STUDIES I N LOGIC AND THE FOUNDATIONS OF MATHEMATICS
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STUDIES I N LOGIC AND THE FOUNDATIONS OF MATHEMATICS
Editors
A. H E Y T I N G, Amsterdam A. M 0 S T 0 W S KI, Warszawa A. R O B I N S O N , NewHuven P. S U P P E S, Stanford
Advisory Editorial Board
Y. B A R - H I L L E L, Jerusalem K. L. D E BOUVl?iRE, SuntaChru H. H E R M E S, Freiburg i. Br. J. H I N T I K K A, Helsinki J. C . S H E P H E R D S O N , Eristol E. P. S P E C K E R, Zurich
NORTH-HOLLAND PUBLISHING AMSTERDAM
COMPANY
SET THEORY
K. KURATOWSKI and
A. MOSTOWSKI Members of the Polish Academy of Sciences
1968 PWN - POLISH SCIENTIFIC PUBLISHERS - WARSZAWA NORTH - HOLLAND PUBLISHING COMPANY - AMSTERDAM
Copyright by PAfiSTWOWE WYDAWNICTWO NAUKOWE (PWN-POLISH SCIENTIFIC PUBLISHERS) Warszawa, Poland 1966 and 1968
This book is a translation of the original Polish “Teoria mnogoici’’ published by PWN-Polish Scientific Publishers, 1966 in the series “Monografie Matematyczne” edited by the Polish Academy of Sciences. Translated from the Polish by M. Mrlnydski The English edition of this book has been publis.hed jointly by PWN and NORTH - HOLLAND PUBLISHING COMPANY
Library of Congress Catalogue Card Number 67-21972.
PRINTED IN POLAND
PREFACE
The creation of set theory can be traced back to the work of XIX century mathematicians who tried to find a firm foundation for calculus. While the early contributors to the subject (Bolzano, Du Bois Reymond, Dedekind) were concerned with sets of numbers or of functions, the proper founder of set theory, Georg Cantor, made a decisive step and started an investigation of sets with arbitrary elements. The series of articles published by him in the years 1871-1883 contains an almost modern exposition of the theory of cardinals and of ordered and wellordered sets. That the step toward generalizations which Cantor made was a difficult one was witnessed by various contradictions (antinomies of set theory) discovered in set theory by various authors around 1900. The crisis created by these antinomies was overcome by Zermelo who formulated in 1904-1908 the first system of axioms of set theory. His axioms were sufficient to obtain all mathematically important results of set theory and at the same time did not allow the reconstruction of any known antinomy. Close ties between set theory and philosophy of mathematics date back to discussions concerning the nature of antinomies and the axiomatization of set theory. The fundamental problems of philosophy of mathematics such as the meaning of existence in mathematics, axiomatics versus description of reality, the need of consistency proofs and means admissible in such proofs were never better illustrated than in these discussions. After an initial period of distrust the newly created set theory made a triumphal inroad in all fields of mathematics. Its influence on mathematics of the present century is clearly visible in the choice of modern problems and in the way these problems are solved. Applications of set theory are thus immense. But set theory developed also problems of its own. These problems and their solutions represent what is known as abstract set theory. Its achievements are rather modest in comparison
vi
PREPACE
to the applications of set theoretical methods in other branches of mathematics, some of which owe their very existence to set theory. Still, abstract set theory is a well-established part of mathematics and the knowledge of its basic notions is required from every mathematician. Last years saw a stormy advance in foundations of set theory. After breaking through discoveries of Godel in 1940 who showed relative consistency of various set-theoretical hypotheses the recent works of Cohen allowed him and his successors to solve most problems of independence of these hypotheses while at the same time the works of Tarsk showed how deeply can we delve in the domain of inaccessible cardinals whose magnitude surpasses all imagination. These recent works will certainly influence the future thinking on the philosophical foundations of mathematics. The present book arose from a mimeographed text of Kuratowski from 1921 and from an enlarged edition prepared jointly by the two authors in 1951. As a glance on the list of contents will show, we intended to present the basic results of abstract set theory in the traditional order which goes back still to Cantor: algebra of sets, theory of cardinals, ordering and well-ordering of sets. We lay more stress on applications than it is usually done in texts of abstract set theory. The main field in which we illustrate set-theoretical methods is general topology. We also included a chapter on Borel, analytical and projective sets. The exposition is based on axioms which are essentially the ones of ZermeloFraenkel. We tried to present the proofs of all theorems even of the very trivial ones in such a way that the reader feels convinced that they are entirely based on the axioms. This accounts for some pedantry in notation and in the actual writing of several formulae which could be dispensed with if we did not wish to put the finger on axioms which we use in proofs. In some examples we use notions which are commonly known but which were not defined in our book by means of the primitive terms of our system. These examples are marked by the sign #. In order to illustrate the role of the axiom of choice we marked by a small circle O all theorems in which this axiom is used. There is in the book a brief account of the continuum hypothesis and a chapter on inaccessible cardinals. These topics deserve a more thorough presentation which however we could not include because of lack of space
PREFACE
vii
Also the last chapter which deals with the descriptive set theory is meant to be just an introduction to the subject. Several colleagues helped us with the preparation of the text. Dr M. Mqczyriski translated the main part of the book and Mr R. Kowalsky collaborated with him in this difficult task. Professor J. EoS wrote a penetrating appraisal of the manuscript of the 1951 edition as well as of the present one. His remarks and criticism allowed us to eliminate many errors and inaccuracies. Mr W. Marek and Mr K. Wibiewski read the manuscript and the galley proofs and helped us in improving our text. To all these persons we express our deep gratitude. KAZIMIERZ KURATOWSKI ANDRZEJMOSTOWSKI
ERRATA Page, lin :
For
Read
18’’
Awl -A into subset
Awl =I onto subsets
232’ 324’
K. Kuratowski and A. Mostowski, Set Theory
CHAPTER I
ALGEBRA OF SETS
0 1.
Propositional calculus
Mathematical reasoning in set theory can be presented in a very clear form by making use of logical symbols and by basing argumentation on the laws of logic formulated in terms of such symbols. In this section we shall present some basic principles of logic in order to refer to them later in this chapter and in the remainder of the book. We shall designate arbitrary sentences by the letters p , q,r, ... We assume that all of the sentences to be considered are either true or false. Since we consider only sentences of mathematics, we shall be dealing with sentences for which the above assumption is applicable. From two arbitrary sentences, p and q, we can form a new sentence by applying to p and to q any one of the connectives: and,
or,
if
... then ... ,
if and only if.
The sentence p and q we write in symbols p A q. The sentence p A q is called the conjuncrion or the logical product of the sentences p and q, which are the components of the conjunction. The conjunction p A y is true when both components are true. On the other hand, if any one of the components is false then the conjunction is false. The sentence p or q, which we write symbolically p v q, is called the dis&nction or the logical sum of the sentences p and q (the components of the disjunction). The disjunction is true if either of the components is true and is false only when both components are false. The sentence i f p then q is called the implication of q by p , where p is called the antecedent and q the consequent of the implication. Instead
2
I. ALGEBRA OF SETS
of writing if p then q we write p + q. An implication is false if the consequent is false and the antecedent true. In all other cases the implication is true. If the implication p -+ q is true we say that q follows from p ; in case we know that the sentence p is true we may conclude that the sentence q is also true. In ordinary language the sense of the expression “if ..., then ...” does not entirely coincide with the meaning given above. However, in mathematics the use of such a definition as we have given is useful. The sentence p if and only if q is called the equivalence of the two component sentences p and q and is written p = q. This sentence is true provided p and q have the same logical value; that is, either both are true or both are false. If p is true and q false, or if p is false and q true, then the equivalence p = q is false. The equivalence p = q can also be defined by the conjunction
(P
+
4) A (4 -+ PI
-
The sentence it is not true that p we call the negation of p and we write l p . The negation l p is true when p is false and false when p is true. Hence l p has the logical value opposite to that of p. We shall denote an arbitrary true sentence by V and an arbitrary false sentence by F; for instance, we may choose for V the sentence 2.2 = 4, and for F the sentence 2.2 = 5. Using the symbols F and V , we can write the definitions of truth and falsity for conjunction, disjunction, implication, equivalence and negatiom in the form of the following true equivalences: (1)
F A F = F,
FAVGF,
VAFGF,
VAVEV,
(2)
F v F r F,
FvVrV,
V V F r V,
VVVZV,
(3) ( F + F ) r V ,
(F+V)=V,
(Y+F)rF,
(V+V)EV,
(4) ( F r F ) = V ,
(F=V)=F,
(V=F)=F,
(V=V)=V,
(5)
1 F r V,
1 V = F.
Logical laws or tautologies are those expressions built up from the letters p, q, r, ... and the connectives A , v ,-t, =, 1which have the
1. PROPOSITIONAL CALCULUS
3
property that no matter how we replace the letters p , q, r, ... by arbitrary sentences (true or false) the entire expression itself is always true. The truth or falsity of a sentence built up by means of COM~Ctives from the sentences p , q, r, ... does not depend upon the meaning of the sentences p , q, r, ... but only upon their logical values. Thus we can test whether an expression is a logical law by applying the following method: in place of the letters p , q, r, ... we substitute the values F and V in every possible manner. Then using equations (1)-(5) we calculate the logical value of the expression for each one of these substitutions. If this value is always true, then the expression is a tautology. Example. The expression ( p A q) + ( p v r) is a tautology. It contains three variables p , q and r. Thus we must make a total of eight substitutions, since for each variable we may substitute either F or V. If, for example, for each letter we substitute F, then we obtain ( F A F) -+ ( F v F ) , and by (1) and (2) we obtain F + F, namely V. Similarly, the value of the expression ( p A q) + ( p v q ) is true in each of the remaining seven cases. Below we give several' of the most important logical laws together with names for them. Checking that they are indeed logical laws is an exercise which may be left to the reader. ( P v 4 ) = (4 V P ) KP v 4) rl = CP v (4 v r)l ( P 4) = (4 A P)
*
"
[ P A (q A r)l= K P A 4 ) A rl
[ p A (q v r)] = [ ( p A q) v ( p A r)] [ p v (q A r)] = [ ( p v q ) A ( p v r)] (PAP)EP (PvP)=P, ( p A F ) = F, ( p v F) "p,
( p A V )= p ( p v V )= V
law law law law
of commutativity of disjunction,
of associativity of disjunction, of commutativity of conjunction, of associativity of conjunction,
first distributive law, second distributive law,
laws of tautology, laws of absorption.
In these laws the far reaching analogy between propositional calculus and ordinary arithmetic is made apparent. The major differences occur in the second distributive law and in the laws of tautol-
4
I. ALGEBRA OF SETS
ogy and absorption. In particular, the laws of tautology show that in the propositional calculus with logical addition and multiplication we need use neither coefficients nor exponents.
[(p -,q ) A (q r)] (p + r) (P v 1PI = v (PA 1 P ) G F P”l-7P 1 ( P v 4 ) = (1 P A 14 ) l ( P A 4 ) = ( 1P v 1 4 ) (P 4 ) = (1 4 1P) (P-,q)‘(lPVq), F+P, P+P, P + V . --f
+
-+
+
law law law law
of the hypothetical syllogism, of excluded middle, of contradiction, of double negation,
de Morgan’s laws, law of contraposition,
Throughout this book whenever we shall write an expression using logical’ symbols, we shall tacitly state that the expression is true. Remarks either preceding or following such an expression will always refer to a proof of its validity. $2. Sets and operations on sets
The basic notion of set theory is the concept of .set. This basic concept is, in turn, a product of historical evolution. Originally the theory of sets made use of an intuitive concept of set, characteristic of the so-called “naive” set theory. At that time the word “set” had the same imprecisely defined meaning as in everyday language. Such, in particular, was the concept of set held by Cantor’), the creator of set theory. Such a view was untenable, as in certain cases the intuitive concept proved to be unreliable. In Chapter 11, $2 we shall deal with the antinomies of set theory, i.e. with the apparent contradictions which appeared at a certain stage in the development of the theory and l) Georg Cantor (1845-1918) was a German mathematician, professor at the University of Halle. He published his studies in set theory in the journal Mathematische Annalen during the years 1879-1897.
2. SETS AND OPERATIONS ON SETS
5
were due to the vagueness of intuition associated with the concept of set in certain more complicated cases. In the course of the polemic which arose over the antinomies it became apparent that different mathematicians had different concept of sets. As a result it became impossible to base set theory on intuition. In the present book we shall present set theory as an axiomatic system. In geometry we do not examine directly the meaning of the terms “point”, “line”, “plane” or other “primitive terms”, but from a well-defined system of axioms we deduce all the theorems of geometry without resorting to the intuitive meaning of the primitive terms. Similarly, we shall base set theory on a system of axioms from which we shall obtain theorems by deduction. Although the axioms have their source in the intuitive concept of sets, the use of the axiomatic method ensures that the intuitive content of the word “set” plays no part in proofs of theorems or in definitions of set theoretical concepts. Sometimes we shall illustrate set theory with examples furnished by other branches of mathematics. This illustrative material involving axioms not belonging to the axiom system of set theory will be distinguished by the sign #= placed at the beginning and at the end of the text. The primitive notions of set theory are “set” and the relation “to be an element of”. Instead of x is a set we shall write Z ( x ) , and instead of x is an element of y we shall write x E y l). The negation of the formula x ~y will be written as x non EY, or x 4 y or 7(x E Y ) .To simplify the notation we shall use capital letters to denote sets; thus if a formula involves a capital letter, say A, then it is tacitly assumed that Ais a set. Later on we shall introduce still one primitive notion: x T R y (x is the relational type o f y ) . We shall discuss it in Chapter 11. For the present we assume four axioms:
1. AXIOMOF EXTENSIONALITY: r f the sets A and B have the same elements then they are identical. l) The sign E, introduced by G. Peano, is an abbreviation of the Greek word dart (to be).
6
I. ALGEBRA OP SETS
Al). AXIOMOF UNION: For any sets A and B there exists a set which contains all the elements of A and all the elements of B and which does not contain any other elements. B I). AXIOMOF DIFFERENCE: For any sets A and B there exists a set which contains only those elements of A which are not elements of B. Cl). AXIOMOF EXISTENCE: There exists at least one set. The axiom of extensionality can be rewritten in the following form: $, for every x, x E A t x E By then A = B, where the equality sign between the two symbols indicates that they denote the same object. It follows from axioms I and A that for any sets A and B there exists exactly one set satisfying the conditions of axiom A. In fact, if there were two such sets C, and C,, then they would contain the same elements (namely those which belong either to A or to B ) and, by axiom I, C1 = C,. The unique set satisfying the conditions of axiom A is called the sum or the union of two sets A and B and is denoted by A v B. Thus for any x and for any sets A and B we have the equivalence (1)
X E Au B ~ ( x E A ) v ( x E B ) .
Similarly, from axioms I and B, it follows that for any sets A and B there exists exactly one set whose elements are all the objects belonging to A and not belonging to B. Such a set is called the difference of the sets A and B and denoted by A-l?. For any x and for arbitrary sets A and B we have (2)
X EA-B
= ( X E A ) A ( x $B).
By means of de Morgan’s law and the law of double negation (0 1, p . 4) it follows that (3)
1( X E A -
B ) f 1( X E A ) v ( X E B),
i.e. x is not an element of A -B if x is not an element of A or x is an element of B. l)
In Chapter I1 these axioms will be replaced by more general ones.
