Chapter I Algebra of Sets

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 n A )-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 (B AA) . (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

<

X = [(XI n A,) u ( X z n A,) u

... u (Xnn A,,)] u (-Al n -Az n ... n -An).

Therefore

-X = [(-Xl u - A , ) n (-1, u --A2) n ... n (-X,, u -An)] n (A1 u ... u An). Since the sets Ai are disjoint, we have X n A ,

= X, nA,

and --X

n A , = -1, n A,. By applying (iii) we obtain (i).

Next we assume that for all r, s (1 < r, s < n) either A , = A , or A , n A , = 0. We shall designate conditions (i) by W,,Wz, ... ,W,,,V1,V 2, ...,V,,. We shall show that if A , = A , then W, -+ W,, or W, -+ W r y or else W, A W, = F. Indeed, if i, = is = 0, then W, --t W, if p r < p , , and W,-,W, if p , < p , . If i, = 1 and is = 0, then W, + W, in case p r p s , and in case p , < p , , W, A W, 3 F. Finally if i, = is = 1 then W, + W, for p r = p , and otherwise W, A W, = F. Similarly it can be shownthat either V , + V,, or V , --t V,, or V , A V , G F. We conclude that either the conjunction of (i) is false or else we may omit from (i) certain components and obtain an equivalent conjunction in which none of the sets A, occurs more than once. Thus this case is reduced to the preceding case. Now we shall reduce the general case to the case in which the sets A , are either identical or disjoint. For this purpose we note that if M n N = 0, then rj(MV N)

rj(M) V [rj-l(mA r ! ( N ) ]V [r:-2(WA r ; ( N ) ]V ... V [rh(kf) A r;(N)],

25

7. CONSTITUENTS

r b ( M U N ) 3 [ r ; ( M ) A r i ( N ) ] V [ r i - ~ ( M )A r : ( N ) ]V

... V [rA(kf) A rb(N)]. By induction, if the sets S,, ...,s h are pairwise disjoint, then the condition of the form I';(S, u ... u &) can be expressed equivalently as a disjunction of conjunctions, where each conjunction has the form I$:(Sl) A

... A F i ( S h ) .

Represent the sets A, as unions of constituents; then according to the above remark, each of the conditions (i) can be expressed as

a disjunction of conjunctions each of which has the form

r ; : ( x n S,)A

... A . F : ~ ( x ~&),

or respectively,

r,",'(--Xn

s,) A ... Ar,",h(-Xn

&).

Applying the distributive law for conjunction over disjunction, we express the conjunction of conditions (i) as a disjunction, each of which n S,> is a conjunction whose components have either the form or the form I';(-X n S,). Sets occurring in each such conjunction are either identical or disjoint. Thus the general case is reduced to the preceding one. Example. We shall find necessary and sufficient conditions for the existence of a set X satisfying the conditions

X n A n B f 0, XnAQB,

-Xn A n B # 0 , XnB+A.

These conditions can be expressed equivalently as the conjunction of the following six conditions:

r f ( X n A n B), r?(--Xn A n B),

I'f(Xn --A n B),

ry(Xn A n -B), T,"(-Xn A n -B),

r t ( - X n -A n B).

Hence we obtain the desired condition n B ) A Pf(A n - B )

A

I'f(--A n B).

26

I. ALGEBRA OF SETS

In other terms, A n -8 and B n -A have to be non-empty and A n B has to contain at least two elements I). Esercisas 1. Assuming that the set 1 is infinite and that A l , ...,A, are finite, describe a method of obtaining necessary and suflicient conditions for the existence of a finite set X satisfying the conjunction of conditions (i). 2. Let Z be the unit n-dimensional cube, that is, the set of sequences (xl, ...,x,) such that 0 < xi Q 1 ( i = 1,2, ...,n). Let Zm consist of those sequences (xl, ...,x,) E I where 1/2 < x, < 1. Show that the sets Zl,...,Z, are independent. Give a geometrical interpretation for n = 2 and n = 3.

0 8. Applications of the algebra of sets to topology *) In order to illustrate applications which the calculus developed in the preceding sections has outside of the general theory of sets, we shall examine the axioms of general topology and apply the algebra of sets to establish several results. In general topology we study a set 1, called the space, whose elements are called points. We assume, moreover, that to every set A contained in 1 there corresponds a set 2also contained in 1 and called the closure of A. The space 1 is called topological if it satisfies the following axioms (see also p. 119) (1)

AUB=AUZ

(2)

z=A,

(3)

ACA,

(4)

0

-

= 0.

