0, 3)] = [(2. 5)] = [(6. ())]. as expected. The rest of the theory is now straightforward. One can prove that ( Z. <) has no endpoints and that {x E Z I a < x < b}, a, bE Z. a < b, has a finite number of elements. Also, every nonempty set of integers bounded from abov(' has a greatest element, and every nonempty set of integers bounded from below has a least element. One can define addition and multiplication of integers bv [(a, b)] + [(c, d)] = [(a+ c, b +d)], [(a, b)]· [(c,d)J = [(ac + bd,ad +be)] and prove that these operations satisfy the usual laws of algebra (commutativity, associativity, and distributivity of multiplication over addition), and that for those integers that are natural numbers, addition and multiplication of integers agree with addition and multiplication of natural numbers. We define subtraction by [(a, b)]- [(c, d)]= [(a, b)]+ ( -[(c, d)]), where -[(c,d)] = [(d,c)] is the opposite of [(c,d)J. Notice that -[(n,O)] = [(O,n)] = -nand -[(O,n)] = [(n,O)] = n, in agreement with our previous notation. Absolute value of an integer a. Ia I, is defined by

Jaj =

{a-a

~fa ~ 0; 1f a< 0.

173

1. INTEGERS AND RATIONAL NUMBERS

Additional properties of these concepts can be proved as needed. Addition, subtraction, and multiplication are defined for all pairs of integers. There remains the inconvenience that division cannot always be performed. We say that an integer a is divisible by an integer b if there is a uniquP integer x such that a = b · x; this unique x is then called the quotient of a and b. We would like to extend the system of integers so that any a is divisible by any band all useful arithmetic laws remain valid in the extended system. Now. if 0 · x = 0 is to be valid in the extended system for all x, we see immediately that no number a can be divisible by 0; the equation a = 0 · x has either none or many solutions, depending on whether a f:- 0 or a = 0. The best wP can hope for is an extension in which for all a and all b f:- 0 there is a unique x such that a= b · x. 2 I b f:- 0}. We call Q' the set of X (Z- {0}) = {(a,b) E Z Let Q' = fractions over Z and write a/bin place of (a,b) for (a, b) E Q'. We define an equivalence ~on the set Q' by

z

if and only if a · d = b · c. Let Q

= Q' / :::::be the set of equivalence classes of Q modulo:::::. Elements of

Q are called rational numbers; the rational number represented by a/b is denoted [a/b]. [Later, the brackets are habitually dropped and we do not distinguish between a rational number and the (many) fractions that represent. it.] There is an obvious one-to-one mapping i of the set Z of integers into the rationals: i(a) =

[i].

(Later, we identify integers with the corresponding rationals.) We now define addition and multiplication of rationals:

[ ~] + [ ~)

=

[a ·~ ~db c] . ·

[~]. [~] = [;."~]. In order to lay the foundations satisfactorily, one should prove the following: (a) Addition and multiplication of the rationals are well defined (i.e .. indPpendent of the choice of representatives). (b) For integers, the new definitions agree with the old ones: i.e., i(a + /1) = i(a) + i(b) and i(a ·b)= i(a) · i(b) for all a,b E Z. (c) Addition and multiplication of rationals satisfy the usual laws of algebra. (d) If A E Q, BE Q, and B f:- [0/1], then the equation A= B ·X has a unique solution X E Q. Thus division of rational numbers is defined, as long as the divisor is not zero; we denote this operation -:-: X =A-:- B. Finally, we extend the ordering of integers to the rationals. First, notice that each rational can be represented by a fraction a/b where the denominator b is greater than 0:

[-ab] = [ --ab]

and either

b

> 0 or - b > 0.

CHAPTER 10. SETS OF REAL NUMBERS

174

We now define the natural ordering of rationals: Ifb>Oandd>O,let

[*] <

[~]

ifandonlyifa·d
\Ve again leave to the reader the proof that the definition does not de~JPnd on the choice of representatives as long as b > 0 and d > 0. that < is n·all~· a linear ordering, that for a. b E Z, a < b if and only if [a/ 1] < [b / 1]. and that t In• usual algebraic laws (such as: if a < b then a+ c < b + c, etc.) hold. Wt> have again

1.2 Theorem The set of rationals Q is countable. Proof. Theorem 3.13 in Chapter 4. Theorem 3.12 in Chapter 4 giv<'s another proof. i .

Tlw next result has been referred to in Chapter 4.

1.3 Theorem (Q, <) is a dense linearly or·dered set and has no endpoints. In 7' E Q there exists n E N such that r < n.

fact. for every

Pmof. Q is infinite: since ajb - 1 < ajb < ajb + 1. Q has no endpoint;;. If r :::; 0 we can take n = 1. If 7' > 0. we write r = ajb when• a > 0. /J > 0. a. bE N. and taken =a+ 1. It remains to show that (Q. <)is dense. Let 7', s be rationals such that r < s: assume that r =a/band.~= cjd. where b > 0 and d > 0. Now we !Pt

:r = [i.P .. x = (r

+ s)/2].

Then r < x <

a·d+b·c 2-b·d 8.

\\'e conclude this section with a few remarks on decimal (or, in general. base p) t>Xpansions of rationals. Every integer p > 1 can serve as a basf' of a numlwr system. The one generally used has p = 10. Another useful case is p = 2.

1.4 Lemma Given a mtional n-umbe7' 7', there i8 a unique integer c:::; r < e + l. We call e the integer part ofr, e = [r].

f'

sur:h that

Pr·oof. Let r = ajb. b > 0. Assume that a ':::" 0. 1 :::; b. so a :S o · /J a11d r = ajb:::; a E Z. It now follows that S = {x E Z I x:::; r} <;;; Z ha..~ all up~wr

z.

bound a in If a < 0. then 0 is all upper bound on S. Therefon•. S has a greatest element e. Then e :S r < f' + 1. and dearly r is the uniqup integer with this property. C~ The expansions of integers in base pare sufficiently well known. By LPmiua 1.4. r = [r] + q where [r] is an intq?;er and q E Q. 0 :::; q < 1. WP t·ow·ent.rat<' on the t>xpansion of q. Construct a sequence of digits 0. 1, .... p - 1 by recursion. as follows:

2. REAL NUMBERS

175

Find a 1 E {0, ... ,p- 1} such that ai/p::;: q < (a1 + 1)/P (let a1 = [q · p]). Then find a2 E {0, ... ,p-1} such that ai/p+a2jp2 ::;: q < a1/P+ (a 2 + l)/p 2 (let a2 = [(q - ai/p) · p 2]). In general, find ak E { 0, ... , p - 1} such that

We call the sequence (a, I i E N) the expansion of q in base p. When p == 10, it is customary to write q = O.a1a2a3 .... One can show (a) There is no i such that aj = p- 1 for all j ~ i. (b) There exist n E N and l > 0 such that an+l = an for all n :2: no (the expansion is eventually periodic, with period l). Moreover, if q == afb, then we can find a period l such that l ::; lbl. Conversely. each sequence (a; I i E N) with the properties (a) and (b) is an expansion of some rational number q (0::; q < 1). Exercises

1.1 Prove some of the claims made in the text of this section. 1.2 If r E Q, r > 0, then there exists n E N- {0} such that ljn Use the fact mentioned in Theorem 1.3.]

2.

< r. [Hint:

Real Numbers

The set R of all real numbers and its natural linear ordering < have been defined in Chapter 4, Section 5, as the completion of the rationals. In particular. every nonempty set of real numbers bounded from above has a supremum, and every nonernpty set of real numbers bounded from below has an infimum. Here we consider the algebraic operations on real numbers. We first prove a useful lemma. 2.1 Lemma For every x E R and n E N- {0} there exist r, s E Q such that r < x ::; s and s- r ::;: 1/n.

Proof. Fix some ro, s 0 E Q such that r 0 < x < s 0 , and some k E N- {0} such that k > n(so - r 0 ). Consider the increasing finite sequence of rational numbers (r,)~=o where r, = r 0 + i/n. Let j be the greatest i for which r, < .r: notice that j < k. Now we have Tj < x::; Tj+I and Tj+l - Tj = lfn. It suffiet's to taker== Tj, s = Tj+l· 0

We now define the operation of addition on real numbers.

CHAPTER 10. SETS OF REAL NUMBERS

176

2.2 Definition Let x, y E R. We let x+y = inf{r+s I r, sEQ, x::; r. y::; s }. (The symbol +on the right-hand side refers to the addition of rational numbers.) We note that the infimum exists, because the nonempty set in question is bounded below (by p + q, for any p, q E Q, p < x, q < y). It is also clear that if both x and y are rational numbers, then x + y, under the new definition. is equal to x + y, where + is the previously defined addition of rationals. 2.3 Lemma Let x, y, z E R.

{i}

X+ y = y +X.

{ii) (x+y)+z=x+(y+z). (iii)

X+

0

=X.

(iv) There exists a unique w E R such that x the opposite of x. {v) If x < y then x

+w

= 0.

We denote w = -:r,

+ z < y + z.

Proof. We use some simple properties of suprema and infima (see Exercise 2.1). (i), (ii), and (iii) follow immediately from the corresponding properties of rational numbers. (iv) We recall that x = inf{s E Q I x ::; s} = sup{r E Q I r < x}. This suggests letting w = inf{ -r IrE Q, r < x}. We have x + w = inf{s- r I r,s E Q, x :S s, r < x}. As r < x, x :S simply s-r > 0. it follows x + w ~ 0. Assume x + w > 0; by density of Q in Rand Exercise 1.2. then• exists n EN- {0} such that 1/n < x + w. But, Lemma 2.1 guarantees existence of some r, s E Q, r < x :S s, such that s - r ::; ljn. ThereforP x + w :S 1/n. a contradiction. This shows the existence of a w with the property x + w = 0. If also x + v = 0, we have (using (i), (ii). and (iii)) W = W + 0 = 0 + W = ( V + X) + W = V + (X + W) = V + 0 = V, SO 11! is uniquely determined. (v) If x < y then x + z ::; y + z is immediate from the definition of addition (and Exercise 2.1). If x + z = y + z, we have x = x + 0 = x + (z + (- z)) = (x+z)+(-z) = (y+z)+(-z) = y+(z+(-z)) = y+O = y, a contradiction. 0

The operation of multiplication on positive real numbers can be defined in an analogous way. We let n+ = {x E R I X> 0}. 2.4 Definition Let

X,

y E n+. We let

X.

y = inf{r. sIr, s E Q,

X ::;

r . .II ::; s}.

2. REAL NUMBERS

177

2.5 Lemma Let x, y, z E R+.

+ z)

(vi')

X·

(y

(vii')

X·

y = y ·X.

X •

1=

y

+X· Z.

= x · (y · z).

(viii') (x · y) · z

(i:i)

=X·

X.

(x) There exists a unique wE R+ such that x · w = 1. We denote w = 1/x, the reciprocal of x. (xi') If x < y then x · z < y · z. Proof. Entirely analogous to the proof of Lemma 2.3. For (x') we let w=inf{1/rlrEQ, O
It is now a straightforward matter to extend multiplication to all real numbers. We first define the absolute value of x E R: lxl =

{X -X

and notice that for x

~f X~ 0;

If

X<

0

# 0, lxl E R+ (if x < 0 then 0 = x + (-x) < 0 + (-x) =

-x). 2.6 Definition For x, y E R we let

lxl · IYI x ·Y =

-(lxl·lyl)

{0

if x > 0, y > 0 or x < 0, y < 0; if x > 0, y < 0 or x < 0, y > 0; if x = 0 or y = 0.

We leave to the reader the straightforward but tedious exercise of showing that this definition agrees with the definition of multiplication of rationals from Section 1, and proving the following lemma.

2.1 Lemma Let x, y, z E R.

+ z)

(vi)

X·

(y

(vii}

X-

y = y

(viii) (x · y) · z (ix)

X·

=X·

y

+X· Z.

·X.

= x · (y · z).

1 =X.

(x) For each x E R, if x # 0, then there exists a unique w E R such that x-w=l.

CHAPTER 10. SETS OF REAL NUMBERS

178

(xi) If x < y and z > 0 then x · z < y · z. As before. we denote the unique win (x) by 1/x. We also clefirw diviswn ll\· a nonzero real number x: y-:- x = y · (1/x). A structure 21 = (A,<,+.·, 0, 1) where < is a linear ordering. + and · are binary operations, and 0, 1 are constants, such that all the properties ( i) ( \') and (vi)-(xi) are satisfied is called an ordered field in algebra. The content~ of Lemmas 2.3 and 2.5 can thus be summarized by saying that the real numbers (with the usual ordering and arithmetic operations as defined abov<') an· an ordered field. As the ordering of the real numbers is complete. they arP a complete ordered field.

2.8 Theorem The structure 9l = (R, <, +. ·,0, 1) is a complete orderPd jiFld. It is possible to prove that the complete ordered field is unique. i.<> .. if 2l =

(A,<,+,·, 0, 1) is also a complete ordered field, then 2l and 9l are isomorphic. As the proof requires a fairly heavy dose of algebra, we do not present it her!' (but see Exercise 2.5). We conclude this section by mentioning a well-known fact about real numbers that we use in Section 6 of Chapter 4. We leave the proof as an exercise.

2.9 Theorem (Expansion of real numbers in base p) Let p ~ 2 be a nat1tral number. For every real number 0 s u < 1 thrre is a uniqur sPquntcf' of rml numbers (an)~ 1 such that (a) 0 San < p, for each n = 1, 2, ... {b) There zs no no such that an = p - 1 for all n > no. (c) a1/P + · · · + an/Pn Sa< aJ/p +···+(an+ 1)/pn, for each n?:: 1. The real number a is rational if and only if (an)~=! is eventually periodic.

Exercises

s

2.1 Let A<;; B <;; R, A f. 0. If B is bounded from below then inf B inf A If B is bounded from above then sup A sup B. If, in addition, for every b E B there exists a E A such that a b, then inf B = inf A. Similarly for sup. 2.2 For every x E n+ and 11 E N- {0} there exist r, .s E Q such that 0 < r < x s and 1 < s/r 1 + 1/n. [Hint: Fix r 0 E Q such that 0 < ro < x, and kEN- {0} such that kr0 > n. Use Lemma 2.1 to find r, s E Q, ro < r < x s, so that s- r < 1/k. and estimate sjr.J 2.3 Prove that if a, b E R, b =/= 0, then the equation a = b · x has a unique solution. 2.4 Prove that for every a E n+ there exists a unique X E n+ such that X·X =a. 2.5 Prove that every complete ordered field is isomorphic to R. [Hints:

s s

s

s

s

3. TOPOLOGY OF THE REAL LINE

179

1. Given a complete ordered field 2l = (A,<,+, ·,0, 1) consider the closure C of {0, 1} under + and ·. Let I! be the structure obtained by restricting the ordering and the operations of 2l to C. Show that I! is (uniquely) isomorphic to Q = (Q, <, +, ·, 0, 1). This is the

"algebraic" part of the proof. 2. Show that for every a E A there is c E C such that a :::; c. (If not. then S = {a E A I a> c for all c E C} is nonempty and bounded below, but does not have an infimum, because a E S implies a- 1 E S.) 3. Use 2. to show that Cis dense in (A,<). 4. Use 3. to show that the isomorphism between Q and I! extends uniquely to an isomorphism between~ and 2l.] 2.6 (Complex Numbers) Let C = R x R, and let us define addition and mult.iplication on C as follows: (a,,a2)

+ (b,,b2)

= (a1

+ b1,a2 + b2):

(a,,a2) · (b1,b2) =(a,· b,-

Show that

+ and

a2 ·

b2,a1 · b2

+ a2

· b!).

· satisfy (i)-(iv) of Lemma 2.3 and (vi)-(x) of Lemma

2.7.

3.

Topology of the Real Line

The main result of the preceding section is a characterization of the real number system as a complete ordered field. This is the usual departure point for the study of topological properties of the real line. We give some basic definitions and theorems of the subject in this section. Our objectives are to justify the claim that set theory, as we developed it so far, provides a satisfactory foundation for analysis, to show some consequences of the previously proved results. and to provide a convenient reference for some concepts and theorems used elsewhere. Readers who studied advanced calculus should be familiar with most of this material. In any case, one can skip this section and refer to it only a.s needed. We begin with a definition of a familiar concept.

3.1 Definition Let (P, <) be a linear ordering, a, b E P, a < b. A (bounded) open interval with endpoints a and b is the set (a, b) = { x E P I a < x < b}. A (bounded) closed interval with endpoints a and b is the set [a, b] = { x E P I a :::; x :::; b}. We also define unbounded open and closed intervals (a. x). ( -oo, a), [a, oo), ( -oo,a], and ( -oo, oo) = P in the usual way: for example. (a, oo) = { x E P I a < x}. We employ the standard notation (a, b) for open intervals, even though it clashes with our notation for ordered pairs; the meaning should always be clear from the context.

180

CHAPTER 10. SETS OF REAL NUMBERS

The next theorem is a consequence of the existence of a countable dense subset in R.

3.2 Theorem Every system of mutually disjoint open intervals in R is at most countable. Let P = Q n US where S is a system of mutually disjoint open Proof. intervals in R and Q is the set of all rational numbers. As Q is dense in R, each open interval in S contains at least one element of P. As elements of S are mutually disjoint, each element of P is contained in a unique open interval from S. The function assigning to each element of P the unique open interval from S to which it belongs is a mapping of an at most countable set Ponto .'i. Hence S is at most countable, by Theorem 3.4 in Chapter 4. D This is a convenient place to mention a famous problem in set theory dating from the beginning of the century: Let (P, <) be a complete linearly orderPd set without endpoints where every system of mutually disjoint open intervals is at most countable. Is (P, <) isomorphic to the real line? This is the Suslin's Problem. Like the Continuum Hypothesis, it remained unsolved for decades. With the help of models of set theory, it has been established that it, like the Continuum Hypothesis, can be neither proved nor refuted from the axioms of Zermelo-Fraenkel set theory. The reader can find out more about these mattprs in Chapter 15.

3.3 Definition A system of sets S has the finite intersection property if evPfy nonempty finite subsystem of S has a nonempty intersection. An example of a system of sets with the finite intersection property is tht· range of a nonincreasing sequence of nonempty sets, i.e., S = {An I n E N} where An of. 0, An ~An' for all n, n' E N. n::; n'. The next theorem is a consequence of the completeness of the real line.

3.4 Theorem Any nonempty system of closed and bounded intervals in R with the finite intersection property has nonempty intersection. Let S be such a system. Let A= {x E R I [x, y] E S for some y E Proof. R} be the set of all left endpoints of intervals in S. We show that A is boundt•d from above. Fix [a, b] E S, then certainly x :::; b, because in the opposite cas<' we would have a < b < x < y and the intervals [x, y], [a, b] would be disjoint. violating the finite intersection property for the 2-element subset {[1:. y]. [a. b]} of S. So b is an upper bound on A. By the completeness property, A has a supremum a. If [a, b] is an interval in S, we have a ::; a because a E A. and a ::; b because b is an upper bound on A. So a E [a, b] for any [a, bj E S and S

n

is~~~

0

Completeness of the real line is also the reason why the notion of limit works as well as it does. We recall some very familiar definitions from calculus.

3. TOPOLOGY OF THE REAL LINE

181

3.5 Definition Let (an I n E N) be an infinite sequence of real numbers. (a) (an) is nondecreasing if an :::; an' holds for all n, n' E N such that n < n'. It is increasing if an < an' holds whenever n < n'. (b) (an) is bounded from above if its range {an I n E N} is bounded from above. Similarly for bounded from below. (an) is bounded if it is bounded from above and below. (c) (an) has a limit a [converyes to a] if for every E > 0, E E R, there is no E N such that, for all n ~ no, ian - al 0. E E R, there is no E N such that, for all m, n ~ no, lam -ani 0 be given. Since a - E < a, and a is the least upper bound on {an I n E N}, there is n 0 such

that a- E
a+ E, 0

3.7 Definition Let (kn I n E N) be an increasing sequence of natural numbers. The sequence (ak .. I n E N) is called a subsequence of (an I n E N). (Note that it is just the composition of (an) o (kn).) 3.8 Theorem Every bounded sequence of real numbers has a convergent subsequence. Assume that (an I n E N) is bounded. Let bn = inf {ak I k ~ n} Proof. and observe that the sequence (bn I n E N) is nondecreasing and bounded from above (every upper bound on (an) is an upper bound on (bn)). By Theorem 3.6, (bn) has a limit a. We construct a subsequence of (an) with limit a by recursion: ko = 0, i.e., ak., = ao; given kn, let kn+l be the least kEN such that k > ko and -

1

-

1

a - - - < ak
n+1

We have to show that such k's exist. But a = sup{ bn I n E N} (see the proof of Theorem 3.6) and so there is i > kn such that

1 n +1

a - - -
182 Also, b,

CHAPTER 10. SETS OF REAL NUMBERS =

inf {ak

Ik

::=: i} and so there is k ::=: i such that

b, :::;

ak-

< a+

1 --. n+1

We now have k > kn with 1

1

a- - < ak < a+ --. n+1 n+1 as required. The subsequence (ak.. I n E N) of (an) is such that for PVery n ::=: 1. lak .. - al < 1/n. From this it follows easily that (akJ converges to a (st>P Exercise 3.4). r~

3.9 Theorem Every Cauchy sequrmr:e of real number·s converge:;. Proof. Every Cauchy sequPIW<' is bounded: If (an) iii a C'auchv SP(jliPilct•. let c; = 1 in Definition 3.S(d); t.h<" result is no E N such that for all n ::=: 11 11 • ian., -ani < 1 (we also let m =no). So an., - 1
Mt = max{an, ... ,an .. -t.an.,

Ah

+ 1}

= min{an, ... ,an.,-!, an., -1}

produces the desired upper and lower bounds on {an I n E N}. By Theorem 3.8, (a,.) has a subsequence (ak .. ) converging to a E R; we provt• that (an) itself converges to a. So lPt c; > 0 be given; since (ak .. ) converges to a. there is n 0 E N such that for all n ::=: n 0 , iak .. - ai < c/2. Since (a") is a Cauchy sequence. there is n 1 E N such that for all m. n "2' n 1 • In"' -a, I < :- /'2. Let nz = max{ n 0 • n1}. If n "2' nz. then Ian - al :::; Ian - ak .. ,l + lak ... -· ul < c/2 + c/2 = c; and we haV(' established that (an) has a limit a. n \Ve now turn our attention to continuous functions and open and dosPd :;Pts. 3.10 Definition A function f : R __. R is continuous at a E R if for Pverv > 0 there is 6 > 0 such that for all x E R, if lx- ai < 6, then 1/(:r)- /(a)l < c. f is continuous if it is continuous at every a E R. A set A t;;; R is open if for every a E A there is b > 0 such that lx - ai < li implies that :r E .4. [Or. in other words, the open interval (a- tl. a+ b) <:;; A.] A st>t B is closed if its rdativ" complement R - B is open. c;

Continuous funt.:tions are particularly simple and are a main object of stud,· in mathematical analysis. For our purposes, we need only a result showing that a continuous function is uniquely determined by its values on n dense subs<>t of R (such as the rationals). 3.11 Theorem Let D <:;; R be dense in R. If I continuous. then f =g.

rD

= g

r D when' I and g

(lJ"('

3. TOPOLOGY OF THE REAL LINE

183

Proof. Suppose that f =J g; then f(a) =J g(a) for some a E R. Let c = tf(a)- g(a)t/2. As both f and g are assumed continuous at a, there exist 61 ,6 2 > 0 such that if tx- at xistence of .r E D such that lx- at < min{o1, 62}. But now If( a) - g(a)t S [f(a)- f(x)[ + lf(.r) g(x)[ + tg(x)- g(a)l < c + 0 + c =If( a)- g(a)t. a contradiction. [We haw used the fact that f(x) = g(x) for x E Din replacing tbe middle term by 0.] 0 Similarly, open and closed sets are relatively simple and well-behaved. It is obvious from the definition that a set is open if and only if it is a union of a system of open intervals. Consequently, the union of any system of open sets is open and the intersection of any system of dosed sets is closed. Open intervals are the simplest examples of open sets; closed intervals, finite sets. and {1/n In EN- {0}} U {0} are typical examples of closed sets. Other examples can be obtained by using the next lemma. 3.12 Lemma The intersection of a finite system of open sets is open. union of a finite system of closed sets is closed.

The

Let A 1 and A 2 be open sets. If a E A1 n Az, there are 6 1.6 2 > 0 Proof. such that lx - a[ < 61 implies x E A1 and [x - at < 62 implies :r E Az. Put t5 = min{ 61 • 6:!}; then t5 > 0 and tx- at < 6 implies .r E A 1 n Az. This argument proves that the intersection of two open sets is open. Tlw claim for any hnitP system of open sets is proved by induction, and the daim for dosed set:-; by taking relative complements in R and using the DP l'dorg;an Laws (Section 4 in Chapter 1). 0 3.13 Theorem The following statements about f : R ___. R (a) f is continuous. (b) f- 1 [A] is open whenever A<;;; R is open. (c) f - 1 [ BJ is closed whenever B <;;; R is closed.

an~

eqaivalmt:

r

Proof. (a) implies (b). Assume that f is continuous. If (l E I [A]. f(a) E A and so there is c > 0 such that IY- f(a)t < E implies y E A (b!'causP A is open). By definition of continuity there is 6 > 0 such that [.r - al < 6 implies lf(x)- j(a)t < c. So we found o > 0 such that fx- a[ < 8 implit>s 1[A], showing j-l [A] open. j(x) E A, i.e., x E (b) implies (a). Let a E R and E > 0. The assumption (b) implies that I [(!(a) - E, f(a) +c)] is open (and contains a). Thus there is 6 > 0 such that lx- a/ < 6 implies x E /- 1 [(/(a)- c,/(a) +c)], i.e., f(x) E (J(a) -E. f(a) +E). establishing the continuity off at a. The equivalence of (b) and (c) follows immediately from Exercise 3.G(b) in Chapter 2. C

r

r

In the rest of this section we prove some results about open and closed sets. The following lemma is used in Chapter 5 to determine the cardinality of the system of all open sets.

CHAPTER 10. SETS OF REAL NUMBERS

184

3.14 Lemma Every open set is a union of a system of open intervals unth rational endpoints. Let A be open and let S be the system of all open intervals with Proof. rational endpoints included in A. Clearly, US~ A; if a E A, then (a a+J) ~ A for some a > 0 and we use the density of Q in R to find r 1 , r 2 E Q such that a- a< rt
-o,

US=A.

0

We have defined closed sets as complements of open sets. It is useful to haw a characterization of them in terms of the behavior of their points.

3.15 Definition a E R is an accumulation point of A ~ R if for every c5 > 0 there is x E A, xi- a, such that lx- ai < J. a E R is an isolated pomt of A ~ R if a E A and there is c5 > 0 such that lx- al < J, xi- a, implies x 1- A. It is easy to see that every element of A is either an isolated point or au accumulation point, and that there may be accumulation points of A that do not belong to A.

3.16 Lemma A to A.

~

R is closed if and only if all accumulation points of A belon_q

Proof. Assume that A is closed, i.e., R - A is open. If a E R - A. then there is > 0 such that lx- al < implies x E R- A, so a is not an accumulation point of A. Thus all accumulation points of A belong to A. Conversely, assume that all accumulation points of A belong to A. If a E R- A, then it is not an accumulation point of A, so there is o > 0 such that there is no x E A, x i- a, with lx - al < J. But then lx - al < c5 implies x E R- A, soR-A is open and A is closed. 0

a

o

As an application, we generalize Theorem 3.4.

3.17 Theorem Any non empty system of closed and bounded sets with thP. finite intersection property has a nonempty intersection. Proof. Let S be such a system. We note that the intersection of any nonempty finite subsystem T of S is a nonempty closed and bounded set. So S= T 1 Tis a nonempty finite subsystem of S} is again a nonempty system o! closed and bounded sets with the finite intersection property, and the additional property that the intersection of any nonempty finite subsystem of S actuallv belongs to Since obviously s = S, it suffices to prove that i- 0. The proof closely follows that of Theorem 3.4. Let A = { inf F I F E S}. WP show that A is bounded from above. Fix G E S; then for any F E S we haw FnG E Sand infF ~ inf(FnG) ~ sup(FnG) ~ supG, so A is bounded by supG. By the completeness property, A has a supremum a. The proof is concluded by showing a E F for any FE S.

{n

s.

n

n

ns

3. TOPOLOGY OF THE REAL LINE

185

Suppose that a¢ F. For any a > 0 there is H E S such that a- c5 < inf H '5 a, by definition of a. Since H n FE Sand inf H '5 inf(H n F), we have also a- a < inf(H n F) '5 a< a+ o. It now follows from the definition of infimum that there is X E H n F for which inf(H n F) '5 X 0, there exists x E F such that x f. a (we assumed that a 1. F) and lx- al < J. But this means that a is an accumulation point ofF, and as F is closed, a E F, a contradiction.

0

Closed sets contain all of their accumulation points; in addition, they may or may not contain isolated points. An example of a closed set without isolated points is any closed interval. [0, l] U N is an example of a closed set with (infinitely many) isolated points. Nonempty closed sets without isolated points are particularly well-behaved; they are called perfect sets. We use them in the next section as a tool for analyzing the structure of closed sets. We conclude this section with an example showing that, their good behavior notwithstanding, perfect sets may differ quite substantially from the familiar examples, the closed intervals. 3.18 Example Cantor Set. LetS= Seq({O, l}) = UnEN{O, l}N be the set of all finite sequences of O's and l 's. We construct a system of closed intervals D = (D. I s E S) as follows:

Do = [0.1]; 1 D(o) = [0, 3J. D(O,O)

=

[0,

1

8

g-J.

Dp,!) = [ ,1],

9

etc.

()

0

()

--

D(II,J,n)D(u,l,l)

In general, if D(so, ... ,s,_I) = [a,b], we let D{so, ... ,s,._ 1 ,o) = [a,a + 1(b- a)] and D(so, ,s,._,,l) = [a+ ~(b- a), b] be the first and last thirds of [a, b]. The existence of the system D is justified by the Recursion Theorem. although it takes some thought to see: we actually define recursively a sequence (Dn I n E N), where each Dn is a system of closed intervals indexed by {0, 1 }n: D?l = [0, 1]; having defined Dn we define vn+l so that for all (so .... 'Sn-1) E { 0, 1} n' I

CHAPTER 10. SETS OF REAL NUMBERS

186

D(.~,

,,o) [a, a+ i(b- a)) and D(.~. _ ,s.,_,,l) = [a+ ~(b- a). h) w!wn• [a,b] = D~o .... s,._,, and then let D = UnENDn. Now let Fn = U{Ds Is E {0, l}n}, so that Fo = [0, 1], F1 = [0. j] U [~.1). etc. Notice that each Fn is a union of a finite system of dosed intervals. and hence closed (see Lemma 3.12). The set F = nnEN Fn is thus also closed: it is called the Cantor set. For any f E {0, 1}N we let DJ = nnEN Dnn- We note that Df ~ F and Df i- 0 (Theorem 3.4). Moreover, the length of the interval Dnn is 1/3". so the infimum of the lengths of the intervals in the system (Dnn I n E N) is zero. It follows that D f contains a unique element: D f = { d f}. Conversely. for any a E F there is a unique f E {0, 1}N such that a = dr let f = U{s E Seq({0,1}) I a ED,}. (Compare these arguments with Exercise 3.13 in Chapter 2_) d = (d f I f E { 0, l}N) is thus a oue-to-one mapping of {0. 1} N outo F, and we conclude that the Cantor set has the cardinality of the continuum 2N". 1

1

,s.,.

\Ve prove two out of many other interesting properties of the set F.

( 1) The Cantor set is a perfect set.

