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  • 1. 5.1CHAPTER 5CardinalityHow many elements are in the setA = in, 28, ji,!, -3, fl, cr, o}?After a short pause, you said "eight." Right? Consider for a moment how you ar-rived at that answer. You probably looked at n and thought "I," then looked at 28and thought "2," and so on up through 0, which is "8." What you have done is set upa one-to-one correspondence between the set A and the "known" set of eight ele-ments {I, 2, 3, 4, 5, 6, 7, 8}. Counting the number of elements in sets is essentiallya matter of one-to-one correspondences. This process will be extended when we"count" the number of elements in infinite sets in this chapter. Here is anothercounting problem.A certain shepherd has more than 400 sheep in his flock, but he cannot countbeyond 10. Each day he takes his sheep out to graze, and each night he brings themback into the fold. How can he be sure all the sheep have returned? The answer isthat he can count them with a one-to-one correspondence. He needs two containersand a pile of pebbles, one pebble for each sheep. When the sheep return in theevening, he transfers pebbles from one container to the other, one at a time for eachreturning sheep. Whenever there are pebbles left over, he knows there are lostsheep. The solution to the shepherds problem illustrates the point that even thoughwe have not counted the sheep, we know that the set of missing sheep and the set ofleftover pebbles have the same number of elements-because there is a one-to-onecorrespondence between them.Equivalent Sets; Finite SetsTo determine whether two sets have the same number of elements, we see whetherit is possible to match the elements of the sets in a one-to-one fashion. This idea maybe conveniently described in terms of a one-to-one correspondence (a bijection)from one set to another.195
  • 2. 196 CHAPTER 5 CardinalityDEFINITION Two sets A and B are equivalent iff there exists a one-to-one function from A onto B. A and B are also said to be in one-to-onecorrespondence, and we write A =B.If A and B are not equivalent, we write A* B.Example. The sets A = {5, 8, ep} and B = {r, p, m} are equivalent. The functionf: A --+ B given by f(5) = r, f(8) = p, and f(ep) = m is one of six such functionsthat verify this.Example. The sets C = {x, y} and D = {q, r, s} are not equivalent. There are ninedifferent functions from C to D. An examination of all nine will show that none ofthem is onto D and so none of them is a bijection. Thus C is not equivalent to D.Example. The set E of even integers is equivalent to D, the set of odd integers.To prove this, we employ the function f: E --+ D given by f(x) = x + 1. The func-tion is one-to-one, because f(x) = fey) implies x + I = y + 1, which yields x = y.Also, f is onto D because if z is any odd integer, then w = z - 1 is even andf(w) =w + I = (z - 1) + I = z.Example. For a, b, c, dE IR, with a < band c < d, the open intervals (a, b) and(c, d) are equivalent.Proof. Let f: (a, b)--+ (c, d) be the linear function pictured in figure 5.1. Thefunction is given by:d-cf(x) = b _ a(x - a) + cThis is simply the linear function through the points (a, c) and (b, d) restricted to thedomain (a, b). We leave the proof that f is a bijection to exercise 4. •yc /d--r-----~------~--~xa bFigure 5.1
  • 3. Theorem 5.1Theorem 5.25.1 Equivalent Sets; Finite Sets 197We now know that any two open intervals are equivalent, even when the inter-vals have different lengths. Also, the interval (5, 6) is equivalent to (1, 9), eventhough it is a proper subset of (1,9).Example. Let gji be the set of all functions from N to {O, I}. This set is sometimesdenoted {a, I}N. We will show gji =QJI(N), the power set of N.Proof. (This proofwill show that gji is precisely the set ofall characteristic func-tions with domain N, which is in a one-to-one correspondence with the subsets ofN.) To show gji =QJI(N), we define H: gji -+ QJI(N) as follows:for g E gji, H(g) = {x E N: g(x} = I}.(Note that under the function H, everyfunction in gji has an image in QJI(N).)To show H is one-to-one, let g" g2 E gji. Suppose gl *" g2 Then there existsn E N such that g1(n) *" g2(n). Both gland g2 have codomain {O, I}, so we may as-sume gl (n) = 1 and g2(n) = 0. (The case gl (n) = °and g2(n) = 1 is similar.) Butthen n E {x E N: gl(x) = I} and n ~ {x E N: g2(X} = I}. Thus H(gl} *" H(g2}To show that H is onto QJI(N), let A E QJI(N). Then A k N. We note thatXA: N --+ {a, I} and thus XA E gji. Furthermore,HCxA) = {x E N: XA(X) = I} = A (.by definition of XA- Thus H is onto.Because H is a bijection, gji =QJI(N). •The relation = is reflexive, symmetric, and transitive. Thus = is an equivalencerelation on the class of all sets.Proof. Exercise I. •The next theorem will be particularly useful for showing equivalences of sets.Suppose A, B, C, and D are sets with A = C and B = D.(a) A X B = C X D.(b) IfA and B are disjoint and C and D are disjoint, then A U B =CUD.Proof. Since A = C and B =D, there exist one-to-one correspondences h: A -+ Candg: B-+D.(a) Let f: A X B-+ C X D be given by f(a, b) = (h(a), g(b)). We leave it as exer-cise 2 to show thatfis a one-to-one correspondence. Therefore, A X B =C X D.(b) By Theorem 4.15, hUg: AU B-+ CUD is a one-to-one correspondence.Therefore, A U B =CUD. •
  • 4. 198 CHAPTER 5 CardinalityLemma 5.3We shall use the symbol Nk to denote the set {I, 2, 3,..., k}. Think of Nk as thestandard set with k elements against which the sizes of other sets may be compared.DEFINITIONS A set S is finite iff S = 0 or S is equivalent to Nk forsome natural number k. A set is infinite iff it is not finite.We say that the empty set has cardinal number 0 (or cardinality 0),and if S = Nk we say that S has cardinal number k (or cardinality k).The set X = {98.6, c, n} is finite and has cardinal number 3. Exhibiting anyone-to-one function from X onto N3 will prove this. One such correspondencef: X -+ N3 is given by f{J8.6) = 1, f(c) = 2, andf(n) = 3.We use the symbol Sto denote the cardinal number ~ a set S. At this time Sisdefined only for a tinite set S. In the previous example, X = 3. For the empty set,0= 0 and clearly, Nk = k for every natural number k. The setA = {8, 7, 3, 7, 2} isfinite and has cardinality 4, since it is equal to the set {8, 7, 3, 2}. Since A = N4,A=4.The symbol Ais used for the cardinality ofA so as to distinguish what we provehere from the more informal results in section 2~, where we used the symbol # A.We shall show that for finite sets the cardinality Aof A corresponds to our intuitivenotion of the number of elements in A.The empty set is finite by definition. Also, each set Nk is finite and has cardinalnumber k because the identity function IrJk is a one-to-one function from Nk ontoNk • In addition, any set equivalent to a finite set must also be finite. Suppose A is afinite set and A =B. If A = 0, then B = 0 (see exercise 3). Otherwise, A =Nk forsome k and thus B =Nk by transitivity of =.In either case B is finite.The remaining theorems of this section give other properties of finite sets. Onegoal is to make sure that the notion of cardinality for a finite setA corresponds to ournotion of the number of elements in A in the following important way: The cardinal-ity ofafinite set should be unique. That is, ifA has cardinality m (that is, A =Nm) andA has cardinality n (A = Nn), then m = n.You are asked to show this in exercise 14, using a property of finite sets calledthe pigeonhole principle, discussed later in this section. Our immediate goal is thekey theorem that every subset of a finite set is finite. Our proof of this theorem usestwo lemmas.If S is finite with cardinalityk and xis any object not in S, then S U {x} is finite andhas cardinality k + I.Proof. If S = 0, then S U {x} = {x}, which is equivalent to NI and thus finite. Inthis case S has cardinality 0 and S U {x} has cardinality 0 + 1. IfS =1= 0, then S =Nkfor some k. Also, {x} = {k + I}. Therefore, by Theorem 5.2, S U {x} = Nk U {k + I}= Nk+ I This proves S U {x} is finite and has cardinality k + 1. •
  • 5. Lemma 5.4Theorem 5.5Theorem 5.65.1 Equivalent Sets; Finite Sets 199f1-1, ontofiT1-1, ontoFigure 5.2Every subset of Nk is finite.Proof. Let A be a subset of Nk• (We prove A is finite by induction on the num-ber k.)Let k = 1. Then either A = 0 or A = NI, both of which are finite.Assume all subsets of Nk are finite and let A !: Nk+ I Then A - {k + I} is a sub-set of Nk and is finite by the hypothesis of induction. If A = A - {k + I}, then A isfinite. Otherwise, A = (A - {k + I}) U {k + I}, which is finite by Lemma 5.3. -Every subset of a finite set is finite.Proof. Assume S is a finite and T !: S. If T = 0, then T is finite. Thus we may as-sume T =1= 0 and hence S =1= 0. Since S = Nk for some kEN, there is a one-to-onefunction J from S onto Nk. Then the restriction off, J IT, is a one-to-one functionfrom Tonto J(T). Therefore, T is equivalent to J(T) (see figure 5.2). But J(T) is asubset of the finite set Nk and is finite by Lemma 5.4. Therefore, since T is equiva-lent to a finite set, T is finite. -At this point you may think that Lemmas 5.3 and 5.4 and Theorem 5.5 are a lotof hard work to prove the very obvious result that subsets of finite sets are finite.You may be right. The value of these results lies in the reasoning and in the use offunctions to establish facts about cardinalities. This work will be helpful when wedeal with infinite sets because there our intuition often fails us.The next result of this section is that the union of a finite number of finite setsis finite. To this end the next theorem is a special case: the union of two disjoint fi-nite sets is finite. Its proof is a rigorous development of the sum rule (Theo-rem 2.17), which states that if A has m elements, B has n elements, and A n B = 0,then A U B has m + n elements.If A and B are finite disjoint sets, then A U B is finite and A U B = A + B.Proof. Suppose A and B are finite sets and A nB = 0. IfA = 0, then A U B = B;if B = 0-,- th~n. A U B = A. In either case A U B is finite, and since 0 = 0,A U B = A + B. Now suppose that A =1= 0 and B =1= 0. Let A = Nm and B = Nn ,
  • 6. 200 CHAPTER 5 CardinalityCorollary 5.7Lemma 5.8Theorem 5.9and suppose that f: A --+ Nm and g: B --+ Nn are one-to-one correspondences. LetH = {m + 1, m + 2,..., m + n}. Then h: Nn --+ H given by hex) = m + xis aone-to-one correspondence, and thus Nn = H. Therefore, B = H by transitivity. Finally, byTheorem 5.2, A U B = Nm U H = Nm +m which proves that A U B is finite and thatA UB=m+n. -(a)(b)IfA and B are finite sets, then A U B is finite.nIfAI> A2, ...,An are finite sets, then U Ai is finite.i=1Proof. We prove part (a) and leave part (b) as an exercise in mathematical induc-tion (exercise 6). Assume that both A and B are finite. Since B - A ~ B, B - A is fi-nite. Thus by Theorem 5.6, A U B = A U (B - A) is a finite set. -Lemma 5.3 shows that adding one element to a finite set increases its cardinal-ity by one. It is also true that removing one element from a finite set reduces the car-dinality by one. The proof of Lemma 5.8 is left as exercise 13.Let r E N with r > 1. For all x E N" Nr - {x} =Nr - I •The final property of finite sets we consider is popularly known as the pigeon-hole principle. In its informal version it says: "If a flock of n pigeons comes to roostin a house with r pigeonholes and n > r, then at least one hole contains more thanone pigeon." Ifwe think of the set ofpigeons as Nn and the set of pigeonholes as N"then the pigeonhole principle says any assignment of pigeons to pigeonholes (func-tion from Nn to Nr) is not one-to-one.(The Pigeonhole Principle)Let n, r E N. Iff: Nn --+ Nr and n > r, then f is not one-to-one.Proof. The proof proceeds by induction on the number n. Since n> r, we beginwith n = 2. If n = 2, then r = 1. In this casefis the constant function f(x) = 1,which is not one-to-one.Suppose the pigeonhole principle holds for some n; that is, suppose for all r < n,if f: Nn --+ N" then f is not one-to-one. Let r < n + 1. The case r = 1 is treated justas in the case n = 2 above, so we may assume r > 1. Suppose there is a one-to-onefunction f: Nn+ 1 --+ N" Then fiNn is a one-to-one function. The range of this func-tion may not contain fen + 1), but by Lemma 5.8 there is a one-to-one correspon-dence g: Nr - {J(n + I)} --+ Nr - I • Therefore, the composite g 0 fiNn: Nn --+ Nr - Iis one-to-one. This contradicts the induction hypothesis.By the PMI, there is no one-to-one function f: Nn --+ ~r when n > r. -The pigeonhole principle can be used to prove several important results aboutfinite sets (see, for example, exercise 13). We use it here to give a characterizationoffinite sets.
