t 5 l
EQUIVALENT SETS AND CARDINAL NUMBERS
THESIS
Presented to the Graduate Council of the
North Texas State University in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF SCIENCE
By
Shawing, Hsueh, B. S. Denton, Texas December, 1975 Hsueh, Shawing, Equivalent Sets and Cardinal Numbers.
Master of Science (Mathematics), December, 19 7 5 , 38 pp., biblography, 4 titles. The purpose of this thesis is to study the equivalence relation between sets A and B: A o B if and only if there exists a one to one function f from A onto B.
In Chapter I, some of the fundamental properties of the equivalence relation are derived. Certain basic results on countable and uncountable sets are given.
In Chapter II, a number of theorems on equivalent sets are proved and Dedekind's definitions of finite and infinite are compared with the ordinary concepts of finite and infinite.
The Bernstein Theorem is studied and three different proofs of it are given.
In Chapter III, the concept of cardinal number is introduced by means of two axioms of A. Tarski, and some fundamental theorems on cardinal arithmetic are proved. TABLE OF CONTENTS
Chapter Page
I. BASIC PROPERTIES ...... -.-.-...... - 1
II. EQUIVALENT SETS...... 10
III. CARDINAL NUMBERS...... 29
BIBLIOGRAPHY...... -...-.-.-.....-.-.... 38 CHAPTER I
BASIC PROPERTIES
The purpose of this thesis is to study the equivalence relation between sets A and B: A ru B if and only if there
exists a one to one function f from A onto B. The thesis is
divided into three chapters: "Basic Properties," "Equivalent
Sets," and "Cardinal Numbers."
In Chapter I, some of the fundamental properties of the
equivalence relation are derived. Certain basic results on countable and uncountable sets are given and the Principle of
Mathematical Induction and the Axiom of Choice are introduced.
In Chapter II, a number of theorems on equivalent sets
are proved and Dedekind's definitions of finite and infinite
are compared with the ordinary concepts of finite and infinite.
The Bernstein Theorem is studied in detail and three different
proofs of it are given. Some properties of sets of the power
of the continuum are considered.
In Chapter III, the concept of cardinal number is introduced
by means of two axioms of A. Tarski, and a number of fundamental
theorems on cardinal arithmetic are proved.
1.1. Definition. The Cartesian product of two sets A
and B is the set of all ordered pairs (x,y) such that x c A
and y F B. It is denoted by A x B.
1 2
1.2. Definition. Every subset of the Cartesian product
A x B is called a relation from A to B.
1.3. Definition. The domain DR of a relation R is the set DR = {x : (x,y) 6 R}.
1.4. Definition. The range of a relation R is the set
RR = {y : (x,y) R}.
1.5. Definition. A set f is called a function if and only if f is a relation and for every x,y,z, (x,y) 6 f and (x,z) C f imply y = z.
1.6. Definition. A function f is called one to one, if
and only if for every x,x' 6 Df, f(x) = f(x') implies x = x'
(or if for every x,x', (x,y) 6 f and (x',y) 6 f imply x = x'). 1.7. Definition. If f is a function, Df = A and Rf GB,
then fCA x B and f is said to be a function from A into B.
If Df = A and Rf = B, f is a function from A onto B.
1.8. Lemma. If f is a one to one function from A into B
and g is a one to one function from B into C, then h = g(f) is a one to one function from A into C.
Proof. It is obvious that h = g(f) is a function from
A into C. Choose x,x' 6 A, and assume that g(f(x)) = h(x) =
h(x') = g(f(x')). Since g is one to one, then f(x) = f(x').
Now, since f is one to one, it follows that x = x'. Thus h
is a one to one function from A into C.
1.9. Lemma. If f is a function from A onto B and g is
a function from B onto C, then h = g(f) is a function from A
onto C. 3
Proof. Choose c e C. Since g is from B onto C, there exists some b E B such that g(b) = c. Further, since f is from
A onto B, there exists some a F A such that b = f(a), and therefore, h(a) = g(f(a)) = g(b) = c. Thus h = g(f) is a
function from A onto C.
1.10. Theorem. If f is a one to one function from A onto
B and g is a one to one function from B onto C, then h = g(f)
is a one to one function from A onto C.
Proof. This result follows directly from Lemmas 1.8 and 1.9.
1.11. Definition. Let f be a function from A into B and
let S be a subset of A. Define a new function g from S into B
by g(x) = f(x) for x e S. This function g is called the restriction of f to S and is denoted by fI S.
1.12. Theorem. Let A,A2,B B2 be sets such that A1n A2 = q and B1F)B2 2=. (I)1ff isa one to one function from
A onto B1 and f2 is a one to one function from A2 onto B2'
then f = f1 U f2 (a common extension of f1 and f2) is a one to one function from A = A1U A2 onto B = B1U B2. (II) If f is a
one to one function from A onto B and if f(A1 ) = B1 , then
f, = fjA1 is a one to one function from A onto B1 and f 2 = fjA2
is a one to one function from A2 = A - A1 onto B 2 B - B1 .
Proof. (I) Assume the hypotheses of (I). f(A) = f(A 1 U A2 )
Sf (A) U f (A2) = f1(A) U f2(A2) = B1 U B2 = B. Hence f is a
function from A onto B.
It remains to prove that f is one to one. Choose x1 ,x2 4
A and assume that f(x 1 ) = f(x2 ). Since f(A1 ) Af(A2 ) 1 2
= $, either both x1 and x2 are in A1 or both are in A2'
Using the one to one property of f1 or of f2 x2 =x(2 ) follows.
The proof of (II) is similar and is therefore omitted.
1.13. Definition. If A and B are sets, A is equivalent to B, denoted by A % B, means that there exists a one to one function from A onto B. I 1.14. Theorem. (I) A n A. 1 f f- (II) If A % B then B 'A. f g(f) (III) If A - B and B C then A C.
Proof. (I) and (II) are obvious, and (III) follows by
Theorem 1.10.
1.15. Notation. Let N = {1l,2,3,...,n,... 1, the set of natural numbers, and, for every n N, let N = {x : x 6 N
and 1 x--n} = {1,2,3,...,n}. 1.16. Definition. A set S is finite means that either
S = g or, for some n 6 N, S -v N n 1.17. Definition. A set S is denumerable means that
S N.
1.18. Definition. A set S is countable means that either
S is finite or S is denumerable.
1.19. Definition. A sequence of elements of a set B is
a function from N into B and may be denoted by xn>. The range
of the sequence
inductive if n e S implies that n + 1 S.
1.21. The Mathematic alInduction Principle. If S is an
inductive set and 1 e S, then S is the set of all natural numbers, that is, S = N.
1.22. Theorem. The following are equivalent:
(I) A set S is countable;
(II) A set S is either empty or the range of a sequence.
