Solutions Manual to Naive Set Theory by Paul Halmos

Naive Set Theory: Solutions Manual George Mplikas September 2015

Contents 2 Misc. Exersizes 2.1 Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Logical Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 3 4

3 Unordered Pairs

5

4 Unions and Intersections

6

5 Complements and Powers

10

6 Ordered Pairs

16

7 Relations

20

8 Functions

22

9 Families

23

10 Inverses and Composites

27

11 Numbers

31

12 Peano Axioms

32

13 Arithmetic

33

14 Order

37

15 Axiom of Choice

38

16 Zorn’s Lemma

39

17 Well Ordering

42

18 Transfinite Recursion

44

19 Ordinal Numbers

46

20 Sets Of Ordinal Numbers

47

21 Ordinal Arithmetic

48

22 The Schr¨ oder-Bernstein Theorem

49

23 Countable Sets

51

24 Cardinal Arithmetic

52

1

25 Cardinal Numbers

54

26 Extra Exercises 56 26.1 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2

2 2.1

Misc. Exersizes Logical Operators

Exercise 2.1. We show that ∧, ∨ are commutative. We do this via truth table. Proof. Consider the logical sentences p, q and the following truth table: p T T F F

q T F T F

p∨q T T T F

q∨p T T T F

Thus, p ∨ q ⇐⇒ q ∨ p Proof. Consider the logical sentences p and q and the following truth table: p T T F F

q T F T F

p∧q T F F F

Thus, p ∧ q ⇐⇒ q ∧ p.

3

q∧p T F F F

2.2

Logical Rules

Exercise 2.2. We show that p ∨ p ⇐⇒ p and p ∧ p ⇐⇒ p, via truth table. Proof. Consider the logical sentence p and the following truth table: p∨p T F

p T F Thus, p ∨ p ⇐⇒ p

Proof. Consider the logical sentence p and the following truth table: p∧p T F

p T F Thus, p ⇐⇒ p ∧ p.

Exercise 2.3. We show that p ⇒ p ∨ q via truth table. Proof. Consider the logical sentences p and q and the following truth table: p T T F F

p∨q T T T F

q T F T F

Thus, if p is true, p ∨ q is true, i.e. p ⇒ p ∨ q.

4

3

Unordered Pairs

Exercise 3.1. Consider the sets ∅, {∅}, {{∅}}, etc.; Consider all the pairs such as {∅, {∅}}, formed by two of them; consider the pairs formed by any such two pairs, or else mixed pairs formed by a singleton and any pair; and proceed ad infinitum. Are all the sets obtained in this way distinct from one another? Yes, all sets obtained in this way are distinct. Proof. By the axiom of pairing, we can form A and B as given, and by the axiom of specification, they are the only such sets which contain their elements. Suppose, to the contrary, that A = B. If A = B, then for all a ∈ A, a ∈ B, and for all b ∈ B, b ∈ A. Now, either A and B are singletons, or they are pairs. • A and B are singletons: Say, A = {a}, and B = {b}. Thus, by the Axiom of Extension, we have a = b. Implying that A and B were formed in the same way. • A and B are pairs: Let A = {a, a0 }, B = {b, b0 }. Without loss of generality, suppose that a = b, and a0 = b0 . If this is the case, then by the Axiom of Extension, again, we have that a, b were obtained in the same fashion, as well as a0 , b0 . But in this case, A and B were constructed in the same manner. Either way, we have a contradiction. Thus, A and B are formed distinctly.

5

4

Unions and Intersections

Throughout these exercises we consider a arbitrary non-empty collection C, of sets. For simplicity we will assume that C = {A, B, C} for arbitrary non-empty sets (unless specified) A, B, C and will make liberal use of the notation {x : S S(x)}, in-place of the more precises yet cumbersome notation {x ∈ C : S(x)}. Exercise 4.1. We show the following: A∪∅=A

(1)

A∪B =B∪A

(2)

A ∪ (B ∪ C) = (A ∪ B) ∪ C

(3)

A∪A=A

(4)

A ⊂ B ⇐⇒ A ∪ B = B Proof. ( ⇐⇒ ) Since ∅ has no elements, for all a ∈

(5) S

C, a ∈ / ∅ and we have,

x ∈ A ∪ ∅ = {x : x ∈ A ∨ x ∈ ∅} ⇐⇒ x ∈ A ∨ x ∈ ∅ = {} ⇐⇒ x ∈ A Thus, A ∪ ∅ = A Proof. ( ⇐⇒ )

A ∪ B = {x : x ∈ A ∨ x ∈ B} = {x : x ∈ B ∨ x ∈ A} = B ∪ A

Proof. ( ⇐⇒ )

x ∈ A ∪ (B ∪ C) ⇐⇒ x ∈ A ∨ x ∈ B ∪ C ⇐⇒ x ∈ A ∨ x ∈ B ∨ x ∈ C ⇐⇒ x ∈ {x : x ∈ A ∨ x ∈ B} ∨ x ∈ C ⇐⇒ x ∈ A ∪ B ∨ x ∈ C ⇐⇒ x ∈ (A ∪ B) ∪ C

Proof. ( ⇐⇒ )

6

A ∪ A = {x : x ∈ A ∨ x ∈ A} = {x : x ∈ A} = A

Proof. (⇒) We note that A ⊂ B iff for all x ∈ A, x ∈ B. Thus, we have that x ∈ A ∪ B = {x : x ∈ A ∨ x ∈ B} ⇒ x ∈ A ∨ x ∈ B ⇒ x ∈ B by the comments above. And likewise, x ∈ B ⇒ x ∈ B ∨ x ∈ A ⇒ x ∈ {x : x ∈ A ∨ x ∈ B} ⇒ x ∈ A ∪ B That is, A ∪ B = B. (⇐) Suppose A ∪ B = B. Then, x ∈ A ⇒ x ∈ {x : x ∈ A ∪ x ∈ B} = A ∪ B = B That is, A ⊂ B. Exercise 4.2. We show the following: A∩∅=∅

(1)

A∩B =B∩A

(2)

A ∩ (B ∩ C) = (A ∩ B) ∩ C

(3)

A∩A=A

(4)

A ⊂ B ⇐⇒ A ∩ B = A

(5) (6)

Proof. (1) Suppose that A 6= ∅, then

(⇒) x ∈ A ∩ ∅ = {x : x ∈ A ∧ x ∈ ∅} ⇒ x ∈ ∅ ∧ x ∈ A ⇒x∈∅ (⇐) (∅ ⊂ A) ∧ (∅ ⊂ ∅) ⇒∅⊂A∩∅

Proof. (2) ( ⇐⇒ ) 7

A ∩ B = {x : x ∈ A ∧ x ∈ B} = {x : x ∈ B ∧ x ∈ A} = B ∩ A

Proof. (3) ( ⇐⇒ ) x ∈ A ∩ (B ∩ C) ⇐⇒ x ∈ A ∧ x ∈ B ∩ C = {x : x ∈ B ∧ x ∈ C} ⇐⇒ x ∈ A ∧ x ∈ B ∧ x ∈ C ⇐⇒ x ∈ {x : x ∈ A ∧ x ∈ B} ∧ x ∈ C ⇐⇒ x ∈ A ∩ B ∧ x ∈ C ⇐⇒ x ∈ (A ∩ B) ∩ C

Proof. (4) ( ⇐⇒ ) A ∩ A = {x : x ∈ A ∧ x ∈ A} = {x : x ∈ A} = A

Proof. (5) (⇒) Suppose that A, B 6= ∅. We note that A ⊂ B iff for all x ∈ A, x ∈ B; then, x∈A⇒x∈B ⇒x∈A∧x∈B ⇒x∈A∩B And likewise, x ∈ A ∩ B = {x : x ∈ A ∧ x ∈ B} ⇒ x ∈ A ∧ x ∈ B ⇒ x ∈ A That is, A ∩ B = A. (⇐)Suppose that A ∩ B = A. Then, x ∈ A ⇒ x ∈ A ∩ B = {x : x ∈ A ∧ x ∈ B} ⇒ x ∈ A ∧ x ∈ B ⇒ x ∈ B That is, A ⊂ B. Exercise 4.3. We wish show that (A ∩ B) ∪ C = A ∩ (B ∪ C) ⇐⇒ C ⊂ A: We first show (1) in the following, noting that (2) was proved in the text; From page fifteen, A ∩ (B ∪ B) = (A ∩ B) ∪ (A ∩ C) (1) and, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Proof. (⇒)

8

(2)

Suppose that not all of A, B, C are the emptyset. Then, x ∈ A ∩ (B ∪ C) ⇒ x ∈ A ∧ x ∈ B ∪ C Now, either x ∈ B or x ∈ C: x ∈ B ∧ x ∈ A ⇒ x ∈ A ∩ B ⇒ (A ∩ B) ∪ (A ∩ C) x ∈ C ∧ x ∈ A ⇒ x ∈ A ∩ C ⇒ (A ∩ B) ∪ (A ∩ C) Either way, x ∈ (A ∩ B) ∪ (A ∩ C). To prove the reverse inclusion, we note that similar logic holds; either x ∈ A ∩ B or x ∈ A ∩ C.

Now, we show that (A ∩ B) ∪ C = A ∩ (B ∪ C) ⇐⇒ C ⊂ A: Proof. (⇐)

Suppose that C ⊂ A. Then, for all x ∈ C, x ∈ A; By Exercise 4.1.5 and the proof above, we have the following: (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) = A ∩ (B ∪ C)

(⇒) Suppose that (A ∩ B) ∪ C = A ∩ (B ∪ C). Then, x ∈ C ⇒ x ∈ {x : x ∈ A ∩ B ∨ x ∈ C} = (A ∩ B) ∪ C = A ∩ (B ∪ C) by the assumption. Thus, x ∈ C ⇒ x ∈ A ∩ (B ∪ C) ⇒x∈A∧x∈B∪C ⇒x∈A That is, C ⊂ A.

9

(3)

5

Complements and Powers

Throughout these exercises we assume that ”all sets are subsets of one and the same set E, and that all complements (unless otherwise specified) are formed relative to that E” (Halmos). Likewise, we will often use the more convenient, yet less accurate notation {x : S(x)} in place of {x ∈ E : S(x)} Exercise 5.1. We prove the following: A − B = A ∩ Bc A ⊂ B ⇐⇒ A − B = ∅

(1) (2)

A − (A − B) = A ∩ B

(3)

A ∩ (A − C) = (A ∩ B) − (A ∩ C)

(4)

c

(5)

A ∩ B ⊂ (A ∩ C) ∪ (B ∩ C ) c

(A ∪ C) ∩ (B ∪ C ) ⊂ A ∪ B

(6)

Proof. (1) ( ⇐⇒ )

A − B = {x ∈ E : x ∈ A ∧ x ∈ / B} ⇐⇒ x ∈ A ∧ x ∈ /B ⇐⇒ x ∈ A ∧ x ∈ B c ⇐⇒ x ∈ A ∩ B c

Proof. (2) (⇒) Suppose that A − B 6= ∅, but A ⊂ B; since A ⊂ B, if x ∈ A, x ∈ B. Thus, if x ∈ A − B = {x ∈ E : x ∈ A ∧ x ∈ / B}, x ∈ / B. However x ∈ A, which implies x ∈ B: A contradiction.

