Chapter 5

5.1

CHAPTER 5

Cardinality

How many elements are in the set

A = in, 28, ji,!, -3, fl, cr, o}?

After a short pause, you said "eight." Right? Consider for a moment how you arrived at that answer. You probably looked at n and thought "I," then looked at 28 and thought "2," and so on up through 0, which is "8." What you have done is set up a one-to-one correspondence between the set A and the "known" set of eight elements {I, 2, 3, 4, 5, 6, 7, 8}. Counting the number of elements in sets is essentially a 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 another counting problem.

A certain shepherd has more than 400 sheep in his flock, but he cannot count beyond 10. Each day he takes his sheep out to graze, and each night he brings them back into the fold. How can he be sure all the sheep have returned? The answer is that he can count them with a one-to-one correspondence. He needs two containers and a pile of pebbles, one pebble for each sheep. When the sheep return in the evening, he transfers pebbles from one container to the other, one at a time for each returning sheep. Whenever there are pebbles left over, he knows there are lost sheep. The solution to the shepherd's problem illustrates the point that even though we have not counted the sheep, we know that the set of missing sheep and the set of leftover pebbles have the same number of elements-because there is a one-to-one correspondence between them.

Equivalent Sets; Finite Sets

To determine whether two sets have the same number of elements, we see whether it is possible to match the elements of the sets in a one-to-one fashion. This idea may be conveniently described in terms of a one-to-one correspondence (a bijection) from one set to another.

195

196 CHAPTER 5 Cardinality

DEFINITION Two sets A and B are equivalent iff there exists a oneto-one function from A onto B. A and B are also said to be in one-to-one correspondence, 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 function f: A --+ B given by f(5) = r, f(8) = p, and f(ep) = m is one of six such functions that verify this.

Example. The sets C = {x, y} and D = {q, r, s} are not equivalent. There are nine different functions from C to D. An examination of all nine will show that none of them 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 function 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. The function is given by:

d-c f(x) = b _ a (x - a) + c

This is simply the linear function through the points (a, c) and (b, d) restricted to the domain (a, b). We leave the proof that f is a bijection to exercise 4. •

y

c / d

--r-----~------~--~x

a b

Figure 5.1

Theorem 5.1

Theorem 5.2

5.1 Equivalent Sets; Finite Sets 197

We now know that any two open intervals are equivalent, even when the intervals have different lengths. Also, the interval (5, 6) is equivalent to (1, 9), even though 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 sometimes denoted {a, I }N. We will show gji = QJI(N), the power set of N.

Proof. (This proof will show that gji is precisely the set of all characteristic functions with domain N, which is in a one-to-one correspondence with the subsets of N.) 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, every function 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 exists

n E N such that g 1 (n) *" g2 (n). Both gland g2 have codomain {O, I}, so we may assume gl (n) = 1 and g2(n) = 0. (The case gl (n) = ° and g2(n) = 1 is similar.) But then 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 that XA: 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 equivalence relation 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) If A 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 -+ C andg: 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 exercise 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. •


Lemma 5.3

We shall use the symbol Nk to denote the set {I, 2, 3, ... , k}. Think of Nk as the standard 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 for some 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 any one-to-one function from X onto N3 will prove this. One such correspondence f: X -+ N3 is given by f{J8.6) = 1, f(c) = 2, andf(n) = 3.

We use the symbol S to denote the cardinal number ~ a set S. At this time S is defined 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} is finite and has cardinality 4, since it is equal to the set {8, 7, 3, 2}. Since A = N4,

A=4. The symbol A is used for the cardinality of A so as to distinguish what we prove

here from the more informal results in section 2~, where we used the symbol # A. We shall show that for finite sets the cardinality A of A corresponds to our intuitive notion of the number of elements in A.

The empty set is finite by definition. Also, each set Nk is finite and has cardinal number k because the identity function Ir'Jk is a one-to-one function from Nk onto Nk • In addition, any set equivalent to a finite set must also be finite. Suppose A is a finite set and A = B. If A = 0, then B = 0 (see exercise 3). Otherwise, A = Nk for some 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. One goal is to make sure that the notion of cardinality for a finite setA corresponds to our notion of the number of elements in A in the following important way: The cardinality of afinite set should be unique. That is, if A has cardinality m (that is, A = Nm ) and A has cardinality n (A = Nn), then m = n.

You are asked to show this in exercise 14, using a property of finite sets called the pigeonhole principle, discussed later in this section. Our immediate goal is the key theorem that every subset of a finite set is finite. Our proof of this theorem uses two lemmas.

If S is finite with cardinality k and x is any object not in S, then S U {x} is finite and has cardinality k + I.

Proof. If S = 0, then S U {x} = {x}, which is equivalent to NI and thus finite. In this case S has cardinality 0 and S U {x} has cardinality 0 + 1. If S =1= 0, then S = Nk for 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. •

Lemma 5.4

Theorem 5.5

Theorem 5.6


f 1-1, onto

fiT

1-1, onto

Figure 5.2

Every subset of Nk is finite.

Proof. Let A be a subset of Nk• (We prove A is finite by induction on the number k.)