I) The elimination method given above is due to Skolem, Untersuchungen uber die Axwme des Klassenkalkuls ..., Skrifter utgit av Videnskapsselskapet i Kristiania, I Klasse, No 3 (Oslo 1919). z, For more details on topological calculus developed in this section see K.Kuratowski, Topology Z, Academic Press 1966, Chapt. I. For further investigations on this calculus from an algebraical point of view see the paper of J.C.C. Mc Kinsey and A. Tarski, The algebra of topology, Annals of Mathematics 45 (1944) 141-191. In $8 we apply not only axioms I, A, B, C but also axioms (1)-(4). However, we can deduce all theorems givzn in this section from the full axiom system of set theory given in Chapter 11, treating axioms (1)-(4) as assumptions about the operation of closure.

27

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

In axioms (1)-(3) the letters A and B denote arbitrary subsets of the space 1. # Axioms (1)-(4) are satisfied if, for example, 1 is the set of points of the plane and if the closure operation 2 consists of adding to the set A all points p such that every circle aroundp contains elements of A. This interpretation will be referred to as the natural interpretation of the axioms (1)-(4). # We shall show how, using only laws of the calculus of sets, it is possible to deduce a variety of properties of the closure operation. -

1 = 1.

(5)

PROOF.For every A we have

Ac 1, and by axiom (3),

-

1

c 1.

PROOF. From B u ( A - B ) = A u B applying axiom (1) we obtain B v A - B = 2 u 3.This implies that A c Bu A Y B and thus - - - A - B c ( B uA - B ) - F = A-B-B~ AT, which proves (6).

-

(7)

ACB+ACB.

PROOF.A c B is equivalent to the equation A u B = B. By axiom (l), XU Z= Z, thus A c F (Cf. 8 3, (9,p. 8).

PROOF.Since A n B c A and A n B c B, theorem (7) implies A n B c 2 and A n B c 2, from which it follows that A n B c A n F. -V A = A and B = B, then A n B = A n B. (9) PROOF.In fact, A n B c A n B by axiom (3), But by (8) and by the .hypothesis of (9), A n B c B= A n B. Therefore A n B = A n B. We call a set closed if it is equal to its closure. Theorem (9) states that the intersection of two closed sets is closed, and axiom (3) that the union of two closed sets is closed. We call a set open if it is the complement of a closed set. By de Morgan’s laws it follows that the union and intersection of two open sets is open.

xn

28

I. ALGEBRA OF SETS

# In the natural interpretation of axioms (1)-(4) closed sets are those sets which contain all their accumulation points (cf. p. 32). Open sets have the property: for every point p contained in the open set A there exists a circle with center p entirely contained in A. # The set ') ht(A) = 1-1-A =A"-"

is called the interior of the set A . The interior of any set is clearly an open set. # In the natural interpretation of axioms (1)-(4), the set Int (A) consists exactly of those points p for which there exists a circle with center p entirely contained in A . # Int(A) c A .

(10)

PROOF.By axiom (3), A" c A", from which, by applying equation (7), $ 6 , p. 19, we obtain 1-A"- c 1-A", hence A"" c A""= A. In particular, the relation Int(1nt ( A ) ) c Int (A) is a special case of (10). This relation may be strengthened as follows: Int (Int ( A ) ) = Int (A).

(1 1)

PROOF. It follows from the definition of Int ( A ) that Int ( A ) = A"--", Int (Int (A)) = [ h t (A)]"-"= [ A C - ~ F - C .

By the law of double complementation we may eliminate two consecutive occurrences of the operation AC;we thus obtain Int (Int ( A ) ) = A"--", and because A'-- = A"- by axiom (2), we obtain Int (Int ( A ) ) = A"-" = Int ( A ) . (12)

Int ( A n B ) = Int ( A ) n Int (B).

PROOF.By de Morgan's laws ( A n B)O-" = (Ac u BC)-", l)

Instead of A w e sometimes write A - .

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

29

whence by axiom (1) Int ( A n B ) = ( A n B)c-c = (Ac- u Bc-)c, and a final application of de Morgan's laws gives Int ( A n R) =

n BC-O = Int ( A ) n Int (B).

As a simple consequence of (12) we have: A c B --r Int ( A ) c Int (B). (13) In fact, the assumption A c B gives us A n B = A , from which it follows that Int ( A ) = Int ( A n B ) = Int (A) n Int ( B ) c Int (B).

~-

Int (Int ( A ) ) = Int (A). PROOF.By (10)

Int (Int ( A ) ) c Int (A), whence by (7) and (2) ~-

(14,)

Int(iit(A)) c Int(A).

