text stringlengths 270 6.81k |
|---|
), a(x', y) = J(x, a(x, y), y). The first and second functions involved are initial functions, the third is obtained by composition from them, the fourth is an initial function and, finally, cx is obtained by primitive recursion from the third and fourth. We leave for exercises similar derivations for irmultiplication, the exponential function, and time factorial function. 2.4. The predecessor of x, pd(x), is defined by pd(0) = 0 and pd(x') = x. It is prirnitivc-recursive by virtue of the primitive-recursive derivation: Cot (x) - 0, U; (x, y) = x, Jpd(O) = r,,,,'pd(x') = Ui(x, pd(x)). 2.5. Proper subtraction, =, is defined by x._ 0 = x and x - y' pd(x-y).That is,x=-y=x-yifx>yandx -y=0 ifx<y. This is a primitive-recursive function. To verify this we initially write 6(y, x) for x _ y and obtain the following primitive-recursive derivation for fi: Ui(y, z, x), pd(z), J(.y, z, x) = pcl(tl2'(y, z, x)), 113'0,, Z, x), f S(0, x) = tts(y, Z, x), l.h(y, x) = J(y, 6(y, x), x). Here we have taken a shortcut by listing the predecessor function as an initial function instead of a derivation for it. To ohtain x =- y as the value of _ at (x, y) (instead of (y, x), as in b), three further steps are necessary: 1121(x, y), t1; (X, v), y), U2(X, Y), U (;, y)). The exercises include further instances of primitive-recursive functions. In addition to those listed above and in the exercises, it is possible. to establish as being prirrritive-recursive, for example, the function whose value at n is the (n -}- I)th. prime number |
, and the function JJ (a > 0) whose value at n is the exponent of the nth prime number (in order of increasing magnitude) in the factorization of a into a product of primes, where we regard the factorization a = 2an3a, as extending indefinitely, with all but a finite number of exponents being 0. Further, rn,. 2.2 I Proof and Definition by Induction 77 it is possible to show that all functions which can be obtained by the procedure that is described next arc primitive-recursive. Another form of definition of functions by induction parallels the principle of proof by strong induction. Thus, it is often called definition by strong induction and has, as its distinguishing feature, the possibility of defining a function k on N such that k(n -I-- 1) is specified in., k(n). Specifically, the terms of some, or all, of the values k(0), k(1), circumstances take the following form: There is given a noncmpty set B and a function h, such that for each natural number n, It assigns to each element of Bn-t' an element of B. 'I'hcrr k: N -* B is supposedly defined by the two conditions k(O) = c (a given member of B), k(n + 1) = h(k(O), k(1),, k(n)). As for h, it is convenient to think of it as a function whose range lies in B and whose domain is the set q of all functions j having as domain N,, _ {0, 1,, p} for some p and, as range, a subset of B. For then the intended value of k at n -I- I is simply where AIN,, is the restric, n'. The theorem concerning such circumstances is tion of k to {0, 1,.. TIIEOR EM 2.4. Let B be a noncmpty set, let c be an clement of B, and let h be a function whose range lies in B and whose domain is,) of all functions j having as domain N, for some natural the set number p, and as range a subset of B. Then there exists exactly one function k: N -j- B such that k(0) = c and k(n -1- 1) = for each natural number it. Here kIN,, is the restriction of k |
to the domain N. = 10, 1, 2,..., n}. Proof. We establish first the uniqueness by contradiction. Assume that the functions k, and k2 satisfy the conditions and that k, 7-1 k2Then there exists a first natural number it for which k,(n) k2(n). Since k,(0) = c = k2(0), n > 0 and, hence, it = p -+- I for some /i. But k2(n), a then k,IN,, = k21N,,, so that ki(n) = h(k,IN,) contradiction. Turning to the matter of existence, we propose for the function in question, the union k over the family 3C of all functions.1 in q, such that j(0) - c and, if N is the domain of j, then for each m < n, j(nr + 1) = h(jIN,,,). That k is a function on N satisfying the condi- 78 The Natural Number Sequence and its Generalizations I c:tt AP. 2 tions of the theorem appear below as properties (3), (4), and (5) of k. The proofs of properties (1), (2), and (3) are left as exercises. (1) If jr and j2 are distinct members of 3C, then either jc C 32 or 32Cit. (2) k is a function. (3) k(0) = c. (4) The domain Dk, of k is equal to N. Proof. Clearly At C N and, by (3), 0 C A. To prove equality we assume the contrary. Then there exists a least member of N of the form p -F I which is not a member of Dk. Then / is the greatest member of Dk and k U ((p -F- 1, h(k))} C K. Ilence p ± 1 C At, a con tradic Lion. (5) k(n + 1) = for n in N. Proof. For each n C N, n + 1 C DI, by (4), and hence, for some j in X, n + I C D. Then k(n + 1) = j(n -f- 1) = h(kIN). EXERCISES 2.1. Deduce the principle of strong induction from Theorem 2 |
.1. 2.2. The definition of b" given in the text has the following form. Given g, : B --} B and c C B, there are set forth two conditions f(0) _ c and J(n -1 1) = gi(f(n)) where f is a function which is supposedly being defined. Deduce that these conditions do define exactly one function J: N -p- B as a special case of Theorem 1.2. 2.3. Supply the independent proof of the result in Exercise 2.2 along the following lines. Let us say that a natural number n has property E(f) ill case f is a function on N. into B where f(O) = c -and J(k + 1) = gi(J(k)) for k < n. 't'hen prove that (1) If n has property E(f) and m has property E(g) and n < in, then &) = g(x) for x C N,,. In C NI there exists a function (11) such that n has property E(f,)} = N. Iufer dual f may then he defined by choosing J(n) to be 2.4. An operation * in X is commutative ill a * b = b * a for all a and b in Y. Prove the following general commutative law. Let * be an associative and commutative operation in X. 11'1',=2', then a,*a2*, a' is a rearrangement of 1, 2, 2.5. State and prove a generalization of M4 in Theorem 1.7. 2.3 I Cardinal Numbers 79 2.6. Prove the existence of a function k satisfying the conditions of "Theorem 2.3 by imitating the proof of (1) in Lemma 1.2. That is, define L. to be the intersection of the collection (i of all sets A, that (i) AcNXB", (ii) for each choice of (x2, and (iii) for each choice of (X2,, x,,) in 11" ', (0, x2,, X,,, c(-x2,, C A,., if (X, x2,, Xn, b) C A, then (x', x2,..., x |
n, g(xr, b, x2,..., xn)) C A. 2.7. Show that multiplication in N, the exponential function, and the fac- torial function are each primitive-recursive. 2.8. Give primitive-recursive derivations for the functions designated by each of the following. (a) ruin (x, y). (b) max (x, y). Ix - yl. (c) 0 (`l)sgx- 1 21 ifx0, ifx>0. (e) crn (x, y), the remainder upon division of x by y (see Exercise 1.17). 2.9. Complete tile. proof of Theorem 2.4. 3. Cardinal Numbers In Section 1 we ignored the role of the natural numbers in counting. We considered them merely as a set of objects without intrinsic properties individually and known only through their position in the natural number sequence. In brief, we considered the natural number sequence as an integral system. In this section we shall discuss the concept of a number as a "measure of size" and it will scarcely be unexpected to find that the natural numbers have an application. The intuitive connotation of two exhibited sets having the sank number of urcrnbers is that the members of one can be paired with those of the other. Since a pairing of the nceinbers of two sets is simply it oneto-one correspondence: between theme, our basic definition in connection with the problem at hand assigns a name to two sets so related. Two sets A and B are similar or equinumerous, symbolized A - B, ifi there exists a one-to-one correspondence between A and B. In any nonernpty collection of sets, similarity is an equivalence relation the -reflexivity and symmetry are obvious, and the verification of transitivity 80 The Natural Number Sequence and its Generalizations I CHAP. 2 is left as an exercise. '1'ltus it appears feasible to apply the method of definition by abstraction (Section 1.7) to obtain the concept of a number as a similarity class. The obvious choice for the set to which to apply the relation of similarity is the set of all sets. "Then the concept that is usually called a cardinal number may be defined to be a similarity class. The modifying adjective "cardinal" is included here to distinguish the concept under study from another that will follow, wherein not only magnitude but also order |
is significant. It will prove expedient to anticipate now that the set of all sets has a contradictory character. 'Thus we will be less grandiose and apply the method of abstraction to a set U which we regard as a universal set; that is, all sets which interest its are subsets of U. 'Then similarity is an equivalence relation on 6'(U) and by a cardinal number we shall mean a similarity class. If A C 61((I), tile cardinal number of A, symbolized A or card A, is the cardinal number having A as a meniber. At time expense of a possible loss of generality entailed by the dependence on an underlying universal set, we achieve a degree of precision that is lacking in Cantor's description of a cardinal number as "the general concept which, with the aid of our intelligence, results from a set Ad when we abstract from the nature of its various chcrnents and from the order of their being given." This double abstraction- that is, the abstraction with respect to the nature and the order of the elements- is the origin of his notation M for the cardinal number of Al. There are other definitions of "cardinal number" which have been adopted by differcnt authors. G. Fregc (1884) and 13. Russell (1902) identified the cardinal number M with the set of all sets similar to Al. On the other hand, J. von Ncuinann (1928a) suggested the selection of a fixed set C front the set of all sets similar to Al to serve as cardinal of M. With any one of these definitions one obtains what is essential--that all object is associated in common with those and only those sets which are similar to each other. That is, similar sets and only similar sets have the same cardinal number. It can be successfully argued that is immaterial as to what cardinal numbers are, for mathematics it explicitly, so long as they have the property A = 13 iff A ti B. Indeed, by virtue of this property, we shall find that all questions regarding equality and inequality of cardinals can be reduced to questions of the similarity or nonsimilarity of sets. That is, any property of cardinal 2.3 I Cardinal. Numbers 81 numbers, however they be defined, can be translated into a property about sets and their similarity. In order to compare cardinals we define the notion of domination for sets. If A and B arc sets such that A is similar to |
a subset of B, we shall write A { B and say that A is dominated by B or that B dominates A. Clearly this is a prcordcring relation (see Example 1.11.5) in the power set of the universal set we have adopted for our discussion of cardinal numbers. in the terminology used in the same example, it is obvious that < is indifferent to a pair of similar sets. The converse statement (which is not obvious) is also true; this is the content of Lite following celebrated theorem proved independently by E. Schriider and F. Bernstein in the 1890's. TIIEOREM 3.1 (Schriider-Bernstcin Theorem). If A B and BA,thenA''B. Proof. First we offer some remarks to motivate the proof which follows. The assumptions amount to the existence of a one-to-one map f on A into B and a one-to-one rnap g on I3 into A. To establish the. existence of one-to-one correspondence between A and B, it is sufficient to determine a subset At of A such that g[B - fAt] _ A - At. For then the function h: A B such that for x C A,, hx = fx hx=g'x forxCA - A,, is of the type desired. Concerning a method for constructing any subset Ao of A which could serve as an At, clearly A should include A - gB. Also, since g [B - fAo] should include A - Ao, (g f)Ao should be included in A0. In addition, since we actually want g(B - fAo] = A - Ao, the smallest such All is required. Thus we arc led to consider the collection d of all subsets Ao of A such that (I) A - gB C A0, and (11) (g ° f )Ao C Ao. Since A C a, a is noncmpty. As At we choose (1(t;. Clearly A, satisfies (1). Moreover, since (g o f )A, C (g o f )Ao C Ao for all AO in a, (g o f)A, C net= A,. Hence A, satisfies (11) and is, consequently, a member of a and, indeed, its least member. 82 The Natwal Number Sequence and its General |
izations I C H A P. 2 The proof that the set A, just defined has the required property i' in two parts. First, since A - g13 C A,, A - A, C gB. This implies, since (g ° f)A, rA,, that A - A, C gjB - fA,]. To establish the reverse inclusion we prove first that (A - gB) U (g A,. Since A, C tt, A - gB C A, and (g ° f )A, e A, and hence (A - g13) U (g °f)A, g Al. Also (g -f) [(A - gB) U (g °f)A,l c (g °f)A,, from which it follows that (A - gB) U (g ° f)A, C a. Recalling the definition of A,, the equality in question then follows. But this implies that A. and g[li - fA,] are disjoint, so g(ll - fA,J A - A1. Hence g I B - fA,] = A - A1.This completes the proof in view of our initial remarks. Combining the observations made so far, it follows from the result induces a partial ordering of simobtained in Example 1.11.5 that ilarity classes, that is, of cardinal numbers. We shall symbolize this relation by < ; thus, for cardinal numbers a and b, a < b ifl' there exist representatives A and B of a and b respectively, such that A < B. The strict inequality of cardinals is defined in arithmetical fashion : a <b ill' a < b and a representatives we define b. In order to characterize this relation in terms of A < 13 for sets A and B to mean that A < B and not 13 < A (abbreviating "it is not the case that B;< A" to "not B;< A"). The second of the following two leimuas is the desired result. '1'he proofs are left as exercises. LEMMA 3.1. For sets A amid 13,A < A < B. 1. FM M A 3.2. For cardinal numbers a and b, a < b ifl there exist respective rel)reseutatives A and 13 such that A < 13. |
Is the pal tial ordering relation for cardinal numbers a simple ordering? To analyze this question let a and b be cardinals and assume that A = a and f3 = b. 'I'llcn either A is dominated by 13 or not. Vice versa, 13 is dominated by A or not. Combining these two pairs of either alternatives gives four cases, exactly one of which must apply to A and B. 2.3 I Cardinal Numbers 83 (1) ANBandB<A. (II) A B and not B (III) Not A B and B (IV) Not A < B and not 13 A. A. A. In case (I), a = b according to Theorem 3.1. In case (II), a < b and in case (II1), b < a according to Lemma 3.2. It is only case (IV) that cannot be resolved with results that are available now. Later there is introduced an assumption which rules out this case. At that time we may conclude, as a consequence, that < is a simple ordering relation. Our next objective is to show how the familiar symbols for the natural numbers can be adopted as symbols for the cardinal numbers of certain sets, or, what amounts to the same, we make an application of the natural numbers in which they become cardinals. To this end we make the following definitions. The cardinal number of the empty set we call 0, and the cardinal number of any set A U (c}, where c iZ A, we call.q + 1. It must be proved that these concepts are well defined in terms of our basic principle: A = P if A Clearly 0 is well defined since the only set similar to 0 is itself. That A + I is well defined is shown as a consequence of the following result, whose proof is left as an exercise and c (, A, then B = Bo U Id l where d (4 B0 and B0 A. From this it follows that if B - A U Ic} where c (Z A, then B determines the cardinal number 130 + 1, which is equal to A -J- 1 (that is, A V qccf) since 730 = A. The converse is trivial by virtue of the definition of A + 1. The two definitions just given -together with the understanding that in case A = n, a natural number, then A --A- 1 = n + I should receive its usual symbol, such as 4 |
+ 1 = 5---lead to an assignment of each natural number as a cardinal number for certain sets. The natural numbers in the role of cardinal numbers are the finite cardinals, and sets which have these cardinals are finite sets. In their new role as cardinals, the natural numbers are subject to the ordering of cardinal numbers generally, as defined above, following Cantor; this ordering we write temporarily as <c. In their original role as members of N, the natural numbers possess the familiar ordering, which we write as < N, for the moment. If it were the case that these two orderings did not coincide on N, a confused situation would result. That they are in agreement on N and thereby no awkwardness results, can be proved after two preliminary theorems are established. 84 Die A'a; anal Number Sequence and its Generalizations I C H A P. 2 THEOREM 3.2. For each natural number n, the finite cardinal n is the cardinal of the set of natural numbers which precede n in the natural ordering. Proof. The proof is by weak induction on it. If n = 0, there are no natural numbers preceding n in the natural ordering, and hence the set referred to in the theorem is 0 and 0 = 0 by definition. Thus the theorem is true for it = 0. Assume that n = card 10, 1,, n - 1 } ; to prove that n 1 - card 10, 1, -,n}.Since n(Z {0,,n - 1,n} we have it + I = card 10, 1, 10, 1,,n - 1} U fit),, n} by the definition of A ± 1. THEOREM 3.3. For each natural number n, if it = n, then A is not similar to a proper subset of itself. Proof. We shall prove by weak induction on n that for all n C. N, if A = n then it is false that A r., A, C A. For n = 0, A = 0 and A = 0. Then A has no proper subsets and so no proper subset of A is similar to A. Assume the theorem for any set of cardinal number n; we prove it for a set A with A = n + 1. Our method of proof calls for obtaining a contradiction of the induction hypothesis upon assuming that A does include a proper subset A, such that there exists a mapping f: A -- A,, which is |
one-to-onr and onto. Since A = n -4-- 1, A = B U (b} where 21 = it and b (Z B. There are three possibilities to consider. (I) b (Z At. Then b f(b) and we conclude that the restriction of.j to I3 is a one-to-one correspondence between B and A, - If(b) {_ where the latter set is a proper subset of the former. Since 1) = n, this contradicts the induction hypothesis. (I1) b C A, and f(b) = b. Again f IR yields a contradiction of the induction hypothesis. (III) b C At and f(b) : b. Let f(b) = b, and f-'(b) = a. Then consider the snapping g: A -- Awhich differs from f only in that g(b) = b and g(a) = bi. Clearly g is one-to-one and onto and is the type of correspondence. considered in (11). 'T'hus in all possible cases a contradiction of the induction hypothesis results from time assumption that if A = n + 1, then A is similar to a proper subset of itself. Hence, if the assertion of the theorem is true 2.3 1 Cardinal Numbers 85 for sets of cardinal number it, it is true for sets of cardinal number n+1. TII L' OR.EM 3.4. The natural ordering and the cardinal ordering agree on N. That is, for all natural numbers p and q, p - 1 } and Q = {0, 1, q<Np if q<cp. Proof. We shall prove first that if q <N p, then q < e p. Let P =, q - 1 } (or 0, if q = 0). Then, q - 1 precedes 7' = p and Q = q. Assuming q <N fi, each of 0, 1,, p - 1 }. p - 1 in the natural ordering, so P = {0, 1, 'I-bus Q C P, and since Q - Q we have Q N Q C P. Further, for no Qo is P - Qo S Q, since otherwise it would follow that P is similar to a proper subset of itself, contradicting Theorem 3.3. Nonce q < c p by Lemma 3.2. -, q - |
1,. Next, assume that q < c p; we deduce that q < N p as follows. Since <n simply orders N, it suffices to show that q = p and p < n q are incompatible with our assumption. If q = p, then Q = 1' and so, as cardinals, q = p = Q = P, which is incompatible with q < c p. If p <N q, we apply the first part of the theorem to conclude that p <c q, which is incompatible with the assumption that q < c p. We turn our attention next to the nonfinite cardinals. A nonfinite cardinal is an infinite or transfinite cardinal. If the cardinal number of a set is infinite the set is called infinite. The cardinal number of the set of natural numbers is symbolized by bto. TI!EOREM 3.5. If n is a finite cardinal, then n < bto, n - I } by'f'licoreni 3.2, n < Ko by Proof. Since it = card 10, 1, the definition of < for cardinals. Assume that n = bto. Since n + I is also a finite cardinal, we conclude similarly that it -} 1 < Ko, which with it = Ko gives n + I < n. Since this contradicts the valid result it < n -}- 1, the assumption n = btu is untenable and the remaining alternative n < bto is established. Thus Ku is an infinite cardinal. So, along with the finite cardinals, we now have an infinite one. If it were the case that all infinite sets were similar, so that Ko would be the only infinite cardinal, the theory of cardinal numbers would contribute nothing to mathematics not already known, and hence would scarcely be worthy of mention. That more 86 The Natural Number Sequence and its Generalizations I C i-t A P. 2 than one infinite cardinal exists-that is, that there exist infinite sets which are not similar---is an immediate consequence of the next theorem. TII EOR E M 3.6 (Cantor). For every set A, A < (P(A) or, in other words, f1 < CP(A). Proof. The mapping on A into a'(A), which takes a in A into {a} in?(A), is a one-to-one mapping on A into (P(A). Thus, A < |
cP(A). To prove that A -< a'(A), we show that the assumption A - a'(A) yields a contradiction. Let f : A -} (P(A) demonstrate the assumed similarity of A and a)(A) and consider A, = {a C Aja (Z f(a) }. Since A, C (3)(A), there exists a, in A such that f(a,) = At. Now either a, C A, or a, (Z At. If a, C A,, then a, f(al) and hence a, (Z A,, which yields a contradiction. Similarly, a, (, A, implies that a, C f (a,) oral C A,, and again a contradiction results. Thus we have proved that A < (P(A) and not A' (P(A). This gives the desired conclusion. Cantor's theorem uncovers a hierarchy of distinct infinite cardinals. Just as the set of finite cardinals is unending, so also is the set of infinite cardinals of the form N = Ho, v(N), a'(n'(N)),.... (However, this is not the end of the matter,] as we shall see in Theorem 9.2.) EXERCISES 3.1. Show that similarity is an equivalence relation on any collection of sets. 3.2. Show that < is a preorderiug relation. 3.3. The closed unit interval, written 10, 11, is (x C R10 < x < 1). The open unit interval, written (0, 1), is (x C RI0 < x < 1). The half-often unit intervals, written (0, 11 and 10, 1), are {x C 1110 < x < 1) and (x C RIO < x < 1), respectively. Show that these sets are similar to each other. 3.4. Using the function f: R --)-- R, where f(x)=2(1+1+1xi)' show that R is similar to (0, 1). 3.5. Show that R and R-1 are similar. 3.6. Referring to the proof of Theorem 3.1, what is the function h if A = [0, 1), B = K1, f(x) = x + 1, and g(x) = x/(1 + x |