Let k = 1. Then either A = 0 or A = N I, both of which are finite. Assume all subsets of Nk are fini te 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 is finite. 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 assume T =1= 0 and hence S =1= 0. Since S = Nk for some kEN, there is a one-to-one function J from S onto Nk . Then the restriction off, J I T, is a one-to-one function from Tonto J(T). Therefore, T is equivalent to J(T) (see figure 5.2). But J(T) is a subset of the finite set Nk and is finite by Lemma 5.4. Therefore, since T is equivalent 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 lot of 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 of functions to establish facts about cardinalities. This work will be helpful when we deal 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 sets is finite. To this end the next theorem is a special case: the union of two disjoint finite sets is finite. Its proof is a rigorous development of the sum rule (Theorem 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 n B = 0. If A = 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 = N n ,


Corollary 5.7

Lemma 5.8

Theorem 5.9

and suppose that f: A --+ Nm and g: B --+ Nn are one-to-one correspondences. Let H = {m + 1, m + 2, ... , m + n}. Then h: Nn --+ H given by hex) = m + xis aone-toone correspondence, and thus Nn = H. Therefore, B = H by transitivity. Finally, by Theorem 5.2, A U B = Nm U H = Nm +m which proves that A U B is finite and that A UB=m+n. -

(a)

(b)

If A and B are finite sets, then A U B is finite. n

If AI> A2, ... ,An are finite sets, then U Ai is finite. i=1

Proof. We prove part (a) and leave part (b) as an exercise in mathematical induction (exercise 6). Assume that both A and B are finite. Since B - A ~ B, B - A is finite. 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 cardinality by one. It is also true that removing one element from a finite set reduces the cardinality 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 pigeonhole principle. In its informal version it says: "If a flock of n pigeons comes to roost in a house with r pigeonholes and n > r, then at least one hole contains more than one pigeon." If we think of the set of pigeons as Nn and the set of pigeonholes as N" then the pigeonhole principle says any assignment of pigeons to pigeonholes (function 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 begin with 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 just as in the case n = 2 above, so we may assume r > 1. Suppose there is a one-to-one function f: Nn+ 1 --+ N" Then fiNn is a one-to-one function. The range of this function may not contain fen + 1), but by Lemma 5.8 there is a one-to-one correspondence g: Nr - {J(n + I)} --+ Nr - I • Therefore, the composite g 0 fiNn: Nn --+ Nr - I

is 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 about finite sets (see, for example, exercise 13). We use it here to give a characterization offinite sets.

Corollary 5.10

Exercises 5.1


A 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 the general case as exercise 15. The case k = 1 is trivial, so let k > 1. Suppose A is a proper 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 is one-to-one. But then i 0 I: Nk --+ Nk- 1 is one-to-one, which contradicts the 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-one correspondence). 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 contradictioo. •

Corollary 5.10 says: "If A is a finite set, then A is not equivalent to any of its proper subsets." The contrapositive is: "If A is equivalent to one of its proper subsets, then A is infinite." Earlier we observed that the open interval (0, 2) is equivalent to its proper subset (0, 1). Thus we know that (0, 2) is infinite. There will be many 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 and g: B--+ D are one-to-one correspondences, then I: A X B--+ ex D given by I(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 equivalent

by 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 Earth the set of all positive integer powers of 2 the set of four-letter words in English the set of rational numbers the 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}


Proofs 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 a 2 + b2 = I

6. 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) if A is infinite and A ~ B, then B is infinite.

"* 8. (a) Prove that for all k, mEN, Nk X Nm is finite.

"*

"*

"*

9.

10.

* 11.

12.'

13.

14.

15.

16.

17.

"* "*

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 Bare finite, 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 + B Using th~ meQlods of this section, prove that if A and B are finite sets, then A 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, then m=n.

Complete the proof of Corollary 5.10 by showing that if A is finite and B is a proper 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. Justify assignments of grades other than A. (a) Claim. If A and B are finite, then A U B is finite.

"Proof." If A and B are finite, then there exist m, n E N such that A = Nm , I-I I-I I-I

B = Nn- Let f: AOi1i6Nm and h: BOi1i6 Nn- Then f U h: A U BOi1i6 Nm+m which 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 integer 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 . Thus

S U {x} is finite. -

5.2

Theorem 5.11

5.2 Infinite Sets 203

(c) Claim. If AX 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. Since A is equivalent to a finite set, A is finite. -

Infinite Sets

This section will describe several infinite sets and their cardinal numbers. We will see that there are many infinite sets and that it is not true that all infinite sets are equivalent. Just as we use Nk as our standard set of k elements, we will specify other sets as standard sets for certain infinite cardinal numbers.

In the previous section an infinite set was defined as a nonempty set that cannot be put into a one-to-one correspondence with any Nk . According to the next theorem, the set N is one such set. Since infinite means not finite, our proof, as might be 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 is onto N.)

Let n = f(l) + f(2) + ... + f(k) + 1. Since each f(i) > 0, n is a natural number larger than each f(i). Thus n =I- f(i) for any i E Nk . Hence n t!. Rng if). Therefore, 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 positive integers. The function f: N --+ E+ defined by setting f(x) = 2x is clearly a one-to-one function from N onto E+. (Of course, there are many other one-to-one correspondences; the one given is the simplest.) Thus N is equivalent to one of its proper subsets. Therefore, by Corollary 5.10, the set of natural numbers cannot be finite.

There are many other infinite sets, some-but not all-of which are equivalent to N. Infinite cardinal numbers are written using the symbol X, aleph, the first letter of 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, akin to the integer 0 being the smallest finite cardinal. The set N is our "standard" set for the 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; we write S = Xo.


Theorem 5.12

We 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 the same number of elements (~o) as N. Although our intuition might tell us that only half of the natural numbers are even, it would be misleading to say that N has twice as many elements as E + , or even to say that N has more elements than E +. We must rely on one-to-one correspondences to determine cardinality.

In section 5.4 we will show that every infinite set is equivalent to one of its proper subsets. Together with Corollary 5.10, this will characterize infinite sets: a set 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 equivalent to N.

The set II of integers is denumerable.

Proof. We define f: N ---+ II by

[(xl = {~ ;x if x is even

if 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 1

II = {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 both x Y f dd i-x i-y even, then 2 = 2' and thus x = y. I x and yare both 0 , then -2- = -2-' so

1 - x = 1 - y and x = y. If one of x and y is even and the other is odd, then only one of fix) or f(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 2w is even and f(2w) = 2; = w. If wEll and w ::5 0, then 1 - 2w is an odd natural

i-(i-2w) 2w 7J number and f(l - 2w) = 2 = 2"" = w. In either case, if wE fL, then

wE Rng if). •

Example. The set P of reciprocals of positive integer powers of 2 is denumerable. Writing the set P as

P = {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 a bijection 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 example leads directly to the next theorem.