Proof We know that F is closed and nonempty already. so it remains to prove that it has no isolated points. Let a E F and c5 > 0: we have to hnd x E F, x i- a, such that lx- a! < ;;_ Taken E N for which 1j3n < J. As WP know, a = d1 for some f E {0, 1 }N: let x = d 9 where g is any sequence of (J's and 1's such that g r n = f r n aud f t- g. Then at- X, but a ..r E Dnn = D"'" which has length 1/3n < J, so lx- a! < J, as required. [] (2) The relative complement of the Cantor set in [0, 1] is dense in [0. 1).

Proof. Let 0 :::; a < b :::; 1; we show that (a, b) contains elements not in F. Take n E N such that :I~· < ~ (/J - a). Let k be the least natura.! number for (k+Ll < b. Th e open .mterva I . l1 Fk 2': a,- we t hen h ave a < w l uc _ 3k.. < ~

. third of [k3" , k+1]) (the m1ddle ~

is certainly disjoint from F. and we see that (a, b) contains elements not in F. 0

Exercises

3.1 Every system of mutually disjoint open sets is at most countable. Tlw statement is false for closed sets_ 3.2 Let S be a nonempty system of closed and bounded intervals in R. If inf{b- a! [a, b) E S} = 0. then ns contains at most one element.

3. TOPOLOGY OF THE REAL LINE

187

3.3 Let (P, <) be a dense linearly ordered set where every nonempty systPm of closed and bounded intervals with the finite intersection property ha..;; a nonempty intersection. Then P is complete. (This is a conwrse to Theorem 3.4.) 3.4 (an)~ 0 converges to a if and only if for every k E N, k > 0. therP is no EN such that, for all n 2 n 0 , [an-al < 1/k. 3.5 If (an)~ 0 converges to a and to b, then a = b (the limit is unique). This justifies the usual notation limn_, 00 an for the limit, if it exists. 3.6 Every convergent sequence of reals is a Cauchy :-;equence. 3. 7 The number a = sup{ inf {ak I k 2 n} I n E N} used in the proof of Tlworem 3.8 is called the lower limit of (an) and is denoted lim inf,._, 00 a,. Similarly, b = inf{sup{ak I k 2 n} In EN} is called the upper limit of (an) and denoted limsupn_,oo an· Prove that (a) b exists for every bounded sequence and a::; b. (b) If cis a limit of some subsequence of (an). then a::; c::; b. (c) (an) converges if and only if a= b (and if it docs, a is its limit). 3.8 There is a sequence (an) with the property that for any a E R, (an) has a subsequence converging to a. [Hint: Consider an enumeration of all rational numbers.) 3.9 Show that f : R -> R is continuous at a E R if and only if for every n E N there is k E N such that for all x E R, [x- a[ < 1/k implies lf(x) - f(a)[ < 1/n. 3.10 Let f : R -> R be continuous, a, b, y E R, a < band f(a) ::; y ::; f(b). Prove that there exists x E R, a ::; x ::; b, such that f(x) y (the Intermediate Value Theorem). [Hint: Consider :r = sup{z E [a. b) I f(z) :S: y}.) 3.11 Let f : R _, R be continuous, a, bE R, a 0 there is x E A such that [x- a[ < a. a E R is a closure point if and only if it is either an isolated point of A or an accumulation point of A. Let A be the set of all closure points of A. A is closed if and only if A = A. 3.15 Every open set is a union of a system of mutually disjoint open intervals. [Hint: If A is open and a E A, U{(x,y) I a E (x,y) ~A} is an open interval.) 3.16 Give an example of a decreasing sequence of closed sets with empty intersection. 3.17 Show that Exercises 3.11 and 3.12 remain true for any closed and boundPd set C in place of the closed interval [a, b). 3.18 We describe an alternative construction of the completion of (Q. <). A sequence (an)~=O of rational numbers is a Cauchy sequence if for each

188

CHAPTER 10. SETS OF REAL NUMBERS rational p > 0 there is no such that Ia,. - a"" I < p whenever n ;:;> n 0 . Let C be the set of all Cauchy sequences of rational numbers. We defiut> an equivalence relation on C by (a.. )~=O -;;::: (b,.)~=O if and only if for each p > 0 there is no such that Ia,. - b,.l < p whenever n ;:;> n 0 • and a preordering of C by (an)~=O ~ (b,.)~=O if and only if for each p > 0 thert> is no such that an - bn < p whenever n ;:;> n 0 . Prove that (a) -;;:::is an equivalence relation on C. (b) ~ defines a linear ordering of C I-;;:::. (c) The ordered set C I-;;::: is isomorphic to R.

4.

Sets of Real Numbers

The Continuum Hypothesis postulates that every set of real numbers is either at most countable or has the cardinality of the continuum. Although the Continuum Hypothesis cannot be proved in Zermelo-Fraenkel set theory, the situation is different when one restricts attention to sets that are simple in some sense. for example, topologically. In the present section we look at several results of this type.

4.1 Theorem Every nonempty open set of reals has cardinality 2N". Proof. The function tanx, familiar from calculus, is a one-to-one mapping of the open interval ( -rr12, rr 12) onto the real line R; therefore, the interval (-rrl2,rrl2) has cardinality 2N". If (a, b) and (c,d) are two open intervals. tlw function d-e f(x)=c+-b"(x-a)

-a

is a one-to-one mapping of (a,b) onto (c,d); this shows that all open intervals have cardinality 2N". Finally, every nonempty open set is the union of a nonempty system of open intervals, and thus has cardinality at least 2N". As a subset of R, it has cardinality at most 2N" as well. 0

4.2 Theorem Every perfect set has cardinality 2N". We first prove a lemma showing that every perfect set contains two disjoint perfect subsets (the "splitting" lemma). Let P denote the system of all perfect subsets of R.

4.3 Lemma There are functions G 0 and G, from P into P such that, for each FE P, G0 (F) <;;: F, G 1 (F) <;;: F and G0 (F) n G 1 (F) = 0. Proof We show that for every perfect set F there are rational numbers r. s, r < s, such that F n ( ·-oo, r] and F n [s, +oo) are both perfect. Let o = inf F ifF is bounded from below (notice that in this case a E F) and o = -oo otherwise. Similarly, let (3 = sup F if F is bounded from abow ((3 E F) and (3 = +oo otherwise. There are two cases to consider:

4. SETS OF REAL NUMBERS

189

(a) (o, (J) ~ F. In this case, any r, s E Q such that o < r < s < (3 have the desired property: for example, if o E R, then F n (-oo,r] = [o,rj; otherwise, F n ( -oo, r] = ( -oo, r] (and we noted that all closed intervals are perfect sets). (b) (o, (3) 1:£ F. Then there exists a E (o, (3), a (j: F. The set F is closed (i.e., its complement is open), and thus there is a > 0 such that (a- J, a+ a) nF = 0. We note that o
FE P and each n EN, n

i- 0,

H(F, n)

~

->

P such that. for each ~ 1/n.

F and diam(H(F, n))

Proof Let F E P, n E N, n > 0. There exists m E Z such that F n (m/n, (m + 1)/n) :f 0. (Otherwise, F ~ {m/n I m E Z}, so all points ofF would be isolated.) Take the least m 2 0 with this property, or. if none exists. the greatest m < 0 with this property. Let a= inf F n (m/n. (m + 1)/n) and b = sup F n (m/n, (m + 1)/n). Clearly, mjn ~ a ~ b ~ (m + 1)/11. so b- a ~ 1/n. We define H(F, n) = F n [a, b]. Clearly, F is closed, F :f 0. The definition of infimum implies that a is not an isolated point of H(F. n). and neither is b. An argument similar to the one used in case (b) in the proof of the previous lemma shows that no x E (a, b) ~ (m/n, (m + 1)/n) is an isolated point of H(F, n) either, so H(F, n) is perfect. 0 Proof of Theorem 4.2. This proof is now easy. It closely resembles the argument used in Section 3 to show that the Cantor set has the cardinality 2N". Let F be a perfect set; we construct a system of its perfect subsets ( F,, I s E S), where S = Seq({O, 1}), as follows:

Fo F(so, ... ,• .. -~oO)

F(so, ... ,S .. -),1)

= F;

= H(Go(F(s ,s.,_J)), n); = H(GI(F(so, ... •.. -1))· n). 11 , ••

190

CHAPTER 10. SETS OF REAL NUMBERS

For
4.6 Corollary Every closed set of reals is either at most countable or hw; thf' cardinality 21-!". \Ve give two proofs of Theorem 4.S. The first proof is quite simple: howewr. it uses the fact that the union of any countable system of at most countable sets is at most countable. The proof of this requires the Axiom of Choice (sPe Chapter 8). Our second proof does not depend on the Axiom of Choice at all: it also provides a deeper analysis of the structure of closed sets, which is of independent interest. It uses transfinite recursion (see Chapter 6). Proof of Theorem :1.5. Let A be a set of real numbers. WP wll (1 E R a condensation point of A if for Pwry 15 > 0 the sPt of all :r E A such that !1:- al < J is uncountable (i.e .. HPither finite nor countable). It is obviou~ frolll the definition that any condensation point of A is
Proof. By Lemma 3.16 it suffices to show that every accumulation point of A" belongs to Ac. If a is an accumulation point of Ac and J > 0, theu there is x E Ac such that lx - al < J. Let E. = J - lx - al > 0. Then, art> uncountably many y E A such thnt IY - :cl < E; but our choice of E guarant<'Ps that if IY- xl < E, then IY- ai <: IY- xi+ lx- al
Now let F be a closed set; tlwu pc S:::: F (sP<, Lemma 3.10). observation is

Our m•xt

(2) C = F - Fe is at most countable. Proof. If a E C, then a is not a condensation point of F, so ther<' is J > 0 such that F n (a - J, a + b) is at most countable. The density of the rationals in the real line guarantees the existence of rational numbers r, .s such that a - J < r < a < s < c1 + 6. This shows that for <~ach a E (' th<'r<' i~

4. SETS OF REAL NUMBERS

191

an open interval (r, s) with rational endpoints such that a E F n (r. s) and F n (r, s) is at most countable; in other words, C ~ U{ F n (r, s) I r. s E Q. T < s, and Fn(r,s) is at most countable}. But the set on the right side of the laBt inclusion is a union of an at most countable system of at most countable sets. so it, and consequently C, too, is at most countable. (Here wf' usP the fact dependent on the Axiom of Choice.) 0 (3) If F is an uncountable set, then Fe is perfect. Proof. We know Fe is closed [observation (1)] am! nonernpty [observation (2)]. It. remains to show that Fe has no isolated points. So assunw that 11 E Fr is an isolated point of Fe. Then there is c5 > 0 such that Jx - aJ < c5 •. r f- 11, implies x ~ Fe. But then, for this J, Jx- al < c5 and .r. E F imply .r. = 11 or x E F- Fe, which is an at most countable set by observation (2). Thus there are at most countably many x E F for which lx - al < J, contradicting thP assumption a E Fe. 0

Observation (3) completes the proof of Theorem 4.5.

0

For an alternative proof of Theorem 4.5, and for its own sakP, we lwgin a somewhat deeper analysis of the structure of closed sr~ts. In general. dosed st'ts differ from perfect sets in that they may have isolatPd points. WP first prow 4.1 Theorem Every closed set has at most countably many isolated points.

If a is an isolated point of a closed set F, then there is c5 > 0 such Proof. that the only element of F in the open interval (a - J, a+ J) is a. Using thP density of the rationals in the reals, we can find r. s E Q such that a - li < r < a < s < a+ J, so that a is the only element of F in thf' open intPrval (r. s) with rational endpoints. The function that assigns to each open intPrval with rational endpoints containing a unique element ofF that element as Cl value. is a mapping of au at most countable set onto the set of all isolatPd points of F. The conclusion follows from Theorem 3.4 in Chapter 4. 0 4.8 Definition Let A ~ R; the derived set of A. dPnoted by A'. is the sPt of all accumulation points of A. It is easy to see that a set A is closed if and only if A' ~ A and pPrff'ct if and only if A i- 0 and A' = A. Also, the derived set is itself a closPd set. (See Exercise 4.4 for these results.) Theorem 4. 7 says that for any closPd F. th(' sPt F- F' of its isolated points is at most countable. These consideration::; ::;uggest a strategy for an investigation of the cardinality of dosf'd sets. Let F f- 0 be a closed set. IfF= F', F is perfPct and JFI = 2N". Otherwis('. F = (F- F') U F' and IF- F'l ::; ~o. so it remains to dPtNmine thP cardiualit,v of the smaller closed set F'. If F' = 0 then F = F - F' has only isolated points and IFI :::.= ~o.

192

CHAPTER 10. SETS OF REAL NUMBERS

If F' is perfect, then IF' I = 2~", so IFI = 2N". It is possible, however, that F' again has isolated points; for an example, consider F == N u { m- ~ I m. 11 E N, m 2: 1, n > 1}. In that case we can repeat the procedure and examiuP F" = (F')'. IfF"= 0 or F" is perfect, we get IFI :S No or IF!= 2N". But F" may again have isolated points, and the analysis must continue. We can dPhne an infinite sequence (Fn I n E N) by recursion:

F0 = F; Fn+! = (Fn)'. All Fn are closed sets; if for some n E N Fn = 0 or Fn is perfect, we get ~ No or IFI = 2Nn. Unfortunately, it may happen that all of the Fn 's have isolated points. We could then define Fw = nnEN Fn; this is again a closed set, smaller than all Fn. If Fw is perfect, we get IFI 2 IFwl = 2N", and if F.., = 0. F = UnEN(Fn - Fn+d is a countable union of countable sets, which can bP proved to be countable (without use of the Axiom of Choice). The problem is that even F.., may have isolated points! (Construction of a. set F for which this happens is somewhat complicated, but see Exercises 4.5 and 4.6.) So we> have to consider further Fw+! = F~. Fw+ 2 = F~+!• ... , and, if necessary. PVP!l Fw+w = nnEN Fw+n and so on. As the sets are getting smaller. there is good hope that the process will stop eventually, by reaching either a perfect set or 1/J. This was the original motivation that led Georg Cantor to the development of his theory of transfinite or ordinal numbers.

IFI

4.9 Definition Let A be a set of reals. recursion on a,

For every ordinal a we define. by

A(Ol =A; A(o-+l) =(A("))'=

A
n

AW

the derived set of

A(o);

(a a limit ordinal).

{
4.10 Theorem Let F be a closed set of reals. There exists an at most countable ordinal e such that (a) For every a < 8, the set F(o)- p(o+l} is nonempty and at most countablf'. {b) F. (c) P = p(e) is either empty or perfect. (d) C = F- P is at most countable. In particular, the theorem gives a decomposition of each uncountable clost>d set F into a perfect set P and an at most countable set C. Proof. For each a, the set p
4. SETS OF REAL NUMBERS

193

As long as F(o+l) f. F(o), the transfinite sequence (F<"l) is decreasing (i.e., F(o) :::> F(/3) when a < (3). By the Axiom Schema of Replacement there exists an ordinal number 8 such that F(e+ I) = F( 9 l. [To see this, assume that F(o) - F(o+ll f. 0 for all a and let "' be the Hartogs number of P(F). The function g(a) = F(ol- F(o+l), a<"(, is a one-to-one mapping of"' into P(F), a contradiction.] Let 8 be the least such ordinal. [As a matter of fact, a consequence of the argument given below is that 8 is an ordinal less than Wt.] Let P = F( 9 l. We have ( F(e) )' = F( 9 l, so the set P is either empty or perfect. Let C == F - P. Let (J0 , J 1 , ... , Jn, ... ) be a sequence of all open intervals with rational endpoints. For each a E C, let Oa be the unique a < 8 such that a E p(o) F(o+ll, and let

f(a) ==the least n such that Jn n F(o.,) ={a}. Since a is an isolated point of p(o,.), there is an interval Jn that has only the point a in common with p(o,.), so f is well defined. The function f maps C into w. We complete the proof by showing that f is one-to-one. Thus let a, b E C, and assume that f(a) == f(b) = n. Let us say that Oa::; Ob. Then F(o,,) ~ F(o,.), and bE Jn n F(oi.) ~ Jn n F(o.. ). sob= a. Hence f is one-to-one and soC is at most countable. 0 Every uncountable closed set has a perfect subset. This suggests that there may not be a simple example of an uncountable set of reals that does not have a perfect subset. And indeed, it is necessary to use the Axiom of Choice to find such an example.

4.11 Example An uncountable set without a perfect subset. Using the Axiom of Choice, we construct two disjoint sets X and Y, both of cardinality 2N", such that neither X nor Y have a perfect subset. We construct X andY as the ranges of two one-to-one sequences (x 0 I a < 2N"), (Ya I a < 2N") to be defined below. As proved in Chapter 5, the number of all closed sets of reals is 2N". Thus there are only 2N" perfect sets of reals, and we let (Po I a < 2N") be some one-to-one mapping of 2No onto the set of all perfect reals. To construct (x 0 ) and (y 0 ), we proceed by transfinite recursion (for a< 2N"). We choose xo and Yo to be two distinct elements of P0 . Having constructed (x~ I~< o) and (Yf. I~< a) (where a< 2N"), we observe that the set

has cardinality 2N" (because !Pal ""zNn and lol < 2N") and we choose distinct Yo E Po such that both X 0 and Yo differ from all x~, Y€ (~ < o). The sequences (x{ I ~ < 2N°) and (Y{ I { < 2N") are one-to-one and their ranges X and Y are disjoint sets of cardinality 2N". Neither X nor Y have a perfect subset: if P is perfect, then P = Pa for some a < 2N", and P n X f. 0 f. P n Y because Xa E P n X and Yo. E P n Y. 0 Xa,

CHAPTER 10. SETS OF REAL NUMBERS

194 Exercises

4.1 Every perfect set is either an interval of the form [a, b]. (- x. u]. [IJ. +x). orR= ( -oo, +oo), or it is the union of two disjoint perfect sets. 4.2 Assume that the union of any countable system of countable sets is countable. Prove that every uncountable closed set contains two disjoint uncountable closed sets (a "splitting" lemma for closed t;ets). 4.3 Use Exercise 4.2 to give yet another proof of the fact that ever:v uncountable closed set has cardinality 2N". 4.4 Let A' be the derived set of A ~ R. The set A' is closed. A is closed it and only if A' <;;; A. A is perfect if and only if A' = A and A I= 0. 4.5 The set F = {1} U {1- 1/2" 1 I n1 2 1} U {1- 1/2n 1 - 1/2"•+n 2 I n1 2 1, n2 2 1} is closed, F' = {1} U {1- 1/2n' I n 1 2 1}. F" = {1}. and

F'"

=

0.

4.6 The set F = {1} u {1- 1/2n' - 1/2n,+n 2 - · . · - 1/2n,+n 2 + +u, 1 A- < nt and n, 2 1 for all i:::; k} has Fw = nnEN Fn = {1}. 4.7 The decomposition of a closed set F into PuC where P n C = 0. P is perfect, and C is at most countable, is unique; i.e .. ifF = P 1 U C 1 wlwn· P 1 is perfect and P1 and C 1 are disjoint and IC11 :::; N0 , then P 1 = P and C 1 =C. [Hint: Show that Pis the set of all condensation points of F.]

5.

Borel Sets

Theorem 4.1 and Corollary 4.6 show that open and closed sets are either at. most countable or have cardinality 2N". It is tempting to try to prove similar results for more complicated, but still relatively simple. sets. How do WP obtain such sets'? One idea is to use set-theoretic operations, such a.s unions and intersections. Now, a union or an intersection of a finite system of open .'Wts is again open, so no new sets can be obtained in this way, and the same holds true for closed sets. The next logical step is to consider unions and inters<~ct.ious of countable systems. We define Borel sets as those sets of reals that can lw obtained from open and closed sets by means of repeatedly taking countablP unions and intersections. 5.1 Definition A set B ~ R is Borel if it belongs to every system of sN:-; S ~ P(R) with the following properties. (a) All open sets and all closed sets belong to S. (b) If Bn E S for each n E N, then U~=o Bn and n~=o Bn belong to S. B denotes the system of all Borel sets. In other words, B = n{s c;; P(R) I S has properties (a) and (b)} is the smallest collection of sets of reals that contains all open and closed sets aud is "closed" under countable unions and intersections. [Note that P(R) has properties (a) and (b), so the intersection is defined.] It is possible to prow some properties of Borel sets using this definition. We give one example (seP also Exercises 5.1, 5.2).

195

5. BOREL SETS The notion of a a-algebra of subsets of Sis defined in 2.17, Chapter 8.

5.2 Theorem The collection B of all Borel sets is a a-algebra of subsets of R.

Proof. Let C = {X ~ R I R - X E B}. We show that C has properties (a) and (b) from Definition 5.1. If X is open, then R - X is closed, so R - X E B and X E C. Similarly, X closed implies X E C. This shows C has property (a). If Bn E C for each n E N, then R- Bn E B for each n E N. hence R- U~=o Bn = n~= 0 (R- Bn) E B, and U~=o Bn E C. This, and a similar argument for intersections, shows that C has property (b). We conclude that B ~ C. Hence X E B implies X E C. which implit>s

0

R-XEB.

Theorem 5.2 shows that the collection B of all Borel sets is the a-algebra generated by all open sets (or, by all closed sets); see Exercise 2.5 in Chapter 8. However, for a deeper study of Borel sets it is desirable to have some more explicit characterization of them. Let us consider which sets are Borel once more. First, all open sets and all closed sets are Borel by clause (a). We define

~~ = { B ~ R

IB

is open};

II?= {B ~RIB is closed}. Thus ~~ ~ B, II? ~ B. Next, countable unions and intersections of open s<'ts must be Borel by clause (b). Countable unions of open sets are open (se<' the remark following Theorem 3.11) and do not produce new sets; so we define

n DO

II~= {B ~RIB=

Bn where each Bn

E

~~}.

n=O

Clearly ~~ ~ II~ and it is possible to prove that the inclusion is proper (see Exercise 5.3), so we obtain some new Borel sets in this way. Similarly. one defines DO

~~

= {B

~ R IB

= U B,

where each Bn E II~}

n=O

and proves that II~ C ~~ ~ B. It is also true that c I1~ and ~~ can be continued recursively. We define

rr?

u

c

~~ (see Exercise 5.3). The process

DO

~~ = { B ~ R I B =

B, where each B,

E

ng}

n=O

and similarly

n B, DO

rrg =

{B ~ R \ B =

n=O

where each B, E ~~}.

CHAPTER 10. SETS OF REAL NUMBERS

196 In general, for any k

> 0,

u DO

:E~+l = {B ~RIB=

Bn where each Bn E nZ}

n=O

and

n Bn where each Bn E :EZ}. DO

nz+l

= {B

~RIB=

n=O

By induction, :EZ ~ Band rrZ ~ B, for all k E N, k > 0. It can also be shown (with the help of the Axiom of Choice; we omit the proof) that

:E2 u rrz c :E~+ I and :E2 u n2 c II~+ 1' so the hierarchy produces new Borel sets at each level. One might hope that all Borel sets are obtained in this fashion, that is, that B = 1 :E~ = 1 IT~, but it is not so. There are sequences (Bn I n E N) such that Bn E :E~+ 1 ~ B for each n E N' but u~=O Bn or n~o Bn does not belong to any :E2 or rr~ (although it is of course Borel). We see again the need for a continuation of the construction of the hierarchy into "transfinite."

U:::

U:::

5.3 Definition For all ordinals n < w 1 we define collections of sets of reals :E~~ and Il 0 as follows: (a) :E~ = {B ~RIB is open}, rrJ = {B ~RIB is closed}. (b) :E()[+l = {B ~RIB= u~=O Bn where each Bn En~}. IT~+ = { B ~ RIB= n~=O Bn where each Bn E :E~}. (c) :E~ = {B ~RIB= Bn where each Bn E rrg for some (3 < n}, rr~ = {B ~RIB= Bn where each Bn E :r:g for some {3 < (t} (n a limit ordinal). [In view of the next lemma, one could use clause (c) for successor ordinals as well, instead of (b).]

I

u:=O n:=O

5.4 Lemma

(i) For every n < Wt, :E~ ~ rr~+t and IT~ ~ :E~+t· {ii) For every n < WJ, :E~ ~ :E~+t and IT~ ~ Il~+I· (iii) If BE I:~, then R- BEll~; if BE IT~, then R- B E I:~. (iv} If Bn E :E~ for each n E N, then Bn E Il~. n EN, then

n:=o

U;::"= 0 Bn

E I:~; if Bn E IT~ for each

(v) If n < (3 then :E~ ~ :Eg, :E~ ~IT~, and IT~ ~ rrg, IT~ ~ :r:g.

197

5. BOREL SETS Proof.

(i) Trivial: in Definition 5.3(b), let Bn = B for all n.

(v) Every open set is in

:r:g

(see Exercise 5.3), and similarly, every closed set is in rrg. Hence I:~ ~ :r;g and II~ ~ IIg. Also, II~ ~ :r;g and I:~ ~ IIg. by (i). Then we proceed to prove that the assertion holds for every n < (3, by induction on (3.

(ii) Follows from (v). (iii) By induction on n. (iv) This innocuous statement requires the Countable Axiom of Choice: For each n, if Bn E :E~, then Bn = u:=O Bmn for some countable collection {Bmn I mEN} of sets Bmn E u-y
00

00

U Bn = U U Bmn n=O

n=Om=O

is the union of the countable collection {Bmn I (n. m) E N x N} and thus in :E~. The proof for II~ is similar. 0 It is clear (by induction on n) that each :E~ and each I1~ consists of Borel sets. What is important, however, is that this hierarchy exhausts all Borel sets.

5.5 Theorem A set of reals is a Borel set if and only if it belongs to I:~ for some n < w1 (if and only if it belongs to II~ for some n < WJ}. Proof.

It suffices to show that the set

s=

u :E~ u =

is closed under countable unions and intersections. Thus let { Bn I n E N} bt> a countable collection of sets inS. We show that, e.g., ~::"=o Bn is inS. For each n let On be the least 0 such that Bn E rr()[. The set {On I n E N} is an at most countable set of ordinals less than w 1 and thus (by the Countable Axiom of Choice) its supremum o = sup{on In EN} = U{on / n EN} is an ordinal less than w 1 . By Lemma 5.4 we have Bn E II~ for each n. Consequently, U::"=o Bn belongs to :E~+l and hence to S. 0 We note, without proof, that Theorem 4.5 can be extended to all Borel sets: Every uncountable Borel set contains a perfect subset. In particular, every uncountable Borel set has cardinality 2N". In Chapter 8 we mentioned the aalgebra VR of Lebesgue measurable sets of reals; of course, B ~ VR, but there

198

CHAPTER 10. SETS OF REAL NUMBERS

exist also Lebesgue measurable sets which are not Borel. In fact, one cannot prove in Zermelo-Fraenkel set theory that all uncountable Lebesgue mea.surabl<> sets have cardinality 2N". The construction of the Borel hierarchy is just a special case of a very gen<>ral procedure. We observe that the infinitary unions and intersections used in it are, in some sense, generalizations of the usual operations u and n. \VP d<>htw F is a (countably) infinitary operation on A if F is a function on a subst't of AN (the set of all infinite sequences of elements of A) into A, finition of a set B <;;; A being closed remaius essentially unchanged: if fJ = N WP n·quin· FJ((a, I i EN)) E B for all (a, I i EN) such that a, E B for all i EN and F 1 is defined. The closure C of a set C <;;; A is still the least closed set wntainiug all clements of C. In this terminology, the system of Borel sets is just tlw closure of tlw systPIII of all open and closed sets under the infinitary union and intersection. ~Ion• precisely. we let 2l = (A, F1, F2) where A = P(R), F1 ( (B, I i E N)) = U,:u B, and F2((B, I i EN))= n~o B,, and c = {B <;;;RIB is open orB is closed}. Then B =C. Exercise 5.9 provides another example of a structure with infillitary op<>rations and a closed subset of some mathematical interest. One important general question about the closure is that of its cardiualit~·. \Ve proved the simplest result providing an auswer in Theorem 3.14 of ChaptN 4. There it was assumed that C is at most countable, and all operations an· fiuitary. We now generalize this, first. to arbitrary C. and then to structur<·s with infinitary operations.

5.6 Theorem Let 2l = (A. (Ro, .... Rm- 1 ). (Fo, .... Fn-l)) lw a strur:ture and h't C <;;; .4. Let C be the closun~ of C. If C is finite, then C ts at most ro1Lnta!Jle: if C is mfinite. then ICI = ICI. Proof. We proved in Theorem 5.10 in Chapter 3 that C = U;:;o C, wlwn· c and c,+ I = c, u Fo[C["J u ... u Fn- dC!" "']. If C' is finite. th<'1l Pach (', is finite, and ICI ::; No. If Cis infinite and ICI = N0 ; then we prove by induction that ICkl = N" for all k: Assume that it is true for C,; then IC(' I ::; N~' = N,..

Co =

IF1 [C(']I ::; No. and IC•+ll = N". Hence ICI =No· N" = Nn. When 2l is a structure with infinitary operations, Theorem 5.6 can be gen(•ralized but the cardinality estimate is different. Let 2l be a structure. and for simplicity let us assume that 2l has just one operation F and that F is an infinitary operation (the result can of course be generalized to structures with more than one operation). 2l has universe A, and F is a function on a subsPt of A"' into A, where Aw is the set of all sequences (an I n E N) with values in A.

5. BOREL SETS

199

5.7 Theorem Let Ql = (A, F) be a structure where F is a function on a of Aw into A, and let C ~A. Let C be the closure of C in Ql, i.e.,

.~u/Jset

If C has at least two elements, then ICI :S ICIN". Proof.

We construct an w1-sequence Co

~

cl

~

...

~

Co

~

...

(n < Wt)

of subsets of A as follows: Co=C;

Ca+l

=

Co =

Ca. U F[C~J;

U Ce

if n is a limit ordinal.

~
Let C be the closure of C in Ql and let D = Ua.nce in D, then (an I n E N) E Co for some n < w1: for each n let ~n be the least ~ such that an E C~, and let n = sup{(n In EN}. And if (an In EN) E domF then F((an In EN)) E Ca+l; hence F[Dw] ~ D, soD 2 C. Thus C = Ua
If n is a limit ordinal, and if the estimate is true for

(because

< n, then

lnl :S N0 ). Finally, we have ICI :S

because N1 · 2N" =

ICIN".

all~

N1 · ICINn

· 2No

=

IGIN" · 2N",

2N". And if ICI 2 2, then IGIN" 2 2N", and we have ICI :S 0

Exercises

5.1 If X is Borel and a E R then X+ a= {x +a I x E X} is Borel. [Hint: Imitate the proof of Theorem 5.2.] 5. 2 If X is Borel and f : R -+ R is a continuous function, then f- 1 [X] is Borel.

200

CHAPTER 10. SETS OF REAL NUMBERS

5.3 Prove that every open interval is a union of a countable system of closed intervals and conclude from this fact that '£~ c;;: '£~. Since II~ c;;: '£~. conclude that '£~ C '£~ and II~ C E~. This yields immediatPly that C II~ and E~ C II~. 5.4 Prove by induction that'£~ U II~ c;;: '£~+! n II~+ I for all n E N. 5.5 Prove that for each a, both '£~ and II~ are closed under finite unions and intersections, i.e., if B 1 and B 2 E '£~. then B 1 U B 2 E E~ and B 1 n B2 E '£~, and similarly for II~. 5.6 Let f: R--> R be a continuous function. If BE'£~. then f- 1 [B] E '£:: Similarly for II~. 5. 7 Show that the cardinality of the set B of all Borel sets is 2N". 5.8 Prove that the Borel sets an' the closure of the set of all open intPrvals with rational endpoints under infinitary unions and intersections. This shows that in a structure with infinitary operation:>. the closure of a countable set may be uncountable. 5.9 Let Fn be the set of all functions from a subset of R into R. TltP infinitary operation lim (limit) is defined on Fn as follows: lirn((f, I i •: N)) = f where f(x) = lim;~rx:> f,(x) whenever the limit on the right side exists, and is undefined otherwise. The elements of the closure of the set of all continuous functions under lim are called Baire functions. Show that Baire functions need not be continuous. In particular. show that the characteristic functions of integers and rationals are Baire functions. 5.10 For n < w 1 , we define functions (on a subset of R into R) of class o: as follows: the functions of cla..o;s 0 are all the continuous functions; f is of class n iff= limn~oc fn where each fn is of some dass < n. Show that Baire functions are exactly the functions of class a for some n: < w 1 • 5.11 Show that the cardinality of the set of all Baire functions is '2No

IIY

Chapter 11

Filters and Ultrafilters 1.

