Introduction to Axiomatic Set Theory (1971) - Krivine

INTRODUCTION TO AXIOMATIC SET THEORY

SYNTHESE LIBRARY MONOGRAPHS ON EPISTEMOLOGY, LOGIC, METHODOLOGY, PHILOSOPHY OF SCIENCE, SOCIOLOGY OF SCIENCE AND OF KNOWLEDGE, AND ON THE MATHEMATICAL METHODS OF SOCIAL AND BEHAVIORAL SCIENCES

Editors: DONALD DAVIDSON,

J AAKKO

Rockefeller University and Princeton University

HINTIKKA,

Academy of Finland and Stanford University

GABRIEL NUCHELMANS, WESLEY

University of Leyden

C. SALMON, Indiana University

JEAN-LOUIS KRIVINE


D. REIDEL PUBLISHING COMPANY / DORDRECHT-HOLLAND

THEORIE AXIOMATIQUE DES ENSEMBLES First published by Presses Universitaires de France, Paris

Translated/rom the French by David Miller

Library of Congress Catalog Card Number 71-146965 ISBN-13: 978-90-277-0411-5 DOl: 10.1007/978-94-010-3144-8

e-ISBN-13: 978-94-010-3144-8

All Rights Reserved Copyright © 1971 by D. Reidel Publishing Company, Dordrecht, Holland Softcover reprint of the hardcover 1st edition 1971 No part of this book may be reproduced in any form, by print, photoprint, microfilm, or any other means, without written permission from the publisher

INTRODUCTION

This book presents the classic relative consistency proofs in set theory that are obtained by the device of 'inner models'. Three examples of such models are investigated in Chapters VI, VII, and VIII; the most important of these, the class of constructible sets, leads to G6del's result that the axiom of choice and the continuum hypothesis are consistent with the rest of set theory [1]I. The text thus constitutes an introduction to the results of P. Cohen concerning the independence of these axioms [2], and to many other relative consistency proofs obtained later by Cohen's methods. Chapters I and II introduce the axioms of set theory, and develop such parts of the theory as are indispensable for every relative consistency proof; the method of recursive definition on the ordinals being an important case in point. Although, more or less deliberately, no proofs have been omitted, the development here will be found to require of the reader a certain facility in naive set theory and in the axiomatic method, such as should be achieved, for example, in first year graduate work (2 e cycle de mathernatiques). The background knowledge supposed in logic is no more advanced; taken as understood are such elementary ideas of first-order predicate logic as prenex normalform, model ofa system of axioms, and so on. They first come into play in Chapter IV; and though, there too, all the proofs (bar that of the reduction of an arbitrary formula to prenex normal form) are carried out, the treatment is probably too condensed for a reader previously unacquainted with the subject. Several leading ideas from model theory, not themselves used in this book, would nevertheless make the understanding of it simpler; for example, the distinction between intuitive natural numbers and the natural numbers of the universe, or between what we call formulas and what we call expressions, are easier to grasp if something is known about 1 Numbers in brackets refer to items of the Bibliography, which is to be found on p.98.

VI

INTRODUCTION

non-standard models for Peano arithmetic. Similarly, the remarks on p. 44 will be better understood by someone who knows the completeness theorem for predicate calculus. All these ideas can be found for example in [4] (Chapters I, II, III) or [5] (Chapters I, II, III). The approach of the book may appear a little odd to anyone who thinks that axiomatic set theory (as opposed to the naive theory, for which, perhaps, this is true) must be placed at the very beginning of mathematics. For the reader is by no means asked to forget that he has already learnt some mathematics; on the contrary we rely on the experience he has acquired from the study of axiomatic theories to offer him another one: the theory of binary relations which satisfy the Zermelo/Fraenkel axioms. As we progress, what distinguishes this particular theory from other axiomatic theories gradually emerges. For the concepts introduced naturally in the study of models of this theory are exactly parallel to the most fundamental mathematical concepts - natural numbers, finite sets, denumerable sets, and so on. And since standard mathematical vocabulary fails to provide two different names for each idea, we are obliged to use the everyday words also when referring to models of the Zermelo/Fraenkel axioms. The words are thereby used with a completely different meaning, the classic example of this being the 'Skolem paradox', which comes from the new sense that the word 'denumerable' takes when interpreted in a model of set theory. Eventually it becomes obvious that even the everyday senses of these words are by no means clear to us, and that we can perhaps try to sharpen them with the new tools which are developed in the study of set theory. Had this problem been posed already at the beginning of the study it would have been tempting to shirk it, by saying that mathematics is merely the business of manipulating meaningless symbols. It is an open question how much set theory can do in this field; but it seems likely that what it can do will be of interest beyond mathematical logic itself. Note: This translation incorporates a number of additions and corrections to the original French edition.

CONTENTS

INTRODUCTION

v

Chapter

1

Chapter

I: The ZermelojFraenkel Axioms of Set Theory II: Ordinals, Cardinals

13

Chapter III: The Axiom of Foundation

35

Chapter

IV: The Reflection Principle

48

Chapter

V: The Set of Expressions

56

VI: Ordinal Definable Sets. Relative Consistency of the Axiom of Choice

63

Chapter VII: FraenkelfMostowski Models. Relative Consistency of the Negation of the Axiom of Choice (without the Axiom of Foundation)

70

Chapter VIII: Constructible Sets. Relative Consistency of the Generalized Continuum Hypothesis

81

BIBLIOGRAPHY

98

Chapter

CHAPTER I

THE ZERMELO/FRAENKEL AXIOMS OF SET THEORY

When we formulate the axioms of set theory, we have to rely on an intuitive understanding of sets, in much the same way that we develop the axioms for a vector space from commonsense ideas about three-dimensional space. Despite this, once we have the axioms of set theory written down we are free to study any other structure in which they hold, just as we study many vector spaces over and above the Euclidean space R3. Thus set theory is no different from any other axiomatic theory familiar to the reader. It is, like the theories of groups, rings, fields, vector spaces, lattices, and so on, an abstract theory. A structure satisfying the axioms of set theory is called a universe. A universe tW is basically a collection of objects called sets. We do not say 'a set of objects', since what we are going to call sets are just those objects of which tW is a collection. It is a straightforward precaution when defining vector spaces, for example, not to use the same word both for the vector space and for a vector in it. The same applies here. Also involved in tW is a single undefined binary relation, a relation which links some of the objects of tW with others. It is called the membership relation, and is denoted by E. We read the expression 'x E y' as 'x belongs to y', or 'the set x belongs to the set y', or 'x is a member of y', or 'x is an element of y', or even 'the set y contains the element x'. Since E is a relation on the universe tW, it holds only between sets. The symbol'E' and the words 'belong', 'contain', 'member', and 'element' will be used only when this particular relation is intended; and should we ever use any of these words in their everyday senses, we shall state this explicitly. For example, we will not normally say 'the object xis an element of the collection tW', but' x is in tW'. So a universe can be pictured as a graph of the kind shown in figure 1 ; the arrow indicates that the object at its head is an element ofthe one at its tail. For example, b E a, C E c. Nothing said so far imposes any restriction on the binary relation E. In order to do just this we now begin to list the axioms of set theory.

2


f

e

Fig. 1.

1. AXIOM OF EXTENSIONALITY

No two distinct sets in r1lt have the same elements. We write this as

'v'x'v'y['v'Z(Z E x _

ZE

y)

~

X = y].

This axiom is not satisfied by the binary relation shown in the figure; for band c are different, yet each has c as its only element. A useful 'axiom' to put down now, though we will not number it since it turns out to be a consequence of later ones, is the pairing axiom. Given two sets a and b, this axiom guarantees the existence of a set c containing a and b as its only elements. By the axiom of extensionality there is only one such set c. We write the pairing axiom as

'v'X'v'y3Z'v't[tEz-t=X v t=y]. We write {a, b} for the set c whose only elements are a and b. A special case is when a and b are the same set; it follows from the axiom that there is a set {a} containing a, and a alone.

THE ZERMELO!FRAENKEL AXIOMS

3

If a=/=b we call {a, b} a pair. The set {a} is sometimes called a singleton or unit set. From two sets a and b three applications of the pairing axiom yield the set {{a}, {a, b}}, called the ordered pair of a and b and usually denoted by (a, b). The following theorem justifies this definition. THEOREM: If(a, b)=(a', b') thena=a' andb=b'. PROOF: If a = b then (a, b) = {{a}}, and so (a, b) has only one element. Thus (a', b') is also a singleton, which means that a' =b'. It follows that {{a}} = {{a'}}, and so a=a' ; thus b=b' too. When a=/=b, (a, b) is a pair, and so (a', b' ) must also be one. This can only be true if a' =/= b'. But as {{a}, {a, b}} = {{a'l, {a', b'}}, either

{a}

= {a', b' }

and

{a, b}

= {a'l

or

{a} = {a'l

and

{a, b} = {a', b'}.

The former possibility is ruled out simply because {a} is a singleton whilst {a', b'} is a pair. Thus {a} = {a'l -and so a=a' -and {a, b}={a', b' } -and sob=b'·1 If a, b, c are sets, the set (a, (b, c» is called the ordered triple of a, b, c, and is denoted by (a, b, c). THEOREM: If(a, b, c) = (a', b', c') thena=a ' , b=b', c=c'. PROOF: From the identity of (a, (b, c» and (a', (b', c'» we get a=a' and (b, c)=(b', c'). Thus also b=b' and c=c'.1 In the same way we can define the ordered quadruple (a, b, c, d) by

(a, b, c, d) = (a, (b, c, d», and for any n>O the ordered n-tuple (ao, ai' ... , an-i) by

(ao, a 1, ... , an-i)

= (ao, (a1' ... , an -1»·

The previous theorems generalize. THEOREM:.if (ao, a1' ... ' an-i) = (a~, a~, ... , an-~) then ao =a~, a1=a~, ... , a,,-l =a,,_~. PROOF: Obvious by induction on n.1

4

INTRODUCTION TO AXIOMATIC SET THEOR Y

It should be noted that, given three distinct sets a, h, c in the universe <11, we do not yet have any way of proving the existence of a set d containing just a, h, and c as members. The next axiom fills this gap.

2.

UNION AXIOM (OR SUM-SET AXIOM)

According to this axiom, to every set a there corresponds a set h whose members are precisely the members of the members of a. We write this formally as Vx3yVz [z E y +-dt(t E x A Z E t)]. The set b, called the union of the members of a (or, more briefly, the union of a) and denoted by U x (or Ua), is unique; for any set h' having the xea

same property would necessarily have the same elements as, and so be identical with, b. This axiom solves the problem posed above. For if a, h, and c are sets, the set d = U {{a, h}, {c}} contains a, h, and c, and nothing more besides. We write the set d as {a, b, c}. More generally, any finite number of sets ao, a1, ... , all - 1 can be collected together into a set {ao, a1, ... , an - 1} containing them and them only. This is easily proved by induction on n; we need only observe that U {{a O,a1, ... ,a,,-2}, {all - 1}} has the desired property for n if {a o, a1, ... , all - 2} has it for n-l. If a and b are sets, then the union of {a, h} is called the union of a and h, and written a u b.1t is trivial that a u (b u c)=(a u b) u c=(h u a) u c= = the union of the set {a, b, c}. The same applies for any finite number of sets ao, a1'"'' all - 1; the union of the members of {ao, a1' ... , all -1} is called the union of ao, a1> and ... , and a,,-l and written a o u a 1 u ... u a,,-l . 3.

POWER-SET AXIOM

Let a and b be sets in <11. The statement Vx(x E a -+ x E b) is abbreviated a c: b, and read 'a is included in b' or 'a is a subset of b'. The power-set axiom tells us that, given a set a, there exists a set b whose members are just a's subsets; in symbols it becomes

Vx3y'f/z [z E y

-Ho

Z

c: x].

THE ZERMELO/FRAENKEL AXIOMS

5

By the axiom of extensionality only one set b can have this property. It is called the power-set of a and written gJ(a). It is worth emphasizing at this point that we shall now use the words 'subset' and 'include' only in the sense given them above, to express a relation between two objects of the universe. This sense is quite different from the usual one (when we say, for instance, that a class is included in the universe). We call it the formal sense of the word 'include' (or 'subset'), and the usual one will be called the intuitive sense. To avoid from the start any possible confusion that might arise, let us agree that when we use the words 'subset' and 'include' without qualification, we are using them in their formal senses; any use of these words in their intuitive senses will be noted as such explicitly. For instance: each set a defines a subset (in the intuitive sense) of the universe
6


Starting from the two binary relations x E y and x = y we can, by repeated applications of these rules, obtain relations of any number of arguments. Suppose that R(x, y) and 8(y, z) are two binary relations; then the relation ..., R v 8 is written R -+ 8. The relation ..., (..., R v ..., 8) is written R A 8; it is satisfied by a, b, and c iff R(a, b) and 8(b, c) both hold. Similarly, R +-+ 8 abbreviates (R -+ 8) A (8 -+ R), and is satisfied by a, b, c iff R(a, b) and 8(b, c) either both hold or both fai1. Likewise, the singulary relation ...,3x...,R(x, y), written VxR(x, y) is satisfied by b iff R(a, b) holds for every object ain 0/1. Relations constructed from x E y and x=y by rules I-IV are thus defined by formulas composed (though not in an arbitrary fashion) from the symbols =, E, ..." v, 3, the variables x, y, z, u, v, ... , and objects in Cl/I. It is clear that further relations on 0/1 can be defined by other methods. But we shall not consider them in this book. A relation R(x) of a single argument is usually called a class. A class is a subcollection (intuitively speaking) of the universe Cl/I. For example, the formula

Vu[uEx-+3v[VEX

A

Vt(tEV+-+t=U v tEU)]]

defines a class; the formula

uEx-+3v[VEX

A

Vt(tEV+-+t=U v tEU)]

defines a binary relation R(u, x). So if a is in 0/1, the formula R(a, x), namely

aEx-+3v[VEX

A

Vt(tEV+-+t=a v tEa)],

defines a class. Objects of 0/1 which appear in a formula E are called the parameters of E. R(u, x), for example, has no parameters; R(a, x) has the object a as its sole parameter. A formula of no arguments at all, a closed formula or sentence as it is called, is either true or false in the universe. For example, the formula 3x3y[Vz(z ¢ y) AyE X A VuR(u, x)] (where the relation R is defined above) is a sentence without parameters (it will later turn up in a slightly different form as the axiom of infinity). The theorems of set theory (and the axioms in particular) are all closed formulas without parameters.

THE ZERMELO!FRAENKEL AXIOMS

7

Equivalence Relations: A binary relation R(x, y) is an equivalence relation if, whatever objects a, b, c might be, R(a, b) ~ R(a, a) A R(b, b); R(a, b) ~ R(b, a); R(a, b) A R(b, c) ~ R(a, c). The class R(x, x) is called the domain of the equivalence relation. R(a, b) is also written a",b (mod. R). For any set a in 0/1, the class R(a, y) is called the equivalence class of a.

Ordering Relations (in the weak sense): A binary relation R(x, y) is an ordering relation (or, briefly, an ordering) in the weak sense if for every a,b,andc R(a, b) ~ R(a, a) A R(b, b); R(a, b) A R(b,a)~a=b; R(a, b) A R(b, c) ~ R(a, c). The class R(x, x) is the domain of the ordering. R(a, b) is also written a~b (mod. R); R(a, b) A b¥=a may similarly be abbreviated as a
Ordering Relations (in the strict sense): A binary relation R(x, y) defines astrict ordering on a class D(x) if for all a, b, c, R(a, b) --+ D(a) A D(b); -, (R(a, b) A R(b, a»); RCa, b) A R(b, c) ~ R(a, c). R(a, b) may again be written a
A

R(x, y, z')

--+

z = z'].

8


The binary relation 3zR(x, y, z) is then called the domain of the functional relation R. The class 3x3yR(x, y, z) is called the range of the functional relation R. For any (for example binary) functional relation defined everywhere (in other words, a functional relation with domain x = x A Y =y) we can introduce a new symbol
and

3z[R(x, y, z)

A

E(z, v, w)]

'v'z[R(x, y, z)

-+

E(z, v, w)]

(which are equivalent to one another) can be written simply as

E[
'v't[tezoH-t=x v t=y] is written z = {x, y}; thus the equivalent formulas

3z['v't(tezoH-t=x v t=y) 'v'z [Wet e z oH- t = x v t = y)

A

-+

zea] z e a]

can both be abbreviated by {x, y} Ea. We can now go on to state the other axioms of set theory. 4.

AXIOM SCHEME OF REPLACEMENT (OR SUBSTITUTION)

Suppose a formula E(x, y, ao, ... , ak- 1 ) with parameters a o, ... , ak-l defines a singulary functional relation; and let a be any set. Then we shall postulate that the universe I7It contains a set b whose elements are just the images under this functional relation of those elements of a within its domain. The demand on I7It that it satisfy this condition for any singulary functional relation is known as the axiom scheme of replacement. The scheme is rendered symbolically by the following infinite list of sentences:

'v'xo ... 'v'Xk-l {'v'x'v'y'v'y' [E(x, y, xo, ,." Xk-l) A E(x, y', xo, ... , Xk-l) -+ y = y'] -+'v't3w'v'v[vewoH-3u[uet A E(u,v,xO, ... ,Xk-l)]]}' Here E(x, y, xo, .. " Xk-l) may be any parameter-free formula with at least two free variables x and y.

9


The axioms 1,2,3, the scheme 4, and the axiom of infinity (introduced on p. 29 below) make up the set theory of Zermelo and Fraenkel, usually abbreviated ZF. Easily derived from the scheme of replacement is the following scheme.

Scheme o/Comprehension: If ais a set, and A (x, a o, ... , ak-l) any formula of one free variable (and parameters ao, ... , ak-l)' we can establish that there is a set whose elements are exactly those elements of a for which A holds. This is the content of the comprehension scheme, which consists of the following infinite list of sentences: 'Vxo ... 'Vxt_ 1'Vx3y'Vz [z e y

+-+-

(z e x

A

A(z, xo, ... , Xk-l))]'

Here A(x, xo, ... , Xk-l) is any parameter-free formula of at least one free variable x. To derive any instance of the scheme from the scheme of replacementitisenoughtonotethattheformulay=x A A (x, ao, ... , ak- l ) defines a singulary functional relation F whose domain is the class A(x, ao, ... , ak-l)' By replacement then, there is a set b whose elements are just the images under F of those members of a in F's domain; it is immediate that b is the set whose existence we set out to prove. The set b is usefully represented by the notation

I

{x e a A (x, ao, ... , ak-l)}' THEOREM:

There is one and only one set without elements.

Let a be any set at all. We apply the comprehension scheme to the set a and the formula x::/:x to get a set b={x e a x::/:x}, which is obviously without elements. That there cannot be more than one such set follows at once from the axiom of extensionality .• The set just defined is called the empty set and is denoted by 0. PROOF:

I

Let us give a derivation of the pairing axiom from axiom 3 and the scheme 4. First of all, every subset of 0 is empty, so that &(0) has 0 as its only element; thus {0} exists. This latter set is a singleton, so if a c: {0} then a=0 or a= {0}. So &({0}) has just two distinct elements, 0 and {0}; thus {0, {0}} exists. Now let a and b be any sets you like. It is apparent that the relation (x=0 A y=a) v (x={0} A y=b) is a singulary functional relation, so we can apply the replacement scheme to it and consider

10


the image of the set {0, {0}} thereunder. This turns out to be the set {a, b}, and thus the pairing axiom is proved. We say that a class A (x) corresponds to a set (or even, twisting language a bit, is a set) if there is a set a such that Vx(x E a +-* A (x)). More generally, a relation (of three arguments, for example) A (x, y, z) corresponds to a set, or even is a set, if there is a set a for which we have VxVyVz[
3x3y[z=
1\

xea

1\

yeb].

To write this out in full we would first have to use the definition of the functional relation Z=
3x3y3u3v[z

= {u, v}

1\ U

= {x}

1\ V

= {x, y} A

xEa

A

yeb].

The braces could then be eliminated with the definition of the functional relation Z= {u, v}, leaving

3x3y3u3v[Vt(tez+-*t=u v t=v) A Vt f (tf e u +-* tf = x) A Vtff (tff e v +-* tff = X V tff = y) AxeaAyeb]. So X(z) is just this last formula, and it clearly is the class of all ordered pairs

11

tion of (x,y») (x,Y)E&(&(aub)). But then X(z) is equivalent to X(z) 1\ z e & (&(a u b), so that by the comprehension scheme X is a set. This set X is called the Cartesian (or direct, or cross) product ofthe sets a, b, and it is written a x b. The Cartesian powers a x a, a x (a x a), and so on, are sometimes abbreviated a2 , a3 , •••• A singulary functional relation R(x, y) whose domain is a set is itself a set. For suppose that the domain is a; then by the replacement scheme there is a set b consisting of the images of a under R. Consequently R(x, y) is equivalent to R(x, y) 1\ (x, y) E a x b, and this class is a set f by the comprehension scheme. Such a set f is called a function defined on a, with values in b, or a map from a to b, or afamily of sets indexed by a. If a is a map we will sometimes write dom(a), rng(a) for its domain and range respectively. The formula 'fis a map from a into b', written more compactly f: a -+ b is therefore just the following formula: f c a xb

1\ 1\

'rfx'rfy'rfy' [(x, y) E f 1\ (x, y') E f -+ y = y'] 'rfx[xea-+3y(yeb 1\ (x,y)Ef)].