Theorem 5.13


If 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 cardinal numbers for insight into infinite cardinals. In chapter 2, you were asked to show that if A is a finite set with m elements and B is a finite set with n elements, then A X B has 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 will not pursue the arithmetic of cardinal numbers, but simply remind you that extension of 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, is a 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, denumerable sets are sometimes referred to as "countably infinite" sets.

We have seen examples of countable sets (both finite sets and denumerable sets), but no example, as yet, of a set that is uncountable. Before considering the next theorem, which states that the set of real numbers in the interval (0, 1) is such an example, we need to review decimal expressions for real numbers. In its decimal form, any real number in (0, 1) may be written as 0.ala2a3a4"" where each ai is an integer, 0:::; ai:::; 9. In this form, i2 = 0.583333 ... , which is abbreviated to 0.583 to indicate 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 iff there is no k such that for all n > k, an = 9. For example, 0.82142857 and ~ = 0.40 are in normalized form, but 0.49 is not. Every real number can be expressed uniquely 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 two decimal numbers in normalized/orm are equal iff they have identical digits in each decimal position.

Uncountable Sets

Infinite Sets

Figure 5.3

Finite Sets

Countable Sets


Theorem 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 that f does not map onto (0, I). Writing the images of the elements of N in normalized form, we have

f(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= 5 if 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 decimal place. We conclude there is no one-to-one and onto function from N to (0, I) and hence (0, I) is not denumerable. a

The open interval (0, I) is our first example of an uncountable set. The cardinal number of (0, I) is defined to be c (which stands for continuum) and is the only infinite 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 of cardinality c.

DEFINITION A set S_ has cardinality c iff S is equivalent to the open interval (0, I); we write S = c.

There is nothing special about using the interval (0, I) to define the cardinal number c; every open interval (a, b), where a, bE IR and a < b has cardinality c. In section 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 note that the functionf: (0,1) ---+ A, given by

{2X + 3,

f(x) = 2x +4,

ifO <x <! I ' if -::::; x < I 2

is a one-to-one correspondence.

Theorem 5.15

Exercises 5.2


y

---+--~~--+-~x

Figure 5.4

The 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) -+ ~ by lex) = tan (nx -1)' See figure 5.4. The function / is a contraction and translation of one branch of the tangent function and is one-toone and onto R Thus (0, I) = R •

Because we know that N X N has the same cardinality as N, you may suspect that ~ X ~ has the same cardinality as R Indeed, it is true that ~ X ~ has cardinality c. We omit the proof because the specialized methods needed will not be useful in our current study.

Since N is a proper subset of ~, and N = Xo, ~ = c, and Xo *" c, it is natural to suspect (correctly) that Xo < c. We consider the ordering of cardinal numbers in section 5.4. In the meantime, we delve more deeply into the study of countable sets in 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


(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 even integers 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" A 2 , A3, ... ,An be denumerable sets. (a) Prove thatA I X (A 2 X A3) and (AI X A 2) X A3 are denumerable. (Recall

from 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 denumerable, for every natural number n.

"* 9. Prove Theorem S.13 directly. That is, if A and B are denumerable sets, construct 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 numbers, has cardinality c.

Proofs to Grade

5.3

5.3 Countable Sets 209

12. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

* (a) Claim. The sets E+ of even natural numbers and D+ of odd natural numbers are equivalent. "Proof." E + is an infinite subset of N. Thus E + is denumerable. Likewise D+ is an infinite subset of N and is denumerable. Therefore, E+ =D+. -

(b) Claim. If A 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-one correspondence. Then g: N --+ A U {x}, defined by

{X if t = 1

g(t)= l(t-l) ift>1

is one-to-one and onto A U {x}. Thus N = A U {x}, so A U {x} is infinite. -* (c) Claim. If AU B is infinite, then A and B are infinite.

"Proof." This is a proof by contraposition so assume the denial of the consequent. 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 result is proved. -

* (d) Claim. If a set A is infinite, then A is equivalent to a proper subset of A. "Proof." Let A = {Xl> X2,' .. }. Choose B = {X2, X3""}' Then B is a proper subset of A. The function I: A --+ B defined by I(Xk) = Xk+ 1 is clearly 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 correspondence between Nand N - {I}, so N is equivalent to a proper subset of N. Therefore, by Corollary 5.10, N is infinite. -

Countable Sets

Countable sets have been defined as those sets that are finite or denumerable. Such sets are countable in the sense that they can be "counted" by using subsets of N: A finite non empty set is counted by using exactly the elements of some Nk , while a denumerable set may be counted by using exactly the elements of N. This section will survey some of the important results about denumerable and countable sets. First we 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, or some infinite cardinal in between (the ordering of cardinal numbers is discussed in the next section). You might also suspect that, since there are infinitely many rationals between any two rationals, Q is not denumerable, but this is not the case. Georg Cantor (1845-1918) first showed that Q+ (the positive rationals) is indeed denumerable through a clever rearrangement of Q+.

Every element in Q+ may be expressed as ~ for some p, q E N. Thus the elements of this set can be presented as in figure 5.5, where the nth row contains all the positive fractions with denominator n.