Filters and Ideals

This chapter studies deeper properties of sets in general. Unlike earlier chapters. we do not restrict ourselves to sets of natural or real numbers, and neither do we concern ourselves only with cardinals and ordinals. Here we deal with abstract collections of sets and investigate their properties. We introduce severn! concept.s that have become of fundamental importance in applications of set theory. It is interesting that a deeper investigation of such concepts leads to the theory of large cardinals. Large cardinals are studied in Chapter 13. We start by introducing a filter of sets. Filters play an important role in many mathematical disciplines.

1.1 Definition Let S be a nonempty set. A filter on S is a collection F of subsets of S that satisfies the following conditions: (a) S E F and 0 ¢ F. (b) If X E F andY E F, then X n Y E F. (c) If X E F and X c:;: Y c:;: S, then Y E F. A trivial example of a filter on S is the collection F = { 5} that consists only of the set S itself. This trivial filter on S is the smallest filter on S; it is included in every filter on S. Let A be a nonempty subset of S, and let us consider the collection

F ={X c:;: sIX 2 A}. The collection F so defined satisfies Definition 1.1 and so is a filter on S. It is called the principal filter on S generated by A. If in this example, the set A has just one element. a, i.e., A = {a} where a E S, then the principal filter F ={X c:;: S

201

I a EX}

CHAPTER 11. FILTERS AND ULTRAFILTERS

202

is maximal: There is no filter F' on S such that F' :) F (because if X E F'- F then a¢ X, but {a} E F' as well, and so 0 =X n {a} E F', a contradiction). We shall prove shortly that there are maximal filters that are not principal. To give an example of a nonprincipal filter, let S be an infinite set, and let (1.2)

F

={X~

SIS- X is finite}.

F is the filter of all cofinite subsets of S. (X is a cofinite subset of 8 if 8- X is finite.) F is a filter because the intersection of two cofinite subsets of S is a cofinite subset of S. F is not a principal filter because whenever A E F then there is a proper subset X of A such that X E F (let X be any cofinite subset of A, X =f. A). 1.3 Definition Let S be a nonempty set. An ideal on S is a collection I of subsets of S that satisfies the following conditions: (a) 0 E I and S ¢I. (b) If X E I and Y E I, then XU Y E I. (c) If Y E I and X ~ Y, then X E I. The trivial ideal on S is the ideal {0}. A principal ideal is an ideal of tlw form (1.4)

I= {X

I X~

A}

where A~ S. To see how filters and ideals are related, note that if F is a filter on S then (1.5)

I={S-XIXEF}

is an ideal, and vice versa, if I is an ideal, then (1.6)

F={8-XIXEI}

is a filter. The two objects related by (1.5) and (1.6) are called dual to each other. The filter of cofinite subsets of S is the dual of the ideal of finite subsets of S. In Exercises 1.2 and 1.3 the reader can find other examples of nonprincipal ideals. We recall (Definition 3.3 in Chapter 10) that a nonempty collection G has the finite intersection property if every nonempty finite subcollection {X 1 . . . . • X,.} of G has nonempty intersection X 1 n · · · n Xn =f. 0. It follows from (a) and (b) of Definition 1.1 that every filter has the finit<> intersection property. In fact, if G is any subcollection of a filter F. then G has the finite intersection property. Conversely, every set that has the finite intersection property is a subc:ollection of some filter. 1. 7 Lemma Let G =f. 0 be a collection of subsets of S and let G have the fimt(' intersection property. Then there is a filter F on 8 such that G ~ F.

203

1. FILTERS AND IDEALS

Proof. Let F be the collection of all subsets X of S with the property that there is a finite subset {X 1, ... , Xn} of G such that

Clearly, S itself is in F, and 0 is not in F because G has the finite intersection property. The condition (c) of Definition 1.1 is certainly satisfied by F. As for the condition (b), if X 2 X1 n · · · n Xn for some X 1, ... , Xn E G, and if Y 2 Y1 n- · -nYm for some Y1, ... , Ym E G, then XnY 2 X1 n- · -nXnnYI n- · -nY,,. so X n Y E F. Thus F is a filter. 0 The filter F constructed in Lemma 1.7 is the smallest filter on S that extends the collection G. We say that G generates the filter F. See Exercise 1.6. 1.8 Example Let S be a Euclidean space and let a be a point in S. Consider the collection G of all open sets U in S such that a E U. Then G has the finite intersection property and hence it generates a filter F on S. F is the neighborhood filter of a. 1.9 Example. Density. Let A be a set of natural numbers. For each n E N, let A(n) = lA n {0, ... ,n- 1}1 denote the number of elements of A that are smaller than n. The limit A(n) d(A) = lim - - , n--+00

n

if it exists, is called the density of A. For example, the set of all even numbers has density 1/2 (Exercise 1.7). Every finite set has density 0, and there exist infinite sets that have density 0. (For example, the set {2n I n E N} of all powers of 2 - Exercise 1.8.) Let A and B be sets of natural numbers. If A ~ B then A(n) ::::; B(n) for all n, and so if both A and B have density then d(A) ::::; d(B). In particular, if d(B) = 0 then d(A) = 0. Also, for every n, (AU B)(n) ::::; A(n) + B(n), and if A and Bare disjoint then (AU B)(n) = A(n) + B(n). Hence d(A U B)::::; d(A) + d(B) (provided that the densities exist), and d(A U B) = d(A) + d(B) if A and B are disjoint. If both d(A) and d(B) are zero, then d(A U B) = 0. This gives us an example of an ideal on N, the ideal of sets of density 0:

Id ={A I d(A) = 0}. We have 0 E Id and N ¢ Id (because d(N) = 1). If A ~ B and B E Id then A E Id, and if both A E Id and B E Id then A U B E h Thus Id is an ideal. As noted above, Id contains all finite sets, but also some infinite sets. As a final remark, not every set A C N has density. It is not difficult to construct a set A such that the limit limn~oo A(n)/n does not exist (Exercise 1.9).

204

CHAPTER 11. FILTERS AND ULTRAFILTERS

We conclude this section with the introduction of an important concept that has many applications in analysis: 1.10 Definition A measun~ on a set S is a real-valued function m defizH'd 011 P(S) that satisfies the following conditions: (a) m(0) = 0, rn(S) > 0. (b) If A<;::; B then rn(A) ::; m(B). (c) If A and Bare disjoint then m(A U B)= rn(A) + m(B). (Notice that the density function on P(N) satisfies tl!P conditions. except that it is not defined for all subsets of N.) It follows that m(A) 2': 0 for every A, and that m(S- A) = rn(S)- m(A). PropE>rty (c) is called finite additivity, and clearly m(A1 U · · · U An) = rn(A1)

+ · · · + rn(An)

for any disjoint finite collection {A 1 •... , A"}. At this point. we can only giw two trivial examples of measures: 1.11 Example Let S be a finitP set, and let m(A) = the counting measure on S.

IAI

f(lr

A

<;::;

S. This is

1.12 Example LetS be a nonempty set and let a E 5. Let m(A) = 1 if a E A. and m(A) = 0 if a tf. A. (A trivial measure.) In the next section, we construct nontrivial measures on N.

Exercises 1.1 If Sis a finite nonempty set, then every filter on Sis a principal filter. 1.2 Let S be an uncountable set, and let I be the collection of all X c s· such that lXI ::; No. I is a nonprincipal ideal on S. 1.3 Let S be an infinite set and let Z <;::; S b(' such that both Z and S- Z an• infinite. The collection I = {X <;::; S I X - Z is finite} is a nonprincipal ideal. 1.4 If a set A <;::; S has more than one element, then the principal filt('r generated by A is not maximal. 1.5 If :F is a nonempty set of filters on S, then F I F E :F} is a filter 0!1 S. 1.6 The filter constructed in the proof of Lemma 1. 7 is the smallest filter on S that includes the collection G. 1. 7 Let A be the set of all natural numbers that are divisible by a gi\'('11 number p > 0. Show that d(A) = 1/p. 1.8 Prove that the set {2" I n E N} has density 0. 1.9 Coustruct a set A of natural numbers such that

n{

limsupA(n)/n = 1 and liminf A(n)/n = 0. n~oo

2. ULTRAFILTERS

2.

205

Ultrafilters

2.1 Definition A filter U on Sis an ultrafilter if for every or S- X E U.

X~

S, either X E U

The dual notion is a prime ideal: 2.2 Definition An ideal I on S is a prime ideal if for every X C S. either X E I or S -X E I. 2.3 Lemma A filter F on S is an ultrafilter if and only if it is a maximal filter· on S. Proof. IfF is an ultrafilter, then it is a maximal filter, because ifF' =:> F is a filter on S, then there is X ~ S in F'- F. But because F is an ultrafilter, S- X is in F, and hence in F', and we have 0 =X n (S- X) E F', a contradiction. Conversely, let F be a filter but not an ultrafilter. There is some X C:::: S such that neither X nor S- X is in F. Let G = F U {X}. We claim that G has the finite intersection property. If X I' ... ' X n are in F' then y = X I n ... n X n E F' and y n X f 0 because otherwise (S- X) 2 Y and so S- X E F, contrary to the a.~sumption. Hem·p X I n ... n Xn n X f 0, which means that G = F u {X} has the finite intersection property. Thus there is a filter F' on S such that F' 2 F U {X}. But that means that F is not a maximal filter. S

We have seen earlier that there are principal filters that are maximal. In other words, there exist principal ultrafilters. Do there exist nonprincipal ultrafilters? Let S be an infinite set, and let F be the filter of cofinite subsets of S. If U is an ultrafilter and U extends F, then U cannot bf' principal. Thus. to find a nonprincipal ultrafilter it is enough to find an ultrafilter that extends the filtPr of cofinite sets. Conversely, if U is a non principal ultrafilter, then it extends the filt(•r of cofinite sets because every X E U is infinite (Exercise 2.2). The next theorem states that any filter can be extended to an ultrafilt.Pr. However, the proof uses the Axiom of Choice, and iu fact it is known that th(• theorem cannot be proved in Zermelo-Fraenkel set theory alone. 2.4 Theorem Every filter on a setS can be extended to an ultrafiltr.r on S. The proof uses Zorn's Lemma. In order to apply it we need the following fact: 2.5 Lemma If C is a set of filters on S and if for e1wry Ft. F2 F1 C:::: Fz or F2 ~ F1, then the union of C is also a filtn· on S.

E

C. rithrr·

206

Proof. nition 1.1.

CHAPTER 11. FILTERS AND ULTRAFILTERS This is a matter of simple verification of (a), (b), and (c) in Defi0

Prvof of Theorem 2.4. Let Fo be a filter on S; we find a filter F ;J Fa that is maximal. Let P be the set of all filters F on S such that F ;J F0 , and let us considPr the partially ordered set (P, C). By Lemma 2.5, every chain C in P has an upper bound, namely UC. Thus Zorn's Lemma is applicable, and (P. C) has a maximal element U. Clearly, U is a maximal filter on S and U ;J Fa. By Lemma 2.3, U is an ultrafilter. t~ There is a natural relation between ultrafilters and measures. Let us call a measure m on S two-valued if it only takes values 0 and 1: for Pvery A ~ .':1. either m(A) = 0 or m(A) = 1.

2.6 Theorem (a) If m is a two-valued measure on S then U = {A ~ S m( A) = 1} is an ultrafilter. (b) If U is an ultrafilter on S, then the function m on P( S) defined by m( A) = I if A E U and m(A) = 0 if A ¢. U is a two-valued measure on S. Proof. Compare the definitions of ultrafilter and measure. Note that if A and B are disjoint, then at most one of them is in an ultrafilter (or ha.-; mpa.:-;un·

c:

1).

We now present one of many applications of ultrafilters. This particular application is a generalization of the concept of limits of sequences of real numb(•rs.

2.7 Definition Let U be an ultrafilter on N, and let (an);;-=o be a bound(•d sequence of real numbers. We say that a real number a is the U -limit of the sequence, a= liman,

u

if for every c: > 0, {n llan- al < c:} E U. First we observe that if a U-lirnit exists then it is unique. For. assume that a< band both a and bare U-limits of (an);;'=o· Let c: = (b- a)/2. Then tlw sets {n llan- al < c:} and {n !/an- bl < c:} are disjoint and so cannot both lw in U. If U is a principal ultrafilter, U = {A I no E A} for some no. then limu an =, an"' for any sequence (an)~~o· This is because for every c: > 0, {n iiun- au, I < e:} 2 {no} E U. If (an);;'=o is convergent and limn~oo an = a, then for every nonprincipal ultrafilter U. limu an = a. This is because for every e: > 0, therP exists sorw> k such that { n I \an - a\ < e:} :2 { n I n ~ k}, and {n I n ~ k} E U. The important property of ultrafilter limits is that the U -limit exists for every bounded sequence:

2. ULTRAFILTERS

207

2.8 Theorem Let U be an ultrafilter on N and let (an);;:, 0 be a bounded sequence of real numbers. Then limu an exists. Proof As (an)~=O is bounded, there exist numbers a and b such that a < an < b for all n. For every x E [a, b], let

Ax

= {n

I On < X}.

Clearly, Aa = 0, Ab = N, and Ax ~ Ay whenever x :::; y. Therefore Aa ¢. U, AbE U. and if AxE U and x:::; y, then AyE U. Now let

c=sup{x/Axrf:U}; we claim that c = limu an. As for every c: > 0, A c.-< tf. U while Ac+E E U. and because Ac+c = Ac- ~ U { n I c - c: < an < c + c:}, it follows that {n I Ian - cl <

e:}

0

E U.

As an application of ultrafilter limits, we construct a nontrivial measure on N. 2.9 Theorem There exists a measure m on N such that m(A) every set A that has density. Proof.

d(A) for

Let U be a nonprincipal ultrafilter on N. For every set A t:;; N.

let m(A) =lim A(n)

u n where A(n) = /Ann/. Clearly, if A has density then m(A) = d(A). Also, it is easy to verify that m(0) = 0 and m(N) = 1, and that A t:;; B implies m(A) :::; m(B). If A and B are disjoint then (AU B)(n) = A(n) + B(n), and the additivity of m follows from this property of ultrafilter limits: lim(an u

+ bn) = liman + limbn. u u

We leave the proof, which is analogous to that for ordinary limits, as an exercise 0 (Exercise 2.5). Exercises

2.1 If U is an ultrafilter on S, then P(S)- U is a prime ideal. 2.2 If U is a nonprincipal ultrafilter, then every X E U is infinite. 2.3 Let U be an ultrafilter on S. Show that the collection V of sets X ~ S x S defined by X E V if and only if {a E S I {b E S I (a, b) E X} E U} E U is an ultrafilter on S x S. 2.4 Let U be an ultrafilter on Sand let f : S--> T. Show that the collection 1 V of sets X ~ T defined by X E V if and only if [X] E U is an ultrafilter on T. 2.5 Prove that limu(an + bn) = limu an+ limu bn·

r

CHAPTER 11. FILTERS AND ULTRAFILTERS

208

3.

Closed Unbounded and Stationary Sets

In this section we introduce an important filter on a regular uncountable cardinal -the filter generated by the closed unbounded sets. Although the results oft !tis section can be formulated and proved for any regular uncountable cardinal. ,,.t. restrict our investigations to the least uncountable cardinal N1 . (SeP Exercisp,.; 3.5-3.9.)

3.1 Definition A set C ~ w 1 is closed unbounded if (a) Cis unbounded in w 1 , i.e .. supC = w 1 . (b) Cis closed, i.e., every increa..~ing sequence ao

<

a1

< · · · < On < · · ·

(n

E

w)

of ordinals in C has its supremum sup{ On I n E w} E C. An important, albeit simple, fact about closed unbounded sets is that t!tP,. have the finite intersection property; this is a consequence of the following: 3.2 Lemma If cl and Cz ar-e closed unbounded subsets of WJ' then closed unbounded.

ci

n c2

IS

It is easy to see that cl n C:2 is closed; if Oj < Oz < ... is a Proof. sequence in both cl and Cz, then n =sup{ On I n E w} E cl n Cz. To see that cl n c2 is unbounded, let I < Wj be arbitrary and let us find (\ in C 1 n C:2 such that a > r· Let us construct an increasing sequence O'Q

< f3o <

CtJ

<

fji

< · · · < Un < /3n < · · ·

of countable ordinals as follows: Let a 0 be the least ordinal in C 1 above f. T!H'n let /3o be the least fJ > no in Cz. Then O'J E cl' !31 E Cz. and so OIL Let a be the supremum of {nn}nEw; it is also the supremum of {rin}nE~·· The ordinal a is in both C 1 and C 2 and therefore a E C1 n C 2 . D The set w 1 of all countable ordinals is closed and unbounded. (Howewr. here we use a consequence of the Axiom of Choice, namely the regularity of w 1 : the supremum of every countable sequence of countable ordinals is a cotmtabl!• ordinal.) Another example of a dosed unbounded set is the set of all limit countable ordinals (Exercise 3.2). For other, less obvious examples see Exercise:,; :u and 3.4. As the collection of all closed unbounded subsets of w 1 has the finite int.t>rsection property, it generates a filter on w 1 : (3.3)

F == {X~ w 1 I X 2 C h1r some dosed unbounded C}.

The filter (3.3) is called the closed unbounded filteT on w1 .

3. CLOSED UNBOUNDED AND STATIONARY SETS

209

3.4 Lemma If {Cn }nEw is a countable collection of closed unbounded sets, then n~=O Cn is closed and unbounded. Consequently, if { Xn }nEw is a countable collection of sets in the closed unbounded filter F, then n~=O Xn E F. Proof. It is easy to verify that the intersection C = n~=o Cn is closed. To prove that C is unbounded, let 'Y < w 1 ; we find a E C greater than 'Y. We note that C = n~=o Dn where for each n, Dn = Con· · · nCn; the sets Dn are closed unbounded and form a decreasing sequence: Do 2 D 1 2 D2 2 · · · . Let (an I n E w) be the following sequence of countable ordinals: 'Y < a 0 < c.\' 1 < · · ·, and for each n, O'n+ 1 is the least ordinal in D 11 above On. Let n be the supremum of {O'n }nEw· To show that a E C, we prove that a E Dn for each n. But for any n, a is the supremum of {ak I k 2: n + 1}, and all the O'k for 0 k 2: n + 1 are in Dn because Dn 2 Dn+l 2 · · · 2 Dk 2 · · · (k > n). Closely related to closed unbounded sets are stationary sets. Stationary sets are those sets S ~ w 1 which do not belong to the ideal dual to the closed unbounded filter. A reformulation of this leads us to the following definition. 3.5 Definition A set S

~

w 1 is stationary if for every closed unbounded sPt.

C, S n Cis nonempty. Clearly. every closed unbounded set is stationary, and if S is stationary and S ~ T ~ w 1 , then T is also stationary. Later in this section we present an example of a stationary set that does not have a closed unbounded subset. But first we prove the following theorem. A function f with domain S ~ w 1 is regressive if f(a) < a for all a i= 0. 3.6 Theorem A set S ~ w 1 is stationary if and only if every regressive function f : S -> w 1 is constant on an unbounded set. In fact, f has a constant value on a stationary set. This theorem is a good example of how an uncountable aleph such as N1 differs substantially from N0 . On uncountable cardinals there is no analoguE' of the function f : w -> w defined by f(n) = n- 1 for n > 0, f(O) = 0. which has each value only finitely many times. A consequence of Theorem 3.6 is that on w 1 , a function that satisfies f(a.) w 1 be defined as follows: f(a) = sup(C n a.)

(a. E A).

CHAPTER 11. FILTERS AND ULTRAFILTERS

210

Because f(a) E C (see Exercise 3.1) and CnA = 0, we have f(a) '"Y and so f does not have the same value for uncountably many a's. To prove the other direction of the theorem we need a lemma. 3.8 Lemma Let { c~ set C <;;; w 1 defined by (3.9)

I E < W[}

be a collection of closed unbounded sets. Thr

a E C if and only if a E

q

for· all

E< a

(a E wi)

called the diagonal intersection of the c~' is closed unbounded.

First we prove that C is closed. So let ao < a 1 < · · · < Ltn < · · · Proof. be an increasing sequence of elements of C and let a be its supremum. To prove that a is in c we show that a E cf. for all E < a. Let. E < a. Then there is some k such that E < Uk. and hence E < o, for all n 2: k. Since each an is in C, it satisfies (3.9}, and so for each n 2: k we haVP an E Cf.. But cf. is closed and so U, the supremum of {an }n2':k· is in Cf.. Next we prove that C is unbounded. Let '"Y < w1 and let us find o E C greater than --y. We construct an increasing sequence a 0 < a 1 < o- 2 < · · · as follows: Let ao = -y. The set n~
a" < an+l E

n Cf.. f.
The let a be the supremum of {an }nEw• and let us prove that a E C. We are to show that a E cf. for all E < a. So let E < a. There is k such that E < ak. and for all n:::: k we have an+l E cf., by (3.10). But a is the supremum of {an }n>k and hence is in cf.. 0 Proof of Theorem 3.6. Let S be a stationary set and let f : S --+ w1 lw a regressive function. For each '"Y < w 1 , let A.., = f- 1 [ b}]; we show that for sonw 1 the set A.., is stationary. Assuming otherwise, no A.., is stationary and so for each 1 < WI then• is a closed unbounded set C.., such that A.., n C.., = 0. Let C be the diagonal intersection of the C..,, that is,

(3.11)

a E C if and only if a E C.., for all --y < n.

If a E C..,, then a¢ A.., and so f(a)-/= --y. Thus (3.11) means that if a E C. then f(a) f= 'Y for all 'Y
3. CLOSED UNBOUNDED AND STATIONARY SETS

211

As an application of Theorem 3.6 we construct a stationary set S c:;; w 1 , whose complement is also stationary. Note however that the construction uses the Axiom of Choice. (It is known that the Axiom of Choice is indispensable in this case.) 3.12 Example A stationary set whose complement is stationary. Let C be the set of all countable limit ordinals. For each a E C there exists an increasing sequence X 01 = (x 0171 I n E w) with limit a. For each 11. we let fn: C--> W1 be the function fn(o) = Xan· For each n, because fn(o) < o on C, there exists In such that the sf't Sn = {o E C I /n(o) =In} is stationary. We claim that at least one of the sets Sn has a stationary complement. If not, then each Sn contains a closed unbounded set, so their intersection does too. Hence n~=o Sn contains an ordinal o that is greatPr than the supremum of the set hn}nEw· But in that case the sequence Xa = (x"n In E w) = Un(a) I n E w) = hn I n E w) does not converge to o, a contradiction. D As another application of Theorem 3.6 we prove a combinatorial theorem known as the .6.-lemma. Although it is possible to prove the .6.-lemma directly, the present proof illustrates the power of Theorem 3.6. 3.13 Theorem Let {A, I i E I} be an uncountable collection of finite sets. Then there exists an uncountable J c:;; I and a set A such that for all distinct i,j E J, A; n Ai =A. Proof. We may assume that I = w1, and since the union of N1 finite sets has size N1 , we may also assume that all the A, are subsets of w 1 • Clearly, uncountably many A, have the same size and so we assume that we have a collection { Ao I a < w 1 } of subsets of w 1 , each of size n, for some fixed number n.

Let C be the set of all o < w1 such that max A( < n whenever ~ < o. Tlw set C is closed unbounded (compare with Exercise 3.4). For each k :::; n. let sk ={a E I lAo no!= k}; there is at least one k for which skis stationary. For each m = 1, ... , k, let fm(o) =the mth element of Aa; we have fm(a) < n on Sk· By k applications of Theorem 3.6, we obtain a stationary set T c;; S~,­ and a set A (of size k) such that A, no= A for all o E T. Now when o < (3 are in T, then A, C (3 (because (3 E C) and Aa n a = A13n(3 =A, and it follows that A 01 nA13 =A [because (Aa -a)nA;1 = 0]. Thus {A a I a E T} is an uncountable subcollection of {Ao I o < wl} which satisfies the theorem. D

c

Exercises 3.1 An unbounded set C c:;; w 1 is closed if and only if for every X sup X< w 1 , then supX E C. 3.2 The set of all countable limit ordinals is closed unbounded.

c

C, if

212

CHAPTER 11. FILTERS AND ULTRAFILTERS

If X is a set of ordinals, then a is a limit point of X if for every 1 < ( l there is f3 E X such that 1 < f3 < n. A countable a is a limit point of X if and oulv if there exists a sequence no < a 1 < · · · in X such that sup{ a,. I n E uJ} = o. Every closed unbounded C ~ w 1 contains all its countable limit points.

3.3 If X is an unbounded subset of w1, then the set of all countable liutit points of X is a closed unbounded set. If f : Wt ---> w 1 is au increasing function, then a < w1 is a closuTe point of f iff(()
3.4 The set of all closure points of every increasing function closed unbounded. Let K be a regular uncountable cardinal. A set C supC = l'i and sup(C n a) E C for all a< /'i.

~ K

f : w1

--->

w 1 is

is closed unbounded if

3.5 If C1 and C2 are closed unbounded, then C1 n C2 is dosed unbounclPd. The closed unbounded filteT on bounded sets.

K

is the filter generated by the closed un-

3.6 If A < ti and each Ca, a < A, is closed unbounded. then na<-\ Co is closed unbounded. A set S

~ h

is stationary if S n C i= 0 for every closed unbounded sd C

~

h'

3.7 The set {a < l'i I a is a limit ordinal and d(a) = w} is stationary. If h > N1, then the set {a< hI cf(a) = w!} is stationary. 3.8 If each C'
~

" is rr.gTessive if f(o)

< rr for all

ct E dom f. n

i=

0.

3.9 A set S ~ l'i is stationary if and only if every regressive function on S is constant on an unbounded set.

4.

Silver's Theorem

Using the techniques introduced in Sections 2 and 3 we now prove the following result on the Generalized Continuum Hypothesis for singular cardinals.

4.1 Silver's Theorem Let N-\ be a singular cardinal such that d A > u.1. If for· every a< A, 2l'l'" = No+l• then 2N" = N-\+1· We prove Silver's Theorem for the special case N-\ = Nw,, using the theory of stationary subsets of N1. The full result can be proved in a similar way. using

4. SILVER'S THEOREM

213

the general theory of stationary subsets of"- = cf >. (see Exercise 4.1). Thus assume that 2N .. = No+ I for all o < w 1 • One consequence of this assumption, to be used repeatedly in this section, is that N~ 1 < Nw 1 for all o < w 1 (see Theorem 3.12 in Chapter 9). Let f and g be two functions on w 1 . The functions f and g are almost disjoint if there is some o < w 1 such that f(f3) =I g({3) for all !3 2: o. 4.2 Lemma Let { Aa I o < w!} be a family of sets such that IA,, I every o < w 1 , and let F be a family of almost disjoint functions,

F Then

IFI :S

II

c

:::; Na

for

A,.

Nw,.

Proof. Without loss of generality we may assume that A" <;;; w,, for each o < w 1 . Let So be the set of all limit ordinals 0 < n < w 1 . For f E F and o E S 0 , let J*(o) denote the least {3 such that f(o) < W[J. As f*(o) < n for every o E S 0 , there exists, by Theorem 3.6, a stationary set S C So such that is constant on S. Therefore f r is a function from into W[J. for some {3 < WJ. Let us denote this function f r by VJ(/). Iff and g are two different functions in F, then tp(J) and VJ(g) are also different: even if their domains are the same, sayS, then f r S =/ g r S because f and g are almost disjoint. Thus 'P is a one-to-one mapping with domain F. The values of 'P range over functions defined on a subset S of w 1 into some WfJ < Ww 1 • Thus we have

r

s

s

s

0

A slight modification of Lemma 4.2 gives the following: 4.3 Lemma Let F c that the set T ={a<

no
Aa be a family of almost disjoint functions, such

JIAol :::; Na} is stationary Then IFI:::; N_.;

1 •

Proof. In the proof of Lemma 4.2, let S 0 be the stationary set S 0 = { n E T I o is a limit ordinal}. The rest of the proof is the same. 0

Using Lemma 4.3, we easily get the following: 4.4 Lemma Let f be a function on w1 such that f(o)
WJ.

F1 = {g E F I for some stationan; set T <;;; w 1 , g(o) < f(a) for ail o E T}

CHAPTER 11. FILTERS AND ULTRAFILTERS

214

Proof. For a fixed stationary set T, the set {g E F I g(a) < f(a) for all o: E T} has cardinality at most Nw 1 , by Lemma 4.3. Thus \FI :::; 2N 1 · Nw, = N...,,.

0 We now prove the crucial lemma for Silver's Theorem:

4.5 Lemma Let {Aa I a< wi} be a family of sets such that \Ani :::; Nn+l for every a < w 1 , and let F be a family of almost disjoint functions,

Let U be an ultrafilter on w 1 that extends the dosed unbounded Proof. filter. Thus every set S E U is stationary. Without loss of generality we may assume that A a ~ Wa+ 1 , for each n < w 1 . Let us define a relation < on the set F as follows:

f <

g

if and only if {a< w1

I f(a) < g{a)}

We claim that < is a linear ordering of F. If because

f <

E U.

9 and 9 < h then

{a I f(a) < h(a)} 2 {a I f(a) < g(a)} n {a l9(a) < h(a)}

E

f <

ft.

u.

If f,g E F and f -1- g, then {a I f(o:) = g(a)} is at most countable and therefore not in U, and so one of the sets {a I f(o:) < 9(a)}, {a l9(o:) < f(o:)} belong!:' to U. Thus either f < 9 or 9
Ft = {g E F I for some stationary T. 9(o) < f(a) for all o E T}.

and by Lemma 4.4, 1Ft I :S Nw,· Thus for every f E F. l{g E FIg< f}J:::; N~, As < is a linear ordering of the set F, it follows from Theorem 1.14 in Chapter 8 that IFI :S Nw, +I· 0 Silver's Theorem (for >. = wt) is now an easy consequence of Lemma 4.5:

Proof of Silver's Theorem.

For each a < w1, let An = P(wn); as 2N, = Ww,, let fx E no
Na+l• we have IAal = No+l· For every set X~

function defined by

fx(a) =X nwa. If X -1- Y then fx -1- fy, and moreover, fx and fy are almost disjoint. This is because there is some a < w 1 such that X n we -1- Y n w 13 for all (3 ~ o:. Thus F = {! x I X E P(ww,)} is a family of almost disjoint functions, aud by Lemma 0 4.5, JFJ:::; Nw,+l· Therefore, 2Nw, = Nw,+l·

4. SILVER'S THEOREM

215

Exercises

4.1 Prove Silver's Theorem for arbitrary A of uncountable cofinality. [Throughout Section 4, replace w 1 by "' = cf A, and replace the sequence (t-l 0 I a < w1 ) by a continuous increasing sequence of cardinals (Aa I a < l'i.). Use Exercise 3.9.]

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Chapter 12

Combinatorial Set Theory 1.