On the other hand, by (1 l), (3), and (13) Int ( A ) = Int (Int ( A ) ) c h t (Int (A)), and by (7) it follows that

Inclusions (14,) and (143 imply (14). Replacing Int(X) by X"-" in (14), we obtain AC-C-C-C-

-AC-C-

(15) Moreover, substituting Ac for A and applying the law of double complementation we obtain A-c-c-c= A-c-. 1) (16) Equations (15) and (16) show that if we apply in succession the operations of complementation and closure to the set A, then we obtain l) Formula (16) was given by K. Kuratowski in the paper: Sur I'opkrution d'dnulysis Situs, Fundamenta Mathematicae 3 (1922) 182-199.

d

30

I. ALGEBRA OF SETS

only a finite number of sets. Namely, if we start with the operation of complementation, then we obtain the sets A, AC, A C - , A C - C A C - C - Y A C - C - C AC-C-rAC-c-c-c 9

9

9

The next set in this sequence would be Ac-c-2-c-, but by (15) this set equals AC-"-. If, on the other hand, we start by applying the operation -, then we obtain the sets A', A-C, A-C-, A-c-C A - C - C A-C-L-C Y

3

The next set would be

t u t by (16) it is equal to the set

A-C-

Hence by applying the operations of complementation and closure to an arbitrary set A we obtain at most 14 distinct sets. Formulas (17) and (18) will be used in $9. (17)

Zf B = X+-,

then Int[Int(A-B) n B] = 0.

PROOF.Clearly A -B c Bc, whence by (13) and (7) Int[Int(A-B) n B] c Int[Int(Bc) n B].

Thus it suffices to show that Int[Int(Bc) n B] = 0. Since Int(BC) = BCC-C- = B-C- = X-C--Cformulas (12), (16), and (10) that

=x-C-C-

,

it follows by

Int[Int(Bc) n B] = Int[Int(Bc)] n Int(B) = X-c-c-c-c n BCdC BE-C = BC BC-C - 0. -x-C-c (18)

If A = 2 or B =

-~

then Int(A) u Int(B) = Int(A u B).

PROOF.From theorem (13) we conclude that Int(A)

c Int(A u B)

and Int (B) c Tnt (A u B), which implies that Int (A) u Int (B) c Int (Au B). Applying the closure to both sides of the inclusion we obtain by (1) and (7) -

(181)

___

~-

Int (A) u Int (B) c Int (A u B).

For the proof of the opposite inclusion we suppose that, for instance,

i? = B. We apply the identity AuBu[~-(AuB)]=

1,

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

31

from which by axiom (3) it follows that A u B u 1 -(A u B) = 1 , whence Bul-(AuB)3

1-A.

Applying closure to both sides we obtain (dnce

B= B):

B u 1 -(A u B) 3 1-A, from which it follows that [1-1-A]uBul-(AuB) = 1, whence Int(A) u B u 1-(A u B) = 1.

It follows from this equation that Int(A) u 1-(A u B)

3

1-By

Int(A) u 1-(A u B)

3

1-B.

and thus __

Adding to both sides of this equation the set 1-1-B obtain

= Int(B)

we

Int(A) u Int(B) u 1- ( A u B ) = 1, thus Int (A) u Int(B)

3

1- 1 -(A u B ) = Int ( A u B).

Applying closure to both sides of the inclusion we obtain by (1) and (3)

Int(A)u Int(B)3 Int(A u B).

( 182)

Inclusions (18,) and (18,) prove theorem (18). Exercises 1. Prove that if the set A is open, then for every set X. 2. Let Fr(A) = 2 n Rove that :

I--A (the boundary of A).

(a) Fr(A u B ) u Fr(A n B) u [Fr(A)n Fr(B)] = Fr(A) u Fr(B), [A.H. Stone]

(b) Fr(A) = (A n

m)u @-A),

32

I. ALGEBRA OF SETS

(c) A u Fr(A) = (d) Fr[Int(A)] c Fr(A), (e) Int[Fr(A)] = A n Int[Fr(A)] = Int[Fr(A)]-A.

3. We call the set A boundary if F A = 1. The set A is called nowhere dense if A i s boundary. Prove that (a) the union of a boundary set and a nowhere dense set is boundary; @) the union of two nowhere dense sets is nowhere dense; (c) in order that the set Fr(A) bs nowhere dense it is necessary and sufficient that A be the union of an open set and a nowhere dense set. 4. Let 1 be a space satisfying besides axioms (1)-(4) the following axiom (where {p} denotes the set consisting of the single element p): {PI = {PI. We say that the point p is an accumulation point of the set A if p E A- {p} (for the plane this condition is equivalent t o the condition that p = lim p n , where pn n= m

E A-{p)).