Theorem 5.16

~' //~/~ ll~~ I 2 3 456 - - - - - -

4 5

Figure 5.5

5 3

5 4

5 5

2

6 3

6 4

6 5

7 I

7 2

7 3

7 4

7 5

To show that Q+ is denumerable, Cantor gave a "diagonalization argument" in which the elements of Q+ are listed in the order indicated in figure 5.5. First are all fractions in which the sum of the numerator and denominator is 2 (only t), then those whose sum is 3 (r, ~), then those whose sum is 4, and so on. We omit from the list ~ (= 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 correspondence from N to Q+:

N = {I, 1

Q+ = {I l'

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1 1 1 1 1 1 1 1 1 1 1 1 213143215165 l' 2' l' 3' l' 2' 3' 4' l' 5' l' 2'

14, ... } 1 ~, ... }

This correspondence can be used to establish the following theorem (we omit the details). 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. The next three theorems are analogous results for denumerable sets. An important distinction may be made between finite sets and denumerable sets: adding a finite number of elements or a denumerable number of elements does not change the cardinality of 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 denumerable set. (The function/: N -+ A given by lex) = -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, yields a denumerable set: { ... , -10, -8, -6, -4, -2,2,4, 6, 8}.

Theorem 5.17

I Lobby I


Even adding a denumerable number of elements, say the nonnegative even integers, still yields a denumerable set: { ... , -10, -8, -6, -4, -2,0,2,4,6,8, 1O, ... }.

If A 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. Define g: N ---+ A U {x} by

g(n) = {fen - I) if n = I if n > I

It is now straightforward to verify that g is a one-to-one correspondence between N and 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 by the story of the Infinite Hotel, I' whose rooms are numbered I, 2, 3, 4, .... The Infinite Hotel has Xo rooms and is full to capacity with one person in each room. You approach the desk clerk and ask for a room. When the clerk explains that each room is already occupied, you say, "There is room for me! For each n, let the person in room n move to room n + I. Then I will move into room I, ·and everyone will have a room as before." There are Xo + I people and they fit exactly into the Xo rooms. See figure 5.6.

Figure 5.6

t The Infinite Hotel is one of the topics discussed in Aha! Gotcha: Paradoxes to Puzzle and Delight by Martin Gardner (Freeman, New York, 1981).


Theorem 5.18

Theorem 5.19

Theorem 5.20

Theorem 5.21

Rooms in the Infinite Hotel can also be found for any finite number of additional 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 additional people arrive a finite number of times (Theorem 5.21) or even a denumerable number of times (Theorem 5.22).

If A is denumerable and B is finite, then A U B is denumerable.

Proof. Exercise 3. • If A 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 odd

if n is even.

It is left as exercise 4 to show that h is a one-to-one correspondence from N onto AUR •

Before finishing the theorems about denumerable sets, we note that Theorems 5.16, 5.17, and 5.19 provide a simple proof that the set of all rational numbers is denumerable.

The set Q of all rational numbers is denumerable.

Proof. By Theorem 5.16, Q+ is denumerable. Likewise, the set of negative rationals, Q-, is denumerable. (The function f: Q+ -> ((J)- that assigns f(x) = -x is a bijection.) 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 disjoint denumerable sets. Its proof, as you might expect, is by induction on the number of sets in the family.

Let {A;: i = 1, 2, ... , n} be a finite pairwise disjoint family of denumerable sets. n

Then U Ai is denumerable. i=1

Proof. 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 proved from the properties of sets we have used thus far. The axioms of set theory allow us

Theorem 5.22

Theorem 5.23

Corollary 5.24

Theorem 5.25


to form unions, intersections, power sets, and infinite sets such as N, Z, G, and even IR. However, a new property (The Axiom of Choice) to be introduced in section 5.5 is needed to prove Theorem 5.22. This does not make Theorem 5.22 suspect, for the Axiom of Choice is a widely accepted and useful tool. A complete proof of Theorem 5.22 is given in section 5.5.

Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerable sets. That is, if i E N, then Ai is denumerable, and if i, j E Nand i =1= j, then Ai =1= Aj

and Ai n Aj = 0. Then U Ai is denumerable. i=N

We continue with a sequence of important results about denumerable and countable sets, leading to the fact that the union of any countable collection of coutable 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 bijection 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 smallest integer in feB). For each n E N,f{B} - {g{I}, g{2}, ... , g{n)} is nonempty because feB) 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; thus g(r) =1= g(s). Thus g is an injection. Also, if t Ef(B), and there are k natural numbers less 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 therefore countable. Thus such sets as {k n EN}, G n (0, 1), {~,~, ~}, and Z are countable sets.

A set A is countable iff A is equivalent to some subset of N.

Proof. If A is countable, then A is either finite or denumerable. Thus A = 0, or A = 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. -

If A 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 (Corollary 5.7), where one set is denumerable and the other is finite (Theorem 5.18), and where A and B are denumerable and disjoint (Theorem 5.19). The only remaining case 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 by Theorem 5.23. If B - A is finite, then A U B is denumerable by Theorem 5.18, and if B - A is denumerable, then A U B is denumerable by Theorem 5.19. -


Theorem 5.26

Lemma 5.27

Theorem 5.25 may be extended (by induction) to any finite union of countable sets.

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}. Then

Al = 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 a nEN

denumerable collection of denumerable sets, we cannot use Theorem 5.22 to conclude 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. Let CJ3 be the family {Bn: n EN}, where

B, =AI ={ ... , -2, -I,a, I}, B2 = A2 - A, = {2}, B3 = A3 - (A 2 UA,) = {3}, B4 = A4 - (A3 UA2 UA,) = {4},

and so on. This family is a pairwise disjoint collection. Furthermore, U Bn = nEN

U AI!" Such a family CJ3 may be defined for any denumerable family of sets (in nEN

Lemma 5.27) and used to show that the union of any denumerable family of denumerable 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; - (CJ Ak ). Then {B;: i E N} is a denumerable family of pairwise disjoint sets and k='

UA;= UBi. iEN iEN

Proof. See exercise 8. •

Theorem 5.28

Theorem 5.29


Let {Ai: i E N} be a denumerable family of denumerable sets. Then U Ai IS

denumerable. iEN

Proof. Al ~ U Ai, so the union is infinite by exercise 7(b) of section 5.1. Ifwe let iEN

Bi = Ai - (0 Ak), then by Lemma 5.27 it is sufficient to show that UBi is k=! iEN

denumerable.