Ramsey's Theorems

The classic example of the type of question we want to consider in this section is the well-known puzzle: Show that in any group of 6 people there are 3 who either all know each other or are strangers to each other. We arP implicitly assuming that the relation "x knows y" is symmetric. The argument goes as follows. Consider one of these people. say .Toe. Of the remaining 5 people, either there are at least 3 who know .Joe. or there ar!' at least 3 who do not know Joe. Let us assume that Peter. Paul, and Mary know Joe. If two of them, say Peter and Paul, know each other. then we have 3 people who know each other (Joe, Peter, and Paul). Otherwise, Peter, Paul. and l\hry are 3 people who are strangers to each other. If Peter, Paul, and Mary are 3 people who do not know Joe, the argument is similar. We now restate this problem in a more abstract form that allows for ready generalizations. LetS be a set. For r EN, r -1-0, [sr ={X c;;; S \lXI = r} is the collection of all r-element subsets of S. (See Exercise 3.5 in Chapter 4.) Let. {A,};,;r: be a partition of [sr into s classes (s E N- {0} ); i.e., [Sf = u;,:~ A, and A; nA 1 = 0 fori -1- j. We say that a setH~ Sis homogeneous for the partition if [Hr ~ A; for some i; i.e., if all r-element subsets of H belong to the same class A; of the partition. It may help to think of the elements of the different da..~ses as being colored by distinct colors. We thus have s colors (numbered 0. 1, .... s - 1) and each r-element subset of Sis colored by one of them. Homogeneous set are monochromatic; i.e., all r-element subsets of them are colored by the same color. In this terminology, the introductory example shows that, for a set S with \S\ 2: 6, every partition of [S] 2 into two classes has a homogeneous set H with \H\ 2: 3. (The classes are A0 = { { x, y} E [S] 2 I x and y know each other} and A 1 = {{x,y} E [SJ2\ x andy are strangers to each other}.) Put yet another way: pick a set S of at least 6 points and connect Pach

217

CHAPTER 12. COMBINATORIAL SET THEORY

218

pair of points by a line segment colored either red or blue (but not both). The resulting graph then contains a monochromatic triangle (i.e., all red or all blue). Let K, A be cardinal numbers. We write K -+ (A)~ as a shorthand for tht' statement: for every setS with lSI = 1\. and every partition of [St into s classes. there exists a homogeneous set H with IHI ~ .\. The negation of this stateme11t is denoted K ft (A)~. We proved that 6-+ (3)~; it is an easy exercise to show that 5.;.. (3)~. (See Exercise 1.1.) More generally, K -+ (.\ 1, ... , A,): means: for every set S with lSI = ~-.: and every partition { Ai} :,;;;t of [ into s classes there is some i and a set H ~ S with [Hr <;;; A, and IH,I ~A,. Exercises 1.2 and 1.3 exhibit some very simpk properties of these symbols. The fundamental result about partitions on finite sets is the Finite Ramspy's Theorem. It asserts that if S is sufficiently large, any coloring of its r-elPment subsets will have monochromatic sets of the prescribed size k.

sr

1.1 The Finite Ramsey's Theorem For any positive natural num!Jer·s k. r, s there exists a natural number n such that n-+ (k):. The readers interested primarily in infinite sets can skip the proof without harm.

Proof. We first consider s = 2 and prove that for all r, p, q E N - {0} there exists n E N for which n-+ (p, q)2 by induction on r. Let r = 1. Then it suffices to take n = p + q - 1: it is clear that if IS'I = 11 and [Sjl = 8 = Ao U Ar, we cannot have both IAol 'S p -1 and IArl 'S q- 1. If IAol ~ plet H = Ao; otherwise, let H = A 1 . We now assume that the statement is true for r· (and any p, q) and prove it for r + 1. We denote the least n for which n -+ (p, q)~ by R(p, q; r). The proof now proceeds by induction on (p + q). Consider a set 8 with IS'I = n > 0 (a..'> yet unspecified) and a partition [sr+ 1 = Ao u Ar where Ao n A, = 0. The statement is true if p ::; r or rJ ::; r: if p < r (q < r, respectively) we can take as H any subset of S with IHI = fi (IHI = q, respectively); if p = q = r then either A 0 f=. 0 and any H E A 0 \\"ill do, or A 1 f=. 0 and any H E A 1 will do. So we assume that p > r. q > r and tlH· statement is true for p', q' if p' + q' < p + q. Fix a E S, let sa= 8- {a} and define a partition {Bo,B 1 } of [S"]' hy X E Bo ( B 1, respectively) if and only if {a} U X E Ao (A 1 , respectively) (for

x E [san.

If we choose n so large that n - 1 = R(p', q'; r) (p', q' as yet unspecified) then there will be a set Ha <;;; sa so that either (i) [Hat<;;; Bo and (ii)

[Ha]•·

!Hal~

p'. or

<;;; B1 and !Hal ~ q'.

1. RAMSEY'S THEOREMS

219

Either way, all (r +I)-element sets of the form {a} U X, X E [Hat are thus guaranteed to be colored by the same color, so it remains to concern ourselves with (r +I)-element subsets of Ha. Assume (i) occurs. By the inductive assumption, there exists n for which n -+ (p- 1, q)2+ 1 holds; let p' = R(p- 1, q; r + 1) be the least such n. It then follows that either there isH'~ Ha with JH'J 2" p- 1 and [H't+ 1 ~ A 0 ; in this case we let H = H' u {a} and notice that JHJ ;::- p and [Hr+t ~ A 0 . Or, there isH"~ Ha with JH"J 2" q and [H"]r+l ~At; in this case we let H = H" and notice that JHJ ;::- q and [Hr+l ~ A 1 • The case (ii) is handled similarly, letting q' = R(p, q- I; r + 1), the least n for which n -+ (p, q - 1)2+ 1 holds. In conclusion, if we let n = R(p',q'; r) for p' = R(p- I,q; r + 1) and q' = R(p,q -1;r+ 1) then n-+ (p,q)2+ 1. This concludes the proof for s = 2. In particular, taking p = q = k we have n -+ (k)2 for all k, r. The proof of the Finite Ramsey's Theorem can now be completed by induction on s. The case s = 1 is trivial and s = 2 is proved above. Assume that m has the property m -+ (k)~. Let us consider a partition {A,}i'=o of [sr into (s + 1) classes, where S is a set of an as yet unspecified cardinality n. Then {E0 , Et} defined by s-1

Eo=

UA,,

E 1 = A.

t=O

is a partition of [St into 2 classes. If n = R(l, l; r) (for as yet unspecified I) then there is H' ~ S, JH'J 2" l, such that either [H']r ~ Eo or [H'jT ~ E 1 . We take l = max{ m, k}. In the first case, {A,} :,;;-ci partitions [H't into s classes, so our choice of l guarantees existence of H ~ H' such that JHJ 2" k and [Ht ~ A, for some i E { 0, ... , s - 1}. In the second case, our choice of l allows us to let H = H' and conclude JHJ ;::- k, [HjT ~A •. Either way, we are done. D We remark that the exact determination of the numbers R(p, q; r) is difficult and only a few are known despite much effort and extensive use of computers. Here is a complete list, as of now (1998): R(3, l; 2) for l = 3, 4, 5, 6, 7, 8, 9 has values 6,9,I4,18,23,28,36, respectively; R(4,4;2) = 18, R(4,5;2) = 25. R(3, 3, 3; 2) = 17, and R(4, 4; 3) = I3. They tend to grow rapidly with increasing (p + q) and r. This is a subject of great interest in the area of mathematics known as (finite) combinatorics, but we now turn to analogous questions for infinite cardinal numbers. There the most important result is the (Infinite) Ramsey's Theorem. 1.2 Ramsey's Theorem

N0

-+ (No)~

holds for all1·, sEN- {0}.

In words, if r-elernent subsets of an infinite set are colored by a finite number s of colors, then there is an infinite subset, all of whose r-element subsets are colored by the same color.

220

CHAPTER 12. COMBINATORIAL SET THEORY

Proof. It suffices to consider S = N (see Exercise 1.2). We proceed by induction on r. r = 1: If [N] 1 = N = U:~~ A, then at least one of the sets A, must b(' infinite (Theorem 2.7 in Chapter 4) and we let H be such an A,. r = 2: This is a warm-up for the general induction step. Let [N]2 = u;~(: .4,. We construct sequences (an);,"=O• (in);,"=O• and (Hn)~=O by recursion. W1• set ao = 0 and let B? = {bE N I b f ao and {ao,b} E A,}. Then {B~};~(( is a partition of N- {ao}; we takf' io to be the first i for which B? is infinit(' (refer to the case r = 1) and let Hu = B~,· All the subsequent terms of thP sequence (an) will be selected from Ho, thus guaranteeing that { ao, an} E A,, for all n > 0. We select a 1 to be the first element of Ho and let B,1 = { b c H 0 I b f a 1 and {ao,b} E A,}. Again. {B,l}:~r( is a partition of the infinite sd H 0 - {ad and we let i1 be the first i for which B,1 is infinite. and H 1 = B,\. Proceeding in this manner one obtains the desired sequences. It is obvious from the construction that (a,.);,"=o is an increasing sequencf' and {an, am} E A,,. till· all m > n. The sequence (i,.);,"=o has values from the finite set {0 ..... s- 1 }. hence there exists j and au infinite set M ~uch that i, = j f
We conclude this section with two simple applications of Ramsey's Theorem.

1.3 Corollary Every infinite ordered set (P, :S) contains an infinitP. subsP.t S such that either any two distinct elements of S are comparable (i.e .. 8 L~ a chain) or any two distinct elements of S are incomparable.

2. PARTITION CALCULUS FOR UNCOUNTABLE CARDINALS Proof.

221

Apply Ramsey's Theorem to the partition { Ao, At} of jPj2 where

A 0 = { {x,y} E [P] 2 I x andy arc comparable}, A1 = {{x,y} E [P] 2 I x andy are incomparable}.

1.4 Corollary EvenJ infinite linearly ordered set contains a subset similar to either (N, <) or (N, > ). Proof. Let (P, ~) be an infinite linearly ordered set and let well-ordering of P. We partition [P] 2 as follows:

~

be some

Ao = {{x,y} E [P] 2 1 x < y and x-< y};

A 1 = {{ x, y} E [P] 2 I x < y and x ~ y}. Let H be an infinite homogeneous set provided by Ramsey's Theorem. The relation < well-orders H; let fl be the initial segment of H of order type w. If [Hj2 <:;; Ao then (H, -<)is similar to (N, <);if [H] 2 <:;; A 1 then (H, ~)is similar to (N, <). D Exercises

1.1 Show that 5 f. (3)~. 1.2 Prove that the words "every set S" in the definition of K -+ (.\)~· can bP replaced by "some set S." 1.3 Assume that K -+ (A)~ holds. Prove (a) If n_' 2': "-then K 1 ---> (A);. (b) If A1 ~ A then n_ -+ (.\'):. (c) If s' ~ s then"----> (A):.. (d) If r' ~ r then K---> (A)~'. Prove analogous statements forK---> (A 1 , ... , As)~. Also show that. K-+ (.\1, ... , As)~ holds if and only if K--> (A,( I)> ... , A,(s))~ holds, where rr is any one-to-one mapping of {1, ... , s} onto itself. 1.4 Show that for an infinite cardinal K, K -+ (K)l holds if and only if A < cf(K). 1.5 Let [Sj
2.

Partition Calculus for Uncountable Cardinals

Ramsey's Theorem asserts that any partition of r-element subsets of an infinite set S into a finite number of classes has an infimte homogeneous set. The next natural question is, how large does S have to be in order that every partition has an uncountable homogeneous set. By analogy with Ramsey's Theorem one might expect~~ -+ (~J)~. However, as the next theorem shows, this is false.

222

CHAPTER 12. COMBINATORIAL SET THEORl'

2.1 Theorem 2N" f+ (N!)~. The argument is quite similar to the one used to deduce Corollary Proof. 1.4. Let A = 2N" be the cardincdity of the continuum. The sPt R of all rca! numbers has IRI =A and is linearly ordered by::;. Let~ be some well-ordering of R of order type A. We partition [R]2 as follows:

Ao = {{x,y} E [R] 2 1 x A 1 = { { x, y} E [ R]

2

Ix

-<.y

and x < y},

-< y and x > y}.

Let H be an uncountable l10mogeneous set for this partition. That means that either (i) for all x,y E H, x-< y implies x < y, or (ii) for all x, y E H, x -< y implies x > y. We show that this is impossible. Let us assume (i) holds. Let

H be an isomorphism of (1-L, E) and (H, -<), where 1-L is an ordinal (necessarily, 1-L S A and 1-L uncountable). WP note that ~ < 17 < 1-L implies <;(0 -< p(TJ) and hence <;(0 <
For a slightly different proof see Exercise 2.1. There is a weaker posit.iv<' result.

2.2 Erdos-Dushnik-Miller Partition Theorem

N1

--->

(N 1 , N 0 )~.

In words, for every partition of pairs of elements of an uncountable set into two classes, if one class has no infinite homogeneous set, then the other one has to have an uncountable homogeneous set. Let {A, B} be a partition of [w1]2. For a E w 1 let B(a) = {/3 E Prvof. =/;nand {a, (3} E B}. There are two possibilities. 1. For every uncountable X ~ w 1 there exists some a E X for which IB(u) 11 X I = N 1 . In this case we construct a countable homogeneous set for B by recursion. We let Xo = w 1
w1

I (3

2. PARTITION CALCULUS FOR UNCOUNTABLE CARDINALS

223

the set U,<>.B(a.,) n X of those {3 E X for which some {a,,O} E B (for some 11 < A) is at most countable (it is a countable union of at most countable sets). Therefore, X- U,<>.(B(a,) U {a,}) is uncountable, and we let aA be its first element. Clearly aA -1- a., and {a,, a A} E A. for all v < A. The S<'t H ={a, I 11 < wi} satisfies IHI = Nt and [Hj2 <;;;A. lJ Actually, the theorem holds for any infinite cardinal"' (in place of NJ). i.e., N 0 )~. The proof for regular"' is a straightforward modification of the one just given for"-= N1 , and is left as an exercise. To guarantee that every partition of [5] 2 has an uncountable homogeneous set, the underlying set S must be more than just uncountable. "' ---> ("-,

2.3 Erdos-Rad6 Partition Theorem

(2N" )+ ---.

(N 1 )~".

Assuming the validity of the Continuum Hypotht•sis, this reduces to N2 (NI)~o·

-+

PTOof. We denote (2N")+ by A and let {A,}nEN be a partition of [A) 2 For any pair {a,{J} E [Aj2 we let n(a,{J) be the unique n for which {a, O} E An (n(a,O) is the "color" of{a,{J}). For every a < A we construct a transfinite sequence fa recursively as follows: fo.(O) = 0; If Ua(TJ) IT/<~) is defined (~t B 11 = {a EX I {a,7} E A, for some (or all) 7 EX, a< r}. The collection {B,} is a partition of X and lXI 2: Nt implies that for some n E N, IBn! 2: N1 . Clearly [Bnf <;;; A,, soH= Bn is the desired homogeneous set. D

It is not too hard to generalize this result. First, we define the transfinite sequence of beths, the cardinal numbers obtained by iterated exponentiation.

CHAPTER 12. C'Ofl'lBINATORIAL SET THEORY

2:24

2.4 Definition .:Jo = No; .:Jn+ I = 2:J"; .:JA = sup{.:J" I a < A} for limit A f. 0. A more general version of the Erdos-Rad6 Partition Theorem can now bP stated.

2.5 Theorem (.:Jn)+ --+(Nil~,; I holds for all n EN. The proof is by induction on n, following closely the special case n = 1 in Theorem 2.3. There are analogous results for any infinite caroinal ,. , see Exercise 2.5. Many results of similar nature, both positive and IwgativP. have been obtained by researchers in the part of set theory known as infinita.r.v combinatorics. One problem in particular has played a very important role· in the development of modern set theory. It is the question of whether thert· are any uncountable cardinal numbers 1'\. for which an analogue of Ramsey's Theorem holds.

2.6 Definition An uncountable cardinalK is callecl weakly compact if" --+ holds for all r. s E N- {0}.

(t>:).~

2. 7 Theorem Weakly compact cardinals are strongly inaccessible. Pmof. \Ve have to prove that a weakly compact cardinal a strong limit cardinal.

(i) Regularity: Assume 1'\. = Uv
l'i

and

JP.,.J

h.

is


n~gular

alld

for all v < A.

{a, ,6} E Au if and only if th!.'re is v < A such that cr E Pv and /3 E P,: {n, ,6}

E

A 1 otherwise.

Clearly {Au, A 1 } cannot havP a homogeneous set of cardinality

h..

(ii) Strong limit property: Assume A< 1'\.::::; 2A. By Exercise 2.1. 2A-/-+ (A+):j. hence"-/-+ (A+)~ and, as A+ -:; t>:,"-/-+ (K)~. 0 Exercises

2.1 Prove that 2"-/-+ (1'\.+)~ for any infinite l'i. [Hint: Follow the proof of Theorem 2.1, but replace (R. -:;) by tlw st·t {0, 1}" ordered lexicographically (see Chapter 4, Section 4).] 2.2 Provp that l'i--+ (h. No)~ holds for all infinite regular K..

225

3. TREES

2.3 Prove: If X is an infinite set and ::; and ~ are two well-orderings of X, then there is Y ~ X with JYI = lXI such that Yr ::; Y2 if and only if Yl ~ Y2 holds for all y 1 , Y2 E Y. [Hint: Let A= {{x,y} E [XJ21 x < yandx-< y} and B = {{:r.y} E [X) 2 I :r < y and :r :>- y}. Apply the Erdiis-Dushnik-Miller TbPon'lll and show that B cannot have an infinite homogeneous set.] 2.4 Prow that (.Jn)+---> (N 1 )~,; 1 . [Hint: Induction.] 2.5 For any cardinall'i define exp 0 (1'i) = "-• expn+l("-) = 2exp.,("). Prove: For any infinite l'i, (expn(,..))+--> (,..+)~+ 1 . In particular, (2")+---> (K+)j.

3.

Trees

Like partitions, trees originated in finite combinatorics. but turned out to be of great interest to set theorists as well.

3.1 Definition A tree is an ordered set (T. :.::; ) which has a least elenwnt and is such that, for every x E T, the set {YET I y < x} is well-ordered by::;. SPe Figure 1.

Elements ofT are called nodes. If x, y E T and y < x, we say that y is a predecessor of x and x is a successor of y. The unique least element ofT is tlw root. By Theorem 3.1 in Chapter 6, the well-ordered set {y E T I y < .1:} of all predecessors of :r is isomorphic to a unique ordinal number h(x). the height of .T. The set 7'o ={:rET I h(.r) = n} is the £tth lnwl ofT. If h(.r) is a slln\'ssor ordinal, x is called a successor node, otherwise it is a limit nude. The least o for which Tn = 0 is called the height of the tree T. h(T). A branch in T is a maximal chain (i.e., a linearly ordered subset) in T. ThP order type of a branch b is called its length and is denoted £(b). It is always au ordinal number less than or equal to the height ofT. A branch whose !Pngth equals the height of the tree is called cofinal. A subset T' ofT is a subtree ofT if for all x E T', y E T, y < x implies y E T'. Then T' is also a tree (when ordered by :.::;) and T~ = Tn n T' for all o < h(T'). The set r T13 is a subtree ofT, for any o::.::; h(T). and h(T(<>l) = o. If x E Tn then {y E T y < x} is a branch of y
Finally, a set A ~ T is an antichain in T if any two distinct elements of A are incomparable, i.e., x, y E A, x 'I= y implies that neither x < y nor y < .r. The reader is strongly urged to work out Exercises 3.1 and 3.2, where sol!le simple properties of these concepts are developed.

226

CHAPTER 12.

COMBINATORIAL SET THEORY

v

To Figure 1

It is time for some examples of trees.

3.2 Example (a) Every well-ordered set (W, :S) is a tree. Hence trees can be viewed as generalizations of well-orderings. h( W) is the order type of ll": the only branch is W itself, and it is cofinal. (b) Let A be an ordinal number and A a nonempty set. Define A<>.= Uo<>. A" to be the set of all transfinite sequences of elements of A of length less than A. We letT= A<>. and order it by~; so f:::; g for J,g E T means f \;;; y. i.e., f = g f domf. It is easy to verify that T is a tree. For f E T. h(f) =a: if and only iff E Au: i.e., T 0 =A"'. For a:= f3 + 1 and f E A". f f {3 is the immediate predecessor of f, and all f U { ({3. a)}, u E A. an· immediate successors off. Branches in T are in one-to-one correspondence with functions from A into A: if F E A>. then { F f a: I n < A} is a branch in T. Conversely, if B is a branch in T then B is a compatible system of functions and F = U B E A>.. We note that all branches are co final. (c) More generally, if T ~ A<>. is a subtree of (A<\~). branches in Tare in one-to-one correspondence with those functions F E A<>- U A>. for which F f a: E T for all a: E dom F and either F ¢ T, or F E T and F has no successors in T. We often identify branches and their corresponding functions in this situation. (d) Let A = N, A = w. Let T ~ N f(J) holds for all i < .J < dom f E N. Then T is a subtree of (N.J which has no cofinal branches (see Exercise 2.8, Chapter 3). (e) Let (R, :S) be a linearly ordered set. A representation of a tree (T. ~) /Jy intervals in (R. :S) is a one-to-one function 1> which assigns to t>ach :r E 1·

227

3. TREES

an interval (x) in (R, S) so that, for x, yET, (i) x ~ y if and only if (x) 2 (y);

(ii) x and y are incomparable if and only if (x) n (y) = 0. This implies, in particular, that ([T], 2) is a tree isomorphic to (T. ~). For example, the system (Ds I s E S) constructed in Example 3.18, Chapter 10, is a representation of the treeS= Seq({O, 1}) = {0, l}
{a E T I Cn :$a}= {Cn} U U{a E T I b ~a} bES

where S is the finite set of all immediate successors of Cn (Exercise 3.1(v)). Hence, for at least one bE S, {a E T I b ~a} is infinite, and we let Cn+l be one such b. It is easy to verify that {a E T I a:$ Cn for some n E N} is a branch in T of length w. 0 There is an important point to be made about this recursive construction, and many similar ones to come. The Recursion Theorem, as formulated in Section 3 of Chapter 3, requires a function g, whose purpose is to "compute'' Cn+! from Cn· In the previous proof, we did not explicitly specify such a g. and indeed some form of the Axiom of Choice is needed to do so. For example, let k be a choice function for P(T). Let Sc denote the set of all immediate successors of c in T. If we let g( c, n) = k( {b E Sc I {a E T I b ~ a} is infinite}), then Cn+l = g(Cn, n) defines (Cn I n E N) in conformity with the Recursion Theorem.

228

CHAPTER 12. COMBINATORIAL SET THEORY

It is customary to omit this kind of detailed justification. and we do so from now on. The reader is encouraged to supply the details of the next few applications of the Recursion Theorem or the Transfinite Recursion Theorem as an exercise. Returning to Konig's Lemma: there are several interesting ways to generalizt> it. For example, it is fairly easy to prove that any tree of height " whose levPls are finite has a branch of length l'i (Exercise 3.3). We consider t.lw following question: let T be a tree of height w 1 , each level of which is at most countable: does it have to have a branch of length w 1 ? It turns out that the amwer is .. no:·

3.4 Definition A tree of height w 1 is called an Ar·onszajn tree if all its levels are at most countable and if it has no branch of length w 1 . 3.5 Theorem Aronszajn trees of height w 1 exist.

Proof. We construct the levels T 0 transfinite recursion in such a way that

,

a

<

w 1,

of an Aronszajn trPP bv

(i) T" <;; w"; l1'c.l S No;

f

is one-to-one and (w -ran f) is infinite;

(ii) Iff

E

Tu then

(iii) Iff

E

Tn and f3 < a then f f

fj E

To;

(iv) For any {3 < a, any g E TfJ, and any finite X <;; w- ran g. there is such that f 2 g and ran f n X = 0.

f

E

·1;,

Let us assume this has been done and let us show that T = Uo ITa I $ ~~~ (bv

229

3. TREES

inductive assumption (i)) and the number of finite subsets of w is countablt>. the set T 0 is at most countable, and (i) is satisfied as well. 0 More generally, a tree of height ,. _ (li an uncountable cardinal) is called an A ronszajn tree if all its levels have cardinality less than ,. _ and there are no branches of length l'i. The question of existence of such trees is very complicated and not yet fully resolved. It is easy to show that they always exist when ~'- is singular (Exercise 3.4). Of particular interest are uncountable cardinals for which an analogue of Konig's Lemma holds, i.e., no Aronszajn trees of height"' exist; such cardinals are said to have the tree property. It turns out that strongly inaccessible cardinals with the tree property are precisely the weakly compact cardinals defined in Section 2. Exercises

3.1 Let (T, :::; ) be a tree. Prove (i) The root is the unique r E T for which h(r) = 0. In particular. To I 0. (ii) If a I !3 then Ton TfJ = 0. (1.1..1) T -- T(h(T)) -- Uo
= sup{a +II T I 0} = sup{h(x) 0

+ 11 x E T}.

3.2 Prove: (i) Every chain in T is well-ordered. (ii) If b is a branch in T, x E b, and y < x, then y E b. (iii) If b is a branch in T, then Ibn To I= 1 for a< €(b) and jbn Tnl = 0 for a> €(b). Conclude that e(b):::; h(T). (iv) Show that h(T) = sup{e(b) Ibis a branch in T}. (v) Show that each T 0 (a< h(T)) is an antichain in T. 3.3 Let (T,:::;) be a tree of height ,. _ (li an infinite cardinal), all levels of which are finite. Prove that T has a branch of length l'i. [Hint: Let Ua = {x E Ta I I{Y E T I x :::; y}l = li}. Prove that {IUnll a< !i} <;;; N is bounded; let m be its maximum (m ~ 1). Show that T has exactly m branches of length K.] 3.4 Construct an Aronszajn tree of height Nw. Generalize to an arbitrary singular cardinal ""·

230

4.

CHAPTER 12. COMBINATORIAL SET THEORY

Sus lin's Problem

WP- proved in Chapter 4, Theorem 5.7, that the real numbers, with their usual ordering, are the unique (up to isomorphism) complete linearly ordered set without endpoints that has a countable dense subset. As an immediate consequence. every collection of mutually disjoint open intervals in ( R, <) is at most countable (Theorem 3.2 in Chapter 10). Suslin asked in 1920 whether the ordering of real numbers is uniquely characterized by this weaker property.

4.1 Definition A Suslin line is a complete linearly ordered set without PHdpoints where every collection of mutually disjoint open intervals is at most countable, but where there is no countable dense subset. The famous Suslin's Hypothesis asserts that there are no Suslin lines. It is now known that Suslin's Hypothesis can be neither proved nor refuted from the axioms of Zermelo-Fraenkel set theory with Choice. In this section we establish a relationship between Suslin lines and certain kinds of trees. In the next section we consider some additional axioms of combinatorial nature that lead to an answer to Suslin's problem. Theorem 3.5 makes it dear that trees of height WJ with countable lewb an• not always "slim enough" to have a cofinal branch. However, we can impose a stronger condition: we can require that all antichains (not just the levels) lw countable.

4.2 Definition A tree of height w 1 is called a Suslin tree if all its antichains are at most countable and there are no branches of length w 1 • Clearly a Suslin tree is an Aronszajn tree, but the converse need not lw true. As the name indicates, there is a close connection between Suslin tr<~es and Suslin lines. It is this connection that we proceed to establish. Let (S, <) be a Suslin line; we recall that every complete ordering is by definition dense. We construct a tree T <;;; N bv open intervals in S using transfinite recursion of length w 1 . More precisely. \lie construct the levels T0 <;;; N° and mappings <1> 0 so that, for each a: < w 1 • (*)

{

/To/ :5 No, y13 is a representation of y
We let To = {0}, <1>(0) = S. Assume T 0 <;;; N° has been constructed and satisfies (*). For each f E T, and n E N - {0} we put fn = f U { ((x, n)} into To+l· Let ifJ 0 ( / ) =(a, b); using density of (S, <)we pick an increasing sequence a = a1 < a2 < a3 < · · · < b and let o+l(/n) = (an, an+ I) for n 2: 1. If ao = sup{an}~=l < b we also put fo = f U {(a:,O)} into T,>+l and set o+ 1 (fo) = (ao,b). It is clear that the condition(*) holds for a:+ 1. Now let a:< w 1 be a limit ordinal. We assume that T11 and <1> 0 satisfying(*) have been constructed for all /3 < c~. It is then clear that T(o) = Ua
4. SUSLIN'S PROBLEM

231

tree, IT(<>) I :-::; No, and (<>) = UIJ13 is a representation of y(cr) by intervals inS. Let f be a branch in y{cr); we can think of it as an element of Ncr. Then (a)(f f !3) = (a 0 , b0 ) is an interval in S and !3 < 1 < a: implies a13 S a-y < b-y S b13 . Put a = supiJcr(/) = (a. b). It is easy to verify that T(<>+l) = y(cr) U Tcr is a tree and (cr+l) = (<>) U n is its representation (for example, if f, g E Tu are incompatible, we consider the first !3 where f (!3) "I g(/3); we then have f f (!3 + 1) f g f (!3 + 1), so (al(J f (!3 + 1)) n (al(g f (!3 + 1)) = 0 by inductive assumption. and so (f) n (g) = 0). The set Tcr is at most countable, because otherwise {{a) (f) I f E Tcr} would be an uncountable collection of mutually disjoint intervals in S. Thus (*) is satisfied at stage a:. This completes the recursive construction. We let T = Ucr = Ucrcr· Clearly Tis a tree and represents it by intervals in (S. <). We claim that T is a Suslin tree. It is clear that T has no uncountable antichains: the image of any antichain in T by the one-to-one function is a collection of mutually disjoint intervals in S and so it is at most countable. We next show that T has no branches of length w 1 • 4.3 Claim Let (T, S) be a tree where each node has at least two immediate successors. lfT has no uncountable antichains then T has no branches of length ~WJ.

Proof. Let (x 01 I a: < WJ) be a chain in T, where Xcr E T 0 • The set of immediate successors of Xcr has at least 2 elements, so let Ya+I be an immediate successor of Xo: different from x<>+ 1 . Then {Ya+ 1 I a: < w!} is an antichain in T. 0 It remains to show that T has height w1 , i.e., that each T"' f 0. We proceed by contradiction. Let o < w 1 be the first ordinal for which T-a = 0; it is dear from the construction that o has to be a limit ordinal. Let C be the set of all endpoints of all intervals (g), g E y(al; then Cis at most countable, and cannot be dense inS. In other words, there is an interval (c, d) disjoint with C. We use it to construct a branch in y(a) as follows: we let F(O) = 0 and note (c, d) <:;; S = (F(O)). Given F(o:) E T 01 such that (c, d) <:;; (F(o:)) = (a, b) and the sequence a= a 1 < a2 < a3 < · · · < a 0 :-::;bused to construct ,>+J• we note that C n (c, d) = 0 implies that there is a unique n E N such that (c, d)<:;; (an, an+ I) (or (c, d) <:;; (ao, b), in case n = 0); we let F(o: + 1) = F(o:) U { (o:, n)} for that n. At o: limit, ni3<<> (F(!3)) 2 (c,d) by the inductive assumption, so g = UIJ<<> F({3) ETa and (g) 2 (c,d), and we let F(a:) =g. The resulting branch f = U<>(f fa:) 2 (c, d), so f E T-a according to the limit step of the recursive construction ofT-a. Hence T-a f 0, a contradiction. The Suslin tree just constructed has two additional properties, both obvious from the proof:

232

CHAPTER 12. COMBINATORIAL SET THEORY

(i) The set of all immediate successors of each node is countable. (ii) IfX,)J E 7',. for limit a:< h(T) and {z E T then x = y.