  • 7. Corollary 5.10Exercises 5.15.1 Equivalent Sets; Finite Sets 201A finite set is not equivalent to any of its proper subsets.Proof. We will show that Nk is not equivalent to any proper subset and leave thegeneral case as exercise 15. The case k = 1 is trivial, so let k > 1. Suppose A is aproper subset of Nk and I: Nk --+ A is one-to-one and onto A.Case 1. Suppose k tE A. Then A ~ Nk - 1 and the inclusion function i: A --+ Nk- 1 isone-to-one. But then i 0 I: Nk --+ Nk- 1 is one-to-one, which contradictsthe pigeonhole principle.Case 2. Suppose k EA. Choose an element y E Nk - A and let A = (A - {k})U {y}. Then A = A (because the function IA-(k} U {(k, y)} is a one-to-onecorrespondence). Thus A = Nb A is a proper subset of Nb and k tE A.This is the situation of Case 1 with Nk and A and again yields a contra-dictioo. •Corollary 5.10 says: "If A is a finite set, then A is not equivalent to any of itsproper subsets." The contrapositive is: "If A is equivalent to one of its proper sub-sets, then A is infinite." Earlier we observed that the open interval (0, 2) is equiva-lent to its proper subset (0, 1). Thus we know that (0, 2) is infinite. There will bemany more interesting infinite sets in the next section.1. Prove Theorem 5.1. That is, show that the relation = is reflexive, symmetric,and transitive on the class of all sets.2. Complete the proof of Theorem 5.2(a) by showing that if h: A --+ C andg: B--+ D are one-to-one correspondences, then I: A X B--+ ex D given byI(a, b) = (h(a), g(b)) is a one-to-one correspondence.3. (a) Show that if A = 0, then A = 0. [See also exercise 12(b), section 4.1.]"* (b) Show that A = A X {x}, for any object x.4. Complete the proof that any two open intervals (a, b) and (e, d) are equivalentby showing that I(x) = (%=~)(x - a) + e is one-to-one and onto (e, d).5. Which of the following sets are finite?* (a)(b)* (c)(d)* (e)(f)(g)(h)(i)(j)(k)the set of all grains of sand on Earththe set of all positive integer powers of 2the set of four-letter words in Englishthe set of rational numbersthe set of rationals in (0, 1) with denominator 2k for some kEN{x E IR: x2 + 1 = o}the set of all turkeys eaten in the year 1620{I, 3, 5} X {2, 4, 6, 8}{x E N: x is a prime}{x E N: x is composite}{x EN: x 2 + x is prime}
  • 8. 202 CHAPTER 5 CardinalityProofs to Grade(I) {x E IR: x is a solution to 4x8 - 5x6 + 12x4 - 18x3 + X2 - X + 10 = o}(m) the set of all complex numbers a + bi such that a2 + b2 = I6. Prove part (b) of Corollary 5.7.7. Let A and B be sets. Prove that* (a) if A is finite, then A n B is finite.(b) ifA is infinite and A ~ B, then B is infinite."* 8. (a) Prove that for all k, mEN, Nk X Nm is finite."*"*"*9.10.*"*"*18.19.*20.(b) Suppose A and B are finite. Prove that A X B is finite.Define BA to be the set of all functions from A to B. Show that if A and Barefinite, then BA is finite.If possible, give an example of each of the following:(a) an infinite subset of a finite set(b) a collection {Ai: i E N} of finite sets whose union is finite(c) a finite collection of finite sets whoseJInio..Q is infinite(d) finite sets A and B such that A U B =1= A+ BUsing th~meQlods of this section, prove that if A and B are finite sets, thenA U B = A+ B- A n B. This fact is a restatement of Theorem 2.17.Prove that if A is finite and B is infinite, then B - A is infinite.Prove Lemma 5.8.Show that if a finite set S has cardinal number m and cardinal number n, thenm=n.Complete the proof of Corollary 5.10 by showing that if A is finite and B is aproper subset of A, then B *A.Prove by induction on n that if r < nand f: Nr ---+ Nm then f is not onto Nn-Let A and B be finite sets with A = B. Suppose f: A ---+ B.(a) If f is one-to-one, show that f is onto B.(b) If f is onto B, prove that f is one-to-one.Prove that if the domain of a function is finite, then the range is finite.Use Corollary 5.10 to show that each of the following sets is infinite.(a) {1O, II, 12, 13, 14, 15,...} (b) {..., -3, -2, -I, o}(c) {3k : kEN} (d) (0, (0)Assign a grade of A (correct), C (partially correct), or F (failure) to each. Jus-tify assignments of grades other than A.(a) Claim. If A and B are finite, then A U B is finite."Proof." IfA and B are finite, then there exist m, n E N such thatA = Nm ,I-I I-I I-IB = Nn- Let f: AOi1i6Nm and h: BOi1i6 Nn- Then f U h: A U BOi1i6 Nm+mwhich shows that A U B = Nm+n- Thus A U B is finite. -* (b) Claim. If S is a finite, nonempty set, then S U {x} is finite."Proof." Suppose S is finite and nonempty. Then S = Nk for some in-teger k.Case 1. xES. Then S U {x} = S, so S U {x} has k elements and is finite.Case 2. x Et: S. Then S U {x} = Nk U {x} = Nk U NI = Nk +l . ThusS U {x} is finite. -
  • 9. 5.2Theorem 5.115.2 Infinite Sets 203(c) Claim. IfAX B is finite, then A is finite."Proof." Choose any b* E B. Then A = A X {b*}. But A X {b*} ={(a, b*): a E A} C A X B. Since A X B is finite, A X {b*} is finite. SinceA is equivalent to a finite set, A is finite. -Infinite SetsThis section will describe several infinite sets and their cardinal numbers. We willsee that there are many infinite sets and that it is not true that all infinite sets areequivalent. Just as we use Nk as our standard set of k elements, we will specify othersets as standard sets for certain infinite cardinal numbers.In the previous section an infinite set was defined as a nonempty set that can-not be put into a one-to-one correspondence with any Nk . According to the next the-orem, the set N is one such set. Since infinite means not finite, our proof, as mightbe expected, is by contradiction. /The set N of natural numbers is infinite.Proof. Suppose N is finite. Clearly, N =I- 0. Therefore for some natural number k,there exists a one-to-one function f from Nk onto N. (We will contradict that f isonto N.)Let n = f(l) +f(2) + ... +f(k) + 1. Since each f(i) > 0, n is a natural num-ber larger than each f(i). Thus n =I- f(i) for any i E Nk . Hence n t!. Rng if). There-fore, f is not onto N, a contradiction. -Here is a second proof that N is an infinite set: Let E+ be the set of even posi-tive integers. The function f: N --+ E+ defined by setting f(x) = 2x is clearly aone-to-one function from N onto E+. (Of course, there are many other one-to-onecorrespondences; the one given is the simplest.) Thus N is equivalent to one of itsproper subsets. Therefore, by Corollary 5.10, the set of natural numbers cannot befinite.There are many other infinite sets, some-but not all-of which are equivalentto N. Infinite cardinal numbers are written using the symbol X, aleph, the first letterof the Hebrew alphabet.The cardinality of N is denoted Xo. The subscript 0 (variously read "zero,""naught," or "null") indicates that Xo is the smallest infinite cardinal number, akinto the integer 0 being the smallest finite cardinal. The set N is our "standard" set forthe cardinal number Xo-a set has cardinal number Xo iff it is equivalent to N.DEFINITION A set S is denumerable iff S is equivalent to N; wewrite S = Xo.