Proof. (I) implies (II). If S is countable, then either
(1) S = $, or (2) St N for some n 6 N, or (3) S N. If (1), n _1 the result is obvious. If (2), N S. Choose s e S and n1 define an extension of f~1 , say h, as follows: -.1 h(m) =f (m) if m c N ; h(m) = s if m 6 N but m i N. 1n Then h(N) = h(Nn) - -(Nn) = S, that is, h is a sequence whose
range is S. If (3), f~1 is a one to one function from N onto
S, and, therefore, f~1 is a sequence whose range is S.
(II) implies (I). Assume S satisfies (II), that is, S is
either empty or S is the range of a sequence. If S is empty,
the result follows immediately. If S is the range of a sequence
f, then S = f(N) = {X1 lx,x...,)Xn 2 ,. where xn = f(n). Clearly S is not empty. Case (1): S has finitely many distinct
elements. In this case, it follows by Definition 1.16 that
S ", N for some n E N. Case (2): S has infinitely many distinct n elements. Define a function g : N + N as follows. Let g(l) = 1.
Assuming g(n) is defined, define g(n+l) = m where m is the
smallest positive integer such that x. x for i<\g(n). :i m 6
Such an integer m exists since S is infinite. By the Principle
of Mathematical Induction, g is defined on N. Now, define a
function h N + S so that if n p N, h(n) = f(g(n)) = xgn)
Then xg(n) i for i< g(n). Hence g(n+l) >g(n) and xg(n+l) Xg(j) x for i and h is one to one. Further, S = {xx 2 ))...,,g( X ,g(n) = h(N), that is, h is a one to one function from N onto S. Hence S is denumerable and therefore countable. 1.23. Theorem. The set I of all integers is denumerable. Proof. I = NUN U{O}, where N = {-n : n c N} is the set of all negative integers. Define a function f from I onto N as follows: f(O) = 1; if n 6 N, f(n) = 2n; if n 6 N, f(-n) = 2n + 1, It is nearly obvious that f is onto N and one to one, so the details will be omitted. 1.24. Remark. If A and B are disjoint denumerable sets then AUB is denumerable. The proof is omitted since it is essentially the same as the proof of Theorem 1.23. 1.25. Lemma. The set Q+ of positive rational numbers is denumerable. Proof. Let E = {m/n : m,n c N}, the set of all rational numerals which represent positive rational numbers. Display the elements of E as follows: 1/1 1/2 1/3 1/4 ... 2/1 2/2 2/3 2/4 ... 7 3/1 3/2 3/3 3/4 ... 4/1 4/2 4/3 4/4 ... Use Cantor's first diagonal process to arrange E in sequence form: <1/IjL/2,2/1,1/3,2/2,3/1,1/4,2/3,3/2,4/1, ...>. Let ay, a2, a3, a 4, a 5,a 6,.. <1/1,1/2,2/1,1/3,2/2,3/1,...>. Delete a if a = a for some m< n. The resulting subsequence n n m is <1/1,1/2,2/1,1/3,3/1, ...>, and its range is Q {l/,l+/2, 2/1,1/3,3/1,... }. By Theorem 1.22, Q is countable. Since Q is not finite, Q is denumerable. 1.26. Theorem. The set Q of all rational numbers is denumerable. Proof. Let Q = Q+UQ U{O}, where Q+ is the set of positive rational numbers and Q~ is the -set of negative rational numbers. By Lemma 1.25, Q+ N. Also Q n Q . Hence, by Theorem 1.14, Q N. By Remark 1.24, Q is denumerable. 1.27. Remark. If A and B are denumerable sets, then A x B is denumerable. The proof is omitted since it is essentially the same as that of Lemma 1.25. 1.28. Theorem. Every subset of a countable set is countable. Proof. Let S = {sn} be a given countable set and let AC S. If A is empty, then A is countable. If A is not empty, choose x _ A, and define a new sequence 1.29. Theorem. If f is monotonic on an interval (ab) then the set of points of (a,b) at which f is discontinuous is countable. Proof. Assume f is increasing. Let E be the set of points at which f is discontinuous. For every x 6 E, there exists a rational number r(x), such that f(x~) Define H : E +-Q by H(x) = r(x). Ifx x2 2 E and x< X2' then f(x+) < f(x ) and thus H(x) H(x2), so H is one to one. The range of H is a subset of rational numbers. By Theorems 1.22 and 1.28, E is countable. 1.30. Definition. Two intervals are said to be non-overlapping if they are disjoint or if they abut. 1.31. Theorem. Every set C of non-overlapping intervals is countable. Proof. If I c C, choose a rational number r = f(I) in I so that r is not an endpoint of I. Let Q = {r r = f(I), for some I F C}. Then Q1 is a subset of rational numbers Q, that is, Q1 C Q. By Theorem 1.28, Q, is countable. Let f : C + Q . It is trivial that f is onto. For, if I ,JO denote, respectively, the interior of I,J, I # J implies I0 and JO are disjoint. Thus, since f(I) E 10 and f(J) 6 J0 f(I) 1 f(J), so f is one to one. Hence C is countable. 1.32. Theorem. If is an infinite sequence of real n numbers in [0,1], there exists a real number x E [0,11 such 9 that x a for every n E N, that is, x F {a } = range of . n n n Proof. Let a1 = 0. a1 1 a1 21a 3 ... aln. a2 =0.a a21 a22a23. a2n a = aa . 3 0.a31 a3 2 a3 3 . a3n ...... a =-O.a a a ...an..., f l Tln2 n3 nn where a. c [0,1], i = 1,2,...,n,...., and a. = 0,1,2,...,9, 1 ij i,j = 1,2,3,...,n,.... Use Cantor's second diagonal process to choose x =0.b b ...b ... [0,1] where b. = 3 if a.. = 2, 1 2S2 n 1i b. = 2 if a # 2. 1 l Hence x f{an = range of 1.33. Theorem. If The proof of this theorem follows from the completeness property of real numbers, and is omitted. It will now be instructive to obtain a second proof of Theorem 1.32 by using Theorem 1.33. 1.34. Theorem. [0,11 is not countable. Proof. Suppose [0,1] is countable, that is, [0,1] = {x ,Xx, ... ,X,. , using Theorem 1.22. Divide [0,1] into 12) 3 n three closed subintervals of equal length. Choose the closed subinterval with the largest right end point which does not contain xl, and denote this interval by I,. Now trisect I, and choose the subinterval of Il not containing x2 with 10 the Largest right end point. Continue this process. A sequence > is defined such that ,IDID... ~DI~k.. and x1 g I, < n1 2 k I , .. x ' I, ..x,. By Theorem 1.33, there0 exists a 2 2 2 n n x [0,) such that x c I for every n c N. unique real number n Since x v I , x x for every n s N. Thus x [o,1J, a n n n contradiction. 