Iz <

x} = {z E T

Iz <

y}.

We call trees with properties (i) and (ii) regular. The arguments just completed prove the following theorem.

4.4 Theorem If a Suslin line exists then there exists a regular· Suslin tree. The theorem has a converse.

4.5 Theorem If a regular Suslin tr,ee exists then there is a Suslin line. The existence of Suslin lines is thus equivalent to the existence of regular Suslin trees. It is in fact equivalent to the existence of Suslin trees, but. we omit the unappealing proof of this fact. Proof. Let (T, ~) be a regular Suslin tree; without loss of generality \\'(' can assume that T ~ A"" 1 for some set A, and ~ is :;;;> (Exercise 4.3). For \'acli f E T the set s, of immediate successors off is countable, and wf• fix a df•nse linear ordering -
b < b'

if and only if

b r (a+ 1) -
r (Cl' + 1)

where n is the least ordinal such that b(a) f. b'((t). (\Ve note that I! C 1/ or 1/ C I! is impossible: branches arP maximal chains.) Arguments similar to t hosP used in the proof of Theorem 4. 7 in Chapter 4 show that < is a linear oniPriug. If b < b' and a is as ubove, there exist g E Sbto such that /J f (n + 1) -. two of its imruediat<' successor:s, say ft - / 1 , 1! 2 ~ f2. \\'t> tlwu haw b1 < b2 , but the interval (b 1 ,b2 ) is dearly disjoint with C. In :summary, (B. <) ha.s all tlw properties of a Sus lin line except compil'tPHess. WP let (B, <) lw the Dedekind completion of (B, <) as constructed iu

5. COMBINATORIAL PRINCIPLES

233

Section 5 of Chapter 4. It is an easy exercise (Exercise 4.4) to verify that (B.<) is a Suslin line. D Exercises

4.1 Let (S, < 1) be a Suslin line, ( {a}, <2) a one-element set with its unique ordering, and (R, < 3 ) the real line. Show that the sum of the ordered sets (S, be as in the proof of Theon~m 4.4: by letting 0 = '"''I in the argument showing that T has height w1 we get a set C of \CI = N1 which is dense in S. Show that each x E S is a limit of a sequence of dements of C; then apply Exercise 3.1 in Chapter 9.j 4.3 Show that a tree has property (ii) from the definition of regularity if and only if it is isomorphic to a tree of transfinite sequences (a.'-> in Example 3.2(c)). 4.4 Let (B, <) be a dense linear ordering without endpoints that ha.-; all the properties of a Suslin line, except completeness. Show that its DPdekind completion (B, <) is a Suslin line. [Hint: If (B, <) had a countable dense subset, then it would be isomorphic to the real line. But every infinite subset of R has a countable dense subset; so B would have a countable dense subset.] 4.5 A regular Suslin tree (T, ~)is normal if for all 0: < {3 < w 1 and all.r E 1:, there exists y E TfJ such that x < y. A Suslin line ( L, ~) is proper if no open interval in L has a countable dense subset. Show that a Suslin tn•<> is normal if and only if the Suslin line constructed from it as in the proof of Theorem 4.5 is proper. 4.6 Prove: If a Suslin line exists, then a proper Suslin line exists. [Hint: Define an equivalence relation on the Suslin line(£,~) by .r ~ y if and only if the interval with endpoints x, y contains a countable dense subset. Show that each [x] contains a countable dense subset. Let L = {[xjl x E L }; let [x] ~ [y] if and only if x ~ y and show that (L, ~) is a proper Suslin line.]

5.

Combinatorial Principles

Suslin's Hypothesis can be neither proved nor refuted from the axioms of ZFC. This was shown in the sixties by constructing models where the axioms of ZFC are satisfied and Suslin's Hypothesis fails, as well as such models where it is true. The study of models of set theory requires familiarity with formal logic and is beyond the scope of this book, except for a brief introduction in Chapter 15. However, early in the development of the subject researchers isolated a number of general principles of combinatorial nature which hold in one or another model of ZFC and have, as a consequence, a definite answer to Suslin's Problem, as well as to a number of other questions. If one is willing to accept such a principle as an additional axiom of set theory, one can then, by purely classical methods,

234

CHAPTER 12. COMBINATORIAL SET THEORY

without deep study of logic, prove from it results unavailable in ZFC alone. This has become the way to proceed in general topology and some parts of abstract algebra. We demonstrate this approach here by two examples. Jensen's PrinciplP Diamond and Martin's Axiom. We want to stress that these axioms do not haw the same epistemological status as the axioms of ZFC. At the present state of set theory there are no compelling reasons to believe that one or another of them is intuitively true. It is merely known that they do not lead to contradictiollC'. assuming ZFC does not. Their significance lies in systematizing the morass of consistency results obtained by thP technique of models. The first combinatorial principle we consider is Jpnsen 's Principle 0.

Principle 0 There exists a sequence (Wa I a < w 1 ) such that. for each a < w 1 , W 0 <;;; P(a), Wa is at most countable, and for any X <;;; w 1 the spt { 0' < Wi I X n 0' E Wa} is stationary. Principle 0 is known to hold in the constructible model (see Chapter 15. Section 2). Here we prove two of its consequences; others can be found in tlw exercises.

5.1 Theorem If 0 holds, then 2N" == N1. Proof. If X <;;; w <;;; W} then Sx == {0' < W} I X n 0' E w Q} is stationary. For 0' E Sx, 0' 2: w this implies X ==X n 0' E Wa. So P(w) <:::: Ua<..;J Wn. but. the cardinality of the last set is at most Ea
5.2 Theorem If 0 holds, then Suslin lines exist. The key idea of the proof is as follows. As usual, we construct T = Un<..;, 1;, by transfinite recursion. We want to make sure that T has no uncountablP antichains. As ITI = N1, there are 2N 1 > N1 subsets ofT. hence more than N1 sets X <;;; T which our construction has to prevent from becoming antichains. The crux of the matter is being able to take care of more than N1 sets in N1 steps. It is here that 0 is used. Roughly speaking, we proceed in such a wav that those A E Wa that are maximal antichains in the subtree T(<>) constructPd before stage a can no longer grow -- they remain maximal antichains for the rest of the construction ofT. 0 implies that when X is a maximal antichain in T, then X n r) so in particular X is at most countable. We now proceed with the details. One minor problem arises because 0 applies to subsets of w 1 , while we plan to construct T as a subset of w<"''. The following lemma takes care of this.

5.3 Lemma 0 implies that there exists a sequence (Za I a < w1) such that, for each a < Wl, Zo. <;;; w<<>, Zu is at most countable, and for any X <;;; w<"' 1 the set {o < w1 I X n w<'-' E Zo} is stationary.

5. COMBINATORIAL PRINCIPLES

Proof.

We note that lw
= I Uo
235 2N"

= N, because

= N1 by Theorem 5.1. Let F be a one-to-one mapping of w
0

onto w 1 . We

]

It is clear that SF is closed. Given f3 < w1 we define recursively: a 0 = {3, an+!= supF[w<""], and set a= sup{an In< w}. It is clear that f3::; a< WJ and a E SF. We now set Zo = {F- 1[A] n w
We construct a particularly nice Suslin tree.

5.4 Definition (see Exercise 4.5) A regular tree (T, ::;) is normal if for all a < f3 < h(T) and all x E T0 , there exists y E Tt3 such that x < y. We construct T0 by transfinite recursion, so that !Tal ::; No and T(u+ll Uf3l). Letting b = UnEw bn we have dom b = sup{ an I n E w} = a, and the other properties of

b required by the Claim are clearly satisfied.

0

Returning to the proof of Theorem 5.2, for each f E T(o) we choose one branch b1 as in Claim 5.5, and we let To = {bJ I f E T("l}. It is clear that T(a+l) is a normal tree and IT(o+l)l ::; No. The key observation is: each Cn remains a maximal antichain in T(<>+Il. This is so because each b E To. is comparable with some g E Cn (in fact, g ~b). This completes the recursive construction. We let T = Ua
236

CHAPTER 12. COMBINATORIAL SET THEORY

5.6 Claim Sx = {a unbounded.

< Wt I X n J'(<>) is

a maximal antichain in T(<>J} i8 closf'fl

Pmof Let /3 < Wt be arbitrary. We construct a sequence (o 11 I n E -..J_) I>\· recursion: n:o = {3. Given n,. the set T(o.,) is at most countable, and for PVerv f E y(n .. l there is some 9! E X comparable with f (otherwisP. X would not a maximal antichain). We let nn+ 1 = sup( { (dom 9!) + 1 I f E 1' 1"'" l} '-J { n 11 f J. If n = sup{ On I n E w} we have 13 :S o < Wt and each f E y!n) is comparahk with some element of X n J'(nl, so X n T!"l is a maximal ant.ichaiu ill 1' 1" 1• This shows S'x is unbounded. It is easy to see that Sx is closed. :_:

IJ;.

We now complete the proof of Theorem 5.2. Lemma 5.3 provides o E Sx. n limit, such that X nw<" E Zn· So X nT(n) is a maximal antichain in 1' 1"l attd belongs to Za. By our construction and the key observation, X n Ti<>J remains a maximal antichain in r+ll and hf'nce f 2 f r n 2 y for some g E X n T(<>l, i.e., f is comparable with some g E X n T(nl. It follow~ that X= X nT(nl, and in particular, lXI<=; IT(")I :S l"{Q. " The second combinatorial principle we study in this section has nonexistPlHT of Suslin lines as one of its many consequences. Before stating it we need to introduce some terminology. 5. 7 Definition Let (P, <=;)be an ordered set. We say that a set C c;;: Pis cofinal in P if for every p E P there is q E C such that p <=; q. A set D c;;: P is dirn·lf'd if for all d 1 ,d'2 ED there is dE D such that d1 <=; d, d2 :S d. A set A<;;; Pis a lower set if a E A, p E P, p <=; a implies p E A. Let C be a collection of cofinal subsets of P. A set G <;;; P is called C -gener-ic if G is a dirPcted lowPr set a11d G n C ¥- 0 for each C E C.

5.8 Example Let (T. :S) be a treP. D c;;: T is directed if and on!~· if D i.~ <~ chain. Assume T is normal. Then 1'(o) = Un::;/1 T6 = {y E T I there t>xist.s .r ·• T, such that x <=; y} is co final in T. for each a < h(T). A set G <;;; T is C-geuNic for C = {T(o) I n < h(T)} if and only if G is a branch of length h(T) iu 'f. The following easy theorem is a fundamental fact about generic sets.

5.9 Theorem Let C be a collection of cofinal subsets of P with for every pEP there exists a C-generic set G such that pEG.

/CI

:S

l"{o·

Thn1.

Proof. Let C = {Cn I n E N}. We construct a sequence (Pn I n E N) recursively. We let Po = p; given Pn. we let Pn+l = q for some q E Cn such that fin :Sq. Finally we let G = {r E P lr <=; Pn for some n E N}. This makes G a lower set and G is clearly directed and C-generic. iJ As au application we prove a special ease (for R) of the Baire Category ThPorem.

237

5. COMBINATORIAL PRINCIPLES

5.10 Corollary The intersection of any at most countable collection of open dense sets in R is dense.

Proof. Let 0 be an at most countable collection of open dense sf?ts in R. Let (a, b) be an open intervaL We consider thf? set P of all closed intervals [a, ;3] t;;; (a, b), a, {3 E R, a < {3, and order P by reverse inclusion ~- If 0 C: R is open dense, we let Co = {[a, {3] E P I [o. {3] <:::: 0}; it is easy to verify that Co is cofinal in P. We let C ={Co 1 0 E 0}; by Theorem 5.9, there is a C-generic set G. In particular, G is directed by 2- It follows that G is a collection of nonempty closed and bounded intervals with the finite intersection property. By Theorem 3.4 in Chapter 10, G 1- 0. Clearly, any X E G belongs to

n

(a,b)nno.

n

o

The interesting question is whether C-generic sets exist for uncountable C. The next example shows they do not, in generaL

5.11 Example LetT = wfw, ordered by <:::: as usual. So T is the tref? of all finite sequences of countable ordinals. For each n < i.Ut we let Cn = {f E T I n E rani}. Cn is cofinal in T: iff E T, domf = n. tlwn g = f U { (n. It)} E C,. and J <:::: g. Let G be C-generic for C = { Ca I a < w 1 }. G is a collection of finite sequences and, being directed, any two finite sequences in G are compatible. Therefore F = U G is a function from a subset. of i.U into Wt. As I ran Fl :S l'{o. there exists 'Y E w 1 - ran F, but this contradicts C-r n G 1- 0. As in our study of trees, we have to restrict ourselvPs to sufficiently ··slim" ordered sets to have a hope of positive results.

5.12 Definition Let (P. :S) be an ordered set. We say that p.q E Par<' wmpatible if there is r E P such that p ::; r, q :'::: r; otherwise they are incompat.iblr. An antichain in P is a subset A <:::: P such that p. q E A, p 1- q implies p and q are incompatible. We note that when (P, S) is a tree, p, q E P are compatible if and only if they are comparable, and so our definition of antichaius agrees with the previous one for trees. An ordered set (P, :':::) satisfies the countable antichain condition if PVPry antichain in P is at most countable. Let "-be an infinite cardinal. We can now state Martin's Anom for"Martin's Axiom MA,.. If (P, :':::) is an ordered set satisfying t.hl' countnult' antichain condition, then, for every collection C of cofinal subsets of P \\'ith ICI :'::: K, there exists a C-generic set. WP note that MANo is true by Theorem 5.9, so MAN, is the first interesting case. We prove two of its consequences here, and consider some further exampks in the exercises.

238

CHAPTER 12. COMBINATORIAL SET THEORY

5.13 Theorem MA"' implies that the intersection of any collP.ction 0 of opPn dense sets in R of 101 ::; "- is dense. Proof. Exactly the same as the proof of Corollary 5.10, with the reference to Theorem 5.9 replaced by a reference to MA"'. P satisfies the countable antichain condition on account of Theorem 3.2 in Chapter 10. 0

5.14 Corollary MA"' implies 2N" > "-·

0

Proof. For each x E R, 0. = R - {x} is open dense. If 2N" -:; " the11 {Ox I x E R} is a collection of open dense sets in R with 101 :::; K, but 0 is not dense in R. D

=

no=

In particular, MAN 1 implies 2N" > N1 , the negation of the Continuum Hypothesis.

5.15 Theorem MAN 1 implies that there are no Suslin lines. LetT be a regular Suslin tree of height w 1 . We let T = {t E T I { r E Proof. T I t :::; r} is uncountable}. It is clear that tis a subtree ofT and t satisfies the condition from Definition 5.4 (although need not be regular, even if T is!); in particular, ITI = N,. For each a< w 1 , T(a) ={yET l.r:::; y for some x E i~.} is cofinal in f. Let G be C-generic for C = {T(a) I a < wi}. Then G is a hence through T, contradicting the assumption that cofinal branch through T is Suslin. C

t

t.

Martin's Axiom (MA) is the statement that MA"' holds for all infinite " < 2N". If 2N" = N1 , then"-< 2N" implies"-= No and MA holds. However. it is consistent to have MA and 2N" > N 1 . Thus MA can be regarded as a generalization of the Continuum Hypothesis. From Theorem 5.13 we see immediately that, if MA holds, then the intersection of any collection 0 of open dense subsets of R of [01 < 2No is dense. MA also implies that the union of less than 2N" sets of Lebesgue measure 0 has measure 0, the Lebesgue measure is K-additive for all "- < 2N" (not just countably additive), and many other results about R. topological spaces, and sets in general. Exercises

5.1 Show that 0 is equivalent to the statement: There exists a sequenn' (W~ I a < w 1 ) such that, for each a < w 1 , w:, ~ a". w:, is at most countable, and for any J : WJ -> W! the set {a < WJ I J r (} E w~} is stationary. 5.2 Assuming 0 show that there is a Suslin tree (T, ::;) such that the only automorphism of (T, 'S) is the identity mapping (a rigid Suslin trPe). [Hint: Imitate the proof of Theorem 5.2: at limit stages a make surP that no automorphism h of T(n) such that hE Zu can be extended to an automorphism ofT(<>+ I).]

5. COMBINATORIAL PRINCIPLES

239

5.3 Assuming 0 show that there are 2N 1 mutually non-isomorphic normal Suslin trees. 5.4 Assuming 0 show that there exist 2N' stationary subsets of w 1 such that the intersection of any two of them is at most countable. [Hint: Consider Sx ={a< WJ I X n a E Wa} for all X~ WJ.] 5.5 Show that MA"' is equivalent to the statement: If (P, ::;) is an ordered set satisfying the countable antichain condition. then for every collection C of maximal antichains with ICI ::; "- there exists a directed set G such that, for every C E C, there exist c E C and a E G so that c.::; a. 5.6 Show that there exist 2N" subsets of w such that the intersection of any two of them is finite. [Hint: For J E {O,l}w let At= {f r nInE w} ~ w
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Chapter 13

Large Cardinals 1.

The Measure Problem

Theorem 2.14 in Chapter 8 shows that there exists no cr-additive translation invariant measure J.L on the cr -algebra of all subsets of R such that J.L( [a. b]) = b- a for every interval [a, b] in R. This raises the question whether there exists any cr-additive measure on P(R), or for that matter Oil P(S) for any infiHitP set S. Of course, the counting measure, which allows th~· value J.L(S) = x. is a trivial example, so we formulate the problem differently. We allow a measure to have only finite values, and we further require (without loss of generality) that J.L(S) = 1.

1.1 Definition LetS be a nonempty set. A (nontrivial probabilistic cr -additive) measure on Sis a function J.l: P(S)--> [0, 1] such that (a) J.L(0) = 0, J.L(S) = 1. (b) If X<;;; Y, then J.L(X):::; J.L(Y). (c) If X andY are disjoint, then fJ-(X U Y) = J.L(X) + J.L(Y). (d) J.L({a}) = 0 for every a E S. (e) If {Xn} ~=O is a collection of mutually disjoint subsets of S, then fJ-(

00

00

n=O

n=O

U Xn) = L

J.L(Xn).

A consequence of (d) and (e) is that every at most countable subset of S has measure 0. Hence if there is a measure on S, then S is uncountable. It is clear that whether there exists a measure on S depends only on the cardinality of S: If S carries a measure and IS'I = lSI then S'' also carries a measurP. In Section 2 of Chapter 11 we constructed a nontrivial finitely additive mea.<>ure on N, a function that satisfies (a)-(d) in Definition 1.1. The measur<' problem is the question whether there exists such a function on some S that is a-additiw. The measure problem is related to the question whether the LebesguP measur<' can be extended to all sets of reals. Precisely: Does there exist a a-addit.iw 241

CHAPTER 13. LARGE CARDINALS

242

measure flo : P( R) --> [0, oo) U { oo} such that for every Lebesgue measurable set X, flo( X) is the Lebesgue measure of X. If there is such an extension J.L of tht• Lebesgue measure then the restriction of J.l to P([O, 1]) is a nontrivial measure on S = [0, 1] (satisfying (a)-(e) in Definition 1.1). Conversely, it can be proved that if a nontrivial measure exists on a set S of cardinality 2N" then the Lebesgue measure can be extended to a cr-additive measure J.L: P(R)--> [O,oo) U {oo}. The measure problem, a natural question arising in abstract real analysis. is deeply related to the continuum problem, and surprisingly also to the subject we touched upon in Chapter 9 - inaccessible cardinals. This problem has beconw the starting point for investigation of Large cardinals, a theory that we Pxplon"' further in the next section.

1.2 Theorem If there is a measure on 2N", then the Continuum Hypothesz.~ fails. Proof Let us assume that 2N" = NI and that there is a measure flo on thP setS= WI, a function on P(S) satisfying (a)-( e) of Definition 1.1. Let I denot.P the ideal of sets of measure 0: I= {X

c;;

s I J.L(X) = 0}.

This ideal has the following properties:

(1.3)

(I.4)

For every xES, {x} E I.

If Xn E I for all n EN, then

U Xn E /. n=O

(1.5)

There is no uncountable mutually disjoint collectionS

c;; P(S)

such that X ¢ I for all X E S.

Property (1.4) is an immediate consequence of Definition l.l(e). As for {l.G), suppose that S is such a disjoint collection. For each n, let I

Sn = {X E S I J.L(X) 2: - }. n Because J.L(S) = I, each Sn can only be finite, and because S = U::"=o Sn. Sis at most countable. We now construct a "matrix" of subsets of S (Aan I a E WI, nEw). For each ~ < w 1 , there exists a function h on w such that ~ c;; ran h. Let us choose one h for each ~, and let Aan={~
(a
The matrix (Aan) has the following properties: (1.6)

For every n, if a # {3, then Aun n A11n = 0. 00

{1.7)

For every a, S-

U Aan is at most countable. n=O

243

1. THE MEASURE PROBLEM

We have (1.6) because~ E AannApn would mean that f~(n) =a and h(n) = (3. The set in ( 1. 7) is at most countable because it is included in the set a u {a}: If~¢ Ann for all n, then a¢ ranf~ and so~ :S n. Let a < w1 be fixed. From (1.7) and (1.4) WP conclude that not all the sets Aarn, nEw, are in the ideal I: Wi is the union of the sets Aun and an at most countable set, which belongs to I by (1.3) and (1.4). Thus for each a < w1 there exists some no E N such that A.w., ¢ I. Because there are uncountably many a < w 1 and only countably many n E N, there must exist some n such that the set {a I no = n} is uncountable. Let S = {A on I no = n}.

This is an uncountable collection of subsets of S. By (1.6), the sets in S arf? mutually disjoint, and Aan ¢ I for each Aan E S. This contradicts (1.5). We have a contradiction and therefore the assumption that 2N" = l'{ 1 must be false. This proves the theorem. 0 The proof of Theorem 1.2 can be slightly modified to obtain the following result.

1.8 Theorem If there is a measure on a setS, then some cardinal

K ::;

lSI

is

weakly inaccessible.

1.9 Corollary If there is a measure on 2No, then 2No ~ ~>- for some weakly inaccessible cardinal ~>.. To prove Theorem 1.8 we follow closely the proof of Theorem 1.2. We assume that for some set S there is a function J1: P(S) -+ [0, 1] that satisfies Definition 1.1. We have shown that then there exists an ideal I on S that satisfies (1.3). (1.4), and (1.5). Let (1.10)

"' = the least cardinal such that for some S of size "' there exists an

(1.11)

ideal I on S with properties (1.3), (1.4), and (1.5) and I is an ideal on S

= K with properties (1.3),

(Of course, if such an ideal exists on some S of size

(1.4), and (1.5). K,

then there is one on

K.)

1.12 Lemma For every A< then

UtJ
~>-,

if {X.,}.,
XtJ E I.

Proof. Otherwise, there exists some A < "'and { XTJ }tJ
J

= {Y ~

A1

U xt) E I}. tJEY

244

CHAPTER 13. LARGE CARDINALS

We verify that J is a ideal on ,\ (e.g., ,\ ¢ J because U110 Xr1 ¢ I). For each E ,\, {7J} E J because Xr1 E I: thus J satisfies (1.3). Similarly. onP verifi<>s that J satisfies (1.4) and (1.5) as well. [For (1.5) we use tlw assumption that the X 11 are mutually disjoint.] Thus J is an ideal on,\< K with properties (1.3), (1.4), and (1.5), contrary to the assumption (1.10). L: TJ

1.13 Corollary If XC 1.14 Corollary

K.

K

and [XI <

then X E I.

t;.,

is an uncountalile regular cardinal.

K is uncountable because every at most countable subset of li Proof. belongs to I. And K is regular, because otherwise K would be the uaion of l('ss than t;. sets of size < t;., and therefon• would belong to the id(•al. a coutradictiou LJ

We can now finish the proof of Theorem 1.8. 1.15 Theorem

K

is weakly inaccessible.

Prnof In view of Corollary 1.14 it suffices to prove that K is a limit cardinal. Let us therefore assume that K is a successor cardinal, K = ~v+ 1 . For each ( < K. we choose a function f~ on w, such that ( <:;; ran f~ and l<>t

As in the proof of Theorem 1.2, one shows that the matrix (Anry) has tlw following properties: ( 1.16) (1.17)

For every

1],

if u

I /J,

For every cr. [K.-

then Aor1 n A1311 = 0.

U Am

1[ :::;

~,.

Still following the proof of Theorem 1.2, but using Lemma 1.12 in place of (LI). we show that for each o < Wv+l there exists some 1] < w, such that An.1 E/c I And the same argument as before produces a collection S, of size ~~/+ 1 ( t hen·fon· uncountable), of mutually disjoint sets ¢ I. This contradicts the property ( l.S). and therefore K. cannot be a successor cardinal. A measure p. is two-valued if it takes only two values, 0 and 1. We return to two-valued measures in the next section, as a starting point for the theory of large cardinals. 1.18 Definition Let p. be a measure on S. A set A <:;; S is an atom if p.(A) > 0 and if for every X <:;; A, either ft(X) = 0 or p.(A - X) = 0. The measure Jl is atomless if there exist no atoms.

1. THE MEASURE PROBLEM

245

We conclude this section with the proof of the following dichotomy due to Stanislaw Ularn. 1.19 Theorem If there exists a measure then either there exists a two-1•alued measure or there exists a measure on 2 11 ". 1.20 Lemma Let p. be an atomless measure on S. (a) For every E: > 0 and every X <:;;; S with Jt(X) > 0 there is a Y that 0 < p.(Y) ::; E:. (b) For every X<:;;; S there is a Y <:;;;X such that p.(Y) = ~11(X).

<:;;;

X .mrh

Proof. (a) Let Xo = X, and for each n, we find an Xn+l C Xn such that 0 < p.(Xn+d ::; ~p.(Xn)- This is possible because Xn is not an atom and therefore there is an X C Xn such that p.(X) > 0 and p.(Xn- X) > 0. As p.(Xn) = p.(X) + {t(Xn- X), it follows that either Xn+i =X or Xn+i = Xn -X has the desired property. Clearly, 0 < p.(Xn) ::; 2~. Jt(X) for ev<'r:'> n, and part (a) follows. (b) Let X <:;;; S be such that p.(X) = m > 0. By transfinite recursion on cr < w 1 we construct a disjoint family of subsets Yo of X as follows: Lt>t Yo C X be such that 0 < p.(Yo) ::; m/2; if p.(Utku Yo) < m/2, we choos<' Yu C X - U.e
p.(u{x, If E Z})

(Z

<:;;;

{0, 1}"').

As p. is a measure, it is easily verified that v has properties (a). (b). (c) and (e) of Definition 1.1. Property (d) follows from the fact. that Jl( X f) = 0 for t•ach /E {0,1}w. 0 Thus if there exists a measure then either there exists one on 2 11 " in which case 211 " ~ K for some weakly inaccessible cardinal t>.. or else there exists a two-valued measure, an alternative that we investigate in the next section.

CHAPTER 13. LARGE CARDINALS

246

2.

Large Cardinals

In the preceding section we proved that if there exists a nontrivial a-additiw measure, then there exists a weakly inaccessible cardinal. This result is just one example of a vast body of results, known as the theory of large cardinals. In this section we give some more examples of large cardinal results and introducP the most studied prototype of large cardinals - measurable cardinals. 2.1 Lemma If p. is a two-valued measure on S, then

u = {X c;; s I p.(X) = 1} is a non principal ultrafilter on S, and

n ()Q

(2.2)

if {Xn}~=o is such that Xn E U for each n, then

X, E U.

n=O

Proof. An easy verification. U is nonprincipal because p. is nontrivial. and satisfies (2.2) because p. is a-additive. 0 Property (2.2) is called a-completeness. The converse of Lemma 2.1 is also true: if U is a a-complete nonprincipal ultrafilter on S, then the function p. : P(S)---. {0, 1} defined by

p.

( X) = { 1 0

if X E U, if X¢ U

is a two-valued measure on S. Thus the problem whether two-valued measures exist is equivalent to the problem of existence of non principal a-complete ultrafilters. We now investigatP this question. First we generalize the definition of a-completeness.

2.3 Definition Let K be an uncountable cardinal. A filter F on Sis K-completl' if for every cardinal>. < K, if X a E F for all 0: < >., then X a E F. An ideal I on S is K-complete if for every cardinal >. < K, if X" E I for all 0: < >., then Ua<..\ X a E I. A filter is K-complete if and only if its dual ideal is K-complete. An N1 r.omplete filter is also called a-complete or countably complete: it means that n~=o Xn E F whenever all Xn E F. Similarly for ideals.

nu<).

2.4 Lemma If there exists a nonprincipal (J-complete ultrafilter then there cJ·ists an uncountable cardinal K and a nonprincipal "--complete ultmfilter on "-·

Proof. Let "- be the least cardinal such that there exists a nonprincipal (Jcomplete ultrafilter on "-• and let U be such an ultrafilter. We wish to show that U is "--complete (note that because U is (J-Complete, "-must be uncountable).

2. LARGE CARDINALS

247

Let I be the ideal dual to U: I= P(K-)- U. I is a nonprincipal a-complete prime ideal on ~~:; we show that I is K-complete. If not, then there exists >. < 11: and {X11 } 11 <.>. such that X 11 E I for each T/ but U11 d X 11 ¢ J. We may assume that the X 11 are mutually disjoint. Let

J = {Y ~>.I

u Xry

E

I}.

ryEY

J is a a-complete prime ideal on >.; it is nonprincipal because X 11 E T/ E >.. The dual of J is a nonprincipal a-complete ultrafilter on >..

I for each

But >. < 11: and that contradicts the assumption that 11: is the least cardinal on which there is a nonprincipal a-complete ultrafilter. Thus U is ~~:-complete.

0 2.5 Definition A measurable cardinal is an uncountable cardinal there exists a nonprincipal ~~:-complete ultrafilter.

11:

on which

The discussion leading to Definition 2.5 shows that measurable cardinals are related to the measure problem investigated in Section 1. The existence of a measurable cardinal is equivalent to the existence of a nontrivial two-valued a-additive measure. We devote the rest of this section to measurable cardinals. 2.6 Theorem Every measurable cardinal is strongly inaccessible. Proof. We recall that a cardinal is strongly inaccessible if it is regular. uncountable, and strong limit. Let 11: be a measurable cardinal, and let U be a nonprincipal ~~:-complete ultrafilter on 11:. Let I be the prime ideal dual to the ultrafilter U. Every singleton belongs to I, and by ~~:-completeness, every X C 11: of cardinality < 11: belongs to I. If 11: were singular, the set 11: would also have to belong to I, by ~>:-completeness. Hence 11: is regular. Now let us assume that 11: is not strong limit. Therefore, there is >. < 11: such that 2.>. ;:=: 11:, so there is a set S ~ {0, 1 }.>. of cardinality 11:. On S there is also a nonprincipal ~~:-complete ultrafilter, say V. For each o < >., exactly one of the two sets

(2.7)

{!

E

S

I f(o) = 0}

and

{!

E

S

I f(o)

= 1}

belongs to V; let us call that set Xa. Thus for each o <>.we have Xo E V, and by /\:-completeness, X = na<.>. X a is also in v. But there is at most one function f in S that belongs to all the Xa: the value off at o is determined by the choice of one of the sets in (2.7). Hence lXI ::; 1, a contradiction since V is nonprincipal. Hence K is a strong limit cardinal, and therefore strongly inaccessible. 0 One can prove more than just inaccessibility of measurable cardinals. We give several examples of properties of measurable cardinals that can be obtained by elementary methods.

248

CHAPTER 13. LARGE CARDINALS

Let us recall (Section 3 in Chapter 12) that an uncountable cardinal K ha..-; the tree property if there exists no Aronszajn tree of height /{,. The following theorem shows that every measurable cardinal has the tree property.