We leave to the reader the task of eliminating the defined symbols c, x, and ( ). If a and b are sets the class X of all maps from a into b is again a set. For a map from a to b is a subset ofax b, and therefore a member of &(a x b). SO XU) is equivalent to XU) 1\ fE &(a x b). Comprehension yields the desired result. This set of maps is written abo Union, Intersection, and Cartesian Product of a family of sets: Let a be a family of sets indexed by I; a is of course just a map whose domain is I, and to indicate this we will usually write a family as (ai)iEI' The union ofthefamity (ai)iEl> written Ua i, is the union of the range iEI of the map a. By the replacement scheme and union axiom it is a set b, and we have 'rfx[ x E b ~ 3i(i E I 1\ X E ai)]' Likewise the intersection of the family (aJiEl is the class X(x) defined by 'rfi(i E I -+ X E a;). If 1=0, then X is the class of all sets, and so not a set itself. Otherwise, take any io E I; then X(x) is equivalent to x E aio 1\

12


Vi(i E I - X E ai), and X is seen to be a set, by comprehension. We write it ai • ieT Now take the class X of map sf: 1- Ua i such thatf(i) E ai for every

1\

n

ieI

i

E I. Such a map belongs to the set I( U aJ, so XU) is equivalent to

XU)

1\

f

E I(

U a)

ieI X is thus a set, called the Cartesian (or direct, or

ieI cross) product of the family (ai)ieb and written

X ai (or sometimes IT aJ ieT

ieT

Note that for the rest of the book we shall be using the words 'map' and 'function' in the sense defined above, which is by no means their everyday mathematical sense (since here, for example, all maps f: a -4 b are in the universe). As we proliferate definitions the same thing will happen to nearly all the familiar words of mathematical language. If we ever use the words in their normal senses (and this will not happen often), we will, as we decided before, say so explicitly. For example: let a and b be two sets, and A, B be the corresponding parts of the universe. If f maps a into b there will correspondingly be a map (in the intuitive sense) from A to B. But there may well be, intuitively speaking, maps from A to B to which there are no corresponding maps from a into b.

CHAPTER II

ORDINALS, CARDINALS

Well-ordering Relations: Let a be a set, all of whose elements lie in the domain of some linear ordering R(x, y). We say that a is well ordered by R if every non-empty subset of a has a smallest element (mod. R). It is immediate that if a is well ordered by R then every subset of a is also well ordered by R. If rca x a then the pair (a, r) is said to be a well-ordered system if a is well ordered by the relation (x, y) E r. Suppose that the set a is well ordered by the relation R. A subset s of a is said to be an initial segment of a if XES and y ~ x (mod. R) imply YES, for any x, yEa. For each Xo in a we write Sxo(a, R) (or Sxo(a) if there is no danger of ambiguity) for the set {x E a I x
This is easily proved by noting that if s is an initial segment of a, but not a itself, then the set a""'-s = {x E a I x rf: s} is non-empty. Consequently, it has a least element xo; so if x
If R is a well-ordering with domain D, and T any non-empty subclass of D (that is, Vx( T(x) ~ D(x») /\ 3xT(x») , then Thas a least element (mod. R). For Tis non-empty, so take Xo in T. Either Xo is the least element of T, or else the class T /\ Sxo(R) is non-empty. This class is however a set (since Sxo(R) is a set), and indeed a non-empty subset of Sxo(R), which is well ordered. It therefore has a least element, which is obviously also the least element of T.

14


(Our talk here of elements of the class T is to be understood as using 'element' in an informal way only.) The Class of Ordinals: A set a is called transitive if every element of a is also a subset of ct; that is, if\fx(x E a --+ x c a). A set ct is called an ordinal if, in addition to being transitive, it is strictly well ordered by the membership relation x E y. The class 'a is an ordinal' is written On(a); spelt out more or less in full this becomes \fx\fy [x E a AyE ct --+ X ~ Y v Y ~ x] A \fx\fy\fZ[XEct A YEa A ZEa A xEy A YEZ --+ XEZ] A \fz[z c: ct A Z #0 --+ 3X(XEZ A \fY(YEZ --+ XEy V x=y))] A \fx\fy [x E ct AyE X --+ YEa] . (In this extensive formula there is no explicit mention of the fact that the ordering x E y is linear; however, no such clause is needed, since we have stated that every non-empty subset of a contains a smallest element, and a fortiori this applies to subsets of two elements.) Examples of ordinals that are easily shown to be such are 0, {0},

{0, {0}}. Let ct be an ordinal. Then the initial segments of ct are a itself, and the elements ofct. For a proper initial segment of ct is S~(a) for some E a; and S~(a)= = {11 E ct 111 < = {11 E ct 111 E = n ct. But since c a, this last term reduces to and the result is proved. Every element of an ordinal is an ordinal. Furthermore, Take E a; then c ct, so membership well-orders if y E X and x E then x E a (since c ct); so x c ct, so yEa. Now membership strictly orders the elements of ct, so from y E X and x E we get y E Thus x E implies that x c and the two conditions for to be an ordinal are established. For every ordinal ct, ct ~ ct. If is any element of a, then ~ since E strictly orders ct. So if a is an element of a we must have a ~ ct. Thus ct ~ ct. For every two ordinals a, p, either ct = p, or p E ct, or ct E p,' moreover, the three cases are mutually exclusive.

e,

e

e.

e

e}

e} e

e

e

e

e.

e

e

e,

e e,

e

e

e

15

ORDINALS, CARDINALS

Put e=(X n p. Then e consists of all members ohthat are

e

e

then (X = P. then rx e p . and e rx ; then p E rx . and e e p ; then e (X n p, so by the definition e But this is a contradiction, since is an ordinal.

e = (X and e = p ; e = (X and e e p ;

e= p e ee (X of e, e e.

e

e

Furthermore, (X = P and rx E p together would imply that PEP; similarly, if (X e p, then rx c p, so if P E (X were also true, we would have PEP again. But p ¢ p. Thus the three cases are mutually exclusive. The membership relation on the class On (that is, the relation x E y 1\ 1\ On(x) 1\ On (y») isawell-ordering. We have just shown the relation in question to be a strict linear ordering. So we need only note that since
e

e

"ea

of p, then for at least one rxo E a we will have x n (xo non-empty. Since rxo is well ordered, its subset x n rxo will therefore have a least element, which is clearly also the least element of x. Since x was an arbitrary non-empty subset of p, it follows that Pis well ordered bye. Now suppose

16


that y E X and x E p. Then x E rx for some rx E a, so y E rx since rx is an ordinal. We conclude that YEP, and this establishes that P is transitive. It is thus an ordinal. Moreover rx ~ P for any IX E a, so P is an upper bound of a. And if IX ~ Y for any IX E a, then IX c y for any IX E a, so P c y, so P~ y. Thus Pis the least upper bound of a. THEOREM: Let rx, P be ordinals, and / an order-preserving isomorphism from rx on to p. Then rx=P, and/is the identity map. PROOF: Suppose / is not the identity, and let ~ be the least member of rx for which /(~)=I~. Then 11 E ~ -+ /(11)=1], so ~ c p. Being an ordinal, ~ is an initial segment of p; but it is not Pitself, for in that case /would map a proper initial segment of rx on to the whole of p, which is not possible if/is an isomorphism. Thus ~ E p. Since / is order-preserving, 11
e. e

e

e

THEOREM: For each well-ordered system u=
(order-preserving) isomorphism/rom a on to an ordinal. PROOF: (Uniqueness) Suppose/maps a one-one on to IX, and g does the same on to p. Then go/ -1 is an isomorphism from IX on to p, and it can only be the identity. So/ = g. (Existence) Put b = {x E a S.,(a) is isomorphic to an ordinal}. By uniqueness no such Sx(a) can be isomorphic to more than one ordinal, so let P(x) be the ordinal that Sx(a) is isomorphic to. If YEb and x
I

e

ORDINALS, CARDINALS

17

Our existence proof will be complete if we can show that b=a. And this is simply done. For suppose b=l=a; since b is an initial segment of a, it must be true for some Xo E a that b=Sxo{a). We have just proved that b is isomorphic to an ordinal, so it follows from the definition of b that Xo E b; but it is impossible that Xo E Sxo(a), by the very definition of initial segment. Thus b = a after all. I Consider now any well-ordering R(x, y) whose domain D is not a set; we shall define a functional relation J which establishes an isomorphism between D and On. The functional relation J{x)=rt is defined by the formula 'rt is an ordinalisomorphic to the segment Sx(R)'. This is a functional relation all right, and its domain is D; for whatever x we pick in D there is one and only one ordinal isomorphic to Sx{R). It is also an order-preserving isomorphism. For if x
VOl: [VP{P < rt holds generally.

-+ E(P») -+ E(rt)]

18

INTRODUCTION TO AXIOMA TIC SET THEORY

The proof falls out at once. If E(a.) is false for any ordinal at all, we let a be the least such ordinal. Then E({3) is true whenever (3
THEOREM: Let a be an ordinal,' A a class,' M the class of all maps defined on ordinals less than a, and taking values in A; and H a functional relation with domain M, and taking values in A. Then there exists one and only one map f defined on afor which f ({3) = H(J t (3)Jor every (3 E a. PROOF: (Uniqueness) This is routine. Suppose distinct functions f and 9 both did the job required, and let (3 be the least ordinal in a where f({3)=f:g({3). Then f(y)=g(y) for y<{3, so f t {3=g t {3. But this means thatf({3)= H(J t (3)=H(g t (3)=g«(3), which is impossible. (Existence) Let't' be the set of all {3
ORDINALS, CARDINALS

19

map f,. defined on IX such that, for all fJ
F (fJ) = H (lp) = H (I,. t fJ) = I,. (fJ) for all fJ
The axiom of choice (AC for short) is the following assertion: If a is a set 01 pairwise disjoint but non-empty sets, then there is a set whose intersections with elements ofa are always singletons. Spelt out more fully this becomes

x #:0) A 'v'x'v'y(x e a Aye a -+ (x = y v x n y -+ 3b'v'x3u(x e a -+ b n x = {u})}.

'v'a{['v'x(xea

-+

= 0»)]

Given the other axioms the following statements are equivalent to the axiom of choice. ACt: For any set a there is afunction h mapping the set of all non-empty subsets of a into a in such a way that hex) e x for all non-empty

xc a. A C": The Cartesian product of a family of non-empty sets is itself nonempty. ACt -+ AC: Let b= U a. We apply ACt to b. Every element x of a, being non-empty, is a non-empty subset of b. So take the function h that ACt guarantees, and consider {hex) x e a}. This set has exactly one element from each x e a, and is the set needed to establish A C.

I

20


AC --+ AC": Let (ai)iEI be a family of non-empty sets, and putbi = {i} x ai. Then the family (bi)iEI consists throughout of pairwise disjoint non-empty sets. Let ~ be a set with just one element in each bi(i E I). Then ~ E X ai' isI by the definition ofX. AC" --+ AC': Take any set a and form the product X x. By AC" this xCa x",0

is not empty, so choose an element
Every set can be well ordered. PROOF: AC' follows at once from this. For take some r c a 2 that wellorders a and define a map h: &(a)".{0} --+ a by putting h(x)= the least element of x (mod. r). Conversely, suppose that AC' is true and that the map ZERMELO'S THEOREM:

h:&(a)". {0}

a

--+

satisfies the requirement that hex) E x for every non-empty x c a. Define a second map

g:&(a)". {a}

--+

a

by putting g(x)=h(a".x). Then this satisfies the requirement that g(x)¢:xforanyproperx c a. Let () be some object that is not an element of a (since the class of all sets is not itself a set, there is no difficulty about finding such a 0). We define by recursion a functional relation F: On --+ a u {O} by decreeing

F(IX) =

I

I

I

g({F(f3) f3 < IX}) if {F(f3) f3 < IX} C a (that is, if it is in the domain of g); '" ootherwise.

Suppose that Fnever takes the value 0; in other words, that F(IX) E a for every ordinal IX. Then {F(f3) I f3
I

I

F «(X) = g ( {F (f3) 13 < (X}) ¢ {F (13) 13 < (X} by the definition of g. Thus F is an injection; On is its domain, and all its values are in a. Inverting F, therefore, and using replacement, we can

21

ORDINALS, CARDINALS

infer that On is a set. Since this is false we must conclude that F(IX) is not alwaysina. Thus F(IX)=O for some ordinal IX, and we may suppose IXo to be the least such. In this case F(P) e a for every P
I

I

I

I

BP(a)"'-{0}

-+

afor which hex) E xfor every non-empty x c a.

If R is an ordering and T a subclass of the domain D of R, then an object x of D is called an upper bound (respectively, a strict upper bound) of Tif, for every yeT, we have x~y (respectively, x>y). If Thas a strict upper bound we call it dominated. If the class of upper bounds of T has a smallest element under R, then this element is called the least upper bound of T. An object x in D is called a maximal element of D if there exists no greater element of D (under the ordering R). DEFINITION:

We now prove yet another equivalent of the axiom of choice.

LEMMA: Let u be an ordered system, every well-ordered subset of which has an upper bound. Then u has a maximal element. PROOF: Put u=(a, r) where r c a Z is an ordering. Assuming the axiom of choice in the form AC' we have a map h: BP(a)"'-{0} -+ a for which h (x) E X for every non-empty x c a. Let c be the set of all dominated subsets of a (subsets with strict upper bounds). We define amapmfrom cinto a by putting ZoRN'S

m (x) = h (the set of strict upper bounds of x) . Then for x E c, m(x) itselfis a strict upper bound of x, and so m(x) cannot be a member of x. Choose again some 0 which is not a member of a. As in the proof of

22


Zermelo's theorem, we can define a functional relation F: On ~ a u {O} as follows: m {F (P) fi < a} if {F (P) P < a} E C ; F(a) = 0 otherwise (that is, if {F(P) P < a} has no strict upper bound or is not included in a).

I

1

I I

Again we show that F must take the value 0 at least once. For, were it otherwise, andF(a) E a for every ordinal a, we would have {F(P) P
I I

I

I

I

I

Conversely, we can show that a seemingly weaker statement than Zorn's lemma is sufficient to entail the axiom of choice. CONVERSE: Suppose that there is a maximal element in every ordered system whose linearly ordered subsets all have least upper bounds. Then the axiom of choice is true. PROOF: Let a be a set of non-empty pairwise disjoint sets; and let b = U a. Put X equal to the set of all subsets of b which do not have more than one element in common with any x E a; then X is ordered by set inclusion. Suppose Y is any linearly ordered subset of X. We shall show that Y has a least upper bound in X. Now the least upper bound of Y (under the c-ordering) is U y: so we will show that U y E X. yeY

For any x in a, then, the set x n

U

yeY

yeY

y must be proved to have fewer

23

ORDINALS, CARDINALS

than two elements. There are two cases to be considered. In the first, x n y = 0 for every ye Y; here we have at once x n U y=0, and nothing yeY

remains to be proved. In the second, x n y is a singleton for one or more y e Y. But however many y e Y yield a singleton when intersected with x, it will always be the same singleton. For suppose x n Yo ={u} and x n Y1 = ={v}. Then since Y is linearly ordered, Yo u Yl is one or the other of Yo, Yb which means that x n (Yo u Yl) = {u} u {v} is a singleton; and this means that u = v. It follows at once from all this that, in this second case, x n U y is itself a singleton, so that U Y e X, as was required. yeY

yeY

By supposition, then, X has a maximal element Yo. We claim that Yo n x has exactly one element for each x e a. Were it otherwise, indeed, Yo n x would have to have no elements for some x e a (since Yo e X). In this case we could take any element e of x and form the set Yo u {e}. This, however, would equally be in X, and would destroy Yo's maximality. I CARDINALS

(In this section the axiom of choice will be used.) Two sets a and b are termed equipollent (or equinumerous, or sometimes just equivalent) if there is a one-one map (a bijection) from a on to b. It is to be observed that two sets a, b can be equipollent in an intuitive sense without being so according to this definition; for if A, B are the parts of the universe corresponding to a, b, there may be a bijection, intuitively speaking, from A on to B. But unless this bijection itself corresponds to an object of the universe there is no reason to suppose that a and b are therefore equipollent. It is obvious enough that the relation 'x is equipollent to y' is an equivalence relation whose domain is the class of all sets. With the axiom of choice, however, we can say more than this; namely that every set is equipollent to an ordinal. For if a set can be well ordered, it is certainly equipollent to the ordinal of its well-ordering. The least ordinal equipollent to a set a, written ii, or sometimes card(a), is called the cardinal of the set a. It is trivial that a and b are equipollent if and only if ii = r;. We write Card for the class of all cardinals. It is defined by the formula

24


Card(rt.): 'rt. is an ordinal not equipollent to any smaller ordinal'.

LEMMA: Let rt. be an ordinal and ~ a subset of it. Then the ordinal of the well-ordering induced on ~ by the ordering of rt. is less than or equal to rt.. PROOF: Let Pbe the ordinal in question, and f: ~ -+ Pthe isomorphism between ~ and p. We show that for y E~, f(y)t;;;;,y, which immediately gives pert., and so p t;;;;, rt. as required. Suppose, on the contrary, thatf(y»y for some y E~. There will be a least such, Yo. For y E ~ we have then

y < Yo

-+

fey)

y ~ Yo

-+

fey) ~ f(yo) > Yo·

t;;;;,

y < Yo;

and also As in an earlier theorem,! clearly misses out on Yo; but since it does take greater ordinals as values (for example, f(yo»Yo), it cannot after all have an ordinal for its range; and this contradicts the hypothesis .• THEOREM: Let a and b be non-empty sets. Then thefollowing conditions are equivalent. (1) There is a one-one map from a into b. (2) There is a map from b on to a. (3) Zit;;;;,b.

PROOF: (1) -+ (2» Let j be an injection from a into b. We define a surjection s in the opposite direction as follows. Take any Xo in a, and let s(y) be Xo for any y in b that is not in the range ofj; and for elements y of b that are in the range ofj we can take s(y) as the inverse image of y underj, sincejis one-one on its range. (2) -+ (1» Conversely, suppose s: b -+ a is a surjection. AC gives us a map h: .9(b)""-{0} -+ b such that hey) E Yfor every non-empty Y c b. So for x E awe putj(x)=h({y E b s(y)=x}). This is manifestly one-one. (3) -+ (1» IfZit;;;;,b we have two bijectionsf: a -+ a and g: b -+ b; and, sincea c b, anidentityinjectioni:a -+ b. Composing these into g-l 0 i oJ, we produce what is clearly an injection from a into b. (1) -+ (3» Supposej: a -+ b is an injection. Then, withfand g defined as above, the map g 0 j 0 f -1 is an injection k from a into b. The range of k is some subset c jj which, according to the previous lemma, is

I

e

ORDINALS, CARDINALS

25

isomorphic to some ordinal p~fj. Thus ais equipollent with p, and since it is a cardinal, a~p; so a~b.1 CoROLLARY (CANTOR/BERNSTEIN THEOREM): For two sets a, b to be equipollent it is necessary and sufficient that there be injections from each one into the other. PROOF: For given the two injections we get at once a~ band b ~ a, and thusa=b.1 (Although the axiom of choice was used in this theorem, and thereby in the proof of the corollary, it could, in the latter case, have been dispensed with.) THEoREM (CANTOR): Foreveryseta,a
I

THEOREM: For no set a does there exist a map from a on to flJ(a); and, a fortiori, no injection offlJ(a) into a. THEOREM: The class ofcardinals is not a set. PROOF: Suppose it were, x for example. Then x would be a set of ordinals with a least upper bound A= U IX. Since every ordinal is equipollent "'ex

with some cardinal or another, every ordinal is equipollent with some subset of A. Now let Ybe the class of all well-orderings defined on subsets of A. Every r E Y is a subset of A2 ; so Y( r) is equivalent to the formula r E flJ(12) A Y( r), which establishes that Y is a set (by the scheme of comprehension). But every r in Y is associated with a unique ordinal (namely, the ordinal of the well-ordering r). Consequently, we can write down a functional relation performing this association, and the range of this functional relation is On. Since Y is a set, this contradicts the replacement scheme. I Let a, b be two disjoint sets, and a', h' two more, respectively equipollent

26


with a and b. Then a u b and a' u b' are equipollent too; for from the first to the second there is a bijection whose restrictions to a and bare just the given bijections from a to a' and from b to b'. So let ct, /3 be cardinals. There certainly exist disjoint sets a, b with these cardinalities, for we can, if we like, take ct x {O} for a and /3 x {I} for b. Whatever disjoint a and b we actually choose, we define the cardinal sum ct + /3 as the cardinal of a u b; it is easily seen that ct + /3 is quite independent ofthe choice of a and b. Amongst the properties easily proved to hold for cardinal addition are commutativity (ct + /3 = /3 + ct) and associativity (ct + (/3 + ')') = (ct + /3) + I'). To get at cardinal multiplication we can drop the disjointness conditions on a, band a', b' that were imposed above. Iff: a -+ a' and g: b -+ b' are some appropriate bijections, then we can define a bijection h: a x b -+ a' x b' by writing h«x,y»=(f(x),g(y» for all xea and yeb; thus these Cartesian products are equipollent. So if ct, /3 are cardinals we define the (cardinal) product of ct and /3, written for brevity ct/3, or ct· /3, as the cardinal of ct x /3. It should be clear that if a has cardinal ct and b has cardinal /3, then a x b has cardinal ct/3. If ct, /3, I' are all cardinals we can easily check that the commutative (ct/3 = /3ct) and associative (ct (/3')' ) = (ct/3) I') laws of multiplication hold; there is no more difficulty involved in checking the distributive law (ct(/3 + ')') = =ct/3+ct,),); we simply pick three sets, a, b, c, the latter two disjoint, and observethatax(b u c)=(axb) u (axe). More generally, let (ai)ieI> (bi)ieI be families of sets each indexed by I. Suppose that for every i e I we have lii = 1)1' Then X ai and X bi are ieI

ieI

equipollent. This is proved by noting that for any i e I the set Bi of bijections from ai on to bi is non-empty; by AC", therefore, the product of the family (Bi)ieI is not empty. So there is a family (CPi)ieI of maps, each CPi being a bijection from ai on to bi' We now define the required bijection fromXaj on to Xb i by associating each element (Xi)ieI of the former set ieI

ieI

with (CPi(Xi)ieI in the latter. If (ct i); e I is a family of cardinals we call the cardinal of the set X cti the

product of the family, and write it

ncti. It is easily checked that this cardinal ieI

ie I

is the cardinal of the Cartesian product of any family (ai)ieI indexed by I and satisfying lii = cti for all i E I.