1.35. Axiom of Choice. For every non-empty family F of non-empty sets, there is a function f defined on F such that f(A) E A for each set A F. The function f is called a choice function. 1. 36. Countable Axiom of Choice. Every countable family of non-empty sets has a choice function. 1. 37. Axiom of Choice for One Set. Every family of one non-empty set has a choice function. Obviously, 1.36 implies 1.37. CHAPTER II EQUIVALENT SETS 2.1. Definition. If A and B are sets, A is equivalent a one to one to B, denoted by A ' B, means that there exists function from A onto B. 2.2. Theorem. If A % A', B % B', AnB = , and A'nB' = then AUB A'u B'. Proof. This result follows immediately from the definition of equivalence relation and Theorem 1.12. 2.3. Theorem. Let A and B be sets, A = U A where the n=l n00 A are disjoint subsets of A, n = 1,2,..., and B = U B where n n=l n the B are disjoint subsets of B, n = 1,2,.....If An n B for nn n every n E N, then A o B. f Proof. Now An %B for every n 6 N. Define h : A-+ B by h(x) = fn(x) if xe An, n = 1,2,3,.....It is easy to prove h that h is one to one and onto, and thus A % B. 2.4. Theorem. If A % A' and B o B' then A x B r, A' x B'. f Proof. By assumption A % A' and B r B'. Define E A x B. h : A x B + A' x B' as h((x,y)) = (f(x),g(y)) where (x,y) Then h is one to one and onto, and, therefore, A x B A' X B'. B 2.5. Definition. A ={f f B + A). 2 ={f f B + {0,1}}. 2.6. Theorem. 2A % P(A) for every set A, where P(A) is the power set of A. 11 12 A Proof. 2 = {f : f : A + {0,}} and P(A) ='{B B CA}. Let BCA and fB(x) = 1 if x B, fB(x) 0 ifEx A - B. Define a function g P(A) + 2A by g(B) = fB. It may be proved that g is one to one and onto. Hence 2 - P(A). B B' 2.7. Theorem. If A % A' and B B', then A A' f g B B Proof. Since A % A' and B P B', if h c A , then foh e A' -1 B'B B' -1 and fohog~ 1 A's. Define : A + A' by e(h) = fohog B B for h A . e is one to one, since h1 ,h2 c A and e(hi) -1 -1 e(h2) means foh og = foh2og , and this implies h1 = h2. B' - B For every h' 6 A' , define h = f oh'og. Then h c A and e(h) = fohog~ 1 = fof~ oh'ogog = h', so that e is onto. B B' This implies that A A' BUC B C 2.8. Theorem. (I) If B l C = p, then A o A x A C C C (II) (A x B) %A x B BC B x C (III) (ABC A BUC Proof. (I) Let f c A . If x c BUC, f(x) s A. Define fIB the restriction of f to B, that is, if x c B, fIB(x) = f(x); likewise, define fIC, so that if x c C, fjC(x) = f(x). Then BB C fjB c AB, and hence (fjB,ffc) AB x AC. Define e(f) = (fjB,fjC); BUC B C then e is a function from A into A x A Now suppose f,g c ABUC with f # g. Thus there exists some x0 c BUC, so that f(x0) 0g(x). It follows that, if X0 F B, fjB(x0 ) # gjB(x0), and if x0 6 C, fjC(x0 ) gjc(x0 ). In either case, e(f) = (fIB,fIC) # (gfB,gjC) = e(g). Thus e is one to one. To see that e is a function from A onto A x A . 13 B C choose an element (f ,f ) c A x AC. Define a function f from b c BU C into A as follows: f(x) = fb(x) if x s B; f(x) = f (x) if x E C. c It follows readily from the definition of e that e(f) = (fjB,fjC) = (f ,f b'c~). (II) If f c (A x B) , then f is a function from C into A x B. Thus, if x F C, f(x) = (fA(x),fB(x)) where fA(x) A C and fB (x) c B. Thus if f c (A x B) , define e(f) =(fA'fB C C C C where fA A and f s B . Hence, e(f) E A x B , that is, A B C C C o is a function from (A x B) into A x B . Now choose f,g (A x B) with f g; then there exists some x0 F C, such that f(x) g(x0). That is, (fA( ) (X)0 (gA(0 'gB(0)) and this means either fAA(x0) gA(x 0 ) or (gAg B) =e(g). fB (x0) g(x). Hence e(f) = (fA fB A Thus e is one to one. Choose an element (fa' f) A x B . Then f = (fa' b) is a function from C into A x B. It follows from the definition of o that e(f) = (f ,f) = (fa ,f) So e is a function from C C C (A x B) onto A x B (III) The proof is similar to that of (I) and (II) and is omitted. 2.9. Definition. A set S is ordinary finite if S is empty or there exists a natural number n such that S n Nn. Note that ordinary finite is called finite in Chapter I. 2.10. Definition. A set S is ordinary infinite if S is not ordinary finite. 14 2.11. Definition. A set S is Dedekind infinite if it has a proper subset S'CS such that S U'' . 2.12. Definition. A set S is Dedekind finite if S is not Dedekind infinite. 2.13. Lemma. If a set A is denumerable, then there exists a proper subset A' of A such that A " A'. Proof. Let A = {a a2 , .... ,an,...} be a denumerable set. Then A' = {a 2 ,a3 ,...,an,...}cjA and A' # A. Define f : A + A' by f(a.) = ai+ 1 , i = 1,2,3).... It is obvious that f is f onto and one to one; hence A u A'. 2.14. Theorem. S is Dedekind infinite if and only if S has a denumerable subset. Proof. Assume S has a denumerable subset A. Then S = A U(S - A). By Lemma 2.13, A n A' where A' is a proper subset of A. It follows by Theorem 2.2 that S = A U(S - A) - A' U(S - A). Hence S is Dedekind infinite. Now assume S is Dedekind infinite, and let f be a one to one function from S onto a proper subset S' of S and let a1 5, Sz S'. Let AI1 = {a1 }. Let a2 = f(a1). Since a2 S', a2 # ay. Let A2 = {aa2}. Then A2 2%N2 Let a3=3 f(a) Since a3 36S', a3 # a1. Also, since a2 # a1 and f is one to one, a3 = f(a2) # f(a1 ) = a . Let 2 A3 = A2 U{a 3 } = {a1 ,a2 ,a3 }- Then A3 consists of distinct elements of S and A3 N . 