(⇐) Suppose that A − B = ∅ and that there exists some x ∈ A which is not in B, i.e. A 6⊂ B. However this is a contradiction since x∈A∧x∈ / B ⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ / B} = A − B and yet A − B = ∅. Proof. (3) ( ⇐⇒ ) 10

By the proof above, A − (A − B) = A − (A ∩ B c ) = {x ∈ E : x ∈ A ∧ x ∈ / A ∩ Bc} ⇐⇒ x ∈ A ∧ x ∈ / A ∩ Bc ⇐⇒ x ∈ A ∧ x ∈ / Bc ⇐⇒ x ∈ A ∧ x ∈ B ⇐⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ B} = A ∩ B

Proof. (4) ( ⇐⇒ )

x ∈ A ∩ (B − C) ⇐⇒ x ∈ A ∧ x ∈ (B − C) ⇐⇒ x ∈ A ∧ x ∈ {x ∈ E : x ∈ B ∧ x ∈ / C} ⇐⇒ x ∈ A ∧ x ∈ B ∧ x ∈ /C ⇐⇒ x ∈ / {x ∈ E : x ∈ A ∧ x ∈ C} ∧ x ∈ A ∩ B ⇐⇒ x ∈ / A∩C ∧x∈A∩B ⇐⇒ x ∈ {x ∈ E : x ∈ / A ∩ C ∧ x ∈ A ∩ B} ⇐⇒ x ∈ (A ∩ B) − (A ∩ C)

Proof. (5) (⇒) Consider C ⊂

S

C, as stated above.

x ∈ A ∩ B ⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ B} ⇒x∈A∧x∈B

(1)

⇒ (x ∈ A ∧ x ∈ B) ∧ (x ∈ C ∨ x ∈ C c )

x ∈ C ∧ (1) ⇒ x ∈ A ∩ C ∧ x ∈ B ∩ C

(2)

x ∈ C c ∧ (1) ⇒ x ∈ A ∩ C c ∧ x ∈ B ∩ C c

(3)

Thus, either way, x ∈ (A ∩ C) or x ∈ (B ∩ C c ), i.e. x ∈ (A ∩ C) ∪ (A ∩ C c ) Proof. (6) (⇒)

11

x ∈ (A ∪ C) ∩ (B ∪ C c ) ⇒ x ∈ A ∪ C ∧ x ∈ B ∪ C c ⇒ (x ∈ A ∨ x ∈ C) ∧ (x ∈ B ∨ x ∈ C c )

(1)

We note that either x ∈ C, or x ∈ C c : x ∈ C ∧ (1) ⇒ x ∈ / C c ∧ (x ∈ B ∨ x ∈ A) ⇒x∈A∪B x ∈ C c ∧ (1) ⇒ x ∈ / C ∧ (x ∈ A ∨ x ∈ B) ⇒x∈A∪B

(2)

(3)

Either way, x ∈ A ∪ B. Exercise 5.2. We show that P(E) ∩ P(F ) = P(E ∩ F ): Proof. Let E, F be sets such that E, F ⊂ E ; via Axiom of Powers, P(E), P(F ); specifically, from the discussion in Halmos1 , P(E) = {X : X ⊂ E}, and P(F ) = {X : X ⊂ F }. Then, we have the following: X ∈ P(E) ∩ P(F ) ⇐⇒ X ∈ P(E) ∧ X ∈ P(F ) ⇐⇒ X ∈ {Y : Y ⊂ E ∧ Y ⊂ F } ⇐⇒ X ⊂ E ∧ X ⊂ F ⇐⇒ ∀x(x ∈ X ⇒ (x ∈ E ∧ x ∈ F )) ⇐⇒ ∀x ∈ X, x ∈ E ∩ F ⇐⇒ X = {x ∈ E : x ∈ X} ⊂ E ∩ F ⇐⇒ X ∈ {Y : Y ⊂ E ∩ F } = P(E ∩ F ) Exercise 5.3. We show that P(E) ∪ P(F ) ⊂ P(E ∪ F ): Proof. Let E, F be given as in the discussion of proof 5.2.0. We first claim that if A, B are sets, then (x ⊂ A ∨ x ⊂ B ⇒ x ⊂ A ∪ B). Indeed; if x ⊂ A, then x ⊂ A ∪ B and likewise, x ⊂ B implies x ⊂ B ∪ A. With this in mind, we have the following: X ∈ P(E) ∪ P(F ) = {X : X ∈ P(E) ∨ X ∈ P(F )} ⇐⇒ X ∈ P(E) ∨ X ∈ P(F ) ⇐⇒ X ∈ {Y : Y ⊂ E} ∨ X ∈ {Y : Y ⊂ F } ⇐⇒ X ∈ {Y : Y ⊂ E ∨ Y ⊂ F } ⇒ X ∈ {Y : Y ⊂ E ∪ F } = P(E ∪ F ) 1 Halmos,

P. ”Naive Set Theory”. pg 19

12

T

Exercise 5.4. We show that

X∈C

P(X) = P(

T

X∈C

X):

Proof. Let C be a non-empty collection of sets. (⇒) z∈

\

P(X) ⇒ (∀X ∈ C)(z ∈ P(X))

X∈C

⇒ (∀X ∈ C)({z} ⊂ P(X)) ⇒ (∀X ∈ C)(z ∈ X) \ ⇒z∈ X X∈C

⇒ {z} ⊂

\

X

X∈C

⇒ z ∈ P(

\

X)

X∈C

(⇐) z ∈ P(

\

X) ⇒ z ∈ {Y : Y ⊂

X∈C

\

X}

X∈C

⇒z⊂

\

X

X∈C

⇒ (∀X ∈ C)(z ⊂ X) ⇒ (∀X ∈ C)(∀v ∈ z)(v ∈ X) ⇒ (∀X ∈ C)(∀v ∈ z)({v} ⊂ P(X)) ⇒ (∀X ∈ C)(∀v ∈ z)(v ∈ P(X)) ⇒ (∀X ∈ C)(z ∈ P(X)) \ ⇒z∈ P(X) X∈C

And this completes the proof. Exercise 5.5. We show that

S

X∈C

P(X) ⊂ P(

S

X∈C

X):

Proof. Let C be a collection of sets. A common fact used in the following proof is that if x ∈ P(X), then x ⊂ X for any set X. This is indeed true, since

13

P(X) = {Y : Y ⊂ X}. With this in mind, we have [ z∈ P(X) ⇒ (∃Y ∈ C)(z ∈ P(Y )) X∈C

⇒z⊂Y ∈C ⇒ z ⊂ {x : x ∈ X(∀X ∈ C)} =

[

X

X∈C

⇒ (∀s ∈ z)(s ∈

[

X)

X∈C

⇒ (∀s ∈ z)(s ∈ P(

[

X))

X∈C

⇒ z ∈ P(

[

X)

X∈C

Which completes the proof. Exercise 5.6. We show that

T

X∈P(E)

X = ∅:

Proof. Let E be a set and consider P(E) = {Y : Y ⊂ E}. We note that this is non-empty because ∅ ∈ P(E), T since ∅ is a subset of every set. We also note that the statement (x ∈ ∅ ⇒ x ∈ X∈P(E) X) is vacuously satisfied since ∅ has no elements. Thus, we only show the forward inclusion: (⇒) x∈

\

X ⇒ (∀X ∈ P(E))(x ∈ X)

X∈P(E)

⇒ x ∈ ∅ ∈ P(E) And this completes the proof. Exercise 5.7. We show that (E ⊂ F ⇒ P(E) ⊂ P(F )): Proof. Suppose that E ⊂ F for sets E, F . Then, we have the following: e ∈ P(E) ⇒ e ⊂ E ⇒ (∀x ∈ e)(x ∈ E) ⇒ (∀x ∈ e)(x ∈ F )

(by Assumption)

⇒e⊂F ⇒ e ∈ P(F )

(by Definition)

Which completes the proof. Exercise 5.8. We show that E =

S

P(E) =

S

X∈P(E)

Proof. Let E be a set. Then, we have the following: 14

X:

(⇐) x∈

[

P(E) ⇒ (∃Y ∈ P(E))(x ∈ Y ) ⇒x∈Y ⊂E ⇒x∈E

(⇒) x ∈ E ⇒ {x} ∈ P(E) ⇒ x ∈ {x} ∈ P(E) [ ⇒ x ∈ {X : X ∈ P(E)} [ = P(E)

Which completes the proof. ´S ¯ E is a set that includes E, typically, as a Exercise 5.9. We show that P proper subset. ´S ¯ E = P(E) = {X ∈ E : X ⊂ E}. If E is non-empty, Proof. Indeed, P then it is clear that E Ĺ P(E). However, if E = ∅, then P(E) = ∅ and E = P(E).

15

6

Ordered Pairs

Throughout this section, we will consider A, B, C, X, Y to be any sets, and will use the notation {x : S(x)} in place of the more accurate, yet cumbersome S notation {x ∈ C : S(x)}. Exercise 6.1. We show the following: (A ∪ B) × X = (A × X) ∪ (B × X)

(1)

(A ∩ B) × (X ∩ Y ) = (A × X) ∩ (B × Y )

(2)

(A − B) × X = (A × X) − (B × X)

(3)

(A = ∅ ∨ B = ∅) ⇐⇒ (A × B = ∅) (A ⊂ X) ∧ (B ⊂ Y ) ⇐⇒ (A × X ⊂ X × Y )

(4) (5)

Proof. (1) (⇒) (x, y) ∈ (A ∪ B) × X ⇐⇒ (x, y) ∈ {(a, b) : a ∈ A ∪ B ∧ b ∈ X} ⇐⇒ x ∈ A ∪ B ∧ y ∈ X ⇐⇒ (x ∈ A ∨ x ∈ B) ∧ y ∈ X (by Distributivity)

⇐⇒ (x ∈ A ∧ y ∈ X) ∨ (x ∈ B ∧ y ∈ X) ⇒ (x ∈ A ∪ X ∧ y ∈ A ∪ X) ∨ (x ∈ B ∪ X ∧ y ∈ B ∪ X)

(by Remarks in Text)

⇒ {x, {x, y}} ⊂ P(A ∪ X) ∨ {x, {x, y}} ⊂ P(B ∪ X) ⇐⇒ (x, y) ∈ P(P(A ∪ X)) ∨ (x, y) ∈ P(P(B ∪ X))

(by Remarks in Text)

⇐⇒ (x, y) ∈ (A × X) ∪ (B × X)

(⇐) (x, y) ∈ (A × X) ∪ (B × X) ⇐⇒ (x, y) ∈ (A × X) ∨ (x, y) ∈ (B × X) (by Definition)

⇒ (x ∈ A ∧ y ∈ X) ∨ (x ∈ B ∧ y ∈ X) ⇐⇒ (x ∈ A ∨ x ∈ B) ∧ y ∈ X ⇐⇒ x ∈ A ∪ B ∧ y ∈ X ⇒ x ∈ (A ∪ B) ∪ X ∧ y ∈ (A ∪ B) ∪ X ⇒ {{x}, {x, y}} ⊂ P((A ∪ B) ∪ X) ⇐⇒ {{x}, {x, y}} ∈ P(P((A ∪ B) ∪ X)) ⇒ (x, y) ∈ (A ∪ B) × X

Which completes the proof. 16

Proof. (2) (⇒) (c, d) ∈ (A ∩ B) × (X ∩ Y ) ⇐⇒ c ∈ A ∩ B ∧ d ∈ X ∩ Y ⇐⇒ (c ∈ A ∧ c ∈ B) ∧ (d ∈ X ∧ d ∈ Y ) ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ B ∧ d ∈ Y )

(by Commutativity)

⇒ (c ∈ A ∪ X ∧ d ∈ A ∪ X) ∧ (c ∈ B ∪ Y ∧ d ∈ B ∪ Y ) ⇒ {{c}, {c, d}} ⊂ P(A ∪ X) ∧ {{c}, {c, d}} ⊂ P(B ∪ Y ) ⇐⇒ (c, d) ∈ P(P(A ∪ X)) ∧ (c, d) ∈ P(P(B ∪ X)) ⇐⇒ (c, d) ∈ A × X ∧ (c, d) ∈ B × X ⇐⇒ (c, d) ∈ (A × X) ∩ (B × X)