ORDINALS, CARDINALS

27

Similar considerations for addition lead us to the following. Let (al)ieI and (bi)ieI be families as above which satisfy the further condition that i¥:-j ~ ai II aJ=bi II bJ =0. Then Uai and Ubi are equipollent. As ieI ieI before, we prove this with the axiom of choice, which supplies us with a family (lPi)leI of maps, each lPi being a bijection of al on to bi' Then if x e U ai it can only be in one of the a" say ai (x), and so we can couple it ieI with lPi (x) (x) in U b" thus establishing the desired correspondence between leI Ualand Ubi' tel ieI Given any family of cardinals (lXi)ieI we can construct a family of disjoint sets (ai)ieI which are equipollent to them for every i e I, by writing ai = IX; x {i}. The cardinal of Ua" which is quite independent of leI the choice of the disjoint family, is called the sum of the family (lXi)lel> and written LOCI' leI _ To define exponentiation of cardinals, let a= {l and b = 5'. Then ba (the set of maps from b into a) is equipollent with b'a'. For supposef: a ~ a' and g: b ~ b' are appropriate bijections. Then to each IP e ba we can associate the map l/I e b'a' defined by l/I = f 0 IP 0 g-l . This clearly establishes a bijection between the two sets of maps. So if 0(, P are cardinals, the cardinality of the set Poc is called the pth power of 0(, and written ocp• If a, b have cardinalities oc, p respectively, it is easily checked that ba has cardinality ocp• The usual laws of exponentiation for cardinals, that O(P+Y=ocp• O(Y and that (O(P) Y= O(Py are easily established from this definition. FINITE ORDINALS

(This section does not use the axiom of choice.) We recall that the next greatest ordinal after a given ordinal 0( is its successor 0( U {O(} (or IX + 1 for short; but this + is different from the + of cardinal addition above). If p is the successor of 0( we say that 0( is the predecessor of p. An ordinal oc is called finite if every p ~ 0( (apart from 0) has a predecessor. The formula '0( is a finite ordinal' is easily expanded into

28


On(a) A Vp[On(f3) A 13 c: a A 13#0 ~ 3y(p=y u {y})J. Finite ordinals are also known as natural numbers. It should be obvious that if a is a finite ordinal and 13 ~ a, then 13 is finite; and that if a is finite, so is a + 1. Mathematical Induction: Let P be a class such that P(O) is true, and for every finite ordinal a, Pea) ~ P(a+ 1) is also true. Then Pea) is true for every finite ordinal a. For if not, take the smallest finite ordinal ao not in P. Since 0 is in P, ao # O. Thus it has a predecessor 13 o. As ao was the smallest finite ordinal not in P, we must have P(Po). We are given that P(Po) ~ P(Po+ 1). So P(ao), which is impossible. It is worth pointing out again that the definitions above of the words 'finite' and 'natural number' are not intended to provide for them their usual senses. In future, therefore, everyday uses of the words will be explicitly recorded as such. It is easily seen that what we would intuitively say was an ordinal a with a finite number of elements is indeed finite according to the above definition. But it could happen that a finite ordinal 13 had, intuitively speaking, infinitely many elements; in this case since 13 c: a is ruled out, 13 would be greater than a. In addition, the part of the universe made up of these finite ordinals (in an intuitive sense) could not then be a class; for supposing P(x) to define the class, we would have -,P(f3), and so could prove the existence of a least ordinal Po for which -,P(f3o). Then if Po were equal to Yo + 1, we would have P(Yo), showing that, intuitively speaking, Yo had finitely many elements; but with f3o=Yo u {Yo} this would hardly be possible.

Every finite ordinal is a cardinal. This is proved by induction. 0 is obviously a cardinal. So suppose that a is a finite ordinal which is a cardinal, whilst a + 1 = a u {a} is not. Then there must exist an ordinal y < a + 1 and a bijection f from a + 1 on to y. As a+ 1 #0, y cannot be 0; thus it is 13 u {f3} for some 13, and since y
ORDINALS, CARDINALS

29

does map Ct one-one on to p, by setting g(x)=f(x) for all x in Ct except for 1'[; and g(1'[)=~. Since Ct is a cardinal, we have a contradiction.• We call a cardinal finite if it is a finite ordinal. Let us note the following consequence of the theorem above. If Ct is a finite ordinal, then its successor Ct + I is indeed the cardinal sum of oc and I. Thus within the domain offinite ordinals our notation is unambiguous. THEOREM: ljCt, P are finite cardinals, so too are oc+ p, ocP, ocp• PROOF: Each proof is done by induction on p. It is trivial that if oc is finite and Pis 0 then oc+P is finite. Moreover, since oc+(P+1)=(oc+P)+ I, a finite oc + Pmeans a finite Ct + (P + I). This proves the first part. The proof for multiplication is similar. Since oc(P + 1) = ocp + oc, the assumption that ocp is finite when oc, P are, together with the previous result, yields at once that oc (P + 1) is finite. Again, the identity ocP+ 1 = ocP• oc supplies all that is needed to establish that ocP is finite .• We are now in a position to state the one remaining axiom of Zermelo/ Fraenkel set theory. 5. AXIOM OF INFINITY This axiom says, in effect, that there is an ordinal which is notfinite. On the strength of the axiom we write 0) (or, in some cases, N) for the first ordinal which is not finite. Thus 0) is the set of finite ordinals. For if oc is a finite ordinal we do not have 0):::;; oc (which would make 0) finite); so, necessarily, oc <0) or, equivalently, oc E 0). Conversely, if YEO) we have Y < 0), and so y is finite, by the definition of 0). Thus the axiom of infinity can be stated in the form: the class offinite ordinals is a set. For the class in question is an initial segment of On and so, if a set, is an ordinal 0); but it is not a finite one, since no ordinal belongs to itself, and in particular 0) ¢ w. An ordinal oc#O without a predecessor is called a limit ordinal; it is thus a non-zero ordinal for which P
P
For if oc is not a limit it is y+ 1, which means that sup P=y. If it is a P
limit, put sup p=y; since y:::;;oc anyway, suppose y
30


of limit, then, y + 1 < ex, and so y is not the supposed supremum after all, since y+ 1 :$')'. So another way of formulating the axiom of infinity is: there is a limit ordinal. For, given the axiom, 00 could only have a predecessor if that predecessor were finite; which would make OJ finite too, directly contrary to its definition. Since a limit ordinal is necessarily not finite, the converse holds too. A fourth way of expressing the axiom of infinity is by

3x[Oex A'v'y(yex-+yu{y}ex)]. Given the axiom we can take 00 itselffor x. Conversely, suppose the above sentence holds, and choose any a satisfying it. Let ex be the least ordinal not in a. It cannot be 0, and if ex=P u {P} we get P
(Unless otherwise stated, the axiom of choice will be allowed in this section.)

A set a is called finite if its cardinal is finite, infinite otherwise; if a~ OJ, we say that a is denumerable. Every infinite set includes an infinite denumerable subset (one, that is, equipollent with OJ) • For if a is infinite and / a bijection of a on to a, the range of/ trois a subset of a equipollent to OJ. A set a is infinite iff it is equipollent with one 0/ its proper subsets. If a is finite, and b a proper subset, take x E a"-.b. The cardinality of a is certainly not zero, but it is finite, so of the form ex+ 1=ex u {ex}. There is no difficulty in finding some bijection / from a on to ex u {ex} which maps x to ex; in this case/ t b maps b one-one into ex. Consequently 1j ~ ex, so 1j < and b is not equipollent with a. If a is infinite, a-;:,OJ. Writing/for some bijection from a on to a, we define an injection g from a into itself by setting THEOREM: PROOF:

a,

g(x) = x, if f(x) -;:. 00; g (x) = y, if f(x) < OJ and f(y)

=

f(x) + 1.

31

ORDINALS, CARDINALS

The range of g is a proper subset of a since it does not contain f thus g is a bijection from a on to a proper subset.•

-1 (0);

We will write Card' for the class of infinite cardinals; it is a subclass of On, but not a set (for then Card would be too, since OJ, the class of finite cardinals, is a set). So Card' is just a well-ordered class, and therefore there must be a functional relation y = ~(IX) establishing an isomorphism between On and Card'. For simplicity we write the infinite cardinal ~(IX) as ~I%; the relation y= ~I% then abbreviates 'y is an infinite cardinal, and the set of infinite cardinals that are less than y is isomorphic, as a well-ordered set, to IX'. We have ~o =OJ; and, for each IX, ~I%+1 is the first cardinal > ~I%'

U

IflXisalimitordinal, ~I%=

To show that yE

U

P
U

~I/

1/<11.

~I/' so Y E ~I/ for

1/<1%

is a cardinal, suppose that y <

C

since IX is a limit ordinal.

U

I/
~I/ >

U

1/<11.

1/<11.

~II.

~I/'

so

U

~I/

~I/'

y and that there is

no bijection from

on to any smaller ordinal y. Since this cardinal U ~I/ ~ ~I/ for all {3
U

I/
some {3
Y::;;~P<~I/+l

This proves that

~I/'

for all {3
U

I/
~I/ c ~rz'

~ ~II.'

1/<11.

But con-

and this proves the result.

By Cantor's theorem we have 2N~~~rz+h since 2N~ is the cardinal of &(~",).

The continuum hypothesis (CH) is the sentence 2No=~1; alternatively,

&(OJ) can be so well ordered that every strict initial segment is denumerable. To show that this second formulation entails the first (the converse is clear), note that the ordinal of such a well-ordering could not in any case be denumerable, since, by Cantor's theorem, &(OJ) is not. By the specification of the well-ordering, it follows that the ordinal in question is the first non-denumerable ordinal, that is ~1' The generalized continuum hypothesis (GCH) is the sentence: 2N • = ~
32


Write On 2 for the class of all ordered pairs of ordinals (ex, P). On this class we can define a well-ordering relation R as follows: (ex, P):;;;' (y, (» (mod. R) if and only if sup (ex, P) < sup(y, (5) v sup (ex, P) = sup(y, (5) 1\ ex < y v sup (ex, P) = sup(y, (5) 1\ ex = y 1\ P:; ;, (). Every segment is seen to be a set. For if S~(R) is an initial segment, and = (ao, Po), and (ex, P) < (ao, Po) (mod. R), then

e

sup (a, P):;;;, sup (exo, Po). Putting Yo = sup (exo, Po), we note that every pair (a, P) in S~(R) is a member of the set (Yo+1)2; so S~(R) is a subclass of that set, and thus a set itself. On the other hand, take a non-empty set X all of whose elements are in On 2 ; the set of all sup (a, P) for (a, P) E X has a smallest element Yo; the set of all ex such that for some P we have both (ex, P) E X and sup(a, P)=Yo has a smallest element exo; and the set of P for which sup(ex o, P)=Yo and (exo, P) E Xhas a least element Po. From all this it drops out that (ao, Po) is the least element of X (mod. R); and so R is a well-ordering. Since On 2 is therefore a well-ordered class, but not a set, a relation y=J(a, P) can be found which establishes an isomorphism between On 2 and On. In the definition of J, be it noted, there is no need for the axiom of choice. THEOREM: For each ordinal ex, ~; = ~... PROOF: If the theorem is false, it fails first at some ordinal p. To obtain a contradiction we need only show that ~p' ~p:;;;' ~P' since there is no trouble with ~; ~ ~p. Restricting J to ~p x ~p we get a one-one map, and so the proof reduces to showing that for a, PE ~P' J(a, P) E ~p. Now J(a, P) E ~p +-+ J(a, P) <~P' so, ~p being a cardinal,

J(ex, P) E ~p +-+ I (a, P) <

~p.

So if we prove that Jea, P)<~pfor a, PE ~p, we are finished. Since J(a, P) is the set of ordinals less than J(a, P), it is the range of J restricted to the set of all (ex', P') less than (a, P) (mod. R). Writing y

33

ORDINALS, CARDINALS

for sup (ct, P), we easily see that this latter set is included in (y + 1)2 (since (ct ' , P') «ct, P) ~ ct'
(y+ 1)2=y+ 1 <~p. EachcaseseparatelyyieldsJ(ct, P)<~P' and so the theorem is proved.• COROLLARY: ~:= ~(Jor every natural number n ~ 1, and every ordinal ct. PROOF:

Immediate, by induction on n.•

COROLLARY: ~,.. ~p=sup(~", ~p). PROOF: If~,,<~p,

~p

then clearly

= l'~p ~ ~,,'~p ~ ~p.~p = ~P'

and so ~,,' ~p = ~P' as required .• COROLLARY: Let (K1)leI be a family of cardinals, all non-zero. If i, or any Kj , is infinite, then l: K1=SUp(i, SUPKJ leI

PROOF;

Recall that

leI

L KI is defined as the cardinal of the set X = U K/ X {i}.

leI

leI

U {OJ x {i}, that is that X includes leI which is equipollent with I. Thus X ~ i. It is clear that X ~ K for

Since each KI~O, we have that X:::>

{OJ X I, every i E I, and soX~sup(i, SUpK j ).

j

ieI

For the reverse inequality, put K = sup K i • Then X c: Thus X ~K' j which, since either

iel

K

or

U

ieI

K

x {i} = K x I.

i is infinite, is equal to sup(i, K) ••

An isomorphism between the class of finite sequences of ordinals and On: (To establish this we do not need the axiom of choice.) For a class C, afinite sequence s of objects of C is, by definition, a map whose domain is a natural number, and whose values are in C. This natural number, the domain, is otherwise called the length of the sequence, and is written I(s). We write (i(C) for the class of all finite sequences of objects of C. An ordering R can be defined on (i(On), the class of finite sequences of

34


ordinals, by setting s < t (mod. R) if and only if sup{s) < sup{t) v sup{s) = sup{t) A I{s) < let) v sup{s) = sup{t) A l{s) = l{t) A s:f= t A sen) < ten) for the first n where

s(n):f= ten).

(Here sup(s) stands for the greatest value taken by the function s.) We show that R, defined in this way, is a well-ordering. This involves, in the first place, that the class S,(R) of sequences I. If sup(s)=y, however, and l(s):f=n, we have in the same way l(s»n=l(f); and so s> I. Finally, if sup(s)=y and l(s)=n, let i I (i), so once more s > I. But a well-ordered class that is not a set is isomorphic to On; and there is consequently a functional relation s=J(rx.) which establishes an isomorphism between On and q (On).

CHAPTER III

THE AXIOM OF FOUNDATION

The axiom offoundation (AFfor short) is the sentence 'v'x[x¥=0~3Y(YEX A

ynx=0)];

in other words, the axiom states that every non-empty set has an element which is disjoint from it. From the axiom it follows that there is no infinite sequence (that is, no map with domain (1)) (un)ne;ro for which Un +l E Un for all n E (1); for if x were the set of terms of such an infinite descending sequence (its range, in other words), it would fail to contain an element disjoint from itself. In particular, therefore, the axiom implies that for every x, x ¢; x; and, more generally, that there are no E-cycles - finite sequences xo, ... , Xn -l thatis,satisfyingxo EXl E ... EXn -l E Xo. For, again, the set {xo, ... , Xn -l} would contradict AF. We can take advantage of the axiom to provide a more simple definition of ordinals than the one given earlier. This is established in the following theorem. THEOREM: For X to be an ordinal it is necessary and sufficient that the two conditions

'v'U'v'v [u E X

A V

E X ~ (u E V

'v'U[UEX

U c X]

V

U= v v

V

E u)]

and ~

be satisfied.

PROOF: The necessity of both conditions is immediate. Conversely, we note first that E imposes a strict linear order on any X satisfying them. For by AF we cannot have U E v and v E U together; moreover, when U v, WE X, and U E v and v E w, the possibility U= W is ruled out (for then U E v and v E u), and the possibility W E U is ruled out (for then we get the cycle U EVE W E u); and so by the first condition all that is left is U E W. To show that this ordering is a well-ordering, we take any non-empty Y c X; by AF there is some U E Y satisfying un Y=0. But this means

36


that if v e Y then v ¢ u, and therefore that u is the desired least element (mod. e) of Y.I Without necessarily supposing the axiom of foundation to hold in the universe IYI, we can define by recursion a functional relation y= VIZ by setting Vo =0; VIZ = U &,(Vp). P
It is clear enough from this that ot~ot' ~ VIZ C V"'. Consequently, since Voc+l = U &,(Vp), we have V,,+l =&,(V,,)jor every ordinal a. p",,,

On the other hand, for limit a we have p< a. ... p+ 1 < a., so V,,= U &'(Vp)= U Vp+1 = U Vpforeverylimitordinala.. P
U &'(Vp) = V"' p<" from which a e V,,+1' contrary to hypothesis. Likewise, let a be the first ordinal, if there is one, such that a e V". Then a e U &,(Vp), so a is in &,(Vp) for some p
U

p<"

Vp+ 1 =

37

THE AXIOM OF FOUNDA TION

THEOREM: The axiom of foundation holds if and only if V is the whole universe; or, briefly, AF +-+ Vx Vex). PROOF: Supposing VxV(x) to hold, we pick a non-empty set a, and from its elements pick some set b of minimum rank !Y.. Then b's elements all have rank
U

x ef
f(n). Then it is obvious that Xc Y; and further, Y is

new

transitive. For if x E Y, then x ef(n) for some nero, so xc f(n+ 1), so x c Y. Now if Z is a transitive superset of X, we prove by induction that every f(n)c Z; n=O is immediate; if x e fen) c z, x e Z, so x c Z, becauseZistransitive;thereforef(n+l)= U xc Zasrequired. xef(n)

This completes the inductive step, and we can conclude that

Y=

U

new

f(n) c Z .•

Returning to the stranded theorem, suppose that AF holds, but that some object a is not in V. Let b be a transitive set including a, and b' the set of elements of b that are not in V. Then b' is non-empty; for a is not in V, so has an element Xo not in V; and since a c b, Xo is also in b'. But for every x E b' we also have b' () x non-empty. For if x eb' , it is not in V, so has an element y not in V. Since y e x e b, and b is transitive, y e b; and thus y E b' , by the definition of b'. But then y e b' () x, which therefore cannot be empty. Thus b' contradicts AF, and the theorem is established .• Let us note one property oftransitive closures.

38


THEOREM: ~(X)=X u

U

yeX

~(y).

PROOF: ~(X) certainly includes X. If y E X then y c ~(X), and as ~(y) is included in every transitive set including y, ~(y) c ~(X). Thus Xu U ~(y) c ~(X). Conversely, if Z=Xu U ~(y), then Xc: Z, yeX

yeX

and, moreover, Z is transitive. For if v E u E Z, then either u is in some ~ (y) with y E X, in which case v E ~ (y); or u E X, in which case v E U c ~(u). Either way, v E U ~(y), so V E Z, as desired. Thus yeX

Xc Z, and Z is transitive; so 'tf(X) c Z, and the proof is complete.

I

Well-foundedRelationsandExtensionaIRelations: Let r be a binary relation on a set a (so r c a 2 ); r is said to be well founded on a if for every nonempty X c a there exists an x E X such that with domain a is called collapsing for r if for all x E a we have q>(X) = {q>(y) lYE a

1\

Evidently, if q> is collapsing the range of q> is a transitive set, since each element of q>(x), for x E a, is of the form q>(y). THEOREM: Ifr is a well-founded relation on a there exists one and only one map with domain a which collapses for r. Moreover, the range of this map is in the class V. PROOF: (Uniqueness) Let q>, tjJ be two such collapsing maps, and write X for {x E a q>(x);6tjJ(x)}. If q>;6tjJ, X must be non-empty, so there will be an Xo E X for which Vy E X( (xo);6tjJ(xo), and so without loss of generality we can assume a u that is in q>(xo) but not in tjJ(xo). Now q> is collapsing, so there is ayE a for which q>(y)=u and E r. By the specification of xo, y cannot be in X; so q>(y)=tjJ(y)=u. But tjJ collapses too, so from tjJ(y)=u and E r we get u E tjJ(xo), which is a contradiction. (Existence) A subset Z of a is called r-transitive if x E Z 1\ has domain a and is collapsing for r then q> t Z is collapsing for r (') Z 2 when Z is an r-transitive subset of a. If Z, Z' are both r-transitive subsets of a, so too is Z (') Z'; and so if d>, d>' are col-

I


39

lapsingmapswithdomainsZ,Z'respectively,!l> t Z II Z'and!l>' t Z II Z' are also collapsing maps, both with domain Z () Z'; and so, by the uniqueness part of the theorem, they are identical. So let qJ be the union of all the collapsing maps !l> that are defined on any r-transitive subset of a. From our discussion above it follows that qJ is also a map, and that its domain Y is also an r-transitive subset of a (since it is the union of a set of r-transitive subsets). Now qJ is obviously collapsing; and so Y must be the largest r-transitive subset of a which is the domain of a collapsing map. Is Y equal to a itself? If not, take some Xo e a"'.Y such that Yy e a"'. Y( (y, xo> ¢ r); that such an Xo would exist under these conditions follows at once from the well-foundedness of r on a. We now define a map t/J on Yu {xo} such that t/J and qJ are identical on Yand t/J(xo)= {qJ(y) lye a 1\ (y, xo> e r} (this definition is a correct one since (y, xo> e r ~ ye Y). But now it falls out at once that Yu {xo} is an r-transitive subset of a and that t/J is collapsing. And this is impossible, since the domain of t/Jis Yu {xo}, which properly includes Y. Consequently Y = a; and qJ is a collapsing map with domain a. It remains to show that rng(qJ) is in V, or, what is the same thing, that all its elements yare. Suppose otherwise, and put X = {x e a I -, V(qJ(x»}; X is then non-empty. As r is well founded there must be an Xo e X such that YyeX«y,xo>¢r). Thus -,V(qJ(xo», which means that there is au e qJ(xo) such that -, V(u). On the other hand, as qJ is collapsing, there exists aye a such that qJ(Y)=u and (y, xo> e r. Thus y ¢ X, in view of the choice of Xo; so V(qJ(y», so V(u); and this contradicts the conclusion of the previous paragraph.• A binary relation r on a set a is called extensional if, whatever elements x, y may be of a, ifYzea[(z,x>er-(z,y>er] then x=y; or, more succinctly, if the set a, together with the relation r, satisfies the axiom of extensionality. A set a is called extensional if the membership relation is extensional on a; that is, if x () a=y () a ~ x=yforany x,yina. Observe that every transitive set is extensional; for if a is a transitive setandxeathenx () Q=X.