3 3' Continue this process. Assume A = {a1 ,a2 ,. . . ,a} such that a = f(ak-),2 = {al, a ,... .a }. It is immediate that ak = f(a ) for 1' 2 n+l k-I 2< k< n+l. Also an+l s S' and a rn+lVA . In the first place, a n a . Secondly a j a for 2< k< n. For, if an+l = a n+l 1 n+l k +1 k for some k, 2< k< n, then f(a ) = f(a ) and hence a = a . -- n k-i n k-i This means that a = a for some h, l< h< n - 1, a contradiction. n h Therefore, by the Mathematical Induction Principle 1.21, the set A = {a1,a,a,...,a ,...} is defined and is denumerable 1 23 n subset of S. 2.15. Lemma. If A v B, A' is a proper subset of A and A' r A then there exists some proper subset B' CB such that B ' B'. Proof. Consider the following diagram: f A+ B -1 g + + hogof = A' B' Define the function h, the restriction of f to A', that is, if x c A, h(x) = f (x) . Choose B' = {h(x) : x e A'}. Then 0 B o B'. 2.16. Theorem. There exists no Nk such that if Sk is a proper subset of Nk then Nk Sk Proof. Let G be the subset of N for which this theorem holds. If k = l, S1C N and S1 N1 then S =$. It is trivial that S N . Thus 1 E G. 1 1 Assume k e G and k + 1 V G; then there exists S c N with Sk+1 Nk+1 and Sk+1 v Nk+l. Consider the following diagram: N : 1 2 3 ... k k+1 f +k+l + + + + + S : s s s ... s s k+Ii 1 2 3 k k+l 16 where f(i) = s , i = 1,2,3,...k+l. i Case (1): If sk+l = k+ 1, then Skk = {ss1 2 ,. ., skk c Nkk S N and Nk S , and this contradicts the assumption k k' kk the k e G. Case (2): If s = i, where 1< i< k, exchange k+l to the position of s. and s.4 ~ in the diagram, and go back case (1). So it is impossible for k + 1 v G. Hence if a natural number k c G, then k + 1 6 G. By the Mathematical Induction Principle 1.21, G = N. The theorem follows. 2.17. Theorem. If a set S is Dedekind infinite, then S is ordinary infinite. Proof. If S is ordinary finite, then S = $ or S - Nn for some n c N. By Theorem 2.16, S is Dedekind finite. 2.18. Theorem. If a set S is ordinary infinite, then S is Dedekind infinite. Proof. Let S be an ordinary infinite set. Then by definition $, by 1.37, there S # p and S /4Nk for every k e N. Since S # exists a function f1 such that f1 (S) = x1 e S. Further, Sl therefore S x N , a S - {x1 } f $, for otherwise S = {x1 } and 1 contradiction. Again by 1.37, there exists a function f2 such that f2 (S1 ) = S 1. Then x2 i, and S2 ='-2 , a contradiction. . For otherwise, S = {x1 ,x 2 1} and. S n N2 distinct Continuing this process, assume x1 ,x2 ,.. . ,xk e S are all elements. ThenSk = S - {x 1 x2 ' . xk} # 4, for otherwise By Theorem S = {xx2' ... xkx} and S Nk a contradiction. 1.37, there exists a function fk+l such that 17 (S ) = x E S . Using the Mathematical Induction f k+1 k k+l k '''kxk+1'*.1, Principle 1.21, the denumerable set A = {x11 x2, . of distinct elements xk of S is defined. By Theorem 2.14, it follows that S is Dedekind infinite. 2.19. Definition. If A and B are sets then A 4 B means that A ' B for some B CB and B r A1 for every A1 CA. 2.20. Remark. In the sequel A 'B 1 for some B CB will be written A ' B1C B. 2.21. Definition. If A and B are sets, then A ( B means either A B or A B. 2.22. Theorem. If A2 AC A0 and A 0 A2' hen A0 A). Proof. By the hypothesis, A f A2 , that is, there exists a function f : A ~* A2 which is one to one and onto. Since f A AO, there exists a subset A3C A2 such that A1 1%A3. In fact, A3 = f(A). This may be abbreviated by A A3C A2 f It follows by 1.12 that CO = A0 - A1 ' A2 - A3. By the Mathematical Induction Principle 1.21, the sequence of relations follows: A0 A2 A 1 A3C A2 CO A0-A1 A2 -A3 C2 A ACA C =A -AoA-A =C 2 4 3 1 1 2 3 4 3 A 2 ACA 3C= A - A2 A - A G=C 3 5 4 2 2 3 4 5 4 A % A C A C =A -A A-A = n-i n+1 n n-2 n-2 n-i n n+1 n A A CA C = A - A n A - A = C n n+2 n+1 n-l n-l n n+l n+2 n+l ...... 18 00 Let D =n A Then n=0 n A = DUC 0 UC1 UC2 U...UC UC 2 UC+U.-.. and 0 0 1 2 2n-I 2n 2n+l A = DUC1UC2U...UC UC U C Un---. 1 1 2 12n 2n+l 2n+2 Now, D o D and C b C , n = 0,1,2,..., where I is the 2n+l f2n+l identity function; C f C , n = 0,1,2, ..... Let g = I on 2n % 2n+2 D and C 2 , g = f on C2, n = 0,1,2, ..... By Theorem 2.3, A2n+12n'g A0 A1 . 2.23. Theorem. If A and B are sets, the relations A nu B, A - B, and B -' A (or A 'rB) are mutually exclusive. Proof. The result follows immediately from the definitions of the relations % and-'. 2.24. Theorem. (I) If A o B then A B; (II) If A 4 B and B 4 A, then A r B. Proof. (I) This follows directly from the definitions of A % B and AAB. (II) follows from Theorem 2.23. 2.25. Theorem. The following are equivalent: (I) If A and B are sets and if A B CB and B % A c A, then A CB. (II) If A cA1CA and if ACA2, then A % A . (III) If A1 CA, then A 1 -A. Proof. (I) implies (II). Assume A2C A1C A and A A2 Then A1 % A CA. By assumption, A o A2 cA .'By (I) A 1 (II) implies (I). Assume A % B1c B and B % A1C A. Then B A CA CA. Using Theorem 1.14 (III), A A . Using the 1 2 1 2 result (II), it follows that A1n A. Using Theorem 1.14 (III) 19 again, B r A; thus A , B. (I) implies (III). Assume that A1 CA. Then A1 A1C A. there exists some set A CA so that A - A then by (I) If 2.. 1 2'0 it follows that A1u A and hence A1 A. Otherwise, for every that A2 CA1 , A r A2. In this case, it follows by definition A A and thus A A. (III) implies (I). Assume (III) and let A A B ,B be sets such that (1) A B. C B and (2) B - A CA. By (III) and (1), B B, and since ArB , A ( follows by (II) of Theorem 2.24 that A o B. 2.26. Bernstein Theorem. If A and B are sets, A nu. B CB and B % A1CA, then A B. First proof. The theorem follows immediately by Theorem 2.22 and Theorem 2.25. f Second proof. Let A J B1 C B and B A1 C A. Suppose that neither f nor g is onto; for otherwise, the theorem follows. Now B1 = f(A), A1 = g(B), f is a one to one function from A onto 1 B1 , and g is a one to one function from B onto A1 . Thus, f -l is a one to one function from B = f(A) onto A and g is a one to one function from A1 = g(B) onto B. The theorem will follow if a function H: A B can be defined which is onto and one to one. For the purpose of defining such a function H, consider the following decomposition of A into three disjoint sets, AVAaAb' where the union is A (Birkhoff and Maclane, p.340). 