)? 3.7. Prove Lemmas 3.1 and 3.2. 3.8. Deduce from Theorem 3.1 that if A, B, and C are sets such that A B?CandA.C,then A-B'C. 2.4 1 Countable Sets 87 3.9, Prove that if A?Band B -BUC,thenA - AUC. 3.10. Prove that if A, B, C, and D are sets such that A 1) C, B C I), and C U D ^ C, then 3.11. Prove the result stated in the text that if A U {c} ti B and c (Z A, then B = Bu U {d} where d (Z Bo and Be N A. 3.12. Let X be a set which is similar to none of its proper subsets. Without using any theorems in this section, prove that the same is true of X U (a) where a (z X. 3.13. Of the given definitions of a finite set and of an infinite set, we have taken the former as primary; the infinite sets are the nonfinite sets. Dedekind proceeded in the complementary way by initially defining an infinite set as one that is similar to a proper subset of itself; then the finite sets are the noninfinite ones. Show that a set which is infinite in the sense of Decekind is infinite in our sense. The converse can be proved, using the axiom of choice; see Corollary 2 of 't'heorem 9.1. 3.14. Tarski (1924) has given the following definition of a finite set: The set S is finite if in each nonempty collection 3 of subsets of S there exists one mernber 7' such that no proper subset of T belongs to 3. Prove that this definition is equivalent to that in the text. 3.15. Let If be a nonempty collection of sets such that for each A in CF there exists it 11 in 9 with B > A. Prove that card U;F > A for each A in F. 4. Countable Sets The smallest infinite cardinal number that we have turned up so far is No. A surprisingly large variety of sets in mathematics have this cardinality, as we will proceed to show after introducing some further terminology. A set is denumerable if it has cardinal number No. |
neoaq sirtulou letuw(Xlod 2 13gwnu a JO'I < xaput : j u U3n1dc 31iuy 33111S 3110 (l0d 1e1I1 Aluo 2.4 1 Countable Sets 89, in f(x) = 0, where f(x) has index i. By assigning to i the values 2, 3, 4, turn, and listing the new algebraic numbers obtained at each step in sonic order, there results an enumeration of the distinct algebraic numbers. No algebraic nurn'.ber escapes being assigned to a natural number since every polynomial has an index. Beginning terms in one such enumeration are 0, -1, 1, -2, -a, of 2, -3, 12 (1 - 51/2), -2h12, -a. 2",.... As in the case of Q', this type of argument makes it plausible that A is denumerable. To put such matters on a firm footing, we need to prove some theorems. THEOREM 4.1. A subset of a countable set is countable. Proof. By definition, if A is countable and not denumerable, then it is finite. If B C A, then 71 < fl, so V is finite and B is a finite and hence a countable set. Next, suppose that A is denumerable; let f : N -} A be one-to-one and onto. If B C A, then the restriction off to f-'[B] is a one-to-one mapping on a subset of N onto B. If we can f--'[BJ is countable, then a one-to-one mapping onto B show that can be constructed by composition. Thus the proof reduces to showing that an arbitrary subset C of N is countable. To prove this, let g(O) be the first member of C. Proceeding inductively, for n in N, let g(n + 1) be the least member of C different from, g(n). If this is impossible for some n, then g is a funcg(0), g(1),, n) onto C, and C is finite. Otherwise, according tion on 10, 1, there is a function g on N into C such that for to Theorem 2.4,, each n in N, g( |
n) is the least number of Cdiferent fromg(0), g(l), g(n - 1). Clearly g is one-to-one. It remains to show that its range is C. For this we point out that g(n) > n for all n; the proof is left as an exercise. Consequently, each number c in C is one ofg(0), g(1),, g(c). Thus g is onto C. TIIEOREM 4.2. If the domain of a function is countable, then its range is also countable. Proof. As in the case of the preceding theorem, the proposition now at hand can be reduced to the case of a function whose domain is a subset of N. So consider f: A -. 13 where A _C N and B is the range off. It is to be shown that B is countable. Let C be the set of all memf(y). That is, C bers x of A such that if y E A and y < x, then f(x) consists of the least member of each of the sets f ''(b) for b in B. Then fIC maps C onto B in a one-to-one manner. Since C is countable by Theorem 4.1, so is B. 90 Die Natural Number Sequence and its Generalizations I C 1.1 A P. 2 TI-1 E0 RE M 4.3. N X N is denumerable. Proof. t Let f : N X N ->- N, where f (m, n) = (m + n) (m + n + 1) { 2m. Then f is one-to-one. For assume that f(m, n) = f(m, n); that is, (m +n)(m +n + l ) +2m = (m +n)(m +n + 1) + 2m. Setting x = m + n and z = m + n, this becomes x(x + 1) + 2m = z(z + 1) ± 2m. Now exactly one of x < r, x = x, x > z holds. Assume that x < T. By A4 of 'T'heorem 1.5, x = x + d + I for some d in N. After substitution in the above equation and simplification, one finds that z(z+ |
1) =x(x+1)+2x(d+1)+(d2+3d+2), and hence z(x + 1) > x(x + 1) -f 2x. Since x = m + n implies that x > in, it follows that x(z -I- 1) > x(x + 1) -I- 2m, and hence R(x+1)+2m>x(x+1)+2rn, contrary to the assumption that f(rn, n) = f(m, n). The assumption that z < x leads to a similar contradiction. It follows that x = z, and hence, in turn, m = m, n = n, and (rn, n) = (m, Ti) as desired. Let A he the range off. "Then f -I maps A onto N X N in a one-toone manner. By 't'heorem 4.1, A is countable and hence, by Theorem 4.2, N X N is countable. Since N X 101 is a denumerable subset of N X N, it follows that N X N is denumerable. If X is a denumerable set, then so is X X X. CORO1,1,AItY. More generally, if n is a natural number, then X"'' is den"InCrable. The proof is left as an exercise. f 'T'he proof is adapted from Rosser's Logic for Mathematicians, p. 439. if we felt free to use rational numbers in our proofs at this point we would set f (rn,n) = -z(nt -I- n)(m + n + 1) + in. This is precisely the enumeration of IN X N which results fioni arranging the elements in a sequeucc by proceeding clown the successive diagonals in the display (0, 0) (1,0) (2.0) (3,0) (0, 1) (1, 1) (2, 1) (3,1) (0, 2) (1, 2) (2, 2) (3,2) (I), 3) (1, 3) (2, 3) (3, 3)............ 2.4 I Countable Sets EXAMPLE 91 4.4. Using the preceding theorem, the denumerability of the |
positive rationals can he established. Clearly, the function f, such that f(m, n) Hence Q} is countable by Theorem (m -} 1)/(n 4.2. Since 0,+ has a denumerable subset, it is not finite, and hence it is denumerable. 1) maps N X N onto THEOREM 4.4. If a is a nonempty finite collection of denumerable sets, then Ua is denumerable. If (t is a nonempty finite collection of countable sets, then Ua is countable. Proof. The proof of the first statement is by weak induction on n where n + 1 is the number of members of a. For n = 0 the result is U A. is denumerable when each obvious. Assume that Ao U At U Ai is denumerable and consider Ao U Ai U U A. U where each Ai is denumerable. Since AoU A,U... UA.f.r = (A0UA,U... UA.) UA.+j, the induction step is established as soon as it is known that if A and B are denumerable then so is A U B. We prove this now. Assume that A and B are denumerable. 't'hen there exists a function f. mapping N X {0} onto A in a one-to-one manner and a function f, mapping N X11 ) onto B in a one-to-one manner. Then f = Jo U f, maps N X 10, 1) onto A U B. As a subset of N X N, N X 10, 11 is countable, and hence A U B is countable by Theorem 4.2. Since A C A U B, the union is actually denumerable. The proof of the second statement in the theorem is left as an exercise. There is a rather obvious possibility for extending the method used above to prove that if A and B are denumerable then so is A U B. This is to prove that the union over a denumerable collection of denumerable sets is denumerable, and proceeds as follows. If a is a denumerable collection of denumerable sets, then there exists a function g on N onto a. Since, for each n in N, g(n) is denumerable, there exists a function f on N X fit } onto g(n) for each it in N. Then f = U { f I n C N } is |
a function on N X N onto Ua. Applying Theorem 4.3 and Theorem 4.2 we conclude that Ua is countable and, actually, denumerable. But there is an argumentative point involved in this reasoning. To explain it we define F,,, for n in N, as the set of all functions on N X { n } onto g(n). By hypothesis, F. is nonempty for each n in N. Also, the Fn's 92 The Naluial Number Sequence and its Generalizations I C n A p. 2 are disjoint in pairs. 'T'hus, in (fnI n C N) we have a set of functions with exactly one from each member of the family ( F jn C N). Now---and this brings us to the heart of the matter C N) is not a defined set until f, is uniquely specified for each n. In general, there is no procedure to attain uniqueness. In effect, what we did to arrive at. (j In C N) was to assume that one can choose a unique member of each of the That is, we employed the. (denunicrablc) axioirt of choice, a principle which, beginning in Section 8, we officially accept. In Section 8 the principle and its ramifications are discussed in detail. It will involve us in no circular arguments and there will incur certain advantages to accept the above proof now. It is left as an exercise to supply the necessary modifications for the version that we state next. THEOREM 4.5. If a is a countable collection of countable sets, then U a is countable. Our principal reason for the acceptance of this theorem now is that it implies the following result: Any set which can be formulated as the union of a countable collection of countable sets is countable. Hence, any set which can be divided into a denumerable collection of countable sets is dcnurnerable. For instance, the set of all real algebraic numbers is denumerable. Again, the sarric is true of the collection of all finite subsets of N. It should be at least mentioned, however, that the dcnumcrability of each of these sets can be established without using the axiom of choice. Further examples of denum crable sets appear among the exercises for this section. To say that a set is uncountable means that it is infinite and nondcnumcrablc. We have |
already noted, as an application of Thcorens 3.6, that card 61(N) > No, so 6'(N) is an uncountable set. We shall denote. card 6'(N) by According to the result stated in Example 1.8.7, 6'(N) ti 2H, the set of all functions on N into 10, 1) or, in other words, the set of all infinite C N} where an = 0 or a = I for each n. Consequently, sequences 2N is an uncountable set. It is instructive to give a direct proof of this fact by the classical Cantor "diagonal" procedure. This is a process for, deriving, tinder the assumption that an enumeration of a set is given, a.. member different from all those in the enumeration. '[Isis, of course;. renders it absurd that the given enumeration is one of all members of the set. So assume that an enumeration of 2"- is given. Since each member of 2.4! Countable Sets is a sequence of 0's and l's, it is possible to indicate all in a table of the form, 2r-' 2N 93 of arn,, a,0, ago, am, all, all, a02, a12, a22, a;j = 0 or I Now consider the sequence {1 - an C NJ. Clearly this sequence is in 2'ti and is different from each of the above, since it differs from the ntl member in the nth place. Thus the enunicration is riot exhaustive and this is the desired contradiction. We state this result as our next theorem and also provide a compact proof. THEOREM 4.6. 2N- is an uncountable set. Proof. Assume to the contrary that there exists a one-to-one correspondence f between N and V. Then a:N -* 10, 11, where a _ 1 - f (n),,, is a member of 2". But a 7 f (n) for all n because a,, Thus the enumeration does not exhaust 2". EXAMPLE 4.5. We assume as known the following property of the real number system. Each real number x; 0 < x < 1, has an r-adic expansion Fi n;r-', where r is a natural number greater than 1 and n; is an integer such that 0 < n; < r |
- I; this expansion is unique unless x is a nonzero number of the form nr ', in which case there exists exactly two expansions, one finite and the other infinite. For r = 2 we obtain the dyadic expansion of x: x =.a1a2, a; - 0 or 1. This representation is unique if we agree to always use the infinite one whenever a choice exists. Defining C to be (x C i 0 < x < 1), it follows that cc 211 and hence C<bt. On the other hand, using the decimal expansion (r = 10) for the numbers of C, the subset of C consisting of those numbers having expansions composed exclusively of 3's and 4's is similar to 2N. Thus bt < C. By Theorem 3.1 it follows that Each finite cardinal n has an successor, n -}- I. This fiuggcsts the question of whether too also has this property. One candidate for the immediate successor of tjo is rt. The question of whether H is the 94 The Natural Number Sequence and its Generalizations I C II A r. 2 smallest cardinal greater than No is known as the continuum problem. This terminology suggests that < simply orders any set of cardinals, which we shall prove later, and that the answer is not as yet known, which is a fact. Since N is the cardinal number of 11 (this appears among the examples for this section), the problem may be formulated as the query: Has every infinite subset of R either the cardinal number No or N? It has been discovered that a number of theorems, some of them important, can be based on the hypothesis that the answer to the continuum problem is in the affirmative. This conjecture is known as the continuum hypothesis. EXERCISES 4.1. Supply the missing part of the proof of Theorem 4.1-n arnely, that g(n) > n for all n. 4.2. Show explicitly that the proof of 't'heorem 4.2 can be reduced to that of a function whose domain is a subset of' N. 4.3. Prove the Corollary to 't'heorem 4.3. 4.4. Prove the second statement in Theorem 4.4. 4.5. Assuming that the union of a denumerable family of denumerable sets is denumerable, prove'I'heorem 4.5. 4.6. Show that N |
can be represented as a union of a denumerable family of denumerable disjoint sets. 4.7. Give an intuitive proof that any infinite set includes a denumerable subset. From this deduce that (1) No is the least infinite cardinal, and (II) an infinite set is similar to a proper subset of itself. 4.8. Show that any set of circles, no two of which overlap and all located within a fixed circle in the plane, is countable. 4.9. Show that U?-,(ZI)" (the set of all ordered n-tuples of positive integers for n = 1, 2,.. ) is denumerable by mapping (ri, r2, pk, where Pk is the dth prune., Ti) onto 2" 3r' 4.10. Show that the set of all complex algebraic numbers (tile roots of poly- nomial equations having integral coefficients) is denumerable. 4.11. Show that the set of real numbers in the closed unit interval having a decimal expansion which ends in an infinite sequence consisting solely of 9's is denumerable. 4.12. Prove that the set of all infinite sequences of natural numbers is un- countable. 4.13. Prove that the set of all finite sequences of rational numbers is de- numerable. 4.14. Show that if 1) is a denumerable set of points in a coordinate plane, then D is the union of two sets Ds and 1.)., let us say, such that the intersection 2.5 I Cardinal Arilhrnelic 95 of D= and each line parallel to the x-axis is a finite set, and the intersection of D,, and each line parallel to the y-axis is also a finite set. 4.15. Prove that the set of all straight lines in a coordinate plane, each of which passes through at least two distinct points with rational coordinates, is denumerable. 4.16. Let A, B, and C be sets such that C C A, A fl B = 0, and B and C are denumerable. Prove that A U B ti A; 4.17. Deduce from the preceding exercise that a set which contains a denumerable subset is similar to the set obtained by adding to it a denumerable set. 4.18. Deduce from the preceding result that if A is uncountable and B is denumerable, then A - B is uncountable |
. 5. Cardinal Arithmetic In this section we shall define the operations of addition, rnttltiplication, and cxponentiation for arbitrary cardinal numbers, and sketch briefly the properties of each. It will be left to the reader to convince himself that these definitions, when applied to the finite cardinals for which we have found it possible to use the natural numbers as labels, are in agreement with those fir addition, multiplication, and exponentialion of natural numbers. Thus the definitions given below are extensions of those for natural numbers. The sum, it 4- v, of the cardinal numbers u and v is A U 13, where A and I3 are disjoint representatives of u and v, respectively. (The required disjointness can always be realized by replacing, if necessary, given representatives A and I1 by A X {01 and B X 111.) It is iinnit-diate that the definition of the sum it + v is independent of the choice of representatives for it and v. Moreover, it is an easy matter to verify the properties stated in the following theorem. TI I E O It E M 5.1. For cardinal nutnl)ers u, v, and rv, (I) u+v =V+u, (11) u+(z+w) = (u+v)+w, (III) u < v itnplies u + w < v + rv. The proof is left as an exercise. The product, uv, of the cardinals u and v is A X B, where A and B are representatives of u and v, respectively. This definition is independent 96 The Natural Number Sequence and its Generalizations I CHIAP. 2 of the choice of representatives for u and v. Multiplication has the propertics listed below. TIIEOR EM 5.2. For cardinal numbers u, v, and w, (I) uv=vu, (I.I) u(vw) = (uv)w, (III) u <vimplies uw <vw, (IV) (u-}-v)w = 11w-1-vw. The proof is left. as an exercise. Properties of addition and multiplication of infinite cardinal numbers lose much of their interest as a consequence of a theorem which will be proved later, using the axiom of choice. This theorem asserts that the sum or product of two cardinals, at least one of which |
is infinite, is simply the greater of the two. Important instances of this result can be proved without appeal to this axiom, however. Several results of this kind follow. EXAMPLES 5.1. If n is a finite cardinal, then since n -l- Ro = card {0, 1, n -l, n - 1} -I- card In, n Ko A- No = MO + 1,... } _ N ; also, since card {1, 3, 5, } + card {0, 2, 4,. } = N. 5.2. n4o = Mo and MoNo = Ro are true by virtue of 'I'heercin 4.4 and Theorem 4.3. It is left as all exercise to show that, similarly, 5.3. The relation may be established as follows. Using the open unit interval S, as a representative of M (see Exercise 3.4), the product MR may be represented as the set Sz of all ordered pairs (x, y) of real numbers x and y, such that 0 < x, y < 1, that is, the interior of the unit square in the plane. If x and y are written in decimal notation (where, to achieve uniqueness, the infinite expansion is chosen when there is a choice), then the correspondence (x, y) = (.x,x2... yj y2...) -.x,y,x2y2... 2.5 I Caidinal A,itlvnetic 97 is a one-to-one mapping on S2 into S1, so S2 < St. But,Si <,S2 by virtue of the correspondence x --s` (x, Z). Thus the assertion follows. 't'urning to the operation of exponentiation, if u and v are cardinals, the Uth power of u, in symbols u", is card A° where A and B are representatives of it and v, respectively. The independence of this definition upon the representatives for u and v is easily verified, as are the following properties of exponentiation. THEOREM 5.3. For cardinal numbers u, v, and w. (1) u"u'° = u" I-, (II) (UV)" = u"'Ui°, (III) (IV) u' = u and 1u = 1, (V) u<vimplies w"<w", |
(VI) u < v implies u'° < U"'. u Again the proof is left as an exercise. Since al(A) ti 2A and 2" is an abbreviation for (0, 11 A, it follows from the definition of exponentiation that Thus, in particular, we may now write a'(A) = 21. I,'t = 2u0. Also, if u is a cardinal and n is a finite cardinal, 4" may be given its familiar interpretation, since, from (I) and (IV) above, u" = U1 III u' = uu... u. (n factors) In particular, recalling the definition of A" in Section 1.8, An = (j)n EXAMPLE 5.4. The following are sample computations which may be carried out with cardinal numbers, using exponentiation. bt2 = (2N°)2 = 22tt° = 2't° = I't Wt°_ (2N')tt°=2't°'=2et°_Lt H = 2't° < t'tu ° < Mtt° = (`t, and hence t`tu ° = Id. 98 The Natural Number Sequence and its Generalizations I C tt A P P. 2 EXERCISES Note: The results of Exercises 1.9.13 1.9.16 should be used whenever pos- sible in the proofs required in these exercises. 5.1. Show that the definition of the suin u + v of two cardinals u and v is independent of the representatives used. l'i ove Theorem 5.1. 5.2. 5.3. Prove 't'heorem 5.2. 5.4. Prove that nK = &rj = tj for n C N and n 0 0. 5.5. Show that the definition of u° for cardinals u and v is independent of the representatives used. 5.6. Prove Theorem 5.3. 6. Order Types The theory of similarity of sets and cardinal numbers ignores the possible existence of an ordering relation on sets under consideration. Indeed, in the language of Cantor, one arrives at the notion of a cardinal number by an abstraction from the nature of the elements of sets and from any existing ordering. Taking orderings into account, similar sets may show much diversity. This contention is illustrated by the similar sets N and Q |
