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For example, in Exercise 32, we will want R 0 R for an arbitrary relation R. Example Assume that G is some one-to-one function. Then by Theorem 3H, G- 1 0 G is a function, its domain is {x E dom G I G{x) E dom G- 1} = dom G, and for x in its domain, (G- 1 0 G)(x) = G-1{G{x)) = x by Theorem 3G. Thus G- 1 Similarly one can show that Go G- 1 is Iran G (Exercise 25). 0 G is I dom G, the identity function on dom G, by Exercise 11. Theorem 31 For any sets F and G, {F 0 Gt 1 = G- 1 0 F- 1 • Proof Both (F 0 G)-l and G- 1 0 F- 1 are relations. We calculate: <x, y) E (F 0 G)-l <=> <y,x)EFoG yGt & tFx <=> <=> xF-1t&tG-1y <=> <x, y) E G- 1 0 F- 1. for some t for some t -I 48 3. Relations and Functions In a less abstract form, Theorem 31 expresses common knowledge. In getting dressed, one first puts on socks and then shoes. But in the inverse process of getting undressed, one first removes shoes and then socks. Theorem 3J Assume that F: A --+ B, and that A is nonempty. (a) There exists a function G: B --+ A (a "left inverse") such that Go F is the identity function I A on A iff F is one-to-one. (b) There exists a function H: B --+ A (a" right inverse") such that F 0 H is the identity function I Bon B iff F maps A onto B. Proof (a) First assume that there is a function G for which G 0 F = I A. If F(x) = F(y), then by applying G to both sides of the equation we have x = G(F(x)) = G(F(y)) = y, and hence F is one-to-one. For the converse, assume that F is one-to-one. Then F- 1 is a function from ran F onto A (by The |
orems 3E and 3F). The idea is to extend F- 1 to a function G defined on all of B. By assumption A is nonempty, so we can fix some a in.11. Then we define G so that it assigns a to every point in B - ran F: In one line, G(x) = {:-l(X) if x E ran F if x E B - ran F. G = F- 1 u (B - ran F) x {a} (see Fig. lOa). This choice for G does what we want: G is a function mapping B into A, dom(G 0 F) = A, and G(F(x)) = F-1(F(x)) = x for each x in A. Hence G 0 F = I A • The converse poses a difficulty. We cannot take H = F- 1 (b) Next assume that there is a function H for which F 0 H = lB. Then for any y in B we have y = F(H(y)), so that y E ran F. Thus ran F is all of B., because in general F will not be one-to-one and so F- 1 will not be a function. Assume that F maps A onto B, so that ran F = B. The idea is that for each y E B we must choose some x for which F(x) = y and then let H(y) be the chosen x. Since y E ran F we know that such x's exist, so there is no problem (see Fig. lOb). Or is there? For anyone y we know there exists an appropriate x. But that is not by itself enough to let us form a function H. We have in general no way of defining anyone particular choice of x. What is needed here is the axiom of choice. Functions 49.a A F- 1 (a) B-ranF ran F H(y) = x • ___ --+-------------~~F·y x' • B A x" • Fig.lO. The proof of Theorem 3J. In part (a), make G(x) = a for x E B - ran F. In part (b), H(y) is the chosen x for which F(x) = y. (b) Axiom of Choice (first form) For any relation R there is a function H ~ R with dom |
H = dom R. With this axiom we can now proceed with the proof of Theorem 3J(b); take H to be a function with H ~ F- 1 and dom H = dom F- 1 = B. Then H does what we want: Given any y in B, we have <y, H(y) E F- 1 ; hence <H(y), y) E F, and so F(H(y)) = y. 1 In Chapter 6 we will give a systematic discussion of the axiom of choice. It is the only axiom that we discuss without using the marginal stripe. 50 3. Relations and Functions Theorem 3K The following hold for any sets. (F need not be a function.) (a) The image of a union is the union of the images: F[A u B] = F[A] u F[B] and F[Ud] = U{F[A] I A Ed}. (b) The image of an intersection is included in the intersection of the images: F[A 11 B] ~ F[A] 11 F[B] and for nonempty d. Equality holds if F is single-rooted. (c) The image of a difference includes the difference of the images: F[A] - F[B] ~ F[A - B]. Equality holds if F is single-rooted. Example Let F: ~ -+ ~ be defined by F(x) = x 2 • Let A and B be the closed intervals [-2,0] and [1,2]: A={xl-2~x~0} and B={xI1~x~2}. Then F[A] = [0,4] and F[B] = [1,4]. This example shows that equality does not always hold in parts (b) and (c) of Theorem 3K, for F[A 11 B] = F[0] = 0, whereas F[A] 11 F[B] = [1, 4]. And F[A] - F[B] = [0, 1), whereas F[A - B] = F[A] = [0,4].. Proof To prove Theorem 3K we calculate Y E F[A u B] <=> (3x E A u B)xFy <=> (3x E A)xFy or (3x E B |
)xFy <=> y E F[A] or y E F[B]. This proves the first half of (a). For intersections we have the corresponding calculation, except that the middle step (3x E A 11 B)xFy = (3x E A )xFy & (3x E B)xFy is not always reversible. It is possible that both Xl E A with Xl Fy and x 2 E B with x 2 Fy, and yet there might be no x in A 11 B with xFy. But if F is single rooted, then Xl = x 2 and so it is in A 11 B. Thus we obtain the first half of(b). The second halves of (a) and (b) generalize the first halves. The proofs follow the same outlines as the first halves, but we leave the details to Exercise 26. Functions 51 For part (c) we also calculate: y E F[A] - F[B] ¢> (3x E A)xFy & I (3x E A - B)xFy => ¢> yE F[A - B]. (3t E B)tFy Again if F is single-rooted, then there is only one x such that xFy. In this case the middle step can be reversed. -j Since the inverse of a function is always single-rooted, we have as an immediate consequence of Theorem 3K that unions, intersections, and relative complements are always preserved under inverse images. Corollary 3L For any function G and sets A, B, and d: G- 1[Ud] = U{G- 1[A] I A Ed}, G- 1[n.W'] = n{G- 1 [A] I A E d} G- 1 [A - B] = G- 1 [A] - G- 1 [B]. for d"# 0, We conclude our discussion of functions with some definitions that may be useful later. Our intent is to build a large working vocabulary of set theoretic notations. An infinite union is often "indexed," as when we write U. I A.. We can give a formal definition to such a union as follows. Let I be a set, called the index set. Let F be a function whose domain includes I. Then we define IE I U F(i) = U |
{F(i) liE I} i E I = {x I x E F(i) for some i in I}. For example, if I = {O, 1, 2, 3}, then U F(i) = U{F(O), F(1), F(2), F(3)} i E I = F(O) u F(1) u F(2) u F(3). Similar remarks apply to intersections (provided that I is nonempty): n F(i) = n{F(i) liE I} i E I = {x I x E F(i) for every i in I}. If we use the alternative notation Fi = F(i), 52 3. Relations and Functions then we can rewrite the above equations as and iEI iEl = {x I x E Fj for some i in I} = {x I x E Fi for every i in I}. For sets A and B we can form the collection offunctions F from A into B. Call the set of all such functions AB: AB = {F I F is a function from A into B}. If F: A --+ B, then F £ A x B, and so FE &(A x B). Consequently we can apply a subset axiom to &(A x B) to construct the set of all functions from A into B. The notation AB is read "B-pre-A." Some authors write BA instead; this notation is derived from the fact that if A and B are finite sets and the number of elements in A and B is a and b, respectively, then AB has b" members. (To see this, note that for each of the a elements of A, we can choose among b points in B into which it could be mapped. The number of ways of making all a such choices is b· b····· b, a times.) We will return to this point in Chapter 6. Example Let w = {O, 1, 2,... }. Then "'{O, 1} is the set of all possible functions f: w --+ {O, 1}. Such an f can be thought of as an infinite sequence f(0),f(I),f(2),... of O's and 1's. Example For a nonempty set A, we have A0 = 0. This is because no function could have a nonempty domain and an empty range. On the |
other hand, 0 A = {0} for any set A, because 0: 0 --+ A, but 0 is the only function with empty domain. As a special case, we have 00 = {0}. Exercises. )11. Prove the following version (for functions) of the extensionality principle: Assume that F and G are functions, dom F = dom G, and F(x) = G(x) for all x in the common domain. Then F = G. 12. Assume thatfand 9 are functions and show that j£ 9 ¢> domf£ dom 9 & (\1'x E domf)f(x) = g(x). Functions 53 13. Assume thatfand 9 are functions withf~ 9 and dom 9 ~ domf. Show thatf = g. 14. Assume thatfand 9 are functions. (a) Show thatf n 9 is a function. (b) Show that fu 9 is a function iff f(x) = g(x) for every x in (domf) n (dom g). 15. Let d be a set of functions such that for any f and 9 in d, either f ~ 9 or 9 ~ f. Show that U d is a function. 16. Show that there is no set to which every function belongs. 17. Show that the composition of two single-rooted sets is again single rooted. Conclude that the composition of two one-to-one functions is again one-to-one. 18. Let R be the set {<o, 1), <0, 2), <0, 3), (1, 2), (1, 3), (2, 3)}. Evaluate the following: R 0 R, R r {1}, R - 1 r {1}, R[{ 1}], and R - 1 [{ 1}]. 19. Let A = {<0, {0, {0}}), <{0}, 0)}. Evaluate each of the following: A(0), A[0], A[{0}], A[{0, {0}}], A- l, A 0 A, A r 0, A r {0}, A r {0, {0}}, UUA. 20. Show that F r A = F n (A x ran F). 21. Show that (R 0 S) 0 T = R 0 (S |
0 T) for any sets R, S, and T. 22. Show that the following are correct for any sets. (a) A~B=>F[A]~F[B]. (F 0 G)[A] = F[G[A]]. (b) (c) Q r (A u B) = (Q r A) u (Q r B). 23. Let I A be the identity function on the set A. Show that for any sets Band C, and 24. Show that for a function F, F-l[A] = {x E dom F I F(x) E A}. 25. (a) Assume that G is a one-to-one function. Show that Go G- l is Iran G' the identity function on ran G. (b) Show that the result of part (a) holds for any function G, not necessarily one-to-one. 26. Prove the second halves of parts (a) and (b) of Theorem 3K. 27. Show that dom(F 0 G) = G-l[dom F] for any sets F and G. (F and G need not be functions.) 54 3. Relations and Functions 28. Assume that f is a one-to-one function from A into B, and that G is the function with dom G = f!IJ A defined by the equation G(X) = f[ X]. Show that G maps f!IJ A one-to-one into f!IJ B. 29. Assume that f: A --+ B and define a function G: B --+ f!IJ A by G(b) = {x E A If(x) = b}. Show that iff maps A onto B, then G is one-to-one. Does the converse hold? 30. Assume that F: f!IJ A --+ f!IJ A and that F has the monotonicity property: X £ Y £ A => F(X) £ F(Y). Define B=n{X£AIF(X)£X} and C=U{X£AIX£F(X)}. (a) Show that F(B) = Band F(C) = C. (b) Show that if F(X) = X, then B £ X £ C. INFINITE CARTESIAN PRODUCTS 2 We can form something |
like the Cartesian product of infinitely many sets, provided that the sets are suitably indexed. More specifically, let I be a set (which we will refer to as the index set) and let H be a function whose domain includes I. Then for each i in I we have the set H(i); we want the product of the H(i)'s for all i E I. We define: X H(i) = {Ilfis a function with domain I and (Vi E I)f(i) EH(i)}. 'E I Thus the members of X, E I H(i) are" I -tuples" (i.e., functions with domain I) for which the "ith coordinate" (i.e., the value at i) is in H(i). The members of Xi E I H(i) are all functions from I into U, E I H(i) and hence are members of I(UiEI H(i)). Thus the set X'EI H(i) can be formed by applying a subset axiom to I (U, E I H(i)). Example X'EIH(i)=IA. If for every i E I we have H(i) = A for some one fixed A, then Example Assume that the index set is the set w = {O, 1, 2,.,.}. Then X'Ero H(i) consists of "w-sequences" (i.e., functions with domain w) that have for their ith term some member of H(i). If we picture the sets H(i) as shown in Fig. 11, then a typical member of X, H(i) is a "thread" that selects a point from each set.,E ro 2 This section may be omitted if certain obvious adjustments are made in Theorem 6M. Equivalence Relations 55 If anyone H(i) is empty, then clearly the product XieI H(i) is empty. Conversely, suppose that H(i) "# 0 for every i in I. Does it follow that XieI H(i)"# 0? To obtain a member! of the product, we need to select some member from each H(i), and put!(i) equal to that selected member. This requires the axiom of choice, and in fact this is one of the many equivalent ways of stating the axiom. Ax |
iom of Choice (second form) For any set I and any function H with domain I, if H(i)"# 0 for all i in I, then XieI H(i)"# 0. H(O) H(I) H(2) H(3) H(4) Fig. 11. The thread is a member of the Cartesian product. Exercise 31. Show that from the first form of the axiom of choice we can prove the second form, and conversely. EQUIVALENCE RELATIONS Consider a set A (Fig. 12a). We might want to partition A into little boxes (Fig. 12b). For example, take A = w; we can partition w into six parts: {a, 6, 12,... }, {1, 7, 13,... }, {5, 11, 17,... }. By "partition" we mean that every element of A is in exactly one little box, and that each box is a nonempty subset of A. Now we need some mental agility. We want to think of each little box as being a single object, instead of thinking of it as a plurality of objects. (Actually we have been doing this sort of thing throughout the book, when ever we think of a set as a single object. It is really no harder than thinking of a brick house as a single object and not as a multitude of 56 3. Relations and Functions bricks.) This changes the picture (Fig. 12c); each box is now, in our mind, a single point. The set B of boxes is very different from the set A. In our example, B has only six members whereas A is infinite. (When we get around to defining "six" and "infinite" officially, we must certainly do it in a way that makes the preceding sentence true.) The process of transforming a situation like Fig. 12a into Fig. 12c is common in abstract algebra and elsewhere in mathematics. And in Chapter 5 the process will be applied several times in the construction of the real numbers. Suppose we now define a binary relation R on A as follows: For x and y in A, xRy ¢> x and yare in the same little box. • • • • • • (c) (a) (b) Fig. 12. Partitioning a set into six little boxes. Then we can easily see that R has the following three properties. 1. R is reflexive on A, by |
which we mean that xRx for all x E A. 2. R is symmetric, by which we mean that whenever xRy, then also yRx. 3. R is transitive, by which we mean that whenever xRy and yRz, then also xRz. Definition R is an equivalence relation on A iff R is a binary relation on A that is reflexive on A, symmetric, and transitive. Theorem 3M If R is a symmetric and transitive relation, then R is an equivalence relation on fld R. Proof Any relation R is a binary relation on its field, since R ~ dom R x ran R ~ fld R x fld R. What we must show is that R is reflexive on fld R. We have X E dom R => xRy => xRy & yRx => xRx for some y by symmetry by transitivity, and a similar calculation applies to points in ran R. Equivalence Relations 57 This theorem deserves a precautionary note: If R is a symmetric and transitive relation on A, it does not follow that R is an equivalence relation on A. R is reflexive on fld R, but fld R may be a small subset of A. We have shown how a partition of a set A induces an equivalence relation. (A more formal version of this is in Exercise 37.) Next we want to reverse the process, and show that from any equivalence relation R on A, we get a partition of A. Definition The set [X]R is defined by [X]R = {t I xRt}. If R is an equivalence relation and x E fld R, then [X]R is called the equivalence class of x (modulo R). If the relation R is fixed by the context, we may write just [xl The status of [X]R as a set is guaranteed by a subset axiom, since [X]R ~ ran R. Furthermore we can construct a set of equivalence classes such as {[X]R I x E A}, since this set is included in &(ran R). Lemma 3N Assume that R is an equivalence relation on A and that x and y belong to A. Then [X]R = [Y]R iff xRy. Proof (=» Assume that [X]R = [y]R' We know that y E |