2.8 Theorem Let K be a measurable cardinal. If T is a trF.I' of hFight " such that each node has less than K unmediate successors, then T has a brunch of length K. Proof For each o: < K, let To be the set of all s E 1' of height n. Since " is a strongly inaccessible cardinal. it follows, by induction on o:. that 11',..1 < J.; tor all o: < K. Hence 11'1 'S K, and therefore IT! = K. Let U lw a nonprincipal K.-complete ultrafilter on T. W~· find a branch of length K as follows. By induction on o we show that for each o: there exists a unique s 0 E T" such that {t E T

(2.9)

I S 0 'S t} E U

and that S 0 < SiJ when o: < (3. First, let so be the root of T. Given s,. and a...;;suming (2.9), we note that the set {t E T I Su 'S t} is th<> disjoint union of {S 0 } and the sets {t E 1' I u 'S t} where u ranges over all imm<>diatP successors of Su. Since U is a !{,-complete ultrafilter and s" has fewer than " immediate successors, there is a unique immediate successor u = so+ 1 such that {t E T I So+l 'S t} E U. \\'hen 11 is a limit ordinal less than li and {Sn }o
s=

n{

t E T

I Su

-::::

t} E

u.

(\:<1}

because u is K-complete. The set sf] = { s E Tt] I s -:::: t for some t E s} is nonempty and of size less than /{,; it follows that there is a unique s E 8,1 for which {t E T I s 'S t} E U. and we let st7 be this s. It is clear that /1 = {s., I ct < li} is a branch in 1'. of length K. C::. In Section 2 of Chapter 12 we introduced weakly compact cardinals. \Ve statP without proof the following equival!'nce that gives another charact<•rizat.ion of weakly compact cardinals.

2.11 Theorem The following ar·p equivalent. for pvenJ uncountable caniinal": (a) /{,-+ (KH. (b) ,.. -+ ( K) ~ for all positive integers r and s. (c) li 2s stmngly inaccessible and has the tr·ee pmperty. Since mea..~urable cardinals an· strongly inaccessible and have the tree property, it follows that every measurable cardinal is weakly compact. This ('an also bt~ proved directly, by showing K -+ (/{,)~. We conclude this introduction to large· cardinals by presenting the proof of the special case for r = s = 2.

2. LARGE CARDINALS

249

2.12 Theorem If li is a measurable cardinal, tlwn every partition of [h] 2 into two sr.ts has a homogeneous sr.t of cardinality h.

Proof. Let { P 1 , P2} be a partition of [~~'f To find a homogeneous set, let U be a nonprincipal /\:-complete ultrafilter on h. For each cr E li:, let

I {3 # {/3 E 1\: I {3 #

s~ = {/3 E

s; =

K

Q

and {(t. J} E pl}.

Q

and { 0. !1} E P:z}.

Exactly one of S~, S~ belongs to U. Let

Z1 ={a 1 s~ E U},

Z2 =

{rt 1 s~

E

U}.

Since Z 1 U Z 2 = li:, either Z 1 or Z 2 is in U. L<'t us assume that Z 1 E U and let us find H c;; 1\: of cardinality 1\: such that [Hj2 c;; P 1 . We construct H = { cr~ I ~ < K} by recursion. At step "f. we have constructed an increasing sequence (cr~ I~<"!)

of elements of Z 1 such that for each ( < T/ < "f,

(2.13) Because S~( E U for all ( < "f, the set

Z1 n

n

S~(

~<1'

is in U (by /\:-completeness) and therefore contains some cr greater than all

( < "'· Let cr" be the least such cr. Then (2.13) holds for all ( <

<

n~.

+

1. Let H = {a~ I~< li:}. If~< 'f/, then o,l E s,~( and so {cr~.n,l} E Pl. 17

[Hj2 c;; P 1 and therefore His homogeneous for tlw partition.

1

0

Exercises A strongly inaccessible cardinal/\: is a Mahto cardinal if the sPt of all rC'gular cardinals < h is stationary. 2.1 If 1\: is Mahlo, then the set of all strongly inaccessible cardinals < ~> is stationary. [Hint: The set of all strong limit o < li is closPd unbounded.] Let 1\: be a measurable cardinal. A nonprincipall\:-complete ultrafilter U on is normal if every regressive function f with dom f E U is constant on some set A E U. In Exercises 2.2-2.5, let U be a non principal h-complete ultrafiltPr on ~>. For f and gin li:", let 1\:

f =9

if {cr <

K

I f(cr)

= g(cr)} E U.

cardinals by presenting the proof of the special case for r

= s = 2.

CHAPTER 13. LARGE CARDINALS

250 2.2 =is an equivalence relation on

K".

Let W be the set of all equivalence classes of = on K"; let [!] denote t.h<> equivalence class of f. Let

[!] < [g] if {cr <

K

I f(cr)

< g(cr)} E U.

2.3 < is a linear ordering of W. 2.4 < is a well-ordering of W. [Hint: Otherwise, there is a sequence Un I n EN) such that [in] > [fn+d· Let Xn = {cr I fn(cr) > fn+t(n)} and let X= n::=o Xn. If cr EX, then /o(cr) > ft(cr) >···.a contradktiou.] 2.5 Let h : K -+ K be the least function (in (W, <)) with the property that for all1 < K, {cr < K I h(cr) > 1} E U. Let V ={X c;; K I h- 1 [X] E U}. Show that V is a normal ultrafilter. Thus for every measurable cardinal K, there exists a normal ultrafilter on In Exercises 2.6-2.9, U is a normal ultrafilter on K.

K..

2.6 Every set A E U is stationary. [Hint: Use Exercise 3.9 in Chapter 11 and the definition of normality.] 2.7 Let>. < K be a regular cardinal and let E>. = {cr < K I d(cr) = ..\}. The set E>, is not in U. [Hint: Assume that E>, E U. For each cr E E>,. let {xa~ I ~ < .X} be an increasing sequence with limit o. For <'adt ~ < >. there is y~ and A~ E U such that Xu( = y~ for all cr E A(. L<'t A = nG<>- A~. Then A E U, but A contains only one element, nanwl_v sup{y~ 1 ( < ,\};a contradiction.] 2.8 The set of all regular cardinals < /'\. is in U. [Hint: Otherwise, tlw set S = {cr < K I cf(cr) < a} is in U, and because cf is a regressive function on S, there exists>.< K such that {cr < K I cf(cr) =.X} E U: a contradiction. J 2.9 Every measurable cardinal is a Mahlo cardinal. [Hint: Exercises 2.6 and 2.8.]

Chapter 14

The Axiom of Foundation 1.

Well-Founded Relations

The notion of a well-ordering is one of the key concepts of set theory. It emerged in Chapter 3, when we introduced natural numbers; the fact that the natural numbers are well-ordered by size is essentially equivalent to the Induction Principle. We studied well-orderings in full generality in Chapter 6, and we saw many applications of them in subsequent chapters. It is thus of great interest to consider whether the concept allows further useful generalizations. It turns out that, for many purposes, the "ordering" stipulation is unimportant; it is the "well-" part, i.e., the requirement that every nonempty subset has a minimal element, that is crucial. This leads to the basic definition of this section.

1.1 Definition Let R be a binary relation in A, and let X <:;;: A. We say that a E X is an R-minimal element of X if there is no x E X such that xRa. R is well-founded on A if every nonempty subset of A has an R-minimal element. The set {x E A f xRa} is called the R-extension of a in A and is denotl:'d extR(a). Thus a is an R-minimal element of X if and only if extR(a) n X = f/J

1.2 Example (a) The empty relation R = 0 is well-founded on any A, empty or not. (b) Any well-ordering of A is well-founded on A. In particular, Eu is wellfounded on a, for any ordinal number a. (c) Let A= P(w); then EA is well-founded on A. (d) EA is well-founded on A if A= Vn (n EN) or A= Vw (see Exercise 3.3 in Chapter 6). (e) If (T, :S) is a tree (Chapter 12, Section 3) then < is well-founded on T.

The next two lemmas list some simple properties of well-founded relations. 251

252

CHAPTER 14. THE AXIOM OF FOUNDATION

1.3 Lemma Let R be a well-founded relation on A. (a) R is antireftexive in A. i.e., aRa is false, for all a E A. (b) R is asymmetric in A, i.e., aR/i implies that bRa does not hold. (c) ThcTe is no finite sequence (ao. a1, . . . . a,) such that a 1 Ra 0 • a~ Ra 1 . UnRan-1. aoRan. (d) There is no infinite sequenre (a, a,+ 1 Ra, holds for all i E N.

(a) (b) (c) (d)

I

i E N) of Flements of A s1U·h that

Proof If aRa then X = {a} f= f/J does not have an R-minimal element. If aRb and bRa then X == {a, I!} f= 0 does not have an R-minimal element. Otherwise, X = { a0 , . ..• a 11 } has no R-rninimal element. X= {a, I i EN} would have no R-minirnal clement if (d) failed.

n 1.4 Lemma* Let R be a binary r·elation in A such that ther·e is no infinite sequence (a, I i E N) of elements of A joT which a,+ 1 Ra, holds for all i E N. Then R is well-founded on A.

This is a converse to Lemma 1.3( o). In this chapter we do not a._<;sume the Axiom of Choice. The few results where it is needed are marked by an ast!'fisk. Under the assumption of the Axiom of Choice, a relation R is wdl-foundecl if and only if therf' is no infinite ·'R-decreasing" sequenc<' of den!f'nt.s of .4. Pmof. If R is not well-founded on A, then A has a nonempty subset X with the property that, for every a E X. tl!f're is some b E X such that bRa hold;;. Choose a 0 E X. Then choose a 1 E X such that a 1 Rao. Given (ao .... . a,). choose a,+ 1 EX so that a,+ IRa,. The resulting sequence (a, I i E N) satisfies a,+ 1 Ra, for all i EN. 0

The importance of well-founder! relations rests on the fact that, like wellorderings, they allow proofs by induction and constructions of fumtions bv recursion.

1.5 The Induction Principle

Let P be some property. Assume that R a well-founded TClation on A and, for all x E A,

(*)

1s

iJP(y) holds for ally E extR(:r), then P(:r).

Then P(x) holds for all x E A. Pmof. Otherwise, X = {:z: E A I P(x) does not hold} f= f/J. Let a llP an R-minimal dement of X. Tlwn P(a) fails, but P(y) holds for all yRa. contradicting (*).

1. WELL-FOUNDED RELATIONS

253

1.6 The Recursion Theorem Let G be an operation. Assume that R is a well-founded relation on A. Then there is a unique function f on A such that. for all x E A, f(x) == G(f r extR(.r.)). The proof of Theorem 1.6 uses ideas similar to those employed to prow ot lwr Recursion Theorems, such as the original one in Section 3 of Chapter 3, and Theorem 4.5 in Chapter 6. It is convenient to state some definitions and prow a simple lemma first.

1.7 Definition A set B <;; A is R-transitive in A if extR(x) <;; B holds for all x E B. In other words, B is R-transitive if it has the property that :r E B and yRx imply y E B. It is clear that the union and the intersection of any collection of R-t.ransiti\'<' subsets of A is R-transitive.

1.8 Lemma For every C <;; A there is a smallest R-tmnsitive B <;; A such that C<;;B. Proof. Let Bo == C, Bn+l ={YEA lyRxholdsforsomcx E B,}, B = U~=o Bn. It is clear that B is R-transitive and C <;; B. tvloreover, for any R-transitive B', if C <;; B' then each Bn <;; B', by induction, and hence Bc:;;B'. 0 Pmof of the Recursion Theorem. Let T == {g I 9 is a function. dom g is R-transitive in A, and g(x) = G(g r extR(x)) holds for all X E domg}. We first show that T is a compatible system of functions. Consider 9 1 ,92 E T and assume that X= {.r. E domg1 n domg2 I gJ(.r.) # 92(x)} # 0. Let a be an R-minimal element of X. Then yRa implies y E dom.r/1. y E dom92, and 91(Y) == 92(y). Hence 9l(a) = 91 f extR(a) = 92 f extR(a) = 92(a), a contradiction with a EX. We now let f = Clearly f is a function, domf = U{dom9 I 9 E T} is R-transitive, and, if x E domf, then x E dom9 for some 9 E T. 9 c:;; f. so f(x) = 9(x) = G(9 f extR(x)) = G(J f extR(x)). It remains to prove that dom f = A. If not, there is an R-minimal elPment. a of A- domf. Then extR(a) c:;; domf and D = domf u {a} is R-transit.ive. We define 9 by

ur.

g(x) = f(x)

9(a)

for x E dom f;

= G(f f extR(a)).

Clearly 9 E T, so 9 <;; f and a E dom j, a contradiction. Uniqueness off follows by the same argument that was used to prow that T is a compatible system. 0

254

CHAPTER 14. THE AXIOM OF FOUNDATION

In Chapter 6 we used transfinite induction and recursion to show that evPrv well-ordering is isomorphic to a unique ordinal number (ordered by E). In the rest of this section we generalize this important result to well-founded relations. We recall Definition 2.1 in Chapter 6: a set T is transitivr. if every e!Pmeut ofT is a subset ofT. 1.9 Definition A transitive set Tis well-founded if and only if the relation ET is well-founded on T, i.e. for every X ~ T, X :;f 0, there is a E X such that an X= 0.

Ordinal numbers, P(w), Vn (n E N), and Vw are examples of transitive well-founded sets (Exercise 1.2). The next theorem contains the main idea. 1.10 Theorem Let R be a well-founded relation on A. There is a unique function f on A such that

f(x) = {f(y)

I yEA and yRx}

= f[extR(x)]

holds for all x E A. The set T = ran f is transitive and well-founded. Proof The existence and uniqueness of Recursion Theorem; it suffices to take G{ z) = {ran z 0

f follow immediately from the

if z is ~ function; otherwise.

T =ran f is transitive. This is easy: if t E T then t = f(x) for some :r E A. so sEt= {f(y) I yRx} implies s = f(y) for some yEA, so sET. T is well-founded. If not, there is S ~ T, S :;f 0, with the property that for every t E S there is s E S such that s E t. Let B = f- 1 [S]; B :;f 0 because f maps onto T. If x E B then f(x) E S so there is s E S such that s E f(x) = {J(y) I yRx}. Hence s = f(y) for some yRx, andy E f- 1 [S] =B. We have shown that for every x E B there is some y E B such that yRx. a [j contradiction with well-foundedness of R. The mapping f is in general not one-to-one. For example, f(x) = 0 whenever x is an R-minimal element of A (i.e., extR(x) = 0). Similarly, f(x) = {0} whenever all elements of extR(x) :;f 0 are R-minimal, and so on. 1.11 Definition R is extensional on A if x :;f y implies extR(x) :;f extR(Y). for all x, y E A. Put differently, R is extensional on A if it has the property that. if for all z E A, zRx if and only if zRy, then x = y. 1.12 Theorem The function f from Theorem 1.10 is one-to-one if and only if R is extensional on A. If it is one-to-one, it is an isomorphism between (A. R) and (T, Er).

1.

WELL-FOUNDED RELATIONS

255

Proof Assume that R is extensional on A but f is not one-to-one. Then X = {x E A I there exists y E A, y =J x, such that f(x) = f(y)} =J 0. Let a be an R-minimal element of X and let b =J a be such that f(a) = f(b). By extensionality of R, there exists c E A such that either eRa but not eRb, or cRb but not eRa. We consider only the first case; the second case is similar. From eRa it follows that f(c) E f(a) = f(b). As f(b) = {!(z) I zRb}, there exists some dRb for which f(e) = f(d). We have e =J d because cRb fails. But this means that c E X, contradicting the choice of a as an R-minimal element of X. Assume that R is not extensional on A. Then there exist a, b E A, a =J b. for which extR(a) = extR(b). We then have f(a) = f[extR(a)] = f[extR(b)] = f(b) showing that f is not one-to-one. Finally we show that f one-to-one implies f is an isomorphism. Of course aRb implies f(a) E f(b), by definition of f. Conversely, if f(a) E f(b) then f(a) = f(x) for some xRb. As f is one-to-one, we have a= x, so aRb. D The corollary of Theorems 1.10 and 1.12 asserting that every extensional well-founded relation R on A is isomorphic to the membership relation on a uniquely determined transitive well-founded set T is known as Mostowski"s Collapsing Lemma. It shows how to define a unique representative for each class of mutually isomorphic extensional well-founded relations, and thus generalizes Theorem 3.1 in Chapter 6 (see Exercise 1.5). Exercises

1.1 Given (A, R) and X<;;;: A, say that a EX is an R-least element of X if a is an R-minimal element of X and aRb holds for all bE X, b =J a. Show: if every nonempty subset of A has an R-least element then {A. R) is a well-ordering. 1.2 Prove that EA is well-founded on A when A= P(w), A= Vn for 11 EN. and A= Vw. 1.3 Let (A, R) be well-founded. Show that there is a unique function p, defined on A and with ordinal numbers as values, such that for all x E A, p(x) = sup{p(y) + 11 yRx}. p(x) is called the rank of x in {A. R). 1.4 (a) Let A = a, R =Eo-, where a is an ordinal number. Prove that p(x) = x for all x E A. (b) Let A = Vw, R =EA. Prove that p(x) = the least n such that x E Vn+I, i.e., Vn = {x E Vw I p(x) < n}. 1.5 Let (A, R) be a well-ordering and let f : A --> T be the function from Theorem 1.10. Prove that T =a is an ordinal number and f = pis an isomorphism of (A, R) onto (a, Ea-)· 1.6 (a) Let A be a transitive well-founded set and let R =EA. Iff: A ...... T is the function from Theorem 1.10 then T =A and f(x) = :r for all X

EA.

(b) If A and Bare transitive well-founded sets and A =J B then (A. EA) and (B, Ea) are not isomorphic.

CHAPTER 14. THE AXIOM OF FOUNDATION

256

2.

Well-Founded Sets

In our discussion of Russell's paradox in Section 1 of Chapter 1 we touched upon the question whether a set can be an element of itself. We left it unanswer<>d and, in fact, an answer was never needed: all results we provPd until now hold. whether or not such sets exist. Nevertheless, the question is philosophically interesting, and important for more advanced study of set theory. We now haw the technical tools needed to examine it in some detail. We start with a simple, but basic, observation. 2.1 Lemma For any set X ther·e e:rists a smallest transitive set containing X as a subset; it is called the transitive closure of X and is denoted TC(X).

Proof. We Jet Xo =X, Xn+l = UXn = {y I y EX for some X Ex .. } and TC(X) = U{Xn I n E N}. It is dear that X <;;; TC(X) and TC(X) is transitive. It T is any transitive set such that X <;;; T, we see by induction that X n <;;; T for all n E N, and hence TC(X) <;;; T. 1:1 2.2 Lemma y E TC(X) if and only if there is a finite sequence (x 0 ,x 1 •...• :r.,) such that Xo =X, x,+l Ex, jo1· i = 0.1. ... ,n- 1, and .T 11 = y.

Proof. Assume y E TC(X) = U::":o Xn and proceed by induction. If y E Xo it suffices to take the sequence (X, y). If y E Xn+I WP have y E .r. E X, for some x. By inductive assumption, there is a sequence (xo, ..... r,.) wlH•r(• xo = X, x,+l E x, fori = 0, .... n- 1, and Xn = .T. We let .Tn+l = y and consider (xo .... . X 11 ,Xn+J). Conversely, given (x 0 , . . . . :r:,) where xo =X and x,+ 1 E .r, fori= 0.1 .. n- 1. it is easy to see, by induction, that x, E X, <;;; TC(X), for all i :S n. LJ 2.3 Definition A set X is called well-founded if TC{X) is a transitive wdlfounded set.

For transitive sets, this definition agrees with Definition 1.9 because TC(X) = X when X is transitive. The significance of well-foundedness is revealed by tlw following theorem.

2.4 Theorem (a} If X is well-founded then then'. is no sequence (X n I n E N) such that Xo =X and Xn+l E Xn fm· all n EN. (b)* If there is no sequence (Xn In E N) such that X 0 =X and Xn+l E X,. for all n E N then X is well-founded. In particular, a well-founded set X cannot be an element of itself (let Xn = X for all n) and one cannot have X E Y and Y E X for any Y (consider tlw s<>quence (X, Y. X, Y, X, Y . ... , ) ) or any other ''circular'' situation.

257

2. WELL-FOUNDED SETS

Proof (a) Assume X is well-founded and (Xn In EN) satisfies Xo =X, Xn+l E Xn for all n EN. We have X 0 =X~ TC(X) and, by induction, Xn E TC(X) for all n :=:: 1. The set {Xn I n :=:: 1} ~ TC(X) does not have an E-minirnal element, contradicting we!l-foundedness of TC(X). (b) Assume X is not well-founded, so TC(X) is not a transitive well-founded set as defined in 1.9. This means that there exists Y ~ TC(X). Y f- 0. with the property that for every y E Y there is z E Y such that z E y. Choose some y E Y. By Lemma 2.2 there is a finite sequence (Xo ..... X,) such that X 0 =X, Xt+ 1 EXt fori= 0, .... n- 1. and Xn = y. W extend it to an infinite sequence by recursion, using the Axiom of Choice. Choose Xn+! to be some z E Y such that z E y = Xn (so Xn+l E Xn n Y). Given Xn+k E Y choose Xn+k+l E Xn+k n Y similarly. The resulting ~o;equence (X n I n E N) is as required by the Theorem.

0 It is time to reconsider our intuitive understandin!!; of sets. We recall Cantor's original description: a set is a collection into a whole of definite. distinct objects of our intuition or our thought. It seems reasonable to interpret this a .'i meaning that the objects have to exist (in our mind) before they can be collected into a set. Let us accept this position (it is not the ouly possible om>, as W<' S('E' in Section 3), and recall that sets are the only objects we are interested iu. Suppose we want to form a set "for the first time." That means that there are as yet no suitable objects (sets) in our mind, and so the only collection we can form is the empty set 0. But now we have something! 0 is now a definite object in our mind, so we can collect the set {0}. At this stage there are two objects in our mind. 0 and {0}, and we can collect various sets of these. In fact. tl1('re are four: 0, {0}, { {0}}, and {0, {0} }. Next we form sets made of these four objects {there are eight of them), and so on. Below, we describe this procedure rigorously, and show that one obtains by it precisely all the well-founded sets.

2.5 Definition (The cumulative hierarchy of well-founded sets.)

Vo

=

0;

Va+l = P(Va)

Va =

U V13

for all a; for all limit a f- 0.

13<<>

v3

We note that VI = {0}, Vz = {0, {0}}, = {0, {0}, { {0} }, {0. {0}}} and \/11 for n ::=; w have been defined in Exercise 3.3 in Chapter 6, and some of their properties stated in Exercises 3.4 and 3.5 in Chapter 6 (see also Exercises 1.2 and 1.4(b) in this chapter).

258 2.6 (a) (b) (c)

CHAPTER 14. THE AXIOM OF FOUNDATION Lemma Jfx EVa andy Ex then y EVa for some (3 < If (3 < ll' then Va ~ Va. For all ll', V0 is transitive and well-founded.

ll'.

Proof (a) We proceed by transfinite induction on ll'. The statement is clear wlwn ll' = 0 or ll' -1 0 limit. If x E Va+l then x ~ Va, soy E x implies y E \l,. and we Jet (3 = ll'. (b) Again we use transfinite induction on ll'. Only the successor stage is nontrivial. We show that Va ~Vex+!· By (a), if x EVa then x ~ Ua. showing that Va is transitive. We next show that each Va is well-founded. Let Y c;;; V0 , Y -#0. Let (3 be' the least ordinal for which Y n Va -1 0; clearly (3 :S ll'. Take any x E Y n \111 : by Lemma 2.6(a), y E x implies y E V,. for some 1 < (3, hence y rj: Y. This shows that x is an E-minimal element of Y. As V0 is transitive. it is well-founded. 0

2. 7 Theorem A set X is well-founded if and only if X E Va for some ordinal (}',

Proof. 1) Assume that X is well-founded. It is easy to check that TC(X) U {X} = TC( {X}) is a transitive well-founded set. Let Y = { x E TC( {X}) I :r E Va for some ll'}; it suffices to prove that Y = TC( {X}). If not, there is an E-minimal element a of TC( {X}) - Y. For each y E a we then have y E Y: let f(y) be the least ll' such that y E v;,. The Axiom Schema of Replacement guarantees that f is a well-defined function on a. We let "! = sup f[a]. Th£•11 y E Vf(y) c;;; V,. holds for each y E a, so a c;;; V,. and a E V-y+ 1, contradietin~ a rj: Y. 2) If X EVa, we have X c;;; Va by transitivity of V0 , so also TC(X) c;;; v;,. Any Y ~ TC(X), Y -1 0, has an E-minimal element, because Va is transitive and well-founded. We conclude that TC(X) is well-founded, and, by definition. w~X.

~

Earlier we proposed an intuitive understanding of the word "collection" that assumes that the objects being collected exist "previously." There is thus a process of collecting more and more complicated sets stage by stage, described rigorously by the cumulative hierarchy V0 , and yielding all weJI-founded sets. On the other hand, a set which is not well-founded does not fit this understanding, because the existence of a sequence (Xn I n E N) where X o 3 X 1 3 X 2 3 · · implies an infinite regress in time. These considerations suggest that, from our present position, only well-founded sets are ''true" sets. The reasonableness of

2. WELL-FOUNDED SETS

259

this attitude can be further confirmed by observing that all axioms of set theory we have considered (so far) remain true if the word "set" is replaced everywhere by the words "well-founded set." We illustrate this by a few examples, and leave the rest of the axioms as an exercise.

2.8 Example (a) The Axiom of Extensionality. All elements of a well-founded set are wellfounded (X E Va implies X ~ Va)· Therefore, if well-founded sets X and Y have the same well-founded elements then X = Y. (b) The Axiom of Powerset. The powerset P{S) of a well-founded set S is well-founded. (If S E Vo then S ~ Va, so P(S) ~ P(Vo) = V<>+l and P(S) E Va+Z is well-founded.) Thus for any well-founded set S there is a well-founded set P ( = P(S)) such that for any well-founded X, X E P if and only if X~ S. (c) The Axiom Schema of Comprehension. Let P(x) be any property of x (in which the word "set" has been everywhere replaced by "well-founded set'"). If A is well-founded, {x E A I P(x)} ~A is also well-founded. These and similar arguments can be converted into a rigorous proof of tht> fact that all the axioms of set theory we have adopted so far are consistent with the assumption that only well-founded sets exist. Doing so requires a rigorous formal analysis of what is meant by a property, a subject of mathematical logic that we cannot pursue here. Nevertheless, this fact. the sharpened intuitive understanding of the way sets can be collected in stages that we described above, and an observation that all particular sets needed in mathematics art> well-founded (see Exercise 2.3), make most set theorists include the statement that all sets are well-founded among their axioms for set theory.

The Axiom of Foundation are well-founded.

(Also called the Axiom of Regularity.) All sets

It should be stressed that, whether or not one accepts the Axiom of Foundation, makes no difference as far as the development of ordinary mathematics in set theory is concerned. Natural numbers, integers, real numbers and functions on them, and even cardinal and ordinal numbers have been defined, and their properties proved in this book, without any use of the Axiom of Foundation. As far as they are concerned, it does not make any difference whether or not there exist any non-well-founded sets. However, the Axiom of Foundation is very useful in investigations of models of set theory (see Chapter 15).

Exercises 2.1 For any set X, TC({X}) is the smallest transitive set containing X as an element. 2.2 (a) Va E Vo+l- Va. (b) {3
CHAPTER 14. THE AXIOM OF FOUNDATIO:\'

260

2.3 Assume that X andY are well-founded sets. Prove that {X. Y}. (X. Y). Xu Y, X n Y, X- Y, X X Y, xY, domX. ami rauX an• wdlfounded. Prove that N, Z. Q, and Rare well-founded sets. 2.4 Show that the Axioms of Existence, Pair. Union, Infinity. and Choice. as well as the Axiom Schema of Replacement, remain true if the word "set" is replaced everywhere by the words ''well-founded set." [Hint: UsP Exercise 2.3.] 2.5 An Induction Principle. Let P be some property. Assume that. for all well-founded sets x,

ux.

(*)

if P(y) holcb for ally E

1.·.

then P(x).

Conclude that P(x) holds for all well-founded x. 2.6 A Recursion Theorem. Lt>t G be an operation. Ther<' is a uuiqll<> opl·ration F such that F(x) = G(F f x) for all well-founded :r. F(:r) = f/J fell all non-well-founded x. 2. 7 Use Exercise 2.6 to show that there is a unique operation p (rank) Sll('h that p(x) = sup{p(y) + 1 I y E x} for well-founded x. p(:r) = { {f/J}} otherwise. Prove that Va ={xI p(x) E a} holds for all ordinals o. 2.8 The Axiom of Foundation cau be used to give a rigorous definition of order types of linear orderings (Chapter 4, Section 4). If 21 = (A, <) is a linear ordering, Jet a be the least ordinal number such that then' is a lint>ar ordering 21' = (A',<') isomorphic to 21, and of rank p(21') = o. (Cf. Exercise 2.7.) We define T(21) = {21' I 21' is bomorphic to 21 and p(21') = n} This is a set becaust' T{21) c:; Vn+l· Prow that 21 is isomorphic to iJ3 = (B.-<) if and only if T(21) = T(iJ3). Isomorphzsm typf'S of arbitrary structures can be defined in the same way.

3.

Non-Well-Founded Sets

For reasons discussed in the previous section, most set theorists accept the Axiom of Foundation as part of their axiomatic system for set theory. Nevertheless. alternatives allowing non-well-founded sets are logically consistent. have a ('Prtain intuitive appeal, and lately have found some applications. We diseuss two such "anti-foundation" axioms in this section. An alternative intuitive view of sets is that the objects of which a sd is composed have to be definite and distinct when the set is "finished.'' but not necessarily beforehand. They may be formed as part of the same process that leads to the collection of the set. A priori, this does not exclude the possibility that a set could become one of its own elements, perhaps even its only element. How could such sets be obtained? It turns out that natural generalizations of results in Sections 1 and 2, in particular of Theorems 1.10 and 1.12, lead to non-well-founded sets. It is convenient to restate these results using some IH'W terminology.

3. NON- WELL-FOUNDED SETS

261

3.1 Definition A graph is a structure (A, R) where R is a binary relation in A. A pointed graph is (A, R,p) where (A, R) is a graph and p E A f- 0. A decoration of (A, R) or (A, R,p) is a function f with dom f =A such that. for all x E A, f(x) = {f(y) I yRx}.

Iff is a decoration of a pointed graph (A, R,p), the set f(p) is its value. In this terminology, Theorems 1.10 and 1.12 imply. respectively: Every well-founded graph has a unique decoration. Every well-founded extensional graph has an injective (one-to-one) decoration. Moreover, a set is well-founded if and only if it is the value of a decoration of some well-founded graph. To prove this last remark, note that, if (A, R, p) is a well-founded graph and f a decoration, f(p) E ran f = T where T is transitive and well-founded by Theorem 1.10, so f(p) ~ Tis well-founded. Conversely, given a well-founded set X, let A= TC({X}), R =EA, p =X, and f =IdA. It is easy to verify that (A, R,p) is a well-founded extensional pointed graph, f is an injective decoration of it, and f(p) =X (see Exercise 2.1). Let us now consider decorations of non-well-founded graphs. In the picture of (A, R, p), elements of A are represented by dots, aRb is depicted as an arrow from b to a, and the "point" p is circled. 3.2 Example (a) See Figure 1(a). Let A = {a} be a singleton, R = {(a, a)}, and p = a. Let S be the value of some decoration f of (A, R. p). Then S = { S}. (b) See Figure 1(b). Let A = {a, b} where a f- b, R = {(a, b), (b, a)}, and p = a. If Tis the value of some decoration of (A, R,p) then T = { {T} }. (c) For a more complicated example of a non-well-founded set, and a hint of possible applications, we consider a task that arises in the study of programming languages. One has a set D of "data" and a set P of "programs." A program rr E P accepts data as "inputs" and returns data as '·outputs": mathematically, it is just a function from D to D. This is straightforward when D n P = 0, but interesting complications arise when programs accept other programs (or even themselves) as input, a situation quite common in programming. As a simple specific example, we let P = { rr}, D = { 0. rr}, and we stipulate that the program rr accepts any input from D and returns it as output. without any modification; i.e., rr(0) = 0, rr(rr) = rr. We then have rr = {(0, 0), (rr, rr)} = { { {0} }, { {rr}} }. Such a set can be obtained as the value of any decoration of the graph depicted in Figure 1 (c).