2. SETS AM) OPERATIONS ON SETS
7
Using the operations u and - we can define two other operations on sets. The intersection A n B of A and B we define by A nB
= A-(A-B).
From the definition of difference we have for any x
X E A ~ B ~ ( ~ E A ) A ~ ( ~ E A - B ) , from which, by means of (3) and the first distributive law (see p. 3), it follows that x E A n B 3 ( x E A ) h [ l ( xE A ) v ( x E B)]
= [ @ € A )A l ( X E A ) ]
v [(XE A ) A (XE B)]
EFv[(xEA)A(xEB)]E[(xEA)~(xEB)],
and finally (4)
x E A n B = ( x E A ) h ( x E B).
Hence the intersection of two sets is the common part of the factors; the elements of the intersection are those objects which belong to both factors. The symmetric difference of two sets A and B is defined as (5)
A - B = ( A - B ) u (B-A).
The elements of the set A l B are those objects which belong to A and not to B together with those objects which belong to B and not to A . Exercises 1.Definetheoperations u , n , - b y m e a n s o f : ( a ) - , n , (b)-,u,(c) -,A. 2. Show that it is not possible to define either the sum by means of the intersection and the difference, or the difference by means of the sum and the intersection.
$3. Inclusion. Empty set A set A is said to be a subset of a set B provided every element of the set A is also an element of the set B. In this case we write A c B or B 13 A and we say that A is included in B. The relation c is called the inclusion relation. The following equivalence results from this definition {for every x ( X E A+ X E B ) } = A c B. (1)
8
I. ALGEBRA OF SETS
Clearly from A = B it follows that A c B, but not conversely. If A c B and A # B we say that A is a proper subset of B. If A is a subset of B and B is a subset of A then A = B, i.e.
( A c B) A ( B c A ) + ( A = B). To prove this we notice that from the left-hand side of the implication we have for every x
XEB+XEA,
X E A + X E B and
from which we obtain the equivalence x E A = x E B, and thus A = B by axiom I. It is easy to show that, if A is a subset of B and B is a subset of C , then A is a subset of C: ( A c B) A ( B c C ) + ( A c C),
(2)
i.e . the inclusion relation is transitive. The union of two sets contains both components; the intersection of two sets is contained in each component: (3)
A c A v B,
B c A v B,
(4)
A n B c A,
A n B c B.
In fact, from p
+ ( p v q)
x EA
it follows that for every x -P
[(xE A ) v (xE B)],
from which, by (I), 92, p. 6, X E A+ x E ( A v B), and by (1) we obtain A c A v B. The proof of the second formula of (3) is similar, the proof of (4)follows from the law ( p A q ) + p . From (2), $ 2 it follows that
A - B c A. Thus the difference of two sets is contained in the minuend. The inclusion relation can be defined by means of the identity relation and one of the operations u or n. Namely, the following equivalences hold (5)
(A c B) =(A u B
= B)
= ( A n B = A).
9
3. INCLUSION. EMPTY SET
In fact, if A c B then for every x , x E A + x E B; thus by means of the law (P 4) 4 [(PV 4) 41, [(x E A ) v ( x E B)] 4 (x E B), +
which proves that A u B c B. On the other hand, B c A u B and hence AuB=B. Conversely, if A u B = By then by (3) A c B. The second part of equivalence ( 5 ) can be proved in a similar manner. It follows from axiom B that if there exists at least one set A then there also exists the set A-A, which contains no element. There exists only one such set. In fact, if there were two such set Z1and Z 2 , then (for every x ) we would have the equivalence XEZ,
=XEZ2.
This equivalence holds since both components are false. Thus, from axiom I, Z1= Z 2 . This unique set, which contains no element, is called the empty set and is denoted by 0. Thus for every x X$O,
i.e. ( ~ € 0= ) F. The implication x E 0 -+ x E A holds for every x since the antecedent of the implication is false. Thus 0 t A, i.e. the empty set is a subset of every set. Formula (l), $2, p. 6 implies that X E
( A u 0) = (XE A ) v (xEO)
=( X E A ) v F =
XEA ,
because p v F 3p . From this we infer AuO=A, and from
7F E V A-0 = A .
The identity A n B = 0 indicates that the sets A and B have no common element, or -in other words -they are disjoint.
10
I. ALGEBRA OF SETS
The equation B-A = 0 indicates that B c A . The role played by the empty set in set theory is analogous to that played by the number zero in algebra. Without the set 0 the operations of intersection and subtraction would not always be performable and the calculus of sets would be considerably complicated. 84. Laws of union, intersection, and subtraction
The operations of union, intersection, and subtraction on sets have many properties in common with operations on numbers: namely, union with addition, intersection with multiplication, and subtraction with subtraction. In this section we shall mention the most important of these properties. We shall also prove several theorems indicating the difference between the algebra of sets and arithmetic I). The Commutative laws: A u B = B u A , A n B = B n A. (1) These laws follow directly from the commutative laws for disjunction and conjunction. The associative laws: (2) A u ( B u C ) = (A u B) u C,
A n ( B n C ) = ( A nB) nC.
Again, these laws are direct consequencer of the associative laws for disjunction and conjunction. Formulas (1) allow us to permute the components of any union or intersection of a finite number of sets without changing the results. Similarly, formulas (2) allow us to group the components of such a finite union or intersection in an arbitrary manner. For example: A u (Bu [CU (Du A')]} = [ Au ( D u C)]u (Bu E ) = (Eu C )u [Bu ( Au D)].
In other words, we may eliminate parentheses when performing the operation of union (or intersection) on a finite number of sets. The distributive laws: A n ( B u C ) = ( A n B ) u ( A n C), (3) A u ( B nC ) = ( A u B) n ( A u C ) . l) The theorems given in $ 4 are due to the English mathematician G. Boole (1813-1864), whose works initiated investigations in mathematical logic.
4. LAWS OF UNION, INTERSECTION AND SUBTRACTION
11
The proofs follow from the distributive laws for conjunction over disjunction and disjunction over conjunction, given in 5 1. The first distributive law is completely analogous to the corresponding distributive law in arithmetic. Similarly, as in arithmetic, from this law it follows that in order to intersect two unions we may intersect each component of the first union with each component of the second union and take the union of those intersections: ( A u B u ... v H ) n ( X u Y u ... u T ) =(An
X ) u (A nY )u
... u ( B n T ) u
... u ( A n T ) u ( B n A') u ( B n Y ) u ... u ( H n X ) u ( H n Y ) u ... u ( H n T ) .
The second distributive law has no counterpart in arithmetic. The laws of tautology: A uA =A,
(4)
A nA =A.
The proof is immediate from the laws of tautology ( p v p ) = p and
(PAP) =P. We shall prove several laws of subtraction. A u (B-A) = A u B.
(5)
PROOF.By means of (1) and (2), 92, p. 6 we have X E [ Au (B-A)]
A ) v [(XE B ) A ~ ( X A)], E from which, by the distributive law for disjunction over conjunction
X E [ Au (B-A)]
(XE
[ ( x E A )v (XE B)] A [ ( x E A )v ~ ( x E A ) ]
= ( X E A ) v (XE B), since (x E A ) v l ( x E A )= V , and Y may be omitted as a component of a conjunction. Thus XE [Au
(B-A)]
XE (A
u B),
which proves (5). From ( 5 ) we conclude that the operation of forming difference of sets is not the inverse of the operation of forming their union. For example, if A is the set of even numbers and B the set of numbers divisible by 3 then the set A u (B-A) is different from B, for it contains all even numbers.
12
i. ALGEBRA OF SETS
On the other hand, in case A P. 8,
c By we have
by (5) and (9,0 3,
A u (B-A) -- B,
as in arithmetic. A-B=A-(AnB).
(6)
PROOF. XEA-(A n B ) = ( x e A ) ~T ( ~ E A ~ B ) = ( X E AT)[A( X E A ) A ( X E B ) ] =(XEA)A[l(XEA)V l(XEB)] ~[(xEA)A~(XEA)]V[(XEA)A~(XEB)]
.
= P V [ ( X E A ) A 1( X E B)] = [(X E A ) A 7( X E B)] ~xEA-B.
The distributive law for union over subtraction has in the algebra of sets the following form A n (B-C) = ( A n B)-C. (7) This law follows from the equivalence X EA
n(B-
c)
[(X E A ) A ( X
EB) A
1( X E c)]
E[(xEA~B)A~(~EC)] n B)-C.
From (7) it follows that A n(B-A) =( A nB ) - A
= (B nA)-A
=B
n (A-A)
= B n 0 = 0.
Thus A n (B-A)
= 0.
De Morgan’s laws for the calculus of sets take the following form (8)
A - ( B n C ) = (A-B) u ( A - C ) , A - ( B u C)= (A-B) n (A-C).
In the proofs we make use of de Morgan’s laws for the propositional calculus. The following identities are given without proof.
(9) (10) (1 1)
( A u B ) - c = ( A - C ) u (B-C), A - ( B - C ) = (A-B) u ( A n C ) , A-(B u C ) = ( A - B ) - c .
4. LAWS OF UNION, INTERSECTION, A N D SUBTRACTION
13
The following formulas illustrate the analogy between the inclusion relation and the “less than” relation in arithmetic: (12)
( A c B ) A (c c 0)4 (A u C c B u D),
(13)
( A c B ) A (C c 0)4 ( A n C c B n D),
( A C B ) A (cC 0)-b (A-D C B-c). (14) From (14) it follows as an easy consequence that
(C c 0)+ ( A - D c A-C), (15) which is the counterpart of the arithmetic theorem: x Q y 4 2 - y Q z-x. Exercises 1. Prove the formula: N ( A u B) = N(A)+ N ( B ) - N ( A n B), where N ( X ) denotes the number of elements of the set X(under the assumption that X is finite). Hint: Express N ( A - B ) in terms of N ( A ) and N ( A n B). 2. Generalize the result of Exercise 1 in the following way
N ( A , u A2 u
... u An) = c N(Ai) I
N(Ai n Aj) 1.i
+ i Jc. k N(Ai n A j n Ak)- ...,
where the indices of the summations take as values the numbers from 1 to n, and they are different from each other. 3. Applying the result of Exercise 2 show that the number of integers less than n and prime to n is given by the formula
where p , ,p 2 , ...,p r denote all different prime factors of n.
0 5. Properties of symmetric differenceI) The symmetric difference A - B was defined in Q 2, p. 7 by the formula : A-B = (A-B) u (BAA). (0) I) The properties of symmetric difference were extensively investigated by M.H. Stone. S e e his The theory of representations for Boolean Algebras, Transactions Of the American Mathematical Society N(1936) 37-111. See also F. Hausdorff, Mengenlehre, 3rd edition. Chapter X.
14
I. ALGEBRA OF SETS
The operation (1)
is commutative and associative: A-B
(2)
=B
A-(B-I-C)
IA,
= (A-r-B)'C.
Formula (1) follows directly from (0). To prove (2) we transform the left-hand and right-hand sides of (2) by means of (0): A - ( l e C ) = A - [ ( B - C ) u (C-B)] = {A-[(B-C)
u (C-B)]} u {[(B-C) u ( C - B ) ] - A } .
Using (8), (9), (lo), and (ll), $4, p. 12, we obtain A-(B-C) = {[A-((B-C)]n[A-(C-B)]} = {[(A-B)
u [(B--)--A]
u [(C-@--A]
u ( A n C)] n [ ( A - C ) u ( A n B ) ] }u [B-(C u A)]
u [C-(Bu A)] = [ ( A - B ) n ( A - C ) ] u [ ( A - B ) n B] u [(A- C ) n C ] u ( A nB n C )u[(B-(C u A)] u [C-(B u A)] = [A-(B
u C)]u [B-(C u A)] u [C-(A u B)] u ( A n B n C ) .
Thus the set A - ( B - C ) contains the elements common to all the sets A, B, and C as well as the elements belonging to exactly one of them. To transform the right-hand side of (2) it is not necessary to repeat the computation. It suffices to notice that by means of (1) (A-B)-c
= C-(-A..-l?),
from which (substituting in the formula for A - ( B - L C ) the letters C, A , B for A , B, C respectively) we obtain (ALB)-C = [C-(AuB)]u[A-(BuC)]u[B-(CuA)]u(CnAnB)
= [A-(B
u C)]u [B-(Cu A)]u [C-(A u B)] u ( A n B n C ) .
Thus the associativity of the operation has been proved. It follows from (1) and (2) that we may eliminate parentheses when performing the operation on a finite number of sets.
15
5. PROPERTIES OF SYMMETRIC DIFFERENCE
The operation of intersection is distributive over
A,
that is
A n (BAG') = ( A n B)-(A n C ) . (3) In fact, it follows from (6) and (7), $4,p. 12 that A n ( B I - C ) = A n [(B-C) u (C-B)] = [(A n B ) - C ] u
[ ( An C ) - B ]
= [B n (A-C)]
u [C n (A-B)]
= ( B n [A-(A n C ) ] }u { C n [ A - ( A n B)]} = [(A n B)-(A n C)] u [(A n C ) - ( A n B)] = ( A n R)-I-(An C ) .
The empty set behaves as a zero element for the operation ,:
that is
(4) AAO = A . In fact, (A-0) u (0-A) = A u 0 = A . The theorems which we have proved so far do not indicate any essential difference between the operations and u. However, a difference can be seen in the following theorems. A - A = 0. In fact, A A A = (A-A) u (A-A) = 0. The operation of union has no inverse operation. In particular, we have seen that the operation of subtraction is not an inverse of the union operation. However, there does exist operation inverse to the operation :: for any sets A and C there exists exactly one set B such that A A B = C, namely B = A - C . In other words: (5)
(6)
(7)
A - ( A - C ) = c, A-B= C + B = A - C .
In fact, (2), (4) and (5) imply A - ( A I - C ) = ( A A A ) A C = OLC = c-0 = c,
which proves (6). If A - B = C then AA(A'-B) = A - C and hence B = A'C by means of (6). Thus (6) and (7) indicate that the operation A does have an inverse: the operation A itself. In algebra and number theory we investigate systems of objects usually called numbers with two operations and (called addition and
+
-
16
I. ALGEBRA OF SETS
multiplication). These operations are always performable on those objects and satisfy the following conditions: X+Y = y+x, 0) (ii) x+dv+z) = ( X + Y ) + Z , there exists a number 0 such that x+O = x , (iii) (iv) for arbitrary x and y there exists exactly one number z = x-y (the diflerence) such that y+z = x , (9 x * y =y . x , ( 4 xe(y.2) = (x.y).z,
(vii)
X.(Y+Z)
= (x.y)+(x.z).
Such systems are called rings (more exactly: commutative rings). If there exists a number 1 such that for every x (viii) x . 1=x , then we say that the ring has a unit element. The algebraic computations in rings are performed exactly as in arithmetic. For, in proving arithmetic properties involving addition, subtraction and multiplication, we make use only of the fact that numbers form a commutative ring with unit. Formulas (1)-(7) show that sets form a ring (without unit) if by “addition” we understand the operation A and by “multiplication” the operation n.A peculiarity of this ring is that the operation “subtraction” coincides with the operation “addition” and, moreover, “square” of every element is equal to that element. Using A and n as the basic operations, calculations in the algebra of sets are performed as in ordinary arithmetic. Moreover, we may omit all exponents and reduce all coefficients modulo 2 (i.e., 2kA = 0 and ( 2 k f l ) A =A). This result is significant because the operations u and - can be expressed in terms of A and n. Owing to this fact the entire algebra of sets treated above may be represented as the arithmetic of the ring of sets. In fact, it can easily be verified that: (8)
A u B = A L B - ( A n B),
(9)
A--B = A - ( A n B ) .