[Y]R (because yRy), consequently y E [X]R (because [X]R = [y]R)' By the definition of [X]R' this means that xRy. (<=) Next assume that xRy. Then t E [Y]R => yRt => xRt because xRy and R is transitive => tE[X]R' Thus [Y]R ~ [X]R' Since R is symmetric, we also have yRx and we can reverse x and y in the above argument to obtain [X]R ~ [Y]R' -j Definition A partition n of a set A is a set of nonempty subsets of A that is disjoint and exhaustive, i.e., (a) no two different sets in n have any common elements, and (b) each element of A is in some set in n. Theorem 3P Assume that R is an equivalence relation on A. Then the set {[X]R I x E A} of all equivalence classes is a partition of A. Proof Each equivalence class [X]R is nonempty (because x E [x ]R) and is a subset of A (because R is a binary relation on A). The main thing that we 58 3. Relations and Functions must prove is that the collection of equivalence classes is disjoint, i.e., part (a) of the above definition is satisfied. So suppose that [X]R and [Y]R have a common element t. Thus But then xRy and by Lemma 3N, [X]R = [y]R' xRt and yRt. If R is an equivalence relation on A, then we can define the quotient set AIR={[x]RlxEA} whose members are the equivalence classes. (The expression AIR is read "A modulo R.") We also have the natural map (or canonical map) cp: A -+ AIR defined by cp(X) = [X]R for x EA. Example Let w = {O, 1,2,... }; define the binary relation ~ on w by m ~ n ¢> m - n is divisible by 6. Then ~ is an equivalence relation on w (as you should verify). The quotient set wi ~ has six members: [2], [0], [3], [1], [5 |
], [4], corresponding to the six possible remainders after division by 6. Example The relation of congruence of triangles in the plane is an equivalence relation. Example Textbooks on linear algebra often define vectors in the plane as follows. Let A be the set of all directed line segments in the plane. Two such line segments are considered to be equivalent iff they have the same length and direction. A vector is then defined to be an equivalence class of directed line segments. But to avoid the necessity of dealing explicitly with equivalence relations, books use phrases like "equivalent vectors are regarded as equal even though they are located in different positions," or "we write PQ = RS to say that PQ and RS have the same length and direction even though they are not identical sets of points," or simply "we identify two line segments having the same length and direction." Example Let F: A -+ B and for points in A define x ~ y iff F(x) = F(y). The relation ~ is an equivalence relation on A. There is a unique one-to-one function F: AI ~ -+ B such that F = F 0 cp (where cp is the natural map as Equivalence Relations 59 ·1.. F Fig. 13. F factors into the natural map followed by a one-to-one function. shown in Fig. 13). The value of F at a particular equivalence class is the common value of F at the members of the equivalence class. The last problem we want to examine in this section is the problem of defining functions on a quotient set. Specifically, assume that R is an equivalence relation on A and that F: A -+ A. We ask whether or not there exists a corresponding function F: AIR -+ AIR such that for all x E A, F([X]R) = [F(X)]R (see Fig. 14). Here we are attempting to define the value of F at an equivalence class by selecting a particular member x from the class and then formin~ [F(X)]R' But suppose Xl and x 2 are in the same equivalence class. Then F is not well defined unless F(x l ) and F(x 2 ) are in the same equivalence class. F ---------- ·1 ·1 F... Fig. 14. This diagram is said to be commutative if F 0 <p = <p 0 F. 60 3. Relations and |
Functions Example Consider wi ~ where m ~ n iff m - n is divisible by 6. Three functions from w into ware defined by In each case we can ask whether there is F.: wi ~ --+ wi ~ such that for every n in w: I F 3(n) = 2n. It is easy to see that if m ~ n, then 2m ~ 2n. Because of this fact F 1 is well defined; that is, there exists a function F 1 satisfying the equation F1([n]) = [2n]. No matter what representative m of the equivalence class [n] we look at, we always obtain the same equivalence class [2m]. (For further details, see the proof of Theorem 3Q below.) Similarly ifm ~ n, then m2 ~ n2, - n2 = (m + n)(m - n). Consequently F 2 is also well for recall that m2 defined. On the other hand, F 3 is not well defined. For example, 0 ~ 6 but 2° = 1 + 64 = 26. Thus although [0] = [6], we have [20] "# [26]. Hence there cannot possibly exist any function F 3 such that the equation F 3([ n]) = [2n] holds for both n = 0 and n = 6. In order to formulate a general theorem here, let us say that F is compatible with R iff for all x and y in A, xRy => F(x)RF(y). Theorem 3Q Assume that R is an equivalence relation on A and that there exists a unique is compatible with R, then F: A --+ A. If F F: AIR --+ AIR such that F([X]R) = [F(X)]R (-{:r) If F is not compatible with R, then no such F exists. Analogous results apply to functions from A x A into A. for all x in A. Proof First assume that F is not compatible; we will show that there can be no F satisfying (-{:r) The incompatibility tells us that for certain x and y in A we have xRy (and hence [x] = [y]) but not F(x)RF(y) (and hence [F(x)]"# [F(y)])' For (-{:r) to hold we would need both F([x]) = [F(x)] and |
F([y]) = [F(y)]. But this is impossible, since the left sides coincide and the right sides differ. Now for the converse, assume that F is compatible with R. Since (-{:r) demands that the pair <[x], [F(x)]) E F, we will try defining F to be the set of all such ordered pairs: F = {<[x], [F(x)]) I x E A}. Equivalence Relations 61 The crucial matter is whether this relation F is a function. So consider pairs <[x], [F(x)]) and <[y], [F(y)]) in F. The calculation [x] = [y] => xRy => F(x)RF(y) by Lemma 3N by compatibility => [F(x)] = [F(y)] by Lemma 3N shows that F is indeed a function. The remaining things to check are easier. Clearly dom F = AIR and ran F ~ AIR, hence F: AIR --+ AIR. Finally ({:r) holds because <[x], [F(x)]) E F. We leave it to you to explain why F is unique, and to formulate the -j "analogous results" for a binary operation (Exercise 42). Exercises 32. (a) Show that R is symmetric iff R - 1 ~ R. (b) Show that R is transitive iff R 0 R ~ R. 33. Show that R is a symmetric and transitive relation iff R = R - loR. 34. Assume that d transitive relation. is a nonempty set, every member of which is a Is the set nd a transitive relation? Is Ud a transitive relation? (a) (b) 35. Show that for any R and x, we have [X]R = R[{xH 36. Assume that f: A --+ B and that R is an equivalence relation on B. Define Q to be the set {<x, y) E A x A I <f(x), f(y) E R}. Show that Q is an equivalence relation on A. 37. Assume that n is a partition of a set A. Define the relation Rn as follows: xRn y ¢> (3B E n)(x E B & y E B). Show that Rn is an equival |
ence relation on A. (This is a formalized version of the discussion at the beginning of this section.) 38. Theorem 3P shows that AIR is a partition of A whenever R is an equivalence relation on A. Show that if we start with the equivalence relation Rn of the preceding exercise, then the partition AI Rn is just n. 39. Assume that we start with an equivalence relation R on A and define n to be the' partition AIR. Show that Rn, as defined in Exercise 37, is just R. 62 3. Relations and Functions 40. Define an equivalence relation R on the set P of positive integers by mRn ¢> m and n have the same number of prime factors. Is there a function/: PIR --+ PIR such that/([n]R) = [3n]R for each n? 41. Let IR be the set of real numbers and define the relation Q on IR x IR by <u, v)Q<x, y) iff u + y = x + v. (a) Show that Q is an equivalence relation on IR x IR. (b) equation Is there a function G: (IR x IR)/Q --+ (IR x IR)/Q satisfying the G([<x, Y)]Q) = [<x + 2y, y + 2x)]Q? 42. State precisely the" analogous results" mentioned in Theorem 3Q. (This will require extending the concept of compatibility in a suitable way.) ORDERING RELATIONS The first example of a relation we gave in this chapter was the ordering relation {<2, 3), (2, 5), (3, 5)} on the set {2, 3, 5}; recall Fig. 7. Now we want to consider ordering relations on other sets. In the present section we will set forth the basic concepts, which will be useful in Chapter 5. A more thorough discussion of ordering relations can be found. in Chapter 7. Our first need is for a definition. What, in general, should it mean to say that R is an ordering relation on a set A? Well, for one thing R should tell us, given any distinct x and y in A, just which one is smaller. No x should be smaller than itself. And furthermore if x is less than y and y is less than z, then x should be less than z. The following definition captures these ideas. Definition Let A |
be any set. A linear ordering on A (also called a total ordering on A) is a binary relation R on A (i.e., R £:: A x A) meeting the following two conditions: (a) R is a transitive relation; i.e., whenever xRy and yRz, then xRz. (b) R satisfies trichotomy on A, by which we mean that for any x and y in A exactly one of the three alternatives holds. xRy, x =y, yRx Ordering Relations 63 To clarify the meaning of trichotomy, consider first the special case where x and yare the same member of A (with two names). Then trichotomy demands that exactly one of xRx, x = x, xRx holds. Since the middle alternative certainly holds, we can conclude that xRx never holds. Next consider the case where x and yare two distinct members of A. Then the middle alternative x = y fails, so trichotomy demands that either xRy or yRx (but not both). Thus we have proved the following: Theorem 3R Let R be a linear ordering on A. (i) There is no x for which xRx. (ii) For distinct x and y in A, either xRy or yRx. In fact for a transitive relation R on A, conditions (i) and (ii) are equivalent to trichotomy. A relation meeting condition (i) is called irrejlexive; one meeting condition (ii) is said to be connected on A. Note also,that a linear ordering R can never lead us in circles, e.g., there cannot exist a circle such as This is because if we had such a circle, then by transitivity Xl Rx l'contradicting part (i) of the foregoing theorem. Of course" R" is not our favorite symbol for a linear ordering; our favorite is "<." For then we can write "x < y" to mean that the pair <x, y) is a member of the set <. If < is a linear ordering on A and if A is not too large, then we can draw a picture of the ordering. We represent the members of A by dots, placing the dot for x below the dot for y whenever x < y. Then we add vertical lines to connect the dots. The resulting picture has the points of A stretched out along a line, in the |
correct order. (The adjective "linear" reflects the possibility of drawing this picture.) Figure 15 contains three such pictures. Part (a) is the picture for the usual order on {2, 3, 5}. Parts (b) and (c) portray the usual order on the natural numbers and on the integers, respectively. (Infinite pictures are more difficult to draw than finite pictures.) In addition to the concept of linear ordering, there is the more general concept of a partial ordering. Partial orderings are discussed in the first section of Chapter 7. In fact you might want to read that section next, before going on to Chapter 4. At least look at Figs. 43 and 44 there, which contrast with Fig. 15. 64 3. Relations and Functions 1 -2 -3 (a) (b) (e) Fig. 15. Linear orderings look linear. Exercises 43. Assume that R is a linear ordering on a set A. Show that R - 1 is also a linear ordering on A. 44. Assume that < is a linear ordering on a set A. Assume that f: A --+ A and that f has the property that whenever x < y, then f(x) < f(y). Show thatf is one-to-one and that whenever f(x) <f(y), then x < y. 45. Assume that < A and < B are linear orderings on A and B, respectively. Define the binary relation <L on the Cartesian product A x B by: <al' b1 ) <L <a 2, b2 ) iff eithera 1 <A a2 0r (a 1 = a2 & b1 <B b2) Show that < L is a linear ordering on A x B. (The relation < L is called lexicographic ordering, being the ordering used in making dictionaries.) Review Exercises 46. Evaluate the following sets: (a) nn<x, y). (b) nnn{<x, y)r 1. 47. (a) Find all of the functions from {O, 1, 2} into {3, 4}. (b) Find all of the functions from {O, 1, 2} onto {3, 4, 5}. Ordering Relations 65 48. Let T be the set {0, {0}}. (a) Find all of the ordered pairs, if any, in & |
T. (b) Evaluate and simplify: (&Tt 1 0 (&T r {0}) 49. Find as many equivalence relations as you can on the set {O, 1, 2}. 50. (a) Find a linear ordering on {O, 1, 2, 3} that contains the ordered pairs <0, 3) and <2, 1). (b) Now find a different one meeting the same conditions. 51. Find as many linear orderings as possible on the set {O, 1, 2} that contain the pair <2,0). 52. Suppose that A x B = C x D. Under what conditions can we conclude that A = C and B = D? 53. Show that for any sets Rand S we have (R u st 1 = R- 1 u S-l, (R n st 1 = R- 1 n s-l, and (R - st 1 = R- 1 54. Prove that the following equations hold for any sets. - S-l. (a) A x (B n C) = (A x B) n (A x C). (b) A x (B u C) = (A x B) u (A x C). (c) A x (B - C) = (A x B) (A x C). 55. Answer "yes" or "no." Where the answer is negative, supply a counterexample. (a) (b) Is it always true that (A x A) u (B x C) = (A u B) x (A u C)? Is it always true that (A x A) n (B x C) = (A n B) x (A n C)? 56. Answer "yes" or "no." Where the answer is negative, supply a counter example. (a) (b) Is dom(R u S) always the same as dom R u dom S? Is dom(R n S) always the same as dom R n dom S? 57. Answer "yes" or "n6." Where the answer is negative, supply a counterexample. (a) (b) Is R 0 (S u T) always the same as (R 0 S) u (R 0 T)? Is R 0 (S n T) always the same as (R 0 S) n (R 0 T)? 58. Give an example to show that F[F |
-1[S]] is not always the same as S. sets Q r (A n B) = (Q r A) n (Q r B) and 59. Show Q r (A - B) = (Q r A) 60. Prove that for any sets (R 0 S) r A = R 0 (S r A). (Q r B). for any that CHA.PTER 4 NATURAL NUMBERS There are, in general, two ways of introducing new objects for mathe matical study: the axiomatic approach and the constructive approach. The axiomatic approach is the one we have used for sets. The concept of set is one of our primitive notions, and we have adopted a list of axioms dealing with the primitive notions. Now consider the matter of introducing the natural numbers! 0, 1, 2,... for further study. An axiomatic approach would consider "natural number" as a primitive notion and would adopt a list of axioms. Instead we will use the constructive approach for natural numbers. We will define natural numbers in terms of other available objects (sets, of course). In place of axioms for numbers we will be able to prove the necessary properties of numbers from known properties of sets. 1 There is a curious point of terminology here. Is 0 a natural number? With surprising consistency, the present usage is for school books (through high-school level) to exclude 0 from the natural numbers, and for upper-division college-level books to include O. Freshman and sophomore college books are in the transition zone. In this book we include 0 among the natural numbers. 66 Inductive Sets 67 Constructing the natural numbers in terms of sets is part of the process of "embedding mathematics in set theory." The process will be continued in Chapter 5 to obtain more exotic numbers, such as,j2. I NDUCTIVE SETS First we need to define natural numbers as suitable sets. Now numbers do not at first glance appear to be sets. Not that it is an easy matter to say what numbers do appear to be. They are abstract concepts, which are slippery things to handle. (See, for example, the section on "Two" in Chapter 5.) Nevertheless, we can construct specific sets that will serve perfectly well as numbers. In fact this can be done in a variety of ways. In 1908, Zermelo proposed to use 0, {0}, {{0}},... as the natural numbers |