with their familiar orderings; N has a least member and (Q does not. Further, each member of N has a covering element while no member of 'Q has this property. The notion of ordinal similarity to be discussed in this section is that of similarity applied to simply ordered sets (or chains) with the respective ordering relations being taken into account. To case the notation fir such sets, we shall suppress the symbol for the ordering relation and speak of "the chain X." When explicit mention of the ordering relation is called for, the symbol " <" will be used. In particular, this symbol will designate possible different orderings in different contexts. Two chains X and 1' are called ordinally similar, symbolized X = Y, if they are isomorphic ordered sets. This means (Section 1.11) that there exists a one-to-one correspondence, say f, on X onto Y such that both f and J'are order-preserving. In the case of chains, the condition that J-' be order-preserving can be dropped, since it follows from the others. Both an isomorphism and its inverse preserve < and, conversely, a function that maps one chain onto another is an isomorphism if it and its inverse preserves <. Ordinal similarity is an equivalence rela- 2.6 I Order 'T'ypes 99 Lion on any collection of chains. An equivalence class under ordinal similarity is called an order type. Order types of infinite sets will generally be designated by lower case Creek letters. If the ordered set A is a representative of the order type a, we shall say that A is of order type a and sometimes write A for a. With regard to our definition of an order type, remarks corresponding to those accompanying that of a cardinal number are appropriate. What an order type is, is immaterial so long as it is an object associated in common with those and only those simply ordered sets which are ordinally similar. That is, ordinally similar simply ordered sets, and only such, have the same order type: Ordinal similarity implies similarity, and hence A = B implies :4 = B. In particular, therefore, for finite chains A and B, A = B implies A = B. Conversely, for finite chains A and B, A = B implies that each is, n - 1 } for some natural number n and, indeed, similar to 10, 1,, n - 11 with its natural ordering. That is ordirally similar |
to 10, 1, is, f1 = B implies 71 = B, and hence two finite chains are of the same order type if they have the same cardinal number. Thus there is but one order type corresponding to any set of n elements (n a natural number) and this order type will also be designated by n. As a trivial consequence of this, a given finite set determines a single order type. This is not true of an infinite set, which admits of a simple ordering; relative to various orderings there will correspond different order types. For the purpose of illustrating this remark, as well as for later examples, it is convenient to employ our notation (...) for an ordered n-tulle to indicate the ordering intended for a given set. For example, (0, 1, 2, will serve to denote the natural number sequence. We denote the order type of N with its natural ordering by co. As our first example of an infinite set which, relative to different. simple orderings, represents different order types, we take N. The chains (0,1,2,,n, - ) and ( -, n,,2,1,0) are not ordinally similar. Indeed, assume the contrary and let the mapping f on the first onto the second demonstrate their ordinal simf(k) = 0 and hence, ilarity. Then there exists k in the first set such since k + I > k (according to the first ordering), f(k + 1) must be greater than 0 according to the second ordering. This is a contradiction, 100 The Natural Number Sequence and its Generalizations I C H A P. 2 since 0 is the greatest clement under the second ordering. Thus the chains in question are not ordinally similar. Again, (0,1,2,..-,n,...) and (1,3,5,...,0,2,4,...) arc not ordinally similar, since the contrary assumption implies the existence in the first set of infinitely many elements less than k, the clement which maps onto 0 in the second set. This is a contradiction. There is an arithmetic for order types- one that is more interesting than that for cardinals. Let A and B be disjoint sets of order types a and P. Then the sum, a + a, of a and a is the order type of A U B, simply ordered as follows. Pairs in A and pairs in R are |
ordered according to the simple orderings of A and B respectively, and each a in A precedes each b in B. The product, afi, of a and P is the order type of A X B ordered by (a, b) <(a',b')ifb <b',orb = b' and a < a'. It is left as an exercise to show that both of these definitions are independent of the representatives used in their formulation. Although the commutative law is valid for addition and multiplication of finite order type, the same is not true in general. For example, n -}- w but n - 1 ) --}- \_n, n -}- 1,.. = 0, 1,..., n,- = w, w -F- n = n, n + 1, :: -) -f- i,,-.. rt - 1- (n, it } since the representative of w 4 n has a greatest element but that of n + w does not. Also, w2=(0,1, (0, b), (1, b), co while 2w = - ( - b,-, 0), ), (b, l ), --. = w. 2w, since the member (0, b) of the representative of w2 is 't'hus w2 preceded by infinitely many elements unlike any member of the represen.tative of 2w. The exercises for this section call for proofs of the associative laws for both addition and multiplication of order type. The general associative law for each operation then.follows by Theorem 2.2. The distributive law a(f3 + Y) = a# A- ay 2.6 1 Order 7 y/pes and consequently its general form holds, but general, as shown by the example 101 y)a 5-4- pot -1- y(X in (w -I- 1)2 = (w + 1) + (w -I- 1) = co -{- [(1 -1 w) + }- 1 ] =w+(w+1) =w2-{-1 p!5 w24-2. We shall pursue the study of the arithmetic of arbitrary order types 110 further. However, additional properties will be obtained for the arithmetic of a restricted class of order types, namely, those which are represented by well-ordered sets. These appear in the next section after the derivation of |
some fundamental properties of such sets. EXERCISES 6.1. Show that the definitions of a -I- /3 and a/3 for order types a and (3 are independent of the representatives used. 6.2. Prove the associative law for addition and for multiplication of order types. 6.3. Prove the distributive law a((3 -I- 'y) = a/3 1- ay for order types. 6.4. Supply the details of the example in the text which demonstrates that -1- y)a 5-' (3a -1- ya for all order types a, 0, y. 6.5. Let A be a chain. A subset I3 of A such that if b C B, a C A, and a < b, then a C B -is a segment of A. Clearly, 0 and A are segments of A; other segments are proper segments. Prove that if R and C are segments of A, then one is a subset of the other. 6.6. Prove that a chain is of order type co if it is infinite and every proper segment is finite. 6.7. From a chain (A, <), a chain (A, <*) can be derived upon defining a <* b if b < a. If the original set has order type a, that of the new chain is denoted by a*. For example, (., 2, 1, 0) is of order type w*. Give two reasons, each of which is adequate, to justify the assertion that w + co* 76 w* + w. 6.8. Prove that for any order types a and 0, (a -I- 8)* _ (3* + a*. 6.9. Prove that for an order type a, a = a* if a has the form (3 -I- /3* or p -1- 1 + 6* for some order type (3. 6.10. Given I -I- a = a for an order type a iff a = co + 0 for some (3, prove that a + 1 = a iff a = (3 + w* for some 0. 6.11. The order type of the set Q of rational numbers with its familiar ordering is designated by r). It can be proved that a chain is of order type n iff it (i) has neither a first nor a last element, (ii |
) is dense (that is, between each pair of distinct elements there is a further element of the set), and (iii) is denuwerable. Use this result to prove that (a) if a and b are rational numbers with a < b, then the set of all rationals between a and b has order type n, and 102 The Natural Number Sequence and its Generalizations ( c t-r A P I. 2 (b) if from an ordered set of order type rt there is removed a finite number of elements, the remaining set has order type r,. 7. Well-ordered Sets and Ordinal Numbers We recall that a well-ordered set is a partially ordered set such that each nonempty subset has a least (or, first) element. Such a set is a chain, among other things. To ensure that a relation p well-orders a set A, it is sufficient that p (restricted to A) be antisymmetric and that in each nonempty subset A, there exist an element a, such that a,pb for every b C A,. The proof is left as an exercise. EXAMPLES 7.1. The empty set is well-ordered relative to any simple ordering. Every subset of a well-ordered set is well-ordered relative to the same ordering relation as for the original set. 7.2. It is left as an exercise to prove that a simply ordered set which is ordinally similar to a well-ordered set is itself well-ordered. It follows that any simply ordered set of order type w is well-ordered. 7.3. The simple ordering of the set of natural numbers given by is a well-ordering. 'thus, sets of order type w + w are, 0, 2, 4, (1, 3, 5, well-ordered. 7.4. The ordering of the set of positive rationals (1/1, 2/1, 3/1,..., 1/2, 3/2, 5/2, 1/3, 2/3, 4/3,...) is a well-ordering, since any uonempty subset contains fractions with a smallest denominator, and among these is one fraction with smallest numerator. 't'his fraction is the least member of the subset. The order type of this ordered set is found to be ww or w2, following the conventional abbreviation. Thus, sets of order type w2 are well-ordered. 7 |
.5. A set of order type w* (see Exercise 6.7) is not well-ordered since it has no first element. It is left as an exercise to show that any infinite chain, no infinite subset of which has a first element, is of order type w*. 7.6. Any simple ordering of a finite set is a well-ordering. 7.7. Well-ordered sets have many properties in common with the natural number sequence. However, N has two properties not shared by well-ordered sets in general. First, N has no last element-as does, for instance, the wellordered set (0, 1, 2, - -, oo). Second, each element n of N, apart from its first element, has an immediate predecessor n - 1. This is not true of the element in the set above; nor is it true of the element 0 in Example 7.3. Both proofs and definitions by induction can be carried out in any well-ordered set. Their respective formulations are entirely analogous 2.7 I Well-ordered.Sels and Ordinal Numbers 103 to those of proofs and definitions by strong induction, given in Section 2. The principle of proof by transfinite induction is as follows, where, as earlier, 1'(x) stands for "the clement x has the property P." if P(x)), where xu is the first element of the well-ordered set X, and if for all z in A', P(y) for ally < z implies P(z), then P(x) for all x in X. f A demonstration of this principle is simply a repetition of the argumerit used to substantiate the earlier statement of proof by strong induction. 'This result is seldom used since, in practice, it is usually as easy to carry out the proof of the principle for the case at hand as to apply the principle. In order to state the generalization to any well-ordered set of our earlier result ('1'heorern 2.4) concerning definition by induction, it is convenient to introduce two auxiliary concepts. If A is a well-ordered set and if x C A, then (a C Ala < x} is called the initial segment determined by x; we shall denote it by As. 1f B is an arbitrary nonempty set, then by a sequence of type x in B we shall mean a function on A. into B. Then the principle of definition by transfinite induction may be |
stated as follows. Let A be a well-ordered set having ao as its least element, let B be a set, and let c be a nicrnber of B. If h is a function whose range is included in 13 and whose domain is the set,j of all sequences j of type x in B for spine x ° ao, then there exists exactly one function k: A--} B such that k(ao) = c and k(x) = h(klA.,) for each x in A other than ao. The proof is left as an exercise. It involves a slight modification of that given for 'T'heorem 2.4 because of the possible variance of an arbitrary well-ordered set A with N, noted in Example 7.7. We turn our attention next to the derivation of structural features of well-ordered sets. Three basic results (Theorems 7.2-7.4) in this category follow easily from the next theorem. TIl E0 R L M 7.1. If A is a well-ordered set and f is an isomorphism of A into itself, then a < f(a) for each a in A. Proof. Assume that for some element a in A we have a > f(a). I;et B be the subset of A of all such elements and b its least rtrenther. Since b > f(b) it follows chat f(b) > f(f(b)). Thus f(b) E B, which is a contradiction. f The hypothesis that xo have property P is redundant, for it is that instance of the second hypothesis which results upon choosing z as xo. 104 The Natural Number Sequence and its Generalizations I c.Ii A P. 2 THEOREM 7.2. A well-ordered set is not ordinally similar to any of its initial segments. Proof. Assume, to the contrary, that f is an isomorphism of the wellordered set A and one of its initial segments A. Theorem 7.1 is applicable and, consequently, x < f(x). Thus, f(x) V A=, which is a contradiction. COROLLARY. IfA is a well-ordered set and ifA. = As,, then x = y. Proof. We assume that A, = A along with x 7` y and derive a contradiction. If x 0 y, then either x |
< y or y < x. Suppose that x < y. Thet, A. is an initial segment of the well-ordered set A and is ordinally similar to Ar. This is a contradiction of Theorem 7.2. The assumption that y < x yields a similar contradiction. THEOREM 7.3. If A and B are ordinally similar well-ordered sets, then there exists exactly one isoinorphisni between them. Proof. Assume that g and h are isomorphisnis of A onto B. Then f = g-'o It is an isornorphisin of A onto itself. According to Theorem 7.1 this implies that a < (g'-' o h) (a) for each a in A and, conscquently, that g(a) < h(a). Reversing the roles of g and h, we may also conclude that h(a) < g(a) for all a in A. Hence, g = h. THEOREM 7.4. If A and B are well-ordered sets, then exactly one of the following hold: A is ordinally similar to B, A is orclinally similar to an initial segment of B, B is ordinally similar to an initial segment of A. Proof. The conclusion is trivially trite if A or B is empty. So assume that neither set is empty and that neither is ordinally similar to an initial segment of the other. We shall prove that, in these circumstances, A = B. Let x be a member of A distinct front the least element, ao, of A, and let j be a sequence of type x in B. If the range of j has an upper bound in 11, let h(j) be its least upper bound, while if the range of j has no upper bound, let h(j) = bo, the least element of B. Finally, let h(ao) = bo. Then It is a function of the type described in our formulation of the principle of definition by Iransfinite induction. Hence, there exists a function k on A into 13 such that k(ao) = bo and k(x) = h(kjA7) for each x in A other than ao. It is left as an exercise to prove by transfiiiite induction that, for each x in A, the function k maps the initial segment determined by x in A onto the initial 2.7 |
I Well-ordered Sets and Ordinal Numbers 105 segment determined by k(x) in B in a one-to-one fashion. i'lien it is another easy exercise to conclude that k is an isomorphism of A onto B. CORC)LI.ARY. For well-ordered sets A and B, exactly one of A = B, A < 1, Ii < A holds. In other words, any two cardinal numbers which have well-ordered sets as representatives are comparable. The order type of a well-ordered set is called an ordinal number, or simply an ordinal. Since a chain which is ordinally similar to a wellordered set is itself well-ordered, every representative of an ordinal is well-ordered. Among the specific order types mentioned so far, some may be classified as ordinals. 't'his is true of the order types represented by finite simply ordered sets, since such an ordering is automatically a well-ordering. 't'hus the natural numbers may henceforth be called ordinal numbers. The ordinals which are not natural numbers are called transfinite ordinals. ']'he one-to-one correspondence between finite cardinal and ordinal numbers is due to the fact that, not only can any finite set A be simply ordered, but that. all orderings of A are similar (and, indccd, well-orderings). In contrast, one cannot expect that infinite cardinals can serve as ordinals, because a given infinite set which can be well-ordered, can be at least simply ordered in a variety of ways and, consequently, determines a variety of order types, some of which may be different ordinals. The state of affairs is adequately illustrated by the set of natural numbers. Below are indicated seven simple orderings of N, each having a different order type: (0, 1, 2, 3,... ) (..., 3, 2, 1, 0), (1,3,5,...,0,2,4,...), (...,4,2,0,...,513, 1), (1,3,5,...,4,2,0), (...5,3,1,0,2,4,...), (0,3,6,9,...1,4,7,10,...,2,5,8,...). Of these, the first, third, and last determine ordinals (namely w, w2, |
and w3 respectively). This suggests what is indeed the case: the infinite ordinals are much more abundant than the infinite cardinals. If a and P are ordinals, we shall say that a is less than 03, symbolized a < /3, 106 The Natural Number Sequence and its Generalizations I C II A P P. 2 if there exists a representative of a which is ordinally similar to an initial segment of one for f3. Certainly this relation is transitive and, ac_ cording to 'T'heorem 7.2, is irreflexive. Hence, the relation < partially orders any set of ordinals. Theorem 7.4 yields the further result that this partial ordering relation is actually a simple ordering; that is, any two ordinals are comparable. The still stronger result, that the ordering relation for ordinals well-orders any set of ordinals, can be proved. As a preliminary to the proof of this fundamental result we derive a special case of it. TI I EOREM 7.5. The set s(a) of all ordinals less than the ordinal a is a well-ordered set of ordinal number a. Proof. Let f3 C s(a). Then there exist representatives A of a and B of f3 such that B = Al for some x in A. The element x is uniquely determined by [3 in view of the Corollary to 'T'heorem 7.2. Hence, a mapping f: s(ee) -} A is defined by setting f(/3) = x. Clearly f is one-to-one. Moreover, f is onto A, since given y in A, if we set [3 = Av, then [3 C s(a) and f(a) = y. Finally, it is clear that f is order-preserving. 't'hus, we have proved that s(a) -= A; since A is well-ordered, so is.s(a) and s(a) = A = a. This result gives a certain "normal" representation for ordinal numbers; whenever it is permissible to replace a set A by one that is ordinally similar to it, we may use s(a) if a = A. "I' I I E () R Is M 7.6. Any set of ordinals is well-ordered. It roust be shown that each nonenrpty set A of ordinals has |
Proof. a least member. Let a C A. If a is not the least member of A, then A n s(a) -x 0. 'T'hen A n 3(n), as a subset of s(a), is well-ordered, and thus has a first member j3. If b C A, then h < 0 implies b < a and, hence, b C A n s((.Y), which contradicts the choice of ft. As ne of the simply ordered set A, (3 and S are comparable, so we may conclude that / < b for all b in A. 'T'hus 0 is the least member of A. 'I H L' ORE M 7.7. If A is any set of ordinals, then there exist ordinals greater than any ordinal of A. Indeed, there exists a smallest such ordinal. Proof. Let A' = A U U { s(a) ja C A I. By Theorem 7.6, A' is well 1 Well-ordered Sets and Ordinal Numbers 2.7 107 ordered. If a C A, s(a) C A' and s(a) = A.. Hence a = -s-(-a) < O'. Thus a < A' for each a in A. Since ordinal numbers are order types, the sun, a + /3, of ordinals a and /3 is defined. It is left as an exercise to show that this sum is an ordinal number. That the product, af, of a and 0 as order types is an ordinal is proved as follows. If either of the ordinals a and /3 is equal to zero, then so is a/ and the proof is complete. So assume that a > 0 and /3 > 0. Let A and B be well-ordered sets such that fl = a and B _ /3. Then, by definition, a$ = A _X W if in A X B we define (a, b) < (a', b') iff b < b' or b = b' and a < a'. It suffices to prove that A >( B is well-ordered by this relation. Let C be a nonernpty subset of A X B and let (ar, br) be some member of C. Let bo be the least member of (b C BI(ar, b) C C} and, in turn, let an be the |
least member of is an easy exercise to prove that (ao, bo) is the (a C AI(a, least element of C, and this completes the proof. C C1. I t Our first result relating addition and ordering of ordinals is THEOREM 7.8. If a and /3 are ordinals and 0 > 0, then a + 0 > a. Proof. Let C be a well-ordered set of ordinal number a + P. Then C=AUBwhereA= a, R =/3, and B- 0, since/ -X 0. Hence A is a segment of C. It follows that r > A; that is, a + /3 > a. The above result implies that a + I > a for every ordinal a. It is left as an exercise to prove further that a + I is the successor of a; that is, there is no ordinal f; such that a < i; < a -f- 1. We have already seen, in contrast to this, that there are ordinals having no predecessor. Ordinals having it predecessor are ordinal numbers of the first kind and those having no predecessor are ordinal numbers of the second kind. For example, 5, w -+- 2, and w2 + 3 are ordinals of the first kind while w, w2, and w2 are ordinals of the second kind. THEOREM 7.9. Let a and /3 be ordinals with a < /3. Then there exists exactly one ordinal -y > 0 such that a -}- y = 0. Proof. Let A and B be well-ordered sets with A = a and B = /3. The assumption that a < /3 implies that A = B. C B. Let C = B - B. and -y = C. Then B,UC= B and B lC=0,so a+y =/3. To prove the uniqueness, suppose that both a + yr = /3 and a + y2 = 0, where yl 0 y2. Let us say that yr < rye. Then there 108 1'lze Natural Number Sequence and its Generalizations I C,ttA1'. 2 exists a b > 0 such that 72 = 7r + S. I-fence 0 = a + 72 = a + (-y' + S) = ((x + 7t) + S = a -{- S |
, which is a contradiction of 'I'heorern 7.8. EXAMPLES 7.8. According to Theorem 7.9, the equation a + > = 13 has a unique solution for given ordinals a and fl with a < 13. This solution is denoted by 14 - a. On the other hand, the equation t -{- a = Q may not have any solution for given a and 6 with a < B. An example is + I = w. 7.9. As with order type generally, addition of ordinals is not a commutative operation. Indeed, it may be, for ordinals a and 0, that a + 0 < 0 + a or that a+0>/3 1- a. For example, w > 1and1+w<w-I-Iwhile w-1.2>w+I and (w -I- 1) -}- (w -I- 2) > (w + 21 - (w A- 1). 7.10. According to '['heorcrn 7.5, if the well-ordered set A is of order type a,.c(a), the set of ordinals less than a. Ilence this set of ordinals may be then A used to index the members of A. That is, we may describe A as {aelt < a} where at < a ifT < +t. In this connection it is desirable to note the first members of a set s(a) for sufficiently large a. First come the natural numbers; their ordering, as dictated by their role as ordinals, coincides with that in their original role as members of the natural number sequence. After the set of all finite ordinals, occurs the first transfinite ordinal. According to Theorem 7.5 it -), that is, w. There follows its successor w l- 1, is the order type of (0, 1, 2, then w + 2, and so on. So far we then have 0,1,2,...,w,w-l- 1,w+2, This sequence has order type co + w - w2, so the number following these is w2. Continuing in this fashion we arrive at the sequence 0,1,2,.,w,w4-1,w-I-2,,w2,(o24-1,,w3,w3+1 |