40


THEOREM: Let r be an extensional relation which is well founded on a. Then

there exists a unique isomorphism from a, together with the binary relation r, on to a transitive set (together with the membership relation); and this transitive set is in V.

PROOF: Any isomorphism such as that required by the theorem is evidently a collapsing map; and so by the previous theorem we have uniqueness. To prove that there is any such isomorphism at all, we consider that one map qJ defined on a which collapses for r, and show that it fills the bill. Let b be the range of qJ; we have already seen that b is transitive and in V. Firstly, qJ is injective. For otherwise there would be an element d of b of minimal rank for which there were two inverse images x, yEa; that is to say, x, yEa, x¥:Y, and qJ(x)=qJ(y)=d. As x, yare distinct and r is extensional, we can assume, without loss of generality, that we have some u E a for which (u, x) E rand (u, y) ¢ r. The former yields qJ(u) E qJ(x)since qJ is collapsing; so qJ(u) E qJ(Y) too. Thus for some v E a we have (v, y) E rand qJ(v)=qJ(u); and as (u, y) ¢ r, we also have v:l:u. But qJ(u)=qJ(v) e d, whence qJ(u), qJ(v) are of rank less than d, in contradiction to the definition of d. Secondly, (y, x) e r -+ qJ(Y) e qJ(x), since qJ is collapsing. And lastly, the converse of this is proved as follows. If qJ(Y) e qJ(x) then there is a z e a such that (z, x) e rand qJ(z)=qJ(Y) (since qJ is a collapsing map). But qJ is also injective, so Z= y. Thus (y, x) e r. This concludes the proof that qJ is indeed an isomorphism .• Note that this theorem is a generalization of the theorem on pp. 16f. For if r is a strict well-ordering on a it is easily seen to be both extensional and well founded. What is more, the range set of the collapsing isomorphism is transitive and well ordered by E; an ordinal, no less. The following corollary is proved with the help of the axiom offoundation. It will be used in Chapter VIII. COROLLARY: Every extensional set is isomorphic, under a unique isomorphism, to a transitive set. PROOF: Let a be extensional, and r the restriction of the membership relation to a; thus r= {(x, y) E a 2 1 x e y}. It follows at once from AF that r is well founded; it is extensional, by hypothesis; and so the theorem above gives the result.•


41

THE HEREDITARILY FINITE SETS

A set a is called hereditarily finite if it is in V"" or, in other words, if it is in Vn for some natural number n. Without having to use the axiom of choice we are able to define a functional relation without parameters that maps V", one-one on to ill. LEMMA:

lin is a natural number, n{o, I} is equipollent with a unique natural

number.

The uniqueness is trivial since no natural number is equipollent with any ordinal apart from itself. The rest is a simple induction on n. Given that n{o, I} is equipollent with k, say, n+l{o, I} is equipollent with k x {O, I}, and thus with 2k. And that is all .• We now define recursively on the natural numbers a sequence of bijections CPn from Vn on to a natural number Vn• For n=O, CPn=0. So suppose that we are presented with CPn: Vn ~ V n; our object is to define CPn+l: Vn+1 ~ Vn+1 • In terms of CPn we can define a bijection I/In: &'(Vn) ~ Vn{o, I} (that is, from Vn+1 on to Vn{O, I}) by PROOF:

I/In (X)(i) = 1 if cP;; 1 (i) EX, I/In (X)(i) = if cP;; 1 (i)¢= X,

°

where X c Vn and 0:::; i
42


on Vto by setting x
rn(x) < rn(y) v [rn(x) = rn(y) = n

A

x < y(mod, rll) ] ,

Every initial segment under r is isomorphic to a natural number; for the set of r-predecessors of Xo is a subset of VII where n is rn(xo); and so its ordinal is finite; This shows that the well-ordering r on V", is isomorphic to the wellordering of en; and the isomorphism K between en (ordered by membership) and the well-ordered system V"" r> is the functional relation sought, since it has been defined without parameters,

<

RELA TIVIZED FORMULAS

Let X be some class, and E(xo, .. " XII-l) some formula in (at most) the variables xo, .. " XII-l' whose parameters are all objects of X, We write EX(xo, .. "XII_l) for the relativization of the formula E to X, namely that formula defined in the following way by (informal) recursion on the length of E, (Since the formulas are not objects of the universe, there is no question here of a recursion on the integers of the universe, but one on intuitive objects,) (1)

(2) (3) (4)

E is of the form x E y, X = y, X E a, x = a, a E x, a = x, a E b, a = b (a, b, objects of X), E is in this case called atomic, and is its own relativization. E is -, F ; then EX is -, FX. E is F v G; then EX is F X v GX• E is 3xF(x, xo, .. ,' X,.-i); then EX is 3x[X(x) A FX(x,xo, ... ,X,._I)]'

It follows at once from (2) and (4) that if Eis VxF(x, xo, .. " XII-l)- that is, at length, -,3x-,F(x, xo, .. " XII-I) - then EX is Vx[X(x) - FX(x, xo, "" XII-l)]'

Note that if the class X qualifies as a set, then the relation X(x) takes the form X E X, Consider any n objects ao, .. " a,.-l of the class X, Then it is apparent that the formula EX(ao, .. " all-l) is true in the universe iff the formula E(ao, .. " an-I) is true in the class X (augmented by the binary relation E t X), As an example, a set X is extensional if and only if the formula EX is true (in the universe), where E itself is one form of the axiom of


43

extensionality, 'v'x'v'y[x = y+-+ 'v'Z(ZEX +-+

ZE

y)].

RELATIVE CONSISTENCY OF THE AXIOM OF FOUNDATION

In this section we shall show that the deployment of the axiom of foundation cannot of itself lead to contradiction within ZF, by assuming a model for ZF - that is, a universe 0/1 - and from it constructing a model for ZF + AF; this latter model will be a set (in the intuitive sense), augmented by a binary relation, for which all the ZFaxioms hold, and the axiom of foundation too. More precisely, we show that if ZF has a model, so also has ZF + AF; this is what is meant by relative consistency. In fact, ifd/t is a universe, then the ZFaxioms and the axiom offoundation all hold in the class V constructed within d/t. This amounts to proving that for each axiom A of ZF+AF, A V holds in 0/1. Axiom of Extensionality: If x, yare in V, so too are all their members; so if they have the same members in V, they have the same members all told. Thus x = y, since the axiom of extensionality holds in 0/1. Union Axiom: If x is in V, anything in the lemma on p. 36, Ux itself is in V.

Ux is obviously also in V, so by

Power-set Axiom: If x is in V and y E 8i'(x), then y c: x, and so all y's elements are in V; thus y is too, and by the same lemma, 8i'{x) belongs

toY. Scheme of Replacement: Let a be a set in V, and R{x, y) some formula (whose parameters are in V) which, within the class V, defines a functional relation. Clearly this functional relation is defined in d/t by the formula V{x) A V(y) A R V (x, y); since replacement holds in 0/1, we can apply it to this functional relation, and obtain in 0/1 a set b made up of the images of a under it. Since everything in b is in V, b itself must be, and the proof is complete.

44


Axiom of Infinity: Every ordinal is in V, ill in particular. Since 0 E 1\ 'v'x[x E ill ~ X U {x} E illJ, the axiom of infinity holds in V.

ill 1\

Axiom of Foundation: Take a non-empty set a in V. Let b be some element of a of minimal rank. But every element of b is of lower rank than b, and therefore cannot appear in a. Thus b n a = 0.

This proof of relative consistency - like all others - can be looked at from a different angle. Without reference to models, we can treat the theory ZF as a formal system, in the sense that its axioms and, more generally, its parameter-free formulas are regarded as mere finite sequences of symbols. The rules of inference (which need not be spelt out here, but can be found in [4J for example) are then formal rules allowing the derivation of certain finite sequences from certain others. A proof is just a finite sequence of formulas A o, ... , An-I' each of which, if not an axiom, is derivable from earlier formulas in the sequence by one of the rules of inference. A theorem of the theory is any sentence which is the last formula of some proof, and the theory is called consistent (or non-contradictory) if and only if 0 # 0 is not a theorem. The relative consistency of the axiom of foundation can then be expressed as: if 0#0 is not a theorem of ZF, then it is not a theorem of ZF+AF. We can prove this straightforwardly by supposing a proof of 0#0 in ZF+AF, say the sequence Ao, ... , A n- 2 , 0#0, and constructing from it a proof of 0 # 0 in ZF alone. To do this we note that every parameter-free sentence E which we have shown to hold in any universe 0If has in effect been proved from the axioms of ZF by means of the formal rules of inference. We can thus consider the sequence of relativized formulas A~, ... , A~_2' (O#ot. If Ai is an axiom of ZF, we have just shown that Ar is a theorem of ZF. Thus this sequence furnishes a proof in ZF of (0 # ot, and therefore one of (0 # 0) itself. We give two more simple examples of relative consistency proofs. CONSISTENCY OF THE NEGATION OF THE AXIOM OF INFINITY

We shall show here that if ZF is consistent then so also is the theory T obtained by striking out the axiom of infinity from the list of axioms, and


45

admitting its negation in its place. In fact, if 0/1 is a universe, the set Vw (together with the membership relation) satisfies every axiom ofT. Since Vw is a transitive set, the extensionality axiom automatically holds. To show the union axiom holds, take a E Vw ; for some nEW, then, a E V", in which case Ua c Vn for sure, so Ua E Vn+ 1 • Thus Ua EVa>' Likewise, every subset of a is in Vn + 1 , so that PJ1(a) c Vn + 1 , which gives PJ1(a) E Vn + 2 ; and this shows that the power-set axiom holds. Let a be in Vw , and take an arbitrary formula R(x, y) with parameters in Vw , which, interpreted in Vro , defines a functional relation. In 0/1, then, the functional relation can be defined as x E Vw /\ Y E Vw /\ RV",(x, y) - indeed, since its domain is a set, this functional relation is a map f with values in Vw ' Let
In this section we suppose that the axiom of choice holds in the universe 0/1. A cardinal Ais called (strongly) inaccessible if it has the following three properties: (1) A>w; (2) If fJ is a cardinal < A, then 2/1 < A; (3) If (fJt)ieI is a family of cardinals less than A, indexed by a cardinal 1<,1., then sup fJ i < A. leI A cardinal that is not inaccessible is called accessible. Note that nothing prevents a cardinal less than an accessible cardinal from being inaccessible; for example, if A is inaccessible, 2'\ though strictly greater than A, is accessible. LEMMA: If Ais an inaccessible cardinal then VA = A; a set a is in VA iff a c VA anda
46


PROOF:

Since A. c V;. we certainly have

inequality we first show that if

0(

=

V..t~A..

To obtain the reverse

ordinal
I

p<"

&(Vp) = sup(Ci, sup 2Vp ), P
0(

be the first

V,,= U

t?i'(Vp)~ P
then Vp
So V"
0(

If a E V;. then a E Va, so a c Va for some O«A., so a~ Va; thus a
Conversely, suppose a c V;. and a
If A.

is inaccessible then in V;. hold all the axioms of ZF, together with AF

andAC. It is easy to check that the axioms of extensionality, union, power-set, foundation, and infinity are satisfied. For the axiom of choice, let a E V;. be a family of disjoint non-empty sets. Then (since AC is here imagined to hold in °It) there is a set b c U a whose intersection with each x E a is a singleton. Since a c VII. for some O«A., b c VII. too, so bE Vd1 , so bE V;.. For the replacement scheme we take a E V;., and pick any formula R(x, y), with parameters in V;. only, that within V;. defines a functional relation. Within o/t the functional relation is defined by the formula x E V;. /I. Y E V;. /I. RV.;>.(x, y). As its domain is a set, it corresponds to


47

some map f, and the range off is a set b c: VA' Since trivially Ii ~ a < ..t, it follows that b EVA' The axiom of accessibility Ace denies the existence of inaccessible cardinals: All cardinals are accessible. We can prove the relative consistency of ACC (relative, that is, to ZF+AC) as follows. Take a universe Cft where AC holds. Either ACC does too - in which case nothing needs to be proved - or there is an inaccessible cardinal in 0/1, the first such being n say. Then, by the above, V" satisfies ZF + AF+A C. We shall see that it satisfies A CC too. The ordinals of V" are the ordinals < n. Forifods an ordinal of V"' it is a transitive element of V" linearly ordered by E. But then it is an ordinal in 0/1; since it must be in V" it is an ordinal less than n. The cardinals of V" are the cardinals < n. If 0( is a cardinal < n, we have 0( E V,,; and within V" there is no one-one map of 0( on to a smaller ordinal, since there is not one even in 0/1 itself. Thus 0( is a cardinal of V". Conversely, if 0(
(1') 0( ~ (j); here 0( must be accessible in V". (2') There is a cardinal P< 0( for which 2';;?; 0(; then PE V"' 2' E V"' and so 0( is accessible in V". (3') 0( ~ U Pi> where I is a cardinal less than 0( and every PI is leI

likewise a cardinal <0(. The map i --+ PI is a subset of Ix 0( and therefore of 0( x 0(; so it belongs to V". Thus 0( is accessible in V". This suffices to establish that the axiom of accessibility holds in V".

CHAPTER IV

THE REFLECTION PRINCIPLE

Let X be a class and E(xo, ... , Xk- 1) a formula with all parameters confined to X. We shall say that X mirrors the formula E (or is a mirror for E) if, for arbitrary a o, ... , ak-l in X, E holds if and only if EX also holds. That is, 'the class X mirrors the formula E' is none other than

VXo· .. VXk-l [X (xo) /\ ... /\ X(Xk-l) ~ (E(xo, ... , Xk-l) -- EX(xo, ... , Xk-l»)]' Note that if E is quantifier-free then EX is E itself, and so any class whatever is a mirror. A formula E is said to be prenex (or in prenex normal form) if it is N oN1 ".Nr - 1 E', where each Ni either is "1 or is 3 followed by some variable x, and E' is free of quantifiers. For prenex formulas E, F we say that F is a truncation of E if it is obtained from the latter by removing some of the "1 's and 3x's at the beginning. So if E is N oN1 ... Nr - 1 E', for somep>O the formulaFis of the form N p Np +1".Nr - 1E'. It is well known, and will not be proved here, that every formula A is equivalent to a prenex formula E containing exactly the same parameters and exactly the same free variables (see [4] or [5J). Let E be a prenexformula without parameters, and let (Xn)nero be an increasing sequence of sets whose limit (that is, U Xn) is X. If each new Xn is a mirror for E and all its truncations, then X too is a mirror for E and all truncations. PROOF: We prove this lemma by informal (metamathematical) induction on the number of "1 's and 3x's at the front of E. When Eis quantifier-free there is nothing to prove since either E has no truncations or all its truncations are quantifier-free as well; in either case every set is a mirror. If Eis "1 Fsuppose that each Xn mirrors E and all its truncations; then it obviously mirrors F and all its truncations, and so by the induction LEMMA:

49


hypothesis X mirrors P and all truncations. Consequently, VXo ••• VX k- 1

[X (xo)

X(Xk-l) (P(Xo, ... , Xk-l) +-+ pX (Xo, ... , Xk-l))] '

A .•• A

-+

and since E is ..., P, EX is ..., pX, and so X mirrors E. Likewise, if E is 3xP(x, x o, ... , Xk-l) and every Xn mirrors E and its truncations, then every Xn mirrors P and its truncations; so (induction hypothesis) X mirrors P and its truncations. It remains to show that X is a mirror for E. Pick ao •...• ak-l in X. If EX(ao, ...• ak-l) is true we have 3x(x E X A A pX(x. ao •...• ak-l)). and so there is ana E X for which pX (a, ao •... , ak-l)' As X mirrors p. this gives pea. ao •...• ak-l) and so 3xP(x. ao•...• ak-l)' But this is E(ao, ... , ak-l), which is thereby established. For the converse suppose that E(ao, ...• ak-l)' By taking n sufficiently big we can ensure that each of ao •...• ak-l is in XII (since they are in X). As Xnis a mirror for Ewe get EXn(ao •... , ak-l), or

Thus there is an element a of Xn for which pXn(a. ao, ... , ak-l)' Now XII mirrors P, so it follows from this that F(a. ao •...• ak-d. And X also mirrors p. so pX (a. ao, ... , ak-l) too. But this entails that 3x(x E X A A pX (x. ao, ... , ak-l)) or. what is the same thing, EX (ao, ...• ak-l)' And this proves the lemma. I Note that this lemma is in effect a lemma scheme. since for each formula E it yields the proof of a true formula depending on E. The reflection principle is another theorem scheme derivable (if AP is assumed) within the theory ZP. According to the principle. for every

parameter-free formula E(xo •...• Xk-l) there exists an arbitrarily large limit ordinal /3 for which Vp mirrors E. More formally. we state it as follows. THEOREM: Por every parameter-freeformula E(xo •... , Xk-l) we have Vrx3/3 > rxVxo ... VXk-l [/3 is a limit ordinal A

(xo

E

Vp

-+ (E (xo •... ,

A

•.. A

Xk-l)

Xk-l E Vp (xo. "0, Xk-l»))] 0

+-+ E V "

50


PROOF: We suppose E to be in prenex normal form, and prove by informal induction on its length that for each ordinal ex there is a greater limit ordinal Psuch that Vp mirrors E and all truncations. If E is quantifier-free we take for P the first limit ordinal >ex, since every set - and therefore Vp - is a mirror for E. If E is I Fwe can find a limit P> ex such that Vp mirrors F and all truncations of F. So for ao, ... , ak-1 in Vp we get

and so

F(ao, ... , ak-1) ~ FVP (ao, ... , ak - 1),

thus Vp also mirrors E. So suppose E is 3xF(x, xo, ... , x k- 1); according to the induction hypothesis, whatever ordinal ex is there is a larger limit P such that Vp mirrors F and all its truncations. We define the functional relation y= 4i (xo, ... , Xk-1) of k arguments to mean that 'y is the set of all x of minimal rank for which F(x, Xo,' .. , Xk-1) holds, if there are such x, and the empty set otherwise'. It is clear from the definition that

3xF(x, xo, ... , Xk-1) ~ 3x [x E 4i (xo, ... , Xk-1) 1\ F (x, x o, ... , x k- 1)] . We now proceed to define an ro-sequence of ordinals, (Pn)ne(JJ> as follows. Po is the first ordinal > ex such that VPo is a mirror for F and all its truncations. When Pi is defined for each i.,,;;,2n we put P211+1 equal to the first ordinal> P211 such that for every k-tuple ao, ... , ak-1 drawn from Vp2n • (Since

U {4i(ao, ... , ak-1) I (ao, ... , ak-1) E V~2J

is a set, it is included in some Vy , and so P2n+1 is well defined.) P2n+2, on the other hand, is put equal to the first ordinal Y>P2n+1 for which we get a mirror Vl' for F and all truncations. The sequence PH is strictly increasing, as is plain enough. So if P= sup Pm P is a limit ordinal and Vp = U VPn = U Vp2". But, by nero

II E

ro

nero

51


stipulation, each VP2n mirrors F and all truncations; and therefore, by the lemma, Vp is also a mirror for F. It remains to prove that Vp is a mirror for E. Take, then, any ao, ... , ak-l E Vp, and fix n large enough to fit all these objects into Vp2n. If E(ao, ... , ak-l) is true we have 3xF(x, ao, ... , ak-l) and so 3X(XE cli(ao, ... , ak-l) 1\ F(x, ao,···, ak-l»' This gives anaE cli(ao, ... , ak- 1) for which F(a, ao, ... , ak- 1). As each of ao, 00., ak-l is itself in Vp2n , a qualifies for Vp2n + l ' so a E Vp. Vp, however, mirrors F, so FVp(a, ao, ... , ak- 1), so 3x(x E Vp 1\ FVp(x, ao, 00', ak - 1», so EVp(ao, ... , ak-l) as required. Conversely, given the latter fact there must exist an a E Vp verifying FVP(a, ao, 00', ak- 1). Since Vp mirrors F, we have then F(a, ao, 00., ak-l) for this a, and so E(a o, 00', ak-l)' which concludes the proof.• To see the significance of this theorem in a special case, note first that if E is a closed formula, a sentence, then to say that Vp is a mirror for it is just to say that it is true in the universe iff it is true in Vp. Thus if E is a sentence which is true in the universe, there are arbitrary large limit ordinals f3 such that E is true in Vp. In later parts of the book we shall make use of a somewhat more general principle of reflection. Consider afunctional relation y = W" with domain On, increasing (rt.~f3 ~ W" c Wp) and continuous (for limit rt., W,,= U Wp). p<"

Let W be the union (in the intuitive sense) of all the W" - that is, the class defined by 3rt.[On(rt.) 1\ x E W,,]. Then, provided E(xo, 00., Xk- 1) has no parameters, there is,for every ordinal rt., a greater limit ordinal f3 such that 'Vxooo.'VXk- 1 [XoEWp 1\ ... 1\ Xk-1EWp ~ (EW (xo, 00', Xk-l) ~ EWP (xo, ... , Xk-1))] . No real modification of the proof is required to establish this form of the principle; for, apart from its continuity, the only fact about the functional relation y = V" used above was that it was increasing. In this more general case, the expression 'the set X mirrors the formula E' can be taken as shorthand for the assertion that every member of X is in the class Wand 'Vxo ... 'VXk -l [xo EX

1\ ... 1\

Xk- l EX

~ (EW (xo, 00., Xk-l) ~ EX (xo, ... , Xk-1»]'

52


It should be observed that the axiom of foundation is no longer needed for the more general form of the reflection principle; its only use previously was in guaranteeing that every set belonged to some Va' COMPARISON OF THE THEORIES OF ZERMELO AND ZERMELO/FRAENKEL

Let rx be a limit ordinal >w (rx is then not less than w+w). There is no difficulty in showing that in the set V" (enriched with the membership relation) all the following axioms hold. (1) Axiom ofExtensionaIity. (2) PairingAxiom:\fx\fy3zVt[tEz~t=x v t=yJ. (3) Union Axiom. (4) Power-set Axiom. (5) Axiom ofInfinity: 3x[0 E x /\ Vy(y E X -t y U {y} E x)J. (6) Scheme of Comprehension: for every formula A (x, xo, ... , Xk-l) without parameters, the axiom VXO.·.VXk_1Vx3yVz[ZEY~[ZEX /\ A(z,x o, ... ,Xk-l)]]'

(1) and (3) hold because V" is transitive; (2) and (4) because rx is a limit ordinal- for if x, y E Va, we get x, y E Vp for some /J and .9(X)EVp+1> where /J+l w), as we have just seen, all the axioms of Z hold. The ordinals of Vro+ro - the sets, that is, that satisfy the relativization of On(x) to Vro + ro - are simply the ordinals
I


53

well-ordering of m, being a subset of m x m, is in Vw + w' Among these wellorderings is one whose ordinal is m + m; but this ordinal is not an ordinal of Vw + w , so the well-ordering in question is isomorphic to no ordinal. Thus Vw + w fails to satisfy the sentence mentioned, which therefore cannot be derived in Z. Note that the foundation axiom also holds in V.. (IX being a limit ordinal beyond m). We now intend to prove that no consistent extension of ZF + AF can be obtained by adding finitely many new axioms to Z + AF; put another way, this means that whatever sentence, consistent with Z + AF, A may be, we can find aformula B which is a consequence ofZF+AF+A, but not ofZ+ AF+ A. The formulaB is 'there exists a limit ordinal P>w such that AVp,. This is certainly a consequence of ZF + AF+ A, since A is a consequence of that theory. We need only apply the reflection principle with w for IX. Now if IX is a limit ordinal, it is easily checked that V.. is a mirror for all the following formulas (whose free variables are just e, x, y). (1) On( e), when it is written out as V'x(x e e ~ x c: e) A V'xV'y[x e e Aye e ~ xey v x=y V yex]. (2) 'eisalimitordinal',thatis, On(e) A e~O A V'x[e~x u {x}]. (3) y=&J(x), that is V'z[z e y ~ z c: xl (4) y= Ux, that is, V'u[u e y +-dz(z e x A u e z)]. (5) x e V~, that is, On(e) A 3f[fis a map with domain e+ 1 such thatV'l1 e e+l(j(f/)= U &J(f(C))) A xef(e)]. 'ell

Taking these as proved we can proceed to the following lemma.