By A' we denote the set of all accumulation points of the set A, called the derivative of the set A. Prove the formulas: (A U B)'= A' U Be, A'- B' c (A -B)', A" c A', A = A U A', A' = A'. 5. Let 1 denote the space considered in exercise 4. We call the set A dense in itself if A c A'. Prove that (a) if the space 1 is dense in itself, then every open set is also dense in itself; (b) if sets A and 1-A are boundary, then 1 is dense in itself; (c) the sets Int [Fr(A)] and A n Int [Fr(A)] are dense in themselves. 6. Conditions (1)-(3) are equivalent t o the condition

AuAuZ=A~B

.

[Iseki]

0 9. Boolean algebras We shall conclude this chapter with certain considerations of axiomatic character. If we examine the theorems of $52-8, we notice that the symbol E does not occur in the majority of them, though of course it does appear in the definitionsand proofs. This suggests developing aseparate theory to cover that part of the calculus of sets which does not make reference to the E relation. In this theory we shall speak only

9. BOOLEAN ALGEBRAS

33

of the equality or inequality between objects and terms resulting from these objects by performing certain operations on them. We shall base this theory on a system of axioms, from which we shall be able to prove all theorems of the preceding sections in which the E symbol does not occur. This theory, which is called Boolean algebra, has applications in many areas of mathematics'). Let K be an arbitrary set of elements, A and A operations of two arguments always performable on elements of K and having values in K. Finally, let o denote a particular element of K. We say that K is a Boolean ring or Boolean algebra with respect to these operations and to the element o if for arbitrary a, b, C E Kthe following equations hold (axioms of Boolean algebra): (1)

aAb

=b

A a,

(4)

aAa=o,

(5)

aAb=bAa,

(6) (7)

a A (b A c ) = (a A b) A c,

(8)

a A a = a,

(9)

a A (6 A c ) = ( a A 6 ) A ( a A c).

aAo=o,

We define the sum and the difference of elements of K by the equations

A [6 A (a A b)], = a A ( a A 6).

av b =a

a-b

We call a A 6 the symmetric difference of a and b, a A b theproduc, of a and b, and o the zero element'). l) There is a number ?f books with exposition of Boolean algebras. Among them let us mention R. Halmos, Lectures on Boolean Algebras (Princeton 1963) and the monograph of R. Sikorski, Boolean Algebras, 2nd edition (Berlin 1964). 2, The fact that we are using the same symbols for operations in Boolean algebra and for logical operations should not lead to misunderstanding.

34

I. AujEBRA OF SETS

An example of a Boolean algebra is the family of all subsets of a given fixed set 1 where the operations A and A are the set-theoretical operations of symmetric difference and intersection and where o denotes the empty set. We dealt with this interpretation of axioms (1)-(9) in 56l). More generally, instead of considering all the subsets of the space 1, we may limit ourselves to the consideration of any family of subsets K of 1 where the symmetric difference and intersection of two sets belonging to K also belong to K. Such a family is a Boolean algebra with respect to the same operations as in the preceding example. Each Boolean algebra of the type just described is called a jield of sets. We introduce Boolean polynomials. Let x1,x,, ... be arbitrary letters. The symbols (9 0 , (ii) x1,x,, ... are polynomials; iff and g are polynomials then the expressions (iii) ( f ) n (g), (iv) ( f ) A (g) are polynomials. A polynomial is to be understood as a sequence of symbols. Let us suppose that K is a Boolean algebra and that to every letter xi there corresponds a certain element n j E K. We define inductively the value of a polynomial with respect to this correlation. The value of polynomial (i) is the zero element of the algebra K, the values of polynomials (ii) are the corresponding elements in K; if the values of f and g are the elements a and by then the value of polynomial (iii) is a A b and the value of (iv) is a A b. The value of the polynomial .f is denoted by f K ( a , ,a,, ...); clearly f&,, 4 ,...) EK. Let the polynomial f have the form ... (h’) A (h”)..., and the poly’) Similarly as in $8, our exposition is based not only on the axioms of set theory but also on the axioms of Boolean algebra and, in part, also on topological axioms. As a matter of fact, we can deduce all theorems from the axioms of set theory given in Chapter 11, treating the axioms of Boolean algebra as assumptions about the operations A, A and the element o and the axioms of topology-as assumptions about the closure operation. Similar remarks apply to § 10.