If all the Bi, i E N, are denumerable, then by Theorem 5.22, U Bi is denumeriEN

able. Likewise, if all the Bi , i E N, are finite, then U B; is denumerable (exercise 9). ;EN

Let ~I = {Bi: Bi is finite} and ~2 = {B;: B; is denumerable}. Both ~I and ~2

are 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 1

nite (by Theorem 5.7) and U B is denumerable by Theorem 5.22. If~1 is infinite, BE!'Jl2

then U B is denumerable (exercise 9 again). If ~2 is finite, then U B is denu-BE!'Jll BE!'Jl2

merable (by Theorem 5.21); otherwise U B is denumerable by Theorem 5.22. BE!'Jl2

Thus U B; is either the union of a finite set and a denumerable set or the union of ;EN

two denumerable sets. In either case, U B; is denumerable. • iEN

We can pull all these results together in one simply stated theorem:

Let .il be a countable collection of countable sets. Then U.il is countable.

Proof. You should check to see that every case is considered, either as a theorem or an exercise. •

Theorem 5.29 provides another means to prove that the positive rational Q+ is denumerable. Q+ is certainly infinite, and Q+ may be written as U Am where

nEN

An = {~: a EN}. For each n E N, An is denumerable. By Theorem 5.29, Q+ is denumerable.


Exercises 5.3

Example. Each set Nb kEN is a finite set and therefore countable. Their union, U Nk = N, is of course countable.

kEN

Example. For each n E N, let Bn = {-n, n}. The countable union of these countable sets is 71 - {O}, which is countable.

We will give one more example of a denumerable set, one with particular significance for computer programmers. A computer program is written in a given programming language and consists of a finite sequence of symbols. These symbols are selected from a finite set called an "alphabet" (typically all 26 upper and lowercase letters, the 10 digits, a blank space, certain punctuation marks, arithmetic operations, etc.). For any particular language, let PI be all programs of length I, let P2 be all programs of length 2, P3 be all programs of length 3, and so on. We note that in most programming languages, the first few P; are empty sets. An example of an element of Pso in the language BASIC is

Program Xl (10 symbols) Let X = 1 (9 symbols) Let Y = 2 (9 symbols) Print "X + Y is II • X + Y (19 symbols) I

End (3 symbols)

Since any computer program is finite in length, every program is an element of Pn for some n. Thus, the set of all possible programs is

By Theorem 5.29, this countable union of countable sets is countable. Hence only a countable 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. This means that there are many functions from N to {O, I} for which there can be no computer programs to compute them! Put a different way, there are not enough solutions (programs) for all the possible problems (functions).

1. What is the 28th term in the sequence of positive rationals given in the discussion of Theorem 5.16?

2. Explain how a diagonalization argument can be used to show that the set of rationals {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 defined is one-to-one and onto A U B.


*: 5. Prove Theorem 5.21: The union of a finite pairwise disjoint family of denumerable 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)

20

U ((j) n (n, n + I») is denumerable. n=l

U ((j) n (n, n + I») is denumerable. nEN

U {nk: kEN} is denumerable. nEN 2

7. 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 distinct finite sets, then U Bi is denumerable.

iEN

10. The Infinite Hotel is undergoing some remodeling, and consequently some of the rooms will be taken out of service. Show that, in a sense, this does not matter as long as only a "few" rooms are removed. That is, (a) prove that if A is denumerable and x E A, then A - {x} is denumerable. (b) prove that if A is denumerable and B is a finite subset of A, then A - B is

denumerable.

*: 11. Use an argument similar to the diagonalization argument for Theorem 5.16 to prove Theorem 5.13 (if A and B are denumerable sets, then A X B is denumerable).

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 is

finite.

13. (a) Let S be the set of all sequences of O's and I's. 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. Prove that S is uncountable.

*: (b) Let Tbe the set of all sequences of 0' sand l' s where all but a finite number of terms are O. Use a diagonalization argument, like that in Theorem 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.)


Proofs to Grade

5.4

15. Prove that if A 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. Justify 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. If A and B are denumerable, then A X B is denumerable. "Proof." Assume A and B are denumerable, but that A X B is not denumerable. Then A X B is finite. Since A and B are denumerable, they are not empty; therefore, choose a E A and bE B. By exercise 3(b) of section 5.1, A = A X {b}, and by an obvious modification of that exercise, B = {a} X B. Since A X B is finite, the subsets A X {b} and {a} X B are finite. Therefore, A and B are finite. This contradicts the statement that A and B are denumerable. We conclude that A X B is denumerable.

-(c) Claim. (Theorem 5.16) The set Q+ of positive rationals is denumerable. "Proof." Consider the positive rationals in the array in figure 5.5 (Theorem 5.16). Consider the order formed by listing all the rationals in the first row, then the second row, and so forth. Omitting fractions that are not in lowest terms, we have an ordering of Q+ in which every positive rational 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, ... } and B = {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-one and onto B, so A = B. _

The Ordering of Cardinal Numbers

The theory of infinite sets was developed by Georg Cantor over a 20-year span, culminating primarily in papers that appeared during 1895 and 1897. He described a cardinal number of a set M as "the general concept which, with the aid of our intelligence, results from M when we abstract from the nature of its various elements and from the order of their being given." This definition was criticized as being less precise and more mystical than a definition in mathematics ought to be. Other definitions 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

Theorem 5.30

5.4 The Ordering of Cardinal Numbers 219

of sets under the relation = , and then to call this set the cardinal number of each set in the class. Under such a procedure we would think of the number 0 as being the empty 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. For our purposes the essential point is that the cardinal number A of a set A is an object ass~iated with all sets equivalent to A and with no other set. The double overbar on A is suggestive of the double abstraction referred to by Cantor. Another common notation for the cardinal number of set A is IA I.