(⇐) (c, d) ∈ (A × X) ∩ (B × X) ⇐⇒ (c, d) ∈ (A × X) ∧ (c, d) ∈ (B × X) ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ B ∧ d ∈ X) (by Commutativity)

⇐⇒ (c ∈ A ∧ c ∈ B) ∧ d ∈ X ⇐⇒ (c ∈ (A ∩ B)) ∧ d ∈ X ⇒ c ∈ (A ∩ B) ∪ X ∧ d ∈ (A ∩ B) ∪ X ⇒ {{c}, {c, d}} ⊂ P((A ∩ B) ∪ X) ⇒ (c, d) ∈ P(P((A ∩ B) ∪ X)) ⇐⇒ (c, d) ∈ (A ∩ B) × X

Which completes the proof. Proof. (3) (⇒) (c, d) ∈ (A − B) × X ⇐⇒ c ∈ (A − B) ∧ d ∈ X ⇐⇒ (c ∈ A ∧ c ∈ / B) ∧ d ∈ X ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B ∧ d ∈ X) (since c ∈ / B)

⇒ (c, d) ∈ A × X ∧ (c, d) ∈ / B×X ⇐⇒ (c, d) ∈ (A × X) − (B × X)

17

(⇐) (c, d) ∈ (A × X) − (B × X) ⇐⇒ (c, d) ∈ (A × X) ∧ (c, d) ∈ / (B × X) ⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B∨d∈ / X) (since (d ∈ X ∧ d ∈ / X, →←))

⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B) ⇐⇒ (c ∈ A ∧ c ∈ / B) ∧ d ∈ X ⇐⇒ c ∈ (A − B) ∧ d ∈ X ⇒ {{c}, {c, d}} ⊂ P((A − B) ∪ X) ⇐⇒ {{c}, {c, d}} ∈ P(P((A − B) ∪ X)) ⇐⇒ (c, d) ∈ (A − B) × X

Which completes the proof. Proof. (4) (⇒) Suppose, without loss of generality, that A = ∅. Then, A contains no elements. Consider A × B = {(a, b) : a ∈ A ∧ b ∈ B}. Then, since there are no a ∈ A, {(a, b) : x ∈ A ∧ x ∈ B} = ∅. (⇐) Suppose to the contrary that A 6= ∅ and B 6= ∅, but A × X = ∅. Since A 6= ∅ and B 6= ∅, there exists some a ∈ A and b ∈ B. And so, {{a}, {a, b}} ⊂ P(A ∪ B) and further we have (a, b) ∈ A × B. However, this is a contradiction since we assumed A × B = ∅. Proof. (5) (⇒) We show that if A ⊂ X and B ⊂ Y , then A×B ⊂ X ×Y . Since A ⊂ X, for all a ∈ A, a ∈ X and likewise for B. (c, d) ∈ A × B ⇐⇒ c ∈ A ∧ d ∈ B (by Assumption)

⇒c∈X ∧d∈Y ⇒c∈X ∪Y ∧d∈X ∪Y ⇒ {{c}, {c, d}} ⊂ P(X ∪ Y ) ⇒ (c, d) ∈ P(P(X ∪ Y )) ⇐⇒ (c, d) ∈ X × Y

18

(⇐) Suppose that A × X 6= ∅. And thatA × B ⊂ X × Y . We show that A ⊂ X and B ⊂ Y . a ∈ A ∧ b ∈ B ⇒ {{a}, {a, b}} ∈ P(P(A ∪ B)) ⇐⇒ (a, b) ∈ A × B ⇒ (a, b) ∈ X × Y ⇒ (a, b) ∈ {(x, y) : x ∈ X ∧ y ∈ Y } ⇒a∈X ∧b∈Y

Which completes the proof.

19

7

Relations

Exercise 7.1. We show that for each of symmetric, reflexive, and transitive properties, there exists a relation that does not have that property but does have the other two. Proof. Let X be a set. We first show that if X = ∅, then any relation in X is an equivalence relation: Lemma. Let X = ∅. Then, by a previous theorem, we have X × X = ∅. And so, if R is a relation in X, R = ∅. However, R is an equivalence relation, since all statements are vacuously satisfied. Likewise, suppose that X = {a}; i.e. X is a singleton. We claim, also that any relation in X is an equivalence relation: Lemma. Let X = {a}. Then, by the comments of the previous section, we have X ×X = {(a, a)}. Thus, since R is the set {(a, b) : a ∈ X ∧ b ∈ X} it follows that R = {(a, a)} for any relation.2 It is easy to verify that the transitive condition is vacuously satisfied for R. And, it is symmetric, since (a, a) = (a, a). And, clearly, R is reflexive. Thus, R is an equivalence relation. We also show that for X = {a, b} each of which is distinct, there does not exist a relation, which is symmetric, reflexive, but not transitive: Lemma. Let X = {a, b}. Then, the relation R in question must contain {(a, a), (b, b)} = I. However, it is easy to verify that this is an equivalence relation.3 Thus, we must add an element of X × X which maintains its symmetry, yet is not transitive. However, this is not possible since either way (a, b), (b, a) must be in R and so we would have R = {(a, a), (b, b), (a, b), (b, a)}.And, this is an equivalence relation since (x, x) ∈ R for all x ∈ X and, by the comments above, R is symmetric, as well as transitive.4 Thus, let X = {a, b, c....}, for distinct elements. We first show the relational properties which include reflexivity: 2 see

Halmos pg. 27 the above lemma 4 this is left to the reader 3 see

20

Let R be a relation in X which is reflexive. Then, we must have I = {(x, x) : x ∈ X} ⊂ R, by definition. Thus, we show a representation of R which is transitive, but not symmetric, and symmetric but not transitive: • Consider R = I ∪{(a, b), (b, a), (b, c), (c, b)}. Then, this relation is symmetric, since (x, x) = (x, x) for all x ∈ R and {(a, b), (b, a), (b, c), (c, b)} ⊂ R. However, R is not transitive, since (a, b) ∈ R and (b, c) ∈ R and yet (a, c) ∈ / R. • Consider R = I ∪ {(a, b)}. Then, R is transitive; note, we only need to check (a, b), (a, a), (b, b) for transitivity, since they are the only ordered pairs which share a first, or second coordinate, where only (a, a), or (a, b) come first. Correctly, (a, b) ∈ R and (b, b) ∈ R implies (a, b) ∈ R; (a, a) ∈ R and (a, b) ∈ R implies (a, b) ∈ R. However, R is not symmetric: (a, b) ∈ R, however, (b, a) ∈ / R. Finally, we show the relation in X which is not reflexive. Let X be given as above: • Consider R = {(a, b), (b, a), (a, a), (b, b)} ⊂ X ×X. This is symmetric since (x, x) = (x, x) for all x ∈ X and {(a, b), (b, a)} ⊂ R. It is, also, reflexive since (a, b) ∈ R and (b, a) ∈ R implies (a, a) ∈ R; likewise, (b, a) ∈ R and (a, b) ∈ R implies (b, b) ∈ R.5 However, R is not reflexive since c ∈ X, yet (c, c) ∈ / R.

Exercise 7.2. We show that the set of all equivalence classes, X/R, is indeed a set. Proof. Let R be an equivalence relation on a set X. We show that X/R ⊂ P(X): z ∈ X/R ⇒ (∃x ∈ X)(z = {x/R} = {y ∈ X : (x, y) ∈ R} ⊂ X) ⇒ z ∈ P(X) ⇒ {z} ⊂ X

Which completes the proof.

5 the

others follow

21

8

Functions

Exercise 8.1. We show the special cases in which the projections are 1-1. Proof. Without loss of generality, we only consider g : X × Y → Y , the range projection. • We first note that if X, Y have a pair or more of distinct elements, then g is not 1-1: Lemma 8.1. Suppose X, Y , have more than a pair of distinct elements, then g is not 1-1: To see this, let {a, b} ⊂ X, and {c, d} ⊂ Y , then {(a, c), (a, d), (b, c), (b, d)} ⊂ X × Y . If we suppose g is one-to-one, then g((a, c)) = g((b, d)) implies (a, c) = (b, d). But this is a contradiction, since a = b and yet a, b are distinct. • Next, if X = {x} and Y = {y}, then we claim g is 1-1. Lemma 8.2. Let X = {x} and Y = {y}; then, X × Y = {(x, y), (y, x)}. Upon inspection, we see that g is 1-1: if (x, y) 6= (y, x), then y 6= x and y = g((x, y)) 6= x = g((y, x). We note that this covers the case in which x = y. • Next, we show that if either X or Y is a singleton and the other is an arbitrary set, g is one-to-one. Lemma 8.3. Suppose, without loss of generality, that X is the singleton. Then either Y = ∅ or Y 6= ∅. If Y = ∅, then X × Y = ∅, by previous exercise; which is trivially 1-1. Else, we assume Y has more than a pair of distinct elements. Since X = {x} for some x, it follows that {(a, b) : a ∈ X ∧ b ∈ Y } = {(x, b) : b ∈ Y }, which means g is 1-1. • Lastly, if X = Y = ∅, then all conditions are vacuously satisfied, and so g is 1-1.

Exercise 8.2. We show that Y ∅ has exactly one element, namely ∅, whether Y is empty or not; i.e. Y ∅ = {∅}. Proof. Suppose that Y is a set. Consider f = Y ∅ . From page thirty, and by a previous theorem, we have f ⊂ P(∅ × Y ) = P(∅) = {∅}, where f is a set of elements which are ordered pairs. From the comments in section seven, ∅ is a set of ordered pairs. It follows that f = {∅}. Exercise 8.3. We show that if X 6= ∅, then ∅X is empty; i.e. ∅X = ∅. Proof. Suppose that there is some function f ∈ ∅X . Then, for all x ∈ X, there exists some b ∈ ∅ such that (x, b) ∈ f . However, this is not true, since ∅ contains no elements. Thus, ∅X = ∅.