  • 10. 204 CHAPTER 5 CardinalityTheorem 5.12We showed above that E+, the set of positive even integers, is equivalent to N.Therefore, E+ is denumerable. Even though E+ is a proper subset of N, E+ has thesame number of elements (~o) as N. Although our intuition might tell us that onlyhalf of the natural numbers are even, it would be misleading to say that N has twiceas many elements as E +, or even to say that N has more elements than E +. We mustrely on one-to-one correspondences to determine cardinality.In section 5.4 we will show that every infinite set is equivalent to one of itsproper subsets. Together with Corollary 5.10, this will characterize infinite sets: aset is infinite iff it is equivalent to one of its proper subsets.The next theorem will show that ll, the set of all integers, is also equivalentto N.The set II of integers is denumerable.Proof. We define f: N ---+ II by[(xl = {~;xif x is evenif x is odd.Thus f(l) = 0, f(2) = 1, f(3) = -1, f(4) = 2, f(5) = -2, and so on. Pictorially,f is represented as a one-to-one matching:N = {I, 2, 3, 4, 5, 6, 7,...}1 1 1 1 1 1 1II = {O, 1, -1, 2, -2, 3, -3,...}We will first show that f is one-to-one. Suppose f(x) = f(y). If x and yare bothx Y f dd i-x i-yeven, then 2 = 2 and thus x = y. I x and yare both 0 , then -2- = -2- so1 - x = 1 - y and x = y. Ifone ofx and y is even and the other is odd, then only oneof fix) orf(y) is positive, so f(x) *-f(y). Thus whenever f(x) = f(y), x = y.It remains to show that the function maps onto ll. If wEll and w > 0, then 2wis even and f(2w) = 2; = w. If wEll and w ::5 0, then 1 - 2w is an odd naturali-(i-2w) 2w 7Jnumber and f(l - 2w) = 2 = 2"" = w. In either case, if wE fL, thenwE Rng if). •Example. The set P of reciprocals of positive integer powers of 2 is denumerable.Writing the set P asP = {dk: kEN}exhibits the one-to-one correspondence f: N ---+ P given by f(k) = 1/2k.Example. The set N X N is denumerable. To see this, we must show there is abijection f: N X N ---+ N. The function F in section 4.3 given by fern, n) =2m- 1(2n - 1) is one such function. Thus N X N has cardinality ~o. This exampleleads directly to the next theorem.
  • 11. Theorem 5.135.2 Infinite Sets 205If sets A and B are denumerable, then A X B is denumerable.Proof. Since A = Nand B = N, A X B = N X N by Theorem 5.2(a). Since N X N= N, A X B = N. Therefore, A X B is denumerable. -Theorem 5.13 is a surprising result if you rely on properties of finite cardinalnumbers for insight into infinite cardinals. In chapter 2, you were asked to show thatif A is a finite set with m elements and B is a finite set with n elements, then A X Bhas mn elements. There is no direct corresponding result for sets with cardinality Xo.In fact, Theorem 5.13 can b~ construed as saying that "Xo times Xo is Xo." We willnot pursue the arithmetic of cardinal numbers, but simply remind you that extensionof results for finite cardinals may not be straightforward.DEFINITIONS A set is countable iff it is finite or denumerable;otherwise the set is uncountable.The set N3 = {I, 2, 3} is countable because it is finite. The set of integers, Z, isa countable set because Z is denumerable. The concepts of countable, uncountable,and denumerable along with the ideas of finite and infinite are related in figure 5.3.Because denumerable sets are those that are countable and infinite, denumerablesets are sometimes referred to as "countably infinite" sets.We have seen examples of countable sets (both finite sets and denumerablesets), but no example, as yet, of a set that is uncountable. Before considering thenext theorem, which states that the set of real numbers in the interval (0, 1) is suchan example, we need to review decimal expressions for real numbers. In its decimalform, any real number in (0, 1) may be written as 0.ala2a3a4"" where each ai is aninteger, 0:::; ai:::; 9. In this form, i2 = 0.583333..., which is abbreviated to 0.583 toindicate that the 3 is repeated. A block of digits may also be repeated, as in~ = 0.82142857. The number x = 0.ala2a3" is said to be in normalized form iffthere is no k such that for all n > k, an = 9. For example, 0.82142857 and ~ = 0.40are in normalized form, but 0.49 is not. Every real number can be expresseduniquely in normalized/orm. Both 0.49 and 0.50 represent the same real number~,but only 0.50 is normalized. The importance of normalizing decimals is that twodecimal numbers in normalized/orm are equal iffthey have identical digits in eachdecimal position.UncountableSetsInfinite SetsFigure 5.3FiniteSetsCountable Sets
  • 12. 206 CHAPTER 5 CardinalityTheorem 5.14 The interval (0, I) is uncountable.Proof. We must show that (0, I) is neither finite nor denumerable. The interval(0, I) is not finite since it contains the infinite subset g, t, *,...}.(See Theorem 5.5.)Suppose there is a function f: N ---+ (0, I) that is one-to-one. We will show thatf does not map onto (0, I). Writing the images of the elements of N in normalizedform, we havef(l) = 0.a11 a I2a 13a I4a I5···f(2) = 0.a21 a 22a 23a 24a 25···f(3) = 0.a31a32a33a34a35···f(4) ~ 0.a41a 42a 43a 44a 45···if au =1= 5if au = 5.(The choice of 3 and 5 is arbitrary.)Then b is not the image of any n E N, because it differs fromfi,n) in the nth decimalplace. We conclude there is no one-to-one and onto function from N to (0, I) andhence (0, I) is not denumerable. aThe open interval (0, I) is our first example of an uncountable set. The cardinalnumber of (0, I) is defined to be c (which stands for continuum) and is the only in-finite cardinal other than Xo that will be mentioned by name. We add the interval(0, I) to our list (Nk and N) of standard sets and will use (0, I) as the standard set ofcardinality c.DEFINITION A set S_has cardinality c iff S is equivalent to the openinterval (0, I); we write S = c.There is nothing special about using the interval (0, I) to define the cardinalnumber c; every open interval (a, b), where a, bE IR and a < b has cardinality c. Insection 5.1, we showed that any two open intervals are equivalent. In particular,(0, 1) = (a, b). Thus, (a, b) has cardinal number c.Example. The set A = (3, 4) U [5, 6) has cardinal number c. To see this, we notethat the functionf: (0,1) ---+ A, given by{2X + 3,f(x) =2x +4,ifO <x <!I if -::::; x < I2is a one-to-one correspondence.
  • 13. Theorem 5.15Exercises 5.25.2 Infinite Sets 207y---+--~~--+-~xFigure 5.4The set of all real numbers also has cardinality c, as the next theorem shows.The set ~ is uncountable and has cardinal number c.Proof. Define/: (0, I) -+ ~ bylex) = tan (nx -1) See figure 5.4. The function /is a contraction and translation of one branch of the tangent function and is one-to-one and onto R Thus (0, I) =R •Because we know that N X N has the same cardinality as N, you may suspectthat ~ X ~ has the same cardinality as R Indeed, it is true that ~ X ~ has cardinal-ity c. We omit the proof because the specialized methods needed will not be usefulin our current study.Since N is a proper subset of ~, and N = Xo, ~ = c, and Xo *" c, it is natural tosuspect (correctly) that Xo < c. We consider the ordering of cardinal numbers insection 5.4. In the meantime, we delve more deeply into the study of countable setsin section 5.3.1. Prove that th~ following sets are denumerable.* (a) D+, the odd positive integers(b) T+, the positive integer multiples of 3(c) T, the integer multiples of 3(d) {n: n E Nand n > 6}"* (e) {x:xE£:andx<-12}(f) N - {5, 6}(g) {(x, y) E N X ~: xy = I}(h) {x E £:: x == 1 (mod 5)}2. Prove that the following sets have cardinality c."* (a) (1,00)"* (b) (a, 00), for any real number a
  • 14. 208 CHAPTER 5 Cardinality(c) (-00, b), for any reai number b"* (d) [1,2) U (S, 6)(e) (3, 6) U [10, 20)(f) IR - {O}3. State whether each of the following is true or false.(a) If a set A is countable, then A is infinite.(b) If a set A is denumerable, then A is countable.(c) If a set A is finite, then A is denumerable.(d) If a set A is uncountable, then A is not denumerable.(e) If a set A is uncountable, then A is not finite.(f) If a set A is not denumerable, then A is uncountable.* 4. (a) Give an example of a bijection g from N to the set E+ of positive evenintegers such that g(l) = 20.(b) Give an example of a bijection h from N to E+ such that h(l) = 16,h(2) = 12, and h(3) = 2.5. Which sets have cardinal number Xo? c?* (a) IR - [0, 1)(b) (S, (0)* (c) {k: n E N}(d) {2x: x E N}* (e) {(P,q)EIRXIR:P+~=I}(f) {(p,q)ElRxlR:q= l-p2}(g) {(x,y)EIRXIR:x,yEZ}6. Give an example of two denumerable sets A and B such that(a) A U B is denumerable and A n B is denumerable.(b) AU B is denumerable and A - B is denumerable.(c) AU B is denumerable and A - B is finite and nonempty.7. Give an example of two sets A and B each with cardinal number c such that(a) A - B is empty.(b) A - B is finite and nonempty.(c) A - B is uncountable.8. LetA" A2, A3,...,An be denumerable sets.(a) Prove thatA I X (A2 X A3) and (AI X A2) X A3 are denumerable. (Recallfrom exercise 16 of section 3.1 that these sets are not equal.)"* (b) Let B = {(a" a2, a3): ai E Ai, i = 1,2, 3}. Prove that B is denumerable.(c) Let Cn = {(ai, az,..., an): ai EAi, i = 1,...,n}. Prove that Cn is denu-merable, for every natural number n."* 9. Prove Theorem S.13 directly. That is, if A and B are denumerable sets, con-struct a function from A X B to N that is one-to-one and onto N.10. GiveanotherproofofTheoremS.lSby showingthat!(x) = (x - !)/[x(x - I)]is a one-to-one correspondence from (0, I) onto IR.11. IR X IR has cardinality c. Use this fact to prove that the set C of complex num-bers, has cardinality c.