Cardinal numbers may be ordered (compared) in the following manner:

DEFINITIONS

A = B if and only if A = B; otherwise A oF B. = = 1-1 A :S B if and only if there exists f: A--+B.

A < B if and only if A :S B and A oF 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 A 1:. B and A "$ B to denote the denials of A < ~ anQ.A :S B, respectively.

Usually a proof of1::s §. will involve constru£.tin~a one-to-one function from A to B, while a proof of A < §. wi!! have a proof of A :S B together with a proof, generally by contradiction, that A oF B. Once we have developed some properties Qf ca!:::, dinal inequalities those facts can be used to prove statements of the form A :S B without resorting to the construction of functions.

Since 1, 2, 3, ... are cardinal numbers, the natural numbers may be viewed as a subset 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 extensions 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) If 1: = §. and §. = ~, then 1: = ~. !f A :§ B a,!!d B...? C..1.. the.!! A :S C. A :S B iff A < B or A = B. If A 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 and 1-1

(e)

[[ B-=---C. Since the composite go f: A -+ C is one-to-one, we conclude A :sC. Let A C B. We note JhatJhe inclusion map i: A -+ B, given by i(a) = a, is one-to-one, whence A :S B. •


Theorem 5.31

Example. k < Xo for every finite cardinal number k. From Nk ~ N we have k :::; 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 name remains attached particularly to the next theorem, which states that the cardinality of a set A is strictly less than the cardinality of its power set. We already know the result to be true if A is a finite set with n elements, since QJ>(A) has 2n elements (by Theorem 2.4) and n < 2n for all n E N. The proof given here holds for all sets A.

(Cantor's 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 exists g: A ~~td QJ>(A). Let B = {y E A: y t!. g(y)}. Since B ~ A, BE QJ>(A), and since g is onto QJ>(A), B = g(z) for some z EA. Now either z E B or z t!. B. If z E B, then z t!. g(z) = B, a contradiction. Similarly, z t!. B implies z E giz), which implies z E B, again a contradiction. We conclude A :f;QJ>(A) and hence A < QJ>(A). •

Cantor's Theorem has some interesting consequences. First, there are infinitely many infinite cardinal numbers. We know one, Xo, which corresponds to N. By Cantor's Theorem, Xo < QJ>(N). Since QJ>(N) is a set, its power set QJ>(QJ>(N)) has a strictly greater cardinality than that of QJ>(N). In this fashion we may generate a denumerable 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 up later in this section. It is also an immediate consequence of Cantor's Theorem that there 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} is equivalent to QJ>(N). Since N < QJ>(N), we know there are uncountably many functions from N to {O, I}. Since a function from N to {O, I} is a sequence of 0' sand 1 's, Cantor's 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 ~~ in figure 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. Cantor solved 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 independently by Ernest SchrOder in 1896 and two years later by Felix Bernstein.

Theorem 5.32


A

F

C = Rng (G) G

Figure 5.7

((!nto!-S(h!iid~-Berl!Ste~ Theorem) If A ::5 Band B ::5 A, then A = B.

B

Proof. We may assume that A and B are disjoint, for otherwise we could replace A and B with the equivalent disjoint sets A X {a} and B X {I}, respectively. Let

1-1 1-1 F: A---+B, with D = Rng (F), and let G: B---+A, with C = Rng (G). If B = D we already have A = B, so assume B - D =1= 0.

Define a string to be a function f: N ---> A U B such that

f(1) EB - D, fen) EBimpliesf(n + 1) = G(j(n», and fen) E A implies fen + 1) = F(j(n».

We think of a string as a sequence of elements of A U B with first term in B - D, and such that thereafter the terms are alternately in C and in D. Each element of B - D is the first term of some string. See figure 5.8.

G

F

G

F

G

Figure 5.8 String I: 1(1), 1(2), 1(3), 1(4), 1(5), 1(6), ...


H

Figure 5.9

Let W = {x EA: x is a term of some string}. We note that We Rng (G) and that x E W iff x = f(2n) for some string f and natural number n. Let H: A -+ B be given by

_ {F(X) H(x) - G-1(x)

if x ff. W ifxEW.

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 both

be in W or both in A - W. (We will use a proof by contradiction to show this.) Suppose x E A - Wand yEW. (The symmetric situation where yEA - Wand x E W produces a similar contradiction.) In this case, H(x) = H(y) is F(x) = G-1(y), so y = 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 G is one-to-one). If n = 1, then f(1) = F(x), which implies f(I) E Rng (F), a contradiction to the definition of string f Thus n:::=:: 2. But then f(2n - 2) = x, since f(2n - 1) = F(x). This implies that x is a term in the string!, a contradiction to x 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). Therefore, x = y since G- 1 is one-to-one. Likewise, if x, yare both in A - W, then F(x) = F(y) and x = y because F is one-to-one. In either case, we conclude x = y and H is one-to-one.

Next we show that H is onto B. Let bE B. We must show that b = H(x) for some 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 of some string and G(b) is the second element of that string, a contradiction to G(b) ff. W.) Since bE Rng (f), there exists x E A such that F(x) = b. Furthermore, x ff. w. (If x E W, then x is


a term in some string and therefore F(x) and G(F(x)) are the next two terms of the same string. But this is a contradiction, since G(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 by Theorem 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-SchraderBernstein Theorem and thus (0, I) = [0, I].

Example. The Cantor-Schrader-Bernstein Theorem can be used to determine the relationship of c, the cardinal number of the open interval (0, I), to the increasing sequence of cardinal numbers

N < rJ>(N) < rJ>(rJ>(N)) < rJ>(rJ>(rJ>(N))) < ....