22

9

Families

Exercise 9.1. After the comments on page thirty-five, we formulate and prove a generalized version of the commutative law for unions. Proposition 9.1. Let K be an arbitrary, non-empty set, and let p ∈ Aut(K K ) be an element of the set of one-to-one and onto functions from K to K. We note that in the current notation this means that p is the family and {pi } denotes the values of p evaluating some S i ∈ K. SLet {Ak } be a family of functions with domain K. We claim that k∈K Ak = i∈K Api . Proof. (⇒) x∈

[

Ak ⇒ (∃k0 ∈ K)(x ∈ Ak0 )

k∈K

(since p is onto)

⇒ (x ∈ Ak0 ) ∧ (∃s ∈ K)(k0 = ps ) ⇒ x ∈ Ak 0 = Ap s [ ⇒x∈ Ap i i∈K

(⇐) [

x∈

Api ⇒ (∃s ∈ K)(x ∈ Aps )

i∈K

(since ps ∈ K and onto)

⇒ (x ∈ Aps ) ∧ (∃k0 ∈ K)(ps = k0 ) ⇒ x ∈ Aps = Ak0 [ ⇒x∈ Ak k∈K

Which completes the proof. (

1 i=0 ; it 0 i=1 is easy to show that p is 1-1 and onto. From the statement above, we have [ Ak = A0 ∪ A1 (1) To illustrate the previous proof, consider K = {0, 1} and pi =

k∈K

[

Api = Ap0 ∪ Ap1 = A1 ∪ A0

i∈K

23

(2)

Exercise 9.2. We show that if both {Ai } and {Bi } are families of sets, with domains I, J, then ´[ ¯ ´[ ¯ [ Ai ∩ Bj = (Ai ∩ Bj ) (1) i

´\

j

¯ Ai ∪

i

Where the symbols such as

´\

i,j

¯ Bj

\ = (Ai ∪ Bj )

j

T

i,j

mean

(2)

i,j

T

(i,j)∈I×J

Proof. (1)

( ⇐⇒ ) x∈

´[

¯ ´[ ¯ ´[ ¯ ´[ ¯ Ai ∩ Bj ⇐⇒ x ∈ Ai ∧ x ∈ Bj

i

j

i

j

⇐⇒ (∃i0 ∈ I)(x ∈ Ai0 ) ∧ (∃j0 ∈ J)(x ∈ Bj0 ) ⇐⇒ (∃(i0 , j0 ) ∈ I × J)(x ∈ (Ai0 ∩ Bj0 )) [ ⇐⇒ x ∈ (Ai ∩ Bj ) i,j

Proof. (2)

( ⇐⇒ ) x∈

´\ i

¯ ´\ ¯ ´\ ¯ ´\ ¯ Ai ∪ Bj ⇐⇒ x ∈ Ai ∨ x ∈ Bj j

i

j

⇐⇒ (∀i ∈ I)(x ∈ Ai ) ∨ (∀j ∈ J)(x ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(x ∈ Ai ∨ x ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(x ∈ Ai ∪ x ∈ Bj ) ⇐⇒ (∀(i, j) ∈ I × J)(x ∈ (Ai ∪ Bj )) \ ⇐⇒ x ∈ (Ai ∪ Bj ) (i,j)∈I×J

24

Exercise 9.3. Let {Ai }, {Bj } be families of sets with domains I 6= ∅, J = 6 ∅, respectively. We show that ´[ ¯ ´[ ¯ [ Ai × Bj = (Ai × Bj ) (1) i

j

´\

¯ Ai ×

i

´\

i,j

¯ Bj =

j

\

(Ai × Bj )

(2)

i,j

Proof. (1)

( ⇐⇒ ) (a, b) ∈

´[

¯ ´[ ¯ [ [ Ai × Bj ⇐⇒ a ∈ Ai ∧ b ∈ Bj

i

j

i

j

⇐⇒ (∃i0 ∈ I)(∃j0 ∈ J)(a ∈ Ai0 ∧ b ∈ Bj0 ) ⇐⇒ (∃(i0 , j0 ) ∈ I × J)((a, b) ∈ Ai0 × Bj0 ) [ ⇐⇒ (a, b) ∈ (Ai × Bj ) (i,j)∈I×J

Proof. (2) Suppose that for each i ∈ I, Ai 6= ∅ and for each j ∈ J, Bj 6= ∅.

( ⇐⇒ ) (a, b) ∈

´\ i

¯ ´\ ¯ \ \ Ai × Bj ⇐⇒ a ∈ Ai ∧ b ∈ Bj j

i

j

⇐⇒ (∀i ∈ I)(a ∈ Ai ) ∧ (∀j ∈ J)(b ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(a ∈ Ai ∧ b ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)((a, b) ∈ Ai × Bj ) ⇐⇒ (∀(i, j) ∈ I × J)((a, b) ∈ Ai × Bj ) \ ⇐⇒ (a, b) ∈ Ai × B j i,j

Exercise 9.4. Suppose that for each i ∈ I, Xi T is non-empty. SConsider the family of sets {Xi } with domain I. We show that i Xi ⊂ Xj ⊂ i Xi for each index j ∈ I.

25

Proof. Let {Xi } be stated as above, and let j ∈ I be arbitrary. \ x∈ Xi ⇒ (∀i ∈ I)(x ∈ Xi ) i

⇒ x ∈ Xj

(since j ∈ I)

Which proves the first inclusion. For the second, suppose j ∈ I is arbitrary, then we have the following: [ x ∈ Xj ⇒ x ∈ Xj j

⇒x∈

´[

⇒x∈

[

[

Xj ∪

j

¯ Xz

z∈I−J

Xi

i

Which prove the second inclusion. Exercise 9.5. We show the minimality condition of intersection S and unions: (1) That is, if Xj ⊂ Y for each index T j and some set Y , then i Xi ⊂ Y and that it is unique; (2) Likewise, if Y ⊂ j Xj for each index j and some set Y , T then Y ⊂ i Xi and it is unique. Proof. (1) Suppose that {Xi } is a family of sets with domain I, such that j ∈ I, and let Y be a set. Then, we have the following: [ (∀j)(Xj ⊂ Y ) ⇒ Xj ⊂ Y j

⇒

[

Xi ⊂ Y

i

Proof. (2) Suppose that {Xi } is a family of sets with domain I, such that j ∈ I. And let Y be a set. Then, we have the following: \ \ Y ⊂ Xj ⇒ Y ⊂ Xi j

i

26

10

Inverses and Composites

Exercise 10.1. Suppose that f : X → Y . Let {Ai } be a family of subsets of X. Define f 0 : P(X) → P(Y ), as the mapping from each Ai to the image subset, f 0 (Ai ), as stated in the text. We claim that ´[ ¯ [ f0 Ai = f 0 (Ai ) i

i

Proof. After the comments on page thirty-four, we note that A : I → P(X), for some index set I 6= ∅. Then, we have the following: (⇒) x∈

[

(since Ai0 ⊂

[

f 0 (Ai ) ⇒ (∃i0 ∈ I)(x ∈ f 0 (Ai0 ))

i

Ai )

⇒ x ∈ f0

´[

i

¯ Ai

i

(⇐) x ∈ f0

´[

¯ Ai ⇒ (∃i0 ∈ I)(x ∈ f 0 (Ai0 ))

i

´

¯ [ f 0 (Az ) ∪ f 0 (Ai0 ) = f 0 (Ai )

[

⇒x∈

i

z∈I−{i0 }

And the result follows. Exercise 10.2. We show that, in general, ´\ ¯ \ f0 Ai = f 0 (Ai ) i

i

does not hold. The following proof was suggested by David. H.. Proof. Consider X = {a, b, c} and Y = {d, e} and suppose all the elements are

27

distinct. Then define f 0 : P(X) → P(Y ) as the following: f 0 (∅) = ∅ f 0 ({a}) = {d} f 0 ({b}) = {e} f 0 ({c}) = {d} f 0 ({a, b}) = {d, e} f 0 ({b, c}) = {d} f 0 ({a, c}) = {e} 0

f ({a, b, c}) = ∅

Then, let {Ai } be a family of subsets of X with domain I = 2, such that A0 = {a} and A1 = {a, c}. Then, we have the following: ´\ ¯ f0 Ai = f 0 ({a}) = {d} i

\

f 0 (Ai ) = f 0 (A0 ) ∩ f 0 (A1 ) = {d} ∩ {e} = ∅

i

Which completes the proof. Exercise 10.3. Let f : X → Y . We show that f maps X onto Y iff for every non-empty subset B of Y , f −1 (B) 6= ∅. Proof. (⇒) If f maps X onto Y , then for every y ∈ Y , there exists an x ∈ X such that f (x) = y. Since this is the case, if B is a non-empty subset of Y , then for any b ∈ B, b ∈ Y . And we have that there exists an x0 ∈ X such that f (x0 ) = b; therefore, f −1 (B) = {x ∈ X : f (x) ∈ B} = 6 ∅. (⇐) Suppose that f −1 (B) 6= ∅ for any B ⊂ Y . In particular, for each y ∈ Y , {y} ⊂ Y and f −1 ({y}) = {x ∈ X : f (x) = y} = 6 ∅. The result follows immediately. Exercise 10.4. In the context of page forty-one, R := ”son”, S := ”brother”, what do R−1 , S −1 , RS and R−1 S −1 mean? Proof. (R−1 ): From the comments on page forty-one, xRy implies that that x is son-related to y. That is, y is the father of x. Thus, yR−1 x means that y has the father relation to x. (S −1 ): Likewise, xSy means that x bears the brother-relation to y. That is, x is the brother of y. Thus, yS −1 x implies y is the brother of x. 28

(T = R ◦ S): Also, by the previous comments, if xT z, then there exists y such that xSy and yRz. Thus, x is the brother of y and z is the son to y. That is, x is the uncle of y. (T = R−1 ◦ S −1 ): Likewise, if xT z, then there exists a y such that xS −1 y and yR−1 z. Thus, x is the brother of y and y is the father to z. Thus, x is the uncle of z. Exercise 10.5. Assume that for each of the following f : X → Y . We show the following: 1. If g : Y → Z such that g ◦ f is the identity on X, then f is 1-1 and g maps Y onto X. 2. f (A ∩ B) = f (A) ∩ f (B) for all A, B ⊂ X iff f is 1-1. 3. f (X − A) ⊂ Y − f (A) for all subsets A of X iff f is 1-1. 4. Y − f (A) ⊂ f (X − A) for all subsets A of X iff f maps X onto Y . Proof. (1): Suppose that g : Y → Z such that g ◦ f is the identity on X. Since g ◦ f is the identity on X, for all x ∈ X, g(f (x)) = x. Thus, if f (x0 ) = f (x1 ), then x0 = g(f (x0 )) = g(f (x1 )) = x1 and f is 1-1. To show that g maps Y onto X let x ∈ X, then x = g(f (x)) where f (x) ∈ Y . Thus, g maps Y onto X. Proof. (2): (⇒) By assumption, we have that f (A ∩ B) = f (A) ∩ f (B) for all A, B ⊂ X. Suppose that f (x) = f (y) for some x, y ∈ X. If x = y, then we are done. So, suppose that x 6= y. Letting A = {x}, B = {y}, we have that f (A ∩ B) = ∅ 6= {f (x) = f (y)} = f (A) ∩ f (B), a contradiction. (⇐) Suppose that f is 1-1. If y ∈ f (A ∩ B), then y = f (x) for some x ∈ A ∩ B. Thus, x ∈ A and x ∈ B. This implies that f (x) = y ∈ f (A) and f (x) = y ∈ f (B). Thus, y ∈ f (A) ∩ f (B). Now take y ∈ f (A) ∩ f (B). Then y ∈ f (A) so there is some a ∈ A with f (a) = y. Also, y ∈ f (B), so there is some b ∈ B such that y = f (b). Now f (a) = y = f (b), so f being 1-1 now implies a = b. Thus, a ∈ A ∩ B and f (a) = y ∈ f (A ∩ B). Proof. (3):

29

(⇒) Suppose that f (x) = f (y) for some x, y ∈ X. Let A = {x, y}. Then, f (X − {x, y}) ⊂ Y − f ({x, y}) = Y − {f (x) = f (y)} = Y − {f (x)}.By the comments on page thirty-nine, we have that f −1 f ((X − A)) ⊂ f −1 (Y − {f (x)}) ⇒ X − A = X − f −1 ({f (x)}) ⇒ X − {x, y} = X − {x}

It follows that {x, y} = {x} i.e. y = x. (⇐) Suppose that f is 1-1. And let y ∈ f (X − A) for any non-empty subset A ⊂ X. Then, there exists an x0 ∈ X − A such that f (x0 ) = y. Suppose to the contrary that y ∈ / Y − f (A). Then, f (x0 ) = y ∈ (Y − f (A))c = f (A). This implies that there exists some a ∈ A such that f (x0 ) = y = f (a). However, this is a contradiction since f being 1-1 implies a = x0 ∈ A. Proof. (4): (⇒) Let y ∈ Y and consider an arbitrary subset A of X. Now, either y ∈ f (A) or it is not. If it is, we are done. So suppose that it is not. Then, y ∈ Y − f (A) ⊂ f (X − A) and so, there exists some x0 ∈ X − A such that y = f (x0 ); and the result follows. (⇐) Let A ⊂ X and suppose that y ∈ Y − f (A). Since f maps X onto Y , there exists some x ∈ X such that y = f (x) ∈ Y − f (A). Thus, x ∈ / A and so x ∈ Ac = X − A. It follows from he definition that y = f (x) ∈ f (X − A) and the result follows.