LEMMA: Let E(xo, ... , Xk-l) be a parameter-free formula, and IX a limit ordinal. Then V.. mirrors EV;(xo, ... , Xk-l) (whose free variables are xo, ... , Xk-l)' PROOF: We use metamathematical induction on the length of E, again assumed to be in prenex normal form. If E has no quantifiers there is nothing to prove, and if Eis -,Fthen EV~is -,Fv;;so V.. mirrors EV;iff it mirrors FVc (which it does, by the induction hypothesis).

e,

54


So suppose that E(xo, ... , Xk-l) is 3xF(x, xo, ... , Xk- 1 ), F itself being a formula for which the lemma holds. Then E V" is the formula 3x [x E V~ /\ /\ FV,,(x, xo, ... , Xk-l)] and so [Ev«xo, ... , Xk-l)Y" is 3x[x E V" /\ /\ (x E v~t" /\ (FVIi(X, Xo,"" Xk- 1 )t"]. We have noted above (in (5» that V" mirrors the formula of two variables x E V~. As a consequence, if x, (E V"' we get (x E V~)V" +-+ X E V~; furthermore, since ( is in V" it is less than IX, so x E V" /\ X E V~ +-+ X E V~. And again, since V" mirrors F V " (induction hypothesis), we have for x, Xo,"" Xk-l E V" that (FV~ (x, xo, ... , Xk-l»)Y" +-+ FVI;' (x, xo, ... , Xk-l)'

Gathering together all these remarks, we can conclude that for (, xo, ... , Xk-l E V" we have

[E V" (xo, ... , Xk-l)t" +-+ 3x [x E V~ /\ FVIi(X, XO, ... , Xk-l)]; and this latter being EVIi(XO' ... , Xk-l), the result follows .• So let IX be the first limit ordinal greater than w for which V" mirrors the sentenceA; since A is true in 0lI, IX is the least limit ordinal > wsuch that A V ". By the lemma V" mirrors the formula A V " (of one free variable only, ().Bisjust3([(isalimitordinal> w /\ A V"],soBv"is3( [(E V", /\ g is a limit ordinal >wt" /\ (AV"t"J. Since V", mirrors both A V' and '( is a limit ordinal >w', BV " is equivalent to 3([( E V", /\ (is a limit ordinal >w /\ AVIiJ. However, as IX was chosen to be the least limit above w for which A V ", we cannot have, for a limit (, both A V " and ( E Vo;; and, as we can see, BV " requires both to be true. Thus BV " is false. This proves that, though V" satisfies all the axioms of Z, AF, and A (by the choice of IX), the formula B does not hold in it. So B cannot be a consequence of Z+AF+A. As a corollary to the above result we may note that if ZF is consistent, it is not finitely axiomatizable. For the assumption of consistency allows us to assume ZF+AF consistent; so if ZF were equivalent to a single sentence A, ZF + AF would be equivalent to Z + AF + A. Note that even if A is an arithmetical formula (by this we mean one relativized to Vro ), B is not. G6del's second incompleteness theorem provides us with another proof of the above result, but one which associates


55

with every A consistent with ZF + AF an arithmetical B' following from ZF+AF+A, butnotfromZ+AF+A. The following, however, remains an open problem: if ZFis consistent, does there exist a sentence A, consistent with Z, such that Z + A is equivalent to ZF + A? From what we have just shown it is clear that such a sentence A would have to contradict the axiom of foundation; for otherwise Z + AF + A would be consistent and equivalent to ZF + AF + A.

CHAPTER V

THE SET OF EXPRESSIONS

In the first two chapters we constructed, within each universe 011, a sort of replica for several of the fundamental ideas of mathematics; the idea of a mapping, for example, or that of a natural number. And we agreed thenceforth to use these words in the senses we had given them in 011, and not at all in their everyday senses. We are now going to carry out the same 'reconstruction' for the idea of a formula of set theory. In this case, however, we shall continue to use the word 'formula' in its intuitive sense only, as we have used it up to now, and will take advantage of the word 'expression' in order to refer to the kind of objects that we shall define in the universe. Choose from 011 any five distinct sets, to be denoted by v, "', 3, e, and =; these sets might be, for instance, the sets 0, 1, 2, 3,4. Choose also a denumerable set "Y whose elements, called variables, differ from all five sets chosen above - "Y might consist of the natural numbers ~ 5. If x is a variable we shall write the ordered pair <3, x) as 3x. We define by recursion a map n -+ Iffn with domain co: Iffo is the set of ordered triples e, x, y) and =, x, y) where x, y are in "Y; more formally,

<

<

The elements of Iff0 are called atomic expressions. Further, Iffn+1 = Iffn U [{ ""'} x Iffn] U [{v} x Iff;] U

[({3} x "Y) x Iffn] ;

it consists therefore of everything in Iffn' together with ordered pairs and triples of the form <"', CP),
It is easy to check by induction on n that, in virtue of our choices for v, ""',3,8, =, and "Y, every Iffn c: V"" and so Iff c: V",; every expression is a hereditarily finite set.


Given an expression

depth of(P.

57

(p we shall call the first n for which (p E Cn the

LEMMA: For each expression (P one and only one 0/ the following possibilities is realized:

(P is atomic; (P iso/the/orm (-, P); (P is o/the/orm (v, P, PI); (P is o/the/orm (:Ix, P). Furthermore, P and P' are expressions determined by (P, and strictly lower in depth than (P itself. PROOF: Bearing in mind that the ordered triple (a, b, c) is defined as the ordered pair (a, (b, c», it is clear that each expression is an ordered pair. Also clear is the fact that the first element of such a pair is bound to be 8, =, -, v, or :Ix (where x E 1); and these objects are distinct from one another, as that is the way we chose them. Ifthe first element of a pair is 8 or = we must have (P E Co; if it is -, the second element of the pair is an expression P. Writing n+ 1 for the depth of (P we have (-, P) E C n+ 1 and (-, P) ¢ Cn; so by the definition of Cn+ 1 we get P E Cn; and therefore the depth of P is less than n+ 1. On the other hand it is obvious that P is determined uniquely by (P. Exactly the same goes if the first element of the pair is :Ix. If, however, it is v, then the second element takes the form (P, PI), P and P' being determined by (P, obviously enough. From (v, P, P') E Cn + 1 and (v, P, PI) ¢ Cn it again follows at once that P, P' E Cm and therefore P and P' have depth less than n + 1.1 In the sequel the expressions (8, x, y), (=, x, y), (-, (P), ( v, (P, P), and (:Ix, (P) will be shortened to x B y, x=y, -(P, «(p) v (p), and :Ix (P respectively; the sets 8, =, -, v will be called the membership sign, the equals sign, the negation sign, and the disjunction sign respectively; and :I is the existential quantifier. The expressions (,..,(p) v (p), -(P) v (- P», -:lx( -(P) will be still further shortened to «(P) ~ (P), «(P) A (P), Yx«(P) respectively; and the expression «(P ~ P) A (P ~ (P) will be written «(P) ~ (P). We now define recursively on the depth of an expression a map w from C into the set of all finite subsets of r; w((P) will be called the set

-«

58


offree variables of the expression iP. DEFINITION:

IfiP=xBY or iP=x=y then w(iP) = {x,y}; w(", iP) = w(iP); w(iP v P) = w(iP) u w(I['); w(3xiP) = w(iP)"'-.. {x}. An expression iP is called c1osedifw(iP) =0. It should be noted that each parameter-free formula A corresponds in an

obvious way to an expression, an expression we shall denote by r AI. But the converse is false. Were there to exist in ott a natural number n with, intuitively speaking, infinitely many elements, then there would be an expression with n as its depth, but no formula could be expected to correspond to it. In fact, one can readily see that an expression corresponds to a formula if and only if its depth is a natural number with only a finite number of elements (in the intuitive sense of the words). We can also define (recursively on the depth of the expression iP) a binary functional relation Y = Val (iP, X) whose domain is 'iP is an expression and X is a set'. Y is called the value of the expression iP in the set X, and it is a subset of w(
(1)

(2) (3)

(4)

If iP is x B y (respectively, x = y) then Val(iP, X) = {b E (x,y}X 1 b(X) E bey)} (respectively {<5 E {x, y) X <5 (x) = <5 (y)}) ; If iP = '" P then Val(iP, X) = w( b(<5' E W('P) X 1\ <5' E Val(P, X»} (remember that in this case w(iP) = w(P)"'-.. {x}).

I

I

If iP is the expression corresponding to the formula A (xo, ... , Xk-l) it is simple to see that Val(iP, X) is the set of maps <5 from {xo, ... , Xk-l} into


59

Xfor which AX (c5(xo), ... , c5(Xk-l» holds; this is easily proved by informal metamathematical induction on the length of the formula A. We shall write an expression
Yo, ... , YI-l (and they are all different), and the domain of 11 is the subset {xo,"" Xk-l} of w(
60


subset Y c: X such that every expression in d is satisfied in Y, Y";;;; P + ~o, andP c: Y. PROOF: The axiom of choice is needed to ensure a map 9 : f!IJ(X)"'. {0} ~ X such that for every non-empty U c: X, 9( U) E U. Using the map () we define recursively an (V-sequence Pn of subsets of X: Po is just P itself; and Pn +1 is defined from Pn in the following manner. Let @n be the set of monadic Pn-expressions 4>(x, ao, ... , ar -1) for which 3x4>(x, ao, ... , ar - 1) is satisfied in X. Let Uw= g E X 14>(e, ao, ... , ar -1) is satisfied in X}. Then Pn+1is defined as the set of all 9(Uw) for 4> in @n' Thus Pn + 1 contains, for each monadic 4> E @n, at least one 'satisfier' of 4> in X, namely 9(Uw). Observe first of all that Pn+1 :::> Pn; for if a E Pn then the expression x=a belongs to @m and, for this expression, Uw={a}; thus 9(Uw)=a, and a E Pn + 1 • Secondly, since the set of expressions with parameters in Pn has cardinality P,. + ~o, we have @n";;;; Pn + ~o. But the map sending the expression 4> to 9(Uw) is by definition a surjection, a map from @n on to

Pn+1. Thus Pn+1";;;;Pn+~0' By induction, therefore, Pn";;;;P+ ~o. We set Y=

U

nero

Pn' ThenP c: Y; and Y";;;;

L Pn";;;;~o·(P+~o)=P+~o.

new

Now let 4>(ao . ... , ak-1) be any closed expression whose parameters are in Y. We show by induction on its depth that 4> is satisfied in X if and only if it is satisfied in Y. F or atomic cI> the result is obvious. If 4> is ,.., 'P (or 'P v 'P'), 4> is satisfied in Yiff 'P is not (or one or other of 'P, 'P' is); the induction hypothesis then tells us that cI> is satisfied in Y iff 'P is not satisfied in X (or either 'P or 'P' is satisfied in X); so that 4> is satisfied in Y iff it is satisfied in X. If 4> (a o, ... , ak-1)= 3x'P(x, ao, ... , ak-1) we suppose to start with that 4> is satisfied in X. Then, for a big enough n, we will have ao, ... , ak-1 all in Pm and therefore 'P(x, ao, ... , ak-1) E @n. Then

say, where a E Pn+1> and so 'P (a, ao, ... , ak-1) is satisfied in X (by the definitions of 9 and U'l')' The depth of 'P(a, ao, ... , ak-1), however, is strictly less than that of 4>, and so, by the induction hypothesis, 'P(a, ao,

THE SET

OF

EXPRESSIONS

61

..• , ak-I) is satisfied in Y; from this it follows at once that 3xP(x, aQ, •.• , ak-I), whichisjustIP(aQ , .. " ak_I),issatisfiedin Y. Conversely, if 3xP(x, aQ, .. " ak-I) is satisfied in Y, there must be an a E Y for which pea, aQ, , .. , ak-I) is satisfied in Y. By the induction hypothesis pea, aQ, .. " ak-I) is also satisfied in X, so 3xP(x, aQ, .. " ak-I)

is too. In particular we have proved that every expression in d is satisfied in Y .• The relativization of an expression to a set: Suppose we are presented with an expression IP(xQ, , .. , Xk-I, aQ, .. " ai-I) whose parameters aQ,"', a l - I are all in a. We define, recursively on the depth of IP, another expression IPa(xQ," " Xk-I, aQ, ... , al-d known as the relativization of IP to the set a.

DEFINITION: If IP is atomic then IP a= IP; If IP = '" P then IP a= '" pa; If IP= PI V P z then IP a= P~ v P~; IfIP=3xP then IP a=3x(xea A pa). Clearly enough the expressions IP, IP a have the same free variables. The parameters of IP a, however, are a, aQ, ... , a/-I' We shall use the following theorem in Chapter VIII. THEOREM: Suppose that a E b and a c: b. If IP is an expression with parameters in a, then Val(lP, a)= w(c;I)a () Val(IP", b). PROOF: The proof is by induction on the depth of IP. For atomic IP the result is obvious. For molecular IP, suppose IP= '" P (the proof is more or less identical, mutatis mutandis, for IP= PI v P 2)' Then Val ( IP, a)= W(c;I)a"'-Val(P, a) = W(
= {(j E

W(c;I)

a 13(j'

:::>

(j(c5' E W(2') (a) A (j' E Val(P, a)}.

62


The induction hypothesis gives Val ('1', a) equal to w('P) a r\ Val ('1'", b), so w(0) a I 3.5' ~ .5 (.5' e w('P) a A .5' e Val ('1'", b))} = {.5 e w(0) a I .5 e Val (:Jx (x 8 a A '1'a), b)} = w(0) a r\ Val(4)", b). 1

Val ( 4>, a) = {.5 e

CHAPTER VI

ORDINAL DEFINABLE SETS Relative Consistency of the Axiom of Choice

Notation: Let 4i(x) be a monadic expression whose parameters are all in some set X; the value of this expression in X is then a subset of {x) X. By the

canonical bijection of {x) X on to X (that which sends {(x, u)} to u, for every u E X) there corresponds to this subset a subset of X itself, written val(4i, X), and also called, by a slight abuse of language, the value of 4i inX. We now consider a universe
make use of the reflection principle to obtain an ordinal IX not only greater than lXo, ... , IXk - 1 but also large enough for a to be in Va, and such that Va mirrors the formula A (x, lXo, ... , IXk-l)' Then a is the only member of Va for which the formula AV",(x, lXo, ... , IXk-l) holds, and so the value of the expression r A(x, lXo, ... , IXk-tY"' in the set Va is {a}. Thus a is ordinal definable .• CONVERSE: Let a be ordinal definable. Then there is a formula A(x, Do) of one free variable and one parameter (the ordinal Do) which holds for a and a alone.

Consider the parameter-free formulas S=J(IX) and x=K(n) which establish, respectively, bijections from On on to 0'( On) (the class of finite sequences of ordinals), and from (J) on to VO)' These formulas have been discussed above on pp. 33 and 41 respectively. With the help of J one can easily define a parameter-free functional

PROOF:

64


relation 17 = J' (0() which is a surjection from On on to the class of functions 17 with a finite domain included in "f'" and with values in On, Since a is ordinal definable, there are an expression cPo(x, 0(0' "" 0(,-1) with parameters in On, and an ordinal Yo, such that val(cP o, VYo)={a}, We therefore consider the following quaternary parameter-free formula E(x, n, y, 13): 'n is a natural number, 13, yare ordinals, the ordered pair (K(n), J' (13» is an expression cP whose parameters are ordinals less than y, and val(cP, Vy)={x}'. Writing then no for that natural number such that K(no) is the parameterfree expression cPo (x, xo, .. " X,-1) and Po for that ordinal such that J'(Po) is the function 170 ={ (xo, 0(0), .. " (x'-l> O(,-1)}' it is apparent that the formula E(x, no, Yo, Po) holds for a and for nothing else; this formula, however, has three parameters, the ordinals no, Yo, and Po' To reduce the number of parameters to one, we write ao for the inverse image under J of the sequence no, Yo, Po; then the formula A (x, ao): 'there are a natural number n and two ordinals y, 13, such that J(a o) is the sequence {(O, n), (I, y), (2, p)} and E(x, n, y, 13)' has ao as its sole parameter, and holds for a and a alone.• A formula like A (x, ao) is said to be a definition of a in terms of ordinals, or an ordinal definition of a. Note that what we have actually produced above is a parameter-free formula A (x, y) of two free variables such that if a is ordinal definable there is an ordinal 80 for which A(x, 80 } provides a definition of a. In other words, the following theorem: 'ltx[OD(x) ~ 38(On(a) " 'ltu(A(u, 8) ~ u=x»]. The class OD is not necessarily a transitive class, for there may be ordinal definable sets a not all of whose elements are ordinal definable. However, we can readily isolate a subclass HOD of OD, the hereditarily ordinal definable sets, which is transitive, by the formula HOD(x): 'Every element of ~({x}) is ordinal definable'. (Remember that ~(a) is the transitive closure of a, the smallest transitive set, that is, which includes a; and ~({x})= {x} u ~(x) - see p. 38.) It should be observed that HOD has been defined by a formula without parameters. LEMMA: A set a is hereditarily ordinal definable if and only if it is ordinal definable and all its elements are hereditarily ordinal definable.

ORDINAL DEFINABLE SETS

65

PROOF: The condition is obviously necessary. So suppose that every element of a is in HOD, and that a is in OD. We must show then that aisinHOD. From the properties of transitive closures investigated on p. 38 we havethat~({a})={a} u ~(a)={a} u au U ~(y),andso

~({a})={a} u =

{a} u

U ({y} u ~(y») U ~({y}).

yea

yea yea

By assumption every element of ~({y}) is in OD, whatever element y may be of a. Since a is in OD, this shows that every element of ~({a}) is in OD; and so HOD (a). I Our main task in this chapter will be the proof of the relative consistency of the axiom of choice. We do this by constructing from a model 0/10 of ZF another model, one for ZF + AF + A C. Moving off from 0/10 we first construct a model 0/1, as we have done before (p. 43), for ZF + AF. We then show that not only ZF+ AF but also ACholds in the class HOD constructed within 0/1. Axiom of Extensionality: Obvious, since if a, b are in HOD, so are all their elements. Union Axiom: If a is in HOD, let b= U a. It is clear that every element of b is in HOD. According to the above lemma, all that remains to be shown is that b is ordinal definable. Now a is certainly ordinal definable, so is the only set for which a certain formulaA(x, oc) holds; as b is the only set for which B(y, a): \fz(z E y ~...du(u E a A Z E u»)

holds, we can combine the two into 3x[A(x,oc)

A

B(y, x)]

which is a formula of one free variable, and one ordinal parameter. It obviously holds just for b, so by the lemma on p. 64, b is ordinal definable. Power-set Axiom: Suppose a to be in HOD, and b the set of all hereditarily ordinal definable subsets of a. We want to show HOD (b), and since we have at once HOD (x) for every x in b, we need only check that OD(b).