20 Let a F A. Then a is called the zeroth ancestor of itself. If g~1(a) F B exists, then g (a) is called the first ancestor of a. If f~g~ 1 (a) exists, then f~1 g~1 (a) 6 A is called the second -l -1-1 -l -1 -1 ancestor of a. If g~ f~ g (a) exists, then g1f g (a) F B is called the third ancestor of a. Continue this process. Either (1) a has an infinite number of ancestors, or (2) a has a last (nth) ancestor. If (1) holds, let a F A . If a has a ancestor, let a c A or a e A according as this last last a b ancestor is in A or in B. Now, let b c B, b is its zeroth ancestor, and let f~1(b), if it exists, be the first ancestor of B. Continuing in a similar fashion to the last paragraph, decompose B into disjoint sets BBaBb. Define a function H : A + B as follows: H(a) (a) if a 6 A U A = f~ I a' H(a) =gg1(a) if a c Ab* Notice that the restriction of f (and H) to A is a function onto BI and is one to one since f is one to one. The restriction f (and H) to A is onto B and is one to one since f is one of a a to one. Now the restriction of g~ (and H) to Ab is into Bb. -l Also this restriction is onto Bb and one to one since g is a one to one function from A onto B. Hence by Theorem 1.12, H H is a one to one function from A onto B, that is, A % B. Third proof. In order to show a derivation of the proof, it is broken into three parts (1), (2) and (3). In each of the f parts, the hypothesis of the theorem will be assumed: A % B CB and B % A1 CA. 21 (1) If there exists a set AOCA such that g(B - f(A0)) -A - A then A nuB. Assume A CA and g(B - f(A )) = A - A Define a function 0 0 0 0 h : A-+ B as follows: h(x) = f(x) for xs6 A ' h(x) = g~ (x) for x E A - A0 ' Since f is a one to one function from A onto f(A) C B, then the restriction of f and h to A is 0 a one to one function from A0 onto f(A0 ). Further, since g is a one to one function from B onto g(B)CA and g(B - f(AO)) = A - A0 the restriction of g- and h to A - A0 is a one to one function from A - A0 onto B - f(A0). By Theorem 1.12, h is a one to one function. from h A onto B, that is, A B. (2) There exists a smallest subset A' of A which satisfies the properties (I) A - g(B)cA', (II) (gof) (A') CA'. It is readily seen that a set A' which satisfies the hypothesis (1) necessarily satisfies properties (I) and (II). For, if g(B - f(A')) DA - A', then g(B)2A - A', and hence A - g(B)C A'. Further, since g(B - f(A')) = g(B) - g(f(A')), g(B - g(f(A')) DA - A', and taking complements, (A - g(B)) Ug(f (A')) CA' and thus g(f(A')) CA'. Now, let G be the collection of all sets A'c A which satisfies (I) and (II) and AO = f A' for A' c G. Certainly A0 satisfies (I). For every A' c G, g(f(A0))cg(f(A'))CA'. Hence, g(f(A))c A0' and A0 satisfies (II). Thus, A0 satisfies (I) and (II), and is the smallest such set. 22 (3) A as defined in (2) satisfies the hypothesis (1), that is, A0C A and g(B - f(A )) =A - A' The following set identity will be first verified: (A - g(B)) U (gof)(AA) =A0 . Since A0 sG, A- g(B)cA 0 and (gof)(A 0 )cA0 ,and thus (A - g(B)) U (gof)(A)CA0 . Thus (I) A - g(B) c (A - g(B)) U (gof) (A0), and, (II) (gof) I(A - g(B)) U (gof) (A0)J c (gof) (AO) c (A - g(B)) U (gof) (A0). Hence, (A - g(B))U (gof) (AO) E G, and, therefore, (A - g(B)) U(gof)(A0)DA0. The above set identity follows: (A - g(B))U (gof)(A0 ) = A0 . Now, using one of De Morgan's theorems, (A - g(B))fn(gof)(A0 ) =A'0 ; g(B) r)(go f) ) = A - A0 g(B) - (gof)(A0) = A - A0 ; and therefore since g is one to one, g(B - f(A0 )) = A - A0 ' 2.27. Theorem. If A is a set and A1 c A, then Aj A. Proof. The theorem follows immediately by Theorem 2.26 and Theorem 2.25. 2.28. Theorem. The following are equivalent: (I) If A and B are sets then A 4 B, that is, A * B C B and B 4 A for every A1C A. (II) If A and B are sets, A n B1C B and A 4 B. Proof. (I) implies (II). A 4 B means A B C B and for every A1C A, B l A Take A = A Thus A z B. (II) implies (I). Assume (II) and not (I). Then A % B1C B, A L B, and there exists some set AC1 A for which B % A11 . 23 By Theorem 2.26 it follows that A % B. This is a contradiction, and the theorem follows. 2.29. Theorem. (I) If A A B and B A C, then A-- C. (II) If A %,Band B 4 C, then A A C. (III) If A A B and B ' C, then A C. (IV) If A - B and B C, then A i C. Proof. (I) Assume A A B and B 4 C. Then A r' B CB, B ' C1 C C, and C 4 B' for every B' C B. Hence, by 1.14 and 1.12, A u C2 CC CC. Also, C ', B1 and therefore C i, A. By Theorem 2.28, AR4 C. (II) Using (I), it is sufficient to prove that if A % B and B - C, then A- C. Assume A * B and B -A C. By Theorem C CC and 2.28, B % C CC and B rZ C. By Theorem 1.14, A 1 A l C. Thus, again by Theorem 2.28, A ' C. (III) The proof of (III) is similar to that of (II) and is omitted. (IV) follows by (II), (III), and 1.14. 