30

11

Numbers

Exercise 11.1. Let {Ai } be a family of non-empty successor sets for some non-empty domain I. We show that \ Ai i

is a successor set. Proof. set, and so 0 ∈ Ai . This implies that T For all i ∈ I, Ai is a successor T 0 ∈ i Ai . Likewise, suppose that n ∈ i Ai ; then n ∈ Ai for all iT∈ I and so n+ ∈ Ai for each i ∈ I; from the definition of intersection n+ ∈ i Ai . This completes the proof.

31

12

Peano Axioms

Exercise 12.1. We prove that if n ∈ ω then n 6= n+ . Proof. Suppose, to the contarary, that n = n+ for some n ∈ ω. Then, from the axiom of extension we have that n+ ⊂ n. However, then we would have that n+ = n ∪ {n} ⊂ n implying that n ∈ n which is a contradiction. Exercise 12.2. Let n ∈ ω. We show that if n 6= 0, then n = m+ for some m ∈ ω — We note that this show any element n ∈ ω has a predecessor. Proof. We proceed by mathematical induction: Let S be that subset of ω for which if n 6= 0, then n = m+ for some m ∈ ω. We note that 1 ∈ S, since 1 = 0+ and 0 ∈ ω. Suppose that for some n ∈ ω, n = m+ for some m ∈ ω; indeed, if n+ = m for some m ∈ ω we have (n+ )+ = m+ = z for some z ∈ ω by the induction hypothesis. This completes the proof. Exercise 12.3. We prove that ω is transitive. That is, if x ∈ y and y ∈ ω, then x ∈ ω. Proof. We proceed by mathematical induction. Let S be the set of all z ∈ ω such that if x ∈ z, then x ∈ ω. We note that {} = 0 ∈ S, vacuously. Thus, let y ∈ S. We show that y + ∈ S. That is, let x be any element in y + : x ∈ y + = y ∪ {y} . Then, either x ∈ y or x = y. In the former cases, we that x ∈ ω, by assumption; in the later case we have that y = x ∈ S. Thus, ω = S and so ω is transitive. Exercise 12.4. We show the following: if E is non-empty subset of some natural number, then there exists an element k ∈ E such that k ∈ m whenever m is an element of E and distinct from k — This says that any finite set has a lower bound, namely a greatest lower bound. Proof. We proceed by mathematical induction. Let S be the set of z ∈ ω such that the above holds. We note that {} = 0 ∈ S, vacuously. Suppose that n ∈ S for some n ∈ ω; we show that n+ ∈ S. Suppose that E is a non-empty subset of n+ = n ∪ {n}. If E is a singleton, then we are done; the above is vacuously satisfied. We note that if E = n, then we are done, by the induction hypothesis. Thus, there exists some k ∈ E, as specified above. We claim that this k, so specified, will work even if n ∈ E; Indeed from a previous proof, we know that no element is equal to its successor and so it follows that for any a ∈ E, a ∈ n. In particular, this is true for k. Therefore, S = ω.

32

13

Arithmetic

In many of these exercises, we use induction. As pointed out in the Axiom of Substitution, and the conversation I had with S. Bleiler, and later sections, it is still valid to start induction at some x ∈ ω; we have used this fact a couple times. Exercise 13.1. Prove that if m < n for some natural numbers m, n, then m + k < n + k for any k ∈ ω. Proof. We proceed by mathematical induction. Suppose that for some natural numbers m, n, m < n. Let S be the set of all z ∈ ω such that m + z < n + z. We show that S = ω. Indeed; we note that since m + 0 = m and n + 0 = n, it follows that m = m + 0 < n + 0 = n, and so, 0 ∈ S. Now, suppose that k ∈ S for some k ∈ ω. We show that k + ∈ S. Note, sn (k + ) = (sn (k))+ = sn (k) ∪ {sn (k)}, and by the definition of <, and the induction hypothesis, we have that {sn (k)} ⊂ sm (k) ⊂ (sm (k) ∪ {sm (k)}) = sm (k + ). This implies that sn (k + ) = sn (k) ∪ {sn (k)} ⊂ sm (k) ∪ {sm (k)} = sm (k + ). That is, S = ω Exercise 13.2. Prove that if m < n for some m, n ∈ ω, and k 6= 0, then m · k < n · k. Proof. We proceed by mathematical induction. Suppose that m < n for some m, n ∈ ω. Let S be the set of z ∈ ω, which satisfy the following statement: if z 6= 0, then m · z < n · z. We note that 1 ∈ S. Suppose that k ∈ S for some k ∈ ω; we show that k + ∈ S. Indeed: by definition of product, we have the following: pm (k + ) = pm (k) + m = sm (pm (k))

(1)

< sm (pn (k))

(2)

= pn (k) + m < pn (k) + n

(3)

+

= pn (k ) We only note that (1),(2), and (3) follow from the inductive hypothesis and the previous exercise. Therefore, S = ω. Lemma 13.1. For any n ∈ ω, n is transitive. Proof. Suppose that E = n ∈ ω and y ∈ E. Since y ∈ E, y is a proper subset of E. Now, if x ∈ y, then x is a proper subset of y and it follows directly that x ⊂ y ⊂ E. That is, x is a proper subset of E. That is x ∈ E. Lemma 13.2. If m, n ∈ ω such that m 6= n, then m ∈ n iff m ⊂ n Proof. Let m, n be stated as above. 33

(⇒) Suppose that m ∈ n. We show that m ⊂ n: let x ∈ m, then by the lemma above, x ∈ n. Thus, m ⊂ n. (⇐) Suppose that m ⊂ n. If m = n, then we are done. Otherwise, m 6= n. If this is the case, then n ∈ m cannot happen (for them m would be a subset of one of its elements), and therefor m ∈ n; this is by the comparability of and x, y ∈ ω. Exercise 13.3. We prove that if E is a non-empty set of natural numbers, then there exists an element k ∈ E such that k ≤ m for all m ∈ E. — This statement says that any non-empty subset in ω has a minimum. Proof. Let E ⊂ ω such that E 6= ∅. Since E is non-empty, we can form the intersection of elements in E. Consider \ z= x x∈E

It follows directly from the formulation of z that z ⊂ x, for all x ∈ E: z ≤ x. Thus, it is only necessary to prove that z ∈ E. From the previous lemma we see that since z ∈ x for all x ∈ E, and x ∈ E, that z ∈ E. The result follows. Exercise 13.4. A set E is called finite if it is equivalent to some natural number; otherwise E is infinite. We use this definition to prove that ω is infinite. Proof. Suppose, to the contrary that ω is finite. Then, there exists a 1-1 and onto function, f , from ω to n for some n ∈ ω: f ⊂ ω × n. Since f is 1-1 and onto, for ever y ∈ n, there exists a unique x ∈ ω such that (x, y) ∈ f . Let g be the union of all such x, which is an element of ω. Then, g + = g ∪ {g} is in ω but not in g. Since f is a function, we must have (g + , y1 ) ∈ f for some y1 ∈ n; likewise since f is onto, there exists some x ∈ g such that (x, y1 ) ∈ f . From here we see that we must have (g + , y1 ) = (x, y1 ), implying g + = x; which is a contradiction since g + ∈ / g, whereas x ∈ g. Exercise 13.5. If A is a finite set, then A is not equivalent to a proper subset of A. The contrastive of this statement is that if A is equivalent to a proper subset of A, then A is not finite (infinite). We use this definition to show that ω is infinite. Proof. We show that ω ∼ A, where A = ω − {0}. First, it is clear that A is a non-empty proper subset of ω. Consider the map f ⊂ ω × ω, given by f (n) = n+ . First, we show that f is indeed a function, and then that it is 1-1 and onto: • Well-Defined : Suppose that x = y for some x, y ∈ ω. From a previous theorem, we have x+ = y + . This shows that f is well-defined.

34

• Everywhere-Defined : This follows directly from the definition of ω as being a successor set. This shows that f is everywhere defined. • 1-1 : Suppose that n+ = m+ . Again, from a previous exercise (page 46), we have n = m. This shows that f is 1-1. • Onto: Suppose that n ∈ A = ω − {0}. Then no n ∈ A is 0. Thus, by a previous theorem, we have that n = m+ = f (m) for some m ∈ ω. This implies that f is onto.

Exercise 13.6. We prove that union of a finite set of finite sets is finite. Proof. We proceed by mathematical induction. Let S be a finite set, S such that #(S) = n, for some n ∈ ω. Let B be the set of all N S ∈ ω such that S is finite. Note 0 ∈ B since, if #(S) = 0, then S = {} and {} = {} (which is finite). Next, suppose that k ∈ B: that is, if S is a finite set such that #(S) = k, then S S is finite. We show that k + ∈ B. Suppose that #(S) = k + . Then, there exists some 1-1 and onto function f i.e. f : k + ↔ S. And we have the following: ´[ ¯ [ [ S= f (i) = f (i) ∪ f (k) i∈k+

i∈k

From the comments on page 53, and the induction hypothesis, we see that k + ∈ B. Thus, B = ω. Exercise 13.7. We prove the following: If E is finite, then P(E) is finite and, moreover, #(P(E)) = 2#(E) . Proof. It is only necessary to show that if #(E) = n for some n ∈ ω, then # P(E) = 2#(E) ; we conclude that 2#(E) ∈ ω by a previous exercise. We proceed by mathematical induction. Let S be the set of all N ∈ ω such that #(E) = N and # P(E) = 2#(E) . Note that 0 ∈ S, since #(E) = 0 implies E = {} and #(P(E)) = #({∅, {∅}}) = 2 = 20 . Next, assuming that k ∈ S, we show that k + ∈ S: Suppose that #(E) = k + = k + 1, and consider the set E ∗ = E − {0}. Note, #(E ∗ ) = k. We note that either 0 is in a subset of E, or it is not. If it is not, then we are forming subsets of E ∗ . From the inductive hypothesis, there are ∗ 2#(E ) = 2k subsets of E ∗ . By similar logic, there are 2k subsets which include 0. It follows that there are 2k + 2k = 2 · 2k = 2k+1 = 2#(E) subsets of E. Thus, k + 1 = k + ∈ S. That is, S = ω. Exercise 13.8. If E is a non-empty finite set of natural numbers, then there exists and element k ∈ E such that m ≤ k for all m ∈ E — This statement says that for any finite set E, non-empty, of natural numbers, there is an upper bound, actually a maximum.

35

Proof. Let E be a finite, non-empty, subset of natural numbers. We proceed by mathematical induction. Let S be the set of all N ∈ ω such that #(E) = N and there exists a k ∈ E such that the above holds. Note that 1 ∈ S; 1 = #(E) implies E = n for some n ∈ ω, and clearly n ≤ n. Next, assuming that k ∈ S, for some k ≥ 1, we show that k + ∈ S: Suppose that #(E) = k + = k + 1. Fix some x ∈ E, and consider E ∗ = E − {x}. We note that #(E ∗ ) = k, and so, it follows from the induction hypothesis that there exists some g ∈ E ∗ such that m ≤ g, for all m ∈ E ∗ . By the comparability of elements in ω, we must have that either x ∈ g, or g ∈ x; we only note that x 6= g, since otherwise we would have #(E) = k. In the former case, if x ∈ g, then x < g, and in particular x ≤ g. In the later case, supposing that g ∈ x, we have g < x; and, again, in particular g ≤ x. But then m ≤ g and g ≤ x for all m ∈ E = E ∗ ∪ {x}, implying m ≤ x for all m ∈ E by the transitivity of ≤. This shows that k + ∈ S, and so S = ω.