. Later von Neumann proposed an alternative, which has several advantages and has become standard. The guiding principle behind von Neumann's construction is to make each natural number be the set of all smaller natural numbers. Thus we define the first four natural numbers as follows: 0=0, 1 = {0} = {0}, 2 = {O, 1} = {0, {0}}, 3 = {O, 1, 2} = {0, {0}, {0, {0}}}. We could continue in this way to define 17 or any chosen natural number. Notice, for example, that the set 3 has three members. It has been selected from the class of all three-member sets to represent the size of the sets in that class. This construction of the numbers as sets involves some extraneous properties that we did not originally expect. For example, and OE1E2E3E'" Os1s2s3s.. ·. But these properties can be regarded as accidental side effects of the definition. They do no harm, and actually will be convenient at times. Although we have defined the first four natural numbers, we do not yet have a definition of what it means in general for something to be a natural 68 4. Natural Numbers number. That is, we have not defined the set of all natural numbers. Such a definition cannot rely on linguistic devices such as three dots or phrases like "and so forth." First we define some preliminary concepts. Definition For any set a, its successor a+ is defined by a+ =au{a}. A set A is said to be inductive iff 0 E A and it is "closed under successor," i.e., (Va E A) a+ E A. In terms of the successor operation, the first few natural numbers can be characterized as 0=0, 1=0+, 2=0++, 3=0+++,.... These are all distinct, e.g., 0 + "# 0 + + + (Exercise 1). And although we have not yet given a formal definition of "infinite," we can see informally that any inductive set will be infinite. We have as yet no axioms that provide for the existence of infinite sets. There are indeed infinitely many distinct sets whose existence we could establish. But there is no one set having infinitely many members that we can prove to exist. Consequently we cannot yet prove that any inductive set exists. We now correct that fault. Infinity Ax |
iom There exists an inductive set: (3A)[0 E A & (Va E A) a+ E A]. Armed with this axiom, we can now define the concept of natural number. Definition A natural number is a set that belongs to every inductive set. We next prove that the collection of all natural numbers constitutes a set. Theorem 4A There is a set whose members are exactly the natural numbers. Proof Let A be an inductive set; by the infinity axiom it is possible to find such a set. By a subset axiom there is a set w such that for any x, x E W ¢> X E A & x belongs to every other ind uctive set ¢> x belongs to every inductive set. (This proof is essentially the same as the proof of Theorem 2B.) Inductive Sets 69 The set of all natural numbers is denoted by a lowercase Greek omega: X E W ¢> x is a natural number ¢> x belongs to every inductive set. In terms of classes, we have W = n{A I A is inductive}, but the class of all inductive sets is not a set. Theorem 4B W is inductive, and is a subset of every other inductive set. Proof First of all, 0 E W because 0 belongs to every inductive set. And second, a E W => a belongs to every inductive set => a+ belongs to every inductive set => a+ E w. Hence W is inductive. And clearly W is included in every other inductive set. -j Since w is inductive, we know that 0 (= 0) is in w. It then follows that 1 (=0+) is in w, as are 2 (=1+) and 3 (=2+). Thus 0,1,2, and 3 are natural numbers. Unnecessary extraneous objects have been excluded from w, since w is the smallest inductive set. This fact can also be restated as follows. Induction Principle for w Any inductive subset of w coincides with w. Suppose, for example, that we want to prove that for every natural number n, the statement _ n _ holds. We form the set of natural numbers for which the desired conclusion is true. If we can show that T is inductive, then the proof is complete. Such a proof is said to be a proof by induction. The next theorem gives a very simple example of this method. Theorem |
4C Every natural number except 0 is the successor of some natural number. Proof Let T = {n E w I either n = 0 or (3p E w) n = p+}. Then 0 E T. And if k E T, then e E T. Hence by induction, T = w. -j 70 Exercise 4. Natural Numbers See also the Review Exercises at the end of this chapter. 1. Show that 1"# 3, i.e., that 0+ "# 0+ + +. PEANO'S POSTULATES2 In 1889, Peano published a study giving an axiomatic approach to the natural numbers. He showed how the properties of natural numbers could be developed on the basis of a small number of axioms. Although he e S(e)......--..-.. \ j e S(e) S2(e)........... ~........... ~ S4(e) • -----t • S3(e) Fig. 16. Any Peano system must behave like (e). (b) (e) attributed the formulation of the axioms to Dedekind, the axioms are generally known as "Peano's postulates." We will first show that the set w we have constructed satisfies Peano's postulates, i.e., the" postulates" become provable when applied to w. Later we will prove that anything satisfying Peano's postulates is, in a certain specific sense, "just like" w. To formulate these results more accurately, we must define the concept of a Peano system. First of all, if S is a function and A is a subset of. dom S, then A is said to be cl()sed uI).der S iff when-ever x E A, then S(x) E A. (This can equivalently be expressed as S[ A] ~ A.) Define a Peano system to be a triple <N, S, e) consisting of a set N, a function S: N --+ N, and a member e EN such that the following three conditions are met:,\ (i) e ¢ ran S. (ii) S is one-to-one. (iii) Any subset A of N that contains e and is closed under S equals N itself. The condition "e ¢ ran S" rules out loops as in Fig. 16a-here the arrows indicate the action of |
S. And the requirement that S be one-to-one 2 The material in this section on Peano systems is not essential to our later work. But the material on transitive sets is essential. Peano's Postulates 71 rules out the system of Fig. 16b. Consequently any Peano system must, in part, look like Fig. 16c. The last of the three conditions is the Peano induction postulate. Its function is to rule out any points other than the ones we expect. We expect the system to contain e, S(e), SS(e), SSS(e),.... The Peano induction postulate replaces the three dots with a precise set-theoretic condition, stating that nothing smaller than N itself can contain e and be closed under S. First we want to show that w (with the successor operation and 0) is a Peano system. In particular, this will show that some Peano system exists. Let (1 be the restriction of the successor operation to w: Theorem 4D <w, (1, 0) is a Peano system. Proof Since w is inductive, we have 0 E wand (1: w --+ w. The Peano induction postulate, as applied to <w, (1, 0), states that any subset A of w containing 0 and closed under (1 equals w itself. This is just the induction principle for w. Clearly 0 ¢ ran (1, since n+ "# 0. It remains only to show that (1 is one-to-one. For that purpose (and others) we will use the concept of a transitive set. Definition A set A is said to be a transitive set iff every member of a member of A is itself a member of A: X E a E A => X E A. This condition can ·also be stated in any of the following (equivalent) ways: UA£A, a E A => a £ A, A £ &A. At this point we have violated a basic rule-we have defined "transitive" to mean two different things. In Chapter 3 we said that A is transitive if whenever xAy and yAz, then xAz. And now we define A to be a transitive set if a very different condition is met. But both usages of the word are well established. And so we will learn to live with the ambiguity. In practice, the context will make clear which sense of "transitive |
" is wanted. Further more when the concept of Chapter 3 is meant, we will refer to A as a transitive relation (luckily A will be a relation), reserving the phrase "transitive set" for the concept defined above. 72 4. Natural Numbers Example The set {0, {{0}}} is not a transitive set. This is because {0} E {{0}} E {0, {{0}}}, but {0} ¢ {0, {{0}}}· Also {O, 1, 5} 4 E 5 E {O, 1, 5} whereas 4 ¢ {O, 1, 5}. is not a transitive set, since Theorem 4E For a transitive set a, U(a+) = a. Proof We proceed to calculate Ua+: Ua+ = U(a u {aD = Ua u U{a} = Uaua =a. by Exercise 21 of Chapter 2 The last step is justified by the fact that Ua S a for a transitive set a. -j Theorem 4F Every natural number is a transitive set. Proof by induction We form the set of numbers for which the theorem is true; let T = {n E win is a transitive set}. It suffices to show that T is inductive, for then T = w. Trivially ° E T. If k E T, then by the preceding theorem whence e E T. Thus T is inductive. We can now complete the proof of Theorem 4D; it remained for us to show that the successor operation on w is one-to-one. If m+ = n+ for m and n in w, then U(m+) = U(n+). But since m and n are transitive sets, we have U(m+) = m and U(n+) = n by Theorem 4E. Hence m = n. -j Theorem 4G The set w is a transitive set. This theorem can be stated as: Every natural number is itself a set of natural numbers. We will later strengthen this to: Every natural number is the set of all smaller natural numbers. Proof by induction We want to show that ('In E w) n S w. Form the set of n's for which this holds: T = {n E win S w}. Recursion on w 73 We must verify that T is inductive. Clearly 0 E T. If k E T, |
then we have and whereupon k u {k} S; w, thus showing that e E T. And so T is inductive and therefore coincides with w. {k} s; w, -j Exercises 2. Show that if a is a transitive set, then a+ is also a transitive set. 3. (a) Show that if a is a transitive set, then f!J>a is also a transitive set. (b) Show that if f!J>a is a transitive set, then a is also a transitive set. 4. Show that if a is a transitive set, then Ua is also a transitive set. 5. Assume that every member of d is a transitive set. (a) Show that Ud is a transitive set. (b) Show that n d is a transitive set (assuming that d is nonempty). 6. Prove the converse to Theorem 4E: If U(a+) = a, then a is a transitive set. RECURSION ON (0 Consider the following guessing game. Suppose I am thinking of a function h: w -.. A. Possibly I am reluctant to tell you directly what the values ofthis function are. Instead I reveal (i) what h(O) is, and (ii) a function F: A ~ A such that h(n+) = F(h(n)) for all nEW. This then gives away all the information; you can compute successively h(O), h(1) = F(h(O)), h(2) = F(h(l)), and so forth. Now for a harder problem. Suppose we are given a set A, an element a E A, and a function F: A ~ A. How can we show that there exists a function h: w ~ A such that (i) h(O) = a, and (ii) h(n+) = F(h(n)) for each nEW. The preceding paragraph tells how to compute h if it exists. But we now want to prove that there exists a set h that is a function meeting the above conditions. Recursion Theorem on W Let A be a set, a E A, and F: A ~ A. Then there exists a unique function h: W ~ A such that h(O) = a, 74 4. Natural Numbers and for every n in w, |
h(n +) = F(h(n)). Proof 3 The idea is to let h be the union of many approximating functions. For the purposes of this proof, call a function v acceptable iff dom v ~ w, ran v ~ A, and the following conditions hold: If ° E dom v, then v(o) = a. (i) (ii) Ifn+ E dom v (where nEw), then also n E dom v and v(n+) = F(v(n)). Let f be the collection of all acceptable functions, and let h = Ufo Thus (u) < n, y) E h iff < n, y) is a member of some acceptable v iff v(n) = Y for some acceptable V. We Claim that this h meets the demands of the theorem. This claim can be broken down into four parts. The four parts involve showing that (1) h is a function, (2) h is acceptable, (3) dom h is all of w, and (4) h is umque. 1. We first claim that h is a function. (Proving this will, in effect, amount to showing.that two acceptable functions always agree with each other whenever both are defined.) Let S be the set of natural numbers at which there is no more than one candidate for h(n): S = {n E W I for at most one y, <n, y) E h}. We must check that S is inductive. If <0, Yl) E hand <0, Y2) E h, then by (u) there exist acceptable Vl and V2 such that vl(O)=Yl and V2(0)=Y2' But by (i) it follows that Yl = a = Y2' Thus ° E S. Next suppose that k E S; we seek to show that e E S. Toward that end suppose that <e, Yl ) E hand <e, Y2) E h. Again there must exist acceptable v 1 and v 2 such that v 1 (e) = Y 1 and v 2(e) = Y 2. By condition (ii) it follows that Yl = vl (e) = F(Vl (k)) and Y2 = v2(e) = F(v2(k)). But since k E S, we have vl (k) = v |
2(k). (This is because <k, vl (k) <k, v2(k) are in h.) Therefore and Yl = F(vl(k)) = F(v2(k)) = Y2' This shows that e E S. Hence S is inductive and coincides with W. Consequently h is a function. 3 This proof is more involved than ones we have met up to now. In fact, you might want to postpone detailed study of it until after seeing some applications of the theorem. But it is an important proof, and the ideas in it will be seen again (in Chapters 7 and 9). Recursion on w 75 2. Next we claim that h itself is acceptable. We have just seen that h is a function, and it is clear from (u) that dom h s;;; wand ran h s;;; A. First examine (i). If 0 E dom h, then there must be some acceptable v with v(O) = h(O). Since v(O) = a, we have h(O) = a. Next examine (ii). Assume n+ E dom h. Again there must be some acceptable v with v(n+) = h(n+). Since v is acceptable we have nE dom v (and v(n) = h(n)) and h(n+) = v(n+) = F(v(n)) = F(h(n)). Thus h satisfies (ii) and so is acceptable. 3. We now claim that dom h = w. It suffices to show that dom h is inductive. The function {< 0, a)} is acceptable and hence 0 E dom h. Suppose that k E dom h; we seek to show that e E dom h. If this fails, then look at v = h u {<e, F(h(k)))}. Then v is a function, dom v S;;; w, and ran v S;;; A. We will show that v is acceptable. Condition (i) holds since v(O) = h(O) = a. For condition (ii) there are two cases. If n+ E dom v where n+ '# e, then n+ E dom hand v(n+) = h(n+) = F(h(n)) = F(v(n)). The other case occurs if n+ = e. Since the successor operation is one-to-one |
, n = k. By assumption k E dom h. Thus v(e) = F(h(k)) = F(v(k)) and (ii) holds. Hence v is acceptable. But then v S;;; h, so that e E dom h after all. So dom h is inductive and therefore coincides with w. 4. Finally we claim that h is unique. For let hl and h2 both satisfy the conclusion of the theorem. Let S be the set on which hl and h2 agree: S = {n E w I hl (n) = h2(n)}. Then S is inductive; we leave the details of this to Exercise 7. Hence S = wand hl = h2 -j • Examples Let 71. be the set of all integers, positive, negative, and zero: There is no function h: 71. --+ 71. such that for every a E 71., 71. = {..., -1,0, 1,2,... }. h(a + 1) = h(a)2 + 1. (For notice that h(a) > h(a - 1) > h(a - 2) >... > 0.) Recursion on w relies on there being a starting point O. 71. has no analogous starting point. For another example, let F(a) = {: + 1 if a < 0, if a;;::: O. 76 4. Natural Numbers Then there are infinitely many functions h: 71. --+ 71. such that h(O) = 0 and for every a in 71., h(a + 1) = F(h(a)). The graph of one such function is shown in Fig. 17. Digression There is a classic erroneous proof of the recursion theorem that people have sometimes tried (even in print!) to give. The error is easier to analyze if we apply it not to w but instead to an arbitrary Peano system <N, S, e). Given any a in A and any function F: A --+ A, there is a unique function h: N --+ A such that h(e) = a and h(S(x)) = F(h(x)) for each x in N. The erroneous proof of this statement runs as follows: "We apply the Peano induction postulate to dom h. First of all, we are Fig. 17. h(a + 1) is h( |