I a < y -I- (3 and conversr'ly; (Il) a < (3 implies a -I- y < (3 -I- y; conversely, a I- y < (3 I- y imnplies a < (3; (III) a < (3 and y > 0 imply ya < y(3; conversely, ya < y(3 implies a < (3; (IV) a < (3 implies ay < (3y; conversely ay < (3y implies a < y; (V) y+a=y-{-i3implies a=(3; (VI) ya = y(3 and y > 0 imply a = (3. The equality signs in (II) and (IV) cannot be dropped; for example, I < 2 but I + w = 2 ± w and 1w = 2w. The equality 1 -I- w = 2 -I- w also illustrates that (V) has no right-hand analogue. There is a righthand analogue of (VI)- if a, (3, and y are ordinal numbers such that ay = fly and y is a number of the first kind, then a = (3. A proof is suggested in an exercise. In Example 7.8 we mentioned subtraction for ordinals. The following result is basic. in formulating the concept and properties of division. However, we shall not pursue this matter. TI1L'OREM 7.1 1. If a and (3 are ordinals and (3 > 0, then a has a unique representation in the forth a=(3k+p where 0<p<(3. Proof. Let (3 > 0. Then 3 > I and hence (3a > 1a = a. If (ice = a, this is a representation of the desired kind. Otherwise fla > a. Now, if a, (3, and y arc ordinals with (3a > y, then y has a unique repre- 110 The Natural Number Sequence and its Generalizations I C11 A P. 2 sentation in the form y = 13a1 + 01 where a, < a and #, < P. The proof is left to the reader. We apply this result with a as y to obtain a = j3a, + #, where /31 < 13. Again we have a representation of the type desired. The proof of uniqueness |
is left as an exercise. EXAMPLES 7.12. Ordinal arithmetic presents a wide assortment of oddities. There follows a sketchy sampling. (a) For n > 1, w"+' = (w" + w)2 - (wn)2. However, it can be shown that,, cannot be represented as a difference of squares of ordinals. (b) The ordinal w2 has infinitely many representations as a difference of squares: w2 = [w(n + 1)]2 - (wn)2 for n = 1, 2,. (c) For every n > 1, (w + 1)"w" is an nth power of an ordinal, namely W2. On the other hand, w"(w + 1)" has no such representation. Indeed, since Wn(w + 1)n _ w2n + W2n 1 +... f Wn (W2 _+. On = W2n + W2n I < Wn(W + 1)n < W2n + In -1 +... / -1- w + 1 = (w2 + W That is, wn(w + 1)" lies between the nth powers of two successive ordinals and hence cannot be an nth Dower. (d) If a = wan + w"-1 and 0 = W2n + w for n > 1, then a2 = A3; yet there is no ordinal y such that a = y3 and Q = y2. 7.13. With S = 2 in Theorem 7.11 we may conclude that every ordinal can be represented either as 2t or 2t + 1, that is, is either even or odd. For example, (w + 1)2 = 2(w2) + 1 is odd! Again, with ;Q = w, we may conclude from Theorem 7.11 that any ordinal a can be represented as a = wE + p where p is finite. If p > 0, then a is a number of the first kind since a = w + (p - 1) + 1. It follows that every ordinal of the second kind is of the form wt. The converse statement is easily established, yielding a characterization of ordinals of the second kind as those having W as a left-hand divisor. EXERCISES 7.1. Prove that if a relation p ( |
restricted to a set A) is antisymmetric and if in each nonempty subset A, of A there exists an element a, such that a,pb for every b E A1, then p well-orders A. 7.2. Prove that a simply ordered set which is ordinally similar to a well- ordered set is well-ordered. 7.3. Show that any infinite chain, no infinite subset of which has a first ele- ment, is of order type w*. 7.4. Establish the principle of definition by transfinite induction given in the text. 2.8 I The Axiom of Choice and Zorn's Lemma 111 7.5. Supply the missing details in the proof of Theorem 7.4. 7.6. In the text we inferred from Theorem 7.4 that any two ordinals are comparable. Give an independent proof of comparability, using Theorem 7.5. 7.7. Prove that the sum of two ordinals is an ordinal. 7.8. Complete the proof of the assertion that the product of two ordinals is an ordinal. 7.9. Find how many different values are assumed by the sum of the ordinals 1, 2, 3, 4, and w in all possible arrangements. 7.10. Determine an arrangement of w, w2 + 1, w3, w5, and w2 for which their sum is w2 + w11 + 1. 7.11. Prove Theorem 7.10 by first proving the forward implications in (I)--(IV). 7.12. Prove that if a and P are ordinals with a > 6, then a + n > fi + n for = 1,2,... 7.13. Prove that 1 + a - a for an ordinal a iff a > w. 7.14. Find ordinals a and fl such that (a - fi) ± f3 P6 a. 7.15. Show that if a, 0, y, and 5 are ordinals such that a > f3 and y > 5, then ay > 06. Use this to prove that if a > f3 and y is an ordinal of the first kind, then ay > fiy. Then show that if cry = 0y and y is of the first kind, a = 0. This is a right-hand analogue of (VI) in Theorem 7.10. 7.16. |
Show that (W' -'- W)5 = (W., + w')2. 7.17. Suppose that a and f3 are positive ordinals with a -{- a = w. What is af3? 7.18. Give an example of two ordinals a and 13 such that a + fi = + a but a2 -l- f32 _ /#2 + a2. 7.19. Prove that for ordinals a and fl, af3 = f3a implies a2f32 = 2a2. 7.20. Prove that an ordinal f3 is of the second kind iff nf3 = 0 for n = 1, 2, 7.21. Prove that the product of two nonzero ordinals is a number of the first kind iff both factors are of the first kind. 7.22. Complete the proof of Theorem 7.11. 8. The Axiom of Choice, the Well-ordering Theorem, and Zorn's Lemma A theorem to the effect that all sets occurring in mathematics can be well-ordered would be extremely valuable. "Then, for instance, definitions and proof's could be fornmulated by induction for all sets, just as for the natural number sequence. In 1904, Zernielo gave a dernonstration of the well-ordering theorem which asserts that every set can be well-ordered. Soon after its publication it was pointed out by E. Borel, that the proof employed a property of sets which may be deduced easily from the well-ordering theorem, thereby making the two properties 112 The Natural Number Sequence and its Generalizations I c 11 A P. 2 equivalent. The property assumed by Zcrmelo has become known as the axiom of choice. One of its formulations is the following. (AC,) If a is a disjoint collection of noncnrpty sets, then there exists a set B such that for each A in (t, B fl A is a unit set. In other words, if a is a disjoint collection of nonernpty sets, then there exists a set which has a single member in common with each member of a. Since a disjoint collection of nonemjrty sets is a partition of the union of the collection, (AC,) is clearly equivalent to: For each partition (a of a set. U there exists a subset of U consisting of exactly one member of each member of a |
. Such a set is called a representative set for the partition as well as for the associated equivalence relation. The restriction in (A(.,) to disjoint collections may be circumvented by formulating it for families of sets. This version reads: If {A,} is a. family of nonempty sets indexed by a nonempty set 1, then there exists a family {xi} with i E I such that xi C Ai for each i C 1. Intuitively, one thinks of arriving at a set B of the type mentioned in (AC,) by a constructive process; one chooses, in turn, an x from each of the sets A; this accounts for the presence of the word "choice" in the name. That it should be named an axiom is simply an indication that no one has been able to infer the existence of such a set, in general (other than from an equivalent property of sets). 1 he axiom of choice has been the subject of serious controversy among totally on such grounds as the utter mathematicians. Some reject it impossibility of making infinitely many selections (needless to say, it is only the case where a is infinite that the axiom injects anything new) or, on the lack of precise definition of a representative set. Others accept the axiom for the case where a is denumerable and reject it in the uncountable case. Many accept it without any reservatin. Of those to whom the plausibility of (AC,) is indisputable, some revise their attitude when propositions which can be proved equivalent are encountered. Several equivalent forms, which are in the nature of more useful working forms, are derived in this section. 'T'here is another category of propositions equivalent to the axiom of choice which might be catalogued as illuminating. For example, in Section 10 it will be shown that it is equivalent to the assertion that the ordering of cardinal numbers, discussed in Section 3, is a simple ordering. This is not a useful version of the choice axiom, but rather serves to point out that someone who "believes" that cardinals should be simply ordered must also "believe" the axiom of choice. Thus, such equivalent 2.8 I The Axiom of Choice and Lorn's Lemma 113 formulations serve primarily to sharpen the delineation between the two schools of thought. Because of the diversity of opinion about the axiom of choice, it is common practice for some present-day authors to point out the occurrences of their us |
ages of it. In cases where a new result rests on some classical mathematics this may amount to merely superficial honesty, since in much of classical mathematics the axiom of choice has slipped into proofs without being noticed, and no one has ever combed through all of it and sorted out the tainted theorems from the untainted. To the best of our knowledge the axiom of choice has been employed in the foregoing only in the proof of Theorem 4.4. Henceforth, we accept the axiom of choice as a valid principle of intuitive set theory and use it without reservation. In this matter we are guided by Cantor, who tacitly accepted the axiom. For intuitive set theory it has the same status as the principles of extension and abstraction (Section 1.2); collectively, these three assumptions serve as a basis for the theory. Turning to the derivation of propositions which are equivalent to the axiom of choice, we present first two variations which are so closely related to (AC,) that. they are also known by the same (AC2) For every set X there exists a function f on the collection, U,(X) - 10 }, of nonempty subsets of X such that f(A) C A. Such a function is a choice function for X. Thus, (ACC2) asserts that every set has a choice function. (AC3) If {A,} is a -family of none,npty sets indexed by a nonempty set I, then X,EIAi is noncmpty. The equivalence of (AC,)--(AC3) is easily established. That of (AC,) and (AC3) follows directly from the formulation of (A(',,) for-a family of sets and the definition of cartesian product. Further, it is clear that (AC,) implies (AC2). To complete the proof of the equivalence of is sufficient to prove that (AC2) implies (AC,). The (AC,)--(AC3) it reader can do this easily. '1'o bridge the gap between the axiom of choice and other useful equivalent forms, we prove a fixed point theorem due to N. Bourbaki (1939). Before tackling its proof, as well as the statement and proof of the theorem that follows it, the reader would do well to refresh his memory with regard to the definitions given at the end of Section 1.11. THEOREM 8.1. Let E be a nonempty partially |
ordered set such that every chain included in E has a least upper bound in E. If 1 14 The Natural Number Sequence and its Generalizations I CHAP. 2 f : E --} E has the property that f (x) exists at least one x in E such that f(x) = x. Proof. Let a be a fixed element of E. A subset A of E will be called admissible (relative to a) if it has the following properties. x for all x in E, then there (1) a C A. (2) J(A) C A. (3) if F is a chain included in A, then lub F C A. Clearly E is admissible. Moreover, it is easily verified that M, the intersection of all admissible subsets, is admissible. Thus M is the smallest admissible subset. It follows that if a subset M0 of M can be shown to be admissible, then MO = M. This technique is used to derive each of three properties of M (designated by Roman numerals), from which the theorem follows easily. (I) The clement a is the first element of M. It is sufficient to prove that the subset A = ;x C Mix > a } of M is admissible. For this we verify, in turn, properties (1), (2), and (3) of an admissible set. (1)' (2)' (3)' Since aCMand a>a,aCA. I.et x E A; to prove f(x) C A. Now x E A implies x E M, and hence, f(x) E M by (2). Also, x C A implies x > a and this, with J(x) > x, yields f(x) > a. Thus f(x) E A. Let to = lub F, where F is a chain included in A. Since A C M, we have F C-: M and hence to C M by property (3) of admissible sets. Also, F C A implies x > a for all x E F, and hence to > a. Thus A is admissible, A = M, and (1) is proved. Before Continuing, we make a definition. An element x of E is said to have property P, in symbols P(x), if y C M and y < x implies f (y) < X. (II) If |
x C M and P(x), then for each z E M either z < x or z > f(x). It is sufficient to prove that the subset B = }z C MIz < x or z > J '(x) } is admissible. (1)" aCIv!anda <x [indeed for each x C M by (I)1,so aCB. (2)" Let z C B; to show that f (z) E B. As in (2)', f (z) E M. Also, z C B implies z < x or z > f(x) by definition. if z = x, then 2.8 1 The Axiom of Choice and Zorn's Lemma 115 f(z) = f(x) so that f(z) > f(x) and hence f(z) C B. If z < x, then since P(x), f(z) < x and f(z) C B. Finally, if z > f(x), then f(z) > z > f(x) and, again, f(z) C B. (3)" Let w = lub F, where b 'is a chain included in B. As in (3)', w C M. Also, for each z C F either z < x or z > f(x). If the first alternative holds for all z C F, then x is an upper bound for F, and hence w < x, so zo C B. Otherwise, there exists a z C F such that z > f(x). Then w > z > f(x) and, again, w C B. Thus B is admissible, B = M, and (II) is proved. (III) Every element of M has property P. It is sufficient to prove that the subset C = [x C MIP(x) [ is admissible. (l)"' a C M and is, moreover, the least element of M. Thus, for no z of M is z < a. Hence, a satisfies I' vacuously and, consequently, is in C. (2) "' Let x C C; to show that f(x) C C. As in (2)', f(x) C M. It remains to prove that f(x) has property P, that is, y C M and y < f(x) imply f(y |
) < f(x). Applying (II) to x we have either y < x or y > f(x). The second possibility cannot hold, for y > f(x) with y < f(x) is impossible. Thus y < x. If y < x, then f(y) < x, using property P for x. This, with x < f(x), implies f(y) < fi(x), as required. Also it is immediate that if y = x the same conclusion holds. Thus f(x) C C. (3)"' Let w = lub F, where F is a chain included in C. As in (3)', w C M. Thus it remains to show that P(ro), that is, y C M and y < w imply f(y) < w. For this we show first that for such a y there exists y, C F such that y < y,. Indeed, if no such y, C F exists, then by (II) [Note: y, C F implies that P(y,) 1, y > f(y,) > y, for ally, C F. Then y is an upper bound for F and hence, y > iv, which contradicts the assumption that y < w. Thus a y, C b with y < y, exists. If y < y,, then by property P for y,, f(y) < y, < w, so that f(y) < iv as required. If y = y,, then P(y) and hence, by (11), either w < y orf(y) < w. The first possibility is excluded and hence, again, f(y) < w. Thus C is admissible, C = M, and (III) is proved. Now for the coup de grace. From (II) and (I1I) it follows that if x, 1 16 '1 he Natural Number Sequence and its Generalizations I CHAP. 2 y E M, then either y < x or y > f(x) > x, so M is simply ordered. Let xo = lub M. Since M is admissible, xo C M and, moreover, f(.to) E M. 'thus f(xo) < xo. But xo < f(xo) by hypothesis. It follows that f(xo) = xo. THEOREM 8. |
2. The following statements arc equivalent to one another. (I) Zcrmclo's axiom of choice: For every set X there exists a function f on the collection of noncmpty subsets A of X into X, such that for each A, f(A) C A. (II) 1-!ausdorff's maximal principle: Every partially ordered set includes a maximal chain, that is, a chain which is not a proper subset of any other chain. (lll) 'horn's lemma: Every partially ordered set in which each chain has an upper bound contains a maximal clement. (IV) Every set can be well-ordered. (I) implies (II). Let (P, <) be a partially ordered set and asProof. sume that (11) is false for it. ']'his means, if. is the family of all subsets X of P which arc simply ordered by <, that for each X in a there exists Y in a. with X C Y. That is, ax - IYC aIXC Y} is noncmpty for each X in U. By (1) there exists a function f on }axlX C a} into a such that f(ax) C ax. "Thus g: a -*- a with g(X) = f(ax) has the property that X C g(X) for all X in U. As such, the partially ordered set (a, C), together with the function g, satisfies the hypotheses. of Theorem 8.1. (It is left as an exercise to show that if a* is a subset of a which is simply ordered by C, then U a* = lub ct*). But X C g(X) for all X in a is a contradiction of the conclusion of 'l'heorcin 8.1. Thus, since (1) and the denial of (11) lead to a contradic-. Lion, (1) does imply (I1). (11) implies (111). Assume that the partially ordered set (1', <) satisfies the hypothesis of (111). By (11), there exists a maximal subset A of 1', simply ordered by <. Let a be an upper bound for A. Then a is a maximal clement for P. Indeed, assume that a < x for some x in 1'. "Then A U {x} is a simply ordered subset which |
properly includes A. This is a contradiction. 2.8'The Axio,n of Choice and Zorn's Lemma 117 (111) implies (IV). Let X be any set. We consider ordered pairs (A, p) where A C X and p simply orders A. Let S be the set of all (A, p) such that p well-orders A. If (A,, pl) and (A2, p2) are members of g, define (A,, p,) < (A2, P2) ill (a) A. S A2, (b) P, C P2, and (c) if a, C A,, a2 C A2, and 02 V A,, then (a,, 02) C P2 In other words, we require that A, be a subset of A2, that the ordering of A2 be an extension of that of A,, and that the elements of A2, not in A,, be greater than the elements of A,, relative to the ordering of A2. It is immediate that < partially orders S. We prove next that (S, <) satisfies the hypothesis of (Ill) ; that is, a chain included in S has an upper bound in S. For a chain e C S, we propose as an upper bound, (A*, p*), where A* = U {AI(A, p) C Cl and p* = U IPI(A, p) C e}. Clearly the only question is whether (A*, p*) C S. To show this we prove that (A*, p*) satisfies the conditions stated at the beginning of Section 7. The proof that p* is antisymmetric is left as an exercise. It remains to prove that if B is a noncnnpty subset of A*, then there exists bo C B such that (b(j, b) C p* for each b C B. For such a B there exists (A,, p,) C e such that B fl A, 7`- 0. In turn, there exists bo C I3 n A, such that (be, b) C p, for all b C 13 () A,. More generally, for each b in 13 there exists p with (A, p) C C such that (bo, b) C p. Indeed, given b in 13, there exists (A, p) C e with b C A. If A e A |
,, then (bo, b) C p,. Otherwise, A D A, and, hence, p D p,. Then (bo, b) C p and so (b,,, b) C p*, as desired. Since the hypothesis of (Ill) is satisfied, we may infer the existence of a maximal element (A, p) of S. The proof will be complete if it can be shown that A = X. To this end assume the contrary, that x C X - A. We now adjoin x to A and extend the ordering p of A to one of A U {x} by defining x to be greater than each element of A. This yields the ordered pair (A', p'), where A' = A U {x} and p' = p U { (a, x)la C A'{. Then p' well-orders A' and hence, (A', p') C S. Moreover, (A, p) < (A', p'), which is a contradiction, since (A, p) is a maximal element. Hence, A = X. (IV) implies (I). Let X be any set. By (IV), Xcan be well-ordered, so we assume that this is given. If A is a noncmpty subset of X, let f(A) be the first element of A. Then f is a choice function for X. 118 The Natural Number Sequence and its Generalizations I C H A P. 2 In mathematics, labeling a proposition with the name of an individual usually indicates his priority to that result. This is not the case in the assignment of names to be found in the literature to the above equivalent formulations of the axiom of choice or in the extensive variety of other formulations which have proved to be useful. We have made what is a relatively common assignment of names. It should be mentioned that prior to the emergence of a variety of statements equivalent to the well-ordering theorem, transfinite induction was a standard proof technique. This has now given way to the use of some form of Zorn's lemma or a maximal principle. f Usually the modern procedure yields a shorter proof. EXERCISES 8.1. Establish the axiom of choice in the form (AC,) for finite collections of sets by proving, by induction, that if A, X A2 X.. X A. is empty, then at least one A; is empty. 8.2. The following is known as Hilbert's |
axiom. If 6' is the set of all properties P such that there exists at least one object having property P, then there exists a function a whose domain is 6' and such that e(P) is an object having property P. Prove that this axiom is equivalent to the axiom of choice (AC2).. the following case of (AC,) : For every disjoint collection (t of n-element sets A, there exists a set B such that, f o r each A in C 3, B (l A has exactly one member. Without using the axiom of choice, prove that [2] implies [4]. 8.3. Following A. Mostowski, let us denote by [n] for n = 1, 2, 8.4. Referring to the proof that (I) implies (II) in Theorem 8.2, show that Ua.* = lub a*. 8.5. In the proof that (I11) implies (IV) in Theorem 8.2, show that p* is antisymmetric. 8.6. Demonstrate the equivalence of the axiom of choice with the following statement: If A and B are nonempty sets and p is a relation with domain A and range B, then there exists a function f : A --b- B such that f C p. 8.7. Show that if p partially orders A, then there exists a simple ordering relation p' such that p' Q p and p' simply orders A. (Hint: Consider the collection of partial ordering relations which include p and use Zorn's lemma.) 9. Further Properties of Cardinal Numbers With the axiom of choice available, some extensions and some simplifications of properties of infinite cardinals are at hand. It will be recalled t Of course, Zorn's leninna is not a substitute for transfinite induction in cases of justifying definitions. 2.9 I Further Properties of Cardinal Numbers 119 that for reasons of expediency one application has already been made to the proof of Theorem 4.4. Another occurs in our next result. THEOREM 9.1. Any infinite set includes a subset of cardinal number No. Proof. Let X be an infinite set. It is sufficient to exhibit a function g on N into X which is one-to-one. Let f be a choice function for X. Then we define g(O) = f(X) and, proceeding inductively |