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14

Order

Exercise 14.1. We express the conditions of antisymmetry and totality for a relation R by means of equations involving R and its inverse — This exercise illustrates that we can determine a partial, or total order, by looking the the co(domains), of a relation. Proof. From the comments on page 54 and the notation in section 10, we adopt the usual assumptions and notation. That is, assume that R is a relation in X. And, we have the following: Antisymmetry: xRy ∧ yRx ⇒ x = y iff

(R(x) = y ∧ R−1 (x) = y) ⇒ R(x) = R−1 (y)

Totality: (∀x, y ∈ X)(xRy ∨ yRx) ⇐⇒ (∀x, y ∈ X)(R(x) = y ∨ R−1 (x) = y)

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15

Axiom of Choice

Exercise 15.1. We prove that every relation includes a function with the same domain. Proof. Suppose that X and Y are sets and let R be a relation from X into Y . By definition, R ⊂ X × Y , and so if X or Y is empty, the X × Y is empty and we are done. Otherwise, there exists at least an element in each of X, Y . Let XR be the subset of X given by the first coordinates of R. Claim: there exists some function f : XR → Y . Let y ∈ Y be arbitrary and consider Ty = {xS∈ XR : xRy}. Let I be the set of z ∈ Y such that Tz 6= ∅. Then, define T = y∈I {Ty } = {Ti }. Assuming the axiom of choice, and by the comments on page sixty, {Ti } is a family of non-empty sets indexed by I and so, there exists a family {ti }, i ∈ I such that ti ∈ Ti for each i ∈ I. Since R is a relation on XR , so defined, it follows that f = {(ti , b) : i ∈ I ∧ b = R(ti )} is a candidate function. Indeed; we verify that f is well-defined and everywhere defined: • Well-Defined : Suppose that ti = tj for some i, j ∈ I. By construction, and the previous supposition, we have that Tti = {x ∈ XR : xRti } = {x ∈ XR : xRtj } = Ttj . Thus, (ti , R(ti )) = (tj , R(tj )) = f (ti ) = f (tj ). • Every-Where Defined : This is clear by construction. Thus, f is a function.

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16

Zorn’s Lemma

Exercise 16.1. We show that Zorn’s Lemma is equivalent to the axiom of choice. Proof. Let X be a set and define F to be the following set of functions:   dom f ⊂ P(X) f ∈ F ⇐⇒ Im f ⊂ X   (∀A ∈ dom f )(f (A) ∈ A) Next, consider the relation ≤, in X, given by the following: for all f1 , f2 ∈ F, f1 ≤ f2 iff f2 is an extension of f1 i.e. f1 ⊂ f2 . As the comments on page 65 point out, ≤ is a partial ordering. We note that (F, ≤) is a partially ordered set. Assuming Zorn’s Lemma, we show that there exists some maximal element f ∈ F: F is already partially ordered. So, let X be a chain in F. Note that XSis a collection of ordered pairs, satisfying the above criterion. S We claim that X is the upper bound for X ; we only need to show that X is a function: S • Function: Suppose that (a, x), (a, y) ∈ X , then (a, x) ∈ f for some f ∈ X and, likewise, (a, y) ∈ g for some g ∈ X . Then, since X is a chain, either f ≤ g, f = g, g ≤ f . Without loss of generality, we S have (a, x), (a, y) ∈ g. Since g is a function, we have x = y. The fact that X is every-where defined is obvious. • The rest of the properties follow by definition. By Zorn’s Lemma, there exists a maximal element in F, call it f . To complete the proof, we show that dom f = P(X) − {∅}: • Suppose that dom f 6= P(X) − {∅}. Then, there exists some a ∈ P (X) such that a ∈ / dom f . Let g = f ∪ {(a, b)}, for any b ∈ X. It is clear that g ∈ F and that g ≥ f . However, this contradicts the minimality of f . We note that using the Axiom of Choice to prove Zorn’s Lemma was shown in text. This completes the proof. Exercise 16.2. We prove that the following is equivalent to the axiom of choice: Every partially ordered set has a maximal chain — This is the Hausdorff Maximality Principal (HMP). Proof. We show that HMP is equivalent to Zorn’s Lemma and use the fact that Zorn’s Lemma is equivalent to the Axiom of Choice to conclude. • (⇒) Suppose the HMP. Let (X, ≤) be the partially ordered set, and A the maximal chain. By the assumptions of Zorn’s Lemma, A has an upperbound, call it a. Now, a ∈ A, since otherwise, we could form B = A ∪ {a}, 39

which is again a totally ordered set, containing A; for all s ∈ B, the elements of A are comparably, likewise, s ≤ a. To complete the proof in this direction, we note that if z ∈ X is any element such that z ≥ a, then z is an upper-bound and must be in A, by the previous discussion. This implies that a ≥ z for all z ∈ X. Specifically, since A is partially ordered, we have a ≤ z and z ≤ a implying a = z. Thus, a is maximal. • (⇐) Suppose Zorn’s Lemma. Let (X, ≤) be a partially ordered set. Consider F be the set of all totally ordered subsets of X. The inclusion order on F is a partial order. Note that if S is a maximal element in F, then it will be a maximal chain in X. We show that any chain A in F has an upper-bound: – (The following was suggested by this example): Indeed, consider A0 = S 0 E∈A E. By definition, A contains all elements of A. We only show 0 that A ∈ F: To show this take x and y, elements of A0 . Then there are sets Ex and Ey , members of A, with x ∈ Ex and y ∈ Ey . Since A is partially ordered by inclusion, Ex ⊂ Ey or Ey ⊂ Ex , that is, either x, y ∈ Ey or x, y ∈ Ex . Since both Ex and Ey are linearly ordered in X then either x ≤ y or y ≤ x. This shows that A0 is a totally ordered set, and so a chain in X i.e. A0 ∈ F. This completes the proof. Exercise 16.3. We show that the following statements are equivalent to the axiom of choice: (i) Every partially ordered set has a maximal chain — This is the Hausdorff Maximality Principal (ii) Every chain in a partially ordered set is included in some maximal chain. (iii) Every partially ordered set in which each chain has a least upper bound has a maximal element. Proof. To do so, we show that i ⇒ ii ⇒ iii ⇒ ZL ⇒ i, noting that Zorn’s Lemma is equivalent to the axiom of choice. Throughout these exercises, we assume that (X, ≤) is the partially ordered set in discussion. • (i ⇒ ii) Suppose that every partially ordered set has a maximal chain. Since (X, ≤) is partially ordered, there exists some maximal chain A ⊂ X. We note that for chains, inclusion is the partial order. Thus, by definition, for all other chains A0 ⊂ X, we must have A0 ⊂ A, since otherwise, A Ĺ A0 , contradicting the maximality of A. • (ii ⇒ iii) Suppose that every chain in a partially ordered set is included in some maximal chain. Let (X, ≤) be such a partially ordered set. Then, there exists some maximal chain A such that A0 ⊂ A for all other chains A0 ⊂ X. We show that each chain A0 of X has a least upper bound: – Let G be the set of all upper-bounds of some chain A0 . Then, we have have A0 ≤ g for all g ∈ G. Likewise G ∩ A0 = A0 implying A0 is the least upper-bound of A0 .

40

From the antecedent in (iii), we see that A has a maximal element in A, call it a. We also note that a is maximal in X: If not, then we can define T = A ∪ {a}, which is a totally ordered set in which A ⊂ T ; contradicting the maximality of A. • (iii ⇒ Zorn’s Lemma) Suppose that every partially ordered set in which each chain has a least upper bound has a maximal element. Let X be such a set. Then for any chain A ⊂ X, A has a l.u.b. call it δA . In particular, δA is an upper-bound. By assumption X has a maximal element. • (Zorn’s Lemma ⇒ i) This was proved in the previous exercise.

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17

Well Ordering

Exercise 17.1. A subset A of a partially ordered set X is cofinal in X in case for each element x ∈ X, there exists an element a ∈ A such that x ≤ a. We prove that every totally ordered set has a cofinal well ordered subset. Proof. Let (X, ≤) be a non-empty totally ordered set. If X has a maximum element, then we are done. Thus, suppose that X has no maximum. Let F be the set of all well ordered subsets of X with the following order: for all f, g ∈ F, f  g iff f is an initial segment of g. We seek to apply Zorn’s Lemma. • From the previous result,  is a partial order on F. • Suppose that S Z is chain in F. We showSthat Z has an upper S bound in F. Consider Z. It is clear that Z  Z. We show that Z is well ordered; Indeed, these are the comments on page 67. We have satisfied the assumptions in Zorn’s Lemma, and so it follows that there exists some maximal element M ∈ F. We claim that M is a well ordered cofinal subset of X. First, it is a subset. Second, it is well ordered by the definition of F. Next, if x ∈ X, then for any m ∈ M , either m ≤ x, x ≤ m or x = m. But, since M is maximal, if m ≤ x, then x ∈ M ; otherwise, we could form N = M ∪ {x} which is a contradiction to the maximality of M . Thus, M is the well ordered cofinal subset of X. Exercise 17.2. Does any such condition apply to partially ordered sets? Proof. Exercise 17.3. We prove that a totally ordered set is well ordered iff the set of strict predecessor of each element is well ordered. Proof. (⇒) Suppose that we have a well ordered set (X, ≤). Note that the set of strict predecessors of some x ∈ X forms a subset. The rest follows from the definition. (⇐) Suppose that (X, ≤) is a partially ordered set and that for any x ∈ X the strict initial segment s(x) is well ordered. Let J be a subset of X. Pick some j0 ∈ J and consider s(j0 ) = {x ∈ X : x < j0 }. If s(j0 ) = ∅, j0 is the least element, and we are done. Otherwise, s(j0 ) 6= ∅. But since s(j0 ) is well ordered, we have J ∩ s(j0 ) ⊂ s(j0 ) and J ∩ s(j0 ) ⊂ J. Implying that J ∩ s(j0 ) has a least element in J, call it z. We note that z is the least element for J; Otherwise, there is some x ∈ J, x 6= z such that x < z < j for all j ∈ J and in particular for j0 . This implies that x ∈ s(j0 ). However, then we must have x ∈ J ∩ s(j0 ) and so x = z; a contradiction.

42

Exercise 17.4. We prove that the well ordering theorem implies the Axiom of Choice. Proof. The following proof was adapted from thisSwebsite: Let C be a collection of nonempty sets. Then S∈C S is a set. By the wellS ordering principle, S∈C S S is well-ordered under some relation <. Since each S is a nonempty subset of S∈C S, each S S has a least member mS with respect to the relation <. Define f : C → S∈C S by f (S) = mS . We show that f is a choice function: • Well-Defined : Suppose that (a, x), (a, y) ∈ f . Then, there exists some S1 ∈ C such that x ∈ S1 . Likewise, there exists some S2 ∈ C such that y ∈ S2 . Without loss of generality, it follows that x, y ∈ S2 , and so x = y = mS2 . • Everywhere Defined : This follows fromt he discussion above. Therefore, f is a choice function. Hence, the axiom of choice holds. Exercise 17.5. We prove that if R is a partial order in a set X, then there exists a total order S in X such that R ⊂ S; in other words, every partial order can be extended to a total order without enlarging the domain. Proof. Let (R, ≤) be a partially ordered subset of X. Let C be the set of all A 6 in R of partial orders, with respect to ≤. Let (C, ⊂) be the partial ordered S set . Let X be the collection of all chains in C. Note that for all y ∈ X y ≤ y∈X y. Thus, every chain has an upper-bound. By Zorn’s Lemma, C has a maximal element. Denote this by c. Claim: c is a total order in X: • Since c ∈ C, c is a partial order in X. Since it is maximal, for all c0 ∈ C, we have c0 ⊂ c. Thus, for all x, y ∈ X, either (x, y), (y, x) ∈ c. Either way, we must have x ≤ y, y ≤ x or x = y.