35

9. BOOLEAN ALGEBRAS

nomial g the form ... (h") A (h') ... where the periods denote sequences of symbols which occur both in f and in g , and where h' and h" are polynomials. In this case we say that the polynomial g j s immediately transformable into the polynomial f by means of axiom (1). Similarly we define immediate transformability by means of the remaining axioms (2)-(9). We say that the polynomial g is transformable into f if there exists a finite sequence of polynomials f =fi ,f2, ... ,& = g such that is immediately transforfor each i (1 < i < k) the polynomial mable into the polynomialfi by means of one of the axioms. In this case we write f - g . Clearly, f - f , f - g - r g - f and f - g - h - r f - h . If f - g then f K ( a , ,a,, ...) = g K ( a l a2, , ...) for every Boolean algebra Kand arbitrary elements a ] E K. Polynomials resulting from the expression f l A f 2 A ... A fk (or from the expression fl ~ f A2... A&) by an arbitrary placement of parentheses are mutually transformable into each other by means of axiom (2) (or axiom (6)). For this reason we shall always omit parantheses when writing such polynomials. Moreover, we shall not take notice of the difference in the order of polynomials to which we apply successively either one of the symbols A or A . THEOREM 1: Every polynomial is either transformable into o or into some polynomial of the form s1 s2 ... A sh, where each of the polynomials sj has the form xi,A xi,A ... A xit (il < i2 < ... < i t , t 2 l), and no two components sj, sk (1 \< j < k h) are identical.') PROOF.The theorem is clear for polynomials (i) and (ii). Assume that it holds for polynomials f and g . Iff o (or g o), then (f)A (g) - g and ( f ) A ( g ) o (or (.f) A (g) f and ( f ) A ( g ) 0). At this point we may assume that f - s , A s2 A ... s h and g - t l A t2 A ... A t k . Thus ( f ) (g)"sl A s2 A Ash t l t2 ..* tk. By applying (3) and (4) we eliminate all redundant occurrences of components and thus obtain (f) ( g ) in the desired form. The theorem, tberefore, holds for formula (iii). In the case of polynomial (iv) we apply axioms (9) and (5) and

<

-

-

N

--

l) Theorem 1, as well as Theorem 2, is a scheme: for each polynomial f we obtain a separate theorem.

36

I. ALGEBRA OF SETS

obtain

(n ( g ) A

sh) A

“(Si

ti]

Sh) A tk1

[($I

A (%IA tk).

(S1 A ti) A A ( s p A tq) By means of ( 5 ) and (8) each of the polynomials spA tq is transformable into the product of individual variables. Omitting as in the previous case redundancies we obtain the desired form. The theorem, therefore, holds for formula (iv). Q.E.D. THEOREM 2: Let K be the &Id of all subsets of the non-empty set 1. I f f is a polynomial such that 1( f o ) , then there exist sets Al ,A Z ,... belonging to K such that fK(A1, A 2 , ...) # 0. PROOF.By Theorem 1 we may limit ourselves to consideration of the case where f has the form s1 As2 ... A sj, and where each of the polynomials sj is a product of letters x i . Let n be the number of distinct letters xi occurring in f. We shall prove the theorem by induction on n. For n = 1, f - x i , thus we may choose any non-empty set for the set A i . Assume that the theorem holds for all numbers less than n and that the polynomial f contains exactly n distinct variables. If one of the variables x, occurs in each of the expressions s j , then f x, A g where g contains less than n variables. By the induction hypothesis there exist sets A l , A z , ... such that gK(A1, A Z ,...) # 0. Replacing A , by 1 we leave the value of g unchanged (because g does not contain xp) and we obtain the set 1 n gK(A1, A z , ...) # 0 as the value off. If none of the letters x p occurs in each of the s j , then we substitute in f the symbol o everywhere for some arbitrary x,. Thus we obtain the polynomial g of fewer variables than f and l ( g 0 ) . Hence in this case the theorem follows from the induction hypothesis. 3 : Every equation f = g which is true for arbitrary sets (and THEOREM even for arbitrary subsets of a given non-empty set) is derivable from axioms (1)-(9). PROOF.If the polynomial f n g has the value o for all Al,A z , ... contained in a non-empty set 1, t h e n f a g - o and thus f - g . Therefore polynomial g arises from f by transformation by means of axioms (1)-(9) and by the general rule of logic which states that equal elements may be substituted for each other. e . 0

..a

-

-

-

31

9. BOOLEAN ALOEBRAS

Theorem 3 shows that the equations derivable from axioms (1)-(9) are identical with the equations true for arbitrary sets. Moreover, this theorem provides a mechanical procedure for deciding when an equation of the form f = g is derivable from axioms (1)-(9). Namely, it suffices to reduce the polynomial f A g by the method given in the proof of Theorem 1 and to determine whether or not it is transformed into 0. We introduce an order relation in Boolean algebra by the definition: a

< b = a ~b

= a.