17
5. PROPERTIES OF SYMMETRIC DIFFERENCE
Formulas (8) and (4) imply the following theorem: (10)
if A and B are disjoint, then A u B = A-B.
The role which symmetric difference plays in applications is illustrated by the following example. Let X be a set and Z a non-empty family of subsets of X,that is, Z is a set whose elements are subsets of .'A Suppose that (Y
(1 1)
Cz)A(ZEz)
3
(YEI),
(YEZ)A(ZEI) + ( Y U Z E z ) .
A family of sets satisfying these conditions is called an ideal. We say that two subsets A , B of X are congruent moduIo I if A - B E Z and we denote this fact by A&B(mod I ) or by A s B if the ideal Z is fixed. Since 0 EZ, it follows from (5) that A IA, i.e. the relation A is reflexive. (1) implies that ( A & B) + (BG A), i.e. the relation =& is symmetric. Finally, the identity A IB = ( A A C )2 (B- C ) implies that A - B c (A'C) u (B-C), because the symmetric difference of two sets is contained in their union. By means of (1 1) we infer that (A&B)h(B&C)
--f
(A-C),
i.e. the relation & is transitive. Replacing the sign = by the sign in the previous definitions we obtain new notions. For example, two sets A and B are said to be disjoint modulo Z provided A n B i O (see p. 9); we say that A is included in B moduIo Z if A - B & 0, etc. Exercises 1. Show that the set A 1 - A 2 z ... -A,, contains those and only those elements which belong to an odd number of sets Ai (i = 1,2,..., n). 2. For A finite let N ( A ) denote the number of elements of A . Prove that if the sets A l , A z , ..., A,, are finite then
N ( A 1 L A 2 L... L A , , ) N ( A i ) - 2 x N(A8 n Aj)+4
= i
i. j
N ( A i n Aj n Ak) i.j.k
N ( A i n Aj n
-8 I, j . k . 1
n Al)+
...
18
I. ALGEBRA OF SETS
3. Show that (A, v A2 u
v An)-(Bi u B2 v
v Bn) c (Ai-Bi) u
(A, n A2 n ... n An)A(Bi n BZn ... n Bn) c ( A 1 - B I ) v
u( A ~ L B ~ ) ,
... u (An'Bn)
(Hausdorf). 4. Show that for any ideal Z the condition A t B implies
A u C 2 B v C,
A n C = B n C,
A -C 1B - C,
C - A = C - B.
5. For any real number t denote by [t] the largest integer < t. Let A t be the set of rational numbers of the form [nt]/n, n = 1,2, ... Prove that if Z is the ideal composed of all finite subsets of the set of rational numbers, then 1( A , G A, (mod Z)) and A, is disjoint (modulo I ) from A , for all irrational numbers x, y > 0, x # y.
.
$ 6 . The set 1, complement
In many applications of set theory we consider only sets contained in a given fixed set. For instance, in geometry we deal with sets of points in a given space, and in arithmetic with sets of numbers. In this section A , By... will denote sets contained in a certain fixed set which will be referred to either as the space or the universe and will be denoted by 1. Thus for every A A c 1, from which it follows that
(1)
Anl=A,
The set 1-A or -A:
Aul=&
1
is called the complement of A and is denoted by Ac -A=A"l-A.
Clearly, (2)
An--A=O,
Au-A=1.
Since - - A = 1-(l-A), we obtain by (lo), $4yp. 12 the following law of double complementation (3)
--A
=A.
Setting A = 1 in de Morgan's laws ((8), $4, p. 12) and substituting A and B for B and C,we obtain (4)
- ( A n B ) = - A u -By
- ( A u B ) = - A n -B.
19
6. THE SET 1, COMPLEMENT
Thus the complement of the intersection of two sets is equal to the union of their complements and the complement of the union of two sets is equal to the intersection of their complements. It is worth noting that the formulas which we obtained by introducing the notion of complementation are analogous to those of propositional calculus discussed in 0 1. To obtain the laws of propositional calculus (see p. 2-4) it sufficesto substitute in (1)-(4) the equivalence sign for the sign of identity and to interpret the letters A , B, ... as propositional variables and the symbols u , n , -,0.1 as disjunction, conjunction, negation, the false sentence and the true sentence, respectively. Conversely, theorems of the algebra of sets can be obtained from the corresponding laws of the propositional calculus simply by changing the meaning of symbols. From this point of view calculations on sets contained in a fixed set 1 can be simplified by using the operations u , n , Subtraction can be defined by means of the operation - and one of the operations u or n . In fact, we have A - B = A n(1-B) = A n -B and A - B = A n - B = - ( - A u B). The inclusion relation between two sets can be expressed by the identity ( A c B ) = ( A n - B = 0). (5) For assuming A c B and multiplying both sides of the inclusion by - B we obtain A n - B c B n - B and since B n - B = 0, we have A n - B = 0. Conversely, if A n - B = 0, then A = A n 1 = A n ( B u -B)
-.
n B ) u ( A n - B ) = ( A n B ) u 0 = A n B c B. Since ( A = B ) = ( A c B ) A ( B c A), it follows from (5) that ( A = B ) = ( A n - B = 0) A ( B n - A = 0), and, since the condition ( X = 0) A ( Y = 0) is equivalent to X u Y = 0, (6) ( A = B ) = [ ( A n - B ) u ( B n -A) = 01 = ( A L B = 0). It follows directly from ( 5 ) that ( A c B ) zz ( - B c -A). (7) (compare with the law of contraposition p. 4). = (A
20
I. ALGEBRA OF SETS
The system of all sets contained in 1 forms a ring where the operation & is understood as addition and n as multiplication. This ring differs from the ring of sets considered in 0 5 in that it has a unit element. The unit is namely the set 1. In fact, formula (1) states that the set 1 satisfies condition (viii), $ 5 , p. 16 characterizing the unit element of a ring. Hence calculations in the algebra of sets are formally like those in the algebra of numbers. Exercise The quotient of two sets is defined as follows A : B = A u -B. Find formulas for A : ( B u C) and for A : ( B n C) (counterpart of de Morgan's laws). Compute A n (B:c).
5 7.
Constituents
In this section we shall consider sets which can be obtained from arbitrary n sets by applying the operations of union, intersection, and difference. We shall show that the total number of such sets is finite and that they can be represented in a certain definite form (normal form). Let A l , A2, ..., A , be arbitrary subsets of the space 1. Throughout this section these subsets will remain fixed. Let A:=I-Al, for i = 1 , 2 , . . . , n . Each set of the form A?nA$n
... n A >
(i,=Oori,=l
f o r k = 1 , 2 ,... , n )
will be called a constituent. The total number of distinct constituents is at most 2", because each of the superscripts ik may have either one of the values 0 and 1. The number of constituents may be less than 2"; for instance, if n = 2 and Al = l-Az, then there are only three constituents: Az = A: n A:. 0 = A: n A: = A: n A:, Al = A: n A:, Distinct constituents are always disjoint. In fact, if and Sz = A i l n A$ n ... n A'," S, = A:' n A p n ... n A$
21
7. CONSTITUENTS
and if for at least one k < n, ik # jk, for instance ik = 0 and j , = 1, then A$ n Aj$ = 0. Hence S1n s2= 0. The union of all constituents is the space 1. It suffices to notice that 1 = (A! u A ; ) n (A! u A:) n ... n (A: u A:). By applying the distributive law of intersection with respect to union on the right-hand side of the equation we obtain the union of all the constituents. The set Ai is a union of all constituents which contain the component A!. If SI, S,, ... sh are all constituents, then 1 = sl u s, u ... s h . Therefore Ai = (Ain S,) u ( A i n S,) u ... u (Ain &). If S , contains the component A:, then Ai n S, = 0 because A in A : = A in (l-Ai) = 0. On the other hand, if S, contains the component A:, then Ai n S, = S,. Thus A iis the union of those constituents which contain the component A:. Q.E.D. We shall now prove the following THEOREM 1: Each non-empty set obtained from the sets Al ,A 2 , ...,A , by applying the operations of union, intersection and subtraction is the union of a certain number of constituents. PROOF.The theorem is true for the sets A l , A2, ... ,A , . It suffices to show that if X and Y are unions of a certain number of constituents then the sets X u Y, X n Y, X-Y can also be represented as the union of constituents (provided X u Y, X n Y, X-Y are non-empty). Assume that X and Y can be represented as unions of constituents:
x= sl u s2 u
u sk,
-
Y = s1 v
-
._
s 2
u
u sl.
It follows that
xu Y = (slu
..a
u sk) u & ( ?!I
u
u
s).
Thus X u Y is a union of constituents. From the distributive law for intersection with respect to union, it follows that XnY=(SlnS,)u(SlnS,)u
... u ( ~ , n S ) ... u u( S n ~ Sj>u
... u ( s k n S,).
I. ALGEBRA OF SETS
22
si n sj = 0 if Si # gj; otherwise Si n S j = Si.Thus X n Y is a union of constituents
X nY
u Si, u
= Si,
... u S i p ,
or else is empty. If among the constituents Sil,Si2, ...,Sip occur all of the constituents SlyS,,
X-Y
..., s k y then
= X-(X n Y)c
u
(s1
... u Sk)-(sl
u
... u s k )
= 0.
Otherwise, let S j l , S j , , ..., S j , be those constituents among S , ,S2, ... , S k which do not occur among the constituents Si,,Si,, ... ,S i p . We have
X-Y
Y) = [(Si, u ... u sip, u ( S j , u ... u Sj*)]-(Si, u ... u Sip) = ( S j , u ... u Sjq)-(Sil u ... u sip, = ( S j , u ... u S j , ) - [ ( S j , u ... u Sj,) n (Si, u ... u Sip,] = sj, u ... u sj,, = X-(Xn
because
(Sjl u
... u Sj,> n (Si, u ... u SiJ = 0.
Thus X u Y,X n Y and X-Y are representable as unions of constituents. Q. E. D. THEOREM 2: From n sets by applying the operations of union, intersection, and subtraction at most 2'" sets can be constructed. In fact, each such set, with the exception of the empty set, is a union of constituents. Because the number of constituents cannot be greater than 2", the number of distinct unions constructed from some (non-zero) number of constituents cannot be greater than 2'"- 1. Of particular importance is the case where all of the constituents are different from 0. In this case, we say that the sets Al , ..., A , are independent ') . l) The notion of independent sets plays an important role in problems connected with the foundations of probability theory. See E. Marczewski, Independence d'ensembles et prolongement de mesures, Colloquium Mathematicum l(1948) 122-132.
7. CONSTITUENTS
23
THEOREM 3 : I f the sets Al , ...,A , are independent, then the number of distinct constituents equals 2". PROOF.If
S =A:' n ... n A? =A(' n ... n A$ and not all of the equations il =j l , ...,in = j , hold, then S = 0. In fact, if for example, ip = 1 and j p = 0, then intersecting both sides of the last equation in (0) with A; we obtain S = 0. Thus if the sets Al , ...,A , are independent then equation (0) holds if and only if il =j l , ... , in =j,. Q. E. D.
(0)
Example. Let the set D, consist of sequences (zl, ... ,z,) such that each zf equals either 0 or 1 but z, = 0. The sets D 1, ... ,D, are independent. In fact, D k consists of those sequences (zl, ...,z,) for which z, = i,. Thus (il , ...,in) E 05 n ... n 0 : . We shall apply the concept of constituents to a discussion of the following problem of elimination. We introduce the abbreviations To(A) = { A contains at least n elements}, Ti@) 3 { A contains exactly n elements}. Let i, , ...,in, jl, ...,j, be sequences of the numbers 0 and 1. Let P I , ... ,Pn , 41, ...,q, be sequences of non-negative integers. We are interested in finding necessary and sufficientconditions for the existence of a set X satisfying the conjunction of the following conditions: Pk(X n Al), rk(X n A J , ... , P k ( X n An), (9 I'~;(--x nA ~ )T , ~ ; ( - xn A ~ ) ,... , T&-x n A,).
We assume at first that n = 1. Writing i, j , p , q, A instead of il, j 1 , PI
,q1,Al , we obtain the solution:
(ii) [(i =j = 1) A I'jtq(A)] v T:,.&A). In fact, if there exists a set X satisfying (i) and i = j = 1; then A is the union of two sets containing respectivelyp and q elements, and in this case A contains exactly p+q elements. If i = 0 v j = 0 then A is the union of two sets, one of which contains at least p elements and the other at least q elements. Therefore A contains at least p+q elements. Conversely, if condition (ii) is satisfied, then it suffices to choose as X any subset of A containing p elements.
24
I. ALGEBRA OF SETS
Assume that n > 1 and Al , ...,A,, are pairwise disjoint. If there exists a set X satisfying (i), then writing X , = X , s = 1,2, ...,n, we conclude that (iii)
T ~ (n x A,) , A T:;(-x,n A,)
s = I, 2,
for
..., n,
and by virtue of (ii)
[(is =is = 1) A r~,+q,(As)lv ~&+q,(As)y s = 1,2, ...,n. (iv) Conversely, if (iv) holds then for every s (1 s < n) there exists a set X , satisfying (iii). Let
l , 'V P ( x ) v Y(X)l = [V @@) v v 9wl. x
(2)
[@(XI
X
(3)
X
X
x
X
Thus the universal quantifier is distributive over conjunction and the existential quantifier is distributive over disjunction.
(5)
a
v X
[@
(x) A y(x)1
-+
[v
@(XI A
X
v Y(x)lX
By means of simple counter-examples we conclude that the converse implications are not, in general, true. The universal quantifier is not distributive over disjunction and the existential quantifier is not distributive over conjunction. (6)
-IA@(x)l = v Cl@(X)l,
(7)
l[\l@(x>l = A [ l @ W l .
Laws (6)and (7) are called de Morgan's laws. It follows from them that: (6')
1A 1 1@(XI1 =
v @W,
(7')
Hence the existential quantifier can be defined in terms of the universal quantifier (and conversely, the universal quantifier in terms of the existential).
48
II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
The first equivalence above shows that the existential sentence
v@(x) X
y@i(x).
can beproved by deriving a contradiction from the assumption X
Such proofs of existential sentences are often used but, in general, are not effective,i.e. they do not give any method of constructing the object satisfying the propositipal function @(x). Sentences can be considered as propositional functions without arguments. Clearly, we have APGP, VP'P.
Moreover, it is easy to check that
A [P v @Wl=[ P v A @(xj1,
(10) (11)
-
v [ P A@(x)l= [P v @Wl. A
X
By means of the law of the propositional calculus [ p + @@)I p v @(x)] it follows easily from (10) that
= [l
A @Wl= A [P
[P
(12) Observe that
[(A@(.I>
(13)
+PI
3
v
+
@Px)l*
[@i(x)
+
PI-
In fact, it follows from (6) that the left-hand side of (13) is equivalent to the disjunction \ l l @ ( x ) vp, and then by (9), to the disjunc-
v p . By (3) we obtain v[l@(x) v p ] or V [ @ ( x ) + p l . X
tion
vl@(x) v X
X
X
X
Propositional functions may involve several variables. For instance: x
>y,
x
E
x,
x2+y2+z2
# 0.
Propositional functions containing two variables are denoted by
@(x,r>,~ ( x , y )*.., .