  • 15. Proofs to Grade5.35.3 Countable Sets 20912. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Jus-tify assignments of grades other than A.* (a) Claim. The sets E+ of even natural numbers and D+ of odd naturalnumbers are equivalent."Proof." E + is an infinite subset of N. Thus E + is denumerable. Like-wise D+ is an infinite subset of N and is denumerable. Therefore,E+ =D+. -(b) Claim. IfA is infinite and x ~ A, then A U {x} is infinite."Proof." Let A be infinite. Then A"= N. Let I: N --+ A be a one-to-onecorrespondence. Then g: N --+ A U {x}, defined by{X if t = 1g(t)= l(t-l) ift>1is one-to-one and onto A U {x}. Thus N = A U{x}, so A U{x} is infinite. -* (c) Claim. IfAU B is infinite, then A and B are infinite."Proof." This is a proof by contraposition so assume the denial of theconsequent. Thus assume A and B are finite. Then by Theorem 5.6,A U B is finite, which is a denial of the antecedent. Therefore, the resultis proved. -* (d) Claim. If a set A is infinite, then A is equivalent to a proper subset ofA."Proof." Let A = {Xl> X2, .. }. Choose B = {X2, X3""} Then B is aproper subset of A. The function I: A --+ B defined by I(Xk) = Xk+ 1 isclearly one-to-one and onto B. Thus A = B. -(e) Claim. The set N is infinite."Proof." The function given by I(n) = n + 1 is a one-to-one corre-spondence between Nand N - {I}, so N is equivalent to a proper subsetof N. Therefore, by Corollary 5.10, N is infinite. -Countable SetsCountable sets have been defined as those sets that are finite or denumerable. Suchsets are countable in the sense that they can be "counted" by using subsets of N: Afinite nonempty set is counted by using exactly the elements ofsome Nk , while a de-numerable set may be counted by using exactly the elements of N. This section willsurvey some of the important results about denumerable and countable sets. Firstwe will deal with the cardinality of the rational numbers Q.Since N ~ Q ~ ~, you should suspect that the cardinality of Q is Xo, or c, orsome infinite cardinal in between (the ordering of cardinal numbers is discussed inthe next section). You might also suspect that, since there are infinitely many ratio-nals between any two rationals, Q is not denumerable, but this is not the case. GeorgCantor (1845-1918) first showed that Q+ (the positive rationals) is indeed denu-merable through a clever rearrangement of Q+.Every element in Q+ may be expressed as ~ for some p, q E N. Thus the ele-ments of this set can be presented as in figure 5.5, where the nth row contains all thepositive fractions with denominator n.
  • 16. 210 CHAPTER 5 CardinalityTheorem 5.16~ //~/~ll~~I 2 3 4 5 6- - - - - -45Figure 5.553545526364657I72737475To show that Q+ is denumerable, Cantor gave a "diagonalization argument" inwhich the elements of Q+ are listed in the order indicated in figure 5.5. First are allfractions in which the sum of the numerator and denominator is 2 (only t), thenthose whose sum is 3 (r, ~), then those whose sum is 4, and so on. We omit from thelist ~ (= t),1 (= r), i (= ~ = t), ~ (= ~), and all other fractions not in lowest terms.The remaining numbers are circled in the array. The result is this one-to-one corre-spondence from N to Q+:N = {I,1Q+ = {Il2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,1 1 1 1 1 1 1 1 1 1 1 1213143215165l 2 l 3 l 2 3 4 l 5 l 214,...}1~,...}This correspondence can be used to establish the following theorem (we omit thedetails). See also the discussion following Theorem 5.29.The set Q+ of positive rational numbers is denumerable.Adding one or any finite number of elements to a finite set yields a finite set. Thenext three theorems are analogous results for denumerable sets. An important dis-tinction may be made between finite sets and denumerable sets: adding a finite num-ber ofelements or a denumerable number ofelements does not change the cardinalityof a denumerable set. Before we proceed with the theorems, we give an example.Example. The setA = {..., -10, -8, -6, -4, -2} of negative even integers is a de-numerable set. (The function/: N -+ A given bylex) = -2x is a bijection, so N = A.)Adding one additional integer, say 6, to A will result in a denumerable set:{..., -10, -8, -6, -4, -2,6}.Adding a finite number of additional integers, for example 2, 4, 6, and 8, yieldsa denumerable set: {..., -10, -8, -6, -4, -2,2,4, 6, 8}.
  • 17. Theorem 5.17I Lobby I5.3 Countable Sets 211Even adding a denumerable number of elements, say the nonnegative even inte-gers, still yields a denumerable set: {..., -10, -8, -6, -4, -2,0,2,4,6,8, 1O,...}.IfA is denumerable, then A U {x} is denumerable.Proof. If x E A, then A U {x} = A, which is denumerable. Suppose that x fl. A.Since N = A, there is a one-to-one function f: N ---+ A that is onto A. Defineg: N ---+ A U {x} byg(n) = {fen - I)if n = Iif n > IIt is now straightforward to verify that g is a one-to-one correspondence between Nand A U {x}, which proves that A U {x} is denumerable. -Theorem 5.16 may be loosely restated as Xo + I = Xo. Its proof is illustrated bythe story of the Infinite Hotel, I whose rooms are numbered I, 2, 3, 4,.... The InfiniteHotel has Xo rooms and is full to capacity with one person in each room. You approachthe desk clerk and ask for a room. When the clerk explains that each room is alreadyoccupied, you say, "There is room for me! For each n, let the person in room n moveto room n + I. Then I will move into room I,·and everyone will have a room as be-fore." There are Xo + I people and they fit exactly into the Xo rooms. See figure 5.6.Figure 5.6t The Infinite Hotel is one of the topics discussed in Aha! Gotcha: Paradoxes to Puzzle and Delight byMartin Gardner (Freeman, New York, 1981).
  • 18. 212 CHAPTER 5 CardinalityTheorem 5.18Theorem 5.19Theorem 5.20Theorem 5.21Rooms in the Infinite Hotel can also be found for any finite number of addi-tional people (Theorem 5.18) or any denumerable number of additional people(Theorem 5.19). In fact, the clerk could find rooms if a denumerable number of ad-ditional people arrive a finite number of times (Theorem 5.21) or even a denumer-able number of times (Theorem 5.22).IfA is denumerable and B is finite, then A U B is denumerable.Proof. Exercise 3. •IfA and B are disjoint denumerable sets, then A U B is denumerable.Proof.1-1 I-I .Letf: N"0iit6A and g: N"0iit6B. Define h: N --+ AU B via_{f(n;l)hen) - g(~)if n is oddif n is even.It is left as exercise 4 to show that h is a one-to-one correspondence from N ontoAUR •Before finishing the theorems about denumerable sets, we note that Theorems5.16, 5.17, and 5.19 provide a simple proof that the set of all rational numbers is de-numerable.The set Q of all rational numbers is denumerable.Proof. By Theorem 5.16, Q+ is denumerable. Likewise, the set of negative ratio-nals, Q-, is denumerable. (The function f: Q+ -> ((J)- that assigns f(x) = -x is abijection.) Therefore, Q+ U ((J)- is denumerable by Theorem 5.19. Finally, Q =(Q+ U ((J)-) U {a} is denumerable by Theorem 5.17. •The next theorem extends Theorem 5.19 to a finite union of pairwise disjointdenumerable sets. Its proof, as you might expect, is by induction on the number ofsets in the family.Let {A;: i = 1, 2,..., n} be a finite pairwise disjoint family of denumerable sets.nThen UAi is denumerable.i=1Proof. See exercise 5. •We can extend Theorem 5.21 to the union of a denumerable number of pair-wise disjoint denumerable sets, but this result (Theorem 5.22) cannot be provedfrom the properties of sets we have used thus far. The axioms of set theory allow us
  • 19. Theorem 5.22Theorem 5.23Corollary 5.24Theorem 5.255.3 Countable Sets 213to form unions, intersections, power sets, and infinite sets such as N, Z, G, and evenIR. However, a new property (The Axiom of Choice) to be introduced in section 5.5is needed to prove Theorem 5.22. This does not make Theorem 5.22 suspect, for theAxiom of Choice is a widely accepted and useful tool. A complete proof of Theo-rem 5.22 is given in section 5.5.Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerablesets. That is, if i E N, then Ai is denumerable, and if i, j E Nand i =1= j, then Ai =1= Ajand Ai n Aj = 0. Then UAi is denumerable.i=NWe continue with a sequence of important results about denumerable andcountable sets, leading to the fact that the union of any countable collection ofcoutable sets is countable.Every subset of a countable set is countable.Proof. Let A be a countable set and let B ~ A. If B is finite, then B is countable.Otherwise, B is infinite, so A is infinite and therefore denumerable. Let f be a bi-jection from A to N. Then fiB is also a bijection from B to f (B).We now define a function g: N ---+ feB) by induction. Let g(l) be the smallestinteger in feB). For each n E N,f{B} - {g{I}, g{2}, ..., g{n)} is nonempty becausefeB) is infinite. Let g(n + 1) be the smallest element in feB) - {g{I}, g{2}, ..., g{n)}.If r, sEN and r < s, then g(r) E {g{I}, g{2}, ... , g{s - I)} but g(s) is not; thusg(r) =1= g(s). Thus g is an injection. Also, if t Ef(B), and there are k natural numbersless than t inf(B), then t = g(k + I). Thus g is a bijection from N tof(B).Therefore, N = feB) = B and so B is denumerable. -We have seen that the set G of all rational numbers is denumerable and thereforecountable. Thus such sets as {k n EN}, G n (0, 1), {~,~, ~}, and Z are countable sets.A set A is countable iffA is equivalent to some subset of N.Proof. If A is countable, then A is either finite or denumerable. Thus A = 0, orA = Nk for some kEN, or A = N. In every case, A is equivalent to some subset of N.If A is equivalent to some subset of N, then A is equivalent to a countable set,since all subsets of (countable) N are countable. Therefore, A is countable. -IfA and B are countable sets, then A U B is countable.Proof. This theorem has been proved in the cases where A and B are finite (Corol-lary 5.7), where one set is denumerable and the other is finite (Theorem 5.18), andwhere A and B are denumerable and disjoint (Theorem 5.19). The only remainingcase is where A and B are denumerable and not disjoint. Write A U B asA U (B - A),a union of disjoint sets. Since B - A ~ B, B - A is either finite or denumerable byTheorem 5.23. IfB - A is finite, then A U B is denumerable by Theorem 5.18, and ifB - A is denumerable, then A U B is denumerable by Theorem 5.19. -
  • 20. 214 CHAPTER 5 CardinalityTheorem 5.26Lemma 5.27Theorem 5.25 may be extended (by induction) to any finite union of countablesets.Let .sil be a finite collection of countable sets. Then .sil is countable.Proof. See exercise 7. •Theorem 5.26 may, in tum, also be extended to any countable union of count-able sets. For example, let.sil = {An: n EN}, where An = {x E Z: x :::; n}. ThenAl = 1...-2, -I, a, I},A2 = ...,-2,-I,a,I,2},A3 = ..., -2, -I, a, 1,2, 3},and so on. Each An is countable and U An = N is countable. Even though .sil is anENdenumerable collection of denumerable sets, we cannot use Theorem 5.22 to con-clude that the union is denumerable because the family .sil is not pairwise disjoint.However, the family .sil does determine another family with the same union. LetCJ3 be the family {Bn: n EN}, whereB, =AI ={..., -2, -I,a, I},B2 = A2 - A, = {2},B3 = A3 - (A2 UA,) = {3},B4 = A4 - (A3 UA2 UA,) = {4},and so on. This family is a pairwise disjoint collection. Furthermore, U Bn =nENUAI!" Such a family CJ3 may be defined for any denumerable family of sets (innENLemma 5.27) and used to show that the union of any denumerable family of denu-merable sets is denumerable (Theorem 5.28).Let {Ai: i E N} be a denumerable family of sets. For each i E N, let B; =A; - (CJAk ). Then {B;: i E N} is a denumerable family ofpairwise disjoint sets andk=UA;= UBi.iEN iENProof. See exercise 8. •
  • 21. Theorem 5.28Theorem 5.295.3 Countable Sets 215Let {Ai: i E N} be a denumerable family of denumerable sets. Then UAi ISdenumerable. iENProof. Al ~ UAi, so the union is infinite by exercise 7(b) of section 5.1. Ifwe letiENBi = Ai - (0Ak ), then by Lemma 5.27 it is sufficient to show that UBi isk=! iENdenumerable.If all the Bi, i E N, are denumerable, then by Theorem 5.22, UBi is denumer-iENable. Likewise, if all the Bi , i E N, are finite, then UB; is denumerable (exercise 9).;ENLet ~I = {Bi: Bi is finite} and ~2 = {B;: B; is denumerable}. Both ~I and ~2are nonempty and UBi = ( U B) U ( U B). If~1 is finite, then U B is fi-;EN BE!Jl 1 BE!Jl2 BE!Jl 1nite (by Theorem 5.7) and U B is denumerable by Theorem 5.22. If~1 is infinite,BE!Jl2then U B is denumerable (exercise 9 again). If ~2 is finite, then U B is denu-BE!Jll BE!Jl2merable (by Theorem 5.21); otherwise U B is denumerable by Theorem 5.22.BE!Jl2Thus UB; is either the union of a finite set and a denumerable set or the union of;ENtwo denumerable sets. In either case, UB; is denumerable. •iENWe can pull all these results together in one simply stated theorem:Let .il be a countable collection of countable sets. Then U.ilis countable.Proof. You should check to see that every case is considered, either as a theoremor an exercise. •Theorem 5.29 provides another means to prove that the positive rational Q+ isdenumerable. Q+ is certainly infinite, and Q+ may be written as UAm wherenENAn = {~: a EN}. For each n E N, An is denumerable. By Theorem 5.29, Q+ is de-numerable.