66


If A (x, IX) is an ordinal definition of a, since b is the only set for which B(y, a):

Vz [z

E

Y - HOD(z) /\ z c: a]

holds, it is clear that 3x[A (x, IX) /\ B(y, x)] holds for b and for nothing else. Thus OD(b). Scheme of Replacement: Let a be some object of HOD, and R(x, y, aQ, ... , ak-i) be a binary relation with parameters in HOD. If, interpreted in HOD, R defines a functional relation, then the formula HOD(x) /\ /\ HOD(Y) /\ RHOD(x, y, aQ,"" ak-i) defining it in the universe o/t also defines a functional relation, Sex, y, aQ,' .. , ak-i) say, for brevity, Let b be the set of images under S of elements of a. Since every element of b is in HOD, we need only prove that aD (b). So let A(x, IX), A(xQ' IXQ), ... , A (Xk-l> IXk-l) be ordinal definitions of a, a Q, • •• , ak - i respectively. Then since b is the only object for which the formula B(y, a, aQ, ... , ak-i): Vz[zEy-3t(tea /\ S(t,z,aQ, ... ,ak_l»]

holds, it enjoys the same distinction with respect to the formula 3x3xQ... 3Xk -i [A (x, IX) /\ A (xa, lXa) /\ ... /\ A (Xk-i, IXk-l) /\ B (y, x, x a, ... , Xk - i )] ,

whose only parameters are the ordinals IX, lXa, ... , IXk-l' Thus OD(b), as required. Axiom of Infinity: If IX is any ordinal it is obviously ordinal defined by X=IX; so every ordinal is in fact in HOD. HOD(w) in particular, and the axiom of infinity holds for w in HOD. Axiom of Foundation: If a non-empty set a is in HOD, and b is one of its elements bearing minimal rank, then b is in HOD and b n a = 0. Axiom of Choice: We can define, by a formula without parameters as follows, an injective functional relation a=e(x) which associates an ordinal with each monadic expression whose parameters are in On. Starting from such an expression
67


Vro' Using our bijection K from Q) on to Vro , we let n be the natural number for which K(n) = 4> (x, Xo, .. " X,-I)' In a similar way we let Pbe the first ordinal for which J' (p)isthefunction {(xo, cxo), ... , (x,-l> cx,_I)}whereJ' is the functional relation defined on p. 64. Then cx= 8 (4) (x, cxo, ... , CX,-I») is taken as that ordinal cx associated with the pair (n, P) by the isomorphism between On 2 and On. Granted 8 then, we proceed to define another functional relation P=D(x), from aD into On, by the formula 'P is the least ordinal to represent a pair of ordinals (cx, 1') where 1'=8(4)) for some monadic expression 4> with parameters in cx and value {x} in V,,'. The formula P=D(x) has no parameters; what is more, it defines an injection, for x =1= x' -+ D(x) =1= D(x'), as is easily checked. It is clear now that the relation R(x, y) defined by the parameterfree formula OD(x) A OD(Y) A D(x)~D(y) is a well-ordering of the class aD. So suppose a to be any hereditarily ordinal definable set. Restricting the well-ordering R to a yields the set b = {(x, y) e a 2 1 D(x) ~ D(y)} ,

and it is this set which we shall show to be a well-ordering of a, and itself in HOD. Taking the latter point first, note that all of b's elements are in HOD, so we have only to prove that OD(b). But since the formula

B(y, a): 'v'z[zey~3u3v(uea

A

vea

z = (u, v)

A A

D(u) ~ D(v»)]

holds only for b, and we may assume that some ordinal definition of a, A(x, cx), holds only for a, it is clear once more that b is ordinal defined by the formula 3x [A (x, cx) A B(y, x)]. So HOD(b). As far as all is concerned, b certainly well-orders a; that this is also the case in HOD is apparent from the fact that all non-empty subsets of a, those lying in HOD in particular, have a smallest element (mod. b). Thus Zermelo's theorem holds in HOD; so AC too. This concludes the proof that A C is consistent relative to ZF. The ordinals of HOD are the same as the ordinals of all; and the natural numbers of all appear unchanged in HOD too. Note that since AFholds

68


in HOD we can write On (x) here as

'v'Y(YEX-+Y ex)

1\

'v'Z[(YEX

1\

-+

ZEX) (z E Y V Y =

Z

v Y E Z)] .

Moreover, Vro is in HOD. For the bijection x=K(n) from W on to Vro provides, for every hereditarily finite set, a definition whose only parameter is a natural number. Thus everything in Vro is in aD, and since Vro is transitive, this means in HOD too. Vro , being definable by the parameterfree formula 'x is the set of hereditarily finite sets', is itself in aD, so too in HOD. Continuing on these lines, we see that if f is the map n -+ Vn defined on w, then HOD (f); for any element of f is a pair
3f3n [! is a map defined on W such that f(O) = 0, f(k + 1) = &(!(k») for all k E W, and nEW and x E f(n)]. This fact, coupled with the relative consistency proof of AC just given, yields the following result. If an arithmetical formula E (one whose quantifiers are relativized to Vw that is) is derivable in the theory ZF + AC, it is derivable already inZF. PROOF: Since (x E Vro)HOD ~ X E Vro we can see without difficulty that E HOD ~ E, if Ehas been relativized to Vro' So if E is a theorem of ZF+AC, there is a proof Ao, ... , A n - 2 , E of it in that theory. We have shown, however, for each axiom A of ZF+AC that A HOD is a theorem of ZF. Thus A~OD, ... , A~!?f, E HOD is a proof of E HOD , and so of E, from ZF alone .• THEOREM:

The Choice Principle: We say that the choice principle holds for a universe cp/ if there is a formula A(x, y) of two free variables and no parameters which defines a well-ordering of the entire universe. In such a case the axiom of choice is easily seen to hold. We should


69

note however that, as it stands, the choice principle is not only not an axiom, it is not even a scheme of axioms; rather, it is something like an 'infinite disjunction' of sentences, of all those sentences, that is, which say, of a binary relation without parameters, that it well-orders the universe <11. Nevertheless, we can establish the following result.

If AF holds in <11, then the choice principle holds in <11 if and only if the axiom YxOD(x) holds in <11. PROOF: If the choice principle holds, some parameter-free formula A (x, y) defines a well-ordering of <11. So there will be generated by this wellordering an isomorphism x = J( a) from On on to <11, and this isomorphism will be definable by a formula free of parameters. Thus x=J(a) will provide an ordinal definition for every element of <11, and therefore YxOD(x) will hold. Conversely, given that YXOD(x) holds, we can proceed as above to write down the formula D(x)=:;;D{y); this is parameter-free, and it defines a well-ordering on <11. I In particular, granted AF, the parameter-free formula D{x)=:;;D(y) has the following property. Ifthere is any parameter-free formula A (x, y) defining a well-ordering of the universe, then D(x)=:;;D(y) defines a wellordering of the universe. In Chapter VIII we will be able to show the relative consistency of the axiom YxOD(x). THEOREM:

CHAPTER VII

FRAENKEL/MOSTOWSKI MODELS Relative Consistency o/the Negation o/the Axiom o/Choice (Without the Axiom 0/ Foundation)

Let R(x, y) be a relation of two arguments that establishes a bijection from the universe %' on to itself; that is, a relation which satisfies the following conditions: VxVyVy' [R(x, y) A R(x, y') ~ Y = y'] ; VxVx'Vy[R(x, y) A R(x', y) ~ x = x']; Vx3yR(x, y) A Vy3xR(x, y).

We shall write this functional relation as y = F( x); and the binary relation x e F(y) will be written x e' y. Likewise, the formula obtained from the formula E(xo, ... , Xk-1) by changing e throughout to e' will be written E'(xo, •.• , Xk-1)' We shall show that every ZFaxiom holds in the class of all sets, enriched with the relation E' (a universe we shall call %,'). This amounts to proving that if A is an axiom of ZF, then A' holds in %'. Axiom 0/ Extensionality: We must show that VxVy [x = y

~ Vz(z E' X ~ Z

e' y)].

So let a, b be two sets in %' such that Vz(z E' a ~ z e' b); by the definition of E' we then have

Vz(z E F(a) ~ Z E F(b»). ThusF(a)=F(b). SinceFis bijective, this means thata=b. Union Axiom: If a is a set then the class 3y [y E' a A x E' y] is one too. For this formula is just an abbreviation of 3y[y e F(a) A x E F(y)], and so is equivalent to x E U F(y). Putting this last set equal to c, we obtain yeF(a) Vx[xEc~3y(ye' a AXE' y)], whence, letting b stand for F -1 ( c), Vx [x

E'

b ~ 3y(y E' a

A

x

E'

which shows that the union axiom holds.

y)],

71

FRAENKEL/MOSTOWSKI MODELS

Power-set Axiom: If a is any set, the formula Vy(y E' x Vy[y EF(x)

~

~

Y E' a) is just

y EF(a)J;

that is, it is equivalent to F(x) c F(a) or, putting c for @J(F(a», to F(x) E c. Defining b by F(b) = {x I F(x) E c}, and remembering that F is bijective, we get F(x) E c equivalent to x E F(b). Thus

x

E'

b - Vy [y

E'

X ~ Y E' a] ,

proving that the power-set axiom also holds.

Scheme of Replacement: Let a be a set and R(x, y) a formula such that R' (x, y) defines a functional relation. Let c be the set of images of elements of F(a) under this functional relation. Then VY[YEc-3x(xEF(a)

i\

R'(x,y»];

so by writing again b for F- 1 ( c) we have

VY[YE' b_3x(XE' a

i\

R'(x,y»].

Thus the replacement scheme holds.

Axiom of Infinity; We define by recursion a mapfwith domain ill :f(O)= =F-l(O); f(71+ 1) is defined by F(J(n+ l»=F(J(n» U {fen)}. Writing 1] for the range off, we then set e=F- 1 (1]). Since Vx(xf F- 1 (0», the set F- 1 (0) must be the empty set 0' ofOll'. Since 0' E 1], we get 0' E' e. Moreover, if x E' e, then x E 1], so X= fen) for some nEW. Now Vz[z E' f(n+ 1) - ZE' fen) v Z= fen)], from the definition of f(n+ 1); consequently,

fen + 1) = X u' {x}', and so Vx[x E' e ~ x u' {x}' E' eJ; thus of infinity in OlI'.

e is a

set satisfying the axiom

If the axiom of choice holds in 0lI, it holds also in OlI'.

For suppose that a is a set for which, in OlI', the formula 'the elements of a are non-empty but pairwise disjoint' holds. Then

'v'x [x and

E'

a

~

VXVy[(xE'a

3y(y E' x)] i\

YE'a

i\ x#y)~Vz(zfx

v zfy)J.

72


I

Put a1 for {F(x) x E F(a)}; then the two sentences above say that the elements of at are non-empty and disjoint in pairs. So if the axiom of choice holds in d/t, we can find a set h1 which intersects each member of at at a single point; that is,

Vx3y[xEal-VZ(Y=Z+-+ZEX /\ zEb l )], or, alternatively,

Vx3y [F(x) E a 1 - Vz(y =

Z +-+ Z E

F(x) /\

Now the definition of al implies that F(x) E a1 iff x if, that is, x E' a. Thus we get

E

ZE

b I )].

F(a); if and only

VX3y[xE'a-Vz(y=Z+-+ZE'X /\ zEb 1 )]. Write now b for F- 1 (b 1 ); we obtain

Vx3y [x

E'

a _ Vz(y =

Z +-+ Z E'

X /\

Z E'

b)],

showing that the axiom of choice holds in d/t'. A set a is called an atom of the universe d/t if a= {a}; or (equivalently) if Vx(x E a +-+ x=a) holds in d/t. No atoms can exist, it is clear, where the axiom of foundation holds. If ZF is consistent so is ZF + 'there is an atom'. To prove this we define a simple bijection F from d/t on to itself by the formula (x=o /\ y= I) v (x= I /\ y=o) v (x:;i:O /\ x:;i: I /\ x=y). In the universe d/t' obtained by applying this bijection to d/t, the empty set of d/t, namely 0, is an atom. For since F(0)={0}, we have at once Vx(x E' 0 +-+ x=0). It is possible, by a suitable choice for the bijection F, to show that several other sentences, stronger than the negation of AF, are consistent with ZF; for example, the sentences asserting the existence of E-cyc1es

aOEat E ... Ean-lEaO' For the rest of this chapter we shall rely on the following example only. If ZF is consistent, so is ZF + there exists a set of atoms that is equipollent to w'. Define F by setting F(n) = {n}, F({n})=n for every natural number n~ I, and F(x)=x everywhere else. This is a bijection all right since n and {p} must be different if nand p are both positive natural numbers. I


73

Since for every n~ 1 we have Vx(x E' n - x=n) we see that n is an atom of 0/1'. We must now show that, within 0/1', this set of atoms is equipollent to w. Given two objects a, b we write (we have already done this above, without explanation) {a, b}' for their pair in 0/1'; it is clear that {a, b}' = =F-1({a, b}). (a, b)' is similarly the ordered pair in 0/1' of a and b, so that (a, b)' =F- 1({F-l({a}), F-1({a, b})}). We can recursively define a function/with domain w by setting Vx [x

E'

fen) - 3i < n (x = f(i»)] ;

thus I(n+ 1)=F- 1({I (O),f (1), ... ,f(n)}) and 1(0)=0'. From this definitionitis clear that when n runs through w, I(n) runs through the set of natural numbers in 0/1'. By putting (x, y)' E' g - (x, y) E f, we can then define a map g in 0/1'. We obtain (x, y)' E F(g) - (x, y) E I, and thus if h is the set of images of elements of/underthe map (x, y) ~ (x, y)', wehaveg=F-1(h). After all this, it is clear that in 0/1' g is a bijection from the natural numbers of 0/1 to the natural numbers of 0/1'. Since all the positive natural numbers of 0/1 are atoms in 0/1', we have shown that the latter universe contains a set of atoms equipollent to w. RELATIVE CONSISTENCY OF THE NEGATION OF THE AXIOM OF CHOICE

In this section we shall provide a model of ZF in which the axiom of choice fails. From what we have just proved we can suppose given a universe 0/1 0 in which there is a set of atoms, X, equipollent to w. We begin by defining a functional relation y= W" on the ordinals by Wo=X; W,,= U @J(Wp)(foralllX#O). p<"

Note that since Xis a set of atoms, Wois transitive. Thus Wo c @J(Wo)= = Wi. For 1 ~ P~ IX, furthermore, it is clear that Wp c W". It follows that whatever ordinals IX, Pare, if P~ IX then Wp c W". In the same way as we proved similar results for V" in Chapter III, we can show that W,,+1 =@J(W,,), and W,,= U Wp when IX is a limit p<"

ordinal. The union (in the intuitive sense) of all the W" we denote by W; W(x), then, is the formula 31X[On(lX) 1\ x E W,,].

74


If a is in W we again call the least ordinal IX for which a E W" the rank ofa. An object a is in W if and only if all its elements are. The rank ofany element of a is strictly below that of a itself, provided only that a does not have rank O. The axioms ofZF hold in the class W. PROOF: All parts of the lemma are proved in the same way as similar results were proved for V in Chapter IlL. LEMMA:

The set X is in W since X E W1 • Moreover, every atom in W is an element of X. For suppose, on the contrary, that there were an atom a of W whose rank were non-zero. Since no limit ordinal is the rank of anything, we can suppose a's rank to be 0(+ 1. Thus a E W,,+l, so a c: W,.; but a E a, so a E W,,' contradicting the definition of rank. Thus a has rank 0, so is in X. Again, the bijection f from X on to w is in W; for f consists only of pairs (a, n) with a E Xandn E w; (a, n)isthereforein Wn + 3 • Although W does not satisfy the axiom of foundation, it does satisfy the sentence Yx[x:#:0-+3Y[YEX

A

(ynx=0 v y={y})]].

For if a non-empty set a is in W, let b be an element of it ofleast rank. If the rank in question is positive, every element of b is of lower rank, so b n a = 0; but if it is zero, b is an atom. Consequently the class W proves the consistency of the axiom system T given by: ZF + the class of all atoms (defined by the formula x = {x}J is a set equipollent to w'+Yx[x:#:0-+3y[yex A (ynx=0 v y= I

= {y})]].

We now consider a universe OU1 satisfying all these axioms; and, writing X for the set of atoms of 0//1 we carry out the above construction within OUt· ThusWo=XandifO(~l, W,.= U L1'(WfI); Wis the class defined by 31X[On{lX) A XE WJ. f/<,. Every object of ou1 is in W; equivalently, 0//1 satisfies the axiom Yx31X[On(lX) A x e W,,]. For suppose the existence of an a not in W; let b be its transitive closure, and c the set of those elements of b not in W. Since a is not in W, some element x of a is not in W; and since this x will

75


be in c, c cannot be empty. So take any y E c; it is not in W, so has an element z not in W. Since y E band b is transitive, z is also in b. Thus z qualifies for c, which means that y n c#O, for every y E c. Moreover, no such y can be an atom either, for in this universe all atoms are in W. The set c thus contradicts the axiom 'v'x[x#0-+3Y[YEX /\ {ynx=0 v y={y})]]. If u is a bijection from X on to itself (a permutation it is also called), we can define by recursion a functional relation (1a. such that (1) if IX is an ordinal (1a. is an E-automorphism of Wa. and (2) (10=(1, and if P' ~P then up is an extension of (1p" For suppose the family ((1p)p
P
(1a. has properties (I) and (2) is entirely straightforward. For IX=P+I we have Wa.=&' (Wp), so can define Ua. by setting (1a.Cu)= ={(1p(x) x E u} for every UE Wa.' Since up is a permutation of Wp, it is clear that (1a. is a permutation of &,(Wp). In addition, (1a. is an extension of (1p; for if UE Wp each x E U belongs to Wp, and since (1p is an E-automorphismofWp, we have X EU+-TUP(X) EUP(U), so that UP(U) = {up(x) x E u}= =(1a.(u). Finally, (1 a. is itself an automorphism of Wa., as can be shown by proving that, for arbitrary x, UE Wa.,

I

I

X E U +-T (1",{x) E (1",(u). But if x E U, we get x E Wp, so up(x) E u",(u) - by the definition of (1a. -, so Ua.{X)E(1a.{U). Conversely, if u..(x) E (1a.(u), then ua.{x) E Wp. As Ua. is a permutation of W", which simply extends the permutation (1 p of WP' we have x E Wp, so (1p(x) E ua.{u), and therefore x E U, by the definition of (1a.' Starting from U 0 = (1 then, the functional relation IX -+ U rt as recursively defined above has all the required properties. From it we can define another functional relation, y = g'.,(x), as the common extension of all the (1 rt; this functional relation, defined by

31X[On(lX) /\ X E W", /\ Y = U..(X)] , has domain 0//1 and, indeed, is an automorphism of 0//1' so that the

76


following are all true in ~1: 'v'x'v'x' [.9"a(X) = .9"a(X') -+ x = X']; 'v'y3x[y = .9"a(x)]; 'v'x'v'x' [x ex' +-+ .9"a(x) e .9".,(x')] .

(The proof of these sentences is immediate from the definition of .9".,.) Thus for each permutation (j of the set X of atoms we have defined a functional relation y = .9"i x) which is an automorphism of the entire universe ~1' For notational simplicity we shall from here on write y=(j(x) for y=.9".,(x); since for any atom x we have (j(x)=.9"a(x), this convention will not precipitate any confusion. LEMMA: Let E(ao, ... , all-i) be a sentence with parameters ao, ... , a,,-1>' and let (j be a permutation of X. Then E(ao, ... , all-i) +-+ E«(jao, ... , (jail-i)' PROOF:

The proof is by informal induction on the length of the formula

E(ao, ... , an-i); if this formula is atomic, that is, ao e a1 or ao=a1 , then the theorem holds because (j is an e-automorphism of the universe ~l ' If E(ao,' '" all-i) is ,F(ao" .. , all-i) we have F(ao, ... , all -1) +-+F((ja o, ... , (jall -1) by the induction hypothesis; and therefore the required equivalence holds also for E. When E(ao, ... , all-i) is F(ao, ... , an-i) V V G(ao, "., a,. -1) the proof is the same. If E(ao, .. " all-i) is 3xF(x, ao, "" all-i), suppose it to be true in ~i; then for some object a we have F(a, ao, .. ,' all-i)' and so F«(ja, (jao, .. " (jail-i) by the induction hypothesis; thus 3xF(x, (jao, ... , (jall -1)' The

converse is handled likewise.• Note that this lemma is another example of a theorem scheme, this time in the theory T defined on p. 74 above. What we have shown is the sentence 'For every permutation (j of the set of atoms, and for every xo, ... , X,.-l> E(xo, ... , X,.-1) +-+ E«(jxo, ... , (jX,.-l)' for an arbitrary parameter-free formula E(xo,"" X n -i)'

LEMMA: Let A (x, ao, ... , ak-1) be a formula of one free variable holding for just one set, the set a; let (j be a permutation of X. Then (ja is the only setfor which theformula A (x, (jao, "., (jak-l) holds. PROOF: By assumption we have 'v'x[x=a +-+ A(x, ao, ... , ak-1)]; so, by

the previous lemma, 'v'x [x = (ja +-+ A (x, (jao, ... , (ja k -

1 )] ••

FRAENKEL/MOSTOWSKI MODELS LEMMA:

77

IflX is an ordinal, and Uis a permutation ofX, then UlX = lX.