These examples show that we obtain non-well-founded sets if we postulat<> that (at least some) non-well-founded graphs have decorations. But first t lwr<> is an interesting question to consider: what is the meaning of equality for nonwell-founded sets? The Axiom of Extensionality, a sine qua non of set theorv.

CHAPTER 14. THE AXIOM OF FOUNDATION

262

u~{S)

n={{T}}

\Jr} (a)

(b)

{{0}}

{{7r}}

{0}

{7r}

0

(c)

Figure 1 tells us that, for any sets X and Y, in order to show X = Y it suffices to establish that X and Y have the same elements. For well-founded sets this is an intuitively effective procedure, because the elements of X and Y haw beP!l formed "previously.'' But if, for example, X = {X} and Y = {Y}, trying to use Extensionality to decide whether X = Y begs the question. Some stronger principle (compatible with the Axiom of Extensionality, of course) is needed to make the determination. The axiom we consider below implies existence of many non-well-founded sets, as well as a principle for deciding their equality. It is a natural generalization of Theorem 1.10, obtained by dropping the assumption that R is well-founded. 3.3 The Axiom of Anti-Foundation

Every graph has a unique decoration.

It is known that this axiom is consistent with Zerme!o-Fraenkel set theory

3. NON- WELL-FOUNDED SETS

263

(without the Axiom of Foundation). 3.4 Definition A set X is reflexive if X= {X}. 3.5 Theorem The Axiom of Anti-Foundation implies that there exists a unique

reflexive set. Let (A, R,p) be the pointed graph from Example 3.2(a). By the Proof Axiom of Anti-Foundation it has a decoration f; X= f(p) is a reflexive set. If Y is also a reflexive set, we define g on A by g(a) = Y. It is clear that g is also a decoration. By uniqueness, f = g and hence X = Y. 0 Next we develop a general criterion for determining whether two decorations have the same value. It involves an important notion of bisimulation, which originated in the study of ways how one (infinite) process can imitate another. 3.6 Definition Let (A 1 ,RI) and (A 2 ,R2 ) be graphs. ForB~ A 1 x A:l we define B+ ~ A 1 x A 2 by: (a 1 , a2 ) E B+ if for every .r 1 E extR, (a I) there exists x 2 E extR 2 (a 2), and for every Xz E extR 2 (a 2 ) there exists Xt E extH, (at). so that (xt, xz) E B. We say that B is a bisimulation between (At, R 1 ) and (A 2 • R 2 ) if B+ ~ B. 3.7 Lemma

(i) If B

~

C

~

A, x Az then B+

~

c+.

(ii) 0 is a bisimulation. (iii) The union of any collection of bisimulations is a bisimulation. Proof (i) and (ii) are obvious. (iii) If B = U,E/ B, and Bi ~ Bt for all i E J, then B,+ ~ B+ for all i E I by (i), soB= U,Ef Bi ~ U,Ef Bt ~ B+. 0 It follows immediately that, for any given {A 1 , RI), (Az, Rz), there exists a largest bisimulation

B=

U{B ~A, x A2 IBis a bisimulation}.

The next lemma connects this concept with decorations. 3.8 Lemma Let JI, fz be decomtions of(A 1 ,R 1 ), (A2.R2), respectively. Set B = {(a 1 ,az) E A 1 x Az I fi(ai) = fz(a2)}. Then B is a bisimulation.

CHAPTER 14. THE AXIOM OF FOUNDATION

264

Proof

We prove that B = B+. We have

(a 1 , a2) E B if and only if fi(ai) = h(a2) if and only if for every Xt E extR, (ai), fi(xi) E h(a2) and vice versa if and only if for every :1:1 E ext R, (a1) there exists x2 E ext R, ( a·2 ) such that f 1 (xi) = fz(x2) and vice versa if and only if for every x 1 E ext R, (a I) there exists x 2 E extR 2 ( a 2 ) such that (xi, x 2 ) E B and vice versa if and only if (a 1 ,a 2 ) E B+. 0

3.9 Definition Pointed graphs (A 1 ,Rt.Pd and (A2,R2.P2) ;ue bisimulation equivalent if there is a bisimulation B between (A 1 , RI) and (A 2 , R 2 ) such that (pt,J!2) E B. (Equivalently, if (Pt,JJ2) E B.) The next theorem provides the promised criterion for equality. 3.10 Theorem Let ft be a decoration of (At, Rt, pi) and let h be a decoratwn of (A2, R 2 ,p2). The Axiom of Anti-Foundation implies that ft and h have thP same value if and only if (At, Rt,JJt) and (A2, R2,J!2) are bisimulation equivalent. 3.11 Corollary Assuming the Axiom of Anti-Foundation, X = Y if and only if the pointed graphs (TC({X}),E,X) and (TC({Y}),E,Y) ar·e bisimulation equivalent.

Before proving Theorem 3.10 we need a technical lemma. 3.12 Lemma (a) For· i = 1,2 let J; be a decoration of (A;,R;,p;). Let A, be the smallest R;-transitive subset of A, s1tch that p, E A,. Let R; = R, n (A, x A,) and f, = f, r A,. Then f, is a decoration of (A;, R,,p,). (b) Let B be a bisimulation between (At,Rt,Pd and (A2,R2,p2). Then B = Bn(A 1 x A2) is a bisimulation between (At, RI) and (A2, R2). lf(p 1 ,p2) E B then domB = A 1 and ranB = A2.

Proof. We recall from the proof of Lemma 1.8 that A; = U~=o C,,n wh!:'n~ C,,o = {p,}, C,,n+l = {y E A 1 I yR,x holds for some x E C,,n} = U{extR,(:~:) I x E C,,n }. In particular, extR, (x) = extR,(x) for x E A,. It follows immediately that J, is a decoration and B is a bisimulation. The last statement implies that dom B is an R 1-t.ransitive subset of A 1 , hence also an Rt-transitive subset of A 1. If (p 1 , p2) E B then PI E dom B, and we conclude that A 1 <;;; dom B. Similarly. A 2 <;;;ran B. 0

3. NON- WELL-FOUNDED SETS

265

Proof of Theorem 3.10. Assume that !I (pi) = f'2(p2)· Then B = {(a I, a2) E A1 x A2 I h(a!) = h(a2)} is a bisimulation by Lemma 3.8, and (p1,p2) E B. Hence (AI, R1, pi) and (A2, R2, p 2) are bisimulation equivalent. Conversely, let B be a bisimulation satisfying (p 1 ,p2 ) E B. From Lemma 3.12 we obtain a bisimulation B between (A-;-, R 1 ) and (A2. R2), and decorations fi and f2. We now define a graph (A, R) as follows:

A=

B=

{(a1,a2)

I a1

EAt, a2 E A2, (a1,a2) E

R = {((bi,b2),(ai,a2)) E Ax A

I (bi,at)

B};

E Rt and (b2,a2) E R:~}.

Vve define functions F 1 and F2 on A by: FI((aL, a2)) = h(aJ) F2((a1,a2)) = h(a2)·

We note that FI((a1,a2)) = fi(ai) = {fi(bi) / b1Rtad = i{FI((bi,b2)) I btR 1 a~, b:~R 2 a:~, (bJ,b:~) E B} = {FL((bl,b2)) I (bJ,b:~)R(aJ,a:~)}, so F1 is a decoration on (A, R). Similarly, F2 is a decoration on (A. R). The Axiom of AntiFoundation implies F 1 = F2, so in particular /J(pt) = fi(pi) = FI((PI,P:>.)) = F2((pt,P2)) = h(P2) = h(P2)· D Other "anti-foundation" axioms have been considered. For example, one can generalize Theorem 1.12 and obtain the following.

3.13 The Axiom of Universality

Every extensional graph has an injective

decoration.

The Axiom of Universality is also consistent, but it is incompatible with the Axiom of Anti-Foundation: as the next theorem shows, it provides many more non-well-founded sets.

3.14 Theorem The Axiom of Universality implies that there exist collections of reflexive sets of arbitrary cardinality. Proof. Let A be any set, and let R = {(a,a) I a E A} be the identity relation on A. If a :/= b then extR(a) = {a} :/= {b} = extR(b), so (A, R) is extensional. Iff is some injective decoration of (A, R) then f(a) = {!(a)} holds for each a E A, and a:/= b implies f(a) :/= f(b), by injectivity of f. Hence {!(a) I a E A} is a collection of reflexive sets of the same cardinality as A. D

Stronger results along these lines can be found in the exercises. Exercises 3.1 Construct pointed graphs whose valueS has the property (a) S = {0. S}; (b) S = (0, S); (c) S=NU{S}.

266

CHAPTER 14. THE AXIOM OF FOUNDATION

3.2 Show that f3+ = f3 holds for the largest bisimulation 3.3 For given graphs (At,RI), (A2,R2), (AJ,R3) show

B.

(i) IdA, is a bisimulation between (At, Rt) and (At. RJ). (ii) If B is a bisimulation between (At, Rt) and (A2. R 2 ) then B-t is a bisimulation between (A:~, R2) and (At, Rt ). (iii) If B is a bisimulation between (At, RI) and (A2, R 2 ), and C bPtWePll (A2, R2) and (A3, R:1), then CoB is a bisimulation betwPen (At. R 1 ) and (A3, R3). Conclude that the notion of bisimulation equivalence is reflexive, symmetric, and transitive. 3.4 Show that any two of the following pointed graphs are bisimulation equiYalent: (i) Example 3.2(a); (ii) Example 3.2(b ); (iii) ( N, {(n

+ 1, n) I n

E

N}, 0);

(iv) (N,>.O). 3.5 Let (A, R) be a graph. Defitw by transfinite recursion:

Wo = 0; Wn+t ={a E A I extR(a) <:;; Wn}; w<> =

u

w{j for limit

Q

:f 0.

6<<>

Show that there exists..\ such that W.\ = WA+t· Prove that (W.\, RnH'f) is well-founded. We call W,\ the well-founded part of (A, R). 3.6 If W is the well-founded part of (A. R) and / 1 , h are decorations ol (A, R) then ft f W = h f W. 3.7 Let (A,R,p) be an extensional pointed graph, Wits well-founded part. and p ¢ W. Assuming the Axiom of Universality, show that then· arP sets of arbitrarily large cardinality, all elements of which are values ol this graph. [Hint: Take the union of an arbitrary collection of disjoint copies of (A, R); identify their well-founded parts, show that the resulting structure is extensional, and consider its injective decoration.)

Chapter 15

The Axiomatic Set Theory 1.

The Zermelo-Fraenkel Set Theory With Choice

In the course of the previous fourteen chapters we introduced axioms which. taken together, constitute the Zermelo-Fraenkel set theory with Choice (ZFC). For the reader's convenience, we list all of them here. There exists a set which has no elements

The Axiom of Existence

The Axiom of Extensionality If every element of X is an element of Y and every element of Y is an element of X, then X = Y.

Let P(:r) be a property of :r. For The Axiom Schema of Comprehension any A, there is B such that x E B if and only if x E A and P(x) holds.

The Axiom of Pair if x = A or x = B.

For any A and B, there is C such that x E C if and only

The Axiom of Union x E A for some A E S.

For any S, there is U such that x E U if and only if

The Axiom of Power Set if X~ S.

The Axiom of Infinity

For any S, there is P such that X E P if and only

An inductive set exists.

267

268

CHAPTER 15. THE AXIOMATIC SET THEORY

The Axiom Schema of Replacement Let P(x, y) be a property such that for every x there is a unique y for which P(x, y) holds. For every A there is B such that for every x E A there is y E B for which P(x, y) holds. The Axiom of Foundation

The Axiom of Choice

All sets are well-founded.

Every system of sets has a choice function.

(The reader might notice that some of the axioms are redundant. For example, the Axiom of Existence and the Axiom of Pair can be proved from the rest.) We have shown in this book that the well-known concepts of real analysis (real numbers and arithmetic operations on them, limits of sequences, continuous functions, etc.) can be defined in set theory and their basic properti<•s proved from Zermelo-Fraenkel axioms with Choice. A similar assertiou can he made about any other branch of contemporary mathematics (except category theory). Fundamental objects of topology, algebra, or functional analysis (say. topological spaces, vector spaces, groups, rings, Banach spaces) a.re customarily defined to be sets of a specific kind. Topologic, algebraic. and analytic properties of these objects are then derived from the various properties of sets. which can be themselves in their turn obtained as consequences of the axioms of ZFC. Experience shows that all tlH:'orems whose proofs mathematicians a.cc:Ppt on intuitive grounds can be in principle proved from the axioms of ZFC. In this sense, the axiomatic set theory serves as a satisfactory unifying foundation tor mathematics. Having ascertained that ZFC completely codifies current mathematicnl pra('tice, one might wonder whether this is likely to be the case also for mathematical pwctice of the future. To put the question differently, can all true mathematical theorems (including those whose truth has not yet been demonstrated) be proved in Zermelo-Fraenkel set theory with Choice? Should the answer he "yes," we would know that all as yet open mathematical questions can be. at least in principle, decided (proved or disproved) from the nxioms of ZFC alorw However, matters turned out differently. Mathematicians have been baffled for decades by relatively easily formulated set-theoretic problems, which they were unable to either prow or disprove. A typical example of a problem of this kind is the Continuum Hypothesis (CH): Every set of real numbers is either at most countable or has the cardinality of the continuum. We showed in Chapter 9 that CH is equivalent to the stateiiH'nt 2~>~" = N1, but we proved neither 2~>~" = N1 nor 2~>~" :f. N1. Another similar problem is the Suslin's Hypothesis, and others arose in topology and measure theory (see Section 3). Persistent failures of all attempts to solve these problems led some mathematicians to suspect that they cannot be solved at all at the current level of mathematical art. The axiomatic approach to set theory makes it possible to formulate this suspicion as a rigorous, mathematically verifiable conjecture that, say, the Continuum Hypothesis cannot be decided from the

1. THE ZERMELO-FRAENKEL SET THEORY WITH CHOICE

269

axioms of ZFC (which codify the current state of mathematical art, as we haw already discussed). This conjecture has been shown correct by work of Kurt Godel and Paul Cohen. First, Godel demonstrated in 1939 that the Continuum Hypothesis cannot be disproved in ZFC (that is, one cannot prove 21-1" :f. N!). Twenty-four years later, Cohen showed that it cannot be proved either. Their techniques were later used by other researchers to show that the Suslin's Hypothesis and many other problems are also undecidable in ZFC. We try to outline some of Godel's and Cohen's ideas in Section 2, but readers more deeply interested in this matter should consult some of the more advanced texts of set theory. The foregoing results put the classical open problems of set theory in a new perspective. Undecidability of the Continuum Hypothesis on the basis of our present understanding of sets as reflected by the axioms of ZFC means that some fundamental property of sets is still unknown. The task is now to find this property and formulate it as a new axiom which, when added to ZFC, would decide CH one way or another. At one level, such axioms are very easy to come by. We could, for example. add to ZFC the axiom

Unfortunately, one could instead add the axiom

and get a different, incompatible set theory. Or how about the axiom ·'21-1" = Nw+ 17"? Cohen's work shows that this, also, produces a consistent set theory, incompatible with the previous two. Furthermore, it is possible to create two subvarieties of each of these three set theories by adding to them either the axiom asserting that Suslin's Hypothesis holds or the axiom asserting that it fails. In the opinion of some researchers, that is where the matter now stands. Similarly as in geometry, where there are, besides the classical euclidean geometry. also various noneuclidean ones (elliptic, hyperbolic, etc.), we have the cantor ian spt theory with 21-1" = N1 as an axiom, and, besides it. various nonc:antoriau set theories, with no logical reasons for preferring one to another. Such an attitude is hardly completely satisfactory. To solve the continuum problem by arbitrarily adding 2Nn = Nw+l7 as an axiom is clearly cheating. It is certainly not the way we have proceeded in the previous chapters. Whenever we accepted an axiom: (a) It was intuitively obvious that sets, as we understand them, have the property postulated by the axiom (some doubts arose in the case of the Axiom of Choice, but those were discussed at length.) (b) The axiom had important consequences both in set theory and in other mathematical disciplines; some of these consequences were in fact equivalent to it. So far, there appears to be very little evidence that the axiom 21-1" = N..,+ 17 (or any other axiom of the form 2N" = N0 ) satisfies either (a) or (b).

CHAPTER 15. THE AXIOMATIC SET THEORY

270

In fact, no axioms that would be intuitively obvious in the same way as the axioms of ZFC have been proposed. Perhaps our intuition in these matters has reached its limits. However. in recent years it has been discovered that a number of open questions, in particular in the field called descriptive s<-'t theory, has an intimate connection with various large cardinals, such as thos<> we mentioned in Chapters 9 and 13. The typical pattern is that such questions can be answered one way assuming an appropriate large cardinal exists. and have an opposite answer assuming it does not, with the answer obtained with the help of large cardinals being preferable in the sense of being much more "natural," "profound," or "beautiful.'" The great amount of research into theiiP matters performed over the last 40 years has produced th1• very rich, oftPn very subtle and difficult theory of large cardinals, whose esthetic app!.'al makes it difficult not to believe that it describes true aspects of the univns!' of sd theory. 'vVe discuss some of these results very briefly in Section 3.

2.

Consistency and Independence

In order to understand methods for showing consistency and independPnu• oft he Continuum Hypothesis with n~sp!>c:t to Zermelo-Fraenkel axioms for set thPorY. let us first investigate a similar, but much simpler, problem. In Chapter 2. WP defined (strictly) ordered sets as pairs (A,<) where A is ;t Sf't a11d < is an asymmetric and transitive binary rdation in A. Equivalently. one might sa~· that an ordered set is a structun• (A. <), which satisfiP.s tlw following axioms. The Axiom of Asymmetry

There are no a and b such that a < b and b <

The Axiom of Transitivity

For all a, b, and c, if a < b and b < r:. then a < c.

11.

\Ve can also say that the Axiom of Asymmetry and the Axiom of TransitivitY comprise an axiomatic theory of order and that ordered sets are models of this axiomatic theory. Finally, let us formulate yet another axiom. The Axiom of Linearity

For all a and b, either a < b or a

=

I> or I> < a.

Let us now ask whether the Axiom of Linearity can be proved or disprowrl in our axiomatic theory of order. First, let us assume that the Axiom of Linearity can be proved in the thPorv of order. Then every model of ti!P theory of order would have to satisfy also the Axiom of Linearity, a logical consequencP of that theory. In a morl' familial terminology, every ordering would have to be a linear ordering. But this is bdsP: an example of a model for the theory of order in which the Axiom of LinearitY does not hold is in Figure 1(a). :\"ext. let us assume that tlw Axiom of Linearity can be disproved in thl' theory of order. Then every model of the theory of order would haw to sat isfv also the negation of the Axiom of Linearity. In other words. I'VPrv ord('ring

2. CONSISTENCY AND INDEPENDENCE

271

{1}

{0, 1}

0

E 0....._----{0}

{0} (a)

(b)

Figure 1: (a) (P({O,l}),C); (b) ({O.{O}},E).

would have to be nonlinear. But this is again false: see Figure 1(b) for an example of a model for the theory of order in which the Axiom of Linearity holds. We conclude that the Axiom of Linearity is undecidable in the theory of order. Let us now return to the question of undecidability of the Continuum Hypothesis in ZFC. By analogy with the preceding example, we see that. in ordPr to show that the Continuum Hypothesis cannot be proved in ZFC, one has to construct a model of ZFC in which the Continuum Hypothesis fails. and. similarly, in order to show that the Continuum Hypothesis cannot be disproved iu ZFC, one has to construct a model of ZFC in which the Continuum Hypot.lwsis holds. In the rest of this section, we outline constructions of just such models. But a technical point has to be clarified first. To describe a model for the theory of order, i.e., an ordered set., we have to specify the members of the model (by choosing the set A) and the meaning of the relation ''less than" (by choosing a binary relation< in A). Similarly, to describe a model for set theory, we have to specify the members of the model and the meaning of the relation "belongs to" in the model. But, in view of the paradoxes of set theory. it is too optimistic to expect that the members of a model for set theory can themselves be gath!:'rt>d into a set. (Actually, such models are not entirely impossible; their existence cannot be proved in ZFC, but can be proved if one extends ZFC by som<' large cardinal axiom - see Section 3). To circumvent this difficulty, models for set theory have to be described by a pair of properties, M(x) and E(x, y). Here M(x) reads "xis a set of the model" and E( x, y) reads "x belongs to y in the sense of the model." A trivial example of a model for set theory is obtained by choosing M(x) to be the property "x is a set" and E(x, y) to be the property "x E y." The model simply consists of all sets with the usual membership relation. The first nontrivial model for set theory was constructed by Kurt Godel in 1939 and became known as the constructible model. Godel wanted to find

CHAPTER 15. THE AXIOMATIC SET THEORY

272

a model in which the Continuum Hypothesis would hold, that is, in which 2N" = NJ. Cantor's Theorem asserts that 2N" ~ N1; in other words N1 is th!:' smallest cardinal to which the cardinality of the continuum can be equal. ThesP considerations suggest a search for a model which would contain as f!:'w sets as possible. Godel constructed such a model by transfinite recursion, in stages. At each stage, sets are put into the model only if their existence is guaranteed b:v one of the axioms of ZFC. To begin with, the Axiom of Infinity and the Axiom Schema of Comprehension guarantee existence of the set of natural numbers, and so we let

Lo =w. Next, if P(n) is any property with parameters from w, then the Axiom Schema of Comprehension postulates existence of the set { n E w I P (n) holds in (w, E)}, and we have to put it into the model. Therefore, we let

L1

={X~

Lo I X= {n E Lo IP(n) holds in (Lo,E)} for some property P with parameters from Lo}.

In order to show that L 1 exists, it is necessary first to give formal definitions of logical concepts, such as "property" and "holds in," in set theory. Such definitions can be found in most textbooks of mathematical logic, but they ar!' beyond the scope of this brief outline. Existence of L 1 itself then follows from the Axiom of Power Set and the Axiom Schema of Comprehension, so L 1 itsPif has to be in the model. Notice that Lo ~ L 1 : We can obtain X = k for an:v k E Lo by taking "n E k" as the property P(n). Next we repeat the argument of the previous paragraph with L 1 in place of L 0 . Since L 1 is in the model, all of its subsets which are definable in ( L 1 • E) h~· some property P must be put into the model, as well as the set L'2 of all such subsets. In this fashion one defines L 3 , L 4 , .... At stage w, we merely take the union of the previously constructed Ln:

(its existence follows from the Axiom Schema of Replacement and the Axiom of Union), and then continue as before. The recursive definition of the operation L thus goes as follows:

Lo =w; La+! ={X~ La I X is definable in (Ln, E) by some property P with parameters from La}; L 01 =

U Lf3 if o is a limit ordinal, o > 0. /3<01

2. CONSISTENCY AND INDEPENDENCE

273

It is easy to see that La ~ Lf3 whenever a < {3. A set is called constructible if it belongs to La for some ordinal a. We are now ready to describe the constructible model. Sets of the model are going to be precisely the constructible sets [that is, M(x) is the property "x is constructible"]. The membership relation in the model is the usual one: If x and y are sets of the model, x belongs to y in the model if and only if x E y [that is, E(x, y) is the property "x E y"]. Of course, it has to be verified that the constructible model is indeed a model for ZFC, that is, satisfies all of its axioms. As an illustration, we show that the Axiom of Pair holds in the constructible model: For any A and B in the model, there is C in the model such that, for all x in the model, x belongs to C in the model if and only if x =A or x =B. Taking into account the definition of the constructible model, this amounts to showing: For any constructible sets A and B, there is a constructible set. C such that, for all constructible x, x E C if and only if x =A or x = B. Let constructible sets A and B be given; we first prove that {A, B} is also a constructible set. Indeed, if A E La, BE Lf3, and 1 = max{a,{3}, then A. BE L"Y and the set {A, B} is definable in (L,, E) as {x E L-y I x =A or x = B}. We can conclude that {A, B} E L"Y+ 1 and is, therefore, constructible. If one now sets C = {A, B}, then C is a constructible set which clearly has the required property. The previous proof actually establishes a stronger result than mere validity of the Axiom of Pair in the constructible model; it demonstrates that the operation of unordered pair, when performed in the model on two sets from the model, results in the usual unordered pair of these two sets. We say that the operation of unordered pair is absolute. A similar detailed analysis of the set-theoretic concepts involved shows that all the remaining axioms of ZFC hold in the constructible model, and that many of the usual set-theoretic operations and notions are absolute. The notion of natural number is absolute (that is, a constructible set is a natural number in the sense of the constructible model if and only if it is really a natural number) and so is the notion of ordinal number. However, some concepts are not absolute; among the most important examples are the operation of power set and the notion of cardinal number. The reason for this is that, say, the power set of w in the sense of the constructible model consists of all constructible subsets of w, while the real power set. of w consists of all subsets of w. So clearly the power set. of w in the model is a subset of the real power set of w, but not necessarily vice versa. Indeed, it is this phenomenon which allows us to "cut down" the size of the continuum in the constructible model. Finally, the notion of constructibility is itself absolute: A set is constructible in the sense of the constructible model if and only if it is constructible. But all sets in the model are constructible! We can conclude that every set in the model is constructible in the sense of the model. This argument proves that the following statement holds in the constructible model.

The Axiom of Constructibility

Every set is constructible.

274

CHAPTER 15. THE AXIOMATIC SET THEORY

To summarize, the constructible model satisfies all axioms of ZFC and. in addition, the Axiom of Constructibility. Consequently, the Axiom of Constructibility cannot be disproved in ZFC, and can be added to it without danger of producing contradictions. It turns out that there are many remarkable t lworems one can prove in ZFC enriched by the Axiom of Constructibility: all ot them then also hold in the constructible model, and thus cannot be disproved in ZFC. Godel himself showed t.hnt the Axiom of Constructibility implies the Generalized Continuum Hypothesis: 2N,.

= ~<>+l

for all ordinals n.

Ronald Jensen used the Axiom of Constructibility to construct a linearly ordered set without endpoints in which every system of mutually disjoint intervals is at most countable, but which has no at most countable dense subset. and thus showed the failure of the Sus lin's Hypothesis in this model. Many other de('P results of a similar character have been obtained since. Here we merely indicatt' how the Axiom of Constructibility is used to show that 2N" = ~ 1 . The proof is based on a fundamental result in mathematical logic. the Skolem-Lowenheim Theorem. This theorem asserts that for any structurP (A; R, ... . F, ... ) where R is a binary relation, F is a unary function. f't<: .. tlwrf' is an at most countable B <;;: A such that any property with parameters from ll holds in (A; R. ... , F, ... ) precisely when it holds in (B; RnB 2 , . . . • F r B .... ). (This is an abstract, generalized version of such theorems as "Every group has an at most countable subgroup," etc. It also generalizes Theorem 3.14 in Chapter

4.) Let now X <;;: w. The Axiom of Constructibility guarantees that X E Lo+J for some, possibly uncountable, ordinal a:. This means that there is a property P such that n E X if and only if P(n) holds in (L 0 , E). By the Skolem-Li.iwenheim Theorem, there is an at most countable set B <;;: La such that (B. E) satisfies the same statements as (L",E). In particular, n EX if and only if P(u) holds in (B, E). Moreover, the fact that a structure is of the form (Lo. E) for some ordinal {3 can itself be expressed by a suitable statement. which holds ill (L 0 .E) and thus also in (B. E). From all this, one can conclude that (B. E) is (isomorphic to) a structure of the form (Lo, E) for some, necessarily at most countable, ordinal {3. Since X is definable in (B, E), we get X E Ld+l. \Ve can conclude that every set of natural numbers is constructed at som(· at most countable stage, i.e., P(w) <;;: Ut.kw, Lo+J· To complete the proof of 2N" = ~~,we only need to show that the cardinality of the latter set is ~ 1 . This in turn follows if we show that L, is countable for all/ < w 1 . Clearly, Lo = _.,· is countable. The set £ 1 consists of all subsets of Lo definable in (£ 0 , E), but there are only countably many possible definitions (each definition is a finite sequence of letters from a finite alphabet of some formalized language, together with a finite sequence of parameters from the countable set £ 0 ), and therefore only countably many definable subsets of L 0 • We conclude that L 1 is countabh· and then proceed by induction, using the same idea at all successor stages. aud the fact that a union of countably many countable sets is countable at limit. stages.

275

2. CONSISTENCY AND INDEPENDENCE

The question of independence of the Continuum Hypothesis from the axioms of ZFC (that is, of showing that it cannot be proved in ZFC) remained opm much longer. Finally, in 1963 Paul Cohen announced the discovery of a method which enabled him to construct a model of ZFC in which the Continuum Hypothesis failed. We devote the rest of this section to an outline of some of his ideas. Let us again consider the universe of all sets as described by Zermelo-Fraenkel axioms with Choice. The onlv information about 2N" we have been able to derive is provided by Cantor's The~rem: 2N" > ~ 0 (more generally, cf(2N") > ~ 0 , see Lemma 3.3 in Chapter 9). In particular, 2N" = ~ 1 is a distinct possibility (and becomes a provable fact if the Axiom of Constructibility is also assumed). In general, it is, therefore, necessary to add "new" sets to our universe in order to get a model for 2N" > N1 . We concentrate our attention on the task of adding just one "new" set of natural numbers X. For the time being X is just a symbol devoid of content, a name for a set yet to be described. Let us see what one could say about it. A key to the matter is a realization that one cannot expect to have complete information about X. If we found a property P which would tell us exactly which natural numbers belong to X, we could set X = { n E w I P(n)} and conclude on the basis of the Axiom Schema of Comprehension that X exists in our universe, and so is not a "new" set. Cohen's basic idea was that partial descriptions of X are sufficient. He described the set X by a collection of ''approximations" in much the same way as irrational numbers can be approximated by rationals. Specifically, we call finite sequences of zeros and ones conditions; for example, 0, (1), (1, 0, 1), (1, 1, 0), {1, 1, 0, 1) are conditions. We view these conditions as providing partial information about X in the following sense: If the kth entry in a condition is 1, that condition determines that k E X. If it is 0, the condition determines that k 1- X. For example, (1, 1,0, 1) determines that 0 EX, 1 EX, 2¢: X, 3 E X (but does not determine, say, 4 E X either way.) Next. it should be noted that adding one set X to the universe immediately gives rise to many other sets which were not in the universe originally. such as w - X, w x X, X 2 , P(X), etc. Each condition, by providing some information about X. enables us to make some conclusions also about these other sets. and about the whole expanded universe. Cohen writes p II- P (p forces P) to indicatP that information provided by the condition p determines that the property P holds. For example, it is obvious that {1,1,0,1) II- (5,3) E w x X

(because, as we noted before, {1,1,0,1) II- 3 EX, and 5 E w is true) or (1, 1, 0, 1) II- {2, 3}

tf

P(X)

(because (1, 1, 0, 1) II- 2 ¢: X). It should be noticed that conditions often clash: For example, (1, 1,0) II- (0, 1) E X

2

,

276

CHAPTER 15. THE AXIOMATIC SET THEORY

while (1,0,1) 11- (0,1) ~

x 2.