2.30. Theorem. A ( B if and only if A % B1 CB. Proof. Assume A B. Then by definition A '' B C B. Assume A% B C B. Using Theorem 2.30, B 4B. By Theorem 2.29 (IV), it follows that A 4 B. 2.31. Theorem. If A is an arbitrary set, A 4 P(A), where P(A) is the power set of A. Proof. Let A' = {{a}}: a E A}. Then A A'cP(A). Thus, by Theorem 2.27, A-, P(A). The theorem will then follow if it can be verified that A ' P(A). Assume that there exists 24 a function f such that A t P(A). Let B = {x :x A and x 1 f(x)}. Since BCA, that is, B 6 P(A), there exists a unique element y F A such that B = f(y). Thus, f(y) = Ix x 6 A and x z f(x)}, that is, if x 6 A, then x 6 f(y) if and only if x z f(x). Let x = y. Hence y E f(y) if and only if y z f(y). This is a contradiction, and the theorem follows. 2.32. Theorem,. If a< b, then fa,b] ' [0,1] and (a,b) i (0,1). Proof. If x e[a,b), define f(x) = (x - a)/(b - a). Then f (a) = 0 and f (b) = 1. For a!5 x < b, it follows that 0 5 x - a < b - a and hence 0< (x - a) /(b - a) <1. Thus, if x c [a,b], 0:5f(x) 51 and f is a function from [a,b] into [0,1]. Choose y c [0,1], that is, 0:5y:1. Then 0 ! (b - a).y < b - a and hence a:5 a + ( b - a)y:5 b. Let x = a + (b - a)y then f(x) = y and f is a function from [a,b] onto [0,1]. Let x1 ,x2 c [a,b] and assume that f(x1 ) = f(x2). Then (x1 - a)/(b - a) = (x2 - a)/ (b - a) and it follows that x1 = x.2 f Thus f is one to one and a,b] '[0,1]. Since f(a) = 0 and f(b) = 1, it follows by Theorem 1.12 that (a,b) % (0,1). 2.33. Theorem. (0,1) 'U R, where R is the set of all real numbers. Proof. By Theorem 2.32, it follows that (-1/2,1T/2) (0,1). Define a function f : (-r/ 2,ur/2)-+ R by f(x) = tan x for every x . (-ff/2,'r/2). Assuming the well known property that f(x) = tan x is increasing on (-f/2,f/2), it follows that f is one to one. And since the range of f(x) = tan x is R, f 25 is onto. Thus (-7T/2,iT/2) % R and by (II) and (III) of Theorem 1.14, it follows that (0,1) R. 2. 34. Theorem. (0,1) (0,1] and (0,1) [0 ,1). Proof. Rather than applying the Bernstein Theorem 2.26, it is instructive to give a direct proof of this theorem. From the standpoint of derivation in a classroom setting, it would be appropriate to first prove this special case and then generalize to Theorem 2.22. (0, 1 = (1/2,1]U(1/3,1/2]U ... U (1/n+l, 1/n] u. . . , (0, 1) = [/2,1) U[1/3,1/2) U. . . U [l/n+l, 1/n) U .... Define a function f : (0,1] + (0,1) as follows: f(x) = x if x 6 (1/n+1,1/n) and f(x) = 1/n+l if x = 1/n, n = 1,2,..,. It is trivial to show that f is one to one and onto. Hence (0,1]% (0,1). Similarly, [0,1)-* (0,1). 2.35. Theorem. [0,1] % (0,1). Proof. [0,1] = [0,1/2)U(1/2,1]U {1/2}, (0,1) = (0,1/2) U (1/2,11 U {1/2}. By Theorem 2.34, [0,1/2) n (0,1/2) and (1/2,1] Tu (1/2,1], {1/2} ,{1/2}, where I is the identity function. By Theorem 2.2, it follows that [0,1] % (0,1). 2.36. Theorem. If a (c,d] -,(c,d). [a,b) -- Ea,b) -u R % [c,d] -u (c,d) % Proof. The theorem follows directly from Theorem 2.32, 2.33,2.34,2.35 and 1.14. 2.37. Definition. A set A is said to be of the power of 26 the continuum if A o R where R is the set of all real numbers. 2.38. Theorem. The interval between any two distinct real numbers a and b is of the power of the continuum. Proof. This theorem follows directly from the definition of the power of the continuum and Theorem 2.36. 2.39. Theorem. The union of two sets which are both of the power of the continuum is again a set of the power of the continuum. Proof. Let A and B be disjoint sets which are both of the power of continuum, that is, A-v R and B u R, AflB = By Theorems 2.36 and 1.14, A [0,1) and B % [1,2) where [0,1)fn[1,2) = p. By Theorem 2.2, A UB % [0,1). Hence, by Theorems 2.36 and 1.14, AUB R. Now let A and B be any two sets such that A R, B R. Let A1 = {(a,l) a c Al and B1 = {(b,2) b c B}. Of course fg A % A B B and A flB = 4. Now let B' = g(B - A). 1' 1 1 1 1 Then B - A rb B'CB . Since A f(B - A) = $ and A fB'c 1 1 1 1 A1lB = p, by Theorem 2.2, it follows that AUJB = A U (B - A) u A U B'. Since A f B = $ and A R, B R, the case of 1 1 1 1 1 1 the first paragraph implies A1 UB R. But A cA UB'cA U B 1 11 11 1 and A1 R % A UB . Then by Theorem 2.22, it follows that 1 1 1 A UB R. Therefore A U B R. 2.40. Theorem. If I = (0,1] and S = {(x,y) x c I and y 6 I}, then I % S. f Proof. First, I S CS. For, if x I, let f(x) = (x,1) C S. Then f (I) = S, = {(x,l) x P I} and f is a one to one 27 function. Now choose (x,y) c S. Then 0< x<1 and O< y<1. Each of x and y can be uniquely expressed as a non-terminating decimal symbol: x = .x1x2*23 n.'. ) 2. 3''' n Let written, g((x,y)) = u where u .x yx2Y2 - n n.... Then u, as is a non-terminating decimal symbol and u P (0,1]. Now choose (x',y') 6 S, (x',y') (xy). Then either x' x or y' y. Let x' = .x 1xx3 ...x.. ,y = -yI yy 3...y. some n 8 N, either x' # x or y' # yn. Hence, Thus for n n n n g((x',y')) = u' where u' =..xy x'yI...X'y'. and u # u'. Thus, g is a one to one function from S into I, that is, S 9 I CI. By the Bernstein Theorem 2.26, I S. 2.41. Theorem. The Cartesian product of two sets which are both of the power of the continuum is of the power of the continuum. Proof. Let A and B be any two sets such that A r\,R and B ) R. By Theorem 2.40, it follows that (0,1] x (0,1] 'x (0,1] o R. Using Theorem 2.4, Ax B (0,1] x (0,1] % R. 2.42. Lemma. If f : N + [0,1], then f is not onto and hence N Z [0,1]. Proof. It follows from Theorem 1.32. 2.43. Theorem. If A and B are disjoint sets with A of has the power of the continuum and B denumerable , then AUB the power of the continuum. Proof. A % [0,1] and B Q1(l,2] where Q is the set of rational numbers. Then AUB r [0,1] U(Q n(l,2 ]) = E. But 28 [0,1] C E C [0,2] and by Theorem 2.36, [0,1] % [0,2]. Hence, by Theorem 2.22, E "' [0,2]. Therefore, AUB [0,2] -uR, and the theorem follows. 2.44. Theorem. If A is a denumerable set, then 2A has the power of the continuum. Proof. Using Theorem 2.6, 2A P(A) for every set A. Let A be a denumerable set. Then A u N = {1,2,3,... ,n,. .