, let g(n + 1) = f(X - 1g(O),, g(u) }). This is possible for each n, since otherwise X would be finite, which is contrary to assumption. According to Theorem 2.4 there exists a function g on N into X such that, for each n in N, g(n) = f(X - {g(0),, g(rr - 1) } ). Now g is s. It is no loss of one-to-one. For consider r and s in N with r generality to assume that r < s and we do so. Then g(r) (, X, g(s - 1) } and hence g(s), as a member of this Ig(0), set, is necessarily distinct from g(r). g(r),. COROLLARY I. If A is an infinite set then fl > Ro. The proof is left as an exercise. Combining Corollary 1 with Theorem 3.5, bto is established as the least infinite cardinal. Theorem 9.1 has another interesting consequence, which places in sharp relief a basic difference between finite and infinite sets. COROLLARY 2. An infinite set is similar to a proper subset of itself. Proo/. Let A be an infinite set. According to Theorem 9.1 we may C NJ and B. 'T'hen the set write A as the union of disjoint sets C N - 1011 U B is a proper subset of A. Moreover, A, = { f: A -- A,, where f(x) = a,, t., if x = a,, and f(x) = x if x C B, is a one-to-one correspondence between A and A,. We infer from this corollary and Theorem 3.3 that a set is infinite iff it is similar to a proper subset of itself. What is for us a characterization of an infinite set, has been taken as the definition of an infinite set in some treatments (Dedekind, 1883). Another simple application of the axiom of choice produces for cardinal numbers the analogue of Theorem 7.7 for ordinals. 120 The Natural Number Sequence and its Generalizations I C II All. 2 TIIEOREM 9.2. If C is a set of cardinal numbers, then there exists a cardinal number greater than each cardinal in C. Proof. A ( |
noncrnpty) set C of cardinal numbers is a disjoint collection of sets. With the axiom of choice it is possible to define a representative set of the form (t =,, = u. Clearly, card U Ct > u for each a in C. Hence card 2" > u for each u in C. C C1 where The hierarchy of infinite cardinals in order of increasing magnitude which was mentioned in Section 3 can now be described with more accuracy. First there is the natural number sequence according to Theorem 3.4. Next we have No, by Corollary I of Theorem 9.1 and Theorem 3.5. 'T'hen we get successively greater cardinals 2k°(= K), by application of 'Theorem 3.6. After all of these we gel. a still 2K, greater one, say p, by application of Theorem 9.2. Then Theorem 3.6 may be applied again to extend the array by 271, 2"°,, and so on. A more profound consequence of the axiom of choice is that every set of cardinal numbers is well-ordered by the ordering relation < introduced for cardinals. To prove this, along with related properties of cardinals, we employ the fact that the axiom of choice implies the wellordering theorem which, in turn, implies that every cardinal number can be represented by a well-ordered set. From this it follows, first, that any two cardinals are comparable in view of the Corollary to Theorem 7.4. Second, it establishes a correspondence between cardinal numbers and ordinal numbers whereby with a cardinal number is associated the (nonempty) set Z(c) of all ordinals having a representative of cardinality c. Specifically, if C is a given set of cardinals, (c, a,)jc C C and a,; is the least nrcrnber of Z. (c) I is a function, f,-with C as do'rnain and a set A of ordinals as range. Clearly, f is one-to-one and consequently a one-to-one correspondence between C and A. Further, it is easily shown that f is an order-preserving map. It follows that the simply ordered sets C amid A are isomorphic and since one is well-ordered, so is the other. We record this result as our next theorem. THEOREM 9.3. Any set of cardinal |
numbers is well-ordered. This property of cardinals is the basis of the following notation for infinite cardinals. If a is the ordinal number of the set of infinite cardinals less than an infinite cardinal u, then u is designated by N. The desig- 2.9 1 Further Properties of Cardinal Nurnliers 121 nation of N as No is an instance of this notation. Again, Mr stands for the immediate successor of No, and consequently the continuum hypothesis may be phrased as the assertion that 2", = 1!ti. This has been extended to the generalized continuum hypothesis which asserts that 2"- = i`t.i I. Another consequence of the axiom of choice is that multiplication and addition of infinite cardinals arc idcmpotent operations; that is, if it is an infinite cardinal, then u2 = u and 2u = u. To prove these results we use the fact that the axiom of choice implies Zorn's lemma. The following proof of the idcmpotcncy of multiplication is due to Zorn (1944). To facilitate the exposition we introduce a temporary definition: To say that a set A has property 9, symbolized i(A), shall mean that A has at least two distinct members and A2 = 1. Additional properties of such sets, as well as some properties of related sets, are derived below. I. If i(A), then A is infinite. The proof is left as an exercise. 12- If #(A) and A > 11, then A + 13 = t1. Proof. Let ao and ar be distinct members of A and suppose that 13NA0CA.Then and it follows that A A<A+ 13=AX njUAo-ark 13 = A. If.9(A), then A + A = A. 13- Proof. The assertion is a corollary of 12- 14- If 4(A) and 0 < lJ < A, then All = A. Proof. Let b E B and let B - A' C A. Then A=AX U <AX 13=AA'=A->-AP«XA=A, which yields the asserted conclusion. I. If,4(A,), Ao C A, and I - Ao < 71o, then 4(A). The proof is left as an exercise. Is. Let A be a set with disjoint subsets Ao and Al such that (i) there exists a |
one-to-one correspondence f between Ao and Au X Ao, (ii) 9(A,), and (iii) lo < A. Then there exists a one-to-one correspondence g between C = Au U Ar and C X C such that g D f. 122 The Natural Number Sequence and its Generalizations f c H A p, 2 Proof. Such a correspondence exists provided there is a one-to-one correspondence between the sets C - Ao = A, and C X C - Ao X Ao = (Ao X A,) U (At X Ao) U (At X A,). In turn, this is the case if Ao X A, + A, X Ao + A, X A, = A1,. This is trivial if A = 0. If Ao - 0, its validity follows from 14 and 13. The next two results are special cases of I6. That is, both assert the existence of a one-to-one correspondence between a subset C of 'A and C X C which properly extends a given mapping of the same variety. L. Assume that Ao is a finite subset of the infinite set A and that f is a one-to-one correspondence between Ao and Ao X Ao (so that As has at most one member). Then there exists a proper extension g of f of the type described in 16. Proof. As an infinite set, A has a denumerable subset arid, therefore, a subset A, such that :1(Ar) (Theorems 9.1 and 4.3). Since Ao is finite, we may assume that Ao (1 A, = 0. Moreover Ao < A,. Thus, 16 may be applied and provides the desired extension. 18. Assume that Ao C A, g (Ao), A - A > A0, and that f : Ao -,. A0 X Ao is a one-to-one correspondence. Then f has a proper extension g of the type described in Is. In view of 16 and 12 it is sufficient to determine a subset A, Proof. of A - A,; which has A0 as cardinal number. Such a set exists by virtue of the assumption that A - Ao > A0. We can now quickly dispose of the principal theorem. THEOREM 9.4. If A is an infinite set, then ii (A). Irr other words, if u is an infinite cardinal, then u2 = it. Proof. Consider the collection if |
of all one-to-one correspondences: f : A' -+- A' X A', where A' C A. This collection is. nonempty since A has a denumerable subset and, as a collection of sets, is partially ordered by inclusion. Each chain e included in if has an upper bound in 9; indeed, U e qualifies. The proof of this is left as an exercise. Hence, by Zorn's lemma, if has a maximal element fo. Now fo is a a subset A0 of A and Ao X Ao one-to-one correspondence be having no proper extension. In view of 17 the set Ao is not finite and, therefore, is idempotent. So, according to I8, it is false that 2.9 I Further Properties of Cardinal Numbers 123 - Ao > Ao, and hence A - Ao < Au according to Theorem 9.3. Since g(Ao), the last inequality implies, using I5, that 9(A). 'T'HEOREM 9.5. If u is an infinite cardinal, then 2u = u. Proof. In view of the preceding theorem this follows from I;,. It is left as an exercise to deduce from this theorem the following, whereby the arithmetic of infinite cardinals is reduced to a triviality. If u and v are infinite cardinals, then u + v TI-IEOREM 9.6. uv = max { u, v I. EXERCISES 9.1. Consider the following three assertions about a set X. (i) X is infinite. (ii) There exists a one-to-one mapping on X onto a proper subset. (iii) There exists a one-to-one mapping on N into X. Show that each of these assertions implies the other two if the axiom of choice may be used. Which of these six implications can be proved without the axiom of choice? 9.2. Expand the proof in the text that any set of cardinals is well-ordered. 9.3. Prove property I,. 9.4. Prove property I5. 9.5. Supply the missing part of the proof of Theorem 9.4. 9.6. Prove Theorem 9.6. 9.7. Extend Theorem 9.6 to the case where only one cardinal is infinite. 9.8. Give a proof, using Zorn's lemma but no properties of well- |
ordered sets, that any two cardinal numbers are comparable. Hint: Recalling the analysis in Section 3, it is sufficient to prove that if A and B are sets, then there exists Bo and either A = A subsets Ao and Bu of A and B respectively such that Ao or Bo = B. Prove this by applying 'corn's lemma to the partially ordered set (1, C), where if is the collection of all one-to-one correspondences f: A' -i- B' with A'CAandB'CB. 9.9. Devise a direct proof that if u is an infinite cardinal, then 2u = u. Hint: Let S = u and 7' = (0, 1). Let be the collection of all pairs (A, fA) where A is a subset of S such that A X T = A and fA is a lixed mapping which demonstrates the similarity of A X T and A. Show that e5' is nonernpty and is partially ordered by the relation <, where (A, fA) < (B, fit) means that A C B and AI A X T = fA. Then deduce that Zorn's lemma may be applied. 9.10. Deduce from the result in Exercise 9.9 that if the set B has an infinite subset A such that A < B, then B - A = B. 124 The Natural Number Sequence and its Generalizations I C H A p. 2 9.11. Use the results of Exercises 9.9 and 9.10 to give another proof of Theo. rem 9.4. Hint: Let u be an infinite cardinal and u. Then consider the collection of all pairs (A, fn), where A is a subset of S such that A X A = A and fA is a fixed mapping which demonstrates the similarity of A X A and A. 9.12. Show that for infinite cardinals r, s, u, and v, if r < s and u < v, then r + u < s + v and ru <.cv. 10. Some Theorems Equivalent to the Axiom of Choice In the preceding section we proved, among other things, that the axiom of choice implies that (i) any two cardinals are comparable, (ii) if u is an infinite cardinal, then u2 = u, and (iii) if u and a are infinite cardinals, then |
u + v = uv. It is a remarkable fact that each of (i), (ii), and (iii) is equivalent to the axiom of choice. As a preliminary to the proofs required to substantiate this statement, we return to the discussion of the relation between cardinal and ordinal numbers which appears prior to Theorem 9.3. If we understand by an aleph a trans. finite cardinal number which has a well-ordered set as a representative, then the first step in the proof of Theorem 9.3 amounts to the observation that the axiom of choice implies that every transfinite cardinal is an aleph. The converse of this implication is easily verified, so the axiom of choice is equivalent to the assertion that every transfinite cardinal is an aleph. Without using the axiom of choice it is possible to prove the following results concerning alephs. TIIEOREM 10.1. To each cardinal number c there corresponds an aleph ti(c), which is not less than or equal to c. If c is a finite cardinal, we may choose Ho for the cardinal Proof. in question. So assume that c is transfinite. We now make a definition. For an ordinal number a, all sets of order type a are similar; we denote the common cardinality of such sets by a and call this the power of a. Now let A be the set of all ordinals cx such that a < c. Then A is an infinite set because every natural number belongs to it and A is well-ordered, being a set of ordinals. I fence the order type of A is a transfinite ordinal l; and & is some aleph, rt(c). We prove that $t(c) is not less than or equal to c by deriving a contradiction from the c. Then, since R (c) = &, we contrary assumption. So assume R(c) have i < c, whence C A. Hence 13 = (01(3 < El C A, since S < l; implies that 0 < E c and E (Z B. Now B = E by 'T'heorem 7.5 and 2.10 I Some Theorems Equivalent to the Axiom of Choice 125 by the definition of. It follows that A is ordirially similar to A one of its initial segments, contradicting Theorem 7.2. THEOREM 10.2. Let c be a transfinitc cardinal |
number and 1`t an alcph. If cR = c -{- K, then either c > b or c < R. Proof. Let C and A be disjoint representatives of c and fit, respectively. By assumption we may take A to be well-ordered. Then CXA=cm =c - Fm, and hence there exist disjoint subsets C, and Ar of C X A such that C,UA,=C'XA, Z;,=c, and fl,=M. Now, either (i) there exists an element b, of Csuch that for all a in A, (b,, a) C C1, or (ii) for every element b of C, there exists an element a of A such that (b, a) V C1. If (i) holds, let Az be {(b,, a)la C A}. Then Az C C1, i = ti, and hence c > t. If (ii) holds, let rp(b) be the least clement of A such that (b, Wp(b)) C A, and let Cz be {(b, (p(b))Ib C C). Then Cz C A,, t:;z = C:, and hence, r. < R. We can now prove the theorems in question. THEOREM 1 0.3 (IIartogs). The axiom of choice is equivalent to the assertion that any two cardinal numbers are comparable. It remains to prove that if any two cardinals are comparable, Proof. then the axiom of choice is valid. Let C be a given set and c = G. In view of Theorem 10.1 and the assumed comparability of cardinals, there exists an alcph R(c) such that c < bt(c). It follows that C is similar to a subset of a well-ordered set, whence follows the existence of a relation that well-orders C. THEOREM 10.4 (Tarski). The axiom of choice is equivalent to the assertion that if u and v are infinite cardinal numbers, then u -l- v = uv. It remains to deduce the axiom of choice from the hypothesis Proof. u + v = uv for infinite cardinals a and v. Let c be an infinite cardinal and K(c) be the aleph of Theorem 10. |
1. Theorem 10.2 is applicable and we conclude that either c > bt(c) or c < KK(c). The inequality c > K(c) is impossible in view of Theorem 10.1. Hence c < 1!I(c), from which it follows that every transfinite cardinal is an aleph. This, in turn, yields the axiom of choice. 126 The Natural Number Sequence and its Generalizations I C H A P P. 2 'I' I I E 0 R E M 10. 5 (Tarski). The axiom of choice is equivalent to the assertion that if It is an infinite cardinal number, then u2 = u. It remains to deduce the axiom of choice from the assumption Proof. that 112 = u for infinite cardinals It. Let c and d be infinite cardinals. Then c2 = c, d2 = d, and (c + d)2 = c -}- d. Since (c + d)2 = c2 + 2cd + d2, it follows that c + d = c +- 2cd -I- d, whence cd < 2cd < c + 2cd A- d = c + d. But we may also set c = c, +- 1 and d = d, -}- l for cardinals c, and d,, to conclude that cd = (c, + 1) (d, + 1) = c,d, + c, + di + 1 > 1 +c,+d,+1 =c+d. Hence our assumptions imply that c + d = cd for infinite cardinals. But this implies the axiom of choice according to the preceding theorem. EXERCISES 10.1. Deduce the axiom of choice from the hypothesis that every transfinite cardinal is an aleph. 10.2. Another of Tarski's results concerning the axiom of choice asserts that it is equivalent to the proposition if2u <it + v, then u <v, while the proposition if 2u > it + v, can be proved without the axiom of choice. Prove this result. then u > v 10.3. Still another of Tarski's theorems states that the axiom of choice is equivalent to the assertion that for any cardinals it, v, and tv, the inequality u + w < v + zv implies that it < v. Prove |
this. 10.4. Prove Tarski's theorem stating that the axiom of choice is equivalent to the assertion that for any cardinal numbers it, v, and tv, the inequality uw < ow implies u < v. 10.5. It is not known whether the formula 2u _ u, which follows from the axiom of choice (Theorem 9.5), implies the axiom of choice. Attempt a proof of this. 11. The Paradoxes of Intuitive Set Theory The theory of sets which has been l)resemcd so far is that used by mathematicians in their daily work. Many theorems which are accepted by a majority of the mathematical couttnuttity, both past and present, 2.11 ( The Paradoxes of Intuitive Set Theory 127 rely on this theory. Unfortunately, it is not free of difficulties. Indeed, as mentioned earlier, it yields contradictions. However, matters are not as bad as this fact might indicate. This is suggested, at least, by its very inclusion in a present-day text. A firm vantage point from which to view the "reliability" of Cantor's theory is one of the axiornatizations which have been devised. The version of axiomatic set theory which we shall discuss later (Chapter 7) is based on the conclusion that the known contradictions of Cantor's theory are associated with "too large" sets. These are not the sort which occur ordinarily in mathematics. Before discussing the best-known contradictions, a preliminary remark is in order. A cornerstone of Cantor's theory is that we are guided by intuition in deciding which objects are sets and which are not. For this reason the name "intuitive set theory" is often applied to it. The implicit faith that individuals have in their intuition seems to be responsible for the contradictions of intuitive set theory commonly being called paradoxes. This is a misnomer, since the connotation of the word "paradox" is that of a seemingly, or superficial, contradiction, whereas the examples in question are bona fide contradictions. As such, they should be labeled "antinomies," which is the correct technical word to describe their status. Few do this, however. The principle of intuitive set theory which asserts that every property determines a set may be regarded as its Achilles' heel. Indeed, when used without restriction, this principle yields at least three sets from which logical contradictions can be derived. The three which we shall discuss are called the Russell paradox, the Cantor paradox |
, and the RuraliForti paradox. The simplicity of the Russell paradox is apparent from the fact that it was possible to mention it as early as Section 1.2. We consider it now in more detail. The formula which Russell considered is xCZx or --i(xCx) where, in the second version, we have used one of the standard symbols for negation (--1). According to the principle of abstraction, this formula determines a set R such that x C R iff --i (x E x). In particular, R C R ill -' (RC R), which is logically equivalent to the contradiction R C R and -(R C R). We note that R (assuming that it exists) is a very large set. For example, its defining property is satisfied by all objects which are not sets, since 128 The Natural Number Sequence and its Genet alizaliuns i (: t-t A P. 2 such objects can have no members and in this event cannot be members of themselves. Moreover, the property is satisfied by most sets; to mention two examples, the set of even integers is not an even integer, nor is the set of all polynomial functions a polynomial function. It is only when one turns to such figments of the imagination as the set of all sets, or the set of all abstract ideas, that violations of the defining property of R can be found. The Cantor paradox, which was discovered by Cantor in 1899 but which was first published only with his correspondence in 1932, is derived from the set defined by the formula x is a set. Let e be the set defined by this formula. Then C is the set of all sets. By Theorem 3.6, W(C) > G. Also, since C is the set of all sets and 6'(C) is a set (the set whose incnibers are the subsets of C), o'(C) C C. Hence, (i'(C) < c or, in other words, it is false that (t'(C) > G. Thus, it follows that both "m((O) > 0" and the negation of this statement are valid. This is a contradiction. The Burali-Forti (1897) paradox, which was known to Cantor as early as 1895, is derived from the set defined by the formula x is an ordinal number. The set 1', which it determines by |
virtue of the principle of abstraction, is that of all ordinal numbers. As a set of ordinals, 1' is well-ordered according to Theorem 7.6, and hence has itself an ordinal number y. By Theorem 7.5, s(y) is a well-ordered set of ordinal number y; hence s(y) is ordinally similar to F. With F as the set of all ordinals, y C 1' and hence s(y) is an initial segment of 1'. Thus, we have proved that I' is ordinally similar tQ one of its initial segments. This is a contradiction of Theorem 7.2. Instead of offering the above paradoxes as proofs of the assertion that the unrestricted use of the principle of abstraction yields a contradictory theory, we may say that if we adhere to ordinary logic, then the paradoxes demonstrate that it is false that corresponding to every property there is a set of objects having that property. Interestingly enough, the converse is also false. That is, it is false that every set has a defining property. The well-known proof of this. is due to Skolcin (1929) and is as follows. It is possible to map the set of real numbers into a collection of sets in a one-to-one fashion. For example, we can assign to a real number x the set of all real numbers less than x. Since the set Bibliogrn/ikical Notes 129 of all real numbers is uncountable, it follows that there exists an uncountable collection of sets. So, if every set has a defining property, the set of defining properties is uncountable. On the other hand, a property (written in English) is a finite sequence of letters of the English alphabet. The set of all such sequences is denumerable so that, in particular, the set of all properties is denumerable. Hence, there exist sets without defining properties. Intuitive set theory with its paradoxes certainly invites a critical examination with the goal of creating a theory which is both consistent and which enjoys as many features of the intuitive theory as is possible. Of the points of departure which may be taken in this matter, that of developing set theory as a formal axiomatic theory has been popular. The present-day status of such axiomatic theories is this: they are flexible enough to permit one to carry on essentially as in intuitive set theory, and they circumvent the classical paradoxes (and thus suggest |