To understand this phenomenon, the reader should think of p II- P(X) as a conditional statement: If the set X is as described by the condition p. then X has the property P. So (1,1,0) II- (0,1) E X 2 means: If 0 EX, 1 EX. and 2 ~ X, then (0, 1) E X 2 • while (1,0, 1) II- (0, 1) ~ X 2 means: If 0 E X. 1 ~X, and 2 EX, then (0, 1) ~ X 2 . If we knew the set X, we would of course be able to determine whether it is as described by the condition (1, 1, 0) or by the condition (1,0, 1) or perhaps by another of the remaining six conditions of length three. Since we cannot know the set X, we never know which of these conditions is "true"; thus we never are able to decide whether (0, 1) E X'2 or not. Nevertheless, it turns out that there are a great many properties of X which can be decided, because they must hold no matter which conditions are "true." As an illustration, let us show that every condition forces that X is infinite. If not, there is a condition p and a natural number k such that p II- "X has k elements.". For example, let p = (1. 0, 1) and k = 5: we show that (1, 0, 1) II- "X has five elements" is impossible. Consider the condition q = (1. 0, 1, 1, 1, 1, 1). First of all, the condition q contains all information supplied by p: 0 E X, 1 ~ X, 2 E X. If the conclusion that X has five elements could be derived from p, it could be derived from q also. But this is absurd becaus(, clearly q II- "X has at least six elements"; namely, q II- 0 E X, 2 E X, 3 E X. 4 E X, 5 E X, 6 E X. The same type of argument leads to a contradiction for any p and k. As a second and last illustration, we show that every condition forces that X is a "new" set of natural numbers. More precisely, if A is any set of natural numbers from the "original" universe (before adding X), then every condition forces that X i= A. If not, there is a condition p such that p II- X = A. Let us again assume that p = (1,0, 1); then p II- 0 EX, 1 ~X, 2 EX. Now there arP two possibilities. If 3 E A, let q 1 = (1, 0, 1, 0). Since q1 contains all intorrnatiou supplied by p, q 1 II- X = A. But this is impossible, because q 1 II- 3 ~ X. when•as 3 EA. If 3 ~A, let q2 = (1,0, 1,1). We can again conclude that q2 II- X= A and get a contradiction from q2 II- 3 E X and 3 ~ A. Again, a similar argument works for any p. Let us now review what has been accomplished by Cohen's construction. The universe of set theory has been extended by adding to it a "new," "imaginary" set X (and various other sets which can be obtained from X by set-theoretic operations). Partial descriptions of X by conditions are available. These (kscriptions are not sufficient to decide whether a given natural number belongs to X or not, but allow us, nevertheless, to demonstrate certain statements about X, such as that X is infinite and diflers from every set in the original universe. Cohen has established that the descriptions by conditions are sufficient to show the validity of all axioms of Zermelo-Fraenkel set theory with Choice in the extended universe. Although adding one set of natural numbers to the universe does not increase the cardinality of the continuum, one can next take the extended universe and.

3. THE UNIVERSE OF SET THEORY

277

by repeating the whole construction, add to it another "new" set of natural numbers Y. If this procedure is iterated ~2 times, the result is a model in which there are at least N2 sets of natural numbers, i.e., 2N" 2: ~ 2 . Alternatively, one can simply add ~2 such sets at once by employing slightly modified conditions. Cohen's method has been used to construct models in which 2N" = ~" for any ~" with cf(Na) > N0 . It can also be used to build models in which Suslin's Hypothesis holds or fails to hold, models for Martin's Axiom MA .. , as well as models for many other undecidable propositions of set theory. The methods described in this section can also be applied to showing that the Axiom of Choice is neither provable nor refutable from the other axioms of Zermelo-Fraenkel set theory. It is not refutable because one can define the constructible model and prove that it is a model of ZFC using only the axioms of Zermelo-Fraenkel set theory without Choice. [A choice function for any system X of constructible sets can be defined roughly as follows: Select from each A E S (A i= 0) that element which is constructed at the earliest stage- that is. belongs to £ 0 or to Lu+l for the least possible a:. Should there be severaL select theE-least element of L 0 or the one whose definition in (L 01 , E) comes first in the alphabetic order of all possible definitions.] On the other hand, the Axiom of Choice is not provable in Zermelo-Fraenkel set theory either, because. as Cohen has shown, it is possible to extend the universe of set theory by adding to it a "new" set of real numbers without adding any well-ordering of this set, and thus obtain a model of set theory in which the set of real numbers cannot be wellordered. Consistency of the Axiom of Foundation with the remaining axioms of ZFC is also established by the method of models. A model for set theory with the Axiom of Foundation is obtained by letting M(x) be the property '·.r is a well-founded set," and E(x, y) be "x E y." The proof that this is indeed a model satisfying the Axiom of Foundation follows the lines of Example 2.8 in Section 2 of Chapter 14.

3.

The Universe of Set Theory

In this last section we give some thought to possibilities of extending ZermeloFraenkel set theory with Choice by additional axioms. Our interest here is in axioms which could with some justification be regarded as true. One possible candidate for such a new axiom has been introduced in Section 2; it is the Axiom of Constructibility. We have seen there that the Axiom of Constructibility has important consequences in set theory; for example. it implies the Generalized Continuum Hypothesis and existence of counterexamples to Suslin's Hypothesis. In recent years it has also been shown that the Axiom of Constructibility can be a powerful tool for other branches of abstract mathematics, and a series of important results, including some of great elegance and intuitive appeal, has been derived from it in model theory, general topology, and group theory. On the other hand, there do not seem to be any good intuitive reasons for belief that all sets are constructible; the ease with which the method of forcing from Section 2 establishes the possibility of existence of

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CHAPTER 15. THE AXIOMATIC SET THEORY

nonconstructible sets might suggest rather the opposite. Moreover, all of thP aforementioned results can be proved equally well from other axioms. some of them weaker than the Axiom of Constructibility, and some of them actual!~· contradicting it. However, the most serious objections to accepting the Axiom of Constructibility at the same level as the axioms of ZFC stem from the fact that it yields som!' rather unnatural consequences in descriptive set theory, a branch of mathPmatics concerned with detailed study of the complexity of sets of real numbers. As descriptive set theory provides some of the most important insights into til!· foundations of set theory, we outline a few of its basic problems and results next. \Ve introduced Borel sets in Section 5 of Chapter 10 as sets of realuumb•·rs that are particularly simple. Their detailed study confirms this by showing that Borel sets exhibit particularly nice behavior: Every Borel spt is Pitlwr at most countable or it contains a perfect subset, and so, in accordance with the Continuum Hypothesis, it has either cardinality ::; N0 or 2N". Similarlr. evE>ry Borel set is Lebesgue measurable, and has many other nice propt>rt.i!'s. Descriptive set theory attempts to extend these results to morE' complex set.:;. How do we obtain such sets? Countable unions and intersections of Borel st>ts are Borel, so we cannot use them to obtain "new'' sets. Continuous fuuctious are well-understood, simple functions, but it turns out that an image of a Bor<>l set by a continuous function (while hopefully still "simple" enough) need not be a Borel set. We define analytic sets as those sets that are images of a Bord set by a continuous function, and coanalytic sets as complements of analytic sets (in R). The usual notation is Ej for the collection of all analytic s<>ts awl II: for the collection of all coanalytic sets. One can then pron~ed furtlwr and define recursively E~+L as the collection of all continuous images of II~, sPts and II~+ 1 as the collection of all complements of sets in E~•+l. This is th•· projective hierar·chy, and sets that belong to some E~, (or II~,) are the JimJt'roti1'1 sets. We would expect projective sets to behave nicely, being still rather simp!!'. and the classical descriptive set tlwory as developed by Nikolai Luzin and bi~ students confirms this to au extent. For example, it is known that all analytic and coanalytic sets are Lebesgue measurable, and that all uncountablt• analytic st•ts contain a perfect subset. As to going further. in ZFC with the Axiom ot Constructibility one gets discouraging results: there exist (or II~) sets that are not Lebesgue measurable, and there exist uncountable coanalytic (IIi) sets without a perfect subset.

E1

Another important classical result is that II: and E~ sets have the so-called r·eduction property and E~ and II~ sets do not have it. (A class of sets 1 is said to have the reduction property if for all A, B E r there exist A'. B' E r such that A' s;; A, B' s;; B, A' u B' = Au B, and A' n B' = 0.) One might expect IIj, E!, II~, etc., sets to have the reduction property, and E~, II!. EA. etr .. to fail to have it. but the Axiom of Constructibility leaves thE' ca.se n = 1 au odd exception and proves instead that all sets for n ~ 2 haw' the reduction property and all II~ sets, n 2: 2, fail to have it.

E:,

3. THE UNIVERSE OF SET THEORY

279

Surprisingly, a more satisfactory answer can be obtained by using some of the so-called large cardinal axioms. The inaccessible cardinal numbers defined in Chapter 9 provide the simplest examples of large cardinals. We remind thP reader that a cardinal number K. > ~ 0 is called inaccessible if it is regular and a limit of smaller cardinal numbers. It is shown in Section 2 of Chapter 9 that the first inaccessible cardinal (assuming it. exists) must be greater than ~~, ~2, ... , Nw, ... , ~N,, ... , NN"', ... , NN"~, ... , etc., hence the adjPctive "large." It is known that the existence of inaccessible cardinals cannot be proved in ZFC; in fact, the construction of a model for ZFC in which there are no inaccessible cardinals is rather simple. If the constructible universe L contains no inaccessible cardinals (in the sense of L). then it is such a model. Otherwise. we take the least inaccessible cardinal {) in L and consider a model, ~wts of which are precisely the elements of L,?, and the membership relation is the usual o1w. It is quite easy to use the inaccessibility of{) and prove that all axioms of ZFC' are satisfied in this model. The fact that.{) is the lea.~t inaccessible cardinal iu L guarantees that there are no inaccessible cardinals in this model. So existence of inaccessible cardinals is independent of ZFC. Unlike the Continuum Hypothesis or the Suslin's problem, however, it is not possible to prove that inaccessible cardinals are consistent with ZFC by way of constructing an appropriate model (doing so would contradict the celebrated Second Incompleteness Theorem of Godel). This means that a set theory with an inaccessible cardinal is esseutiallv stronger than plain ZFC. The assumption of existence of inaccessible cardinals requires a "leap of faith" (similar to that required for arc:eptancP of the Axion1 of Infinity), but some intuitive justification for t.hP plausibility of making it ran be given. It goes roughly as follows. Mathematicians ordinarily consider infinite collections, such as tiH' sl't of natu raJ numbers or the set of real numbers. a.o.; finished. romplet<'d tot alit.ies. On th e other hand, it is not in the power of an ordinary mathematician to regard the collection of all sets as a finished totality. i.e., a set - it would lead to a contradiction. We call the collections which mathematicians ordinarily view as finished the sets of the first order; they are the sets we have been con cerned with until now. Let us now plac<' ourselves in a position of an extraordin ary, more abstract, mathematician, who has <'Xamined tlH' universP of the first-order sets and then collected all those sets into a new finished totality. a second-order set V. Having gotten V, he can, of course. use the met.lwds of Chapters 1· 14 to form various other second-order sets. such as V- w. N 1 x V. P(P(V)), 0 = {a: E V I a: is an ordinal}, 0 + 1, 0 + 0 + w, etc. [Notice that 0 is the (second-order) set of all first-order ordinals.] The intuitive argunwut~ which justified the axioms of ZFC for the first-order mathematician also convincP the second-order mathematician of the validity of these axioms in his '·secondorder" universe of set theory. (Notice that Russell's Paradox is avoided h:v this viewpoint: The second-order mathematician can form the set R of all firstorder sets which are not elements of themselves. but R is not a first-ordPr set, so clearly R ¢ R, and no contradictions appear. Of course, the second-or(kr mathematician still cannot form the "set" of all sPcOIHi-order sets which are not elements of themselves; this is a task for a third-order mathematician.) \Ve now

280

CHAPTER 15. THE AXIOMATIC SET THEORY

claim that 0, the least second-order ordinal, is an inaccessible cardinal in the second-order universe. Certainly 0 > N0 and 0 is a limit of (the first-order) cardinal numbers. If (~~:, I L < a:) is a sequence of ordinals less than 0 having length a:< 0, then both a: and all~~:, are first-order ordinals, and the arguments we used to justify the Axiom Schema of Replacement again convince us thCtt. a first-order mathematician should be able to collect {~~:, I L < a:} into Ct firstorder set. But then sup,
For any real number a,

~1

is inaccessible in L[a],

where L[a] is a model constructed just as L, but starting with L 0 [a] = w u {a} (it is the least model of set theory containing all ordinals and the real number a). Conversely, from the "large cardinal axiom" (*) it follows that all uncountable ll~ sets have a perfect subset. In addition, (*) further implies that all uncountable :E~ sets have a perfect subset (and thus cardinality 2N") and that all :E~ and ll~ sets are Lebesgue measurable (but it does not imply the existence of perfect subsets of uncountable ll~ sets, nor measurability of :E:\ sets). Overall, the axiom (*) produces a mild improvement upon the results obtainable from the Axiom of Constructibility. To get better results. we haw to assume existence of cardinals much larger than mere inaccessibles. But befon· outlining some of them, we have to make another digression. A very important role in modern descriptive set theory is played by a certain kind of infinitary game. Let us consider a game between two players. I and II. described by the following rules. A finite set M of possible moves is given. The players choose moves, taking turns and each moving n times. That is, player I begins by making a move p 1 E M; player II responds by making a move q1 E :\!. Then it is I's turn to make a move p 2 E M, to which II responds by playing q2 E M, etc. The resulting sequence of moves (PI, q1, P2, q 2 , ... , Pn, qn) is called a play. A set of plays S is specified in advance and known to both players. Player I wins if (PI, ... ,qn) E S; player II wins if (p 1, ... ,qn) 1- S. Many intellectual games between two players, including checkers, chess, and go. can be mathematically represented in this abstract form by choosing suitable A/. n. and S (some artificial conventions have to be adopted, e.g., in order to allow for draws). A fundamental notion in game theory is that of a strategy. A strategy for player I is a rule which tells him what move to make at each of his turns.

3. THE UNIVERSE OF SET THEORY

281

depending on the moves played by both players previously. If a strategy for player I has the property that, by following it, I always wins, it is called a winning strategy for I. Similarly, one can talk about a winning strategy for II. The basic fact about games of this type is that one of the players always has a winning strategy; in other words, the game is determined. The reason is rather simple: If there is a move PI such that for every q, there is a move p 2 such that for every

q2 ...

there is a move Pn such that for every qn the play (Pt,q,, ... ,pn,qn) E S, then obviously player I has a winning strategy. If the opposite holds, it means that For every PI there is q, such that for every P2 there is q2 such that ... for every Pn there is qn such that the play (PI, qL, ... , Pn. qn) ¢: S. But in this case, II is the player with a winning strategy. We can now modify the game by allowing each player infinitely many turns. The play now is an infinite sequence of moves: (p 1 , qi, p 2 , q2, ... ) . The payoff set S is a set of infinite sequences of elements of M, and I is the winner if and only if (p 1 , q 1 , p 2 , q2 , ... ) E S. The game played according to these rules is referred to as Gs. It is no longer obvious that such games are determined, and as a matter of fact they need not be. Using the Axiom of Choice one can construct a payoff set S ~ MN for any M with IMI 2 2 such that neither player has a winning strategy in the game Gs. (The argument is quite similar to the one Wf' used to construct an uncountable set without a perfect subset in Example 4.11 of Chapter 10.) The interesting question is whether the game Gs is determined for "simple" sets S. In order to study this question, we take the finite set AI to be a natural number m; infinite sequences of elements of m can then be viewed as expansions of real numbers in base m, and the payoff set S can be viewed as a subset of R. In this way we can talk about payoff sets being Borel, analytic, etc. A profound theorem of D. Anthony Martin states that all games with Borel payoff sets are determined. The situation at higher levels of the projective hierarchy is similar to that discussed in connection with the question of existence of perfect subsets, but the large cardinals involved are much larger. To be specific, the Axiom of Constructibility can be used to construct games with analytic (:ED payoff sets that are not determined. The work of Martin and Leo Harrington showed that determinacy of all games with :E~ (or, equivalently, rr:) payoff sets is equivalent to a certain large cardinal axiom. It would take us too far to try to state this axiom here (it is known technically as "For all real numbers x, x# exists."); for our purposes it suffices to say that this axiom implies Solovay's assumption (*), and much more (for example, it implies the existence of Mahlo cardinals and weakly compact cardinals in the constructible

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CHAPTER 15. THE AXIOMATIC SET THEORY

universe L), and is itself a consequence of the existence of measurable cardinals. So if a measurable cardinal exists then all analytic and coanalytic games an• determined. The existence of measurable cardinals does not suffice to prove determinacy of all games with :E~ (or II~) payoff sets. The work of Martin. John Steel, and Hugh Woodin in the early nineties showed that there arE' largt> cardinals (much larger than the fin.;t mt>asurable one) whose exist.em·e impliPs the Axiom of Projective Determinacy: all games with projective payoff sets arP determined. Conversely, the Axiom of Projective Determinacy implies existE'no· of models of set theory with these large cardinals. What reasons are there for tht> belief that something like the Axiom of Projective Determinacy is true? Besides its own intrinsic plausibility, it st.Pms from the fact that determinacy of infinitary games at a certain level in th(• projE:>ctive hierarchy implies all the nice descriptive set-theoretic properties that the sets at or near that level are expected to have. For example, determinacy of :El games implies that all UnCOUlltable II: and :E1 sets have a perfect sub:·Wt. and that all :E~ and II1 sets are Lt>besgue measurable. The determinacy of :E~ games implies that all uncountabiP II~ and :E~ sets have a perfect subset. aud that all :E~ and IIj sets are Lel)('sguc~ measurable. Moreover, it also geJwralizPs the reduction property correctly to the third level of the projective hit>rarchv; i.e., it implies that IIj sets have the reduction property and :E:\ sets do not. The full Axiom of Projective Determinacy has as its consequences the Lebesgue measurability of all projective sets, the fact that every uncountable projective set has a perfect subset, and the "correct" behavior of the reduction property (II~, :E~, IIi, :E!, ... sets have it, II~, :E~, II!, ... do not). among mauy others not stated here. It is the results like these that are convincing workPrs in descriptive set theory that PD must be true. What emerges from the above considerations is a hierarchy postulating the Pxistence of larger and larger cardinals and providing better and better approximations to the ultimate truth about the universe of sets. Further support for this general picture has come from the work on undecidable st atr>ments in arit hmetic. By arithmetic we mean the theory of Peano arithmetic whosp axio1us were given in Chapter 3; it is easy to establish that Peano arithmetic is equivalent to the theory of finite sets; i.e., the theory that is obtained from ZFC by omitting the Axiom of Infinity (iu the resulting theory, only finite .sets can be proved to exist). It has been known since the fundamental work of Kurt GiidP! in 1931 that there are true statements about. natural numbers (or finite sl'ts) that cannot be decided (either proved or disproved) from the axioms of Pea no arithmetic (or the theory of finitP sets). The statements are true because tlwy can be provE>d in ZFC, but the usP of at least some infinite sets is essential f
:r::,

3. THE UNIVERSE OF SET THEORY

283

infinity is necessary. Another example is a version of the Finite Ramsey's Tllf'orem discussed in Section 1 of Chapter 12. It is possible to determine the size of infinite sets needed to prove a particular statement, and it turns out that hPrt'. too, there is a hierarchy of theories postulating the existence of larger and larger infinite sets and providing better and better approximations to the truth about natural numbers (or finite sets). In most cases these theories are subtheories of ZFC (so few mathematicians doubt that they are true), but there are examples of (rather complicated) statements of arithmetic that cannot be decided even in ZFC (but can be decided, for example, in ZFC with au inaccessible cardinal). so the hierarchy obtained from the study of the strength of arithmetic statPments merges into the hierarchy of large cardinals needed for the study of the strength of statements about real numbers arising in descriptive set theory. with ZFC being just one of the stages. Even the techniques used to prove the results about arithmetic are closely related to those used to study large cardinals; they rely heavily on such concepts as partitions, trees, and games. Most of the work discussed in this section is relatively recent, and by no means complete. Both the theory of large cardinals and the study of the undecidable statements of arithmetic are very active research areas where new insights and interconnections continue to be discovered. As Godel assure~ us in his Incompleteness Theorem, no axiomatic theory can decide all statements of arithmetic or set theory. We can thus feel confident that the euterpris<' of getting closer and closer to the ultimate truth about the mathematical universe will continue indefinitely.

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Bibliography [1] Peter Aczel. Non-well-founded aeta. Stanford University Center for the Study of Language and Information, Stanford, CA, 1988. With a foreword by Jon Barwise [K. Jon Barwise]. [2] Keith Devlin. TM jofl ofaeta. Springer-Verlag, New York, second edition, 1993. Fundamentals of contemporary set theory. [3] F. R. Drake and D. Singh. Intennediate aet theorJI. John Wiley & Sons Ltd., Chichester, 1996. [4] Herbert B. Enderton. Element. ofaet theorJJ. Academic Press [Harcourt Brace Jovanovich Publishers], New York, 1977. [5] Paul Erd&, Andras Hajnal, Attila Mate, and Richard Rado. Combinatori4l aet theort~: partition relationa for cardinals. North-Holland Publishing Co., Amsterdam, 1984. [6] Ronald L. Graham, Bruce L. Rothschild, and Joel H. Spencer. Ramsey theortJ. John Wiley & Sons Inc., New York, second edition, 1990. A WileyInterscience Publication. [7] Paul R. Halmos. Naive aet theorJJ. Springer-Verlag, New York, 1974. Reprint of the 1960 edition, Undergraduate Thxts in Mathematics. [8] James M. Henle. An outline of aet 1986. [9] Thomas Jech. Set

theof'JI. Springer-Verlag, New York,

theof'JI. Springer-Verlag, Berlin, second edition, 1997.

[10] Winfried Just and Martin Weese. Diacovering modern aet theDrJJ. I. American Mathematical Society, Providence, Rl, 1996. The basics. [11] Akihiro Kanamori. The higher infinite. Springer-Verlag, Berlin, 1994. Large cardinals in set theory from their beginnings. [12] Alexander S. Kechris. Claaaical ducriptive aet theortJ. Springer-Verlag, New York, 1995. 285

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286

[13] Kenneth Kunen. Set fheoru. North-Holland Publishing Co., Amsterdam, 1983. An introduction to independence proofs, Reprint of the 1980 original. (14] Yiann.is N. Moechovak.is. Descriptive set Co., Amsterdam, 1980. [15] Yiannis N. Moechovakis. Notes on set 1994.

theorJJ. North-Holland Publishing

theorJJ. Springer-Verlag, New York,

(16] Judith Roitman. Introduction to modem set theof'JI. John Wiley & Sons Inc., New York, 1990. A Wiley-Interscience Publication. (17] Robert L. Vaught. Set theortJ. Birkhliuser Boston Inc., Boston, MA, second edition, 1995. An introduction.

Index absolute, 273 accumulation point, 184 addition, 52, 172, 176 addition of cardinal numbers, 93 addition of ordinal numbers, 119 additive function, 147 aleph, 131 aleph naught No, 78 algebraic number, 101 almost disjoint functions, 213 analytic set, 278 antichain, 225, 237 antisymmetric relation, 33 Aronszajn tree, 228 associativity, 13, 94, 122, 159, 160 asymmetric relation, 33 at most countable, 74 atom, 244 atomless measure, 244 automorphism, 60 Axiom of Anti-Foundation, 262 Axiom of Constructibility, 273 Axiom of Countable Choice, 144, 152 Axiom of Existence, 7 Axiom of Extensionality, 7 Axiom of Foundation, 259 Axiom of Infinity, 41 Axiom of Pair, 9 Axiom of Power Set, 10 Axiom of Regularity, 259 Axiom of Union, 9 Axiom of Universality, 265 Axiom Schema of Comprehension, 8 Axiom Schema of Replacement, 112 axioms,3

Banach, Stefan, 149 basis, 146 beth, 223 binary operation, 55 binary relation, 19 bisimulation, 263 Borel set, 194 bounded sequence, 181 bounded set, 86, 161 branch, 225 Brouwer-Kleene ordering, 85 Cantor set, 185 Cantor's Theorem, 96 Cantor, Georg, 1, 90, 99, 101 Cantor-Bernstein Theorem, 66 cardinal number, 68, 130, 140 cardinality, 65, 68 cartesian product, 21, 57 Cauchy sequence, 181 chain, 34 characteristic function, 91 choice function, 138 closed interval, 179 closed set, 182 closed unbounded filter, 208, 212 closed unbounded set, 208, 212 closed under operation, 55 closure, 60 closure point, 145, 212 coanalytic set, 278 cofinal sequence, 143 cofinality, 163 cofinite set, 202 Cohen,Paul,101, 139,269,275-277 collection, 9 commutativity, 13, 94

Baire function, 200 287

288 comparable, 34 compatible, 26, 237 complete linear ordering, 87 completion, 87 complex number, 179 composition, 21 computation, 48, 115 concatenation, 62 condensation point, 190 condition, 275 connectives, 5 constant, 7, 57 constructible model, 271 constructible set, 273 continuoUB function, 145, 182 continuum, 92 Continuum Hypothesis, 101, 141, 268 convergence, 145 coordinate, 57 countable additivity, 150 countable set, 74 counting measure, 153, 204 cumulative hierarchy, 257 cut, 88 De Morgan Laws, 14 decimal expansion, 86 decoration, 261 decreasing sequence, 144 Dedekind cut, 88 Dedekind finite set, 97 .£\-lemma, 211 dense linear ordering, 83 density, 203 derived set, 191 descriptive set theory, 278 determined, 281 diagonal intersection, 210, 212 diagonalization, 90 difference, 13 digit, 174 disjoint, 14 distributivity, 14, 94, 122, 160 division, 173, 178 domain, 19 double induction, 46

INDEX element, 1 empty sequence, 46 empty set, 6, 8 equipotent, 65 equivalence class, 30 equivalence relation, 29 Erd&t-DUBhnik-Miller Theorem, 222 Erd&-Rad6 Theorem, 223 expansion, 178 exponentuuaon ,54 exponentiation of cardinals, 95 exponentiation of ordinals, 122 extension, 24 extensional relation, 254 factorial, 48 Fibonacci sequence, 50 filter, 201 finite additivity, 204 finite character' 144 finite intersection property, 180, 202 finite sequence, 46 finite set, 69, 72 fixed point, 69 forcing, 275 function, 23 function into, 24 function on, 24 function onto, 24 Gooel, Kurt, 3, 101, 269, 271, 274, 279,282,283 GCH, 165 generic set, 236 Goodstein sequence, 125, 126 graph, 261 greatest element, 35 greatest lower bound, 35 Hahn, Hans, 149 Hahn-Banach Theorem, 148 Hamel basis, 146 Harrington, Leo, 281 Hartogs number, 130 Hausdorff's Formula, 168 hereditarily finite, 114

INDEX Hilbert, David, 101 homogeneoUB set, 217 ideal, 202 identity, 4 image, 20 inaccessible cardinal, 162, 168 inclUBion, 12 incomparable, 34 incompatible, 237 increasing function, 105 increasing sequence, 160, 181 Induction Principle, 42, 44, 114, 260 inductive set, 40 infimum, 35 infinitary operation, 198 infinite product of cardinals, 157 infinite sequence, 46 infinite set, 69 infinite sum of cardinals, 155 initial ordinal, 129 initial segment, 104 injective function, 25 integer, 171 integer part, 174 intersection, 7, 13, 14 inverse, 20 inverse image, 20 invertible function, 25 isolated point, 184 isomorphic structures, 59 isomorphism, 36, 58 isomorphism type, 260 Jensen's Principle, 234 Jensen, Ronald, 274 K-Complete filter, 246 Konig's Lemma, 227 Konig's Theorem, 158 least element, 35 least upper bound, 35 Lebesgue measure, 152 length, 62 lexicographic ordering, 82

289

limit, 160, 181 limit cardinal, 162 limit point, 212 linear functional, 148 linear ordering, 34 linearly independent, 146 linearly ordered set, 34 lower bound, 35 Mablo cardinal, 249 mapping, 23 Martin's Axiom, 237, 238 Martin, Donald Anthony, 281, 282 maximal element, 35 measurable cardinal, 247 measure, 150,204,241 measure problem, 241 minimal element, 35 model, 270 monochromatic set, 217 monotone function, 69 Mostowski's Lemma, 255 multiplication, 54, 172, 177 multiplication of cardinals, 94 multiplication of ordinals, 121 n-ary operation, 57 n-ary relation, 57 n-tuple, 56, 118 natural number, 41 node, 225 nondecreasing sequence, 181 normal form, 124 normal ultrafilter, 2.49 one-to-one function, 25 open interval, 179 open set, 182 operation, 7, 55 order type, 79,113,260 ordered n-tuple, 56 ordered field, 178 ordered pair, 17 ordered set, 33 ordering, 33 ordinal number, 107

290

palindrome, 64 parameter, 5 Paris, Jeffrey, 282 partition, 31 Peano arithmetic, 54 Peano, Giuseppe, 54 perfect set, 185 power set, 10 prime ideal, 205 principal filter, 201 principal ideal, 202 Principle of Dependent Choices, 144 product of an indexed system, 26 projection, 62 Projective Determinacy, 282 projective set, 278 proper subset, 12 property, 1 quantifiers, 5 Ramsey's Theorem, 218, 219 range, 19 rank, 255 rational number, 173 real number, 89 RecU1"8ion Theorem, 47, 115, 260 reflexive relation, 29 regressive function, 209, 212 regular cardinal, 161 relation, 29 restriction, 24 RUBsell's Paradox, 3 RUBsell, Bertrand, 2 sequence, 46 set, 1 set of all sets, 2 set of representatives, 32, 139 set of sets, 2 u-additive, 150, 151, 241 u-algebra, 151 u-complete filter, 246 Silver's Theorem, 212 singleton, 17 singular cardinal, 160

INDEX Skolem-LOwenheim Theorem, 274 Solovay, Robert, 152, 280, 281 statement, 6 stationary set, 209, 212 Steel, John, 282 strategy, 280 strict ordering, 33 strong limit cardinal, 168 structure, 56, 58 subaequence, 181 subset, 6, 10 successor, 40, 51 successor cardinal, 162 successor ordinal, 108 sum of linear orderings, 81 supremum, 35, 109 Suslin tree, 230 Suslin's Problem, 180, 230, 268 Suslin, Mikhail, 230 symmetric difference, 13 symmetric relation, 29 system of sets, 9 ternary relation, 22 Tichonov's Theorem, 153 total ordering, 34 transcendental number, 101 transfinite sequence, 115 transitive closure, 256 transitive relation, 29 transitive set, 107 translation invariance, 150 tree, 225 tree property, 229, 248 trivial filter, 201 trivial ideal, 202 trivial measure, 204 Thkey's Lemma, 144 two-valued measure, 206, 244

u-limit, 206 ultrafilter, 205 unary operation, 55 unary relation, 22 unbounded set, 161 uncountable set, 90

INDEX union, 9, 13 universe, 58 unordered pair, 9 upper bound, 35 value, 23 variables, 4 Venn diagrams, 12 weakly compact cardinal, 224 well-founded relation, 251 well-ordered set, 44, 104 well-ordering, 44 Well-Ordering Principle, 142 winning strategy, 281 Woodin, Hugh, 282 Zermelo, Ernst, 11, 138, 139 Zermelo-Fraenkel Set Theory, 267 Zorn's Lemma, 142

291