a) + 1 when h(a) < 0 and is h(a) when h(a) ~ O. told that h(e) = a, and so h is defined at e, i.e., e E dom h. And whenever x E dom h then immediately h(S(x)) = F(h(x)), so S(x) E dom h as well. Hence dom h'is closed under S. It follows (by induction) that dom h = N, i.e., h is defined throughout N." What is wrong? Well, for one thing, the proof talks about the function h before any such function is known to exist. One might think that a little rewording would get around this objection. But no, a closer examin ation of the proof shows that it does not utilize conditions (i) and (ii) in the definition of Peano systems. The recursion theorem is in general false for systems not meeting those conditions, such as the systems of Fig. 16a or Fig. 16b. So any correct proof of recursion absolutely must make use of conditions (i) and (ii), as well as using induction. (Our proof of recursion on w uses these conditions in part 3.) Our first application of the recursion theorem will be to show that any Peano system is "just like" <w, (1, 0). There are other Peano systems; for example, let N be the set {1, 2, 4, 8,... } of powers of 2, let S(n) = 2n, and let e = 1. Then <N, S, e) is a Peano system. The following theorem expresses the structural similarity between this Peano system and <w, (1, 0). J Recursion on w 77 Theorem 4H Let <N, S, e) be a Peano system. Then <w, a, 0) is isomorphic to <N, S, e), i.e., there is a function h mapping w one-to-one onto N in a way that preserves the successor operation and the zero element h(a(n)) = S(h(n)) h(O) = e. Remark The equation h(a(n)) = S(h(n)) (together with h(O) = e) implies that h(l) = S(e), h(2) = S(S( |
e)), h(3) = S(S(S(e))), etc. Thus the situation must be as shown in Fig. 18. 0 • I ',J 2 • • I.. • 1-----lI.. 1 1 1 S I.. • I S(S(e)).. • I S(S(S(e))).. • I S(e)!----lI.. _ ••• ••• Fig. 18. Isomorphism of Peano systems. Proof By the recursion theorem there is a unique function h: w -+ N such that h(O) = e and for all nEW, h(n+) = S(h(n)). It remains to show that h is one-to-one and that ran h = N. To show that ran h = N we use the Peano induction postulate for <N, S, e). Clearly e E ran h. Also for any x E ran h (say x = h(n)) we have S(x) E ran h (since S(x) = h(n+)). Hence by the Peano induction postulate applied to ran h, we have ran h = N. To show that h is one-to-one we use induction in w. Let T = {n E W I for every m in W different from n, h(m) =F h(n)}. First we claim that 0 E T. Any mEW different from 0 must be of the form p+ (by Theorem 4C). And h(p+) = S(h(p)) =F e since e rt ran S. Hence h(O) = e =F h(p+), and consequently 0 E T. Now assume that k E T and consider e. Suppose that h(e) = h(m). Then m =F 0 by the preceding paragraph, so m = p + for some p. Thus S(h(k)) = h(e) = h(p+) = S(h(p)). Since S is one-to-one, this leaves the equation h(k) = h(p). Since k E T, we have k = p. Hence e = p+ = m. This shows that e E T. So T is inductive, and thus coincides with w. Consequently h is one-to-one. -l 78 4. Natural Numbers Theorems 4D and 4H relate the constructive |
approach to the natural numbers and the axiomatic approach. Theorem 4D shows that Peano's postulates are true of the number system we have constructed. And Theorem 4H shows that the number system we have constructed is, "to within isomorphism," the only system satisfying Peano's postulates. Exercises 7. Complete part 4 of the proof of the recursion theorem on w. 8. Let f be a one-to-one function from A into A, and assume that C E A - ranf. Define h: w --+ A by recursion: h(O) = c, h(n+) = f(h(n)). Show that h is one-to-one. 9. Letf be a function from B into B, and assume that A f; B. We have two possible methods for constructing the "closure" C of A under f. First define C· to be the intersection of the closed supersets of A: C· = n{X I A f; Xf; B& f[X]f; X}. Alternatively, we could apply the recursion theorem to obtain the function h for which h(O) = A, h(n+) = h(n) U f[h(n)]. Clearly h(O) f; h(l) f;... ; define C. to be U ran h; in o~her words C. = U h(i). i EW Show that C· = C •. [Suggestion: To show that C· f; C.' show that f[C.] f; C •. To show that C. f; C·, use induction to show that h(n) f; C·.] In Exercise 9, assume that B is the set of real numbers,f(x) = x 2 10. A is the closed interval [1, 1]. What is the set called C· and C.?, and In Exercise 9, assume that B is the set of real numbers,f(x) = x - 1, 11. and A = {O}. What is the set called C· and C.? 12. Formulate an analogue to Exercise 9 for a function f: B x B --+ B. Arithmetic ARITHMETIC 4 79 We can apply the recursion theorem to define addition and multiplication on w. (Another way of obtaining these operations will be discussed in Chapter 6.) For example, suppose |
we want a function As: w --+ w such that As(n) is the result of adding 5 to n. Then As must satisfy the conditions As(O) = 5, As(n+) = As(nt for n in w. The recursion theorem assures us that a unique such function exists. In general, for each mEW there exists (by the recursion theorem) a unique function A : w --+ w for which m A (0) = m, m A (n+) = A (nt m m for n in w. But we want one binary operation +, not all these little one-place functions. Definition A binary operation on a set A is a function from A x A into A. Definition Addition ( + ) is the binary operation on w such that for any m and n in w, Thus when written as a relation, m + n = A (n). m + = {«m, n), p) 1m E w & nEW & p = Am(n)}. In conformity to everyday notation, we write m + n instead of + (m, n) or +«m, n»). Theorem 41 For natural numbers m and n, (AI) (A2) m+O=m, m + n+ = (m + nt. This theorem is an immediate consequence of the construction of A. m Observe that (AI) and (A2) serve to characterize the binary operation + in a recursive fashion. Our only reason for using the A's is that the recursion theorem applies directly to functions with domain w, not domain W x w. We can now forget the A's, and use + and Theorem 41 instead. m We can now proceed to construct the multiplication operation in much the same way. We first apply the recursion theorem to obtain many functions M m: W --+ W where M m(n) is the result Of multiplying m by n. Specifically, m 4 Readers planning to omit Chapter 5 are permitted to skip this section also. 80 4. Natural Numbers for each mEW there exists (by the recursion theorem) a unique function M : W -4 W for which m Definition Multiplication (.) is the binary operation on W such that for any m and n in w, The theorem analogous to Theorem 41 is the following. m' n = M (n). m Theorem 4J For natural numbers m and n, (Ml) |
(M2) m' 0 = 0, m. n+ = m. n + m. We can now discard the M functions, and use. and Theorem 41 m instead. We could, in the same manner, define the exponentiation operation on w. The equations that characterize exponentiation are (El) (E2) mO = 1, m(n+) = mn. m. Example 2 + 2 = 4 (we would be alarmed if this failed), as the following calculation demonstrates: 2+0=2 2 + 1 = 2 + 0+ = (2 + ot = 2+ = 3, 2+2=2+1+ = (2 + 1)+ = 3+ =4. by (AI), by (A2) by (A2) Having now given set-theoretic definitions of the operations of arithmetic, we next verify that some of the common laws of arithmetic are' provable within set theory. This verification is additional evidence, albeit at an elementary level, that mathematics can be embedded in set theory. Arithmetic 81 Theorem 4K The following identities hold for all natural numbers. (1) Associative law for addition {' m + (n + p) = (m + n) + p. (2) Commutative law for addition m + n= n + m. (3) Distributive law m. (n + p) = m. n + m. p. (4) Associative law for multiplication m. (n. p) = (m. n). p. (5) Commutative law for multiplication m' n= n' m. Proof Each part is proved by induction. This exemplifies a general fact: When a function has been constructed by use of the recursion theorem, then general properties of the function must usually be proved by induction. (1) The proof uses induction on p. That is, consider fixed natural numbers m and n, and define A = {p E wi m + (n + p) = (m + n) + p}. We leave the verification that A is inductive for Exercise 15. (2) It is necessary to prove two preliminary facts, each of which is proved by induction. The first preliminary fact is that 0 + n = n for all nEW. Let A = {n E wi 0 + n = n}. Then 0 E A by (AI). Suppose that k EA. Then 0+ e = (0 + k |
t = k+ by (A2) since k E A, and hence e E A. So A IS inductive. The second preliminary fact is that m+ + n = (m + nt for m and n in w. Consider any fixed mEW and let B = {n E wi m+ + n = (m + nt}. Again o E B by (AI). Suppose that k E B. Then m++e=(m++kt =(m+k)++ = (m + e)+ showing that e E B. Hence B is inductive. by (A2) since k E B by (A2), 82 4. Natural Numbers Finally we are ready to prove the commutative law. Consider any nEW and let C = {m E wlm + n = n + m}. By the first preliminary fact, 0+ n = n = n + 0, whence 0 E C. Suppose that k E C. Then e + n = (k + nt = (n + kt by second fact since k E C by (A2), so that e E C. Hence C is inductive. (3) Consider fixed m and n in wand let A = {p E W 1m' (n + p) = m. n + m. p}. To check that 0 E A, observe that m. (n + 0) = m. n =m'n+O =m'n+m'O Now suppose that k E A. Then m' (n + e) = m' (n + kt = m' (n + k) + m =(m'n+m'k)+m = m. n + (m. k + m) =m·n+m·k+ by (AI) by (AI) by (Ml). by (A2) by (M2) since k E A by part (1) by (M2), which shows that e E A. Hence A is inductive. The reader has no doubt observed that each inductive argument here is quite straightforward. And each is, for that matter, much like the next. (4) Consider fixed m and n in wand let A = {p E W 1m' (n. p) = (m' n). p}. To check that 0 E A we note that m' (n ·0) = m ·0= 0 by (Ml |
), and (m' n) ·0= 0 as well. Now suppose that k EA. Then m. (n' e) = m. (n. k + n) = m. (n. k) + m. n = (m. n). k + m. n = (m. n). e by (M2) by part (3) since k E A by (M2), which shows that e EA. Hence A is inductive. (5) The proof here follows the outline of part (2). There are the analogous two preliminary facts to be proved: O' n = 0 and. The details of the,three inductive arguments are left for Exercise 1 16. Ordering on w Exercises 83 13. Let m and n be natural numbers such that m' n = O. Show that either m = 0 or n = O. 14. Call a natural number even if it has the form 2. m for some m. Call it odd if it has the form (2. p) + 1 for some p. Show that e~ch natural number is either even or odd, but never both. 15. Complete the proof of part (1) of Theorem 4K. 16. Complete the proof of part (5) of Theorem 4K. 17. Prove that mn+ p = mn. mP. ORDERING ON (0 We have defined natural numbers in such a way that, for example, 4 E 7. This may have appeared to be a spurious side effect of our definition, but we now want to turn it to our advantage. We have the following strikingly simple definition of order on w: For natural numbers m and n, define m to be less than n iff mEn. We could introduce a special symbol" <" for this: m < n iff mEn. But the special symbol seems unnecessary; we can just use "E". But it will be necessary to keep in mind the dual role of this symbol, which denotes both membership and ordering. In place of an.:::; symbol, we define iff either mEn or m = n. Observe that pEk+ ¢> p~k, a fact we will use in later calculations. We are now entitled to state the following fact: Any natural number is just the set of all smaller natural numbers. That is, for any n in w, x is a member of n ¢> x |
E W & x is less than n. To verify this, note that we can restate it as XEn ¢> xEw&xEn, which is true because W is a transitive set, and thus x EnE W = X E W. We should show that we do indeed have a linear ordering relation on w, in the sense defined in Chapter 3. The relation in question is the set of ordered pairs E w defined by Ew = {<m, n) E W x wi mEn}. 84 4. Natural Numbers We will prove that this is a linear ordering relation on w, i.e., that it is a transitive relation that satisfies trichotomy on w. Because each natural number is a transitive set, we have for m, n, pin w: mEn & n E p = m E p. That is, our ordering relation on w is a transitive relation. It is somewhat harder to show from our definitions that of any two distinct natural numbers, one is larger than the other. For that result, we will need the following lemma. Lemma 4L (a) For any natural numbers m and n, mEn iff m + E n +. (b) No natural number is a member of itself. Proof (a) First assume that m+ E n+. Then we have mE m+ En. Hence (by the transitivity of n) we obtain mEn. To prove the converse we use induction on n. That is, form T = {n E w I ("1m E n) m+ E n+}. Then 0 E T vacuously. Consider any k E T. In order to show that e E T, we must show that whenever mEk+, then m+Ek++. Given mEk+, we have either m = k (in which case m + = k + E k + +) or mE k. In the latter case (since k E T), m+ E k+ s;: k+ +. So in either case we get m+ E k+ + and thus e E T. Hence T is inductive and coincides with w. Part (b) follows easily from (a). Let T = {n E win ¢ n}. Then 0 E T since nothing k ¢ k = e ¢ e. Hence T is inductive and coincides with w. is a member of O. And by part (a), -l |
(In Chapter 7 we will come to the regularity axiom, which implies among other things that no set is a member of itself. But for natural numbers we can get along without the regularity axiom.) We next use the lemma to prove that for two distinct natural numbers, one is always a member of the other. (It is the smaller one that is a member of the larger one.) Trichotomy Law for w For any natural numbers m and n, exactly one of the three conditions mE n, m= n, nEm holds. Ordering on w 85 Proof First note that at most one can hold. If mEn and m = n, then mE m, in violation of Lemma 4L(b). Also if mEn Em, then because m is a transitive set we again have mE m.'It remains to show that at least one holds. For that we use induction; let T = {n E w I (Vm E w)(m En or m = nor n Em)}. In order to show that 0 E T, we want to show that 0 ~ m for all m. This we do by induction on m. (An induction within an induction!) Clearly o ~ 0, and if 0 ~ k, then 0 E k +. Hence 0 E T. Now assume that k E T and consider k+. For any m in w we have (since k E T) either m ~ k (in which case mE e) or k Em. In the latter case k+ E m+ by Lemma 4L(a), and so k+ ~ m. Thus in every case, either mE k+ or k+ = m or k+ Em. And so k+ E T, T is inductive, and we are done. -l A set A is said to be a proper subset of B (A c B) iff it is a subset of B that is unequal to B. A c B ¢> A f; B & A "# B. Ordering on w is given not only by the membership relation, but also by the proper subset relation: Corollary 4M For any natural numbers m and n, and mEn iff men m ~ n iff m f; n. Proof Since n is a transitive set, mEn = mf;n, and the inclusion is proper by Lemma 4L(b). Conversely assume that men. Then m "# n, and |
n ¢ m lest n E n. So by trichotomy mEn and we are done. -l The above proof uses trichotomy in a typical way: To show that mEn, it suffices to eliminate the other two alternatives. The following theorem gives the order-preserving properties of addition and multiplication. The theorem will be used in Chapter 5 (but not elsewhere). Theorem 4N For any natural numbers m, n, and p, mEn ¢> m + pEn + p. If, in addition, p "# 0, then mEn ¢> m' pEn' p. 86 4. Natural Numbers Proof First consider addition. For the "=" half we use induction on p. Consider fixed mEn E wand let A = {p E W I m + pEn + p}. Clearly 0 E A, and kEA = m+kEn+k (m + kt E (n + kt = = m+k+En+k+ = k+ EA. by Lemma 4L(a) by (A2) Hence A is inductive and so A = w. For the "=" half we use the trichotomy law and the "=" half. If m + pEn + p, then we cannot have m = n (lest n + pEn + p) nor n Em (lest n + p Em + pEn + p). The only alternative is mEn. For multiplication the procedure is similar. For the "=" direction, consider fixed mEn E wand let B = {q E W 1m· q+ En' q+}. (Recall that for a natural number p "# 0 there is some q E w with q+ = p.) It is easy' to see that 0 E B, since m. 0+ = m. 0 + m = m. Suppose that k E B; we need to show that m. k+ + En' k+ +. Thus m·k++=m·k++m E n·k+ +m by applying the first part of the theorem to the fact that m. e En' e. And by again applying the first part ofthe theorem (this time to the fact that mE n), n. k+ + mEn' k+ + n =n·k++. Hence e E B, B is inductive, and B = w. The "=" half then follows exactly as for addition. Corollary |