  • 22. 216 CHAPTER 5 CardinalityExercises 5.3Example. Each set Nb kEN is a finite set and therefore countable. Their union,U Nk = N, is of course countable.kENExample. For each n E N, let Bn = {-n, n}. The countable union of these count-able sets is 71 - {O}, which is countable.We will give one more example of a denumerable set, one with particular sig-nificance for computer programmers. A computer program is written in a given pro-gramming language and consists of a finite sequence of symbols. These symbols areselected from a finite set called an "alphabet" (typically all 26 upper and lowercaseletters, the 10 digits, a blank space, certain punctuation marks, arithmetic opera-tions, etc.). For any particular language, let PI be all programs of length I, let P2 beall programs of length 2, P3 be all programs of length 3, and so on. We note that inmost programming languages, the first few P; are empty sets. An example of an el-ement of Pso in the language BASIC isProgram Xl (10 symbols)Let X = 1 (9 symbols)Let Y = 2 (9 symbols)Print "X + Y is II •X + Y (19 symbols)IEnd (3 symbols)Since any computer program is finite in length, every program is an element ofPn for some n. Thus, the set of all possible programs isBy Theorem 5.29, this countable union of countable sets is countable. Hence only acountable number of programs could ever be written in a given language. However,we have seen that the set of all functions from N to {O, I} is uncountable. Thismeans that there are many functions from N to {O, I} for which there can be nocomputer programs to compute them! Put a different way, there are not enough so-lutions (programs) for all the possible problems (functions).1. What is the 28th term in the sequence of positive rationals given in the dis-cussion of Theorem 5.16?2. Explain how a diagonalization argument can be used to show that the set of ra-tionals {2X /3Y: x, yEN} is denumerable. Omit details of the proof.3. Prove Theorem 5.18 by induction on the number of elements in the finite set B.4. Complete the proof of Theorem 5.19 by showing that the function h as definedis one-to-one and onto A U B.
  • 23. 5.3 Countable Sets 217*: 5. Prove Theorem 5.21: The union of a finite pairwise disjoint family of denu-merable sets {Ai: i = I, 2,..., n} is denumerable.6. Use the theorems of this section to prove that(a) an infinite subset of a denumerable set is denumerable.(b) Q n (1,2) is denumerable.(c)(d)(e)20U ((j) n (n, n + I») is denumerable.n=lU ((j) n (n, n + I») is denumerable.nENU {nk: kEN} is denumerable.nEN 27. Prove Theorem 5.26 by induction on the number of sets in the family si.8. Complete the proof of the set equality in Lemma 5.27.*: 9. Prove that if {Bi: i E N} is a denumerable family of pairwise disjoint distinctfinite sets, then UBi is denumerable.iEN10. The Infinite Hotel is undergoing some remodeling, and consequently some ofthe rooms will be taken out ofservice. Show that, in a sense, this does not mat-ter as long as only a "few" rooms are removed. That is,(a) prove that ifA is denumerable and x E A, then A - {x} is denumerable.(b) prove that ifA is denumerable and B is a finite subset ofA, then A - B isdenumerable.*: 11. Use an argument similar to the diagonalization argument for Theorem 5.16 toprove Theorem 5.13 (ifA and B are denumerable sets, then A X B is denumer-able).12. Give an example, if possible, of* (a) a denumerable collection of finite sets whose union is denumerable.(b) a denumerable collection of finite sets whose union is finite.(c) a denumerable collection of pairwise disjoint finite sets whose union isfinite.13. (a) Let S be the set of all sequences of Os and Is. For example,1 0 1 0 1 0 1..., 0 I I I I 1 I I I... and 0 1 0 I I 0 I I 1... are in S. Provethat S is uncountable.*: (b) Let Tbe the set of all sequences of 0sand ls where all but a finite num-ber of terms are O. Use a diagonalization argument, like that in Theo-rem 5.16, to show that Tis a denumerable subset of S.14. Let A be a denumerable set. Prove that(a) the set of all I-element subsets of A, {B: B ~ A and B= I} is denumer-able.(b) the set of all 2-element subsets of A, {B: B ~ A and II = 2} is denumer-able.(c) for any kEN, {B: B ~A and II = k} is denumerable.(d) thnet of all finite subsets of A, {B: B ~ A and B is finite} is denumer-able. (Hint: Use Theorem 5.22.)