that they are consistent) ; however, no one of them has been proved to be consistent. BIBLIOGRAPHICAL NOTES Section 1. The development of the basic properties of the arithmetic of the natural numbers from the Peano axioms in Dedckind (1888) (reproduced in Dcdekind (1932)] is well worth reading. In H. Wang (1957) there is an interesting account of how Dedekind arrived at his characterization of the natural numbers. The standard classical account of the development of properties of natural numbers and their extension to the real numbers (see the next chapter) is E. Landau (1930). Section 2. An excellent account of both proof and definition by induction is to be found in Rosser (1953). Sections 3-7. For full accounts of the topics considered in these sections, W. Sierpinski (1958) and A. Fraenkel (1961) should be consulted. Sections 8-10. For more complete accounts of consequences of the axiom of choice and propositions which are equivalent to it, Rosser (1953) and Sierpinski (1958) should be consulted. The proposition known as Zorn's lemma appears in M. Zorn (1935). Theorem 10.3 is in F. Hartogs (1914). The various propositions which are equivalent to the axiom of choice and which have been credited to A. Tarski appear in Tarski (1923). CIIAP TER 3 the Extension of the Natural Numbers to the Real.Numbers IN THIS CHAPTER we carry out another variety of extension of (N, -l-,, <), f the systemn discussed in Sections 2.1 and 2.2. Three successive extensions are made, the last of which yields the real number system. The first of these may be described as the completion of N_ with respect to addition-- that is, the minimum enlargement of N to insure the solvability of all equations of the type x + n = nz with n, m C N. The extended set is the set Z. of integers. The second extension amounts to the completion of Z with respect to multiplication--that is, the minimum enlargement of 'Z to attain the solvability of all equations of the form. xb = a with a, b C Z and b s 0. The resulting set is the set Q of rational numbers. The third extension amounts to the |
completion ofO with respect to order---that is, the minimmum enlargement of a which provides least upper bounds for nonempty subsets of Q which have upper bounds. In addition to those theorems which are of permanent interest, any development of the real number system includes a great number of results having just temporary interest (for example, results which justify various definitions). Each statement of the latter sort is labeled a lemma, and if no proof is in evidence the reader can count on his being asked to supply one in an exercise. Finally, we mention that elementary properties of operations and relations for natural numbers are used without explicit reference. 1. The System of Natural Numbers On the basis of definitions and theorems appearing in Sections 2.1 and 2.2, the natural number sequence (N, ', 0) determines the system t It better suits our presentation to adopt <, instead of <, as the basic ordering relation in U. 3.1 1 The System of Natural Numbers 131 of natural numbers, (N, -I-,, <), by which we mean the set N together with the two binary operations and ordering relation which have been defined in this set. Below is a list of those properties of +,, and < and their interrelations upon which this chapter is based. These were all derived as theorems in Sections 2.1 and 2.2 from the assumption that (N, ', 0) is an integral system. Thus, for one who has studied Sections 2.1 and 2.2, this section (which is preliminary to the developments described in the above summary) is simply an abstract of already demonstrated properties of the system of natural numbers. For anyone who, for some other reason, admits the following as valid properties of (N, +,, <), the chapter is self-contained. The properties of (N, +,, <) to which we call attention are the following. A,. x + (y + z) = (x + y) + z. A2. x+y =y+x. A3. O+x=x. A4. x+z=y +zorz+x=z+y implies thatx =y. Ml. M2. Ms. M4. xz = yz or zx = zy, and z D. x(y + z) = xy + xz. x(yz) = (xy)z. xy = yx. 1 |
X = X. 0, imply that x = y. Further, the relation < has the following properties. 01. x < y and y < z imply that x < z (transitivity). 02. For each pair x, y of natural numbers, exactly one of x < y, x = y, y < x hold (trichotomy). (N, <) is a well-ordered set. 03. OAR. x <yimpliesthatx+z <y +z anclz +x <z+y (mon- otonicity of -1- with respect to <). OA2. x -l- z < y + z or z + x < z + y implies that x < y (can- cellation property of + with respect to <). OMI. If z > 0, then x < y implies that xz < yz and zx < zy (monotonicity of with respect to <). OM2. If z > 0, then xz < yz or zx < zy implies that x < y (can- cellation property of with respect to <). A comment about some of the terminology used above is in order. The meaning of the statement that an operation has the cancellation. 132 Extension o f the Natural Numbers to the Real Numbers I C H A P. 3 property with respect to some relation at hand can be inferred immediately from 0A2. (Although 0M2 involves a restriction, no special terminology will be introduced as a reminder of this restriction.) When we state simply that an operation has the cancellation property, we shall mean with respect to the equality relation. 't'hus a binary operation * has the cancellation property ifl' each of x * z = y * z and z * x = z *y implies that x = y. Further, the meaning of the statement that some binary relation is trichotomous or that a binary operation is monotonic with respect to some relation should be clear from the above, examples. Although the less than relation was given a central position in our development of the theory of the natural number system, it can be introduced as an offshoot of the operations of addition and multiplication and the notion of positiveness which stems from the definition of a positive natural number as a nonzero natural number. Indeed, since x < y iff there exists a positive natural number z such that x + z = y, this characterization of less than may be taken as the |
definition of less than in terms of addition and positiveness. Then properties of less than can be derived as consequences of properties of positive elements, properties of addition, and properties of multiplication. In such a treatment, parts As and Mr, of Theorems 2.1.5 and 2.1.7 respectively (which may be stated as "The sum and the product of two positive natural numbers is a positive natural number") occupy a key role. As an illustration we derive 01 within this framework. Assume that x < y and y < z. Then there exist positive natural numbers u and v such that x -1- u = y and y + v = z. Hence x + (u -}- a) = z. Since it positive and v positive imply that u + v is positive, it follows that x < z. We have called the reader's attention to the foregoing approach to the theory of order for the natural numbers because we shall employ it in each of the forthcoming extensions of the system of natural numbers. 2. Differences This section includes the necessary preliminaries for a definition of the integers and a rapid development of their properties, all of which is presented in the next section. In this section the letters "m," "n," "p," and "q" will designate natural numbers. The intuitive motivation for our point of departure is the observation that a solution of x + n = m is determined solely by m and n in a specific order. Thus ordered pairs 3.2 1 Ui(Je)cnccs 133 of natural numbers become the object of study. The way in which one might naively expect such objects to behave in view of their intended role is the source of the succession of definitions which we make. By a difference we shall mean an ordered pair ()n, n). In the set N X N of all differences we introduce the relation ^'d (the subscript is for "difference") by defining (in, n) ',t (p, q) if in + q = p + n. LEMMA 2.1. -,t is an equivalence relation on N X N. We shall call a difference (m, n) positive iff m > n. Two fundamental properties of positive differences arc stated next. LEMMA 2.2. If (in, n) is positive and (m, n) ^'d (p, q), then (p, q) is positive. If (in, n) is positive, |
then there exists a difference (p, 0), with p > 0, such that ()n, n) -,) (/), 0). An operation, which we call addition and symbolize by f-, is defined for differences by (in, n) + (p, q) = (m + p, n + q) Clearly, addition is a binary operation in N X N. The motivation for the definition is the expectation that if x + n = in and y + q = p, then it should follow that (x + y) + (n + q) = in + p. Properties of addition which interest us are given next. LEMMA 2.3. then x+y' 'du+v. If x, y, u, and v arc differences and x -d u and y ^-',i v, LEMMA 2.4. Addition of differences is associative and conunutative. The sum of two positive differences is a positive difference. Further, addition is cancellable with respect to -,r. LEMMA 2.5. If x and y are differences, then there exists a differ- ence z such that z + x Another binary operation in N X N, which we call multiplication and symbolize by, is defined for differences by (in, n). (p, q) = (mp + nq, rnq + )zp). Usually we shall write "xy" instead of "x y" for a product of differences. 134 Extension o f the Natural Numbers to the Real Numbers I C H A P. 3 LEMMA 2.6. then xy ^'d UV- If x, y, u, and v are differences and x -d u and y rvd v, LEMMA 2.7. Multiplication of differences is associative and commutative, and distributes over addition. The product of two positive differences is a positive difference. Further, multiplication is cancellable with respect to -d for differences other than those of the form (m, in). EXERCISES 2.1. Prove Lemma 2.1. 2.2. Prove Lemmas 2.2 and 2.3. 2.3. Prove Lemma 2.4. 2.4. Prove Lemma 2.5. 2.5. Prove Lernma 2.6. 2.6. Prove Lemma 2.7. 3. |
Integers Recalling Lemma 2.1, we define an integer to be a -,,-equivalence class. We shall write [x1 for the equivalence class determined by the difference x. (The new subscript is for "integer.") The set of integers will be symbolized by Z. We shall call an integer positive iff one of its members is a positive difference. It follows from Lemma 2.2 that if [x]; is positive, then every member of [x], is positive. The set of positive integers will be symbolized by Z-1. We consider next a relation from Z X Z into Z : {(([x];, [y],), [x d- y];)l x and y are differences}. According to Lemma 2.3 this relation is a function which, by virtue of its form, is a binary operation in Z. We call this operation addition 'and symbolize it by +. Thus, [x]i + [y], _ [x + y],. LEMMA 3.1. Addition of integers is associative and commutative, and has the cancellation property. Further, the sum of two positive integers is a positive integer. 3.3 I Integers 135 LEMMA 3.2. If x and y are integers, then there exists exactly one integer z such that z + x = Y. From this result it follows that if x is an integer, then there exists exactly one integer, which we call the negative of x and symbolize by -x, such that (-x) + x = x ± (-x) = [(0, 0)1,. Finally, we consider the following relation from Z X Z into Z: {(([x];, [y],), [xy];)l x and y are differences}. According to Lemma 2.6 this relation is a function which, by virtue of its form, is a binary operation in Z. We call this operation multiplication and symbolize it by. Thus [XI. [_fl, = [.ry l,. LEMMA 3.3. Multiplication is associative and commutative, distributes over addition, and has the cancellation property if (0, 0) is not a member of the factor to be canceled. Further, the product of two positive integers is a positive integer. Now let us tidy up our notation for the integers. The first step is the observation that the set Z° of integers of |
the form [(n, 0) J; with n E N and the set of integers of the form [(0, m)], with in C N - {0} are disjoint and exhaust Z. The former statement is obvious. To prove the latter, consider any integer [(p, q)Jj. Exactly one of p > q and p < q holds. In the former case, p = q + it with it C N, and hence [(p, q)]i = [(n, 0)] C Z°. In the latter case, q = p -l- in with in C N - {0) and [(p, q)]i = [(0, rn)J which completes the proof. It is a straightforward exercise to demonstrate that the ordered triple whose coordinates are, in turn, Z°, the map on Z° which takes [(n, 0)]; into [(n + 1, 0)J and [(0, 0)J; is an integral system. 't'heorem 2.1.8 implies that the mapping f on N into Z such that f(n) = [(n, 0)]j is one-to-one, onto Z°, and preserves addition, multiplication, and less than. f We summarize these properties off by calling it an order-isomorphism of N onto Z° and indicate the relationship of L° to N by referring to Z° as an order-isomorphic image of N (or, saying that L° is orderisomorphic to N). Parenthetically we remark that it should be clear t'Tu be precise, Theorem 2.1.8 states that f(x + y) = f(x) -I- f( Y), f(xy) = f(x)f(y), and x < y iflf f(x) < f(y). The last property implies, in turn, that x < y iff f(x) < fly), according to Exercise 1.11.9. 136 Extension of the Natural Numbers to the Real Numbers I c t I A P. 3 that these definitions are applicable to any two systems each of which consists of a set along with two binary operations and an ordering relation in that set. Thus we may apply the definitions to other such pairs of systems. The order-isomorphism of N onto Z° suggests that we call the members of Z° the integers which correspond to the natural numbers as names for them. We |
shall do this, and adopt "Oi," "1 i," "2i," which means we agree that ni = {(n, 0)]t if n C N. Since the remaining integers (that is, the members of _L - Z°) have the form 1(0, in)]i with in C N - {0], and since {(0, m)]i = - [(in, O)]i = -ini, we acquire "-1i," "-2i," integers. Ilcnceforth we may write, therefore, as names for the so-called negative Z = {..., - 2 i, -1 i, 0;, I i, 2 i,...). We summarize our results concerning (Z, +,, Oi, 1 i, Z ), the system of integers, in the following theorem. The theorem does not include all the properties which have been stated. However, in the exercises for this section, the reader is given the opportunity to show that the properties listed in the theorem are complete in the sense that from them follow as logical consequences all others which have been mentioned or might be expected. In particular, it is implied that from the properties listed it may be inferred that for each integer x the equation z + x = Oi in part (4) has a unique solution (which we have already agreed to symbolize by -x). Then the notation "y - x" [which appears in part (14) of the theorem] may be introduced as an abbreviation for "y -i- (-x)." Further, the exercises call for the derivation of all expected properties of less than, as defined in part (14). TIIEOREM 3.1. The operations of addition and multiplication for integers, together with Oi, 1 i, and the set Z+ of positive integers, have the following properties for all integers x, y, and z. (1) x + (y + z) = (x + y) + Z. (2) x + y =y+x. (3) Oi + x = x. (4) There exists an integer - such that z + x = 0i. (5) x(yz) = (xy)z. (6) xy = yx. 3.4 I Rational Numbers 137 (7) 1 ix = x. (8) x(y + z) = xy + xz. (9) x |
z = yz and z 54 Oi imply that x = y. (10) 0i 1 i. (11) x,yEZi imply that (12) x, y E Z+ imply that xy C Z'-. (13) Exactly one of x C Z+, x = 0, -x C Z I- holds. (14) If < i is defined by x < i y if y - x c z+, then < i simply x+yCZ+. orders Z and well-orders (Oil U Z+. EXERCISES 3.1. Prove Lemma 3.1. 3.2. Prove Lemma 3.2. 3.3. Prove Lemma 3.3. 3.4. Prove part (13) of Theorem 3.1. Remark. Exercises 3.5-3.8 arc concerned with proving that from the propertics of (-7,, +,, 0i, 1;, Z+) in Theorem 3.1 can he deduced the other propertics mentioned in this section and the familiar properties of less than. 3.5. From properties (1), (3), and (4) of addition, prove that (i) addition has the cancellation property, (ii) for each x the solution of z -I- x = Oi is unique, and (iii) for each x and y, the equation z + x = y has a unique solution. 3.6. Prove each of the following properties of negatives of integers: -(x -f y) = -x -y, (-x)y = -(xy), (--x)(-y) = xy. 3.7. Using properties of addition and multiplication, prove that (-1i)x = -x. 3.8. Prove each of the following properties of the system of integers. (i) x is positive iff 0 < i X. (ii) The square of a nonzero integer is positive. (iii) < i is transitive. (iv) For each pair x, y of integers, exactly one of x <i y, x = y, y <i x holds. (v) x<iyiffx+z<iy+z. (vi) If 0 <i z, then x <iy if-f xz <i yz. 4. Rational Numbers The steps which precede the definition of a rational |
number parallel those which lead to the definition of an integer. Now we concern ourselves with the solution of equations of the form xb = a where a 1 38 Extension of the Natural Numbers to the Real Numbers I C I3 A P. 3 and b are integers and b 54 0;. So again we consider ordered pairs of the numbers at hand but with the quotient (instead of the difference) in mind as the intended interpretation. Since the formal developments are so similar to those in the two preceding sections, our treatment will be rather summary. The letters "a," "b," "c," and "d" will designate integers in this section. An ordered pair (a, b) with b 0 0; will be called a quotient. The quotient (a, b) will be written as a b The relation -y is introduced into the set of all quotients by defining c a bd iff ad=bc. This is an equivalence relation on the set of all quotients and has the further property that a ac bcb if c O 0;. We shall call a quotient b positive if ab is a positive integer. Further, we introduce operations of addition and multiplication into the set of quotients by way of the following definitions: a c b + d ad -I--bc bd a b c d ac bd Since b 0 0; and d 0 0; imply that bd 34 0 these are operations in the set of quotients. LEMMA 4.1. If x, y, u, and v are quotients and x rvq u and y -y v, then x + y Ny u + v, xy Nq uv and, if x is positive, then u is positive. In summary, this lemma asserts that the equivalence relation defined for quotients has all expected substitution properties. We forego proving for quotients the analogues of the properties derived in Section 3.2 for addition, multiplication, and positive elements. Instead, we turn to the Nq-equivalence classes to obtain the rational numbers. A rational number is a -,equivalence class of quotients. The ra- 3.4 I Rational Numbers 139 tional number having the quotient x as a representative we write, for the moment, as [x],. The letter "s" is intended to refer to "rational"; we do not use "r" since we want to reserve it for real numbers. The set of rational |
numbers will be symbolized by Q. We shall call [x], positive iii it contains a quotient y such that y is positive. It follows from Lemma 4.1 that if [x], is positive then each of its members is positive. The set of positive rationals we symbolize by Q+. The definitions of addition and multiplication for rationals are [x], + [y]n = [x + y13, [XI. [y], = [xy]8. Of course, Lemma 4.1 plays a crucial role in these definitions. Next we make a further definition: a,=[a] for aCZ. Clearly, to-one and, since [ (a, a,) I a C Z) is a function on Z into Q. Further, it is one- a, + b, _ (a -I- b) a,b, _ (ab),, the operations of addition and multiplication are preserved under this mapping. Finally, the image a, of an integer a is a positive rational number ifl' a is a positive integer. This last property implies that if <, is the ordering relation which can be defined in O in terms of its positive elements (see below), then a <; b ill' a, <, b,. Thus, the mapping a -+- a, is an order-isomorphism. The members a, of this order-isomorphic image of Z in 0, we shall call integral rational numbers. There follows one comprehensive theorem concerning properties of (Q, +,, 0 1 (Q 1), the system of rational numbers. THEOREM 4.1. The operations of addition and multiplication for rational numbers, together with 0 1 and the set () i- of positive ratiohals have the following properties for all rationals x, y, arid z. (1) x+ (y+z) _ (x+y) +z. (2) x + y =y+x. (3) 0, + x = x. (4) There exists a z such that z + x = 0,. 140 Extension of the Natural Numbers to the Real Numbers I C 11 A P. 3 (5) x(yz) = (xy)z. (6).y = yx. (7) 1.x = x. (8) If x 7-1 0., there exists a z such that zx = 11. (9) x(y |
-I- z) = xy + xz. (10) 1.7 0.. (11) x,yCQ,+imply that (12) x, y C Q' imply that xy C (13) Exactly one of x C QF-, x = 0., -x C Q'- holds. (14) If P is the intersection of all subsets of Q3 which contain 1, and are closed under addition, then, for each x C Q1', there exist a, b C P such that xb = a. (2;t. x+yCQt. In the exercises the reader is asked to prove the various parts of this theorem [including a more familiar formulation of (14)] and to derive some of the immediate consequences of these properties of the system of rational numbers. Certain results in the latter category are worthy of comment. First, since addition of rationals enjoys the same properties as does addition of integers [properties (1)- (4) of Theorems 4.1 and 3.1, respectively], the results [derived from (1)-(4) of Theorem 3.1 J which appear in Exercises 3.5 and 3.6 hold for rationals. Next, since the basic properties of multiplication [parts (5), (7), and (8) of Theorem 4.11 for nonzero rationals mimic properties (1), (3), and (4) of addition, with 1. in place of 0., we may infer the following multiplicative analogues of the results in Exercise 3.5. (i) xz = yz and z (ii) For each x 0, imply that x = y. 0, the solution of zx = 1, is unique. This solution is called the inverse of x and is symbolized by x-'. (iii) For given x and y with x 5 0., the equation zx = y has a unique solution. Finally, we call attention to the fact that if less than, which we symbolize by <., is defined in Q by x<.y ilf y - xCQ' then it enjoys all of those properties stated in Exercise 3.8 for <c, since the earlier proofs carry over without change. We are now in a position to simplify the notation for rationals. The string of identities CaavJA tiiJ= l 8 = [1J]e L1 tie' = a |
iba 3.4 I Rational Numbers 141 shows that each rational can he written in terms of integral rational numbers. We shall drop the subscript "s" from now on (the context should make plain whether a. or a is the appropriate entity) and further, agree that b or (when convenient) alb is another name for ab-1. In this way we obtain the familiar notation for rationals. In practical terms this means that we agree to adopt names of representatives (that is, members) of rational numbers as names of rational numbers. To clarify this remark, let us consider, for example, the rational number C[(2, Mil «3, 0)]=B By our convention, "2/3" is a name of this rational number. The statement "2/3 = 4/5" means that "4/5" is another name of the sartu: number. This is true iff C3JA = C5Js' which, in turn, is true ill 2 5 = 4 3. Since 2 3, the original statement is false. In general, the same type of analysis yields the following results for rational numbers: 4 5 alb = c/d iff ad = cb, alb + c/d = (ad + bc)/bd, (c/d) = ac/bd. (a/b) We derive next two significant properties for rational numbers. At this point we begin to use elementary properties of rationals without explicit references. THEOREM 4.2. Between any two distinct rational numbers there is another rational number. Proof. Suppose that r, s E Q with r < s. It is sufficient to prove that r < (r + s)/2 and (r + s)/2 < s. To prove the first inequality we start with r < s and infer, in turn, r + r < r + s, 2r < r + s, and r < (r + s)/2. The second inequality is derived similarly. THEOREM 4.3. (Archimedcan property). If r and s are positive rational numbers, then there exists a positive integer n (properly, a positive integral rational number n) such that nr > s. 142 Extension of the Natural Numbers to the Real Numbers I CHAP. 3 Proof. Let r = alb and s = c/d where a, b, c, and d are positive integers. If n is a (rational number corresponding |