THEOREM 4: a < b = a v b = b. PROOF.If a r\b = a, then a v b = a A b A ( a h b) = an b A a = b. Conversely, if a v b = b , t h e n a A b A ( a h b ) = b . T h u s a A b A b A ( a Ab) = b A b = o and a A o A ( a r \ b ) = 0, whence a A a A ( a h b ) = a A o = a and o A (a A b) = a; that is, a A b = a. We call an element i of the Boolean algebra K a unit of K if (10)

aAi=a

for all elements a E K. It is easy to prove that a unit, i f it exists, is unique. In an algebra with unit we define the complement of the element a by the equation: -a = i A a.

Axioms (1)-(9) are very convenient in most calculations but are seldom used to describe Boolean algebra. In the next theorem we shall present a different system of axioms, which is usually taken as the basis of Boolean algebra. We shall limit ourself to the consideration of Boolean algebras with unit. THEOREM 5: If K is a Boolean algebra with unit, then the following equations hold for all a, b, C E K : =bA

(i) a v b = b v a ,

(i’) a A b

a,

(ii) a v (6 v c) = (a v 6) v c,

(ii’) a A (b A c) = (a A b ) A c,

(iii) a v o = a,

(iii’) a A i = a,

(iv) a v -a = i,

(iv’) a A - a

= 0,

(v) a A ( b v c ) = (aAb)v(CtAc), (v‘) a v ( b A c ) = ( a v b ) A ( a v c ) .

38

I. ALGEBRA OF SETS

PROOF.Equations (i), (i’) (ii), (ii’), (%), (v), (v’) are true for arbitrary sets and thus are consequences of axioms (1)-(9). Equation (iii‘) is identical with (10). We establish (iv) and (iv’) as follows:

=o

by the definition of -a, from (9) and (10) from (8) from (4).

=a

A (i A a) A [a A (i A a)] = a A (i A a) A o

by the definition of sum, from (iv)

=(aAa)Ai

from (11, (21, (3) from (4) from (3).

a h - a = a~ ( i n a )

A (aha)

=a

=aAa a v -a

=oAi

= i

A partial converse to Theorem 5 holds: 0 , i E K ,and if v , A , - are operations defined on elements of K satisfying equations (i)-(v’), then K is a Boolean algebra with respect to the operations a A b = ( a A -b) v (b A -a), a A b and the element 0. The proof of the theorem is not difficult and is left to the reader. Equations (i)-(v’) are most often used as axioms for Boolean algebra. In particular, it is worth noting the symmetry of these equations with respect to the operations v and A . We shall conclude this section by giving an interesting example of a Boolean algebra. Let 1 be an arbitrary topological space with a closure operator (see 5 8, p. 26). We call A c 1 a regular closed set if

THEOREM 6: If K is a set,

A = Int ( A ) . By K we denote the family of all regular closed sets contained in 1. Clearly 0 and 1 belong to K since

-

-

Int(0) = 0 = 0

If A E K then

and

-

-

Int(1) = 1 = 1.

A= A , because

- -

A = Int (A) = Int (A) = A .

39

9. BOOLEAN ALGEBRAS

Thus every set belonging to K is closed (see $8, p. 27). By theorem (18), $8, p. 30, it follows that if A, B E K then

- -

A u B = Int(A) u Int(B) = Int(A u B), which proves that A u B EK. For A E K and B EK let

A @ B = Int(A n B),

A’ = Int(-A),

A o B = (A 0 B’) u (BOA‘).

It follows from this definition and from formula (14), $ 8, p. 29 that if A E K and BEK, then A@BEK,A’EKand AoBEK. THEOREM 7: K is a Boolean algebra with unit with respect to the operations o and 0 . PROOF.It suffices to show that the operations u , @ and ’ satisfy axioms (i)-(v’) of Theorem 5. Axioms (i)-(iii) are clearly satisfied. Axiom (i’) follows from the equation A @ B = I n t ( A n B ) = I n t ( B n A ) = BOA. It is equally easy to show that axiom (iii‘) holds: A @ 1 = Int(A n 1) = Int(A)

= A.