  • 24. 218 CHAPTER 5 CardinalityProofs to Grade5.415. Prove that ifA and B are countable, then A X B is countable.16. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Jus-tify assignments of grades other than A.* (a) Claim. If A is denumerable, then A - {x} is denumerable."Proof." Assume A is denumerable.Case 1. Ifx E,lA, then A - {x} =A, which is denumerable by hypothesis.Case 2. Assume x EA. Since A is denumerable, there exists I: N ~~~lA.Define g by setting g(n) =I(n + 1). Then g: N ~:td (A - {x}),so N = A - {x}. Therefore, A - {x} is denumerable. _(b) Claim. IfA and B are denumerable, then A X B is denumerable."Proof." Assume A and B are denumerable, but that A X B is not de-numerable. Then A X B is finite. Since A and B are denumerable, theyare not empty; therefore, choose a E A and bE B. By exercise 3(b) ofsection 5.1, A = A X {b}, and by an obvious modification of that exer-cise, B = {a} X B. Since A X B is finite, the subsets A X {b} and {a} X Bare finite. Therefore, A and B are finite. This contradicts the statementthat A and B are denumerable. We conclude that A X B is denumerable.-(c) Claim. (Theorem 5.16) The set Q+ ofpositive rationals is denumerable."Proof." Consider the positive rationals in the array in figure 5.5 (The-orem 5.16). Consider the order formed by listing all the rationals in thefirst row, then the second row, and so forth. Omitting fractions that arenot in lowest terms, we have an ordering of Q+ in which every positiverational appears. Therefore, Q+ is denumerable. _(d) Claim. If A and B are infinite, then A = B."Proof." Suppose A andB are infinite sets. LetA = {ai az a3, ...}andB = {b J, b2, b3, ...}. Define I: A ---+ B as in the picture{al> a2, a3, a4, ...}! 1 1 1{bl> b2, b3, b4, ..•}Then, since we never run out of elements in either set, I is one-to-oneand onto B, so A = B. _The Ordering of Cardinal NumbersThe theory of infinite sets was developed by Georg Cantor over a 20-year span, cul-minating primarily in papers that appeared during 1895 and 1897. He described acardinal number of a set M as "the general concept which, with the aid of our intel-ligence, results from M when we abstract from the nature of its various elements andfrom the order of their being given." This definition was criticized as being less pre-cise and more mystical than a definition in mathematics ought to be. Other defini-tions were given, and eventually the concept of cardinal number was made precise.One way this may be done is to determine a fixed set from each equivalence class
  • 25. Theorem 5.305.4 The Ordering of Cardinal Numbers 219of sets under the relation =,and then to call this set the cardinal number of each setin the class. Under such a procedure we would think of the number 0 as being theempty set and the number I as being the set whose only element is the number O.That is, 1 = {o}; 2 = {o, I}; 3 = {o, I, 2}, and so on.We will not be concerned with a precise definition of a ~rdinal number. Forour purposes the essential point is that the cardinal number Aof a set A is an ob-ject ass~iated with all sets equivalent to A and with no other set. The double over-bar on A is suggestive of the double abstraction referred to by Cantor. Anothercommon notation for the cardinal number of set Ais IAI.Cardinal numbers may be ordered (compared) in the following manner:DEFINITIONSA= Bif and only if A =B; otherwise A oF B.= = 1-1A :S B if and only if there exists f: A--+B.A< Bif and only ifA:S Band AoF B.A< B is read "the cardinality of A is strictly less tha!!. th~cardi.!!ality of B"while :S is read "Ie~ th~n or ~uaIJo." In addition, we use A1:. Band A"$ Bto de-note the denials ofA< ~ anQ.A :S B, respectively.Usually a proof of1::s §. will involve constru£.tin~a one-to-one function fromA to B, while a proof ofA< §. wi!! have a proof ofA:S Btogether with a proof, gen-erally by contradiction, that A oF B. Once we have developed some properties Qfca!:::,dinal inequalities those facts can be used to prove statements of the form A:S Bwithout resorting to the construction of functions.Since 1, 2, 3,... are cardinal numbers, the natural numbers may be viewed as asubset of the collection of all cardinal numbers. In this sense the properties of :S and< that we will prove for cardinal numbers in the next theorem may be viewed as ex-tensions of those same properties of :S and < that hold for N. Parts (a), (b), (d), and(f) are left as exercise 6.For sets A, B, and C,(a)(b)(c)(d)(e)(f)Proof.A:§A._ (RefkxivJ!y)If1: = §. and §. = ~, then 1: = ~.!fA:§ Ba,!!d B...? C..1..the.!! A :S C.A:S BiffA< Bor A= B.IfA C B, then A :S B.(Transitivity of =)(Transitivity of:S)A :S 13 iff there is a subset W of B such that W = A.(c) Suppose A:s Band B:s C. Then there exist functions f: A..!..=.!.B and1-1(e)[[ B-=---C. Since the composite go f: A -+ C is one-to-one, we concludeA:sC.Let A C B. We noteJhatJhe inclusion map i: A -+ B, given by i(a) = a, isone-to-one, whence A:S B. •
  • 26. 220 CHAPTER 5 CardinalityTheorem 5.31Example. k < Xo for every finite cardinal number k. From Nk ~ N we havek :::; Xo, and since Nk is not equivalent to N, k < Xo, from Theorem 5.30(d).Example. We have implied that Xo < c in previous sections. This is true because(i) Xo :::; c follows from N ~ ~ and (ii) N is not equivalent to ~.Although Cantor developed many aspects of the theory of infinite sets, his nameremains attached particularly to the next theorem, which states that the cardinality ofa set A is strictly less than the cardinality of its power set. We already know the re-sult to be true ifA is a finite set with n elements, since QJ>(A) has 2n elements (by The-orem 2.4) and n < 2n for all n E N. The proof given here holds for all sets A.(Cantors Theo~m)_For every set A, A< QJ>(A).Proof. To show A< QJ>(A), we must show that (i) A:::; QJ>(A), and (ii) A*- QJ>(A).Part (i) follows from the facLthat F: A -+ QJ>(A) defined by F(x) = {x} is one-to-one.To prove (ii), suppose A= QJ>(A); that is, assume A = QJ>(A). Then there existsg: A ~~td QJ>(A). Let B = {y E A: y t!. g(y)}. Since B ~ A, BE QJ>(A), and since g isonto QJ>(A), B = g(z) for some z EA. Now either z E B or z t!. B. If z E B, thenz t!. g(z) = B, a contradiction. Similarly, z t!. B implies z E giz), which impliesz E B, again a contradiction. We conclude A :f;QJ>(A) and hence A< QJ>(A). •Cantors Theorem has some interesting consequences. First, there are infinitelymany infinite cardinal numbers. We know one, Xo, which corresponds to N. ByCantors Theorem, Xo < QJ>(N). Since QJ>(N) is a set, its power set QJ>(QJ>(N)) has astrictly greater cardinality than that of QJ>(N). In this fashion we may generate a de-numerable set of cardinal numbers:Xo < QJ>(N) < QJ>(QJ>(N)) < QJ>(QJ>(QJ>(N))) < ...Exactly where c, continuum, fits within this string of inequalities will be taken uplater in this section. It is also an immediate consequence of Cantors Theorem thatthere can be no largest cardinal number (see exercise 7).In section 5.1 we showed that the set cg; of all functions from N to {O, I} isequivalent to QJ>(N). Since N < QJ>(N), we know there are uncountably many func-tions from N to {O, I}. Since a function from N to {O, I} is a sequence of 0 sand1s, Cantors Theorem provides another proof for exercise 13(a) of section ~3. _It appears to be obvious that if B has at ~ast..ils many elements as A (A :::; Ii),a.Qd A_has at least as many elements as B (Ii:::; A), then A and B are equivalent(A = Ii). The prooL ho~ever,ls n~ obvious. The situation mar_~e represente9 ~~ infigure 5.7. From A :::; Band B:::; A, there are functions F: A--B and G: B----+A.The problem is to construct H: A -+ B, which is both one-to-one and onto B. Cantorsolved this problem in 1895, but his proof used the controversial Axiom of Choice(section 5.5). Proofs not depending on the Axiom of Choice were given indepen-dently by Ernest SchrOder in 1896 and two years later by Felix Bernstein.
  • 27. Theorem 5.325.4 The Ordering of Cardinal Numbers 221AFC = Rng (G)GFigure 5.7((!nto!-S(h!iid~-Berl!Ste~ Theorem)If A::5 Band B::5 A, then A= B.BProof. We may assume that A and B are disjoint, for otherwise we could replaceA and B with the equivalent disjoint sets A X {a} and B X {I}, respectively. Let1-1 1-1F: A---+B, with D = Rng (F), and let G: B---+A, with C = Rng (G). If B = D wealready have A = B, so assume B - D =1= 0.Define a string to be a function f: N ---> A U B such thatf(1) EB - D,fen) EBimpliesf(n + 1) = G(j(n», andfen) E A implies fen + 1) = F(j(n».We think of a string as a sequence of elements ofA U B with first term in B - D, andsuch that thereafter the terms are alternately in C and in D. Each element of B - Dis the first term of some string. See figure 5.8.GFGFGFigure 5.8 StringI: 1(1), 1(2), 1(3), 1(4), 1(5), 1(6), ...
  • 28. 222 CHAPTER 5 CardinalityHFigure 5.9Let W = {x EA: x is a term of some string}. We note that We Rng (G) andthat x E W iff x = f(2n) for some string f and natural number n. Let H: A -+ B begiven by_ {F(X)H(x) - G-1(x)if x ff. WifxEW.See figure 5.9. We will show that H is a one-to-oe correspondence from A onto B.Suppose x, yEA and H(x) = H(y). We will first show that x and y must bothbe in W or both in A - W. (We will use a proofby contradiction to show this.) Sup-pose x E A - Wand yEW. (The symmetric situation where yEA - Wand x E Wproduces a similar contradiction.) In this case, H(x) = H(y) is F(x) = G-1(y), soy = G(F(x)). Since yEW, Y = f(2n) for some string f and some natural number n.Therefore,f(2n - I) = F(x) (because G(j(2n - I)) = f(2n) = y = G(F(x)) and Gis one-to-one). If n = 1, then f(1) = F(x), which implies f(I) E Rng (F), a contra-diction to the definition of string f Thus n:::=:: 2. But then f(2n - 2) = x, sincef(2n - 1) = F(x). This implies that x is a term in the string!, a contradiction tox E A - W. Therefore, we know that x and yare either both in W or both in A - W.If x, yare both in W, then H(x) = H(y) implies that G-1(x) = G-1(y). There-fore, x = y since G- 1 is one-to-one. Likewise, if x, yare both in A - W, thenF(x) = F(y) and x = y because F is one-to-one. In either case, we conclude x = yand H is one-to-one.Next we show that H is onto B. Let bE B. We must show that b = H(x) forsome x EA. There are two cases:Case 1. If G(b) E W, let x = G(b). Then H(x) = H(G(b)) = G- 1(G(b)) = b.Case 2. If G(b) ff. W, then bE Rng (j). (If b ff. Rng (j), then bE B - D. There-fore, b is the first element ofsome string and G(b) is the second elementof that string, a contradiction to G(b) ff. W.) Since bE Rng (f), thereexists x E A such that F(x) = b. Furthermore, x ff. w. (Ifx E W, then x is
  • 29. 5.4 The Ordering of Cardinal Numbers 223a term in some string and therefore F(x) and G(F(x)) are the next twoterms of the same string. But this is a contradiction, sinceG(F(x)) = G(b) and we are in the case where G(b) is not on any string.)From x EA - W we conclude H(x) = F(x) = b.In both cases, H(x) = b, so H is onto B. •The Cantor-Schrader-Bernstein Theorem may be used to prove equivalence be-tween sets in cases where it would be difficult to exhibit a one-to-one correspondence.Example. We will show that (0, I) = [0, 1]. First, note that (0, I) <: [0, I], so byTheorem 5.30(e), (0,1)::5[0,1]. Likewise, since [0,1]<:(-1,2), we have[0, 1] <: (-I, 2). But we know (0, I) = (-I, 2) and thus (0, I) = (-1,2). Therefore,we may write [0, I] ::5 (0, I ). We conclude (0, I) = [0, I] by the Cantor-Schrader-Bernstein Theorem and thus (0, I) = [0, I].Example. The Cantor-Schrader-Bernstein Theorem can be used to determine therelationship of c, the cardinal number of the open interval (0, I), to the increasingsequence of cardinal numbersN < rJ>(N) < rJ>(rJ>(N)) < rJ>(rJ>(rJ>(N))) < ....First, recall that any real number in the interval (0, I) may be expressed in abase 2 (binary) expansion 0.b)b2b3b4..., where each bi is either °or 1. Ifwe excluderepeating 1s, such as 0.01011111 I 1... (which has the same value as 0.01 100000...),then the representation is unique. Thus we may define a function f: (0, I) ---+ rJ>(N)such that for each x E (0, 1),f(x) = {n E N: bn = 1 in the binary representation ofx}.The uniqueness of binary representations assures that the function is defined andis one-to-one. We note that f is not onto rJ>(N) since inductive sets such as {5, 6, 7,8, 9,...} are not images of any x E (0, I). Sincefis one-to-one, (0, 1)::5 rJ>(N).Next, define g:rJ>(N) ---+ (0, I) by g(A) = 0.a)a2a3a4"" where_{2an - 5ifnEAif n I$A.For any set A <: N, g(A) is a real number in (0, I) with decimal expansion consist-ing of2s and 5s. (Any pair ofdigits not 9 will do.) The function g is one-to-one butcertainly not onto (0, 1). Therefore, rJ>(N) ::5 (0, 1).__By the Cantor-Schrader-Bernstein Theorem, rJ>(N) = (0, IL Therefore,~ = c. We can now identify the first two terms of the sequence N < rJ>(N) <rJ>(rJ>(N)) < rJ>(rJ>(rJ>(N))) < ... as being Xo and c.The Cantor-Schrader-Bernstein Theorem is another result in the extension ofthe familiar ordering properties of N to properties for all cardinal numbers. It, intum, leads to others. In the following, parts (a) and (c) are proved; (b) and (d) aregiven as exercise 13.