6 This

has been shown other exercises.

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18

Transfinite Recursion

Exercise 18.1. We prove that a similarity preserves ≺ (in the same sense in which the definition demands the preservation of ≤). Proof. Let (X, ), (Y, ≤) be two partially ordered sets and that there exists some function f : X → Y , 1-1 and onto. f has the property that for all a, b ∈ X with a  b iff f (a) ≤ f (b), for all f (a), f (b) ∈ Y . Suppose, to the contrary, that a ≺ b and f (a) ≥ f (b). However, this is a contradiction, since, by definition, we must have b  a. Thus f preserves strict order. Exercise 18.2. We prove the following statement: A 1-1 function, f that maps one partially ordered set onto another is a similarity iff f preserves < (strict order). Proof. Throughout this exercise, we assume that (X, ) and (Y, ≤) are partially ordered sets and f : X → Y is 1-1 and onto. • (⇒): Suppose that f is a similarity, then the previous exercise prove this conclusion. • (⇐): Suppose that f is as stated above, such that for any a, b ∈ X we have a ≺ b iff f (a) < f (b). To show that f is a similarity, we show that a = b iff f (a) = f (b): – (⇐): Since f is 1-1, then f (a) = f (b) implies a = b. – (⇒): Suppose, to the contrary, that a = b and, (without loss generality) f (a) < f (b). Since f preserves strict order, we must have a ≺ b. However, this is a contradiction. This completes the proof. Exercise 18.3. We show that a subset Y of a well ordered set X is, again, well ordered. Proof. Let (X, ≤) be a well-ordered set, and Y ⊂ X. If Y = X or Y = ∅, then Y is clearly a well-order. Thus, suppose that Y is proper subset of X. Suppose that Y was not well-ordered. Then, some subset, Z of Y does not have a least element. However, Z ⊂ X, and so by definition of a well-ordered set, it has a least element. A contradiction. Exercise 18.4. We prove that each subset of a well-ordered set X is similar either to X, or an initial segment of X.

44

Proof. Suppose that (X, ) is a well-ordered set. If X = ∅, then every subset of X is equivalent to X. So, suppose X 6= ∅, and let S be a non-empty subset of X. Since X ⊂ X, X has a least element, call it x. Likewise, S has a least element, s. Consider the map f given by f (s) = x. And then, since X − {x} ⊂ X it has a least element, call it x0 . Likewise, S − {s} has a least element, call it s0 and define f (s0 ) = x0 , and so on. We show that f is a sequence of type S in X. Now, either this process terminates, or it does not. If it does, then it terminates at some x∗ , s∗ i.e. f (s∗ ) = x∗ . Claim: S ∼ = s(x∗ ). • The fact that f is a function is clear by previous definition. It only need to be shown that f is 1-1, onto and preserves ≺. The fact that f is 1-1 is clear. Thus, if x ¯ ≺ x∗ , then by construction, s(¯ x) = f (t) for some t ∈ S. • f preserves ≺: This follows from the definition of initial segments give on page 65, 78. Otherwise, the process does not terminate. In this case, the claim is that S∼ = X. The fact that f is a 1-1 and onto function is shown similarly as before. Likewise for ≺. Exercise 18.5. We prove that if (X, ≺), (Y, ≤) are well ordered sets, and X ∼ = Y , then the similarity maps the least upper bound (if any) of each subset of X onto the least upper bound of the image of that subset. Proof. Let X, Y be stated as above. And suppose that S ⊂ X has a least upper bound, s∗ . This implies that for all s ∈ S, s ≤ s∗ and for any other upper bound x ∈ X, x ≤ s∗ implies x = s∗ . Suppose, to the contrary, that f (s∗ ) was not the least upper-bound of f (S). Since f is order preserving, we have that f (s∗ ) is an upper-bound for f (S). Let α 6= f (s∗ ) be the l.u.b. of f (S) (since f is onto there exists some x ∈ X such that α = f (x). Thus from here, α = f (x)); for all f (s) ∈ f (S), we have f (s) ≤ f (x), and that for any other upper-bound z of f (S), z ≤ f (x) implies z = f (x), in particular, this implies that s∗ ≥ x. But then, since f is order-preserving, we have that f (s) ≤ f (x) implies that s ≤ x for all s ∈ S. But since s∗ is the l.u.b. for S, we must have s∗ ≤ x. Since X, and consequently S is well-ordered, it is partially ordered, and so s∗ = x, a contradiciton.

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19

Ordinal Numbers

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20

Sets Of Ordinal Numbers

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21

Ordinal Arithmetic

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22

The Schr¨ oder-Bernstein Theorem

Exercise 22.1. Let X, and Y be sets. Suppose that f : X → Y , is a function, and g : Y → X, is a function. We prove that there exist subsets A ⊂ X, B ⊂ Y such that f (A) = B, and g(Y − B) = X − A. (This statement is used to prove the Schr¨ oder-Bernstein theorem.) Proof. Let X, Y , f and g be stated as above. Throughout this exercise, we will use f [A] instead of the more common, yet less accurate, f (A). We look at four cases: • f onto, and g is not: In this case, we have f [X] = Y , and g[Y − f (X)] = g[Y − Y ] = g[∅] = {} = X − X. Thus, choosing A = X ⊂ X and B = {} ⊂ Y , the result follows. • g onto, and f is not: In this case, we have f [∅] = ∅, and g[Y − ∅] = g(Y ) = X = X − {}. Thus, choosing A = {} ⊂ X, and B = {} ⊂ Y , the result follows. • f onto, g onto: Again, in this case, letting A = ∅ ⊂ X we have f [∅] = ∅, and g[Y − ∅] = g[Y ] = X = X − ∅; the result follows. • f not onto, g not onto: We point out that since neither f nor g is onto, there exists some x ¯ ∈ X such that the fiber of x ¯ under g is empty. Likewise, there exists some y¯ ∈ Y such that the fiber of y¯ under f is empty. Let F{} be the set of all such x ¯ ∈ X. For any x ∈ X such that the fiber of x under g is not empty, either g −1 [x] = {f (x)} or f (x) ∈ g −1 [x], and g −1 [x] 6= {f (x)}. In the first case, let F{f (x)} be the set of all x ∈ X such that g −1 [x] = {f (x)} and then let Fx be the set of all x ∈ X for which f (x) ∈ g −1 [x], and g −1 [x] 6= {f (x)}. From the section on relations, we have that X is partitioned by F{} , Fx and F{f (x)} . That is, these sets are pairwise disjoint, and their union is X. These sets have some interesting properties. Namely g[Y − f [F{} ]] = g[f [Fx ] ∪ f [F{f (x)} ]]. This is from the fact that F{} , Fx , and F{f (x)} partition X and that if g is not mapping from to the set F{} , then it must map to the other two. Likewise, g[Y − f [F{f (x)} ]] = g[f [Fx ] ∪ f [F{} ]]. The claim is that the subset A of X, so desired is, F{} ∪ F{f (x)} . We

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compute: ” ı ” ı ” ı g Y − pf [F{} ] ∪ f [F{f (x)} ]q = g Y − f [F{} ] ∪ g Y − f [F{f (x)} ] ” ı ” ı = g f [Fx ] ∪ f [F{f (x)} ] ∪ g f [Fx ] ∪ f [F{} ] ” ı ” ı = g f [Fx ] ∪ g f [F{} ] ” ı = g f [Fx ] ∪ ∅ (1) ” ı = g f [Fx ] = Fx Where the second line follows form the comments above, and (1) is from the construction of F. Indeed, from the comments above, X − (F{} ∪ F{f (x)} ) = Fx .

Exercise 22.2. We prove the Schr¨ oder-Bernstein theorem using the above exercise. Proof. Suppose that X À Y , and Y À X. Then, there exist some f : X → Y such that f is 1-1 and onto a subset Y 0 of Y . Likewise, there exists some g : Y → X which is 1-1 and onto a subset X 0 of X. We note that if either X 0 = X or Y 0 = Y , then we are done. So, suppose that X 0 6= X and Y 0 6= Y . Likewise, if either X 0 or Y 0 was the empty set, then we are done, since it is trivially true that X ∼ Y .So, suppose that X 0 Ĺ X, such that X 0 6= ∅, and Y 0 Ĺ Y and Y 0 6= Y . So, consider f |X−X 0 and g|Y −Y 0 . From the previous theorem, there exists a subset X 00 ⊂ X − X 0 and some Y 00 ⊂ Y − Y 0 such that f |X−X 0 [X 00 ] = Y 00 and g|Y −Y 0 [(Y − Y 0 ) − Y 00 ] = (X − X 0 ) − X 00 . We claim that ∅ = (X − X 0 ) − X 00 i.e. X 00 = X − X 0 ; if not, then g|Y −Y 0 [(Y − Y 0 ) − Y 00 ] 6= ∅ and so there is some g(y) ∈ X 0 such that y ∈ (Y − Y 0 ) − Y 00 . But then, we have f (g(y)) ∈ Y 0 6⊂ (Y − Y 0 ) − Y 00 ; this implies that y ∈ Y 0 , a contradiction. Thus, X 00 = X − X 0 . Since this is the case, f |X−X 0 [X 00 ] ⊂ Y 0 . But then, g|Y −Y 0 [(Y − Y 0 ) − f |X−X 0 [X 00 ]] = g|Y −Y 0 [Y − Y 0 ] = (X − X 0 ) − X 00 = ∅. This implies that Y − Y 0 = ∅, or in other words, Y = Y 0 . A contradiction.

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23

Countable Sets

Exercise 23.1. We prove that the set of all finite subsets of a countable set is countable. Proof. The following proof was suggested by this website: Let A(n) be the set of subsets of A with no more than n elements. Thus, A(0) = {a}, A(1) = A(0) ∪ {{a} : a ∈ A} and for all n ≥ 0: A(n + 1) = {a(n) ∪ a(1) : a(n) ∈ A(n) ∧ a(1) ∈ A(1)} We verify, by induction, that each A(n) is countable. Let A be countable. Then, A(1) is countable, by construction. Suppose A(n) is countable. Then by since the union of countable sets is countable, A(n + 1) is also countable. Denote with Af as the set of finite subsets of A. It is apparent that every finite subset is in some A(n), and so: [ Af = A(n) n∈N

And since the union of countable subsets is countable, Af is countable. Exercise 23.2. We prove that if every countable subset of a totally ordered set X is well ordered, then X itself is well ordered. Proof. Let A ⊂ X be an arbitrary countable subset. Then, A is well-ordered, S and every subset a of A is well-ordered, so they haveSleast elements. Let n An be the union of all countable subsets of X. SinceS n An is the union of wellordered sets, it is well-ordered.S The claim is that n An = X: S S SupposeS otherwise; Since S n An ⊂ X, it follows that X = n An ∪ (X − n An ) = n An ∪ Y . Since n An is the union of all countable subsets of X, Y must be finite; otherwise, we have missed a countable subset. But, since Y is finite, Y is included in some countable subset of X. Thus, [ [ X= An ∪ Y = An n

n

This contradiction concludes the proof.