Laws (1)-( 13) are also valid for propositional functions of several variables. Instead of p in (7)-(13) we can have any function which does not contain the variable x .
49
1. PROPOSITIONAL FUNCTIONS. Q U -
Note the following laws concerning propositional functions of two variables: (14)
A A @(X,Y> = A A @(XtY>,
(15 )
v v @&Y)
X
Y
X
Y
=-
Y
X
Y
X
v v @(XtY).
Thus it makes no difference in which order we write universal or existential quantifiers when they occur together. We usually write instead of A and instead of
A
v
A
XY
X
Y
v v.
XY
X
Y
= A [@(x)
v Yv(Y)l
XY
[V @(XI
(17)
A
v W)l= v v =v X
[@(XI A
YY(Y)l
Y
[@(XI
A
Y(Y)l.
XY
To prove (16) we substitute in (10)
A
Y ( y ) for p and observe that
Y
/\
YQv@(x)
Y
= A [!P(y)v@(x)J by means of the same law (10). Y
The proof of (17) is similar.
PROOF. Using (1) twice we obtain
A @(x,
4
y)
-
@(x, y) and @(x, y )
v @[x, y). These implications hold for any x,y, hence Y
X
By means of (12) and (13) we obtain (18). # As an application of (18) we shall discuss the difference between uniform and ordinary convergence of a sequence of functions. By the definition of limit, the sentence [limfn(x) =f(x)] is equivalent to the
A x
following
n=m
A A v A If,-!-k(x)-f(x)l
s>O
x
k
n
and
(ii) {c> = {a,bl,
or
(iii) {c, d } = {a}
or
(iv) {c, d } = {a,b}.
Formula (ii) holds if a = c = b. Formulas (iii) and (iv) are then equivalent and it follows that c = d = a. Hence we obtain a = c = b = d in which case the theorem holds. Similarly, one can check that the theorem holds for case (iii). It remains to be shown that the theorem holds for cases (i) and (iv). We have then c = a and either c = b or d = b. If c = b then (ii) holds and this case has already been considered. If d = b then a = c and b = d, which proves the theorem. COROLLARY: If (a, b ) = (b, a ) then a = b. By the definition of the set { ~ E A@: ( x ) } axiom VIL implies the following theorem: THEOREM 5: (2)
?€{.YEA:
@(X)}
[@((t)
A
(t € 4 1 .
In particular, if @(x) 3 ( x E A ) (i.e. if the domain of the propositional function @ is limited to A), then t € { x : Q,(x)} = @(t).
(3)
Equivalence (3) leads easily to the following theorems (with the assumption that the domains of Q, and Y are limited to A ) : (4)
{x: Q,(x) v Y(x)} = {x: Q,(x)}u{x: Y(x)},
(5)
{ x : @(x)A Y(x)}= {x: @(x)}n{x: Y(x)},
(6)
{x:
1@(x)} = A + :
@(x)}.
As an example we shall prove (4). For this purpose we apply equivalence (3) to the propositional function @(x)vY(x) and we obtain
(0
t € { x : @ ( x ) v Y ( x ) } 3 [Q,(t) v Y(t)l. According to (3) we have @ ( t ) = t ~ { x : @(x)}
and
Y ( t ) = t ~ { x :Y ( x ) } ,
3. SIMPLE CONSEQUENCES OF THE AXIOMS
61
thus it follows from (i) that t E { x : @(x) v Y ( X ) ) )= t E { x : @ ( x ) } v t E { x : u'(x)} = t E { X : @ ( x ) } u { x : !P(x)},
which proves (4). THEOREM 6: For every non-empty famfly of sets A- there exists a unique set containing just those elements which are common to all the sets of the family A . This set is called the intersection of the sets belonging to the family X. A and is denoted by P ( A ) or
n
XEA
For we have: P(A) = { X E S(A) i ,A, [ ( X EA ) -+ ( X EX)]}. X
If the family A is composed of sets X,,X,, ...,X,,(n finite), then P(A) = X In&n ...nX,,. In case A = 0, the operation P ( A ) is not performable. We conclude this section with a remark on so-called antinomies of set theory. A naive intuition of set would incline us to accept an axiom (stronger than axiom VIL) stating that for any propositional function there exists a set B containing those and only those elements which satisfy this function. The creator of set theory, Cantor, believed (at least at the beginning of his work) that such an axiom was true'). However, it became very early apparent that the axiom formulated in this way leads to a contradiction (to an antinomy). Let us take as an example the propositional function @ ( x ) = (x is a set) A ( x $ x), which leads to Russell's antinomy'). We shall prove THEOREM 7: There exists no set Z such that
A [(X€Z) =@(x)]. X
See G. Cantor, uber unendliche. lineare Punktmannigfaltigkeiten, Mathematische Annalen 20 (1882) 113-121. 2) This antinomy was first published in the appendix to the book of G. Frege, Grundgesetze der Arithrnetik. vol. 2 (Jena 1903). 1)
62
11.
AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
PROOF.If such a set existed, then the equivalence ( x E Z ) = ( X is a set)A(x$x) would hold. From the assumption that 2 is a set we obtain the contradiction 2 E Z = l ( 2 E 2). Interdependence of the axioms. We have shown that axiom A (p. 6 ) follows from Z. Axiom B (p. 6 ) follows from C as well, because A - B = { x E A : l ( x B)}. ~ Axiom C (p. 6 ) follows directly from axiom 11 (the axiom of empty set) or from axiom V (the axiom of infinity). Axiom 11' follows from the other axioms of the system Z. In fact, let A be a family of sets such that O E A and such that there exists at least one non-empty set belonging to A . This family exists by the axiom of infinity. The set {a, b} is the set {@}"A where @ is the propositional function
"
E(x = 0) A (Y = a>l E(x # 0) A (v = bll. Axiom VIL (of subsets) is also a consequence of the other axioms of the system 2. In fact, let A be a set and @(x) a propositional func[(x E A ) -- l @ ( x ) ] , then the empty set satisfies the axiom tion. If
A X
of subsets. Otherwise, let a be an arbitrary element of A such that @ (a). Denote by Y(x, y ) the propositional function [@ ( x ) A (y = x)] v [l@ (x) A (x = a)]. For every x there exists exactly one y such that Y(x, y ) ; namely this element y is x or a, depending on whether @(x) or l @ ( x ) . The set ( Y } " A clearly satisfies the thesis of axiom VI;. Exercises
Show that 1. If X EA , then P(A\ c X c S(A). 2. S(Al u A2) = S(A1) u S(A*). 3. If A l n Az 7t 0. then P ( A l n A*) r> P ( A l ) n P(A&
0 4. Cartesian products. Relations The Cartesian product of two sets X and Y is defined to be the set of all ordered pairs (x, y) such that x E X and y E Y. The existence of this set can be proved as follows. If x EX and y E Y ,
4. CARTESIAN PRODUCTS. RELATIONS
63
then {x, y } c X u Y and { x } c X u Y , whence The set
{t ET
v v (t
xaX YEY
= (x, y))),
where
T = 22x”y~
exists by means of axioms IVY VI‘ and Theorem 2, p. 59. This set contains every ordered pair (x, y ) , where x E X,y E Y , and contains no other elements. Hence this set is the Cartesian product of X and Y. Since there exists at most one set containing exactly the pairs (x, y } , x E X , y e Y , the Cartesian product is uniquely determined by X and Y . This product is denoted by X X Y . If X = 0 or Y = 0 then obviously X X Y = 0. In spite of the arbitrary nature of X and Y,their Cartesian product can be treated in geometrical terms: the elements of the set X x Y are called points, the sets X and Y the coordinate axes. I f z = ( x , y } then x is called the abscissa and y the ord.’nate of z. The fact that the set of points in a plane can be treated as the Cartesian product & x 6 where € is the set of real numbers justifies the use of this terminology. Certain properties of Cartesian products are similar to the properties of multiplication of numbers. Fo; instance, the distributive laws hold:
( x , u x 2 ) x Y =X , X Y U X , X Y , Y X ( X 1 u X,)= Y X X , u Y x X , , (X,-X,) x Y = X , x Y-x, x Y , YX(X,-X,)= rxx,-rxx,. As an example we shall prove the first of these equations: (X,J’) E
(xiU X z ) XY
U X z )A (J’EY) =(XEXiVXEXz)/\(J’EY) = ( x €XIA y E Y ) v ( x EX, A y E Y) (X€X,
= (<X,Y>
x Y ) v ( ( A r>EX2 x Y) 3 (x, y ) E (X, x Y u X,X Y ) . EX1
The Cartesian product is distributive over intersection:
(X,n X , ) x Y
= ( X l x Y )n( X z x Y ) , Y x ( X , n X , ) = ( Y x X , ) n (YxX,).
64
H. AXIOMS OF SET THEORY. RELATIONS. FUNcIlONs
The proof is similar to the previous one. The Cartesian product is monotone with respect to the inclusion relation, that is, (*) If Y # 0, then ( X , c X2) = ( X , X Y c X Z X Y ) = (Y XX,c Y xXz). In fact, let y E Y . Suppose that XI c 1,. Since (for i = 1,2) ( ( x , Y ) E X I X YE) ( x E X ~ ) A ( Y E Y ) , we have the following implication ((X,Y) EX1 x Y ) ((XY u> EX2 x y>; hence X , x Y c XzX Y . Conversely, if X , x Y c X2x Y and y E Y , then (XEX,)+ ( X E X i ) h ( y E Y ) ~ ( ( X , y ) E X ~ X Y ) + ( ( X , Y ) E X z X Y ) ( X EXz) A (J’E y) + ( X EXz), thus Xi cXZ. The proof of the second part of (*) is similar. Using cartesian products we can perform certain logical transformations. For instance, the formulas (p. 49) +
A A @(XY X
Y
X
Y
Y ) =A XY
@iXY Y )
=A @
v v @ ( x ,Y ) = v P ( x , Y ) = v
@(z)
XY
allow us to replace two consecutive universal or existential quantsers by one quantifier binding the variable z = < x , y ) which runs over the Cartesian product X x Y. A subset R of a Cartesian product X X Y is called a (binary) relation. Instead of writing (a, b ) E R, where R denotes a relation, we sometimes write aRb and read: a is in the relation R to by or the relation R holds between a and b. The left domain (DJ(or simply the domain) of a relation R is defined to be the set of all x such that <x,y) E R for some y ; the right domain (0,)-the set of all y such that (x, y ) E R for some x. The right domain of a relation is sometimes called the range, or the counter-domain, or the converse domain. The union F(R)of the left and right domains of R is called the field of R. In geometrical terminology we say that D, is the projection of R on the X axis and D, is the projection of R on the Y axis.
4. CARTFSJAN PRODUCTS. RELATIONS
65
Thus we have
These formulas prove the existence of the sets Dland D,. If the arguments of the propositional function @(x, y ) are limited to the sets X and Y respectively, then the set R = { ( x , y ) : @(x, y ) ] is a relation. Clearly, @(x, y ) = x R y = ( x . y ) E R. Hence it follows from formula (1) that: THEOREM: The projecrion of the set { ( x , y ) : @ ( x ,y ) } on the X-axis is the set ( x :
V@(x,y ) } . Y
The relation ((x, y ) : y R x } is called the inverse of R and is denoted by RQ.Obviously D,(Rc) = D,(R) and D,(Rc) = D,(R). The relation ( ( x , y ) : (xSz A z R y ) } is called the composition of R
v 2
and S and is denoted by R o 5''). Obviously, D,(R o 5') c D, (5') and D r ( R o S) c D,(R). The operation o is associative. In fact, X ( Ro S ) o TY =
V (XTZ
A
Z R o SY)
2
= v/ / ( x T z A z S t A t R y ) z
t
=VV(xTzhzStAtRY) I
E
Z
v [v( x T z ~ z S t ) h t R y ] f
z
=V(xSoTtr\tRy) t
=xRo(SoT)y. Because of the associativity of o we may omit parantheses in expressions of the form R o S o ... oU. We shall prove the formula (ROS)'
=
ScORc,
I) We use this notation instead of the more natural S o R, because composing transformations we normally write on the second place the symbol of the operation which is carried out first (for example, sinoogx)).
66
II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
In fact, ~ ( R S)cy o = y R o SX E
V ( Y S ZA Z R X ) 2
=
(xRCz0 z Soy) z
=xScoRcy. Other properties of the operations and
o are given in the exercises.
Examples and exercises # 1. Let X = Y = C (the set of real numbers). The set {<x, y>: x < y) is that part of the plane which lies above the straight line x = y. The set { <x, y>: y = x') is a parabola, its projection on the Y axis is the set { y : (y = x2)}. #
v X
a family of subsets of X x Y. Let F(Z) denote the projection of the set Z (where Zc X x Y) on the X axis and F(A) the family of all projections F(2). Z EA. Prove that 2. Let A be
F[S('4)1 = SlF(PIl# i.e. the projection of a union is equal to the union of the projections. 3. Give an example showing that the projection of an intersection may be different from the intersection of the projections. 4. Prove the formulas (Ru S)c = Rc u Sc, (Rn S)c = Rc n Sc and ( R c ) c =R. 5. Prove the formulas ( R u s)0 T = ( R 0 T ) u (S 0 T), T O ( R U s)= (TOR)u (T 0 S), (Rn S) o T c (RoT ) n (So T ) , T o (RnS) c ( T O R ) n ( T o S ) . 6. Provr that ( X x Y)' = Y x X. Compute ( X x Y) o ( Z x r).
0 5. Equivalence relations Equivalence relations form an important and frequently encountered class of relations. A relation R is called an equivalence relation if for all x, y , z EF(R),the following conditions are satisfied: xRx (reflexivity), xRy 3 yRx (sYmmetrY) x R y A Y R Z+ x R z (transitivity). # Examples 1. Let x , y be straight lines lying in a plane, Let x R y if and only if x is parallel to y. Then R is an equivalence relation. 9
5. EQUIVALENCE RELATIONS
67
2. Let C be a set of Cauchy sequences ( a,, a,, ...,a,,, ...) of rational numbers. The relation R which holds between two sequences if and only if lim (a,-b,,) = 0 is an equivalence relation. 3. Let X be the set of real numbers x such that 0 < x < 1. The relation R which holds between two numbers a, b EX if and only if the difference a--b is a rational number is an equivalence relation. # 4. Let X be any set, K 5 2' and let Z be an ideal (seep. 17). The relation f which holds between two sets X , Y E R if and only if X 2 Y E Z is an equivalence relation. 5. Example 4 can be generalized in the following way. Let K be an arbitrary Boolean algebra and Z any subset of K satisfying the conditions : a < b E Z - + a E Z , (aEZ)A(bEZ)+(av beZ). Then Z is an ideal of K and the relation & (Example 4) is an equivalence relation. We shall now give several theorems which describe the structure of an arbitrary equivalence relation. Let C be any set. A family A of subsets of C ( A c 2 9 is called a partition of C if 0 $ A , S ( A ) = C and the sets belonging to A are pairwise disjoint (i.e. for any X,Y E Aeither X = Y or X n Y = 0). THEOREM 1 : I f A is a partition of C, then the relation RA defined by the formula
XRAY
v
[ ( X E y) A
YEA
( y E y)]
is an equivalence relation whose field is C. The proof of this theorem is left to the reader. 2: If A and B are two diferent partitions of C, then RA THEOREM # RB. PROOF.Suppose that RA = RB; we shall show that A = B. Because of the symmetry of the assumptions it suffices to prove that A c B. So let Y EA and let y E Y . Since S ( B ) = C, there exists 2 E B such that Y E Z . If X E Y then XRAY and hence x R ~ y Because . Z is the unique element of B containing y, we have X E Z .Similarly we can show that x EZ+ x E Y,which proves that Y = Z and hence Y EB.