4P The following cancellation laws hold for m, n, and p in w: m+p=n+p = m=n "# 0 = m = n. Proof Apply trichotomy and the preceding theorem. Well Ordering of w Let A be a nonempty subset of w. Then there is some mEA such that m § n for all n E A. Ordering on w 87 Note Such an m is said to be least in A. Thus the theorem asserts that any nonempty subset of w has a least element. The least element is always unique, for if mi and m 2 are both least in A, then mi ~ m2 and m2 ~ mi' i = m 2 Consequently m. Proof Assume that A is a subset of w without a least element; we will show that A = 0. We could attempt to do this by showing that the comple ment w - A was inductive. But in order to show that k+ E w - A, it is not enough to know merely that k E w - A, we must know that all numbers smaller than k are in w - A as well. Given this additional information, we can argue that k+ E W - A lest it be a least element for A. To write down what is approximately this argument, let B = {m E W I no number less than m belongs to A}. We claim that B is inductive. 0 E B vacuously. Suppose that k E B. Then if n is less than e, either n is less than k (in which case n ¢ A since k E B) or n = k (in which case n ¢ A lest, by trichotomy, it be least in A). In either case, n is outside of A. Hence e E B and so B is inductive. It clearly follows that A = 0; for example, 7 ¢ A because 8 E B. -l Corollary 4Q There is no functionf: w->w such thatf(n+) Ef(n) for every natural number n. Proof The range off would be a nonempty subset of w without a least -l element, contradicting the well ordering of w. Our proof of the well ordering of w suggests that it might be useful to have a second induction principle. Strong Induction Principle for w Let A be a subset of w, and assume that for every n in w, if every number less than n is in A, then |
n EA. Then A = w. Proof Suppose, to the contrary, that A "# w. Then w - A "# 0, and by the well ordering it has a least number m. Since m is least in w - A, all numbers less than m are in A. But then by the hypothesis of the theorem mEA, contradicting the fact that mEW - A. -l The well-ordering principle provides an alternative to proofs by induction. Suppose we want to show that for every natural number, a certain statement holds. Instead of forming the set of numbers for which the statement is true, consider the set of numbers for which it is false, i.e., the set C of counterexamples. To show that C = 0, it suffices to show that C cannot have a least element. 88 Exercises 18. Simplify: E: 1[{7. 8}]. 4. Natural Numbers 19. Prove that if m is a natural number and d is a nonzero number. then there exist numbers q and r such that m = (d. q) + rand r is less than d. 20. Let A be a nonempty subset of w such that U A = A. Show that A = w. 21. Show that no natural number is a subset of any of its elements. 22. Show that for any natural numbers m and p we have mE m + p+. 23. Assume that m and n are natural numbers with m less than n. Show that there is some p in w for which m + p + = n. (It follows from this and the preceding exercise that m is less than n iff (3p E w) m + p+ = n.) 24. Assume that m + n = p + q. Show that 25. Assume that n E m and q E p. Show that mE p ¢> q En. (m. q) + (n. p) E (m. p) + (n. q). [Suggestion: Use Exercise 23.] 26. Assume that nEW and I : n+ --+ w. Show that ran I has a largest element. 27. Assume that A is a set. G is a function. and 11 and 12 map w into A. Further assume that for each n in w both/l ~ nand 12 ~ n belong to dom G and Show that 11 = 12. 28. Rewrite the proof of The |
orem 4G using. in place of induction. the well ordering of w. Review Exercises 29. Write an expression for the set 4 using only symbols 0. {. }. and commas. 30. What is U4? What is n4? 31. What is UU7? 32. (a) Let A = {I}. Calculate A+ and U(A+). (b) What is U({2} +)? Ordering on w 89 33. Which of the following sets are transitive? (For each set S that is not transitive, specify a member of Us not belonging to S.) (a) {a, 1, {I}}. (b) {I}. (c) <0, I). 34. Find suitable a, b, etc. making each of the following sets transitive. (a) (b) {{{0}}, a, b}. {{{{0}}}, c, d, e}. 35. Let S be the set (1,0). (a) Find a transitive set Tl for which S ~ T. 1 (b) Find a transitive set T2 for which S E T2. 36. There is a function h: w -> w for which h(O) = 3 and h(n+) = 2· h(n). What is h(4)? 37. We will say that a set S has n elements (where nEw) iff there is a one-to one function from n onto S. Assume that A has m elements and B has n elements. (a) Show that if A and B are disjoint, then A u B has m + n elements. (b) Show that A x B has m. n elements. 38. Assume that h is the function from w into w for which h(O) = 1 and h(n+) = h(n) + 3. Give an explicit (not recursive) expression for h(n). 39. Assume that h is the function from w into w for which h(O) = 1 and h(n +) = h(n) + (2. n) + 1. Give an explicit (not recursive) expression for h(n). 40. Assume into w defined by the function h(n) = 5. n + 2. Express h(n+) in terms of h(n) as simply as possible. from w that |
h is CHAPTER 5 CONSTRUCTION OF THE REAL NUMBERS 1 In Chapter 4 we gave a set-theoretic construction of the set w of natural numbers. In the present chapter we will continue to show how mathematics can be embedded in set theory, by giving a set-theoretic construction of the real numbers. (The operative phrase is "can be," not "is" or "must be." We will return to this point in the section on "Two.") INTEGERS First we want to extend our set w of natural numbers to a set 71. of integers (both positive and negative). Here "extend" is to be loosely interpreted, since w will not actually be a subset of 71.. But 71. will include an "isomorphic copy" of w (Fig. 19). A negative integer can be named by using two natural numbers and a subtraction symbol: 2 - 3, 5 - 10, etc. We need some sets to stand behind these names. As a first guess, we could try taking the integer -1 to be the pair <2, 3) of natural numbers used to name -1 in the preceding paragraph. And 1 Other chapters do not depend on Chapter 5. 90 Integers 91 similarly we could try letting the integer - 5 be the pair <5, 10) of natural numbers. But this first guess fails, because - 1 has a multiplicity of names: 2 - 3 = 0 - 1 but <2, 3) =F <0,1). As a second guess, we can define an equivalence relation "" such that <2, 3) "" <0, 1). (Imposing such an equivalence relation is sometimes described as "identifying" <2, 3) and <0, 1).) Then we will have the one equivalence class [<2,3)] = [<Q, 1)], and we can take - 1 to be this equivalence class. Then for the set 7L of all integers, we can take the set of all equivalence classes: 7L = (w x w)/ "". This is in fact what we do. Call a pair of natural numbers a difference; then an integer will be an equivalence class of differences. Consider two differences <m, n) and <p, q). When should we call them equivalent? 0 • • Oz • • Iz 2 • • 2z 3 • • 3z OJ 7L • -2t • -I z Fig. |
19. There is a subset of 7L that looks like OJ. Informally, they are equivalent iff m - n = p - q, but this equation has no official meaning for us yet. But the equation is equivalent to m + q = p + n, and the latter equation is meaningful. Consequently we formulate the following definition. Definition Define "" to be the relation on w x w for which <m, n) "" <p, q) iff m + q = p + n. Thus"" is a set of ordered pairs whose domain and range are also sets of ordered pairs. In more explicit (but less readable) form, the above definition can be stated; "" = {«m, n), <p, q» 1m + q = p + n and all are in w}. Theorem 5ZA The relation "" is an. equivalence relation on w x w. Proof We leave it to you to check that"" is reflexive on w x wand is symmetric. To show transitivity, suppose that <m, n) "" <p, q) and <p, q) "" <r, s). Then (by the definition of "" ). 92 5. Construction of the Real Numbers By use of the cancellation law (Corollary 4P), we obtain m + s = r + n, and thus <m, n) "" <r, s). -l Definition The set 7L of integers is the set (w x w)/ "" of all equivalence classes of differences. For example, the integer 2z is the equivalence class [<2, 0)] = {<2, 0), <3, 1), <4, 2),... }, and the integer - 3z is the equivalence class [<0,3)] = {<O, 3), <1,4), <2, 5),... }. These equivalence classes can be pictured as 45° lines 10 the Cartesian product w x w (Fig. 20). Next we want to endow 7L with a suitable addition operation. Informally, we can add differences: (m - n) + (p - q) = (m + p) - (n + q). This indicates that the correct addition function + z for integers will satisfy the equation [<m, n)] +z[<p, q)] = [<m + p, n + q)J. This equation will serve to define + z' once |
we have verified that it makes sense. The situation here is of the sort discussed in Theorem 3Q. We want to specify the value of the operation + z at a pair of equivalence classes by (1) selecting representatives <m, n) and <p, q) from the classes, (2) -4 o 3 Fig. 20. An integer is a line in OJ x OJ. Integers 93 / / / / / / / / /! <m, n) / 1.......-"""'-.......- <m + p, n + q) i 1 1.....-: 1 1 / 1 /.......-1 1 1 <p, q) Fig. 21. Addition of lines is well defined, operating on the representatives (by vector addition in this case), and then (3) forming the equivalence class of the result of the vector addition, For + z to be well defined, we must verify that choice of other representatives <m', n') and <p', q') from the given classes would yield the same equivalence class for the ~um (Fig, 21), Lemma 5ZB If <m, n) '" <m', n') and <p, q) '" <p', q'), then <m + p, n + q) '" <m' + p', n' + q'), Proof We are given, by hypothesis, the two equations p + q' = p' + q, m + n' = m' + n and We want to obtain the equation m + p + n' + q' = m' + p' + n + q, But this results from just adding together the two given equations, -l This lemma justifies the definition of + z' In the terminology of Theorem 3Q, it says that the function F of vector addition F«m, n), <p, q») = <m + p, n + q) 94 5. Construction of the Real Numbers is compatible with quotient set; P is just our operation + z' It satisfies the equation Hence there is a well-defined function P on the [<m, n)] +z[<p, q)] = [<m + p, n + q)]. In other words, for integers a and b our addition formula is a +zb = [<m + p, n + q)], where <m, n) is chosen from a and <p |
, q) is chosen from b. Theorem 3Q assures us that the equivalence class on the right is independent of how these choices are made. Example We can calculate 2z + z (- 3z )· Since 2z = [(2, 0)] and -3z = [<0, 3)], we have 2z + z (- 3z ) = [(2,0)] + z [<0,3)] = [(2 + 0, ° + 3)] = [(2, 3)] = -lz· The familiar properties of addition, such as commutatlVlty and associativity, now follow easily from the corresponding properties of addition of natural numbers. Theorem 5ZC The operation + z is commutative and associative: a +zb = b +za, (a +zb) +zc = a +z(b +zc). Proof The integer a must be of the form [<m, n)] for some natural numbers m and n; similarly b is [<p, q) 1 Then: a +zb = [<m, n)] +z[<p, q)] = [<m + p, n + q)] = [<p + m, q + n)] = [<p, q)] +z[<m, n)] = b +za. by definition of +z by commutativity of + on w The calculation for associativity is similar (Exercise 4). Let 0z = [<0, 0)]. Then it is straightforward to verify that a + z 0z = a for any integer a, i.e., 0z is an identity element for addition. And the new feature that 7L has (and the feature for which the extension from w to 7L was made), is the existence of additive inverses. Integers 95 Theorem 5ZD (a) Oz is an identity element for +z: a +zOz = a for any a in 71.. (b) Additive inverses exist : For any integer a, there is an integer b such that a +zb = Oz. Proof (b) Given an integer a, it must be of the form [<m, n)]. Take b to be [<n, m)]. Then a +zb = [<m + n, n + m)] = [<0, 0)] = Oz· -l Theorems 5ZC and 5ZD together say that |
71. with the operation + z and the identity element Oz is an Abelian group. The concept of an Abelian group is central to abstract algebra, but in this book the concept will receive only passing attention. Inverses are unique. That is, if a + z b = Oz and a + z b' = 0z' then b = b'. To prove this, observe that b = b +z(a +zb') = (b +za) +zb' = b'. (This proof works in any Abelian group.) The inverse of a is denoted as - a. Then as the proof to Theorem 5ZD shows, -[<m, n)] = [<n, m)]. Inverses provide us with a subtraction operation, which we define by the equation ). We can also endow the set 71. with a multiplication operation, which we obtain in much the same way as we obtained the addition operation. First we look at the informal calculation with differences (mq + np), (m - n). (p - q) = (mp + nq) - which tells us that the desired operation 'z will satisfy the equation [<m, n)] 'z[<P' q)] = [<mp + nq, mq + np)]. (Here we write, as usual, mp in place of m. p.) Again we must verify that the above equation characterizes a well-defined operation on equivalence classes. That is, we must verify that the operation on differences G«m, n), <p, q») = <mp + nq, mq + np) is compatible with "". This verification is accomplished by the following lemma. 96 (1) (2) 5. Construction of the Real Numbers Lemma 5ZE If <m, n) "" <m', n') and <p, q) "" <p', q'), then <mp + nq, mq + np) "" <m'p' + n'q', m'q' + n'p'). Proof We are given the two equations m + n' = m' + n, p + q' = p' + q, and we want to obtain the equation mp + nq + m'q' + n'p' = m'p' + n'q' + mq + np. The idea is take multiples of (1) and (2) that contain the |
terms we need. First multiply Eq. (1) by p; this gives us an mp on the left and an np on the right. Second, multiply the reverse of Eq. (1) by q; this gives us an nq on the left and an mq on the right. Third, multiply Eq. (2) by m'. Fourth, multiply the reverse of Eq. (2) by n'. Now add the four equations we have obtained from (1) and (2). All the unwanted terms cancel, and we are left with the desired equation. It works. -l As for addition, we can prove the basic properties of multiplication from the corresponding properties of multiplication of natural numbers. Theorem 5ZF The multiplication operation 'z is commutative, associa tive, and distributive over + z: a 'zb = b 'za (a 'zb) 'zc = a 'z(b 'zc) a 'z (b + z c) = (a 'z b) + z (a 'z c) Proof Say that a = [<m, n)] and b = [<p, q)]. For the commutative law, we have whereas a 'zb = [<mp + nq, mq + np)], b 'za = [<pm + qn, pn + qm)]. The equality of these two follows at once from the commutativity of addition and multiplication in w. The other parts of the theorem are proved by the same method. Say that c = [<r, s)]. Then (a 'z b) 'z c is [<mp + nq)r + (mq + np)s, (mp + nq)s + (mq + np)r)], where a 'z (b 'zc) is [<m(pr + qs) + n(ps + qr), m(ps + qr) + n(pr + qs)]. Integers 97 The equality of these follows from laws of arithmetic in w (Theorem 4K). As for the distributive law, when we expand a 'z (b + z c), we obtain [<m(p + r) + n(q + s), m(q + s) + n(p + r)], whereas when we expand a 'z b + z a 'z c we obtain [<mp + nq |
+ mr + ns, mq + np + ms + nr)]. Again equality is clear from laws of arithmetic in w. The remaining properties of multiplication that we will need constitute the next theorem. Let lz be the integer [<1,0)]. Theorem 5ZG (a) The integer lz is a mUltiplicative identity element: a'z lz = a for any integer a. (b) Oz =F lz· (c) Whenever a 'z b = 0z' then either a = Oz or b = Oz· Part (c) is sometimes stated: There are no "zero divisors" in 71.. Proof Part (a) is a trivial calculation. For part (b) it is necessary to check that <0, 0) + <1, 0). This reduces to checking that 0 =F 1 in w, which is true. For part (c), assume that a =F Oz and b =F Oz; it will suffice to prove that a 'z b =F Oz· We know that for some m, n, p, and q: b = [<p, q)], a = [<m, n)], a 'zb = [<mp + nq, mq + np)]. Since a =F [<0,0)], we have m =F n. So either mEn or n Em. Similarly, either p E q or q E p. This leads to a total of four cases, but in each case we have mp + nq =F mq + np by Exercise 25 of Chapter 4. Hence a 'z b =F [<0,0)]. In algebraic terminology, we can say that 71. together with + z' 'z' 0z' and lz forms an integral domain. This means that: (i) 5ZD). 71. with + z and Oz forms as Abelian group (Theorems 5ZC and (ii) Multiplication is commutative and associative, and is distributive over addition (Theorem 5ZF). (iii) lz is a multiplicative identity (different from 0z)' and no zero divisors exist (Theorem 5ZG). 98 5. Construction of the Real Numbers There is a summary of these algebraic concepts near the end of this chapter. The value of the concepts stems from the large array of structures that satisfy the various conditions. In this book, however |