}. A N N {1,2,3,...,n,... By Theorem 2.7, 2 r1 2N. P(N) u 2 = 2 2 ' ''is the set of all infinite sequences whose <0,1,0,1,0,1,...>. terms are O's or l's, such as .x 1 x 2*3'n.... Further, if x and y are in [0,1] with x y and if x = .x1x2...x ... and y = .y*y ... yn..., then x y 2yn 1y2. ny n n for some n. On the other hand, certain numbers x in [0,1] can be expressed in the form of two distinct binary symbols, one terminating and one non-terminating. For example, the two symbols .0101111... and .0110000... represent the same number. This set T of numbers in [0,1] with two binary representations is certainly an infinite subset of the rational numbers and hence is denumerable, by 1.28. Let B = 1 UB2, whose B1 denotes the set of non-terminating binary symbols and B2 denotes the set of terminating binary symbols. Then B1 o [0,1] and B2 v A T % N. Thus, by 2.43, B - R, and therefore 2 x R. The theorem follows. CHAPTER III CARDINAL NUMBERS A. Tarski introduces cardinal numbers by means of the followings two axioms (Sierpinski, p.13 5 ): 3.1. Axiom. Every set A is associated with an object which is its cardinal number and every cardinal number is the cardinal number of a set. This assumption is similar to Cantor's original idea. Cantor stated that "every aggregate (set) has a definite power which is also called its cardinal number"(Cantor, p.68). 3.2. Axiom. Two sets A and B have the same cardinal number if and only if they are equivalent, in notation, A A) B. 3.3. Notation. Following Cantor (Cantor, p.86), the cardinal number p of the set A is also denoted by A. The symbol was used to denote a double abstraction from the given set A, namely, the nature of its elements and any order that the elements may possess. Also, following Cantor, the cardinal number of N, the set of all natural numbers, is denoted by N = i0 . Further, if R is the set of all real numbers,, the cardinal number of R will be denoted by R = c. 3.4. Definition. Given two cardinal numbers p and q, there exist two disjoint sets A and B, such that A p and B = q. Then by definition, p + q = AUB. Using Axioms 3.1 and 3.2, it can be seen that the sum p + q of cardinal numbers 29 30 p and q is uniquely defined by the above. For, suppose A'and B'are disjoint sets such that p = N~ q = Wr. Then, by Axiom 3.2, A ' A' and B B'. Using Theorem 2.2, A UB -u A' UB', and hence AU=B = A B'. 3.5. Theorem. Let p,q and r be cardinal numbers. (I) p + q = q + p. (II) (p + q) + r = p + (q + r). Proof. (I) By Axiom 3.1, there are sets A and B so that p = A and q = B. Further, it may be assumed that A fB = For otherwise, replace A with A1= { (a,l) a E Al and replace B with B2 = { (b,2) : b c B}. Then A r\ A, B ' B2, and A1lB = $. But A UB = B UA, and therefore, p + q = q + p. 1 2 (II) This result follows from the associativity of the union operation on sets. 3.6. Definition. Given two cardinal numbers p and q, there exist sets A and B, such that A = p and B = q. Then by definition, pq= A X. Using Axioms 3.1 and 3.2, it can be seen that the product pq of cardinal numbers p and q is uniquely defined by the above. For, suppose A' and B' are sets such that p = K' and q = W. Then by Axiom 3.2, A "-'A' and B 'B'. Using Theorem 2.4, A x B b A' x B', and henceA X B =A'xB' 3.7. Theorem. Let p, q and r be cardinal numbers (I) pq = qp; (TI) (pq)r = p(qr); (ITT) p(q + r) = pq + pr. Proof. (I) follows readily from the relation A x B B x A. 31 for any sets A and B. x C (II) Let A = p, B = q anc C = r. (A x B) x C o A x B o A x (B x C). It follows that (pq)r = p(qr). (III) Let A = p, B = q, C = r, and BflC = p. Now Hence A x (BUC) ov (A x B) U(A x C) and (A x B) n(A x C) = $. p(q + r) = pq + pr. 3.8. Definition. Let p and q be given cardinal numbers q ~~ and A and B be sets such that A = p, B = q. Then p = A Then, by definition, pq is the cardinal number of the set AB in notation, p = AB. Now, to see that pq is thus uniquely defined, suppose A' and B' are sets such that KT = p and 2.7, w = q. By Axiom 3.2, A % A' and B ' B'. Using Theorem B - B' it follows that AB A' . Hence, again by Axiom 3.2, A =A'B 3.9.. Theorem. Let p, pl, p2 ' q, q q 2 be cardinal numbers. Then: M q+ q2=q 1 q 2 (I) p =p .p ; (II) (P )q = P -qP2 (III) (P ) 2 qq2 . Proof. This theorem follows readily by Theorem 2.8. 3.10. Definition. Let p and q be cardinal numbers and A and B be sets such that X = p and B = q. Then p < q means As, before, it is readily that A . B and p < q means that A- B. seen that these definitions are identity of the particular sets A and B. 3.11. Theorem. Let p, q be cardinal numbers. Then 32 (I) p (TI) If p < q and q< p, then p = q. Proof. (I) and (II) follow directly from Theorem 2.24. A. 3.12. Theorem. (I) If A2 C A 1 C A and 2 A, then A1 = (II) If B A1 CA, then B< A. (III) If p, q are cardinal numbers then p + q>p. Proof. (I) The result follows by Theorem 2.22. (II) This follows directly from Theorem 2.30. (III) This result is a corollary of (II). 3.13. Theorem. K Proof. The result follows directly by Theorem 2.31. 3.14. Theorem.40 < C. Proof. Using Theorem 2.36, [0,1] % R. Since N A4, R 40,1], by lemma 2.42, it follows that 0 Proof. The theorem follows directly from Theorem 2.41. 