68
II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
THEOREM 3: For any equivalence relation R with jield C # 0 there exists a partition A of the set C such that R = RA.
PROOF.Let Because of the reflexivity of R the elements of the family A are non-empty and S ( A ) 5 C. If Y EA and 2 E A, then for some y , z E C the following formulas hold : ~ ( U E Y F U R Y )A , (uEZ=URZ). U
U
From the symmetry and transitivity of R we infer that if sets Y and 2 have a common element, then they are identical. This proves that the
family A is a partition of C. We show now that R = R i . Suppose that uRv. Denoting the set {zEC: zRu) by Y u we obtain Y , E A and V E Y , ;hence URAOand R c RA. Now suppose that uRAv; then there exist Y in A and an element y such that ( z R y = zeY) and U E Yand V E Y .Therefore uRy and
A 2
WRY;by the symmetry and transitivity of R,it follows that uRv. This proves that RA c R. Hence R = RA, Q.E.D. It follows from Theorems 1-3 that every equivalence relation with field C # 0 defines exactly one partition A of the set C and vice versa. If R = RA then sets of the family A are called equivalence classes of R. The equivalence class containing an element x is denoted by x/R, the family A itself is denoted by C/R. This family is called the quotient class of C with respect to R. # Examples For the relation R of Example 1 each equivalence class consists of all straight lines lying in the same direction (i.e. mutually parallel). For the relation R of Example 2 each equivalence class consists of all sequences of rational numbers convergent to the same real number. Cantor defined real numbers as the equivalence classes with respect to this relation. #
69
5. EQUIVALENCE RELATIONS
A set of representatives of an equivalence relation with field C is a subset of C which has exactly one element in common with each equivalence class. The existence of a set of representatives for any equivalence relation follows from the axiom of choice. Without the axiom of choice we cannot prove the existence of a set of representatives even for very simple relations. Such is the case for the relation of Example 3 I). Exercises 1. Let I = {x: 0 x < 1); for X c Z let X(r) denote the set of numbers belonging to Z and having the form x + r + n where x E X and n is an integer. Show that if Z is a set of representatives for the relation R of Example 3, then a) Z ( r ) n Z(s) = 0 for all rational numbers r, s (r # s); b) Z = Z ( r ) , where the union is over all rational numbers.
A Y = f ( x z ) - f x ,
= xzl.
71
6. FUNCTIONS
Clearly, this formula is equivalent to (1). The second part of the theorem follows from the formulas for the domain and range of an inverse relation (p. 65).
THEOREM 2: I f f E Y x and g € Z y then the relation g o f is a function Z then X Z Z ) . and g o f € Z X (in other words: if X T Y PROOF.The definition of the composition of two relations implies the equivalences:
v C W Y ) (.Ygz)l = v [ ( f (4 Y ) ( g ( Y )
xg o f z =
A
Y
=
A
=
Y
-g(f(x))
41
= 2;
it follows that
A X, 21.
za
[(xgo f 4 A
(aof2 2 )
-+
z1 = 221
and that every element of X belongs to the domain of g o f . Since the domain of this relation is included in the range of g, g of E Z X . Theorem 2 implies the following formula g of ( x ) =g(f(x))
for
XEX.
THEOREM 3: r f f e y X , g € Z y and the functions f and g are oneto-one, then their composition is also one-to-one.
In fact,
g ( f W ) = s(f(.'I)
-+f(x) =f ( 4
--f
x = x'.
DEFINITION: A one-to-one function whose domain and range are the same set X is called a permutation of the set X . The simplest permutation of X is the identity permutation Ix, that is, the function defined by the formula I,@) = x for all x E X .
THEOREM 4: If f E Y X and f is a one-to-one function, t h e n P o f and f of"= Iy, where Y , is the range o f f .
= Ix
In fact, the equivalence fC(y) = x =f ( x ) = y implies f"(f ( x ) ) == x , thus f" o f = Ix. The proof of the second formula is similar.
72
11. AXIOMS OF SET THEORY. RELATIONS. F"CTI0NS
Let f E Y", g E Z ~q~,E TYand y E Tz.Hence the range of the function 9 o f is contained in T and the same holds for the function y o g . If q~ o f = y o g then we say that the following diagram XT'Y 81
ZTT
1.
is commutative. This diagram shows that starting with an element x E X we can obtain the element q~ of ( x ) = y o g ( x ) in two ways: through an element of the set Y or through an element of the set 2. An example of a commutative diagram will be given on page 79. DEFINITION: A function g is said to be an extension of a function f if f c g . We also say that f is a restriction of g. THEOREM 5 : In order that f c g it is necessary and suficient that D l ( j ) c D l ( g )andf(x) = g ( x ) for all x ~ D , ( f ) . PROOF.Necessity: Suppose that f c g . Then Dlcf)cDl(g), for the projection of a subset is a subset of the projection (see p. 66). If x € D I ( f ) and y =f ( x ) then ( x , y ) EL hence ( x , y ) Eg and y = g(x). Suficiency: Suppose that Qcf) c Dl(g) and f(x) = g(x) for all x e D 1 ( f ) .If (x, y ) Ef then y =f ( x ) = g(x), therefore ( x , y)Eg, which shows that f c g . The restriction f of g for which = A will be denoted by glA. The notion of function should be distinguished from the notion of operation. By an operation we mean a propositional function @(x,y) of two variables satisfying the following conditions:
(w)
A 'V @ (x,v), A X
Y
P(x,Y1) A @(x,uz)
+
@1
= YJl.
X.YI.Yl
These conditions state that for every x there exists exactly one object y such that @(x,y). This object might be denoted by F(x). If, however, the domain of the propositional function @(x,y ) is unlimited, then there may exist no set of all the pairs ( x , y ) such that @(x,y), i.e. there may exist no function f such that @ ( x , y )= [ y = f ( x ) ] . For instance, such a function does not exist if @(x, y ) is the propositional function x = y (see p. 61). On the other hand, the following theorem holds: THEOREM 6: If a propositional function @(x,y ) satisfies conditions (W) and A is an arbitrary set, then there exists a function f A with domain A
6. F U N ~ O N S
13
and such that for arbitrary X E A and arbitrary y LY
=fA(X)l
= @(XX,Y).
Namely, the required function f A is the set { t E A XB:
v Kt
= <X,Y>) A @ ( X , Y ) l )
XY
where B denotes the image of A under the propositional function @ (see p. 55). In particular, if a propositional function @ is of the form ...x ... = y (where on the left-hand side of the equation we have an expression written in terms of the letter x , constants, and operation symbols), then the function fA will be denoted by F [... x . .I. For example, XEA using this notation, Functions of more than one variable
Let X X Y X Z = X x ( Y x Z ) , X x Y x Z x T = X x ( Y x Z x T ) and similarly for any number of sets. If X = Y = 2 then instead of X X X X X we write X3 and similarly for X X X X X X X . Subsets of the Cartesian product of n sets are called n-ary relations. If the domain of a function f is the Cartesian product X x Y , then f is said to be a function of two variables. Similarly, if the domain o ff is the Cartesian product X X Y XZ, then we say that f is a function of three variables. Instead of f ( < x , y ) ) we write f ( x , y ) . ‘THEOREM 7: fl a function f is a one-to-one mapping of the set X x Y onto the set 2, then there exist functions a and By mapping the set 2 onto X and Y respectively, such that f ( a ( z ) , B(z)) = z for every z €2. PROOF.It suffices to take for a the set of pairs ( z , x ) satisfying the [ f ( x ,y ) = z] and for B the set of pairs ( z , y ) satisfying condition
V Y
tile condition
V [ f ( x ,y ) = z]. X
To conclude this section we give a formulation of the axiom of choice using the notion of function. ‘THEOREM 8: If A is a non-empty family of sets and 0 4 A , then there exists a function fE (S(A))” such that f ( X ) EXfor every X EA.
74
11. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
PROOF.Let h = F [{X}Xx]. For X E A we have thus h ( X ) # O XEA
and, moreover, h ( X ) n h ( Y ) = 0 for X # Y . Applying the axiom of choice to the family D,(h) we obtain a set which has exactly one element in common with each set h ( X ) , X EA . As it is easy to show, this set is the required function f. A function with the properties mentioned in Theorem 8 is called a choice function for the family A . Theorem 8 shows that from the axioms 2 it is possible to derive the existence of a choice function for an arbitrary non-empty family of sets not containing the empty set. Conversely, it can be shown that the axiom of choice follows from Theorem 8 and the axioms Z. Exercise 1. Let n
>3
and let
X = A O u... UAn-,,
Bk =Ak+lU
... U A k + n - 1 ,
c k =Ak+lU
where the indices are reduced modulo n. Let tem of functions satisfying the condition fk
( x ) =f k + i ( x )
There exists a function f~ Y k = 0, ...,n-1.
X
for
fk E Y B k ,
... U A k t n - 2 , k = 0 , ...,n - 1,
be a sys-
x Eck+i.
satisfying the equation f k = f l
Bk
for every
$7. Images and inverse images Let A and B be arbitrary sets and R a relation such that R c A x B. For X c A let R'(X) = { y : 'J (xR~)}. X€X
This set is called the image of X under the relation R. Clearly,
R': 2A + 2B. In particular, iff is a function then f ' ( X ) consists of values of the function f on the set X. We shall write f ' ( X ) = {f (x): x E X } . The same symbol will be used for operations, e.g. {(x,y):x~X}, { S ( X ) : X E A } , etc. As we know, there exists neither a function whose value for any x is the pair (x,y> nor a function whose value for any family X is the set S ( X ) . However, every such operation
75
7. IMAGES AND INVERSE IMAGES
determines a function if we limit its domain to an arbitrary given set (see Theorem 6, p. 72). Thus strictly speaking it would be necessary to replace the symbols ( x , y > , S ( X ) , etc., in the formulas {(x,y>: X E X } , { S ( X ) :X E A } by symbols for values of functions with domains Xand A , respectively. It follows from the definition of inverse relation (p. 65) that if Y c B then the image of Y under the relation R" is
R-'(Y) = { x :
v (~R'x)}
=
(x:
Y EY
V(xR~)}. Y €Y
.
This set is called the inverse image of Y under R , If R =f is a function, then f-'(Y) = { x : ( f ( 4= Y ) ) = { x : S ( X ) E Y } ,
v
Y SY
i.e. the following equivalence holds: x Ef-yY) = f ( x ) E Y . If Y reduces to the one-element set { y } , then the set J'-'(Y) is called a coset off determined b p y . Distinct cosets are always disjoint, the union of all cosets is the domain off. We shall now establish several simple properties of images and inverse images. 1: THEOREM (1)
(2)
R c A x B and X,, X , are subsets of A , then R'(X,) u R'(X2) = R'(X1 u Xz), X , c X , + R'(X,) c R'(X2), R1(X, n X,) c R1(Xl)n R1(X2).
(3) PROOF.Formula (1) follows from the equivalence Y E RYXl
u X,)
= v.(C{
EX,)
v ( x E&)l
A
(XRY)}
X
v
[(X
EX')
A
( X R Y ) ]V
X
v [(XEXz)
A
(xR3')l
X
=y E R'(X,) v y E R'(Xz) =y E R'(X,) u R1(X,). To prove (2) it suffices to notice that if ,'A c X , then X , = X , u X,. Thus by means of (1) it follows that R'(X2) = R'(X,) u R1(X2)2 R'(X1).
76
II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
Finally, formula (3) follows from the remark that X , nX2c Xi for i = 1,2, whence, by (2), R'(X1 n XJ c R'(X1) and R'(Xl n X2) c R1(X2);and, in turn, we obtain R'(Xl nX2) c R1(Xl)n R1(X2).
THEOREM 2: r f f ~ and p Y , c B, Y2c B, then
f -yy,u Yz) = f-l(Y1)uf-'(Yz), f - v 1 n Y2) =f-'(YJ n f - 1 ( Y 2 ) , f -l(Y1- Y2)=f --1(Y1)--f-1(Yz).
(4) (5)
(6) PROOF.(4) is a special case of (1). Formula ( 5 ) follows from the equivalence x Ef - l ( Y l n Y2)= f ( x ) E Y 1n Yz = (fWE Yl) A (fWE y2) ( X Ef-I(Y1))A (X Ef-'(YZ)) = ( x ~f-l(Y,) nf-'(Y2)). The proof of (6) is similar. Theorems 1 and 2 show that the operation of forming the image under an arbitrary relation is additive, but it is not multiplicative. On the other hand, the operation of obtaining the inverse image is both additive and multiplicative. THEOREM 3: I f f : A + B and i f f is a one-to-one function, then for any X , , X , c A the .following formulas hold
f1W1 nX2) =f '(X1) nf '(-&I,
f '(X1-XJ
=f YXJ
-f '(&I.
For the proof, substitutef" for f in Theorem 2.
THEOREM 4: I f f : A -+ B, Y c f ' ( A ) and X c A , then
f
l(f-l(v) = y,
f-'(f'(n) = X.
The proof of the first formula can be obtained from the equivalence YEf
yf-Yn)= v [(x E f -'m)(Y =f(41 A
=
v [ ( f (4 y) E
A
(Y =f ( x > ) ]3 0,E
n.
The proof of the second formula follows from the implication (x EX)
+
( f ( 4E f '(XI) = x E f - l ( f
-
l(X>)
77
7. IMAOES AND INVERSE IMAOES
In the formula just proved the inclusion sign cannot in general be replaced by the equality sign. For instance, iff is a function of the real variable x and f ( x ) = x2, then for X = { x : x 2 0} we have f-'(fl(X)) # X. But for one-to-one functions we ooviously have f-'(f'(X)) = X. Finally, let us note the following important THEOREM5 : If S c A x B and R c B x C , t h e n ( R o S ) ' ( X ) = R ' ( $ ( X ) ) for every set X c A. PROOF.
v (XROSY) =v v 'v \/
yLZ(RoS)'(X) =
X€X
X€X
[(XSZ) A
(zRJ91
I
[ ( X S Z ) A (ZRy)]
E
2-
xex
= v [(z E S' (x))A ( Z W ] =y E R' (S' (X)). In particular, it follows from Theorem 5 that if f:A+ B and g : B + C , then (gon'(x>=g'(f'(X)) for every set X c A . Exercises 1. Prove that f1(Xl)-f1(X2) c fl(Xl--Xd and fl(Xnf-'(Y)) = f l ( x ) n Y . 2. If g = f l A , then g - l ( Y ) = A n f - l ( Y ) . 3. A value y of a function f is said to be of order n if the set f-l((y}) consists n if all of its values are of n elements. We say that a yunction f is of order n. of order Prove that if a function f defined on a set X is of order n and A c X, then the restriction f l ( f - ' ( f ' ( A ) ) - A ) is of order n-1. 4. We are given a system of r + l disjoint sets A o , A ] , ..., A , included in X and a function of order < n defined on X (n 2 r). Let B = f ' ( A o ) n ... nfl(Ar). Prove that the restrictionfl ( A i n f - ' ( B ) ) is of order n-r. 5. Images and inverse images are used in topology, in particular to define the notion of a continuous function. Let X and Y be two topological spaces and let f: X + Y . We say that f is continuous if the inverse image of any open set in Y is an open set in X. Prove that the following conditions are necessary and sufficient for a function f to be continuous: (a) inverse images of closed sets are closed,
U P ) =f(x,Y)lR.