, we are concerned with only the most standard cases. Example The calculation [<0,1)] A<m, n)] = [<n, m)] shows that -1 z 'za = -a. Next we develop an ordering relation < z on the integers. The informal calculation m-n<p-q ffi m+q<p+n indicates that ordering < z on 71. should be defined by [<m,n)] <z[<p,q)] iff m+qEp+n. As usual, it is necessary to check that this condition yields a well-defined relation on the integers. That is, we want to define iff m + q E P + n, a < z b where m, fl, p, and q are chosen so that a = [<m, n)] and b = [<p, q)]. But that choice can be made in infinitely many ways; we must verify that we have the same outcome each time. The following lemma does just this. Lemma 5ZH If <m, n) "" <m', n') and <p, q) "" <p', q'), then m + q E P + n iff m' + q' E p' + n'. Proof The hypotheses give us the equations m + n' = m' + n and In order to utilize these equations in the inequality m + q E P + n, we add n' and q' to each side of this inequality: p + q' = p' + q. m+qEp+n - m+q+~+ifEp+n+~+if - ~+n+q+ifE~+q+n+~ - m' + q' E p' + n'. Here the first and third steps use Theorem 4N, while the middle step uses the given equations. -l Theorem 5ZI The relation < z is a linear ordering relation on the. set of integers. Integers 99 Proof We must show that < z is a ~ransitive relation that satisfies trichotomy on 71.. To prove transitivity, consider integers a = [<m, n)], b = [<p, q)], and C = [<r, s)J. Then a<zb&b<zc = m+qEp+n&p+sEr+q = m+q+sEp+n+s&p+s |
+nEr+q+n = m+q+sEr+q+zc. by Theorem 4N Proving trichotomy is easy. To say that exactly one of holds is to say that exactly one of a = b, m+qEp+n, m+q=p+n, p+nEm+q holds. Thus the result follows from trichotomy in w. An integer b is called positive iff Oz < z b. It is easy to check that b <zOz iff Oz <z -b. Thus a consequence of trichotomy is the fact that for an integer b, exactly. one of the three alternatives b is positive, b is zero, - b is positive holds. The next theorem shows that addition preserves order, as does multiplica tion by a positive integer. (The corresponding theorem for w was Theorem 4N.) Theorem 5ZJ The following are valid for any integers a, b, and' c: (a) a<zb-a+zc<zb+zc. (b) If Oz < z c, then a <zb - a 'zc <zb ·zC. Proof Assume that a, b, and care [<m, n)], [<p, q)], and [<r, s)], respectively. The result to be proved in part (a) then translates to the following statement about natural numbers. This is an immediate consequence of the fact that addition in w preserves order (Theorem 4N). 100 5. Construction of the Real Numbers Part (b) is similar in spirit. As in Theorem 4N, it suffices to prove one direction: This translates to = mr + ns + ps + qr E pr + qs + ms + nr. This is not as bad as it looks. If we let k = m + q and I = p + n, then it becomes S E r & k E I = kr + Is E ks + Ir. This is just Exercise 25 of Chapter 4. -l Corollary 5ZK For any integers a, b, and c the cancellation laws hold, a 'z c = b 'z c & c "# Oz = a = b. Proof This follows from the preceding theorem in the same way that the cancellation laws in w (Corollary 4P) followed from the order preserving properties (Theorem 4N). -l Although w |
is not actually a subset of 71., nonetheless 71. has a subset that is "just like" w. To make this precise, define the function E: w -+ 71. by E(n) = [<n, 0)]. For example, E(O) = Oz and E(1) = 1z. The following theorem, in algebraic terminology, says that E is an "isomorphic embedding" of the system <w, +, " Ew) into the system < 71., + z' 'z' < z)· That is, E is a one-to-one function that preserves addition, multiplication, and order. Theorem 5ZL E maps w one-to-one into 71., and satisfies the following properties for any natural numbers m and n: (a) E(m + n) = E(m) + z E(n). (b) E(mn) = E(m) 'z E(n). (c) mEn iff E(m) < z E(n). Proof To show that E is one-to-one we calculate E(m) = E(n) = [<m, 0)] = [<n, 0)] = <m,O) "" <n, 0) = m=n. Parts (a), (b), and (c) are proved by routine calculations (Exercise 8). -l Rational Numbers 101 Finally we can give' a precise counterpart to our motivating guideline that the difference <m, n) should name m - n. For any m and n, [<m, n)] = E(m) - E(m) as is verified by evaluating the right side of this equation (Exercise 9). Henceforth we will streamline our notation by omitting the subscript "Z" on +z, 'Z' <z, 0Z' lz' etc. Furthermore a' b will usually be written as just abo Exercises 1. Is there a function F: 7l. --+ 7l. satisfying the equation F([<m, n)]) = [<m + n, n)]? 2. Is there a function G: 7l. --+ 7l. satisfying the equation G([ <m, n)]) = [<m, m)]? 3. Is there a function H: 7l. --+ 7l. satisfying the equation H([ <m, n)]) |
= [<n, m)]? 4. Prove that + z is associative. (This is part of Theorem 5ZC.) 5. Give a formula for subtraction of integers: [<m, n)] - [<p, q)] =? 6. Show that a ·z Oz = Oz for every integer a. 7. Show that a'z(-b)=(-a)'zb= -(a'zb) for all integers a and b. 8. Prove parts (a), (b), and (c) of Theorem 5ZL. 9. Show that [<m, n)] = E(m) - E(n) for all natural numbers m and n. RATIONAL NUMBERS We can extend our set 7l. of integers to the set iQl of rational numbers in much the same way as we extended w to 7l.. In fact, the extension from 7l. to iQl is to mUltiplication what the extension from w to 7l. is to addition. In the integers we get additive inverses, i.e., solutions to the equation a + x = O. 102 5. Construction of the Real Numbers In the rationals we will get multiplicative inverses, i.e., solutions to the equation r 'Q x = lQ (for nonzero r). We can name a rational number by using two integers and a symbol for division: 1/2, - 3/4, 6/12. But as before, each number has a multiplicity of names, e.g., 1/2 = 6/12. So the name" 1/2" must be identified with the name" 6/12." By a fraction we mean an ordered pair of integers, the second com ponent of which (called the denominator) is nonzero. For example, <1, 2) and <6, 12) are fractions; we want a suitable equivalence relation"" for which <1, 2) "" <6, 12). Since alb = c/d iff a. d = c. b, we choose to {O} of nonzero integers. Then define "" as follows. Let 71.' be the set 71. 71. x 71.' is the set of all fractions. Definition Define "" to be the binary relation on 71. x 71.' for which <a, b) "" <c, d) iff a' d = |
c. b. The set iQl of rational numbers is the set (71. x 71.')/ "" of all equivalence classes of fractions. We use the same symbol "",," that has been used previously for other equivalence relations, but as we discuss only one equivalence relation at a time, no confusion should result. For example, <1, 2) "" <6, 12) since 1. 12 = 6. 2. The equivalence class [<1, 2)] is the rational number" one-half." The rationals zero and one are and These are distinct, because <0, I) + <1, I). Of course we must check that "" is indeed an equivalence relation. lQ = [<1, 1)]. Theorem SQA The relation "" is an equivalence relation on 71. x 71.'. Proof You should verify that the relation is reflexive on 71. x 71.' and is symmetric. As for transitivity, suppose that <a, b) "" <c, d) and <c, d) "" <e.!). Then ad = cb and cf = ed. Multiply the first equation by f and the second by b to get adf= cbf and cfb = edb. From this we conclude that adf = edb and hence (by canceling the nonzero d) af = eb. This tells us that <a, b) "" <e.!). -1 Rational Numbers 103 We can picture the equivalence classes as nonhorizontal lines (in the "plane" 71. x 71.') through the origin (Fig. 22). The fraction (1,2) lies on the line with slope 2; in general, [<a, b)] is the line with slope b/a. We arrive at addition and multiplication operations for iQl by the same methods used for 71.. For addition, the informal calculation a c --+ -= b d ad + cb bd indicates that + Q should be defined by the equation [<a, b)] +Q[<c, d)] = [<ad + cb, bd)J. Note that bd "# 0 since b "# 0 and d "# O. Hence <ad + cb, bd) is a fraction. As usual, we must check that there is a well-defined function + Q on equivalence classes that satisfies the above equation. The following lemma, together with Theorem |
3Q, does just that/ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Fig. 22. Rational numbers are nonhorizontal lines. 104 5. Construction of the Real Numbers Lemma 5QB If <a, b) "" <a', b') and <c, d) "" <c', d'), then <ad + cb, bd) "" <a'd' + c'b', b'd'). Proof We are given the equations ab' = a'b and cd' = c'd. We want the equation (ad + cb)b'd' = (a'd' + c'b')bd, which, when expanded (with the factors in alphabetic order), becomes ab'dd' + bb'cd' = a'bdd' + bb'c'd. T T T T This clearly is obtainable from the given equations. Example Just to be on the safe side, we will check that 2 + 2 = 4 in 0. Let 2Q = [(2,1)] and 4Q = [<4,1)]. Then 2Q +Q2Q = [(2,1)] +Q[<2, 1)] = [(2 + 2,1)] = [<4, 1)] = 4Q, where we use the fact that 2 + 2 = 4 in 71.. The rationals with + Q and OQ also form an Abelian group: Theorem 5QC (a) Addition + Q is associative and commutative: (q +Qr) +Qs = q +Q(r +Qs), r +Qs = S + Qr. (b) OQ is an identity element for + Q: r +QOQ = r for any r in 0. (c) Additive inverses exist: For any r in 0 there is an S in 0 such that r + Q S = OQ. Proof First we verify commutativity. On the one hand, [ < a, b)] + Q [ < c, d)] = [<ad + cb, bd)],. and on the other [ < c, d)] + Q [<a, b)] = [< cb + ad, db)]. Rational Numbers 105 But the right sides of these two equations are equal |
, by known commutative laws for arithmetic in 71.. The verification of associativity is similar. Consider three rational numbers [<a, b)], [<c, d)], and [<e,f)]. Then one grouping for the sum is ([<a, b)] +Q[<c, d)]) +Q[<e,f)] = [<ad + cb, bd)] +Q[<e,f)] = [«ad + cb)f + e(bd), (bd)f)] = [<adf + cbf + ebd, bdf)]. The same expansion for the other grouping is [<a, b)] + Q ([ < c, d)] + Q [ < e,f)]) = [<a, b)] + Q [ < cf + ed, df) ] = [<a(df) + (cf + ed)b, b(df)] = [<adf + cfb + edb, bdf)], which agrees with the first calculation. Part (b) is a routine calculation. We know that r = [<a, b)] for some integers a and b. Then r +QOQ = [<a, b)] +Q[<O, 1)] = [<a' 1 + 0. b, b. 1)] = [<a, b)] = r. Finally for part (c) we select (with r as above) s = [< - a, b)]. Then it is easy to calculate that r + Q S = [< a, b)] + Q [ < - a, b)] = [< ab + ( - a )b, bb)] = [<0, bb)] = 0Q' since <0, bb) "" <0, 1). As in any Abelian group, the inverse of r is unique; we denote it as - r. The above proof shows that - [<a, b)] = [< -a, b)]. For rational numbers, mu.1tiplication is simpler than addition. The informal calculation ac a c - - -= bd b d indicates that 'Q should be defined by the equation [<a, b)]'Q [<c, d)] = [<ac, bd)]. (Notice the close analogy with + z.) This multiplication function is well defined, as the following lemma verifies. 106 5, Construction of |
the Real Numbers Lemma 5QD If <a, b) "" <a', b') and <c, d) "" <c', d'), then <ac, bd) "" <a'c', b'd'), Proof The proofis exactly as in Lemma 5ZB, but with addition replaced -l by multiplication, Example Recall that 1Q = [0, 1)]. We can now check that 1Q is a that r 'Q 1Q = r, We know that r = multiplicative identity element, i,e" [<a, b)] for some a and b, Thus r 'Q 1Q = [<a, b)]'Q [0,1)] =[<a'1,b'1)] = [<a, b)] = r, You should also verify that r'Q OQ = 0Q' Theorem 5QE Multiplication of rationals is associative, commutative, and distributive over addition: (p 'Q q) 'Q r = P 'Q (q 'Q r), q 'Q r = r 'Q q, P 'Q (q + Q r) = (p 'Q q) + Q (p 'Q r), Proof The verification of associativity and commutativity is directly analogous to verification of the same properties for + z'We will proceed to prove the distributive law, We know that we can write P = [<a, b)], q = [<c, d)], and r = [<e,f)] for some integers a, b, c, d, e, and! Then P'Q (r +Qs) = [<a, b)] 'Q ([<c, d)] +Q[<e,f)]) = [<a, b)] 'Q [<cf + ed, df)] = [< acf + aed, bdf)]. On the other side of the expected equation we have (p 'Q r) +Q (P'Q s) = ([<a, b)]'Q [<c, d)]) +Q ([<a, b)] 'Q [<e,f)]) = [<ac, bd)] +Q[<ae, bf)] = [<acbf + aebd, bdbf)]. This agrees with the first calculation because <i, j) "" <bi, bj), The new |
property the rationals have (and that integers lack) is the existence of multiplicative inverses, Rational Numbers 107 Theorem 5QF For every nonzero r in 0 there is a nonzero q in 0 such that r 'Q q = lQ' Proof The given r must be of the form [<a, b)], where a"# 0, lest -l r = 0Q' Let q = [<b, a)J. Then q"# OQ and r 'Qq = [<ab, ab)] = lQ' We can use the existence of multiplicative inverses to show that there are, no zero divisors in 0: Corollary 5QG If rand s are nonzero rational numbers, then r 'Q s is also nonzero, Proof The preceding theorem provides us with rationals r' and s' for which r 'Q r' = s 'Q s' = lQ' Hence (r 'Q s) 'Q (r' 'Q s') = lQ by using commutative and associative laws, But this implies that r 'Q S "# 0Q' because OQ 'Q (r' 'Q s') = OQ "# lQ' -1 We can restate this corollary by saying that the set of nonzero rational numbers is closed under multiplication; i,e" the product of numbers in this set is again in this set. As a result of the foregoing theorems, we can assert that the nonzero rationals with multiplication form an Abelian group, That is, multiplication gives us a binary operation on the nonzero rationals that is associative and commutative, we have an identity element lQ' and we have multiplicative inverses, As in any Abelian group, the inverse of r is unique; we denote it as r- 1, The proof of Theorem 5QF shows that [<a, b)]-l = [<b, a)J. Inverses provide us with a division operation: For a nonzero rational r we can define, S -;- r = S 'Q r -1, Then we have [ < c, d)] + [< a, b)] = [< c, d)] 'Q [ < b, a)] = [<cb, da)], a version of the "invert and multiply" rule for division of fractions, The algebraic concept exempl |
ified by the rational numbers is the concept of a field, To say that < 0, + Q' 'Q' 0Q' lQ) is a field means that it is an integral domain with the further property that multiplicative inverses exist. (Other examples of fields are provided by the real numbers and by the complex numbers,) The method we have used to extend from 71. to 0 can be applied to extend any integral domain to a field, 108 5. Construction of the Real Numbers Next we want to define the ordering relation for the rational numbers. The informal calculation ~ <~. iff ad < cb b d is correct if band d are positive. There is no guarantee that denominators are always positive. But because [ < a, b)] = [< - a, - b) ], every rational number can be represented by some fraction with a positive denominator. (Recall that for nonzero b, either b or - b is positive.) The above informal calculation then suggests that we define < Q so that [<a, b)] <Q[<c, d)] iff ad < cb whenever band d are positive. As with < z, we must verify that this condition yields a well-defined relation. The following lemma accomplishes the verification. Lemma 5QH Assume that <a, b) ~ <a', b') and <c, d) ~ <c', d'). Further assume that b, b', d, and d' are all positive. Then, ad < cb iff dd' < c'b'. Proof The proof is the same as the proof of Lemma 5ZH, but with -l multiplication of integers in place of addition of naturai numbers. This lemma guarantees that when we test to see whether or not r < Q S, it does not matter which fractions with positive denominators we choose from rand s. Example To check that OQ < Q 1Q, we choose fractions <0, 1) E OQ and (1, 1) E 1Q. Then since 0. 1 < 1. 1, we do indeed have OQ < Q 1Q. But we could also have chosen fractions <0, 4) E OQ and 0, 3) E 1Q. Then since o. 3 < 3. 4, we again find, in consistency with the first calculation, that OQ < Q 1Q. Theorem 5Q I The relation < Q is a linear |