UlX#=lX; then u/J=/J for all /JelX, so ulX={u/Jl/JelX}=lX, which contradicts the

PROOF: Otherwise, suppose lX is the first ordinal for which

supposition. I

Within the universe ~1 we can apply the generalized reflection principle (p. 51) to the functional relation y= W,.. Since the universe is exhausted in this case by the (intuitive) union of all the W,.'s, we can conclude that for every formula Ethat is without parameters

\flX3/J > lX\fxo ... \fXk- 1 [xo e Wp A ••• A Xk-l e Wp ~ (E(xo, ... , Xk-1) - EWp(xo, ... , Xk-1»)]. We can proceed to define the class OD1 by the formula OD1 (x), 'There are an ordinallX and a monadic expression 4>(y, lXo, ... , lXr -1, ao, ... , as-t) all of whose parameters are atoms or ordinals < lX, and val ( 4>, W,.) = {x}'. This class will be referred to as the class of sets definable by ordinals and atoms. As in Chapter VI we can prove the following result.

Let A (x, lXo, ... , tXr-1' u) be a singulary formula whose parameters are all ordinals, except for u, which is a finite sequence of atoms (a map, that is to say,from some natural number s into X). Then if a is the only set for which the formula A holds, a is definable by ordinals and atoms. LEMMA:

IXo, ••• , IXr-t, U

PROOF: The generalized reflection principle provides us with an ordinal

and big enough for u and a both to belong to W,., and such that W,. mirrors the formula A. Then the value of the expression r A(x, IXo, .•• , IXr-l> up in W,. is {a}. Write 'l'(x, IXo, •.• , IXr-l> u) for this expression. Since u is a finite sequence ao, ... , as-l of atoms we have IX>IXo, ••• , IXr-1

u = {(O, ao), ... , (8 - 1, as-I)}. Thus the expression

Elz(Yt[t 8 z<-> t = (0, ao) v ... v t = (8 - l,as-l)] A P(x,IXo, ... ,lXr-hZ») gets the value {a} in W" and has only ordinals and atoms for parameters·1 The converse can also be obtained.

78


CONVERSE: If a is a set definable by ordinals and atoms, then there is a formula A (x, bo, u) of one free variable that holds for a and a alone; the parameters of A, moreover, are just two - the ordinal bo and the finite sequence ofatoms u. PROOF: By the specification of a we have an ordinal 1o and an expression !l>(x, 0(0, ••• ,0(,-1> ao, ... , as-t) whose parameters are the atoms ao, ... , a8 -1 and the ordinals 0(0, ••• , O(,-t <1o; and the value of 4> in WyO is {a}. Such an expression specifies uniquely a natural number (the one associated with the parameter-free expression !l>(x, xo, ... , X,-1> Yo,· .. , Y8-t), an ordinal (the one associated with the finite sequence of ordinals 0:o, .•. ,0('-1)' and a finite sequence u=ao, ...• a8 - t of atoms; and therefore it singles out an ordinal Po and a finite sequence of atoms, u. A formula with three parameters E(x, 1o. Po. u). namely 'the expression (whose parameters are ordinals and atoms) associated with the ordinal Po and the finite atom-sequence u has the value {x} in the set WyO ' is therefore at hand. The number of parameters can then be reduced to two by means of the bijection from On 2 on to On; and this yields a formula A (x, bo, u) holding, as required, for a and nothing else. Such a formula A(x, bo, u) will be said to be a definition ofa by ordinals and atoms .• Note that the class OD t has been defined by a parameter-free formula. Also parameter-free is the formula HOD t (x) which defines the class of sets hereditarily definable by ordinals and atoms, HOD t . It is simply 'Every element of~({x}) is in ODt ••

LEMMA: A set a is in HOD t if and only if a is in OD t and every element of aisinHODt · PROOF: The proof is the same as that of the corresponding lemma in Chapter VI .• That the class HOD t satisfies the axioms of ZF can also be proved in the way the same claim for HOD was proved in the last chapter. We can prove that the set X of atoms is in HOD t • For every atom a is definable by the formula x=a, so is in ODt • As ~({a})=a, each atom is in HOD1 • So we need only show that X is in 0 D t ; but since it is the set of atoms, it is definable by 'v'z(z e x +-+ z={z}), so certainly qualifies for ODt • But the axiom of choice is not satisfied in HOD t ; worse, the set of atoms cannot even be linearly ordered.

FRAENKELjMOSTOWSKI MODELS

79

For suppose in HOD 1 there were a set v which imposed a strict linear order on X. Let A (x, 15 o, u) be some definition of v by ordinals and atoms. Then since u is only a finite sequence of atoms ao, ... , as-l> and X is equipollent with co (in 0111 ), there must be two atoms b, c different from all theao, ... , as - 1 ' v linearly orders X, however, so without loss of generality we can suppose (b, c) E v. Let (1 be that permutation which switches band c round, but otherwise leaves X unaltered; then nothing happens to any of the atoms ao, ... , a s - 1 under (1, so (1U=U. Moreover, (115 0 =15 0 , since 15 o is an ordinal. But, by the lemma on p. 76 (1V is defined by the formula A(x, (115 0 , (1u), so (1V actually equals v. Now (b, c) E v by assumption, so «(1b, uc) E UV, so (c, b) E v. Thus v is not a strict order, which contradicts our hypothesis. Let us note two other properties of X which contradict AC and yet are true of Xin HOD1 • Every subset of X either is finite or has a finite complement with respect to X.

Let Y be some object of HOD1 which is a subset of X, but not finite (that is to say, it is not equipollent with any natural number); and suppose A (x, Yo, u) to be a definition of Yin terms of ordinals and atoms. As Y is not finite we can certainly find abE Y which is not among the terms of the finite atom-sequence u. Let c be any other atom ignored by u, and let (1 be that permutation of X which switches band c round whilst keeping the other atoms fixed. We have at once that (1U=U and (1Yo = Yo (since Yo is an ordinal). And as Y is defined by the formula A(x, Yo, u), we must also have uY= Y. As b E Ywe get ub E uY, so c E Y. Thus every atom like c, that is every atom not accounted for in u, is an element of Y. Thus X". Y is finite. X is not finite, but it has no infinite denumerable subset (subset equipollent with co).

As X is equipollent with co within 0111 , it cannot in HOD 1 be equipollent to a natural number. If Y is a subset of X that is equipollent with co, it cannot be finite; and so X".Y is finite according to the previous result. But then X itself is equipollent with co, which means that it can be well ordered. And this, we know, is false. Note that the model of ZF+-,AC just constructed does not satisfy

80


AF, so it does not solve the problem of whether ZF+AF+,AC is consistent. Indeed, the set here which cannot be well ordered is a pathological one, the set of atoms. The results of P. Cohen described in [2] yield far more interesting relative consistency results; for example that of ZF + AF+ 'g;( ill) cannot be well ordered'.

CHAPTER VIII

CONSTRUCTIBLE SETS Relative Consistency of the Generalized Continuum Hypothesis

Given a set X and a subset Y of it we say that Y is a subset of X definable by parameters, if there is a monadic expression 4>(x, ao, ... , ak-I) which has parametersao, ... , ak-I inXand the value Yin the set X. We define a functional relation y= rI(x) on the class of all sets by the formula 'y is the set of subsets of x definable by parameters'. If the axiom ofchoice holds, then ifa is an infinite set, rI(a)=a. For, first of all, every member b of a provides a subset of a, {b}, definable

by parameters by way of the expression x=b; so rI(a)~a. And the set of expressions with parameters in a has cardinality ~o·(1(a)=a (since any such expression can be specified by a parameter-free expression and a finite sequence of members of a); so n(a)~a. This proves that for infinite a, rI (a) is a proper subset of f!jJ(a). Observe however that the functional relation y= rI(x) is not increasing with x; that, in other words, we can have a c b without having rI(a) c rI(b). For if a is a subset of b, but not one definable by parameters, then a e rI (a) (defined as it is by the expression x=x), but a ¢ rI(b). However, if a is also a member of b, this possibility no longer obtains. THEoREM:Ifa c bandaebthenrI(a) c rI(b). PROOF: If c e rICa) then c=val(4), a) for some monadic expression 4>(x) whose parameters ao, ... , ak-I are in a. Thus z e c iff z e val(4), a), that is iff {(x, z)} e Val(4), a). But by the theorem on pp. 61f., Val(4), a)= = Val (4)'', b) (") {x}a; thus z e ciff {(x, z)} E Val(4)'', b) A {(x, z)} e (x}a. Consequently,

z e c +-+ z e val (4)'', b) so c=val( 4>", b) (") a.

A Z

e a,

82


It follows at once that c=val[tP"{x) 1\ x 8 a, b], since in general val(tP, b) is the set of elements of b that satisfy tP. But all the parameters of the expression tP"{x) 1\ x 8 a, namely a, ao, ... , ak-l are elements of b. Thus c E TI{b).1 By setting first Lo = 0 we can define recursively a functional relation on the whole of On by setting thereafter

The class L defined by the formula L{x): 3(X[On{a) A x E L,,], the union (in the intuitive sense) of all the L", is called the class of constructible sets. Appropriately enough, a set a is called constructible if it is in L; if, therefore, it is in L" for some ordinal a. The axiom of constructibility is the sentence 'Every set is constructible', or, equivalently, 'v'x3a[On(a) A x E La]. The main business of this chapter will consist in showing that if ZF is consistent, it remains so on the addition of the axiom of constructibility. We do this by proving that the class L with the usual membership relation satisfies all the ZFaxioms and the axiom of constructibility too. We first prove some results for the L,,'s very similar to ones proved already for the VIZ'S. If a' ~ a then L". c L", since La' = U TI (Lp) c U TI (Lp) = p<,,'

p<"

=L".lfp
the other hand, if a is a limit ordinal,

so that L" =

U

p<"

Lpfor every limit ordinal a.

Corresponding to the definition of rank, we now provide a definition of order. The order of a constructible set x is the first ordinal (X for which x E L". We write it od{x), and note that od(x) is never a limit ordinal; for if a is a limit and x E L" then x is already in Lp for some smaller p.

83

CONSTRUCTIBLE SETS

LEMMA: If a is constructible, so are all its elements; their orders, moreover, are strictly less than the order ofa. PROOF: Let a=p+l be the order of a. Then aELp+ lo so aETI(Lp). Thus a c: Lp, and every element of a must belong to Lp; these elements, therefore, have lower order than a. I It follows that for any ordinal a the set L", is transitive; for if a E L", and b E a, the order of a is ::;; a, so also the order of b.

Every ordinal 0( is constructible, and the order ofa is a+ 1. If there is an ordinal a for which a E L"" let 0(0 be the least. Then 0(0 E U (Lp) and so 0(0 E (Lp) for some PE ao; thus ao c: L p, pe",o giving PE L p, in contradiction with the definition of 0(0' That much done, it remains to prove that 0( E La+1 for every ordinala. Again, let ao be the least ordinal, if there is one, such that 0(0 ¢; L"'o+l' THEOREM:

PROOF:

TI

For every ao c: U

P<"'o

TI

PE 0(0 then, we have PE L p+ 1 , so PE TI(Lp).

TI (Lp); and therefore that 0(0 c: L",o'

It follows that

Consider now the parameter-free expression !V(x) of one free variable VUVV[U8X

A

V8X=>U8V

V

A

U= v

V

V8U]

VUVV[U8X

A

V8U=> V8X].

Let the value of this expression in the set L",o be Y, which is thus the set of elements x of L",o for which the following formula, which is relativized to L",o' holds. VuVv[uEL",o

AVEL",o-+[(UEX AVEX)

-+

(u

A

VuVv[uEL",o

EV V U

= v v

V E

u)]]

1\ VEL",o-+(UEX 1\

VEU)-+VEX)].

Seeing that L",o is transitive, an element x of it will satisfy this formula if and only if VUVV[UEX

1\

VEX-+UEV

V 1\

U=V v VEU] VuVv [u EX 1\ V E U -+ V E x] ;

just, that is, when it is an ordinal. The value of !V(x) in L",o is therefore the set of ordinals belonging to L",o' We have proved that ao anyway is not one of these, and therefore

84


no greater ordinal is (since L"o is transitive). Moreover, as shown above, Clo c Lrzo' As a consequence, 4> (x) takes the vaIueClo inLlZo ' so Clo E n(L"o)=L"o+1> which contradicts the definition of Clo. I We can now proceed to check that all the axioms of ZF + AF hold in the classL.

Axiom of Extensionality: If a is constructible, its elements are too. So this axiom holds. Union Axiom: Let a be constructible, and Cl some ordinal for which a E LIZ' As LIZ is transitive, the set b = U a is a subset of L". Since, moreover, the value in LIZ of the expression Ely(y 8 a A x 8 y) - which has one free variable x, and a single parameter a, in L" by assumption - is clearly b, we can conclude that bE L IZ + 1 , and that b is constructible. Power-set Axiom: Let a be constructible, b the set of constructible subsets of a. The map x -+ od(x) defined on b has some subset of the ordinals for its range, and some ordinal Cl will therefore be an upper bound to this subset. So for ail x E b we will have od(x)~Cl, so that b c LIZ' Nothing prevents us from setting Cl so high that a is a member of LIZ' Take the expression Vu[u 8 x => U 8 a). It has one free variable, x, one parameter a E L". Its value in Lrz is the set of elements x of Lrz which are subsets of a; b then, by the definition of Cl. Thus bE L«+1> and so b is constructible. Scheme of Replacement: Let R(x, y, ao, ... , ak-l) be a formula of two free variables whose parameters ao, ... , ak-1 are in L; and suppose that, when interpreted in L, it defines a functional relation. Within the universe IJlt this functional relation is defined by the formula L (x) A L(y) A A RL(x, y, ao, ... , ak-l); both domain and range are subclasses of L. Let b stand for the set of images of the elements of some constructible set a under this functional relation. We must show b to be constructible. Each element of it certainly is, so pick some ordinal Clo exceeding the order of any element of b and big enough for a, ao, ... , ak-l to be in L"o' Then b c L lZo'

CONSTRUCTIBLE SETS

85

We now take the parameter-free k+2-ary functional relation R(x, y, xo, ... , Xk-l), and apply to it the generalized reflection principle (p. 51), using the functional relation y = Lit' This provides us with an ordinal fJ > lXo such that

'v'x'v'y'v'xo ... 'v'Xk-l [x E Lp 1\ YELp 1\ xoELp 1\ •.• 1\ Xk_1ELp--+(RL(X,y,xo"",Xk_l) ~ RLp (x, y, xo, ... , Xk-l»] . Since ao, ... , ak-l

(*)

E

Lp we have

'v'x'v'y [x E Lp 1\ YELp --+ (RL(X, y, ao, ... , ak-l) ~ RLp(x, y, ao, ... , ak-l»)]'

Now the expression r R(u, v, ao, ... , ak-lrl has two free variables u, v, and parameters ao, ... , ak- 1 E Lp; call it q,(u, v) for short. Thus Val (4), Lp) is the set of mapsJ: {u, v} --+ Lp such that RLP(J(u),J(v), ao, ... , ak-l)' By (*), however, this set is the set ofmapsJ: {u, v} --+ Lp for which RL(J(U), J(v), ao, ... , ak- 1 ). Consequently, Val (3u(u 8 a 1\ 4>(u, v), Lp) is the set of maps J: {v} --+ Lp for which 3x(x E a 1\ RL(X,J (v), ao, ... , ak- 1 »; so val (3u(u 8 a 1\ q,(u, v), Lp) is just the set of elements of L which are images of members of a under the functional relation RL(X, y, ao, ... , ak-l); in fact, b, in view of the choice of fJ. Thus bE L p + 1 , and is constructible.

Axiom oj Infinity: Every ordinal is constructible, as we have seen; so is constructible.

(i)

Axiom oj Foundation: Let a be constructible and non-empty, and b be an element of it of lowest possible order. Every element of b has strictly lower order than b itself, so no element of b is an element of a. Thus b ()

a=0.

We have now proved that all the axioms of ZF + AFhold in L.

Note that the ordinals of L are the ordinals of the universe O/!. For all ordinals IX are constructible, and clearly they all satisfy OnL(a). Conversely, a constructible IX satisfying OnL(IX) is transitive and linearly ordered by E; so an ordinal.

86


THE AXIOM OF CONSTRUCTIBILITY HOLDS IN

L

We must now show that the axiom of constructibility holds in L.

Absolute Formulas: A parameter-free formula E(xo, ... , Xk-l) is called absolute ifit can be obtained by applications ofthe following rilles. (1) A formula without quantifiers is absolute. (2) If A and B are absolute, then A v B and A 1\ B are absolute. (3) If A (x, xo, ... , Xn-t) is absolute then 3xA(x, xo, ... , xn-t)is absolute. (4) If A (x, y, xo, ... , Xn-t) is absolute, then

'Ix [x E y ~ A (x, y, x o, ... , Xn-t)]' a formula with free variables y, xo, ... , X n -l' is absolute. On the basis of this definition we call a relation absolute if it can be defined in IfI by an absolute formula.

Let E(xo, ... , Xk-t) be an absolute formula, Wa transitive class (that is, 'Vx'Vy[W(x) 1\ y E X ~ W(y)]), and W' an extension of W (that is 'Vx(W(x) ~ W' (x»). If ao, ... , ak-1 are objects in W such that E W (ao . ... , ak-1) is true, then EW'(ao, ... , ak-1) is also true. Equivalently,

THEOREM:

'Vxo ... 'VXk-t [W (xo)

1\ ••• 1\

W (Xk-t)

~ (EW(xo, ... , Xk-t) ~ EW'(xo, ... , Xk-1»]' PROOF: The proof is an informal induction on the length of the absolute formula E. For quantifier-free E the result is immediate since E W and E W ' are the same as E. If Eis F v G, takeao, ... , ak-1 in W; by the induction hypothesis

and so

F W (a o, ... , ak-l) ~ F W' (ao, ... , ak-1)

(FW (ao, ... , ak-1) v GW(ao, ... , ak-1» ~ (FW' (ao, ... , ak-1) v GW' (ao, ... , ak-1»,

as required. The same proof works if Eis F 1\ G. If E(xo, ... , Xk-1) is 3xF(x, xo, ... , Xk-1), take any ao, ... , ak-1 in W such that E W(a o, ... ,ak_ 1) holds; thus 3 x [W(x) 1\ FW(x,ao ... , ak-1)].

CONSTRUCTIBLE SETS

87

So for some ain Wwe have FW(a, ao, "" ak_l),Bytheinductionhypothesis therefore F W' (a, ao, .. " ak-l)' whence 3x[W' (x) A F W' (x, ao, .. " ak-l)]' This is E W' (ao, , .. , ak-l)' Lastly, suppose that E(y, xo, , .. , Xk-l) is 'Ix [x E Y --+ F(x, y, xo, .. " Xk-l)], and pick objects b, ao, .. " at - 1 in Wsuch that EW(b, ao, .. " at-I)' Then 'Ix [W(x)

--+

(x

E

b --+ FW(x, b, ao, , .. , ak-l»)]

which is the same thing as

But as b is in W, which is a transitive class, x E b --+ W(x). Moreover, by the induction hypothesis, (W(x) A F W(x, b, ao, , .. , at-I») --+ F W' (x, b, ao, .. ,' at-I)' Thus the last inset formula yields 'Ix [x

E

b --+ F W' (x, b, ao, , .. , at-I)],

and so E W' (b, ao, ... , ak-l)'. Note that the definition of absolute relation given here is equivalent to that ofP. Cohen in [2]; it is not the same as K. Godel's definition in [1].

If the relation G(y, zo, .. " Z,-I) and the functional relation y= F(xo, .. ,' Xk-l) are absolute, then the relation G(F(xo, .. " Xt-l), Zo, ... , Z,-I) is absolute. The relation in question can be defined by 3y [y

= F(xo, ... , Xt-l)

A

G(y, zo, ... , ZI-I)] '

so can be realized by application of rules (2) and (3) to the absolute relations given. In particular, if a is a set for which y=a is equivalent to some absolute formula R(y), and if G(y, Zo, .. " Z,-I) is absolute, then G(a, Zo, .... Z,-I) is absolute. For the rest of this section we shall assume the axiom of foundation to hold for "II, and our efforts will be directed towards proving that the functional relation y=La. is absolute, From this it will follow easily that the axiom of constructibility holds in L.

88


The (singulary) relations On (x) and x = ware absolute. Nothing has to be proved for the former, since AFholds,l and so On(x) can be written 'v'u'v'v [u E x A V E X ~ U E V V U = v v V E u] A 'v'u [u E X ~ 'Iv (v E U ~ V EX)]. For the latter we simply check the absoluteness of the following relations: x={), defined by 'v'y(y E X -+ Y if. x). y=x u {x}, defined by 'v'z(z E Y -+ Z=X v Z E x) A x=w,definedby 'v'Z(ZEX-+ZEY) A XEy. On(x)

'v'y [y E X ~ Y u {y} EX] 'v'y [y E X ~ (y = () v 3z(y = Z U {z}))]. We have shown above that certain compositions of absolute relations are themselves absolute. Since the functional relation y=L", whose absoluteness we are after, is defined by recursion, we shall want to show that functional relations defined recursively from absolute functional relations are themselves absolute. And we shall also want to show that the functional relation y= IT (x) is absolute. These results are the content of the next two lemmas. A () E X A

A

Let y=H(x) be an absolute functional relation whose domain is the class of all maps defined on the ordinals. Then the functional relation F recursively definedfor every ordinal IX py LEMMA:

F(IX) = R(F r IX) is absolute. PROOF: We check successively the absoluteness of a dozen or more relations. z={X,y}:XEZ A yEZ A 'v't(tEZ-+ t=x v t=y). Z= (x, y) : Z= {{x}, {x, y}} (composition of absolute functional relations). y c x: 'v'Z[ZEY -+ ZEX]. z=xuy:x c Z A y e Z A 'v't(tEZ~ tEx V tEY). Z=X n y: Z C X A Z C Y A 'v't[t E X -+ (t E y ~ t E z)]. z=x"",y: Z C x A 'v't[t E x ~ (t if. y ~ t E z)]. Z C xxy: 'v't[tEZ ~ 3u3v[UEX A VEy A t=(u, v)]]. Z::> xxy: 'v'U[U E x ~ 'Iv [V E Y -+ 3t(t E Z A t=(u, v»)]].