3.16. Theorem. (I) If n is a finite cardinal number, then n + = 50' + = +0 (I){ Proof. (I) If n = 1, then there exists a set A = {a} and A = 1, Ar N = c. Define f : AUN + N as follows: f(x) = 1 if x c A; f(x) = x + 1 if x E N. AUN x N. It is trivial that f is one to one and onto, so N, assume Hence AUN = N, that is, 1 +{ }0. Now, - if k c 33 k + = 0. Then (k +1) + 0 = (k + XO) + 1 = X0 + 1 Principl e 1.21, for every 50. By the Mathematical Induction n C, N, n +N0= Io (II) Let I+ denote the positive integers and I- denote the negative integers. I+ fT I N and I~ n N. Define f : I UTI~ N as follows: f(x) = -2x if x 6I~; f(x) = 2x- 1 if x cI It is trivial that f is one to one and onto. So I U I u N, that is, I+UI~ = N. Therefore .0 + 0 O 3.17. Theorem. (I) If n is a finite cardinal number and n 0, then nN 0 { ' (II) 0 1%. Proof. (I) If n = 1, there exists a set A = {a} and A = 1 Define f : A x N -* N by f((a,n)) = n. It is obvious that f is one to one and onto, and therefore A x N o N. Hence N= , that is, -}o. Assume (I) is true for n = k, that is, =k Then (k + 1)j& = k 0 +f0= +N Induction Principle 1.21, for every = }fo. By the Mathematical n N, nX$= 0 KO f 0 (II) This result follows by Remark 1.27. 3.18. Theorem. 4ojc = c. Proof. Let R+ =TI UTIU... UI U ... = U T where 1 2 n n=1 n + R I = (n - 1,n] and I I = $ if n # m. R = (oo,o) % (0,o) = n n m and In R for every n. Since R = 1 UI 2 U'... U In u..., then c = c + C + c + ... = NOC. 3.19. Theorem. If p p + r = q. Proof. Let A = p and B = q. Since p< q, A -< B and this implies A B C B and A 4 B. Then B = B u(B - B1), B - B1 , B n (B-B) B=, B U (B-B- and B=A. So take r = B - B1 > 1. B = B1 + B - B and this means p + r = q. The converse is not true. For, if n is a finite cardinal number, then go = n + gp'o o. 3.20. Theorem. (I) If p< q and q< r, then p< r. (II) If p q and q (III) If p (IV) If p< q and q Sr, then p: r. Proof. These results follow directly by Theorem 2.29. 3.21. Theorem. If p 1 < p2 and q1 < q2 then p1 + ql< p2 + q2 Proof. Let Al= p1 , A2= , B=Bp q , B = q , and A flB =$. 1 12 ~2 1 1 2 q2 2 2 Since p< p ' A1<2, 2 and this means A1Nj A3 CA2 qq1>q 2 ' that is, B1 B2 , and this means B1' B3 GB . Therefore, A U B c A UB and A nB = . It follows by Theorem 3.12 (II) 3 3 2 3 33 that A3 U B true that p1 + q1 < p2 + q2 . For, Hoj< c, c< c, f + c = c, c + c = c, that is, + c< c + c is not true. 3.23. Theorem. If p1 Proof. Let A = p , 2 = q ,1 = q, and B 2 = q2 . Since p1 < p2 , that is=A< A2 it is permissible to assume 1 2 1-B B1 B1 that A1C A . Thus A 1 'CA2 1. By Theorem 3.12 (II), A1 q q that is, p1 1< P2 If = q 2, the theorem is true. If q < q , by Theorem 3.19, there exists some r> 1 such that _'q r q q q+ r = q2 So if q2 q1+r 1 q1 r = 2p 2 22p p22 2 2 2 HenceHenceif ifP2 - q -p2 q 1 3.24. Theorem. 2.9 = c. Proof. This result follows directly from Theorem 2.44. 3.25. Theorem. ( 0 )c = 2c Proof. By Theorem 3.23, 2 c< AJc cc. Using Theorem 3.24, c 9 0 C 0 29 _ c N~O 2 =C, soC = 2c22= 2 c, by Theorems 3.18 and 3.9. Hence c < 2 c and jc > 2 c. Therefore, by Theorem 3.11 (II), Iq 0- 20 (o 0c = 2c. 3.26. Remark. If p1 Proof. Let A = p1 , A 2 = P2 , B1 = q and B2 = q2 . Now assume that A CA and B CB then A x B CA x B2 By 1 2 1 2' 1 BC 2 2' B Theorem 3.12 (III), A x B1 3.28. Remark. If p 1< p2 and q1 < q2 , it is not necessarily true that plq 1 < p2q2 . For, c< c and l< J{ , cl = c, c$ 0 = c, that is, c< c is not true. 3.29. Definition. For every cardinal numbers p and q, the difference p - q exists if and only if there exists a unique cardinal number r such that p = q + r. 3.30. Theorem. If n is any natural number and p is a cardinal number such that p> n, then there exists a difference 36 p - n. Proof. Let n be a natural number and p be a cardinal such that p ,n. By Theorem 3.19, there exists a cardinal number q> 1 such that p = n + q. Suppose r is a cardinal number such that p = n + r. Let A,B,C,M and M' be sets such that p = A, q = B, r = C, N = n = 9r, and BfM= = CfN'. Assume further, without loss of generality, that MUB = A = M'U C. Consider the following decompositions of the sets B and C into two disjoint components: B = A - M = (A - (MUm'))U (M' - M). C=A-M'= (A - (MUM')) U(M - M'). Let u = M-F, V = T' Mand t = M~. Then each of u,v and t is a natural number or zero. Since u + t = N = n and u + t = M~ = n, u + t = v + t. By the cancellation law of the whole numbers (NU{0}), u = v, that is, R - M = F'1-M. Using these two set decompositions together with M M' = M' -, it follows by Theorem 2.2 that q = r. 3.31. Corollary. Let p and q be non-zero cardinal numbers and n c N. If p + n = q + n, then p = q. Proof. Let r = p + n = q + n. Since n e N, r >n. By Theorem 3.30, there exists a difference r - n. Hence p = r - n and q = r - n, and p = q. 3.32. Remark. If n = 90 and p + n = q + n, then p = q is not necessarily true; for instance, 1 + 0 = 2 + 1 and 1 2. 3.33. Theorem. If p is a cardinal number, then p >R0 if and only if p + =p. 37 Proof. Let p be a cardinal number and py f. If 0 and this p= 0 , by Theorem 3.16 (II), 0 +{ 0 %, theorem follows. If P> 0, by Theorem 3.19, there exists r> 1 such that p = .0 + r. Hence p + 40 = (r + + = p. Then r + (J{ 0 +J{ 0 ) = r + ff = p. Now assume p + 0= by Theorem 3.12 (III), p = p+ 0 0 3.34. Theorem. If p and q are cardinal numbers such that p > and q Jjo, and if p + = q +{ then p = q. Proof. If p >J 0 and q >( 0 , by Theorem 3.33, p = p +(T = q + 0 q. 3.35. Theorem. If p is a cardinal number which is not finite, then p > r Proof. Since p is not finite, by Theorems 2.18 and 2.14, there exists a cardinal number r such that p = r + Av' By Theorem 3.12 (III), r +-K >, . Hence, by Theorem 3.20 (IV), p 0' BIBLIOGRAPHY Birkhoff, G. and Maclane, S., A survey of Modern Algeb.ra, New York, Macmillan, 1965. Cantor, M.B., Vorles~ungenkber Geschichte der Mathematik, New York, Johnson Reprint Corp., 1965. Royden, H. L. , Real Analysis, London, The Macmillan Company, 1968. Translated Sierpinski, W., Cardinal and Ordinal Numbers, from polish by Janina~~Smol~E~,sk~Warszawa, 1965. 381, such that 34