to a) positive integer, then nr > s if nad > bc. If for n we choose 2bc, this inequality is satisfied, since ad > 1. We conclude our discussion of rational numbers with the introduction of two functions pertaining to rational numbers. The first has Q as domain and as its value at x, which we symbolize by [x], the greatest integer equal to or less than x. For example, [2] = 2 and [= 2.U The second function has Q as domain. Its value at x, which we symbolize by lxl and call the absolute value of x, is defined as TI I E OR E M 4.4. Ixl=l-x x If x and y are rational numbers, then < 0. (I) 1XI > 0, (Il) Ixyl = 1XI lyl, (II1) Ix + yI <- 1XI + lyl, (IV) lxi - 1,1 <_ lx - A. EXERCISES 4.1. Prove Lemma 4.1. 4.2. Prove that the mapping a --)a. on Z into Q, introduced prior to Theo- rent 4.1, is an order-isomorphism. 4.3. Prove Theorem 4.1. As for part (14) of this theorem, show that P is simply the set of ratitmals which correspond to the positive integers. 4.4. Write a short paragraph to substantiate the assertion made after Theorem 4.1 that the properties of multiplication listed may be inferred without giving new proofs. 4.5. Prove that the relation <. for rational numbers may be characterized as follows: [a] <. [fl iff abd2 <; b2cd. 4.6. Prove Theorem 4.4. 5. Cauchy Sequences of Rational Numbers The set of rational numbers includes nonempty sets which have an upper bound but fail to have a least upper bound. One of these is 3.5 I CaucIty Sequences of Rational Numbers 143 S= (xC(Qt"Ix2<3(, as we proceed to prove. Clearly, any positive rational number whose square is greater than 3 is an tipper bound. On the other hand, no positive rational number whose square is less than 3 (that is, no member of S) is an upper |
bound, since, if s C S, then 3 - s2 s+5-I-2s is obviously greater than s and, as a direct calculation shows, is a member of S. Since there exists no rational whose square is equal to 3, it follows that those positive rationals whose square exceeds 3 exhaust the set of upper bounds for S. Now this set has no least member. Indeed, if u is a positive rational such that u2 > 3, then u -I- 3/u 2 is positive, less than u (since 3/u < u'2/u = u), and its square is greater than 3, since (u+3/2=(-L)2 u-3u,+ 2 J 2 It follows that S has no least upper bound. The failure of the rational number system to include the least upper bound of every nonempty set having an upper bound may be taken as the motivation for the extension of Q that is presented in the next section. We now set the stage for this by developing the theory of Cauchy sequences of rational numbers. We recall that a sequence is a function having '_L° (or, when convenient, Z t) as its domain. The value of the sequence x at n will be denoted by x,,. A sequence of rational numbers is a sequence x such that x E 0, for every n. A Cauchy sequence of rational numbers is a sequence x of rational numbers such that for every positive rational number e there exists a positive integer N such that for every m, n > N EXAMPLES 5.1. The sequence x such that <e. X" _ n + 1 u is a Cauchy sequence (of rational numbers). To prove this we must exhibit for each positive rational number a an integer N such that for m, n > N lx - xml < e. 144 Extension of the Natural Numbers to the Real Numbers i CHAP. 3 Since Ixn - xm! = m_-n _ 1 - l `_1 mn I n ml min if we let N = [1/e] -I- 1, then for all m, n > N, Ixn -x,n1 <_-1 min (rn, n) < N __ 1/e < e. 5.2. The sequence x such that x0 = 0, x, = 1, and xn _ 2 (Xn -t -l- x |
n- 2) a Cauchy sequence. To prove this we note first that is for n > 2 xn}1-xn= 2n \1)n This can be established by induction. Further, from the recursive definition of xn it is clear that for all m > n, x,, falls between xn and xni*.,. So, if a is a positive rational number and we choose N so that 2N > l/e, then, for all m, n > N, Ixm. - xnj Ixnu - xnI < F. We define the operations of addition and multiplication for se- qucnccs of rational numbers in the following way: x --1- y = u where u = x -1- y,, xy = v where Un = Xnyn Clearly, if x and y arc sequences of rational numbers, then sa arc x + y and xy. It is an important fact that if x and y are Cauchy sequences of rational numbers, then so are x + y and xy. In other words, addition and multiplication are binary operations in the set of all Cauchy sequences of rational numbers. The proof for the case of multiplication requires the following preliminary result. 3.5 1 Gauclay Sequences of Rational Numbers 145 LEMMA 5.1. If x is a Cauchy sequence of rational numbers, then there exists a positive rational number 5 such that for every n Ixnl < S. Proof. Corresponding to the positive rational number 1 there exists by assumption an integer N such that for every m, n > N (1) Let (2) IXn - XmI < S = max (Ixol, Ixil,..., IxNI, IAN+II) + 1. Clearly, if n < N + 1, then Ixni < 5. Suppose then that n > N + 1. By virtue of (1), lxn - XN+11 < 1 and, hence, According to (2), IXN_yl + 1 < S. Hence, for all n, Ixtti < S. Ixnl < I XN-, lI + I. LEMMA 5.2. If x and y are Cauchy sequences of rational numbers, then x -{- y and xy arc Cauchy sequences of rational numbers. (Sum.) Let e > 0. By hypothesis there exist |
N, and N2 such Proof. that for all m, n > N, and for all m,n> N2 IXn - xml < E/2, lyn - yml < e/2. Then for all m, n > max (N,, N2) ixn + yn - (Xm + ym) I = I (Xn Xm) + (yn - ym) <IXn-XmI < E/2 + E/2 < E. +lyn - ymi (Product.) Let e > 0. By virtue of the preceding lenuna, there exist positive rational numbers S, and2 such that for all n Ixnl < Si,SIynl < S2. Further, there exist integers NA and N2 such that for all m, n > N, and for all m, n > N2 iXn - xmI < E/(262), lyn - yml < E/(251). 146 Extension of the Natural Numbers to the Real Numbers I C11 A P. 3 Then for all m, n > max (N,, N2) Ixnyn - xym1 = Ixnyn - xmyn + x,,.yn - Iy,.I Ix, - xml + Ix,,nl Iyn /ml US-) 2. ) G E. The basic properties of addition and multiplication may be summarized by the statement that they satisfy properties (1) -(8) of Theorem 3.1, where the distinguished elements in (3) and (7) are taken to be the sequence 0, (whose value is 0 for all n) and the sequence 1 (whose value is 1 for all n), respectively. Again, the results stated in Exercises 3.5- 3.7 hold. The negative of the Cauchy sequence x is the sequence -x such that (-x) = -x,. for all n. We introduce next a relation, which we symbolize by -,, in the set of all Cauchy sequences of rational numbers. If x and y are Cauchy sequences of rational numbers, then x y if for every positive rational number E there is an integer N such that for every n > N, Ixn - ynl < E. _ As an illustration, consider the sequences x and y such that x,, (n + 2)/(n + 1) andy,. = 1 fot all n. These are |
Cauchy sequences and clearly x -,.y, since x,, -y,, = 11(n + 1). It is an easy matter to establish the following property of this relation. LEMMA 5.3. The relation N, is an equivalence relation on the set of all Cauchy sequences of rational numbers. If x is a Cauchy sequence of rational numbers, then x is called positive ifl there is a positive rational number e and an integer N such that for every n > N X. > E. The expected substitution properties of the equivalence relation with respect to addition, multiplication, and positiveness are stated next. LEMMA 5.4. numbers and x if x is positive, then u is positive. If x, y, u, and v are Cauchy sequences of rational u + v, xy -. uv and, u and y -. v; then x + y 3.5 I Gaucliy Sequences of Rational Numbers 147 Proof. The proof that x + y u + v is left as an exercise. Turning to the result concerning multiplication, let e be a positive rational number. By Lemma 5.1 there exist positive rational numbers 51 and 52 such that for every n ly,.l < 5., lung < 52. Since x ti, u, there exists an integer N, such that for every n > N1 Ixn - unl < E/25,, and since y'v, there exists an N2 such that for every n > N2 Then, for every n > max (N1, N2) ly. - v,81 > E/252. unV,, = IXnyn - unyn + unyn - unvnl < lynl Ixn - u,, + 111"1 lyn -- and < < E. 52(e/252) That if x is positive and x -, u, then u is positive is shown as follows. By assumption, there exists a positive rational number 2E and an integer N, such that for every n > N, x > 2E, and there exists an N2 such that for every n > N2 Hence, for n > max (N,, N2), Ixn - E. un> Xn - E > 2E-e > E. LEMMA 5.5. The sum and the product of two positive Cauchy sequences are positive Cauchy sequences. Further, if x is any C |
auchy sequence, then exactly one of the following hold: x is positive, x 'C 06, -x is positive. Proof. We shall prove only the last statement. Clearly, at most one of the three possibilities for x can hold. So we need to prove that at least one holds. Suppose that x is not equivalent to 0,. By the definition of x N, 0, this means that there is a positive rational number 2E such that for every integer N there is an n > N such that (1) I xn1 > 2E. 148 Exten.uon of the Natural Numbers to the Real Nun;be;.c I C uu A P. 3 Since x is a Cauchy sequence there is an integer N, such that if m, n > N1, then Ixn - x+nl < E. (2) From our observation which led to (1), it follows that there exists an integer p > N, such that (3) I x,, l > 2E. Since (3) implies that x 0 0, either x,, > 0 or x, < 0. Suppose that x, > 0. Then xp > 2E by (3), and, as a consequence of (2), for every n > p Hence, for every z1 > p Ixn - x,, < E. E > 2E-E > E. Thus, if.x,, > 0, then x is positive. By a similar argument it can be proved that if x,, < 0, then -x is positive. With the foregoing result available it is easy to prove the following lemma, which is of basic importance when we turn our attention to the -,-equivalence classes of Cauchy sequences (that is, real numbers). LEMMA 5.6. If the Cauchy sequence x is not equivalent to 0, then there is a Cauchy sequence z such that zx N, 1 Proof. The preceding lemma implies that for an x which is not equivalent to 0. there is a positive rational a and an N such that for every n > N Ixnl > E. Consider now the sequence x' such that xn = E if n < N and x;, = x if n > N. Clearly, x' is a Cauchy sequence, x' x, and for all n Ixnl > E. (1) 0 for every it, the sequence z where z |
= l /x, is a sequence Since x' of rational numbers. Further, z is a Cauchy sequence as we proceed to prove. Let 17 be a positive rational number. Since x' is a Cauchy sequence, there exists an N such'that for every m, n > N (2) Ixn - xmI < 'qE2. 3.6 I Real Numbers 149 Further, by virtue of (1), we have I From (2) and (3) it follows that for all m, n > N Izm - zn! < 77, which proves that z is a Cauchy sequence. It is clear that zx' Finally, since x'' x, it follows that zx -, I. 1 EXERCISES 5.1. Prove that the sequence x such that xn=1-3+5-...--2n+1 n is a Cauchy sequence. 5.2. Prove that the sequence x such that xn=1+1 d-I-+...+I is a Cauchy sequence. 5.3. Prove that addition and multiplication for Cauchhy sequences satisfy parts (1)-(8) of Theorem 3.1. 5.4. Prove Lemma 5.3. 5.5. Complete the proof of Lemma 5.4. 5.6. Complete the proof of Lemma 5.5. 6. Real Numbers As promised earlier, we define a real number as a -,equivalence class of Cauchy sequences of rational numbers. The real number having the Cauchy sequence x as a representative we write as [x l r, for the time being. The set of real numbers will be symbolized by R. We shall call a real number positive if it contains a positive Cauchy sequence. In view of Lemma 5.4, if [x]r is positive, then each of its members is positive. The set of positive real numbers we symbolize by R-1. The following definitions of addition and multiplication for real num- bers will scarcely offer any surprise: [XI, -+- [x]r - [[Ylr = [X [y]r = [x! ]r. +ylr, 150 Extension of the Natural Numbers to the Real Numbers + CHAP P. 3 Of course, it is Lemma 5.4 which ensures that these are binary operations in R |
. Next we call attention to a distinguished set of real numbers--those which correspond to rational numbers in a natural way. If a is a rational number, then (a, a, - ) is a Cauchy sequence of rational numbers. Since T.. is a partition of the set of all such sequences, there exists exactly one real number, let us call it ar, which contains (a, a, ). This means that, a,, a, - j {(a, ar)Ia C Q.] into R. It is easily proved that this function is oneis a function on to-one and, moreover, that the operations of addition and multiplication are preserved by this mapping. Finally, a rational number is positive iff its correspondent in B. is positive. Thus, B. includes an order-isomorphic image of Q. Members of this image of Q, will be called rational real numbers. The rational real number corresponding to the rational number 0, we again call zero and symbolize by Or. Thus, Or = ](0=, 0e,..., 0...)]r. The rational real number corresponding to the rational number 1. we again call one and symbolize by J r. Thus, The first major theorem concerning the real number system (g, -1-,, Or, 1 r, B.1) is the following. Its proof relies entirely on those properties of the rational number system appearing in Theorem 4.1, the properties of Cauchy sequences appearing in Lemmas 5.5 and 5.6, and the definitions of addition, multiplication, and positiveness for real numbers. THEOREM 6.1. The operations of addition and multiplication for real numbers, together with Or, 1, and the set of positive reals, have properties (1)-(13) listed in Theorem 4.1. Those further properties of addition and multiplication for rational numbers which are listed immediately after Theorem 4.1 are enjoyed by the corresponding operations for real numbers. If less than, which we symbolize by <r, is defined in $ by x <r y if y- x E 8+, then it has all those properties which <, possesses Also, the definition 3.6 Real Numbers 151 :)f the absolute value function extends to the real numbers and Theorem 4.4 applies. Our results up to this point may be summarized by the statement that the extension of Q to R has resulted in no loss of ground. That a gain has |
been made will be demonstrated when we have proved that every nonempty set of real numbers which has an upper bound has a least upper bound. The proof of this property of the real number system requires some other results which are important in their own rights. The first of these is usually phrased as the statement that the rational numbers are dense in the set of real numbers. THEOREM 6.2.t Between any two distinct real numbers there is a rational real number. Precisely, if x and y are distinct real numbers, then there exists a rational real number z such that if x < y, then x < z < y while if y < x, then y < z < x. Proof. We shall consider the case x < y. Let a C x and b C Y. Then x < y implies the existence of a positive rational 4c and an integer NJ such that for every n > N, fi - a > 4E. (1) Further, since a and b are Cauchy sequences, there exist integers N2 and N3 such that for every rn, n > N2 (2) and for every m, n > N3 Ibn - b,,11 < e. (3) Let N = max (N1, N2, N3) + I and let s be a rational number such that e < s < 2e (see Theorem 4.2). Now consider the real number z corresponding to the rational number aN + s. We contend that x < z and z < y. From (2) we may conclude that for every n > N Ian - a,,,I < E, an - aN < E. Hence, aN -- an > -e and, therefore, for every n > N (aN+S) -an > s This means that the Cauchy sequence e> 0. (aN ± S, aN +s,..., a, - S,...> - a is positive. Since this sequence is a member of z - x, the real number z - x is positive, and hence x < z. t In the remainder of this chapter we shall omit the letter "r" as a subscript for the symbols used in connection with real numbers. 152 Extension of the Natural Numbers to the Real Numbers I C II A r. 3 Using the identity bm - (aN + s) = (bv - aN) + (bm - bN) - s |
, it follows by a similar argument [which employs (1) and (3) ] that y - z is positive, and hence z < y. The following theorem is a generalization of the corresponding property (Theorem 4.3) for the system of rational numbers. THEOREM 6.3 (Archiincdean property). If x and y are positive real numbers, then there exists a positive integer n (properly, a real number n which corresponds to a rational which, in turn, corresponds to a positive integer) such that nx > y. Proof. Let b C y. According to Lemma 5.1, there exists a positive rational number b such that for every n If d is the real number corresponding to (b, b,, a, ), then bn < b. y < d. Also, by assumption, 0 <x. By the preceding theorem there exist rational real numbers s and t such that 0<s<x, y <t <d. By the Archimedean property of rationals (which obviously carries over to rational reals) there exists a positive integer n such that It follows that ns> t. nx>ns>t>y. THEOREM 6.4. A nonempty set of real numbers which has an upper bound has a least upper bound. In the proof which follows, if x is a real number and a is a Proof. rational number such that x < a, we shall abbreviate this to simply "x < a." Let A be a set which satisfies the hypothesis of the theorem. According to Theorem 6.3 there exist integers m and M such that m is not an upper bound of A and M is an upper bound of A. (To obtain 3.6 1 Real Numbers 153 an in, select an element a of A, apply Theorem 6.3 to secure an integer n such that n > -a, and then let m = -n.) Then we may infer the existence of an integer bo such that bo is an upper bound of A while bo - I is not. We now define b inductively as follows: if bn_.1 - 2-n is an upper bound of A, bn _ bn_t - 2_n if bn_t - 2-4 is not an upper bound of A. hn_ t For all n, bn is an upper bound of A and, as may be proved by an induction argument, |
bn - 2-n is not an upper bound. Hence, for every m > n bn - 2-" < b,n. (1) Further, it is clear that for every m > n, (2) Combining (1) and (2) gives b,n < b. Ibn -- bml < 2-n. It follows that if N is a positive integer and m, n > N, then Ibn - bml < 2-N, whence b is a Cauchy sequence of rational numbers. Let u be the real number which it determines. Then by virtue of (1) arid, in turn, (2), for every n (3) (4) bn-2-n <u, u < bn. f We shall now prove that u is an upper bound of A. Assume to the contrary that a > u for some a in A. Then there exists an n such that 2" > (a - u)--l or 2-n<a-u Addition of this to (3) yields the inequality bn < a, a contradiction of the fact that bn is an upper bound of A. Finally we prove that u is the least upper bound of A. Assume to the contrary that v is a smaller upper bound. As above, there then exists an n such that (5) 2-n<u-v. f "That (2) implies (4) is a consequence of the following result. If x and y are Cauchy sequences and there exists an integer N such that for all n > N, x < y,,, then lx], <,[y],. For the contrary implies that the Cauchy sequence z such that z = x - y" is positive and hence there exists an e > 0 and an N, such that for all n > N,, x - y > e. If we choose n = N + N, we are led to a contradiction of the hypotheses. We note further that x < y" does not imply [xJ, <,[y], but only [x], <, [Y],. 154 Extension of the Natural Numbers to the Real Numbers I CHAP. 3 Since b" - 2-" is not an upper bound of A, there exists an a in A such that b - 2-" < a, which implies that b"-2"<v. Addition of this to (5) yields the inequality which contradicts |
(4). b" < u, Later we shall prove that the properties of the real number system stated in 't'heorems 6.1 and 6.4 characterize it to within an orderisomorphism. EXERCISES 6.1. Prove that the system of rational real numbers is order-isomorphic to 6.2. Prove Theorem 6.1. 6.3. Prove the assertion made in the proof of Theorem 6.4 that b" - 2-" is not an upper bound of A. 6.4. Derive as a corollary to 't'heorem 6.4 that a nonempty set of real num- bers which has a lower bound has a greatest lower bound. 6.5. Let f be a real function --that is, a function whose domain and range are each a set of real numbers. Such a function is called continuous at a member a of its domain iff for every e > 0 there exists a S > 0 such that for IhI < S and a -l- h in the domain off I f(a + h) - f(a) I < E. Prove that if f is a continuous at each point of the closed interval [a, b] and f (a) < 0 and J(b) > 0, then there exists a c such that a < c < b and f (c) = 0. Hint: Define c to be the least upper bound of all x between a and b for which f(x) < 0. 6.6. Assume that it has been shown that a real polynomial function is continuous. Let f be the polynomial function such that for all real numbers x, f(x) = x" - a where n is a positive integer and a is a positive real number. Prove that there exists exactly one positive real number c such that f (c) = 0. This number is called the nth root of a and symbolized by Va or a'1" 6.7. If a > 0, b > 0, and ri is a positive integer, prove that 1/a 17'b = ab. 7. Further Properties of the Real Number System A sequence x of real numbers is a sequence such that x" C R for every n. A Cauchy sequence of real numbers is a sequence x of real 3.7 I Further Properties of the Real Number System 155 |
numbers such that for every positive real number e there exists a positive integer N such that for every m, n > N IX'n - xml < E. We define next the notion of limit. This notion is the cornerstone of the calculus and, indeed, of analysis in general. The real number y is a limit of the sequence x of real numbers iff for every positive real number e there exists a positive integer N such that for every n > N The proof of the following lemma is left as an exercise. Ixn - yl < E. LEMMA 7.1. A sequence of real numbers has at most one limit. Thus, if the sequence x of real numbers has y as a limit, then y is its only limit, and we are justified in introducing the following familiar notation for y : lim x,, = y n- or simply lim x = Y. LEMMA 7.2. Let a be a sequence of rational numbers and let x be the sequence of real numbers such that for every n, x = the real number corresponding to a,,. Then x is a Cauchy sequence ifr a is a Cauchy sequence. Further, if a is a Cauchy sequence and y is the real number which it defines, then lim x,, = y. Proof. We shall consider only the second assertion. Let e be a positive real number and let b = (d)r he a rational real number such that 0 < b < E. Since a is assumed to be a Cauchy sequence, there exists an N such that for all m, n > N Ian - a,I < d Since a - a,,, < d it follows that Ran, a,,,.. an,...) - (a,, a2,..., an,...) ]r < b (see the footnote in the proof of Theorem 6.4) or, in other words, that xn - y<S. Similarly, the inequality a. - an < d implies that Hence, for all m, n > N Ixn - yl < b < 156 Extension of the Natural Numbers to the Real Numbers I C I I A P. 3 THEOREM 7.1. (Cauchy convergence principle). A sequence of real numbers has a limit iff it is a Cauchy sequence. It is left as an exercise to prove that if a sequence of real Proof. numbers has a limit, then it is a Cauchy |