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24

Cardinal Arithmetic

Exercise 24.1. We prove that if a, b, c and d are cardinal numbers such that a ≤ b and c ≤ d, then a + c ≤ b + d. Proof. Let a, b, c, d be stated as given. By definition, there exists sets A, B, C, D (without loss of generality, pairwise disjoint) with cardinality a, b, c, d, respectively. Using the stated conditions, we have A À B, C À D. Thus, there exists B 0 ⊂ B, and D0 ⊂ D such that A ∼ B 0 and C ∼ D0 i.e. card A = card B 0 and card C = card D0 . Then, a + b = b0 + d0 ≤ b + d. Exercise 24.2. We prove that if a, b, c and d are cardinal numbers such that a ≤ b and c ≤ d, then ac ≤ bd. Proof. We proceed in a fashion similar to exercise 24.1. Let a, b, c, d be stated as given. By definition, there exists sets A, B, C, D (without loss of generality, pairwise disjoint) with cardinality a, b, c, d, respectively. Using the stated conditions, we have A À B, C À D. Thus, there exists B 0 ⊂ B, and D0 ⊂ D such that A ∼ B 0 and C ∼ D0 i.e. card A = card B 0 and card C = card D0 . Thus, ab = card(A × B) = card(B 0 × D0 ). From a previous exercise and the comments on page 94, we have B 0 × D0 À B × D, and so it follows that card(B 0 × D0 ) ≤ card(B × D). Exercise 24.3. We prove that if {ai } (i ∈ I) and {bi }P(i ∈ I) Q are families of cardinal numbers such that ai < bi for each i ∈ I, then i ai ≤ i bi . Proof. The following proof was suggested to me by Sophie C.. Note, Y X ai = a0 + a1 + . . . ≤ a0 × a1 × · · · = ai i

i

We proceed by mathematical induction. • Initial Case: a0 < b0 — By assumption. Q Q • Hypothesis: Suppose that 0≤i≤n ai < 0≤i≤n bi Q Q • Step: i + an+1 < 0≤i≤n aQ 0≤i≤n bi × bn+1 for an+1 < bn+1 . Therefore, Q a < b i i 0≤i≤n+1 0≤i≤n+1 P Q Q From the above, we have i ai ≤ i ai < i bi . Which completes the proof. Exercise 24.4. We prove that if a, b and c are cardinal numbers such that a ≤ b, then ac ≤ bc . Proof. The following proof was suggested by Sophie C.. We take the definition of ac as multiply a by itself c times. And likewise for bc . We proceed by mathematical induction: 52

• Initial Case: a1 ≤ b1 . Q Q • Inductive Hypothesis: Suppose that i∈I ai ≤ i∈I bi such that card I = C. Q Q Q Q • Step: If i∈I ai ≤ i∈I bi , then i∈I ai × a ≤ i∈I bi × b, since a ≤ b. Q Q Thus, i∈I ai ≤ i∈I bi where card I = C + 1. And so, ac ≤ bc Exercise 24.5. Prove that if a and b are finite, greater than 1, and if c is infinite, then ac = bc . Proof. The following proof was suggested by Sophie C.. We take the definition of ac as multiply a by itself c times. And likewise for bc . If c is countable then, ac , bc are countable, and if c is uncountable, then ac , bc are uncountable. Thus, ac = bc Exercise 24.6. We prove that if a and b are cardinal numbers at least one of which is infinite, then a + b = ab. Proof. The following proof was suggest by Sophie C.. Suppose that a, b are given as stated. Without loss of generality, there exists sets A, B such that card A = a, card B = b, and A ∼ B 0 , for some B 0 ⊂ B. Thus, a + b = card A ∼ card B 0 ∪ card B = card(A ∪ B) ∼ card(B 0 ∪ B) = card B = b = ab. Therefore, a + b = ab. Suppose that a and b are both infinite, then as above, we have a+b = a+a = a and ab = aa = a. Therefore, a + b = ab. Suppose, without loss of generality, that a, and b are both infinite such that card A < card B. Then, a + b = b and ab = b, thus, a + b = ab. Exercise 24.7. We prove that if a and b are cardinal numbers such that a is infinite and b is finite, then ab = a. Proof. The following proof was Q suggested by Sophie C.. Suppose that a, b are given, as above. Then, ab = i∈I ai where card I = b and where ai = a, for all i ∈ I. Then, we have the following: ab = card(A × A × . . . × A)

where card A = a

= a × a × ... × a

b times

= a × a × ... × a

b/2 times, since a × a = a

= a × a × ... × a .. .

b/4 times, since a × a = a

=a This completes the proof.

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25

Cardinal Numbers

Exercise 25.1. We prove that each infinite cardinal number is a limit number. Proof. This proof was inspired from this website. Suppose to the contrary that k is an infinite cardinal which was not a limit number. Then, k = a+ for some ordinal a+ . If a is finite, we are done. Otherwise, we proceed to show that k = a. Not that since k is infinite, ω À k, and in particular, ω À a. Thus, consider φ : a → k given by φ(0) = a, and φ(m+ ) = m (if m ∈ ω), and φ(x) = x (if x ∈ a − ω). We show that φ is a 1-1 and onto function: • Well-Defined : Since no ordinal is equivalent to its successor, it follows that φ is well-defined. • Everywhere Defined : By construction, this is clear. • 1-1 : If φ(x) = φ(y), then in any of the cases, we have x = y. • Onto: Let k1 ∈ k. Then, either k1 ∈ a, or k1 = a. In either case, there exists a ` ∈ a for which φ(`) = a. Thus k = a, a contradiction. Exercise 25.2. We answer a few questions: If card A = a, what is the cardinal number of the set of all 1-1 and onto in A? What is the cardinal number of the set of all countably infinite subsets of A? Proof. If A is infinite, then ω À A and we can consider the infinitely manly maps fi , such that i ∈ ω, given by fi (x) = x + 1 (if x ∈ ω), and fi (x) = x (if x ∈ A − ω). However, if A is finite, then a simple counting argument show that the number is the multiplication of all predecessors of a, except zero. This answer was suggested by this website: ”A has aω countably infinite subsets, and there’s not much more that you can say unless you know something about the cardinal a. For example, 2 ≤ a ≤ 2ω , then aω = 2ω . If cf a = ω, i.e., if a has cofinality ω, then aω > a.” Exercise 25.3. We prove that if α, β are ordinal numbers, then card(α + β) = card α + card β and card αβ = card α card β. We use the ordinal interpretation of the operations on the left side and the cardinal interpretation on the right. Proof. The following proof was suggested by Sophie C.. We compute: • card αβ = card(ord A ord B) = card(ord(A × B)) = card(A × B) • (card α)(card β) = (card(ord A))(card(ord B)) = card(A) card(B) = card(A × B) Thus, card αβ = (card α)(card β) 54

Proof. The following proof was suggest by Sophie C.. We compute: • card(α + β) = card(ord A + ord B) = card(ord(A ∪ B)) = ord(A ∪ B) = ord(A) + ord(B) = card(ord(A)) + card(ord(B)) = card α + card β. Thus, card(α + β) = card α + card β.

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26

Extra Exercises

Many of these exercises were collaborations with other student. They are not marked exercises in the text, however, they are useful.

26.1

Arithmetic

Exercise 26.1. We show that addition for elements of ω is associative. The following proof was suggested by Aidan H. and David H., and I attribute all credit to them. Proof. We proceed by mathematical induction: By the recursion theorem there exists a function sm : ω → ω such that sm (0) = m and sm (n+ ) = (sm (n))+ where m, n ∈ ω and sm (n) = m + n.First note sk+m (n) = (k + m) + n, and similarly sk (m + n) = k + (m + n). Consider the set T = {n ∈ ω | sk+m (n) = sk (m + n)}. Initial case: See that 0 ∈ T since sk+m (0) = k + m = sk (m + 0) = k + (m + 0) = k + m. Inductive hypothesis: Let n ∈ T hold. Then sk+m (n) = sk (m + n). This implies (k + m) + n = k + (m + n). Now sk+m (n+ ) = (k + m) + n+ = (sk+m (n))+ = ((k + m) + n)+ . Similarly, sk ((m + n)+ ) = k + (m + n)+ = (sk (m + n))+ = (k + (m + n))+ . Since sk+m (n) = sk (m + n) by the inductive hypothesis. We obtain (k + m) + n+ = ((k + m) + n)+ = (k + (m + n))+ = k + (m + n)+ Note that sm (n) = m + n, so (sm (n))+ = (m + n)+ = m + n+ = sm (n+ ). Thus, (k + m) + n+ = k + (m + n+ ) Conclusion: Therefore, n+ ∈ T , and thus T = ω, and thus addition is associative. Exercise 26.2. We prove that addition is commutative. Proof. The following proof was suggested by Aidan H. and David H., and I attribute all credit to them. Let B = {m ∈ ω | sm (n) = sn (m)}. Initial case: 0 ∈ B, since s0 (n) = sn (0) = n. Inductive hypothesis: Let b ∈ B hold. That is, sb (n) = b + n = n + b = sn (b). Then sn (b+ ) = (sn (b))+ = (n + b)+ , by definition. Similarly, sb+ (n) = (sb (n))+ = (b + n)+ . Then, by the inductive hypothesis, sb+ (n) = (sb (n))+ = (sn (b))+ = sn (b+ ). Inductive conclusion: b+ ∈ B. Thus, by the induction principle, B = ω. This 56

implies sm (n) = sn (m) ∀m, n ∈ ω, and thus m + n = n + m ∀m, n ∈ ω, and thus addition is commutative. Exercise 26.3. We prove the properties of exponentiation. Proof. The following proof was suggested by Aidan H. and David H., and I attribute all credit to them. First note em : ω → ω is a function defined as em (0) = 1 em (n+ ) = em (n) · m

where em (n) = mn for m, n ∈ ω

I want to show m(n+l) = mn ·ml , that is em (n+l) = em (n)·em (l) for m, n, l ∈ ω. Define Pm : ω → ω such that Pm (0) = 0 and Pm (n+ ) = Pm (n) + m, and Pm (n) = m · n for m, n ∈ ω. First I will show P1 (n) = n ∀n ∈ ω by induction. Consider the set B = {n ∈ ω | P1 (n) = n}. Initial case: Note 0 ∈ B since P1 (0) = 0 by definition. Inductive hypothesis: Let n ∈ B hold, then P1 (n) = n. Then P1 (n+ ) = P1 (n) + 1 = n + 1 = n+ . Inductive conclusion: Thus n+ ∈ B and by principle of mathematical induction, B = ω. Thus, P1 (n) = n ∀n ∈ ω. Now to show m(n+l) = mn · ml . Let F = {n ∈ ω | em (n + l) = em (n) · em (l)}. Initial case: Note 0 ∈ F since em (0 + l) = em (l) and em (0) · em (l) = 1 · em (l) = P1 (em (l)) = em (l). Inductive hypothesis: Let n ∈ F hold, then em (n + l) = em (n) · em (l). Then em (n+ + l) = em ((n + l)+ ) = em (n + l) · m = em (n) · em (l) · m = em (n) · m · em (l) = em (n+ ) · em (l) Inductive conclusion: Thus, n+ ∈ F , and by the principle of mathematical induction, F = ω. Thus, m(n+l) = mn · ml ∀m, n, l ∈ ω. Exercise 26.4. We prove the left distributive law. Proof. The following proof was suggested by Aidan H. and David H., and I attribute all credit to them. We proceed by mathematical induction: Initial case: k · (m + 0) = k · m + k · 0 k·m=k·m+0=k·m Inductive hypothesis: k · (mn) = k · m + k · n Then k · (m + n+ ) = (k · (m + n))+ by definition of addition 57

= (k · m + k · n)+ = k · m + (k · n)+ by hypothesis and addition = k · m + k · n+ by definition of addition Conclusion: k · (m + n+ ) = k · m + k · n+

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