  • 30. 224 CHAPTER 5 CardinalityCorollary 5.33Exercises 5.4For sets A, B, and C,(a)(b)(c)(d)if;i ::5 ~, the~E 1::_A.if;i::5 ~ and ~ < f, then;i < f·if;i < ~ and ~ ::5 f, then;i < f.ifA< Eand E< C, then A< C.Proof.(a) Supp.Qse ~ < A. Then A "* E and E ::5 A. Combining this with the hypothesis(c)Qlat ;i ::5 E, we conclude by the Cantor-Sc.hrod~r-Bernstein Theorem thatA= E, wQich.is a c~tra~ction. !.her~ore, E1:: A.SuppQ§eA_< E andj1::5S. Then A ::5 E; sQJJy "Q1eorel!!..5.21(c), A..? C~Sup­pose JL1:: (, Then A = C, which imp.lies ~ ::5 A. But E ::5 C and C ::5 A im-plies E::5 A. Combining this with... A:§. E, we conclude by th~ C~tor­Schro-.ger-~ernstein Theorem that A = E. Since this contradicts A< E, wehave A<c. _It is tempti~ to~xtend .Qur r~sults even further to include the converse of Corol-lary 5.33ia): JIE 1::j1, t~nA ::5 E." (As far as we know now, for two given sets A andB, both A ::5 E and E::5 A may be false.) The Cantor-Schroder-Bernstein Theoremturned out to be more difficult to prove than one would have guessed from its simplestatement, but the situation regarding the converse of Coroll~ 513(a) is ~en!poreremarkable. This is_disc.!!sseQ. in--.?ectism ~5, where "If E 1:: A, then A::5 E" isrephrased as "Either A < E or A = E or E < A."1. Prove that for any natural number n, n < c."* 2. Prove that QP(N) < QP(IR).3. Prove that if A::5 Eand E= C, then A::5 C.4. Prove that if A::5 B and A= C, then C ::5 B.5. State ~hefuer each of the following is true or false.(a) A::5 E im}21ies that A <;::; B.* (b) 1n!!.::5E.(c) ;i::5 ~ implies ~(AL::5 QP(B).(d) A= Eimplies A::5 E.6. Prove the remaining parts of Theorem 5.30."* 7. Prove that there is no largest cardinal number.8. Arrange the following cardinal numbers i.!!.-order: ~;;;;=;;~* (a) (0,1), [0, 1], {O,T}, 101, ~, Q, 0, IR - g, QP(QP(IR)), 1R(b) Q U {n}, IR - {n}, QP({O, I}), [0,2], (0, 00), E, IR - Il, QP(IR)9. Apply the proof of the Cantor-Schroder-Bemstein Theorem to this situation:A = {2, 3, 4, 5,...},B = H, t, t,···},F: A-+BwhereF(x) = x!6,andG: B-+A
  • 31. Proofs to Grade10.11.5.4 The Ordering of Cardinal Numbers 225where G(x) = i + 5. Note that tand! are in B - Rng (F). Let J be the stringthat begins at t, and let g be the string that begins at !.(a) FindJ(l ),f(2),f(3),f(4).(b) Find g(1), g(2), g(3), g(4).(c) Define Has in the proof of the Cantor-SchrOder-Bernstein Theorem andfind H(2), H(8), H(l3), and H(20).I-I I-I I-ISuppose there exist three functions J: A--B, g: B--C, and h: C--A.Prove A = B = C. Do not assume that the functions map onto their codomains.If possible, give an example of(a) functions J and g such that J: ~N, g: N1.=..!.Q, but neither J nor g isan onto map.* (b)(c)(d)I-Ia functionJ: IR--N.1-1a function J: 0J>~~--N.a functionJ: IR--Q.12. Prove that ifthere is a functionJ: A -+ N that is one-to-one, then A is countable.13. Prove parts (b) and (d) of Corollary 5.33.14. Use a cardinality argument to prove that there is no universal set V of all sets.15. Use the Cantor-SchrOder-Bernstein Theorem to prove the following.(a) The set of all integers whose digits are 6, 7, or 8 is denumerable.(b) IR X IR = IR. Note that this means there are just as many points on the realline as there are in the Cartesian plane.(c) .!fA <: IR and there exists an open interval (a, b) such that (a, b) <: A, thenA=c.Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify as-signments of grades ot~r t~n A._16. (a) Claim. If A::5 Ba.!...d A-== C, t1:!en ~ ::5 B."Proof." Assume A::5 Band A= C. Then there exists a functionJsuchI-I I-I = =thatJ: A--B. Since A-== CJ: C--B. Therefore, C::5 B. -* (b) Claim. If B <: C and B= C, then B = C."Proof." Suppose B oF C. Then B is a proper subset of C. ThusC - B oF 0. This implie~C -_B> O. But C = B U (C - B)~nd,~ince Band C - B are disjoint, C = B+ (C - B). By hypothesis, B= C. Thus(C - B) = 0.1...a c2..ntrad~tio~ -(c) Claim. If A::5 Ba~ B= C, then A::5~."Proof." Assume A::5 B. Then, since B= C, we have A::5 C by sub-stitution. -* (d) Claim. IfA oF 0 and A::5 B, then there exists a function J: Bonto,A.= = I-I"Proof." From A ::5 B, we know there exists g: A--B. Fix a particulara* EA. Define J: B -+ A as follows:(i) For bE B, if bE Rng (g), then g-I({b}) consists of a single elementofA, since g is one-to-one. Definej{b) to be equal to that element ofA.(ii) If bEt: Rng (g), defineJ(b) = a*.The function J is onto A, for if a E A, let g(a) = b. Then bE Bandg-I({b}) = a, soJ(b) = a. -
  • 32. 226 CHAPTER 5 Cardinality5.5 Comparability of Cardinal Numbers and the Axiom of ChoiceTheorem 5.34The Axiomof ChoiceThe goal ofsection 5.4 was to establish results for the relations:::; and < on the classof all cardinal numbers while thinking of these results as extensions of properties of11. One of the most useful ordering properties of 11 is the trichotomy property: ifm and n are any two natural numbers, then m > n, m = n, or m < n. The analog forcardinal numbers is stated in the Comparability Theorem.(The Comparability Theorem)_IfA and B are any two sets, then A< B, A= B, or B< A.Surprisingly, it is impossible to prove the Comparability Theorem from the ax-ioms and other theorems of Zermelo-Fraenkel set theory (see section 2.1). In a for-mal study of set theory one can build up, starting with a few axioms specifying thatcertain collections are sets, to the study of the natural, rational, real, and complexnumbers, polynomial, transcendental, and differentiable functions and all the rest ofmathematics. Still, comparability cannot be proved. On the other hand, it is impos-sible to prove in Zermelo-Fraenkel set theory that comparability is false. Theo-rem 5.34 is undecidable in our set theory; no proof of it and no proof of its negationcould ever be constructed in our theory.At this point we could choose either to assume that Theorem 5.34 is true (or as-sume true some other statement from which comparability can be proved) or elseassume the truth of some statement from which we can show comparability is false.Of course, we have revealed the fact that we want comparability to be true by la-beling the statement as a theorem. It has become standard practice by most mathe-maticians to assume the Comparability Theorem is true by assuming the truth ofthefollowing statement:If .il is any collection of non~mpty sets, then there exists a function F (called achoice function) from .il to U A such that for every A E .il, F(A) EA.AEdThe Axiom of Choice at first appears to have little significance: From a collec-tion of nonempty sets, we can choose an element from each set. If the collection isfinite, then this axiom is not needed to prove the existence of a choice function. It isonly for infinite collections of sets that the result is not obvious and for which theAxiom of Choice is independent of other axioms of set theory.Many examples and uses of the Axiom of Choice require more advancedknowledge of mathematics. The example we present is not mathematical in contentyet has become part of mathematical folklore.A shoe store has in the stockroom an infinite number of pairs of shoes and aninfinite number of pairs of socks. A customer asks to see one shoe of each pair andone sock of each pair. The clerk does not need the Axiom of Choice to select a shoefrom each pair of shoes. His choice rule might be "From each pair of shoes, choosethe left shoe." However, since the socks of any pair of socks in stock are indistin-
  • 33. Theorem 5.22(restated)(1, I) (1,2)(2, I) (2,2)(3, 1) (3,2)(4, I) (4,2)(5, 1) (5,2)5.5 Comparability of Cardinal Numbers and the Axiom of Choice 227guishable, and since there are an infinite number of pairs of socks, the clerk mustemploy the Axiom of Choice to show the customer one sock from each pair.Example. Let.sa = {A: A ~ IR and A =1= 0}. Ifwe are to select one element from eachset A in .sa, then we will need to use the Axiom of Choice. However, if we letZA = {A: A ~ IR, A =1= 0, and A is finite}, then we do not need the Axiom of Choice toselect one element from each set in ZA. Our choice rule might be: for each B E ZA,choose the greatest element in B. Since B is finite, such an element exists for eachBEZA.With the Axiom of Choice, we could, but will not, prove Theorem 5.34. For aproof, see R. L. Wilder, Introduction to the Foundation of Mathematics, 2nd ed.(Krieger, 1980).In section 5.3, we delayed the proof of Theorem 5.22 because it uses the Axiomof Choice. We restate the theorem and give its proof now.Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerablesets. That is, if i E N, then Ai is denumerable and if i, j E Nand i =1= j then Ai =1= Ajand Ai n Aj = 0. Then UAi is denumerable.iENProof. Since each Ai is denumerable, there exist functions /;: N ---+ Ai i E N, eachof which is one-to-one and onto Ai We define the function h: N X N ---+ UAi byiENhem, n) = fm(n). (To see that h is a bijection, refer tofigure 5.10. Infigure 5./O(a),we have an arrangement ofN X N. The union ofthe Ai, i E N, is presented in fig-ure 5./O(b), where the nth row contains the elements ofAi as the range offi Theone-to-one correspondence h is easily seen.) Since all the functions fi are bijections,it is an easy exercise to show that h is a bijection. Therefore, N X N = UAi SinceN X N = N UA- is denumerable. iEN •, IiEN(1, 3) (1,4) (1, 5) AI: fl(l) f,(2) fl(3) f,(4) f,(5)(2,3) (2,4) (2,5) A2: f2(1) f2(2) f2(3) f2(4) fz(5)(3, 3) (3,4) (3, 5) A3: f3(1) 13(2) f3(3) 13(4) 13(5)(4,3) (4,4) (4,5) A4: f4(1) f4(2) f4(3) f4(4) f4(5)(5, 3) (5,4) (5,5) A5: f5(1) f5(2) f5(3) f5(4) f5(5)(a) N X N (b) UAiiENFigure 5.10