1 The formula On (a) as written on p. 14 is patently not absolute. What is more, it can be shown not to be equivalent, in ZF, to any absolute formula.

89

CONSTRUCTIBLE SETS Z Z

=x x y : the conjunction of the previous two relations. maps x into y : Z c x x Y /\

'v'u[uEx-dv3t(VEY /\ tEz /\ t=(u,v»] /\ 'v't'v't''v'u'v'v'v'v' [(t E Z /\ t' E Z /\ U EX /\ V E Y /\ V' /\ t=(u,v) /\ t'=(u,v'»-v=v'].

E

Y

fis a map with domain x: 3y(Jmaps x into y). fis a map: 3x(Jis a map with domain x). g =ft x(a ternary relation):

3x' [x c x' /\ (J is a map with domain x') /\ (g is a map with domain x) /\ g c J] . y=f(x) (a binary functional relation): (fis a map) /\ (x, y) Ef. We can now write the formula y=F(rx) which we are interested in as On(rx) /\ 3f[(I is a map with domain rx) /\ 'v'fJ(f3 E rx - f(f3) = H(J t fJ») /\ y = H(J)]. If the relation y = H(J) is absolute, so too is this .• We shall need also the following absolute relations. fis an injection ofx into y (fis one-one):

(J maps x into y) /\ 'v't'v't''v'u'v'u''v'v [(t E f /\ t' E f /\ U EX /\ U' E X /\ V E

Y /\ t

= (u, v) /\ t' = (u', v» _ u = u'].

fis a surjection from x on to y: (fmaps x into y) /\

'v'v [v E Y - 3u (u EX /\ (u, v) E j)] . h= fo g (a binary functional relation): 3x3y3z [(g maps x into y) /\ (J maps y into z) /\ (h maps x into z)] /\ 'v't [t E h - 3u3v3w [(u, v) E g /\ (v, w) E f /\ t

= (u, w)]] .

Now the relation y = La., whose absoluteness is in question, is a functional relation defined by recursion. We shall therefore apply the above lemma. U f](J(x»), and this we For the relation y=H(J) we take y= xedom(!)

90


must show to be absolute (dom(f) is the domain off). It can be written, however, as

IT

'Ix [x E dom (I) ~ (I(x») c y] "Vz[zey -+3x[xEdom(l) " ZEIT(I(X»)]]. The following lengthy lemma will therefore complete the proofthaty=L~ is absolute. Thefunctional relation y= IT (x) is absolute. PROOF: As the relations x = and y = x u {x} are absolute, it follows that the relations X= 1, x=2, ... are absolute (by successive composition). As "", v, 3, e, =, and "// are respectively 0, 1,2, 3, 4, and the set of natural numbers ;;':5, the relations x= "", X= v, x=3, X=8, X= =, and x="// (this latter being written x=m"'-{O, 1, 2, 3, 4}) are absolute. From this we can derive the absoluteness of several other relations, as follows.

°

LEMMA:

z =
z=

({ e}

x "// x "//) u ({ =} x "// x "//)

(a composition of absolute functional relations). Z=
3f [(I is a map with domain m) "f(O) =
y

=

x u [{ "'} x x] u [{ v} x x x x] u [({3} x "//) x x].

Z E
z E
-+f«3x,y»=f(y)"'-{x}] " y=f(z)].

91

CONSTRUCTIBLE SETS

We now have to show that the binary functional relation y=W(F)X is absolute. Note that the formula y=YX (three variables, y, X, Y) is not absolute; for if Z is a transitive set, (y=YX)Z means that y is the set of maps from Y into X that are elements of Z, and clearly there is no reason for supposing that y is the set of all maps from Y into X. By the theorem on p. 86 then, the formula y = Y X is not absolute. Only because w(F) is always finite can we expect to get the result we want. Some more absolute relations first. k is a natural number and y = k X (a functional relation of two arguments, Xandk): 31 [(I is a map with domain (0) A 1(0) = 0 A 'v'n [n e 00 -+ I(n + 1) = N (n, X,J(n))] A y = I(k)] , where the ternary functional relation Z=N(n, X, Y) is given by the absolute formula nero A w(F)X

A

'v'9 [g e Z

'v'h'v'x[(h e Y

-+

A X

9 maps n + 1 into X] eX) -+ h u {(n, x)} eZ].

c y (a ternary relation with arguments y, X, F):

3ep3n [n e 00

A

(ep is a bijection from w(F) on to n)

A'v'I(lenX-+/oepey)].

y = W (F) X (two arguments, X, F):

w(F)X c y

A

'v'1 [J e y

-+

(J maps w(F) into X)].

y is a closed expression with parameters in X:

3F3ep[Fetf

A

(epmapsw(F)intoX)

A

y=(F,ep»)].

y = t9'~ (y is the set of closed expressions with parameters in X):

'v'x [x e y -+ (x is a closed expression with parameters in X)] A 'v'F[F e tf -+ 'v'ep [ep e w(F)X -+ (F, ep) e y]]. ye x):

tf~

(y is an expression with parameters in X and a single free variable

3F3ep [F e tfAx e w(F) A (epmaps w(F)"'-{x} into X) A y=(F,ep)]. y = tfx:'v'z[z e y -+ Z e tfx] A 'v'F'v'ep [F e tfAx e w(F) A ep e w(F)'{x}X -+ (F, ep) e y].

92


y e is'~ and z is the closed expression obtained by substituting in y the element a of X for the variable x (five arguments, x, y, z, X, a):

3F3cp3t/1 [F e is'

A

X e w(F)

A A

(cp maps w(F)"-{x} into X) (t/I maps w(F) into X) A cp c t/I t/I(X) = a A y=(F,cp) A z=(F,t/I)]. A

4J is a closed expression with parameters in X and O=Val(4J, X) (this is a functional relation of two arguments, 4J, X; 0 takes the values 0, 1): 4J

e

tf~ A 3f [(fmaps is'~ into {O, I}) A Va'v'b[a e X Abe X

«a e b A f(a 8 b) = 1) v (a ¢ b A f(a 8 b) = 0»] A VaVb [a e X Abe X -+ «a = b A f(a = b) = 1) v (a oF b A f(a = b) = 0»] A 'v'I[' [I[' e is'~ -+ f( '" 1[') = 1 - f(I[')] A VI[''v'I['' [I[' e is'~ A 1[" e is'~ -+ f(I[' V 1[") = f(I[') U f(I[")] A VxVI['[x e -r A I[' e is'~ -+ ([3a(a e X A f(I['(a» = 1) A f(3xl[') = 1] v [Va (a eX -+ f(I['(a» = 0) A f(3xl[') = 0])] A 8 = f(4J)].

-+

4J e tf~ A Y = val ( 4J, X) (y is the subset represented by the monadic expression 4J in the set X) :

Va[aey-+(aeX A Val(4J(a),X) = 1)] A 'v'a [a EX -+ (Val(4J(a), X) = 0 v a e y)]. ye

IT (X) (two arguments, y, X):

3x34J[x e -r A 4J e is'~ A Y = val(4J, X)]. Y = IT (X):Vz[z e y -+ ZEIT (X)] A 'v'x'v'4J[xe-r A 4Jetf~-+3z(zey A z=val(4J,X»]. In this waytherelationy= THEOREM:

IT (X) is shown to be absolute.•

The relation y = L« is absolute.

It follows at once that the axiom of consfructibility holds in L. For since ZF + AFholds in the class L, the relation (y = L,,)L is a functional relation. Moreover, according to the theorem on p. 86 we have

VyVa[L(y)

A

L(a)

A

(y = La)L -+ y = La],

CONSTRUCTIBLE SETS

93

since L is transitive and y=La is an absolute formula. So let a be some object in L. For some 0(0 then, a E Lao; and since OCo is an ordinal of L, there is a unique constructible set Yo for which (YO=Lao)L holds; as (yo = Lao)L ~ Yo = Lao' we get Yo = Lao' soa EYo. Thus (a E Lao)L is true, and so [30((a E La)]L is also true. The axiom of constructibility asserts that the three classes 011, V, L are identical (this axiom therefore implies AF). The axiom of constructibility is usually denoted V=L, which is an abbreviation of the formula Vx[V(x) +-+ L(x)] (when this notation is used, AFis understood to hold). What we have shown, therefore, is that if ZFis consistent, then so is ZF + AF+ V = L. The axiom of constructibility has many important consequences. We shall now show that it entails the axiom of choice, in particular - and even the choice principle stated on p. 69-, and the generalized continuum hypothesis. These two sentences are thus satisfied in L, and therefore consistent relative to the remaining axioms.

V= L

IMPLIES

AC

Consider a universe 011 which satisfies the axiom of constructibility. With a parameter-jree formula we shall define on the class of wellorderings in t¥I a functional relation v = F( u) having the following property. If u is a well-ordering of the set X, F(u) is a well-ordering of the set of monadic expressions with parameters in X. Let tff be the set of parameter-free expressions; it is a subset of V"" and we have already defined (on p. 41) an injection, itself without parameters, which maps tff one-one into w. This injection, K- 1, is the inverse of K, suitably restricted. Letj be the isomorphism from u on to its ordinal oc. A monadic expression with parameters in X is a pair (E, 1'/) where E E tff and 1'/ maps some subset of weE) into X. If u well-orders X, each element of 1'/, (x, y), where x E "Y and y E X, can be associated one-one with a pair of ordinals; namely x itself (remember the definition of"Y as w'-.5) and the 'place-number' of y in the well-ordering u of X (that is, the ordinal of the well-ordered initial segment Six, u». This pair of ordinals can itself be associated one-one with an ordinal, by the isomorphism from On2 on to On, defined on p. 33. Thus 1'/ can be associated one-one with a finite sequence of ordinals - namely the sequence

94


which takes in their natural order the ordinals associated with its elements. As a consequence, and using the injection K- 1 noted above, we can find an injection from the set of monadic expressions with parameters in X into (j) x U ( On), aclass which is well ordered by the well-ordering already defined on u( On). In this way we can obtain the well-ordering v of the set of monadic expressions with parameters in X. We can now define recursively on the ordinals a parameter-free functional relation y = B(a) which to each ordinal a associates a well-ordering B(a) of LI1.' such that for p:::;;'a, Lp is an initial segment of LI1. under B{a); and on Lp, B(a) and B(P) are identical. For the definition, suppose B{P) suitably defined for all p
p<11.

common extension of all the B(P) for P< a. If a = P+ I, we use the fact that B(P) well-orders Lp, so that vp = F(B{P» well-orders the set of monadic expressions whose parameters lie in Lp. For the well-ordering B{a) of LI1. = (Lp) we therefore write

TI

x:::;;,y(mod.B{a»-[xeLp A yeLp A x::::;; y(mod. B{P»] v [xeLp A yeLI1."Lp] v [x e LI1. "Lp AyE LI1. "Lp A the first expression defining x in Lp is earlier, in the well-ordering vp of the monadic expressions with parameters in L p, than the first expression defining yin Lp]. It is obvious that the ordering B(a) thus defined on LI1. has all the required properties. The functional relationy = B(a) is thus completely defined by recursion. A binary relation R(x, y) without parameters which well-orders the whole of all is then available from the formula 3a[On(a) A x:::;;'y (mod. B(a»]. This shows that the choice principle holds in all; by a result on p. 69, the axiom 'every set is ordinal definable' holds also in 0/1.

v= L

IMPLIES THE GENERALIZED CONTINUUM HYPOTHESIS

The proof of this result relies on a certain property of absolute relations, dealt with in the next theorem. THEOREM:

Let all be a universe where ZF+AF+AC holds, and E(x, y) an

CONSTRUCTIBLE SETS

95

absolute formula defining in tPJ a functional relation Y=4>(x) of a single argument. Then 4> (x) ~ ~ (x) + Nofor every x in the domain of 4>. PROOF: Let a be in the domain of 4>, and {a) E V"' and such that 'v'x'v'y [x E V" AyE V"

-+

(Ev,,{x, y) +-+ E{x y»)].

That there is such an ordinal is guaranteed by the reflection principle for the formula E(x, y). Let P stand for the transitive closure of {a}, that is 'if ({a}). Since {a} c: V" and V" is transitive, we haveP c: V". Moreover, P=~(a) u {a},

p=

'if (a) + 1. Let ~ be the set of all closed expressions that are satisfied in V" and have all their parameters in P. By the LowenheimJSkolem theorem (pp. 59-61) there is a subset X of V"' including P, satisfying all the expressions

so

in ~, and of cardinality not exceeding P+ ~o. Since V" is a transitive set, the axiom of extensionality holds in it; so the corresponding expression

VxVy [x = y <=> Vz{z 8 X <=> Z 8 y)] is in ~, and is therefore satisfied in X. Thus X is an extensional set. By the corollary on p. 40 we can find an isomorphism j of X on to some transitive set Y; in consequence, Y will satisfy all the expressions in the set ~' which is produced from ~ by replacing each parameter u (inP, and so in X) by j{u). But j(u) = u for every u E P, as can be seen by supposing Uo to be an element of P for whichj(uoh~:uo, and one ofleast rank at that. For then j(v) is equal to v for all v E Uo; for P=~({a}) is transitive, so any such v EP. However,j{uo)={j (v) I v E uo}=uo, which contradicts the definitionofuo' Thus P c: Y, and Y satisfies all the expressions of ~; in addition,

Y=X~P+~o.

We chose V" to mirror the formula E(x, y), so since a and 4>(a) are both in V"' and E(a, 4>(a») is true, we have also EV"(a, 4>(a»). Thus the sentence E(a, 4>(a») is true in V"' and so 3yE{a, y) is also true in V". The expression r3yE{a, y)' is thus satisfied in V"' and has a single parameter

96


a, in P; so it belongs to ~, and is therefore satisfied in Y. Consequently, the formula 3yE(a, y)holdsin Y,andforsomebe YwewillhaveEY(a, b). But as E( x, y) is an absolute formula and Y is a transitive set, we also have EY(a, b) -+ E(a, b). Thus E(a, b), and so b=iP(a). But b was in Y, so 4'(a)e Y, and Ybeing transitive, 4'(a) c Y. Thus iP(a)~:Y~P+~o, and so iP(a)~~(a)+~o.1 As a particular case of this theorem, suppose a is an object of o/t defined by an absolute formula E(y); in other words, that Vy[y=a +-+ E(y)]; thena~~o·

It is enough for our purpose to put x = 0 and to apply the above theorem to a formula E(y) which does not contain x at all. We can deduce in this way that, for example, if ZF + AF+ AC holds in 0lI, then the formula y = gJ( w) is not equivalent to any absolute formula.

= THEOREM: Suppose again that ZF+AF+AC holds in 0lI. Then L(I.=a,for every ordinal (X~w. /fa is a constructible set, od(a)~~(a)+~o. PROOF: We have seen already that (X c L(I.' so that a,~La.' To show the converse inequality for all (X~w we proceed by induction. For w we note that Ln= v" for every new (a trivial induction proves this); thus Lru= Vru ,

U n(Lp), and by the induction hypothesis Lp~P if P<(I. w~f3<(X. Butforinfinite awe have n(a)=a, so Now La.=

An alternative proof of this would apply the previous theorem to the absolute relation y = L(I.' It is in this way that we demonstrate the second part of the theorem. Recall that od(a) is the first ordinal (X for which a e La.; the relation y=od(x) is absolute since it can be written

On(y)

A

xeLy

A

Vz[zey -+ x Et:L.] ,

and the relation y=Lx is absolute. At once then,

od(a)~~(a)+~o.1

CONSTRUCTIBLE SETS

97

With all this achieved, let d/I be a universe satisfying the axiom of constructibility. We have already proved that the axiom of choice holds in CFI, so the theorems above can be applied. Any set a is constructible, so if a c: ~p we have

od(a) ~ since ~(a)

c: ~p.

~(a)

+ ~o ~ ~P'

Thus od(a)<~p+l' so a E L Np +1' But a is any subset of

----

~P' so t1'(~p) c: LNp +1 ' whence 9I'(~P)~LNp+l' Thus 9I'(~P)~~P+l'

Cantor's theorem gives us the reverse inequality, so we can conclude that 9I'(~p) = ~P+ 1> which is the generalized continuum hypothesis.

As far as arithmetical formulas are concerned, we have an analogue of the result obtained in Chapter VI. If E is an arithmeticalformula ( aformula whose quantifiers are relativized to Vro) that is derivable with the help of the axiom of choice and the generalized continuum hypothesis (and, more generally, the axiom of constructibility), then E is already derivable from ZF alone. For Vro=Lro, as we have seen. A formula E with quantifiers relativized to Lro is absolute, and so EL - E is a consequence of ZF. Let, then, Ao, ... , A n - 2 , E be a proof of E from ZF+ V=L. If AI is an axiom, we know that Af is a theorem of ZF. The sequence A~, ... , A~-2' EL is thus a proof of EL from ZF alone. With EL _ E also a theorem of ZF, we see that E itself is a theorem of this theory.

BIBLIOGRAPHY

The two standard works on this subject are [1] Kurt GOdel, 1940, 'The Consistency of the Continuum Hypothesis', Annals of Mathematics Studies 3, Princeton University Press. [2] Paul J. Cohen, 1966, Set Theory and the Continuum Hypothesis, W. A. Benjamin, New York. An exposition of naive set theory can be found in [3] Paul R. Halmos, 1960, Naive Set Theory, D. Van Nostrand, Princeton. The fundamental ideas of elementary logic and model theory are discussed in detail in, for example, [4] Elliott Mendelson, 1964, Introduction to Mathematical Logic, D. Van Nostrand, Princeton. [5] G. Kreisel and J. L. Krivine, 1967, Elements of Mathematical Logic (Model Theory), North Holland Publishing Company, Amsterdam. [6] Roger C. Lyndon, 1966, Notes on Logic, D. Van Nostrand, Princeton. [7] Joseph R. Shoenfield, 1967, Mathematical Logic, Addison-Wesley. This last book contains (amongst much else) an important chapter on set theory. On the subject of ordinal definable sets see [8] John Myhill and Dana Scott, 1967, 'Ordinal Definability', Summer Institute on Axiomatic Set Theory, Los Angeles. The proof given in the text of the generalized continuum hypothesis in L is close to that of [9] Carol Karp, 1967, 'A Proof of the Relative Consistency of the Continuum Hypothesis', in Sets, Models and Recursion Theory (ed. by John N. Crossley), NorthHolland Publishing Company, Amsterdam, pp. 1-32. For the topic of absolute expressions see also [10] Azriel Levy, 1965, 'A Hierarchy of Formulas in Set Theory', Memoirs of the American Mathematical Society 57.

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W. BETH, Aspect of Modern Logic. 1970, XI + 176 pp.

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tPAUL WEINGARTNER and GERHARD ZECHA, (eds.), Induction, Physics, and Ethics. Proceedings and Discussions of the 1968 Salzburg Colloquium in the Philosophy of Science. 1970, X + 382 pp. 00. 65,A. EBERLE, Nominalistic Systems. 1970, IX + 217 pp. Dft.42,PAAKKO HINTIKKA and PATRICK SUPPES, Information and Inference. X + 336 pp. Dft.60,~ROLF

tKAREL LAMBERT, Philosophical Problems in Logic. Some Recent Developments. 1970, VII +176 pp. Dft. 38,tP. V. TAVANEC (ed.), Problems of the Logic of Scientific Knowledge. 1969, XII+ 429 pp. Dft. 95,~ROBERT S. CoHEN and RAYMOND J. SEEGER (eds.), Boston Studies in the Philosophy of Science. Volume VI: Ernst Mach: Physicist and Philosopher. 1970, Vm+295 pp. Dft.38,tMARSHALL SWAIN (ed.), Induction, Acceptance, and Rational Belie/. 1970, VII+ 232 pp. Dft.4O,tNICHOLAS RESCHER et aI. (ed8.), Essays in Honor of Carl G. Hempel. A Tribute on the Occasion of his Sixty-Fifth Birthday. 1969, VII + 272 pp. Dft. 46,tPATRICK SUPPES. Studies in the Methodology and Foundations of Science. Selected Papers from 1951 to 1969.1969, XII +473 pp. Dft.70,tJAAKKO HnmKKA, Models for Modalities. Selected Essays. 1969, IX + 220 pp. Dft.34,-

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tRoBERT S. COHEN and MARX W. WARTOFSKY (eds.), Boston Studies in the Philosophy 0/ Science. Volume IV: Proceedings 0/ the Boston Colloquium for the Philosophy 0/ Science 1966/1968. 1969, VIII + 537 pp. Dfl. 69,tNICHOLAS RESCHER, Topics in Philosophical Logic. 1968, XIV + 347 pp.

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tGONTHER PATZIG, Aristotle's Theory o/the Syllogism. A Logical-Philological Study 0/ Book A 0/ the Prior Analytics. 1968, XVII + 215 pp. Dfl. 45,tC. D. BROAD, Induction, Probability, and Causation. Selected Papers. 1968, XI + 296 pp. Dfl.48,tRoBERT S. COHEN and MARX W. WARTOFSKY (eds.), Boston Studies in the Philosophy 0/ Science. Volume III: Proceedings 0/ the Boston Colloquium for the Philosophy 0/ Science 1964/1966. 1967, XLIX + 489 pp. Dfl. 65,tGUIDO KONG, Ontology and the Logistic Analysis 0/ Language. An Enquiry into the Contemporary Views on Universals. 1967, XI + 210 pp. Dfl. 38,·EvERT W. BETH and JEAN PIAGET, Mathematical Epistemology and Psychology. 1966. XXII + 326 pp. Dfl. 58,·EVERT W. BETH, Mathematical Thought. An Introduction to the Philosophy 0/ MatheDfl. 32,matics. 1965, XII + 208 pp. tPAUL LORENZEN, Formal Logic. 1965, VIII + 123 pp.

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Introduction to Axiomatic Set Theory (1971) - Krivine

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