To show that axiom (ii’) holds we apply (12) and (8) from $ 8 and obtain Int[(AgB) n C ] = Int(A@B) n Int(C), A 0 B = Int (A n B)

= Int (A) n Int (B) c

- -

Int (A) n Int(B)

= A n B;

it follows by (lo), $ 8 that Int[(A@B) n C] c ( A @B) n C c A n B n C c B n C. (*) Thus Int{Int[(AOB) n C]} c Int(B n C); that is (see (1 l), $ 8), Int[(A@B)nC]cInt(BnC) c I n t ( B n C ) = BgC. Since (by (*)) Int[(A @ B) n C] c A, we have Int[(A 0 B) n C] c A n (BO C), whence we obtain Int{Int[(AgB) n C]} c Int[A n (BOC)],

40

I. M E B R A OF SETS

and hence Int[(A@B) n C ] c Int[A n ( B O C ) ] . Taking closure on both sides of the inclusion we obtain Int[(A @ B) n C ] c Int[A n (B@C ) ] , that is, ( A @ B ) @ Cc A @ ( B @ C ) .The opposite inclusion is obtained in an entirely similar manner. Thus we may consider the equation (A @ B) @ C = A @ (BO C) as proved. We examine axiom (v). We have A @ ( B u C) = Int[A n ( B u C ) ] = Int[(A n B) w (A n C)]. The sets A, B and C are closed, thus (see (9) 58, p. 27) A n B = A n B and A n C = A n C. By (18) 58, p. 30 we conclude that Int[(A n B ) ' u (A n C ) ] = Int(A n B) U Int (A n C) = ( A @ B )u ( A @ C ) . Thus A @ ( B u C) = ( A @ B ) u ( A @ C ) . We check axiom (v') as follows. From the definitions,

--

A u ( B O C ) = Int(A) u Int(B n C). By (18) 58, p. 30, the right-hand side of the equation equals Int[Au ( B n C)], which equals Int[(A u B ) n (A u C)], that is, ( A uB) @ ( A u C). Axiom (iv') is an easy consequence of theorem (17)§8, p. 30 and of the fact that every regular closed set has the form X-+. Finally we prove that axiom (iv) holds. By (18) $8, p. 30 we have A u A' = Int(A) u Int(-A) = Int(A u -A), since = A. Thus we conclude immediately that A u A' = Int (1) = 1. Theorem 7 is thus proved. # Let us take the plane as the space 1. Every circle together with its boundary is clearly a regular closed set. Since every non-empty set of the form Int(A) contains some circle, we conclude that the Boolean algebra of regular closed sets in the plane has the following property : If A E K a n d A # 0, then there exists B such that B E K , 0 # B c A and B # A. #

41

9. BOOLEAN ALGEBRAS

Exercises 1. From every equation written in terms of variables, the symbols o and i and the operations v, A and - we obtain a new equation by interchanging the symbols o and i and the operations v and A. If the original equation is true in Boolean algebra, then so is the equation obtained from it in this way (Principle of Duality). 2. Show that axioms (ii) and (ii’) are derivable from axioms (i), (V), (iii)-(v), (iii’)-(v’). (Huntington)

4 10. Lattices ’) The concept of lattice is more general than that of Boolean algebra. Let L be an arbitrary set of elements, upon which are defined the operations v and A . We say that L is a lattice with respect to the operations v and A if the following equations hold (axioms of lattice theory) (1) (2) (3)

ava=a, avb=bva, av(bvc)=(avb)vc, a A (a v b) = a,

(4) We call a lattice distributive if

( 5 ) a A (b v c) = (a A b) v (a A c),

a h a = a, aAb=bAa, alz(br\c)= ( a ~ b ) ~ c , av(aA6)=a.

a v (b A c)

= (a v b) A

(a v c).

We introduce an order relation between elements o f a lattice just as we did for Boolean algebras: a
(6) or, equivalently,

a

(7)


3

a h b = a.

Similarly we define the elements o and i (if they exist in the given lattice) as the elements satisfying conditions (8) for all a

a v o = a, E

ahi =a

L.

*)A detailed exposition of lattice theory is given in the book: G. Birkhoff, Lattice theory, 2nd edition (New York 1948). See footnote 1) on p. 34.