Hence k o f = q o k ' . Example. Let X = K be a field of sets with unit U,Zany ideal in K, R the relation A mod I (see p. 17). The set K/R is denoted by K/Z and is called a factor Boolean algebra. The functions f ( X , Y ) = X u Y , g ( X , Y ) = X n Y , and h ( m = U-X are consistent with the relation G (see Exercise 4, p. 18). The functions induced from f, g, h by the relation iwill be denoted by v , A , and -, respectively. Hence (XIR)A (YlR) = WnV l R , -(X/R) = (U-Xl/R.
(WR)v (YIR) = ( X u Y)IR,
THEOREM 2: The set K/Z is a Boolean algebra with respect to the operations v , A , -, with Q/Rand U/R as the zero and the unit element, respectively.
PROOF.It is sufficient to show that the operations v , A , - and the elements O/R and U/R satisfy axioms (i)-(v'), p. 37. For instance, we check axiom (i). Let a = X/R and b = Y / R ;then a v b = ( X u Y ) / R
80
U. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS
and b v a = ( Y u X ) / R and hence a v b = b v a . The remaining axioms can be checked similarly.
REMARK.The conditions XAO (mod I) and XEZ are equivalent. This proves that O/R = I. The factor algebras K/Z may have properties quite different from those of K. Thus the construction leading from K to K/Z allows us to build new and interesting examples of rings. Exercises 1. Generalize the example given above by taking any Boolean algebra as K and any subset of K satisfying the conditions of Example 5, p. 67, as I. 2. Let K be the field of all subsets of an infinite set U,and let Z be the ideal of all b i t e subsets of U. Show that every non-zero element of the factor ring K/I can be represented as xvy where x # I and y # I.
0
9. Order relations
DEFINITION 1: A relation R is said to be an order relation if it is reflexive, transitive, and antisymmetric. The last condition means that ( x R ~ ) A ( ~ J R x ) +=( yx) .
A relation which is only reflexive and transitive is said to be a quasiorder relation '). Instead of x R y we usually write x G Ry or x y . We also say that the field of R is ordered (or quasi-ordered) without explicitly mentioning R. It is necessary to remember, however, that an ordering is by no means an intrinsic property of the set. The same set may be ordered by different relations.
=S(pl(n, a), n, a), where g E Z A . This is a definition by induction with parameter a ranging over the set A . Schemes (a) and (b) correspond to induction “from n to n+ l”, i.e. pl (n‘) or pl(n’, a) depends upon pl(n) or pl(n, a), respectively. More generally, pl(n’) may depend upon all values y ( m ) where m n (i.e. m en’). In the case of induction with parameter pl (n‘) may depend upon all values q ( m , a), where m < n’; or even upon all values pl(m, b), where m n’ and b E A . In this way we obtain the following schemes of definitions by induction:
0;
r+;+')("+2bf').
2 ("'g'")
> ("+;+'). Likewise we can- derive a contradiction from the assumption that y > b. Therefore and we would obtain
("+!+')
the function J is one-to-one. Now we shall prove that the range Z of J is identical with N. It follows from J(0,O) = 0 and J ( 0 , l ) = 1 that 0 , l EZ. Suppose that n E Z,i.e. that n = J(x, y ) for some x and y. If y > 0 then n+1 = J ( ~ , y ) + l = If y = 0 then
n=
(i)+ x =
Assuming that x
r'"')
ri'),
> 0 we can write
+x+l = J(xf1, y - l ) ~ Z . thus
n+l =
(xi')+ l .
(xi')+1 in the form
+1 =J(l,x-1); hence n+lcZ. Finally, if x = y = O t h e n n = O and n+ 1 = 1. Hence n + l ~ Z .Theorem 1 is thus proved. THEOREM 2: There exist functions K, L mapping N onto N such that J(K(x), L(x)) = x. Moreover, these functions satisfv the inequalities (2)
K(x) < x ,
L(x) < x.
The existence of the functions K and L follows from Theorem 7, p. 73, the inequalities follow from x J ( x , y ) and y < J ( x , y ) .
(zk(elk') = rk(e*lk'))A (ek+l= e?+d
+
--f
--f
(elk'
= e*lk') A
(ek+' = e k l ) + (e = e*).
It remains to be shown that for every number n there exists a sequence e E Nk+' such that zk(e)= n. By the inductive assumption there exists a sequence f € N k + l such that ?&(f)= K(n). We let e to be the sequence whose k f 1 initial terms coincide with those off and whose last term is L(n). For this sequence e the following formula holds by definition : rk+l(e)= J ( r k ( f ) , L(n)) = J(K(n),L(n)) = n. 3. The mapping of the set of all finite sequences of natural numbers onto the set N. Let for eENk+l
% ( 4 = J(k,Z k W ) . This function is a one-to-one mapping of the set of all non-empty finite sequences of natural numbers onto N. We have the following: 4: There exists a one-to-one mapping a of the set S of all THEOREM finite sequences of natural numbers onto N which satisfies the condition a(0) = 0.
MAPPING J OF N X N O N T O N
3.-
99
TO prove this it suffices to let b(e) = l+oo(e) for non-empty sequences e and 4 0 ) = 0. 4. The mapping J' of the set N"+l X N"+l onto the set Nn+l and of the set N N x N N onto the set N N . Let n be a natural number. For e, f E N"+l, let J'(e,f) = F [J(ek,fk)] Y
k j then b k c B k c B j c L ( b j ) ,hence {bk,bj}EG. Letting Y be the set of a11 the terms of b we obtain [ Y ] , c G . The theorem is thus proved, because the set Y is infinite. Example. Let X c N and let G be the set all unordered pairs { x , y } , where x, Y E X and x and y are relatively prime. Applying Ramsey’s theorem we conclude that if X is infinite, then either X contains an infinite subset Y where every two numbers belonging to Y are relatively prime, or else X contains an infinite subset Y where no two numbers belonging to Y are relatively prime.
REMARKS 1. Ramsey’s theorem may be illustrated as follows. Let every edge of the complete graph [XI, where X is an infinite set be coloured either white or black. Then X necessarily contains an infinite subset Y such that all the edges of the graph [yl, are of the same colour. 2. Ramsey’s theorem has a counterpart, also due to Ramsey, in the finite case: For any natural number n there exists a natural number q such that, i f the set X contains at least q elements, then for every graph G c [ X ] , there exists a subset Y c X of n elements such that either [y1, c G or [Yl2 c
G’.
In other words, if G is a graph whose field consists of q elements, then either G or G’ contains a complete subgraph with field consisting of n elements. The theorems given in Exercises 1 and 2 below may also be stated to hold for finite sets. Exercises 1. If G is a graph with infinite field and G = G,uG,U ... UGk, where G i n G j = 0 for i # j , then at least one Gi contains a complete subgraph with infinite field. [Ramsey] 2. Generalize Ramsey’s theorem in the following way. Suppose that X is an infinite set. Let [XI,, be the family of all subsets of X consisting of n elements and let [XI,, = M , u M , u ... u M k where M i n M j = 0 for i # j . Then there exists an infinite subset Z c X such that [Z],cMj for some j .
110
m.
NATURAL NUMLIERS. FINITB AND INFINITE SETS
3. A graph G is said to contain a triangle provided there exist three distinct elements a, b, c such that { a , b } , {b, c } , { a , c } EG. Show that if X has at least 6 elements and G c [ X I z ,then either G or G contains a triangle. 4. Let N =
U Ak
where the sets A k are pairwise disjoint. Show that for any
k€
(x:
v Y
@(x,y)) =
=z E Fy
@(z,y) Y
= V ( Z € F y ) = Z E U F y ~ Z E U { X @: ( x , y ) } . Y
Y
Y
The proof of the second equation in (2) is similar. By (l), the formulas concerning quantifiers given in 0 1, Chapter IT, lead to the following formulas for the generalized operations: (3) (4)
1. GENERALIZED UNION AND INTERSECTION
113
(13)
In formulas (8) and (9) the symbol - denotes complementation with respect to the set X. The proofs of the formulas above follow directly from the respective formulas in $1, Chapter 11. As an example we prove de Morgan’s law (8): XE
-(n Fr) = i nFr) 1[A ( x ~ ~ r ) ] t
E
v
(XE -Ft)
t
3 X E
U(--F,); I
where we apply successively formulas p. 6, (6), p. 47, and (1) above. The diagram on page 50 also leads to formulas for the generalized operations. It suffices to replace the implication sign --r by the inclusion sign c ,and @ by a function F of two arguments having sets as values. In particular, the following important formula holds:
u nF,, n u F,,
(14)
t
s
s
t
This inclusion cannot in general be reversed (see p. 50). THEOREM 1: The union
u Ft is the unique set S satisfying the conditions: I
A (Ft = s)’
(15)
(16) The intersection
A ~[A~~tcaI+(S=X)1. X I nFr is the unique set P satisfying the conditions: A (P= Ft), A { [ A(X=ml ( X 4 . f
(15 ’ ) (16‘)
-+
X
t
114
W. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT
u Ft is the smallest set containing all the
In other words, the union
t
sets Ft and the intersection
nFt is the largest set included in each of I
the sets Ft.
PROOF.It follows from (3) and (13) that the union
uFt satisfies 1
conditions (15) and (16). Conversely, assuming that the set S satisfies Ft c S . Setting these conditions we infer from (15) and (13) that
u X = u Ft in (16) and applying (3), we obtain S u Ft. Hence S u Ft. t
c
t
t
=
t
The proof for intersection is similar.
THEOREM 2: (GENERALIZED ASSOCIATIVE
LAWS.)
If
T=
u H,
UEU
where H is a set-valued function with domain U (i.e. HE(^=)'), then
PROOF.Letting S
=
u Ft
and
S,
tET
=
u F,,
tEHU
we reduce equation (17) to the form (19)
s= lJ s,. UEU
By assumption we have S 2 Ft for every t E T , in particular, for every t E H, . Thus S =I S, by Theorem 1. On the other hand, suppose that X 3 S, for arbitrary U E U. If t E T , then there exists u E U such that t e H,, whence it follows that S, 2 F,, and thus X D Ft. Since t is arbitrary, we conclude that X 2 S. Applying Theorem 1 we obtain (19). The proof of (18) is similar.
115
1. GENERALIZED UNION AND INTERSECTION
THEOREM 3: (GENERALIZED COMMUTATIVE LAWS.) tation of the elements of a set T , then
PROOF.Let S =
u F,+,(t). If
t E T,
If q~
is a permu-
then t = cp(qJ"(t)), and because
t
S 2 Fpr(,,, for arbitrary U E T ,in particular for u = ye(t), we have S 3 F,. Conversely, if X is a set such that X I> Ft for t E T , then X I> Fpr(t),because v(t)E T. Thus X 3 S, which shows that S is the smallest set containing all the sets Ft (i.e. S = U F,). The proof of the second formula is similar. 4: (GENERALIZED DISTRIBUTIVE LAWS.)')If THEOREM M= T,, and K = YE^^: A ( Y n T , , # 0 ) } ,
u
UEU
then (20)
(21)
U d J
n u F f = Yu€ K mi, u 17 Ft = n u F,.
UEUteTu
t€Y
u t U t€TU
Y€Kt€Y
PROOF.Suppose that Y E K and U E U . By the definition of the family K we have Y n Tu# 0, thus there exists toE Y nT,,. This implies by (3) that Since this inclusion holds for any u E U (where Y is constant), we infer from Theorem 1 that n F t c uFt.
n
U E UtcTu
t EY
Since Y is arbitrary, we obtain by (3) the following inclusion: (22)
u n F t c n UF,.
YEK t € Y
UEUtETU
To prove the opposite inclusion, suppose that
UF,.
U E UtETU
Y = { t E M : aEF,}. If U E U then by (23) aE F t . Thus there exists tET, such that
u
tETu
') See: A. Tarski, Zur Grundlegung der Boole'schen Algebra I, Fundamenta Mathematicae 24 (1935) 195.
N. GENERALIZED UNION, INTERSECllON AND CARTESIAN PRODUCT
116
a € F,; hence ~ E Ywhich , proves that Y n T, # 0. By the definition of K we have Y E X . It now follows from (24) that A ( u E F ~ that ); t€Y
is, a E (7 F,. This shows that t €Y
a€ U
(25)
f7 F t .
YER t€Y
This together with (22) gives (20). To prove (211, replace Ft in (20) by S - F t , where S = F t . Then we obtain:
u n u (s-m u n teM
=
U€U t€T"
Y€Kt€Y
whence, by de Morgan's laws (8) and (9) and by - ( - F t ) = Ft, we obtain (21). We shall now generalize formulas (1)--(4), $8, Chapter 11, concerning images and inverse images of finite unions and intersections, to the case of arbitrary unions and intersections. THEOREM 5 : Let F E(231>* and let fE Y X . Then
If the function f is one-to-one, then the inclusion sign in (27) can be replaced b y the identity sign.
PROOF.It follows from the definition of image that
V [b
Y € f l ( U Ft)
E
=
v [v x
U Ft) A (Y
=
v (Y
fW)J
( ( x E Fr) A (Y =f
:
= i j [// ((. t
=
E Fr) A
(41
(u =f))]
X
Ef' (FJ) =Y E
u f' (4) >
which proves (26). Similarly, by means of (18) on p. 49, we obtain the following equivalences:
117
1. GENERALIZED UNION AND INTERSECTION
A v [(x E 4) * (Y =f(41 = A (uEflo;t)) = Y E nm), t
+
f
x
whence it follows that (27) holds. If the function f is one-to-one, then using (27) for the inverse funcwe obtain t i o n y and for the sets fl(Ft)
f-l(nmt))=
(m)) nF,,
nf-1
=
t
and by (2), p. 75, it follows
n
f1
f
n (F,)).
(Ff)= f1(
Since (27) also holds, Theorem 5 is proved.
THEOREM 6 : If G E (2Y)Tand f E Y x then
G,) = nf-'(Gt).
(29)
t
t
The proof can be obtained from the following equivalences, which are consequences of the definition of the inverse image (see p. 75).
(uGf) = f ( y ) u Gf = 'v[fO)GI = v Iv ~f-' (G,)] =y u f-'(Gt); €f-l(nG,) =m nG, = ,\ [mGtl
Y ef-'
E
E
E
f
E
E
=A k ~ f - ' ( G , ) ] r yn ~ f-'(Gt). t
t
Formulas (26) and (28) assert the additivity of the operation of forming images and inverse images. Formula (29) asserts that the operation of forming inverse images is multiplicative. The operation of forming images is multiplicative, however, only for one-to-one functions. Examples Let the set 1 be a topological space (see Chapter I, $8).
118
IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUm
1. r f F is a function whose values are closed sets (see p. 27), then the intersection P = (7 & is also a closed set. t
PROOF. Since P c F,, we have because 8 = Ft. This implies that
c
for every t ; thus Pc F,, 4 = P, hence P= P, for
Pc n t
P c 3 by Axiom (3), 6 8, Chapter I, p. 26. 2. If G is a f l c t i o n whose values are open sets, then the union
S=
u G, is an open set. f
PROOF.The sets I-Gt are closed, thus the intersection n (l-Gt) t is also closed. By de Morgan’s law (9) the set 1-S is closed; hence the set S is open. 3. I f D is afunction whose values are regular closed sets (cf. p. 38), then the set So= D, is a regular closed set containing all the sets Dt. More-
u t
over, every regular closed set containingall the sets Dt also contains the set So.