  • 34. 228 CHAPTER 5 CardinalityTheorem 5.35Why was the Axiom of Choice necessary in the proof? For each i E N, A; is de-numerable, so there are (many) bijections from A; to N. In our proof, we select onebijection (and call itj;) from the set of all bijections from A; to N. We do this an in-finite number of times, once for each i E N. Our collection consisted of sets of bi-jections, and we needed one bijection from each set. There is no way to select the!;without the Axiom of Choice.Many other important theorems, in many areas of mathematics, cannot beproved without the use of the Axiom of Choice. In fact, several crucial results areequivalent to it.t Some of the consequences of the axiom are not as natural as theComparability Theorem, however, and some of them are extremely difficult to be-lieve. One of these is that the real numbers can be rearranged in such a way thatevery nonempty subset of R has a smallest element-in other words, that the realscan be well ordered. Another, called the Banach-Tarski paradox, states that a ballcan be cut into a finite number of pieces that can be rearranged to form two balls thesame size as the original ball. Actually, this "paradox" is hardly more surprisingthan exercise l5(b) of section 5.4-that IR X IR = IR, i.e., there are exactly as manypoints on the x-axis as there are in the entire Cartesian plane, a result that can beproved without the Axiom of Choice.The Axiom of Choice has been objected to because of such consequences, andalso because of a lack of precision in the statement of the axiom, which does notprovide any hint of a rule for constructing the choice function F. Because of theseobjections, it is common practice to call attention to the fact that the Axiom ofChoice has been used in a proof, so that anyone who is interested can attempt to findan alternate proof that does not use the axiom.We present two more theorems whose proofs require the Axiom of Choice. Thefirst is that if there is a functionjfrom A onto B, then A must have at least as manyelements as B. The proof chooses for each b E B, an a E A such that j(a) = b. Thesecond theorem is that every infinite set has a denumerable subset. The axiom isused to define the denumerable subset inductively.If there exists a function from A onto B, then B:::; A.Proof._ I(§ = 0, then B k A, so B:::; A. Let j: A ---> B be onto B where B =1= 0. Toshow B:::; If, we must construct a function h: B ---> A that is one-to-one. Since j is ontoB, for each bE B, bE Rng (f) or, equivalently, j-l({b}) =1= 0. Thus .ii =U-1({b}): bE B} is a nonempty collection of nonempty sets. By the Axiom ofChoice, there exists a functiong from.ii to Uj-l({b}) = A such thatg(j-l({b})) Ej-l({b})forallbEB. hEBDefine h: B--->A by h(b) = g(j-l({b})). It remains to prove h is one-to-one.Suppose h(bl ) = h(b2). Then g(j-l({bl})) = g(j-I({b2})). By definition of g, thiselement is in bothj-l({bl}) and j-l({b2}). Thus j-l({bl}) nj-I({b2}) =1= 0. Fort See H. Rubin and J. E. Rubin, Equivalents a/the Axiom a/Choice (North-Holland, New Amsterdam,1963).
  • 35. Theorem 5.36Corollary 5.37Exercises 5.55.5 Comparability of Cardinal Numbers and the Axiom of Choice 229any x Ef-I({bl }) nf- I({b2}),f(xL= bl1. and f(x) = b2. Therefore, bl = b2. Thisproves that h is one-to-one and that B:5 A. •Every infinite set A has a denumerable subset.Proof. We shall inductively define a denumerable subset ofA. First, since A is in-finite, A* 0. Choose al EA. Then A - {al} is infinite, hence nonempty. Choosea2 E A - {al} Note that a2 *al and a2 EA. Continuing in this fashion, supposeal"" ak have been defined. Then A - {al"" ak} *0, so select any ak+l from thisset. By the Axiom of Choice, an is defined for all n E N. The an have been con-structed so that each an E A and ai *aj for i *j. Thus B = {an: n E N} is a subsetof A, and the functionfgiven by fen) = an is a one-to-one correspondence from Nto B. Thus B is denumerable. •Theorem 5.36 can be used to prove that every infinite set is equivalent to one ofits proper subsets. (See exercise 8.) This characterizes infinite sets because, as wesaw in section 5.1, no finite set is equivalent to any of its proper subsets.Theorem 5.36 also confirms that Xo is the smallest infinite cardinal number.For any set~ wUh infinite cardinality, there is a denumerable subset B ofA. There-fore, Xo = B:5 A.A nonempty ~et A is countable iff there exists a function f: N onloA.Proof. Exercise 9. •We have seen that Xo < c and that the first two terms of the sequenceN < Q]>(N) < Q]>(Q]>(N)) < Q]>(Q]>(Q]>(N))) < ... are the cardinal numbers Xo and c.This does not necessarily mean that c is the next largest cardinal numb~r after Xo.Cantor conjectured that this is so: that is, no set X exists such that Xo < X< c. Thisconjecture, called the continuum hypothesis, is one of the most famous problemsin modern mathematics. The combined work of the logicians Kurt Godel in the1930s and Paul Cohen in 1963 shows that the continuum hypothesis can neither beproved nor disproved in Zermelo-Fraenkel set theory. Like the Axiom of Choice,the continuum hypothesis is undecidable.1. Indicate whether the Axiom of Choice must be employed to select one ele-ment from each set in the following collections.(a) an infinite collection of sets, each set containing one odd and one eveninteger.* (b) a finite collection of sets with each set uncountable.
  • 36. 230 CHAPTER 5 Cardinality(c)(d)(e)(f)(g)* (h)an infinite collection of sets, each set containing exactly one integer.a denumerable collection of uncountable sets.{(a, co): a E JR}{A: A ~ N and both A and N - A and infinite}{A: A ~ JR and both A and JR - A are infinite}{A: A ~ JR and A is denumerable}2. Prove this partial converse of Theorem 5.14 without using the Axiom ofChoice. Let A and B be sets with B =1= 0. If B::5 Athen there exists g: A -+ Bthat is onto B.3. Let A and B be any two nonempty sets. Prove that there existsf: A -+ B thathas at least one of these properties:(i) fis one-to-one or (ii) fis onto B.4. Prove that if f: A -+ B, then Rng if) ::5 A.* 5. Suppose A is a denumerable set and B is an infinite subset of A. Prove A =B.6. Suppose B< Cand B$ A. Prove A< C.7. Let {Ai: i E N} be a collection of distinct pairwise disjoint nonempty sets.That is, if i andj are in Nand i =1= j, then Ai =1= Aj and Ai n Aj = 0. Prove thatUAi includes a denumerable subset.iEN"* 8. Let A be an infinite set. Prove that A is equivalent to a proper subset of A.9. Prove Corollary 5.37: A nonempty set A is countable iff there is a functionf: N -+ A that is onto A.Proofs to Grade 10. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Jus-tify assignments of grades other than A.* (a) Claim. There exists an infinite set of irrational numbers, no two ofwhich differ by a rational number."Proof." For x, y E JR, define the relation S by x S y iff x - y E Q. It iseasy to show that S is an equivalence relation. For the family of equiva-lence classes {xis: x E JR}, choose one element from each equivalenceclass except the equivalence class Q. The set of all such chosen elementsis an infinite set of irrational numbers, no two of which differ by a ratio-nal. (You may accept, without proof, the claim that this set is infinite.) -(b) Claim. Every infinite setA has a denumerable subset."Proof." Suppose no subset ofA is denumerable. Then all subsets ofAmust be finite. In particular A ~ A. Thus A is finite, contradicting the as-sumption. -* (c) Claim. Every infinite set A has a denumerable subset B."Proof." IfA is denumerable, let B = A, and we are done. Otherwise, Ais uncountable. Choose XI EA. If A - {XI} is denumerable, let B=A - {XI}. Otherwise, choose X2 EA - {XI}. If A - {XI X2} is denumer-able, let B = A - {xI X2}. Continuing in this manner, using the Axiom ofChoice, we obtain a subset C = {xI, Xz, . .. } such that B = A - C is denu-merable. -
  • 37. 5.5 Comparability of Cardinal Numbers and the Axiom of Choice 231(d) Claim. Every infinite set has two disjoint denumerable subsets."Proof." Let A be an infinite set. By Theorem 5.36, A has a denumer-able subset B. Then A - B is infinite, because A is infinite, and is disjointfrom B. By Theorem 5.36, A - B has a denumerable subset C. Then Band C are disjoint denumerable subsets of A. -(e) Claim. Every infinite set has two disjoint denumerable subsets."Proof." LetA be an infinite set. By Theorem 5.36, A has a denumer-able subset B. Since B is denumerable, there is a function f: N ~:dBLetC={J(2n):nEN} and D={J(2n-I):nEN}. Then C={J(2),f(4),f(6), ...} and D = {J(1),f(3),f(5),...}are disjoint denumerable subsets~A. -(f) Claim. Every subset of a countable set is countable."Proof." Let A be a countable set and let B ~ A. IfB is finite, then B iscountable by definition. If B is infinite, since B ~ A, A is infinite. Thus Ais denumerable. By Theorem 5.36,1l has a ~enu!!1er~le subset C. Thus{; ~!! ~ A, which implies Xo = C and C:::; B:::; A= Xo. ThereforeA= B= Xo. Thus B is denumerable and hence countable. -