ordering on 0. Proof The proof is the same as the proof of Theorem 5ZI, with multi plication in place of addition. For example, to prove trichotomy, we consider rational numbers rand s. For suitable integers we can write r = [<a, b)] and s = [<c, d)], where band d are positive. Then trichotomy for 7L tells us that exactly one of ad < cb, ad = cb, cb < ad Rational Numbers 109 holds, whence exactly one of r <QS, r = S, holds. One can check that r < Q OQ iff OQ < ~ - r (Exercise 12). Call q positive iff OQ < Q q. Then as a consequence of tnchotomy, we have the fact for any rational number r, exactly one of the three alternatives r is positive, r is zero, - r is positive holds. We can define the absolute value I r I of r by - r is positive, if otherwise. Then OQ :s; Q I r I for every r. Next we prove that order is preserved by addition and by multiplication by a positive factor. Theorem 5QJ Let r, s, and t be rational numbers. (a) r<Qsiffr+Qt<Qs+Qt. If t IS positive, then (b) r < Q s iff r' Q t < Q s 'Q t. Proof Part (b) has the same proof as part (a) of Theorem 5ZJ, but with m.ultiplication in place of addition. To prove part (a), we first write r, s, and t in the form r = [<a, b)], s = [<c, d)], t = [<e,f)], where b, d, and f are positive. Since t is a positive rational, e is also a positive integer. Then ¢> [<af + eb, bf)] < Q [<cf + ed, df)] ¢> {af + eb)df < (cf + ed)bf ¢> adff + bdef < bcff + bdef ¢> ad < be by Theorem 5ZJ as desired. We have already said that the rational numbers form a field; the two preceding theorems state that < 0, + Q' 'Q' 0Q' 1Q, |
< Q) is an ordered field. 110 5. Construction of the Real Numbers Theorem 5QK The following cancellation laws hold for any rational numbers. (a) (b) If r + Q t = S + Q t, then r = s. If r' t = S 'Q t and t is nonzero, then r = s. Q Proof We can prove this as a corollary of the preceding theorem, following our past pattern. But there is now a simpler option open to us. In part (a) we add - t to both sides of the given equation, and in part (b) we multiply both sides of the given equation by t- 1. (This proof works in any Abelian group.) -l Finally, we want to show that, although 71. is not a subset of 0, nevertheless 0 has a subset that is "just like" 71.. Define the embedding function E: 71. --+ 0 by E(a) = [<a, 1)]. This functions gives us an isomorphic embedding in the sense that the following theorem holds. Theorem 5QL E is a one-to-one function from 71. into 0 satisfying the following conditions: (a) E(a+b)=E(a)+QE(b). (b) E(ab) = E(a) ·QE(b). (c) E(O) = OQ and E(1) = 1Q. (d) a < b iff E(a) <QE(b). Proof Each part of the theorem can be proved by direct calculation. First we check that E is one-to-one: E(a)=E(b) = [<a,1)]=[<b,1)] = <a, 1) '" <b, 1) = a=b. Parts (a), (b), and (d) are proved by the following calculations: E(a) + QE(b) E(a) 'Q E(b) [ < a, 1)] + Q [ < b, 1)] [<a + b, 1)] E(a + b), [<a, 1)]'Q [<b, 1)] [<ab, 1)] E(ab), Real Numbers III E(a) <QE(b) [<a, 1)] <Q[<b, 1)].;;..;;. a |
·1<b·1.;;. a < b. Finally part (c) is a restatement of the definitions of OQ and 1Q' -l We also obtain the following relation between fractions and division: [<a, b)] = E(a) -7 E(b). Since b "# 0, we have E(b) "# 0Q' and so the indicated division is possible. Henceforth we will simplify the notation by omitting the subscript "Q" on +,'Q' 0, and so forth. Also the product r. s will usually be written. Q as Just rs. Q Exercises 10. Show that r 'Q OQ = OQ for every rational number r. 11. Give a direct proof (not using Theorem 5QF) that if r 'Q s = OQ' then either r = OQ or s = 0Q' 12. Show that r<QOQ iff OQ<Q-r. 13. Give a new proof of the cancellation law for + z (Corollary 5ZK(a)), using Theorem 5ZD instead of Theorem 5ZJ. 14. Show that the ordering of rationals is dense, i.e., between any two rationals there is a third one: REAL NUMBERS The last number system that we will consider involves the set IR of all real numbers. The ancient Pythagoreans discovered, to their dismay, that there was a need to go beyond the rational numbers. They found that there simply was no rational number to measure the length of the hypotenuse of a right triangle whose other two sides had unit length. In our previous extensions of number systems, we relied on the facts that an integer could be named by a pair of natural numbers, and a rational number could be named by a pair of integers. But we cannot hope to name real number by a pair of rationals, because, as we will prove in Chapter 6, there are too many real numbers and not enough pairs of rationals. Hence we must look at new techniques in searching for a way to name real numbers. 112 5. Construction of the Real Numbers Actually there are several methods that can be used successfully to construct the real numbers. One approach is to utilize decimal expansions, so that a real number is determined by an integer and an infinite sequence of digits (a function from w into 10). This approach may be found in Claude Burrill's book, Foundations of |
Real Numbers, McGraw-Hill, 1967. A more common method of constructing a suitable set IR is to utilize the fact that a real number can be named by giving a sequence of rationals (a function from w into 0) converging to it. So one can take the set of all convergent sequences and then divide out by an equivalence relation (where two sequences are equivalent iff they converge to the same limit). But there is one hitch: The concepts of "convergent" and "equivalent" must be defined without reference to the real number to which the sequence is converging. This can be done by a technique named after Cauchy. Define a Cauchy sequence to be a function s: w -+ 0 such that ISm - snl is arbitrarily small for all sufficiently large m and n; i.e., (V positive sin 0)(3k E w)(Vm > k)(Vn > k) ISm - snl < s. (Here we write sn in place of s(n), as usual for w-sequences.) The concept of a Cauchy sequence is useful here because of the theorem of calculus asserting that a sequence is convergent iff it is a Cauchy sequence. Let C be the set of all Cauchy sequences. For rand s in C, we define rand s to be equivalent (r'" s) iff Irn - snl is arbitrarily small for all sufficiently large n; i.e., (V positive sin 0)(3k E w)(Vn > k) Irn - snl < s. Then the quotient set C; '" is a suitable candidate for IR. (This approach to constructing IR is due to Cantor.) An alternative construction of IR uses so-called Dedekind cuts. This is the method we follow henceforth in this section. The Cauchy sequence construction and the Dedekind cut construction each have their own advantages. The Dedekind cut construction of IR has an initial advantage of simplicity, in that it provides a simple definition of IR and its ordering. But multiplication of Dedekind cuts is awkward, and verification of the properties of multiplication is a tedious business. The Cauchy sequence construction of IR also has the advantage of generality, since it can be used with an arbitrary metric space in place of 0. With these considerations in mind, we choose the following strategy. We will present the Dedekind cut construction, and will prove that |
least upper bounds exist in IR. (This is the property that distinguishes IR from the other ordered fields.) Although we will define addition and multiplication of real numbers, we will not give complete verification of the algebraic properties. The Cauchy sequence construction may be found, among other Real Numbers 113 places, in Norman Hamilton and Joseph Landin's book, Set Theory and the Structure of Arithmetic, Allyn and Bacon, 1961. The idea behind Dedekind cuts is that a real number x can be named by giving an infinite set of rationals, namely all the rationals less than x. We will in effect define x to be the set of all rationals smaller than x. To avoid circularity in the definition, we must be able to characterize the sets of rationals obtainable in this way. The following definition does the job. Definition A Dedekind cut is a subset x of 0 such that: (a) 0;6 x ;6 0. (b) x is "closed downward," i.e., q E X & r < q = rEX. (c) x has no largest member. We then define the set IR of real numbers to be the set of all Dedekind cuts. Note that there is no equivalence relation here; a real (i.e., a real number) is a cut, not an equivalence class of cuts. The ordering on IR is particularly simple. For x and y in IR, define x < R y iff x c y. In other words, < R is the relation of being a proper subset: <R = {<x, y) E IR x IR I x c y}. Theorem 5RA The relation < R is a linear ordering on IR. Proof The relation is clearly transitive; we must show that it satisfies trichotomy on IR. So consider any x and y in IR. Obviously at most one of the alternatives, xc y, x = y, y c x, can hold, but we must prove that at least one holds. Suppose that the first two fail, i.e., that x '* y. We must prove that y c x. Since x '* y there is some rational r in the relative complement x - y (see Fig. 23). Consider any q E y. If r ::; q, then since y is closed downward, Fig. 23. The proof of Theorem SRA. 114 5. Construction of the Real Numbers we would have r |
E y. But r ¢ y, so we must have q < r. Since x is closed downward, it follows that q E x. Since q was arbitrary (and x"# y), we have y c x. -l N ow consider a set A of reals;. a real n urn ber x is said to be an upper bound of A iff y :::; R X for every y in A. The number x itself might or might not belong to A. The set A is bounded (i.e., bounded above) iff there exists some upper bound of A. A least upper bound of A is an upper bound that is less than any other upper bound. First consider an example not in IR, but in 0. The set {r E 0 I r. r < 2} of rationals whose square is less than 2 is a bounded set of rationals that has no least upper bound in 0. (We are stating this, not proving it, but it follows from the fact that y0. is irrational.) The following theorem shows that examples of this sort cannot be found in IR. Theorem 5RB Any bounded nonempty subset of IR has a least upper bound in IR. Proof Let A be the set of real numbers in question. We will show that the least upper bound is just UA. Simply by the definition of UA, we have x <;: UA for all x EA. Furthermore let z be any upper bound for A, so that x <;: z for all x E A. It then follows that UA <;: z; compare Exercise 5 of Chapter 2. The argument so far is not tied to IR; we have only shown that UA is the least upper bound of the set A with respect to ordering by inclusion. What remains to be shown is that UA E IR. Because A is nonempty, it is easy to see that UA "# 0. Also UA "# 0 because UA <;: z where z is an upper bound for A. You can easily verify (Exercise 15) that U A is closed downward and has no largest element. -l The foregoing theorem is important in mathematical analysis. For example, it is needed in order to prove that a continuous function on a closed interval assumes a maximum. And this in turn is used to prove the mean value theorem of calculus. The addition operation for IR is easily defined from addition of rationals. For reals x and y, define: x + RY = {}. |
Lemma 5RC For real numbers x and y, the sum x + R Y is also in IR. Real Numbers lIS Proof Clearly x + R Y is a nonempty subset of 0. To show that x + R Y "# 0, choose some q' in 0 - x and r' in 0 - y. Then'& r < r' = q + r < q' + r' so that any member q + r of x + R Y is strictly less than q' + r'. Hence q' + r' ¢ x + R y. To show that x + R Y is closed downward, consider any p<q+rEx+Ry (where q E X and r E y). Then adding - q to both sides of the inequality, we have p - q < r. Since y is closed downward, we have p - q E y. Thus we can write p as the sum p = q + (p - q) of q from x and p - q from y; this is what we need to have p EX + R y. (Note: Here "p - q" refers to subtraction of rationals, p + (-q). Earlier in this proof" 0 - x" referred to the relative complement of x in 0. If this sort of thing happened often, we would use a different symbol "0 \ x" for complements. But in fact the opportunities for confusion will be rare.) We leave it to you to verify that x + R Y has no largest member -l (Exercise 16). Theorem 5RD Addition of real numbers is associative and commuta tive: (x + R y) + R Z = X + R (y + R z), X +RY = Y +R X. Proof Since addition of rationals is commutative, it is clear from the definition of + R that it is commutative as well. As for associativity, we have (x + R y) + R Z = {s + r 1 SEX + R Y & r E z} = {(p + q) + rip EX & q E Y & r E z}, and a similar calculation applies to the other grouping. Thus associativity of + R follows from associativity of addition of rationals. -l The zero element of IR is defined to be the set of negative rational numbers: OR = {r E 01 r < O}. 116 5. Construction of the Real Numbe,s |
(a) OR is a real number. Theorem 5RE (b) For any x in IR, we have x + R OR = x. Proof (a) It is easy to see that 0"# OR "# 0; for example, -1 E OR and 1 ¢ OR' And it is clear that OR is closed downward. The fact that OR has no largest member follows immediately from the density of the rationals (Exercise 14). For part (b), we must prove that {r + sir E x & s < O} = x. The" <;:" inclusion holds because x is closed downward. To prove the " :;:> " half, consider any p in x. Since x has no largest member, there is some r with p < rEX. Let s = p inclusions hold. r. Then s < ° and p = r + s E OR' Hence both -l x 1 • 1 {rEOI-rEx) {rEOI-r¢x) I[ o I II o o I II 1 I II Fig. 24. Sometimes -x is {r E 01 - r ¢ xl, but not always. -x ~ ________________ ~A~ __________________ \ -, \ x -x __ ----~A~------~ o I (a) o I -.I ~--------------y~------------~ x (b) Fig. 25. In (a) x is negative: in (b) x is positive. Real Numbers 117 Before we can conclude that the real numbers form an Abelian group with + R and OR' we must prove that additive inverses exist. First we need to say just what set - x should be (where x E IR). We think of the real number -- x as the set of all smaller rational numbers. If we draw a picture as in Fig. 24, we might be tempted to think that - x is the complement of {r E 01 - rEx}, or in other words, that - x should be {r E 01 - r ¢ x}. This choice is not quite right, because it may have a largest element. Instead we define - x = {r E 01 (3s > r) - s ¢ x}. In Fig. 25, -x is illustrated as a subset of the rational number line. Theorem 5RF For every x in IR: -xEIR, (a) (b) x + R ( - |
x) = OR. Proof To prove that - x is a real number, we first must show that 0"# -x"# 0. There is some rational t with t ¢ x; let r = -t - 1. Then r E -x because r < - t and - (- t) ¢ x. Hence -x"# 0. To show that -x"# 0, take any pEX. We claim that -p¢ -x. This holds because if s> - p, then -s:<. P E x, whence -s E x. Hence - p ¢ -x and so -x "# 0. It is easier to show that - x is closed downward. Suppose that q < r E - x. Then (3s > r) - s ¢ x. Consequently (3s > q) - s ¢ x, since the same s can be used. Hence q E - x. It remains to show that - x has no largest element. Consider then any element r in -x. We know that for some s > r we have -s ¢ x. Because the rationals are densely ordered (Exercise 14), there is some p with s > p > r. Then p E - x, and p is larger than r. This completes the proof that -x is a real number. Now we turn to part (b). By definition x + R (-x) = {q + r 1 q EX & (3s > r) -s ¢ x}. For any such member q + r of this set, we have r < sand q < - s (lest - s ::;; q EX). Hence by the order-preserving property of addition, q + r < (-s) + s = 0. This shows that x + R (-x) <:::; OR' To establish the other inclusion, consider any p in OR' Then p < 0, and for which so - p is positive. By Exercise 19, q + (- p -;- 2) ¢ x. Let s = (p -;- 2) - q, so that - s ¢ x. Then p is the sum of q (which is in x) and p - q (which is less than s, where -s ¢ x). This makes p a member of x + R (-x). Thus we have both inclusions. -l there is some q E X 118 5 |