PROOF.Clearly So3 Dt, so that Int(So) 3 Int(Dt), thus _______
(9
Int(S0) =3 Int(D,) = Dt.
Since t is arbitrary, we infer by Theorem 1 that Int(S,)
3
u Dt r
and
Int(So) 3
u Dt
= So.
I
On the other hand, Int(So) c So. Thus ~-
Int (So) c So=: So,
which proves that So= Int(SJ. Hence the set So is regular closed. It follows from (i) that So contains each set D,. If 2 is a regular closed set and 2 3 D, for every t, then t
1
4. r f D is a function whose values are regular closed sets, then the set
Po =
is a regular closed set included in each set D*. More-
over, every iegular closed set included in each set D, is also included in Po.
1. GENERALIZED UNION AND INTERSECTION
PROOF.Let X
119
D,. We thus have
=-
t
Po = Int(X) = X"-"- hence
Int (Po)= X"-"-"-"-.
Applying formula (15), $8, Chapter I, p. 29 we have Int(P,,)
=X = Po.
Hence the set Po is regular closed. -~ Since X c D,, we have Int(X) c Int(D,) and Int(X) c Int(D,), that is, Po c Int(D,) = Dt for every t. Finally, if 2 is a regular closed set and 2 c D, for every t, then _ _ _ ~ 2c D,. Hence I n t ( 2 ) c Tnt(X) and Z = Int(2) c Int(X) = Po.
n f
5. As a result of the theorems proved in Examples 1 and 2, it is
possible to define a topological space by taking as primitive notion either that of open set or that of closed set instead of closure. Namely, we may conceive of a topological space as a set with a distinguished family of subsets F. Subsets belonging to the family F a r e called closed sets. We suppose that F satisfies two conditions: (i) I f W c F, then P ( W )E F (that is, the interesection of an arbitrary family of closed sets is closed). (ii) If a family W is finite and W c F, then S(W) E F (that is, the union of a finite number of closed set is closed). We obviously assume that P(0) is the whole space. If we take the notion of open set as primitive, then denoting the family of open sets by G we assume axioms dual to (i) and (Z): (i') If W c G then S ( W ) E G. (ii') If a family W is finite and W c G, then P ( W ) E G. The system of axioms (i)-(ii) is equivalent to the system (1)-(4) given in Chapter I, p. 26. The axioms (1)-(4) are satisfied if we define Aby the formula A= P(W'), where WAis the family of all closed sets containing the set A. Then we have ( A = A> = ( A EF). A similar remark can be made for the system (i')-(Z'). 6. A family R c F is said to be a closed base for the topological space if for every A E F there exists W c R such that A = P(W). A family
120
UNION, INlBRSECI'ION AND CARTESIAN PRODUCT
IV. G E -
R
c F is a closed subbase if the family of all finite unions of the sets belonging to R is a closed base. 7. The notion of open base and subbase can be defined dually replacing P by G (= the family of open sets), intersection by union and union by intersection.
ExerdseS 1. Let F E (2')',
fe
and 35 =
~-I(Y) =
2. Prove that
uFt. Let
=fl
Ft. Prove that
for every Y c 9. UYT'(Y) t
(U Ft)X(UG") t
=
U (FtXG,), t. u
3. Let T be any set and let KC^^. Let the operation DK on Fe(2X)T be defined by the formula XEDK(FE ) {t: x ~ F t } e K . Find K for which the operation DK coincides with the operations of union and intersection discussed above. 4. Show that if Z is an ideal in 2* and K = 2T-Z, then the operation DK is distributive over finite union; that is, 5. Prove that the family of all intervals r < x < s, where r and s are rational numbers, is a base for the space C of real numbers. Prove that the sets { x : r < x ) and ( x : x < r } , where r is rational, form an open subbase for this space. 6. Let X be any set and R be any family of its subsets. Prove that the set 3 can be considered as a topological space with the family R as an open subbase (resp. closed subbase). 7. If X is a topological space and R is an equivalence relation with field X , then X{R becomes a topological space when we assume that a set Uc XIR is open if and only if the union S(U) = UZ is an open set in X. ZEU
8. Prove that
the canonical mapping X -+X/R is continuous if XIR h a the
quotient topology defined in Exercise 7.
8 2. Operations on infinite sequences of sets We shall now consider a special case of the previous operations; namely, where the domain T of the function F coincides with N,that is, where F is an infinite sequence of sets. In analogy with infinite
121
2. OPERATIONS ON INFINlTE SEQUENCES OF SETS
series and products of real numbers, we write m
UFn or UFn n
or Fo u Fl u
...
u F,;
instead of
ntN
n=O
nFn or n
nF,.
00
OFnor Fon Fl n ...
instead of
nEN
n=O
The following formulas follow immediately from formulas (2), Q 1
(j{x: @ ( n , x ) } = (x: v @(n, x)}, w
n=o
n=O
(1)
fi {x: @ ( n , x ) } = ( x : A @ ( n , x ) } , 00
h=O
n=O
where @(n,x) denotes a propositional function of two variables, n is limited to N and x to a given set X. Besides infinite union and intersection we consider the operations Limsup Fn
(limit superior of the sequence Fo,Fly...),
n=m
LiminfF,
(limit inferior of the sequence Fo,Fly...),
n=m
defined as follows m
Limsup Fn n=m
=
w
w
nU
Fn+Ly
LiminfF, = n=m
n=Ok=O
U
w
Fn+k.
n=Ok=O
It is easy to check that LimsupFn is the set of those elements x which belong to Fnfor infinitely many n. Analogously, x belongs to LiminfF, if and only if it belongs to Fn for almost all n; that is, if it belongs to all but a finite number of the F,. It is easily seen that Liminf Fnc Limsup F,. n=w
n=w
(see formula (18), p. 49). If the inclusion sign in (2) can be replaced by the equality sign, that is, if the superior and inferior limits are equal, then their common value is denoted by Lim F,, n=m
122
IV. GENBRALIZBD UNION, O N
ANJJ CARTESIAN PRODUCT
.
and is called the limit of the sequence Fo,Fly ... In this case we also say that the sequence is convergent. This terminology is similar to that used in the theory of real numbers. In order to emphasize this analogy, let us consider the notion of the characteristic function of a given set. Let the set 1 be given and X c 1. The function with domain 1 1 0
(3)
if if
xcX, x~l--X
is said to be the characteristic function of the set Xl). It is easy to show that the sequence Fo,PI, ... of subsets of 1 is convergent if and only if the sequence of the characteristic functions of these sets is convergent to the characteristic function of LimF,,. n=m
It is also easy to show that the following conditions are equivalent2): (4)
Lirn (F,,-A) = 0,
n= m
LimF, = A ,
(4’)
n= w
where the sign A denotes the symmetric difference of two sets. The same equivalence holds for real numbers if we replace F,-A by IF,,-AI. PROOF.Condition (4) is equivalent to the following: every element x belongs to Fn -I A for at most finitely many n. In other words, for every x there exists no such that n > no implies x E F,, = XE A . Suppose that x E Lirn sup Fn, i.e. that x belongs to F,,for infinitely many n. It follows from ( 5 ) that x s A and that x sF,, for all n > no; that is, x E Lirn inf F,,.Thus we have proved that (4) implies
(5)
(6)
Lim sup F,, c A c Lirn inf F,, n=w
n= w
from which (4’) follows by (2). I) See: Ch. de Ia Val& Poussin, Inte‘grales de Lebesgue, fonctions #ensemble, classes de Baire, 2nd ed. (Paris 1936). See: E. Marczewski, Concerning the symmetric direrenee in the theory of sets and in Boolean Akebraq Colloq. Math. 1 (1948) 2W202.
123
6. OPERATIONS ON INFINITE SEQUENCES OF SETS
Conversely, suppose that (6) holds and x E A . Thus x E Lim inf F,, and x E F,, for all n greater than some no. If, on the other hand, x # A , then x $ Lim sup Fn and hence x # F,, from an no on. Hence condition (6) implies that (5) holds for every x if n > n,. We shall still prove several more special laws concerning interchanging the symbols n and v and replacing two operations of the same kind by one such operation. Let F be a function defined on the set N X N , F' a function defined on the set N N x N N and F" a function defined on the set N x N N , where the values of these functions are sets. We use the notation introduced on p. 96-100.
U UFm,n m n
(7)
=
F~(~),~(~,. U F ~ ( ~ ) , ~n (n ~ Fm,n )~ =n P m n P
We shall prove only the first of these formulas. Clearly, Fg(p),L(p) c F,,,,, and thus Fg(p),L(p) c F,, ,,.On the other hand,
uu m
n
if x E U m
u
uu
P
m
n
u Fm,n, then for some m, n we have
x E F,,,, ; hence x E
n
where P = J(m, 4.
FK(P),L(P)
n u Fm,n = U n
(8)
m
n
P,E"
Fm,p(m)*
m
Since q is arbitrary, we have
un
Fm,p(rn)
m
If X E
n u Fm,n,then m
x E F,,
n.
hence x E
c
nU m
Fm,n.
n
for every m E N there exists n such that
n
Letting in) = min ( x E F,, ") we infer that x E
u nFnn, v
m
n
p(m).
n m
Fm,q(m);
N. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT
124
In fact, @'"EN", thus Fi,,(m)
u F i , I , whence it follows that
=
I
F;, m
xE n m
,(m)
n u FZ,,. To prove the opposite inclusion, suppose that
c
u F;,
m
v
and let Zm= {v: x E Fi,,}. Hence Z # 0 for every
I
v
rn E N , and so there exists a choice function h for the family consisting of all the sets 2,. Letting h(Zm) = f ( m ) , we infer that f ( m ) N~ N and x E F i , f(m) for every rn E N . It follows from Theorem 7, p. 99 that there exists a sequence pl E N" such that I$") = f ( m ) for every rn E N . Thus x E Fa, ,(m) for every m E N , that is, xE
n F:,
u nFG,
c
,(m)
m
,,,(m).
~m
Using formulas (7)-(10) we shall obtain a formula to be used in Chapter X. In this formula G is a -function of four variables, with domain N Nx N Nx N x N, whose values are sets.
In fact, replacing F:,I in (10) by
nGv,+,m,n(where p is an arbin
trary but fixed function), we obtain
f3 L' f l Gv,p,m,n m v
=
n
U n fGv,+(m):m,n, l v
thus the left-hand side of (1 1) equals
m
n
u u nnGv,,(m),m,n. ~
p
m
Apply-
n
ing (7) and (9) we obtain (11). Exercises 1. Prove that the characteristic function defined by (3) satisfies the following conditions: (a)
(b) fi(X)
= 0, = 1,
(c) f - X W = 1-fx(x), (d) fA n B ( x ) = fA ( x ) .fB ( X I , fA-B ( X I = fA ( x ) -fA
n
B(XI-
2. Prove that if Fo c F, c F2 c
..., then
m
u Fn
n=O
= LimF,. n=m
125
2. OPERATIONS ON INFINITE SEQUENCES OF SETS
m
3. Prove that if Fo3 F,
..., then
3
F, = LimF,,. n = ~
n=m
4. Prove that if Fo= 1, then m
1 = ~ F ~ - F ~ ~ ~ ~ F ~ - ~ ... ~ u ~ U (7( FFn . ~ - F , ) ~ If, moreover, Fo3 F,
3
F22
n=O
..., then m
( F, - F~ ) u(F ~ -F . )u
... u n=o n F n = l-[(Fo-Fl)u(Fz--F~)u
5. Prove that if k, < kz
1. Let U n n U c, and let U c.,+, U c,,,,.Prove A.
n=l
n=l
8.
= A,-(A,u
=
=
B ~ , =, A =
n
m
c
m
n
n. m
n,m
that
u k=l (D
A
= LhA,,
where
A, =
n=co
... nBk,,)n(CI,kU ... ucn,k).
9. Prove that (a) Liminf(-A,) = -- LimsupA,, (b) Lim(-A,) = -(LimA,), (c) Liminf(A,nB,) = LiminfA,n LiminfB,, (d) Limsup(A,uB,) = LimsupA,u LimsupB,, m
(e)
n A,
uAn, m
c LiminfA, c LimsupA, c
n=l
n=l
( f ) LiminfA,u LiminfB, c Liminf(A,uB,), (9) Limsup(A,nB,) c LimsupA,n LimsupB,, An), (h) A ILiminfA, c Limsup(A A A LimsupA, c Limsup(A A An); show that the opposite inclusions d o not hold in general. _I_
10. A function f from sets into sets is said t o be continuous if for every convergent sequence F,, F2,... the following identity holds: f (LimF.) = Limf(Fn). n=m
n=m
126
N. OENBRALIZED UNION, INTERSEcIlON AND CARTESIAN PRODUCT
Show that the functions X u Y , X n Y ,
-X and generally
uF,, and nF,, are n
n
continuous with respect to each component.
11. Prove the following condition for a sequence F,, to be convergent: for every sequence of pairs < r n i , Q > such that limmi = limni = co, we have I=m
k m
nI (Fmi
Fnj)= 0.
L
[Marczewski]
12. If K is a family of subsets of N such that the complement of every set in K is tinite, DK(F) = LiminfF,; if K is a family of infinite subsets of N then DK(F) = LimsupF, (see 8 1, Exercise 3). Using this result, generalize the operations Limsup, Liminf for the w e where the argument is a function defined on an arbitrary set T (not necessarily on the set N).
5 3. Families of sets closed under given operations Let X be a fixed set and f a function of an arbitrary number of variables, where each variable ranges over the subsets of X. For simplicity let us suppose that f is a function of two variables; that is, the domain off is the cartesian product 2x x 2x. A family R c 2x is said to be closed under a given finction f if
A [WI ER) A (YzE R)
Y1,Yn
4
(f(Y1,Yz)6R)I-
there exists a f h i l y R1 such THEOREM 1: For each fa@Iy R c that: 1 . R c R1 c 2x; 2: the family R1 is closed under the operation f; 3. the family RI is the least family satisfying conditions 1 and 2, that is, if R’ satisfis the following two conditions
then R1 c R’.
PROOF. Let K be the set of all families R’ satisfying (1). K is a non-empty set, for 2x E K. The required family is the intersection R’.
n
R’EK
The family R1 satisfying conditions 1-3 is uniquely determined. In fact, if R2 also satisfies the same conditions, then R1 c R,, since R1is the least such family. Similarly we obtain R2 c R1.Hence R1 = R2 We denote this family by R*.
127
3. FAMILIES OF SETS CLOSED UNDER GIVEN OPERATIONS
THEOREM 2: For arbitrary families R, R1 and R2 the following conditions hold (i) R c R*, (ii) R1 c R2 + RT c R,*, (iii) R** = R*.