42

I. ALGEBRA OF SETS

It is easy to show that o is the smallest element in the lattice and that i is the largest, namely,

(9)

oga
for every a E L. Referring to Theorem 5 , p. 37 we observe that every Boolean algebra with unit is a distributive lattice with zero and unit. The converse does not hold, as is shown by the following counter-example which of itself is important for numerous applications in topology. The family of all closed subsets of an arbitrary topological space is a lattice (with the natural interpretation of the operations: a v b = a u by a A b = a n 6). However, this family is not in general a Boolean algebra, since the difference of two closed sets need not be closed (for example, when the space is the space of real numbers). On the other hand, the following theorem holds. THEOREM: If A is a distributive lattice with o and i and if for every a E A there exists an element -a E A satisfying the equations a v (-a) = i, a A ( - a ) = 0, (10)

then (i) the element -a is unique, and (ii) A is a Boolean algebra with zero and unit with respect to the operations v , A , and -. PROOF.Suppose that the element a’ also satisfies conditions (10). T h e n a ’ = a ’ ~ i = a ’ ~ ( a v - a ) = ( a ‘ r \ a ) v ( a ’ ~- a ) = o v ( a ’ ~ - a ) = a‘ A -a. Similarly, - a = --a A a‘, therefore a’ = -a. For the proof of the second part of the theorem it suffices to show that axioms (i)-(v‘) from p. 37 are satisfied. Axioms (i), (i’), (ii) and (ii’) hold in every lattice, (iii) and (iii’) follow from the assumption that o and i are zero and unit in A, (iv) and (iv’) follow from the assumption that condition (10) holds, and finally (v) and (v’) from the assumption that the lattice is distributive. The concept of Brouwerian lattice l) is intermediate between that of lattice and Boolean algebra. We call a lattice with unit Brouwerian if l) Brouwerian lattices were investigated in detail in the paper: J. C. C. Mc Kinsey and A.-Tarski, On closed elements in closure algebras, Annals of Mathematics 47 (1946) 122-162. The term “Brouwerian algebra” was introduced in this paper because of a close connection between these algebras and logic of Brouwer.

43

10. LA'lTICES

for arbitrary elements a, b 6 L there exists an element of L called the pseudo-diference of a and b and denoted by the symbol a*b such that (u*b c ) = (a b v c).

<

<

The family of closed subsets of a given space considered above is a Brouwerian lattice, where the pseudo-difference of two closed sets A and B is the closed set A - B . We denote by * a the pseudo-complement of a, namely * a = i*u. Notice that, in contrast to the operation of ordinary complementation, the equation (*a) A u = 0 does not hold. This corresponds to the fact that the law of the excluded middle does not hold in intuitionistic logic. In the topological interpretation this means that the nowhere dense set X - A n A , namely the frontier of A , is not necessarily empty. On the other hand, the validity of the equation (*a) v a = i corresponds to the law of contradiction in Brouwerian logic. # Examples

and exercises

1. The set of natural numbers is a lattice with respect to the operation of taking the greatest common divisor as the operation v and the least common multiple for the operation A . The formula a b means that b is a divisor of a. The number 1 is the unit of the lattice, and there is no zero element. 2. We consider euclidean n-space P and the family L, of its linear subsets (points, lines, planes and in general, k-dimensional subspaces where k n) passing through the origin. The family L, is a lattice with respect to the operations v and A defined by: A A B is the intersection of A and B; A v B is the least linear subspace of Lrn containing A and B. For example, if A and B are two planes, then A V B is a 3-dimensional spaceif A n B is a straight line, and is a 4-dimensional space if A n B is a point. The relation is the ordinary inclusion relation. The zero element of the lattice L, is the one-point set consisting of the origin and the unit is the entire space L,. The lattice L, is not distributive but is modular. Nsmely. we call a lattice modular if (a < c) + [a v (b A c) = (a v b) A c].

<

<

<

It is worth noticing that in the lattice L, every increasing sequence a, < a, <... contains at most n + l elements. 3. The set of sentences of an arbitrary mathematical theory is a lattice with respect to the operations of conjunction and disjunction when we identify sentences which are equivalent in terms of the propositional calculus. The zero element of the lattice is the false sentence and the unit element is the true sentence. The formula a < b means that a implies b. #

44

I. ALGEBRA

OF SETS

4 Prove that the formula (a

< C)

-+

[a V (b A

C)

< (U

V

b) A C]

holds in every lattice. 5. Prove that the equations given in (5) are equivalent. 6. Prove that the following fonnulas hold in every Brouwerian lattice (see Mc Kinsey-A. Tarski, op. cit., p. 124): (X S y ) - + ( X - r - Z ~ Y i Z ) A ( Z ' - y ~ Z l , ) ,

(x
A

= ( x l y = o),

y) = (ZLX)

( x v y ) 1.2 = (x 5 2)

V

(Z-f-J2),

" ( y ZZ),

7. The family of compact subsets of a topological space is a lattice with respect to the ordinary set theoretical operations u and n. If the space itself is not compact, then the-lattice does not have a unit; on the other hand, the lattice always has a zero (see p. 139 for the definition of compact space).