First, recall that any real number in the interval (0, I) may be expressed in a base 2 (binary) expansion 0.b)b2b3b4 ... , where each bi is either ° or 1. Ifwe exclude repeating 1 's, 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 and is 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

_ {2 an - 5 ifnEA if n I$A.

For any set A <: N, g(A) is a real number in (0, I) with decimal expansion consisting of2's and 5's. (Any pair of digits not 9 will do.) The function g is one-to-one but certainly 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 of the familiar ordering properties of N to properties for all cardinal numbers. It, in tum, leads to others. In the following, parts (a) and (c) are proved; (b) and (d) are given as exercise 13.


Corollary 5.33

Exercises 5.4

For 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. if A < E and 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 that A = E, wQich.is a c~tra~ction. !.her~ore, E 1:: A. SuppQ§eA_< E andj1::5 S. Then A ::5 E; sQJJy '"Q1eorel!!..5.21(c), A..? C~Suppose JL 1:: (, 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, we have A <c. _

It is tempti~ to ~xtend .Qur r~sults even further to include the converse of Corollary 5.33ia): JIE 1:: j1, t~nA ::5 E." (As far as we know now, for two given sets A and B, both A ::5 E and E::5 A may be false.) The Cantor-Schroder-Bernstein Theorem turned out to be more difficult to prove than one would have guessed from its simple statement, but the situation regarding the converse of Coroll~ 513(a) is ~en '!pore remarkable. This is _disc.!!sseQ. in --.?ectism ~5, where "If E 1:: A, then A::5 E" is rephrased 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 E and 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 = E implies 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

Proofs to Grade

10.

11.


where G(x) = i + 5. Note that t and! are in B - Rng (F). Let J be the string that begins at t, and let g be the string that begins at !. (a) Find J(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 and

find H(2), H(8), H(l3), and H(20). I-I I-I I-I

Suppose 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 is

an onto map. * (b)

(c) (d)

I-I a functionJ: IR--N.

1-1 a function J: 0J>\~~--N. a functionJ: IR--Q.

12. Prove that if there 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 real

line as there are in the Cartesian plane. (c) .!f A <: IR and there exists an open interval (a, b) such that (a, b) <: A, then

A=c. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades ot~r t~n A. _ 16. (a) Claim. If A ::5 B a'.!...d A -== C, t1:!en ~ ::5 B.

"Proof." Assume A ::5 B and A = C. Then there exists a functionJsuch I-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. Thus C - B oF 0. This implie~C -_B> O. But C = B U (C - B)~nd,~ince B and 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 B a~ B =' C, then A ::5~. "Proof." Assume A ::5 B. Then, since B = C, we have A ::5 C by substitution. -

* (d) Claim. If A 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 particular a* EA. Define J: B -+ A as follows:

(i) For bE B, if bE Rng (g), then g-I({b}) consists of a single element of A, 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 Band g-I({b}) = a, soJ(b) = a. -


5.5 Comparability of Cardinal Numbers and the Axiom of Choice

Theorem 5.34

The Axiom of Choice

The goal of section 5.4 was to establish results for the relations:::; and < on the class of all cardinal numbers while thinking of these results as extensions of properties of 1\1. One of the most useful ordering properties of 1\1 is the trichotomy property: if m and n are any two natural numbers, then m > n, m = n, or m < n. The analog for cardinal numbers is stated in the Comparability Theorem.

(The Comparability Theorem)_ If A 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 axioms and other theorems of Zermelo-Fraenkel set theory (see section 2.1). In a formal study of set theory one can build up, starting with a few axioms specifying that certain collections are sets, to the study of the natural, rational, real, and complex numbers, polynomial, transcendental, and differentiable functions and all the rest of mathematics. Still, comparability cannot be proved. On the other hand, it is impossible to prove in Zermelo-Fraenkel set theory that comparability is false. Theorem 5.34 is undecidable in our set theory; no proof of it and no proof of its negation could ever be constructed in our theory.

At this point we could choose either to assume that Theorem 5.34 is true (or assume true some other statement from which comparability can be proved) or else assume 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 labeling the statement as a theorem. It has become standard practice by most mathematicians to assume the Comparability Theorem is true by assuming the truth of the following statement:

If .il is any collection of non~mpty sets, then there exists a function F (called a choice function) from .il to U A such that for every A E .il, F(A) EA.

AEd

The Axiom of Choice at first appears to have little significance: From a collection of nonempty sets, we can choose an element from each set. If the collection is finite, then this axiom is not needed to prove the existence of a choice function. It is only for infinite collections of sets that the result is not obvious and for which the Axiom of Choice is independent of other axioms of set theory.

Many examples and uses of the Axiom of Choice require more advanced knowledge of mathematics. The example we present is not mathematical in content yet has become part of mathematical folklore.

A shoe store has in the stockroom an infinite number of pairs of shoes and an infinite number of pairs of socks. A customer asks to see one shoe of each pair and one sock of each pair. The clerk does not need the Axiom of Choice to select a shoe from each pair of shoes. His choice rule might be "From each pair of shoes, choose the left shoe." However, since the socks of any pair of socks in stock are indistin-

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 227

guishable, and since there are an infinite number of pairs of socks, the clerk must employ the Axiom of Choice to show the customer one sock from each pair.

Example. Let.sa = {A: A ~ IR and A =1= 0}. If we are to select one element from each set A in .sa, then we will need to use the Axiom of Choice. However, if we let 'ZA = {A: A ~ IR, A =1= 0, and A is finite}, then we do not need the Axiom of Choice to select 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 each BE'ZA.

With the Axiom of Choice, we could, but will not, prove Theorem 5.34. For a proof, 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 Axiom of Choice. We restate the theorem and give its proof now.

Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerable sets. That is, if i E N, then Ai is denumerable and if i, j E Nand i =1= j then Ai =1= Aj and Ai n Aj = 0. Then U Ai is denumerable.

iEN

Proof. Since each Ai is denumerable, there exist functions /;: N ---+ Ai' i E N, each of which is one-to-one and onto Ai' We define the function h: N X N ---+ U Ai by

iEN

hem, 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 of the Ai, i E N, is presented in figure 5./O(b), where the nth row contains the elements of Ai as the range offi' The one-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 = U Ai' Since

N X N = N U A- is denumerable. iEN • , I

iEN

(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) UAi iEN

Figure 5.10


Theorem 5.35

Why was the Axiom of Choice necessary in the proof? For each i E N, A; is denumerable, so there are (many) bijections from A; to N. In our proof, we select one bijection (and call itj;) from the set of all bijections from A; to N. We do this an infinite number of times, once for each i E N. Our collection consisted of sets of bijections, 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 be proved without the use of the Axiom of Choice. In fact, several crucial results are equivalent to it.t Some of the consequences of the axiom are not as natural as the Comparability Theorem, however, and some of them are extremely difficult to believe. One of these is that the real numbers can be rearranged in such a way that every nonempty subset of R has a smallest element-in other words, that the reals can be well ordered. Another, called the Banach-Tarski paradox, states that a ball can be cut into a finite number of pieces that can be rearranged to form two balls the same size as the original ball. Actually, this "paradox" is hardly more surprising than exercise l5(b) of section 5.4-that IR X IR = IR, i.e., there are exactly as many points on the x-axis as there are in the entire Cartesian plane, a result that can be proved without the Axiom of Choice.

The Axiom of Choice has been objected to because of such consequences, and also because of a lack of precision in the statement of the axiom, which does not provide any hint of a rule for constructing the choice function F. Because of these objections, it is common practice to call attention to the fact that the Axiom of Choice has been used in a proof, so that anyone who is interested can attempt to find an alternate proof that does not use the axiom.

We present two more theorems whose proofs require the Axiom of Choice. The first is that if there is a functionjfrom A onto B, then A must have at least as many elements as B. The proof chooses for each b E B, an a E A such that j(a) = b. The second theorem is that every infinite set has a denumerable subset. The axiom is used 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. To show B :::; If, we must construct a function h: B ---> A that is one-to-one. Since j is onto B, 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 of Choice, there exists a functiong from.ii to U j-l({b}) = A such thatg(j-l({b})) E j-l({b})forallbEB. hEB

Define 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, this element is in bothj-l({bl}) and j-l({b2}). Thus j-l({bl}) nj-I({b2}) =1= 0. For

t See H. Rubin and J. E. Rubin, Equivalents a/the Axiom a/Choice (North-Holland, New Amsterdam, 1963).

Theorem 5.36

Corollary 5.37

Exercises 5.5


any x Ef-I({b l }) nf- I({b2}),f(xL= bl1. and f(x) = b2. Therefore, b l = b2. This proves 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 of A. First, since A is infinite, A*' 0. Choose al EA. Then A - {al} is infinite, hence nonempty. Choose a2 E A - {al}' Note that a2 *' al and a2 EA. Continuing in this fashion, suppose al"'" ak have been defined. Then A - {al"'" ak} *' 0, so select any ak+l from this set. By the Axiom of Choice, an is defined for all n E N. The an have been constructed so that each an E A and ai *' aj for i *' j. Thus B = {an: n E N} is a subset of A, and the functionfgiven by fen) = an is a one-to-one correspondence from N to B. Thus B is denumerable. •

Theorem 5.36 can be used to prove that every infinite set is equivalent to one of its proper subsets. (See exercise 8.) This characterizes infinite sets because, as we saw 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 of A. Therefore, Xo = B :5 A.

A nonempty ~et A is countable iff there exists a function f: N onlo'A.

Proof. Exercise 9. • We have seen that Xo < c and that the first two terms of the sequence

N < 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. This conjecture, called the continuum hypothesis, is one of the most famous problems in modern mathematics. The combined work of the logicians Kurt Godel in the 1930s and Paul Cohen in 1963 shows that the continuum hypothesis can neither be proved 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 element from each set in the following collections. (a) an infinite collection of sets, each set containing one odd and one even

integer. * (b) a finite collection of sets with each set uncountable.


(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 of Choice. Let A and B be sets with B =1= 0. If B ::5 A then there exists g: A -+ B that is onto B.

3. Let A and B be any two nonempty sets. Prove that there existsf: A -+ B that has 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 < C and 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 that U Ai 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 function f: 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 of which 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 is easy to show that S is an equivalence relation. For the family of equivalence classes {xis: x E JR}, choose one element from each equivalence class except the equivalence class Q. The set of all such chosen elements is an infinite set of irrational numbers, no two of which differ by a rational. (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 of A is denumerable. Then all subsets of A must 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." If A is denumerable, let B = A, and we are done. Otherwise, A is uncountable. Choose XI EA. If A - {XI} is denumerable, let B= A - {XI}. Otherwise, choose X2 EA - {XI}. If A - {XI' X2} is denumerable, let B = A - {x I' X2}. Continuing in this manner, using the Axiom of Choice, we obtain a subset C = {x I, Xz, . .. } such that B = A - C is denu

merable. -


(d) Claim. Every infinite set has two disjoint denumerable subsets. "Proof." Let A be an infinite set. By Theorem 5.36, A has a denumerable subset B. Then A - B is infinite, because A is infinite, and is disjoint from B. By Theorem 5.36, A - B has a denumerable subset C. Then B and 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 denumerable subset B. Since B is denumerable, there is a function f: N ~:dBLet C={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. If B is finite, then B is countable by definition. If B is infinite, since B ~ A, A is infinite. Thus A is denumerable. By Theorem 5.36,1l has a ~enu!!1er~le subset C. Thus {; ~!! ~ A, which implies Xo = C and C:::; B :::; A = Xo. Therefore A = B = Xo. Thus B is denumerable and hence countable. -

Chapter 5

Business

Transcript of Chapter 5