sequence. Turning to the converse, assume that u is a Cauchy sequence of real numbers. The strategy for proving that u has a limit calls for the determination of a sequence a of rational numbers which approximates u sufficiently closely that a is a Cauchy sequence and the real number which it defines is the limit of rc. For each positive integer n, u,, < un + 1/n, and hence (Theorem 6.2) there exists a rational real number x such that un < x < un -i- 1 /n. Let a be a positive real number. Then there exists an integer Nn such that N, > 3/e and, hence, for every n > Nr (1) Further, x is a Cauchy sequence, since tun - xnl < e/3. lxn - Xnel < Ixn - unl + Jun - Umt + June - Xrnl, and for in and n sufficiently large each summand on the right side of the inequality is less than e/3. Let an be the rational number to which xn corresponds. By Lemma 7.2, a is a Cauchy sequence of rational numbers and hence defines a real number y. Further, by Lemma 7.2, urn Xn = Y. Hence, there exists an integer N2 such that for n > N2 (2) We infer from (/) and (2) that for n > Inax (Nn, N2) Ixn - YI < e/2. tun - yI ` Iun - xnl + Ix,, - yj < e/3 + e/2 < e, which establishes that y = Limn un. We establish next the possibility of representing a real number by a nonterminating decimal. A precise formulation of this, generalized to any integer radix greater than or equal to 2, is given in the following theorem. THEOREM 7.2. Let r be an integer greater than or equal to 2e Corresponding to each nonncgative real number x there is a sequellOG 3.7 I Further Properties of the Real Number System 157 - -, d,,, (a, dl, d2, (relative to r) such that - - -) of integers which is uniquely determined by x (i) a = [x], the largest integer less than or equal to x, (ii) 0 < do < r for all |
n, (iii) the sequence whose terms are defined inductively by yo=a, yn+-I = yn + doI l/rnf-1 is a Cauchy sequence and lien yn = x. Proof. Let r be an integer greater than or equal to 2, x be a nonnegative real number, and a = [x]. Then xr = ar + xl for some number xl such that 0 < xl < r. Let dl = lxi ] so xlr = dlr + x2 for some number x2 such that 0 < X2 < r. Let d2 = [x2] so x2r = d2r A- xa for some number x3 such that 0 < xa < r. In general, define xn by xn_Ir = do-lr + xn and set do = [xn]. 't'hen x=a l r --z+..+.-- d2 r d o -n r a Y I>> n.i r where 0 < x,,.1I < r. Hence r r r// 1.11 According to the definition of yn given in (iii), this may be written as 0 <x - yn < r n. It follows that ix - ynI < r '`, whence limy. = x. The proof of the uniqueness (relative to r) of the sequence corre- sponding to x is left as an exercise. If r = 10 in the' preceding theorem we obtain the familiar representation of a nonncgative real cumber as a nontcrminating decimal ;upon writing a = [xj in decimal notation. Of the two possible decimal 10-) where i and j are non'representations of numbers of the form i 158 Extension of the Natural Numbers to the Real Numbers I CHAP P. 3 negative integers, the theorem chooses that one which consists of all zeros after a certain point. The proof of the converse of Theorem 7.2 is left as an exercise. THEOREM 7.3. Let (a, dl, d2, ) be a sequence of nonnegative integers such that for some integer r > 2, 0 < do < r for all n. Then there exists a unique nonnegative real number x such that the sequence whose terms y. are defined inductively by yo = a and yA+, = y + is a Cauchy sequence |
having x as its limit., d,,, We conclude our development of the real number system by calling attention to # common feature of the three extensions whereby R is obtained from N. If the details (see Sections 2 and 3) of the first extension are reviewed, it will become evident that it could be mimicked using integers instead of natural numbers as the initial elements. Suppose this construction is carried out to obtain what might be called the system of superintegers. Then this system has all those properties which the system of integers possesses. Moreover, the system of superintegers has an additional property-one which destroys any further interest in it. Namely, as the reader can readily prove, it is order-isomorphic to the system of integers. In other words, the extension of Z by the method used to extend N to Z yields nothing essentially different from Z. A corresponding result holds for the second type of extension we introduced : the extension of (Q by the method used to extend Z to Q is a system which is order-isomorphic to Q. Finally, let us consider the extension of R, which can be made in terms of Cauchy sequences. We shall call the numbers we get in this way superreal numbers. Thus, a superreal number is an equivalence class of Cauchy sequences of real numbers. Corresponding to the results stated above, it is possible to prove that the system of superreal numbers is order-isomorphic to the system of real numbers. In other words, essentially nothing new results if R is extended by the method used to extend Q to R. To prove this we point out first that, as discussed in Section 6, there is a one-to-one map on the initial system (which is now R) into the extended system SR. This map determines an order-isomorphic image of B. in SR. Suppose now that X is any superreal number. Let x E X, which means that x is a Cauchy sequence of real numbers. According to Theorem 7.1, x has a limit y, whence x is -.,equivalent to (y, y, ), which implies that X = yer. This means that the image of B. in SR exhausts SR or, in other words, that SR is an order-isomorphic image of R., y, Bibliographical Notes EXERCISES 159 7.1. Prove Lemma 7.1. 7.2. Prove the first assertion made |
in Lemma 7.2. 7.3. Prove that if a sequence of real numbers has a limit, then it is a Cauchy sequence. 7.4. In the proof of Theorem 7.2, show that an explicit definition of d" is d" = [xr"] - r[xr"-']. 7.5. Prove Theorem 7.3. 7.6. Prove that the system of superintegers is order-isomorphic to the system of integers. BIBLIOGRAPHICAL NOTES The two basic set-theoretical methods of constructing the system of real numbers from the system of natural numbers are due to Cantor and Dedekind. The difference in these methods appears in the extension of the rational numbers to the real numbers. The extension of the rationals to the reals via Dedekind's method is given in Landau (1930) and in N. FI. McCoy (1960). CHAPTER 4 Logic AS WE SHALL study it, mathematical or symbolic logic has two aspects. On one hand it is logic--it is an analytical theory of the art of reasoning whose goal is to systematize and codify principles of valid reasoning. It has emerged from a study of the use of language in argument and persuasion and is based on the identification and examination of those parts of language which are essential for these purposes. It is formal in the sense that it lacks reference to meaning. Thereby it achieves versatility: it may be used to judge the correctness of a chain of reasoning (in particular, a "mathematical proof") solely on the basis of the form (and not the content) of the sequence of statements which make up the chain. There is a variety of symbolic logics. We shall be concerned solely with that one which encompasses most of the deductions of the sort encountered in mathematics. Within the context of logic itself, this is "classical" symbolic logic. The other aspect of symbolic logic is interlaced with problems relating to the foundations of mathematics. In brief, it amounts to formulating a mathematical theory as a logical system augmented by further axioms. The idea of regarding a mathematical theory as an "applied" system of logic originated with the German mathematician G. Frege (1848-1925), who developed a system of logic for use in his study of the foundations of arithmetic. The Principia Mathematica (1910-1913) of Whitehead and Russell carried on this work of |
Frege and demonstrated that mathematics could be "reduced to logic." In the later chapter treating axiomatic theories some indication will be given of this approach to mathematical theories. 1. The Statement Calculus. Sentential Connectives In mathematical discourse and elsewhere one constantly encounters declarative sentences which have been formed by modifying a sentence with the word not or by connecting sentences with the words and, or, if then (or implies), and if and only if. These five words or combina... 4.1 1 T 1w Statement Calculus: Sentential Connectives 161 tions of words are called sentential connectives. Our first concern here is the analysis of the structure of a composite sentence (that is, a declarative sentence in which one or more connectives appear) in terms of its constituent prime sentences (that is, sentences which either contain no connectives or, by choice, are regarded as "indivisible"). We shall look first at the connectives individually. A sentence which is modified by the word "not" is called the negation of the original sentence. For example, "2 is not a prime" is the negation of "2 is a prime," and "It is not the case that 2 is a prime and 6 is a composite number" is the negation of "2 is a prime and 6 is a composite number." It is because the latter sentence is composite that grammatical usage forces one to use the phrase "It is not the case that" instead of simply the word "not." The word "and" is used to join two sentences to form a composite sentence which is called the conjunction of the two sentences. For example, the sentence "The sun is shining, and it is cold outside" is the conjunction of the sentences "The sun is shining" and "It is cold outside." In ordinary language various words, such as "but," are used as approximate synonyms for "and"; however, we shall ignore possible differences in shades of meaning which might accompany the use of one in place of the other. A sentence formed by connecting two sentences with the word "or" is called the disjunction of the two sentences. We shall always assume that "or" is used in the inclusive sense (in legal documents this is often expressed by the barbarism "and/or"). Recall that we interpreted "or" in this way in the definition of the union of two sets.. From two sentences we may construct one of the form "If., |
."; this is called a conditional sentence. The semtcnce irnthen. ipediately following "If" is the antecedent, and the sentence iuianediately following "then" is the consequent. For example, "If 2 > 3, then 3 > 4" is a conditional sentence with "2 > 3" as antecedent and "3 > 4" as consequent. Several other idioms in English which we shall regard as having the same meaning as "If P, then Q" (where I' and Q arc sentences) are.. P implies Q, only if Q, P is a sufficient condition for Q, Q, provided that P, Q if P, Q is a necessary condition for P. 162 Logic I CHAP. 4 The words "if and only if" are used to obtain from two sentences a biconditional sentence. We regard the biconditional as having the same meaning as P if and only if Q if P, then Q, and if Q, then P; Q is a necessary and sufficient condition for P. By introducing letters "P,".. to stand for prime sentences, a. special symbol for each connective, and parentheses, as may be needed for punctuation, the connective structure of a composite sentence can be displayed in an effective manner. Our choice of symbols for the connectives is as follows: -, for "not," A for "and," V for "or," - for "if.., then. for "if and only if."...," Thus, if P and Q are sentences, then -1P,PAQ,PVQ,P-4Q,P++ Q are, respectively, the negation of P, the conjunction of P and Q, and so on. Following are some concrete examples of analyzing the connective structure of composite sentences in terms of constituent prime sentences. EXAMPLES 1.1. The sentence 2 is a prime, and 6 is a composite number may be symbolized by P A C, where P is "2 is a prime" and C is "6 is a composite number." 1.2. The sentence If either the Pirates or the Cubs lose and the Giants win, then the Dodgers will be out of first place and, moreover, I will lose a bet is a conditional, so it may be symbolized in the form R --> S. The antecedent is composed from the three prime sentences P ("The |
Pirates lose"), C ("The Cubs lose"), and G ("The Giants win"), and the consequent 4.1 1 The Statement Calculus. Sentential Connectives 163 is the conjunction of D ("The Dodgers will be out of first place") and B ("I will lose a bet"). The original sentence may be symbolized in terms of these prime sentences by 1.3. The sentence ((PVC.,) AG)-'(DAB). If either labor or management is stubborn, then the strike will be settled iff the government obtains an injunction, but troops are not sent into the mills is a conditional. The antecedent is the disjunction of L ("Labor is stubborn") and M ("Management is stubborn"). The consequent is a biconditional whose left-hand member is S ("The strike will be settled") and whose right-hand member is the conjunction of G ("The government obtains an injunction") and the negation of R ("Troops are sent into the mills"). So the original sentence may be symbolized by (L V M) -> (S c-), (G A (-1R))). To avoid an excess of parentheses in writing composite sentences in symbolic form, we introduce conventions as in algebra. We agree that +-> is the strongest connective (that is, it is to encompass most), and then follows -+. Next in order are V and A, which are assigned equal strength, and then follows -,, the weakest connective. For example, P A Q-+Rmeans(P A Q)-iR, P +--> Q --+R means P F-' (Q -'R), -, P A Q means (-n P) A Q, and the sentence in Example 1.3 may now be symbolized as L v M -> (S +-> G A -, R). EXERCISES 1.1. Translate the following composite sentences into symbolic notation, using letters to stand for the prime components (which here we understand to mean sentences which contain no connectives). (a) Either it is raining or someone left the shower on. (b) If it is foggy tonight, then either John must stay home or he must take a taxi. (c) John will sit, and lie or George will wait. (d) John will sit and wait, or George will wait. (e) I will go either by bus or by taxi. (f) Neither the North nor the South won the Civil War |
. (g) If, and only if, irrigation ditches are dug will the crops survive; should the crops not survive, then the farmers will go bankrupt and leave. 164 Logic I cnnP. 4 (h) If I am either tired or hungry, then I cannot study. (i) if John gets up and goes to school, he will he happy; and if he does not get up, he will not be happy. 1.2. Let C be "Today is clear," R be "It is raining today," S be "It is snowing today," and Y be "Yesterday was cloudy." Translate into acceptable English the following. (a) C->-,(RAS). (b) Y.-4 C. (c) Y A (C V R). (d) (Y->R) V C. (e) C 4.* (R A -18) V Y. (f) (C H R) A (--,S V Y). 2. The Statement Calculus. Truth Tables Earlier we agreed that by a statement we would understand a declarative sentence which has the quality that it can be classified as either true or false, but not both. That one of "truth" or "falsity" which is assigned to a statement is its truth value. Often we shall abbreviate "truth" to T and "falsity" to F. If P and Q are statements, then, using the everyday meaning of the connectives, each of -,P,PAQ,PV Q,P--3Q,PHQ is a statement. Let us elaborate. On the basis of the usual meaning of "not," if a statement is true, its negation is false, and vice versa. For example, if S is the true stateinent (has truth value T) "The moon is a satellite of the earth," then --S is false (has truth value F). By convention, the conjunction of two statements is true when, and only when, both of its constituent statements are true. For example, "3 is a prime, and 2 + 2 = 5" is false because "2 + 2 = 5" is a false statement. Having agreed that the connective "or" would be understood in the inclusive sense, standard usage classifies a disjunction as false wlicn, and only when, both constituent statements are false. Truth-value assignments of the sort which we are making can he |
summarized concisely by truth tables wherein are displayed the truthvalue assignments for all possible assignments of truth values to the constituent statements. Below are truth tables for the types of composite statements we have already discussed, as well as those for conditional and biconditional statements. 4.2 I The Statement Calculus. Truth Tables Negation P _1 P T F I F T Conjunction 165 Disjunction Condi tional Bicond itio na l P Q I PHQ The motivation for the truth-value assignments made for the conditional is the fact that, as intuitively understood, P -- * Q is true if Q is deducible from P in sortie way. So, if P is true and Q is false, we want 1' - Q to be false, which accounts for the second line of the table. Next, suppose that Q is true. Then, independently of P and its truth value, it is plausible to assert that P-' Q is true. This reasoning suggests the assignments made in the first and third lines of the table. To justify the fourth line, consider the statement P A Q - -) P. We expect this to be true regardless of the choice of I' and Q. But, if P and Q are both false, then P A Q is false, and we are led to the conclusion that if both antecedent and consequent are false, a conditional is true. The table for the biconditional is determined by that for conjunction and the conditional, once it is agreed that P <- Q means the same as (P -+ Q) A (Q ---' P). These five tables are to be understood as definitions; they arc the customary definitions adopted for mathematics. We have made merely a feeble attempt to make them seem plausible on the basis of meaning. It is an immediate consequence of these definitions that if P and Q are statements, then so are each of -, P, P A Q, I' V Q, P ---' Q, and P E--' Q. It follows immediately that any composite sentence whose prime colnponents are statements is itself a statement. If the truth values of the prime components are known, then the truth value of the composite statement can be determined in a mechanical way. 166 EXAMPLES Logic I CHAP. 4 2.1. Suppose that a composite statement is symbolized by PVQ-+(RH-1S) and that the truth values of P, Q, R, and S are T, |
F, F, and T, respectively. Then the value of P V Q is T, that of -1S is F, that of R H -1S is T, and, hence, that of the original statement is T, as a conditional having a true antecedent and a true consequent. Such a calculation can be made quickly by writing the truth value of each prime statement underneath it and the truth value of each composite constructed under the connective involved. Thus, for the above we would write out the following, where, for study purposes, we have put successive steps on successive lines. P V Q -i (R E3 -,S.2. Consider the following argument. If prices are high, then wages are high. Prices are high or there are price controls. Further, if there are price controls, then there is not an inflation. There is an inflation. Therefore, wages are high. Suppose that we are in agreement with each of the first four statements (the premises). Must we accept the fifth statement (the conclusion)? To answer this, let us first symbolize the argument using letters "P," "W," "C," and "I" in the obvious way. Thus, P is the sentence "Prices are high." Then we may present it as follows: P -, W PVC C -' -i I I W To assume that we are in agreement with the premises amounts to the assignment of the value T to the statements above the line. The question posed then can be phrased as: If the premises have value T, does the conclusion have value T? The answer is in the affirmative. Indeed, since I and C -b -,I have value T, the value of C is F according to the truth table for the conditional. Hence, P has value T (since P V C is T) and, therefore, W has value T (since P -- W is T). 2.3. We consider the conjunction (PVC)A(C-'-,I) of two of the statements appearing in the preceding example. In general, the 4.2 ( The Statement Calculus. Truth Tables 167 truth value which such a statement will receive is dependent on the assignments made to the prime statements involved. It is realistic to assume that, during periods of changing economic conditions, the appropriate truth value assignments to one or more of P, C, and I will change from T to F or vice versa. Thus the question may arise as to combinations of truth |
values of P, C, and I for which (P V C) A (C -' -,l) has value T or value F. This can be answered by the examination of a truth table in which there appears the truth value of the composite statement for every possible assignment (211) of truth values to P, C, and I. This is called the truth table for the given statement, and it appears below. Each line includes an assignment of values to P, C, and I, along with the associated value of (P V C) A (C--+ -,l). The latter may be computed as in the first example above. However, short cuts in filling out the complete table will certainly occur to the reader as he proceedsPVC) A (C -- -1I.4. If P is "2 is a prime" and L is "Logic is fun," there is nothing to prohibit our forming such composite statements as PV L,P->L, -,P --*PV L. Since both P and L have truth values (clearly, both are T), these composite statements have truth values which we can specify. One's initial reaction to such nonsense might be that it should be prohibited- -that the formation of conjunctions, conditionals, and so on, should be permitted only if the component statements are related in content or subject. However, it requires no lengthy reflection to realize the difficulties involved in characterizing such obscure notions. It is much simpler to take the easy way out: to permit the formation of composite statements from any statements. On the basis of meaning, this amounts to nonsense sometimes, but no harm results. Our concern is with the formulation of principles of valid reasoning. In applications to systematic reasoning, composite statements which amount to gibberish simply will not occur. EXERCISES 2.1. Suppose that the statements P, Q, R, and S are assigned the truth values T, F, F, and T, respectively. Find the truth value of each of the following statements. 168 Logic I CHAP. 4 (f) P V R 4- R A --,S. (g) S*--+P--(-,PV S). (h) Q A -,S -> (P S). (i) RAS--1(P--), -,Q VS). (j) (a) (P V Q) V R. (b) P V (Q V R). (c) R - |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.