PROOF.Formula (i) follows from Theorem 1 (condition 1). Formula (ii) follows from the fact that R: is a family closed under f and containing R1, thus by miaimality, R t =I RT. Finally, condition (iii) can be proved as follows: (i) implies R* c R**; since R* 3 R* and R* is closed underf, we obtain R** c R* by minimality. Theorems similar to 1 and 2 also hold for the case where there is given not one function f but an arbitrary family of such functions and R* denotes the least family containing R and closed under all these functions. Moreover, the domains of these functions may be sequences of subsets of X. We shall not, however, formulate all of these generalizations. Example 1. Let f denote the union of sets, i.e. f ( Y l ,Y 2 )= Y , u Y , . The least family of sets containing R and closed under f i s denoted by R,. This family consists of finite unions of the form U Y i where k n
# 0 and Y = (Yo,Y 1 ,..., Y n - l ) is a sequence of sets belonging to R ; in other words, Y ER". Similarly, if g is the function defined by g(Yl, Y,) = Y 1 n Y,, then the least family containing R and closed under g is denoted by Rd.
n E N, n
This family consists of all intersections of the form
n Y , where
fen
n E N , nfO, Y E R " .
Example 2. The least family containing R and closed under the operations f and g defined in Example 1 is called the lattice of sets generated by R.
THEOREM 3: The lattice of sets generated by R is identical with the f amih Rsd. More0 ver, RSd= Rd, . PROOF.First we prove the second part of the theorem. Let Z E Rsa, that i s , Z = Y i , where n E N , n # 0 and YiERsfor i < n. We show
n
icn
N. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT
128
by induction that Z E Rd,. For n = 1 we have Z = YoE R, and R, c Rd, because R c Rd. Suppose that the theorem holds for n = k and let Z be the intersection of k+ 1 components Yi belonging to R,. In particT j where m e N , m # 0 and T j e R for j < rn. Let ular, let Yk =
u
j<m
Z'
=
n Yi.By the induction hypothesis, 2' E&, k k
where p
E N,
p # 0 and
2 = 2' n
v h E Rd for
h < p . Since Z
u Ti u (2' n T j ) u u =
i<m
=
thus 2' =
u vh h is called the direct product of the functions ft (resp. of the relations R,) reduced mod Z. Let @(x,y, z) be an arbitrary propositional function. The main problem of the theory of reduced products can be formulated as follows: When the sets { t : @(F,, ft,R,)}are known, under what conditions do the set P/Z, the function y/Z and the relation e / Z satisfy the propositional function @? In order to solve this problem, we consider more general propositional functions @(x, y, z, ul,..., uk) of an arbitrary number of variables. Suppose that e l , ...,ek E P and let A0 = { t : @(Ft7 ft,Ri, el(t)7
*'.,
ek(t))}
(clearly, the set A& depends not only on @ but also on the elements e l , ...,ek; we do not write el, ...,ek in the symbol A , in order to simplify the notation). With this notation, the following theorems (i) - (iv) hold:
A a v y $ Z ~A,$ZV
(9
AygZ.
In fact, A m V y= A,uAp; thus (see formula (ll), p. 17) A ~ P ~ Y E Z A (AYE Z), and (i) follows by de Morgan's laws.
= (&€I)
14
"(ii)
IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT
v @,
If 0 is the propositional function
then
"k
A,$Zr V[Aa$Z]. ek tP
In fact, suppose that A , = { t : @(Ft,ft,Rr,ei(t),* . - , e ~ - i ( t ) ) } # Z .
For t E A, there exists x E F, such that @(Fr,
5,Rt, el(t), - - . , e k - l ( t > 7 x ) .
Let X , denote the set of all these elements x , and let X , = Ft for t # A,. Let ek be a choice function for the family consisting of all the setsX,. We have ekE P and
@(&, 57 R ~el(t)$ ?
(1)
ek(t))
for all t E A,. This implies A , c A , and thus A , #Z. Conversely, if A , these t
# Z then formula (1) holds for all t E A,. Thus for @ ( F t , A, R,, el ( t ) ,
.
*
Y
ek-l(t>).
This implies that t E A@. Hence A , c A , and A , $Z. An ideal Z is said to be prime if for any X c T exactly one of the conditions A E Z, T - X E Z holds. The set {X c T: x $X}is an example of a prime ideal. In Chapter VII we shall prove that every ideal can be extended to a prime ideal. (iii) If 1 is a prime ideal and !P is the formula 7 @, then A&Z= In fact & = T - A , . "(iv) If Z is a prime ideal, then
1(A,$Z).
A,,\~$Z~(A,$Z)A(A~#Z); moreover, if.." is the formula
A uk
Theorem (iv) follows fro= (i)-(iii).
@, then
145
8. REDUCED DIRECT PRODUCTS
A propositional function @(x, y,z, ul, ...,uk) is said to be elementary if it can be constructed from the propositional functions (a) ui = uj, (b) Y(ui,u j ) = uh, (c) € 2 by applying the operations of the propositional calculus and by applying the quantifiers
v, A.
UEX
UEX
The following theorem solves the problem stated above. O THEOREM 3 : If 1 is a prime ideal, @ ( x , y ,z, u l , ..., uk) is an elementary propositional function and el, ...,ek are arbitrary elements of P , then I):
(2)
v/z,@/I,el/z?
@ (p/z,
*
a
Y
ek/z)
PROOF.If @ is one of the propositional functions (a), (b), (c), then (2) holds. In fact, the left-hand side of (2) is equivalent to ej/Z= ej/Z in case (a), to v/Z(ei/Z,eJZ) = eh/Z in case (b), and to (ei/Z, ej/Z)E @ / Z in case (c). The right-hand side of (2) is then equivalent: in case (a) to { t : e j ( t )= e j ( t ) } $ Z , in case (b) to ( t : f t ( e i ( t ) , e j ( t )= ) eh(t)}#Z, in case (c) to { t : (ei ( I ) , e j ( t ) )E R,)$ 1. From the definitions of the set PIZ, the function q / Z , and the relation ,o/Z as well as from the definition of prime idealn it follows that the left-hand and right-hand sides of (2) are equivalent. In turn, it follows from theorems (i)-(iv) that if (2) holds for the propositional functions @ and Y,' then it also holds for the proposi! by applying the operations of tional functions arising from @ and P propositional calculus and by applying the quantifiers and uix
A. UEX
In this way Theorem 3 is proved. 1) Theorem 3 is a scheme, for each propositional function we obtain a separate theorem.
146
IV. QENERALIZED UNJON, INTERSECTION AND CARTESIAN PRODUCT
o
C4: If@D(x, y~, z ) is an elementarypropositional ~ ~ function, ~ then ~
@ ( W ,911, elr) = It : @E, ft ,R,)) 4 1. This corollary follows from Theorem 3 if we assume that the propositional function does not contain the variables 2.4,...,uk. ‘COROLLARY 5: If @(x,y, z) is an elementary propositional function and f o r every t the formula @ ( F c y f t , R tholds, ) then dj(P/I,pl/Z,e/I).
This corollary follows directly from the previous corollary and from the remark that if I is a prime idea1 then T 4 Z. Examples
In the following we suppose that Z is a prime ideal in 2=. 1. If the relations Rt are reflexive, transitive and satisfy the conditions: A (KX, Y> ERtI A Kv,x> ERtI --tx = Y } , x.yaF,
XI
A K<X, Y> ER;) ” ( +,I, YEFt
then the reIation e/I satisfies the same conditions. # 2. If Ft is a field with respect to the operations of addition f, and multiplication g,, then the set PII is a field with respect to the operations q11I and y/Z, where pl and y are the Cartesian products of the operations ft and g, respectively. Similarly, if each of the sets F, is an ordered field with respect to the operations ft and g, and with respect to the order relation R,, then PI1 is an ordered field with respect to the operations y/Zand y / I and the order relation e/Z. These properties follow by Corollary 5 from the remark that propositional functions “ X is a field under the operations D and M” and “X is a field with respect to these operations ordered by a relation R” are equivalent to elementary propositional functions @(x, D, rmp. (X,0,M,R). # These examples show that forming the reduced direct product we can construct from a given family of models of a given system of axioms new models of the same system of axioms. Other applications of reduced direct product will be given in Chapter IX.
~
8. REDUCED DIRECT PRODUCTS
147
Exercises 1. By applying the lemma on p. 139, prove the existence of a prime ideal which contains a given ideal # 2T. 2. Prove that if each of the sets Ft is a Boolean algebra with respect to the operations v t , A t , -* and the elements O t , lf and if for every t the condition
holds, then the set P / I is a Boolean algebra with respect to the operations v / I , AII, --/I and the elements 511, ill, where v, A, - are the Cartesian products of the operations v t , A t and - t , respectively, and l , i are functions such that E(t) = Ot and i(t) = I f for every t. Moreover, P/I satisfies condition (*). Show that the ordinary Cartesian product of Boolean algebras satisfying (*) may fail to satisfy this condition.
$9. Inverse systems and their limits Suppose we are given: Ci) an arbitrary set X, (ii) a set T ordered (or more general, quasi-ordered) by the relation ) ~is , (iii) a function F E ( ~ ~that Ft c X for every t E T;
- = f ( x ) G f W . = T i in the qet A ; then pi 5; x and therefore
v
tET
f(yt)
’4,consisting of all functions f whose values are the numbers 0 and 1 and whose domain is the set A. Each such function is uniquely determined by the set X, of those a for which f ( a ) = 1 (f is the characteristic function of this set, see Chapter IV, p. 122). Under this correspondence to distinct functions fi and fi correspond distinct sets X,, and Xf2. Thus associating with the function f E ( 0 , l}“ the set X, c A, we obtain a one-to-one correspondence between the sets (0, 1jA and 2A.
5 5. Inequalities between cardinal numbers. The Cantor-Bernstein theorem and its generalizations We obtain the “less than” relation beetwen cardinal numbers from the following definition. DEFINITION: The cardinal number m is not greater than the cardinal number n, m d n, if every set of power m is equipollent to a subset of a set of power n. If m < n and m # n we say that m is less than n or that n is greater than m; we write m < n or n > m.
V. THEORY OF CARDINAL NUMBERS
186
For example, (1)
< a, rn < 2"'. n
(2)
For the proof of (2) we notice that m < 2", because every set A of power m is equipollent to the subset of 2A consisting of all singletons of elements of A . Moreover, m # 2m by Theorem 2, 0 3. The following theorem is an interesting consequence of the definition of inequality. O THEOREM 1: r f f is a function defined on the set X and f'(X) = Y , then '.< PROOF.The notion of coset of a functionf (a set of all elements of X which have the same value under f) was defined in Chapter 11, p. 175. Every coset is a set of the form w,= { x E x: f ( x ) = y } .
Since all cosets are distinct and non-empty, there exi ts by the axiom of choice1) a set A containiw exactly one element from every coset. It follows that A is equipollent to the set of all cosets off and thus to the set f'(X). Since A is a subset of X , we conclude that Y Example. The projection of a plane set Q onto an arbitrary straight line has power 2. In this case the cosets of the projection are the intersections Q n L where L is a straight line parallel to the direction of the projection. REMARK: We write m = v " ~ a , - l ( 4 .
We define A z = {aEA: n(a) = N V ( ~ ( ~ ~ E g ' ( B ) ) h ( g Q ( S ( a ) ) $ f l ( A ) ) ) . ,
A1 = A - A , ,
B1 =f'(A,),
B,
= B-BI.
For the proof of the theorem it suffices to show that g'(B2) =,A,, i.e. that
(8)
b e 4 -+g(b)EA*,
(9) A , = g'(B,). PROOFOF (8). Assume that bEB2; that is, that b $ f ' ( A , ) and let a = g(b). If b $f'(A) then a is not extendabIe, s(a) = a, and by definition we have a E A,. If b E ~ ' ( A )then b ~ f l ( A , and ) b =f(a'), where a' E A2. Clearly a' =f"(gc(a')), that is a' = a* ;thus a* E A,. If n(a*) = N then n(a) = N and U E A , ; otherwise s(a*) = $(a) and again ~ E A , . PROOFOF(^). Assume that ~ E A ,If. a is extendable then a = g(f(a*)). If at the same time n(a) is finite, then s(a) = s(a*) and a* € A 2 . The same is true i f n ( a ) = N, because then n(a*) = N as well. Thus in both cases a* $ A', f ( a * ) $ f ' ( A , ) and thus f ( a * ) E B2. It follows then that a = g ( f ( a * ) ) Eg'(B2). If a is not extendable, then s(a) = a and by the assumption that ~ E A we , obtain aeg'(B). If a were an element of gl(B1) then, by the definition of the set B,, a would have the form g ( f ( a ' ) ) and would be extendable contrary to the assumption. Thus a Eg'(B,). Q.E.D.
Y. THEORY OF CARDINAL NUMBERS
190
As a corollary we have the Cantor-Bernstein theorem :
-
THEOREM 3: I f m
< n and n d m
then m = n.
PROOF.Let A = m and B = n. Since m < n, there exists a one-to-one function f from A onto a subset of B. Since II m, there exists, sirnilariy, a one-to-one function g from B onto a subset of A. By Theorem 2, A = Al u A,, where A l and A , are disjoint; and B = B1 v B,, where B1 and B, are disjoint and where f ‘(Al) = B, and g’(B,) = A,. Thus A l N B, and A, N B,; hence A B. The Cantor-Bernstein theorem can be generalized as follows, Let R be an equivalence relation on the family 2A and let R satisfy the following two conditions: P
2'. 5. Show that the closed circle T is equivalent by finite decomposition to the union T u F, where F is an arbitrary line segment disjoint from T l).
N, so
= 2" by Theorem
7 , p. 185.
THEOREM 2: The set of all infinite sequences of natural numbers has power c.
PROOF.This set is NN.Therefore its power is
aa =
c.
#THEOREM 3: The following sets have the power of the continuum: (a) the set of irrational numbers in the interval (0,l); (b) the set of all points in (0,l) ;
6. CARDINALS
a
(c) the see € of all real numbers; (d) the set of all points of the space
AND
c
195
P,where n is a natural number.
PROOF,(a) follows from the remark on p. 138 and from Theorem 2. (b) follows from the observation that the interval (0,l) is the union of the countable set of rationals in (0, 1) and the set of irrationals in
(0, l), which has power c. (c) holds because function y = 1/2
1 +x-arctgx
is one-to-one and maps the set & onto the interval (0, 1). (d) follows from (c) and equation (6). =i+ e
-
c.
THEOREM4: If A = c, B = a and B c A , then A - B = c. PROOF.By (6) we have that A x A N A; therefore it suffices to show that if M is a countable subset of A x A, then the difference A X A --M has power c. The projection onto A of the points in M constitute at most a countable set, which implies that there exists an element of A which does not belong to the projection of M . The set {(a, y ) : y E A} is disjoint from M and has power c, thus the difference A x A --M has power 2 c. O n the other hand, this set has power c as a subset of A x A. Thus the difference A x A --M has power c by the Cantor-Bernstein Theorem.
f , (r2)
9
-
**
,f (r"),
* *
V. THEORY OF CARDINAL NUMBERS
196
Iff and g are distinct then the corresponding sequence?
f
(h)Y
f (4, ,f @"I, ***
*** Y
g(r,), g ( r A *
-
*
Y
g(rn), * * *
are also distinct. In fact, f # g implies that f ( x ) # g ( x ) for some x ; so if r k , is a sequence of rationals converging to x , then it is not true that f(rk,) = g(rk,) for every n, because in that case, by the continuity o f f and g , we would have
f (x) = lim f (rk,) = limg(rkn)= g ( x ) . n- w
,= w
Thus the set of continuous functions of one real variable is equipollent to the set of sequences (2), which has power c by Theorem 4. On the other hand, the set of continuous functions has power >, c because it contains all constant functions. Thus by Theorem 3, § 5 we obtain Theorem 7.