. Construction of the Real Numbers We have now shown that <IR, + R' OR> is an Abelian group. As in any Abelian group, the cancellation law holds: Corollary 5RG For any real numbers. Proof Simply add - z to both sides of the given equation. Next we prove that addition preserves order. Theorem 5RH For any real numbers, x < R Y.;;.. Proof It is easy to prove that (1) x ::; R Y = X + R Z ::; R Y + R Z, because this amounts to the statement that if x S y, then {q + s I q EX & S E Z} ~ {r + sir E Y & s E z}, which is obvious. And by Corollary 5RG we have (2) x;6y = x+Rz;6Y+RZ, which together with (1) gives the "=" half of the theorem. The" <=" half then follows by trichotomy (as in the proof to Theorem 4N). -l We can define the absolute value Ixl of a real number x to be the larger of x and -x. Since our ordering is inclusion, the larger of the two is just their union. Thus our definition becomes Ixl = xu - x. Then by Exercise 20, Ixl is always nonnegative, i.e., OR ::;R Ixl· Consider now the definition of multiplication. For the product of x and Y we cannot use {rs IrE x & s E y} (in analogy to the definition of x + R y), because both x and y contain negative rationals of large magnitude. Instead we use ~he following variation on the above idea. Definition (a) If x and yare nonnegative real numbers, then x 'R y = OR u {rs I ° ::; rEx & ° ::; s E y}. (b) If x and yare both negative real numbers, then x 'RY = Ixl'R IYI· (c) Ifone of the real numbers x and Y is negative and one is nonnegative, then Real Numbers 119 The facts we want to know about multiplication are gathered into the following theorem. Let 1R = {r E 01 r < 1}. Clearly OR <R 1R. We will not give a proof for this theorem; a proof can be found in Appendix F of Number Systems and the Foundations |
of Analysis by Elliott Mendelson (Academic Press, 1973). Theorem 5RI For any real numbers, the following hold: (a) x 'RY is a real number. (b) Multiplication is associative, commutative, and distributive over addition. (c) 0R"# 1R and x 'R 1R = x. (d) For nonzero x there is a nonzero real number y with x 'RY = 1R. (e) Multiplication by a positive number preserves order: IfOR < R z, then x <RY.;;. x 'R Z <RY 'R Z, The foregoing theorems show that, like the rationals, the reals (with + R' 'R' OR' 1 R' and < R) form an ordered field. But unlike the rationals, the reals have the least-upper-bound property (Theorem SRB). An ordered field is said to be complete iff it has the least-upper-bound property. It can be shown that the reals, in a sense, yield the only complete ordered field. That is, any other complete ordered field is "just like" (or more precisely, is isomorphic to) the ordered field of real numbers. For an exact statement of this theorem and for its proof, see any of the books we have referred to in this section, or p. 110 of Andrew Gleason's book, Fundamentals of Abstract Analysis, Addison-Wesley, 1966. The correct embedding function E from 0 into IR assigns to each rational number r the corresponding real number consisting of all smaller rationals. E(r) = {q E 01 q < r}, Theorem 5RJ E is a one-to-one function from 0 into IR satisfying the following conditions: (a) E(r + s) = E(r) + R E(s). (b) E(rs) = E(r) ·RE(s). (c) E(O) = OR and E(1) = 1R. r < s iff E(r) < R E(s). (d) Proof First of all, we must show that. E(r) is a real number. Obviously E(r) is a set of rationals, and it is closed downward. Furthermore o "# E(r) "# 0 because r - 1 E E(r) and r ¢ |
E(r). E(r) has no largest element, because if q E E(r), then by Exercise 14 there is a larger element p with q < p < r. Hence E(r) is indeed a real number. 120 5. Construction of the Real Numbers To show that E is one-to-one, suppose that r "# s. Then one is less than the other; we may suppose that r < s. Then r E E(s) whereas s ¢ E(s). Hence E(r) "# E(s). Next let us prove part (d), because it is easy. If r < s, then clearly E(r) <;: E(s). The inclusion is proper since E is one-to-one. Thus r < s = E(r) c E(s). The converse follows from trichotomy. If E(r) c E(s), then we cannot have r = s nor s < r (lest E(s) c E(r)), so we must have r < s. For part (a), we have E(r) + R E(s) = {p + q I p E E(r) & q E E(s)} = {}. We must show that this is the same as E(r + s), i.e., that {} = {t I t < r + s}. The" <;: " inclusion holds because by Theorem 5QJ. the ";2" inclusion, suppose t) -7- 2; then s > O. Define p = r - sand q = s - t < r + s. Let s = To establish (r + s s. Then p < rand q < sand p + q = t. Thus we can represent t as a sum in the desired form. Hence both inclusions hold. that Finally, we omit the (awkward) proof of part (b), and part (c) is only a -l restatement of definitions. Exercises In Theorem 5RB, show that UA is closed downward and has no 15. largest element. In Lemma 5RC, show that x + R Y has no largest element. 16. 17. Assume that a is a positive integer. Show that for any integer b there is some k in ro with b < a. E(k). 18. Assume that p is a positive rational number. Show that for any rational |
number r there is some k in ro with r < p. E(E(k)). (Here k is in ro, E(k) is the corresponding integer, and E(E(k)) is the corresponding rational.) 19. Assume that p is a positive rational number. Show that for any real number x there is some rational q in x such that p + q ¢x. Summaries 121 20. Show that for any real number x, we have OR ~R IxI21. Show that if x < R y, then there is a rational number r with 22. Assume that x E IR. How do we know that Ixl E IR? x <RE(r) <RY' SUMMARIES In this chapter we have given one way of constructing the real numbers as particular sets. Along the way, some concepts from abstract algebra have naturally arisen. For convenient reference, we have collected in the present section certain definitions that have played a key role in this chapter. Integers Let m, n, p, and q be natural numbers..;;. m + q = p + n, [<m, n)] '" [<p, q)] [<m, n)] +z[<p, q)] = [<m + p, n + q)], -[<m, n)] = [<n, m)], [<m, n)] A<p, q)] = [<mp + nq, mp + np)], [<m, n)] < z[ <p, q)].;;. m + q E P + n, E(n) = [<n, 0)]. Rational numbers Let a, b, c, and d be integers with bd "# O..;;. ad = cb, <a, b) '" <c, d) [<a, b)] +Q[<c, d)] = [<ad + cb, bd)], -[<a, b)] = [< -a, b)], [<a, b)] 'Q[<c, d)] = [<ac, bd)], [<a, b)] <Q [<c, d)].;;. ad < cb, when band d are positive, E(a) = [<a, 1)]. Real numbers. A real number is a set x such that 0 c x c 0, x |
is closed downward, and x has no largest member. x < R Y.;;. X c y, X + R Y = {}, -x = {rE 01 (3s > r) -s¢x}, Ixl =X u - x, Ixl 'R IYI = OR u {rs I ° ~ r E Ixl & ° ~ S E Iyi}, E(r) = {q E 0 I q < r}. Next we turn to the definitions from abstract algebra that are relevant to the number systems in this chapter. 122 5. Construction of the Real Numbers An Abelian group (in additive notation) is a triple 2 <A, +, 0) consisting of a set A, a binary operation + on A, and an element (" zero") of A, such that the following conditions are met: 1. + is associative and commutative. 2. 0 is an identity element, i.e., x + 0 = x. Inverses exist, i.e., "Ix 3y x + y = O. 3. An Abelian group (in multiplicative notation) is a triple <A,., 1) consisting of a set A, a binary operation' on A, and an element 1 of A, such that the following conditions are met:. is associative and commutative. 1. 2. 1 is an identity element, i.e., x. 1 = x. Inverses exist, i.e., "Ix 3y x. y = 1. 3. This is, of course, the same as the preceding definition. A group has the same definition, except that we do not require that the binary operation be commutative. All of the groups that we have considered have, in fact, been Abelian groups. But some of our results (e.g:, the uniqueness of inverses) are correct in any group, Abelian or not. A commutative ring with identity is a quintuple <D, +,., 0, 1) consisting of a set D, binary operations + and, on D, and distinguished elements o and 1 of D, such that the following conditions are met: 1. <D, +,0) is an Abelian group. 2. The operation' is associative and commutative, and is distributive over addition. 3. 1 is a multiplicative identity (x. 1 = x) |
and 0"# 1. An integral domain is a commutative ring with identity with the additional property that there are no zero divisors: 4. If x "# 0 and y "# 0, then also x. )' "# O. A field is a commutative ring with identity III which multiplicative inverses exist: 4'. If x is a nonzero element of D, then x. )' = 1 for some.r. Any field is also an integral domain, because condition 4' implies condition 4 (see the proof to Corollary 5QG). 2 It is also possible to define a group to be a pair <A. +), since the zero element turns out to be uniquely determined. We have formulated these definitions to match the exposition in this chapter. Two 123 An ordered field is a sextuple <D, +, ',0, 1, <) such that the following conditions are met: 1. <D, +, ',0, I) is a field. 2. < is a linear ordering on D. 3. Order is preserved by addition and by multiplication by a positive element: If ° < Z, then x <)' ¢> x + Z < )' + z. x < y ¢> x' Z < Y. z. We can define ordered integral domain or even ordered commutative ring with identity by adjusting the first condition. A complete ordered field is an ordered field in which for every bounded nonempty subset of D there is a least upper bound. The constructions in this chapter can be viewed as providing an existence proof for such fields. The conditions for a complete ordered field are not impossible to meet, for we have constructed a field meeting them. TWO What is a two? What are numbers? These are awkward questions; yet when we discuss numbers one might naively expect us to know what it is we are talking about. In the Real World, we do not encounter (directly) abstract objects such as numbers. Instead we find physical objects: a number of similar apples, a ruler, partially filled containers (Fig. 26). Fig. 26. A picture of the Real World. Somehow we manage to abstract from this physical environment the concept of numbers. Not in any precise sense, of course, but we feel inwardly that we know what numbers are, or at least some numbers like 2 and 3. And we have various mental images that we use when thinking about numbers (Fig. 27). 124 5. Construction of |
the Real Numbers -I o J2 • 2 Fig. 27. Pictures of mental images. Then at some point we acquire a mathematical education. This teaches us many formal manipulations of symbols (Fig. 28). It all seems fairly reasonable, but it can all be done without paying very much attention to what the symbols stand for. In fact computers can be programmed to carry out the manipulations without any understanding at all. Since numbers are abstract objects (as contrasted with physical objects) it might be helpful to consider first other abstractions. Take honesty, for example. Honesty is a property possessed by those people whose utterances are true sentences, who do not fudge on their income tax, and so forth. By way of imitation, we can try characterizing two as the property that is true of exactly those sets that, for some distinct x and y, have as members x and y and nothing else. A slight variation on this proposal would be to eliminate sets and characterize two as the property that is true of exactly those properties that, for some distinct x and y, are true of x and y and nothing else. (This proposal is due to Frege.) t (3x 2 ) 6x Jr~X dx= X fofjx-x 2 17) 35 34 1 6 +8 14 Fig. 28. Pictures of manipulated symbols. Two 125 If we define numbers in terms of properties, someone might ask us what a property is. And no matter how we define numbers, the procedure of defining one concept in terms of another cannot go on forever, producing an infinite regress of definitions. Eventually the procedure must be founded on a commonly agreed upon basis. Properties might form such a basis. For a mathematician, sets form a more workable basis. Actually there is a close connection between properties and sets. Define the extension of a property to be the set of all objects of which that property is true. If a couple of properties have the same extension (i.e., if they are true of exactly the same objects), are they in fact one and the same property? Is the property of being a prime number less than 10 the same as the property of being a solution to X4 - 17x 3 + 101x 2 - 247x + 210 = O? If you answer "yes," then one says that you are thinking of properties extensionally, whereas if you answer "no," then you are thinking of properties intensionally. Both alternatives are legitimate, as long as the choice of alternatives is |
made clear. Either way, sets are the extensions of properties. (All sets are obtainable in this way; the set x is the extension of the property of belonging to x.) We can recast Frege's proposal in terms of sets as follows: Two is the set having as members exactly those sets that, for some distinct x and y, have as members x and y and nothing else. But in Zermelo-Fraenkel set theory, there is no such set. (For the number one, you should check Exercise 8 of Chapter 2.) Our response to this predicament was to select artificially one particular set {0, {0}} as a paradigm. Now {0, {0}} is very different from the property that is true of exactly those sets that, for some distinct x and y, have as members x and y and nothing else, but it serves as an adequate substitute. Then rapidly one thing led to another, until we had the complete ordered field of real numbers. And as we have mentioned, one complete ordered field is very much like any other. But let us back up a little. In mathematics there are two ways to introduce new objects: (i) The new objects might be defined in terms of other already known objects. (ii) The new objects can be introduced as primitive notions and axioms can be adopted to describe the notions. (This is not so much a way of answering foundational questions about the objects as it is a way of circumventing them.) In constructing real numbers as certain sets we have selected the first path. The axiomatic approach would regard the definition of a complete 126 5. Construction of the Real Numbers ordered field as axioms concerning the real number system (as a primitive notion). On the other hand, for sets themselves we have followed (with stripes) the axiomatic method. But what about the Real World and those mental images? And the manipulated symbols? We want not just any old concept of number, but a concept that accurately reflects our experiences with apples and rulers and containers, and accurately mirrors our mental images. This is not a precise criterion, since it demands that a precise mathematical concept be compared with informal and intuitive ideas. And consequently the question whether our concept is indeed accurate must be evaluated on informal grounds. Throughout this chapter our formulation of definitions has been motivated ultimately by our intuitive ideas. Is there any way we could have gone wrong? Yes, we could have gone wrong. In seeking a number system applicable to problems dealing with physical objects and |
physical space, we might have been guided by erroneous ideas. There is always the possibility that lines in the Real World do not really resemble IR. For example, over very short distances, space might be somehow quantized instead of being continuous. (Experimental evidence forced us to accept the fact that matter is quantized; experimental evidence has not yet forced us to accept similar ideas about space.) Or over very large distances, space might not be Euclidean (a possibility familiar to science-fiction buffs). In such events, mathematical theorems about IR, while still true of IR, would be less intere~ting, as they might be inapplicable to certain problems in the Real World. Mathematical concepts are useful in solving problems from the Real World to the extent that the concepts accurately reflect the essential features of those problems. The process of solving a problem mathematically has three parts (Fig. 29). We begin with a Real World problem. Then we need to model the original problem by a mathematical problem. This typically requires simplifying or idealizing some aspects of the original problem. (For example, we might decide to ignore air resistance or friction.) The middle step in the process consists of finding a mathematical solution to the mathematical problem. The final step is to interpret the mathematical solution in terms of the original problem. The middle step in this process is called "pure mathematics," and the entire process is called "applied mathematics." We have, for example, all been given problems such as: If Johnny has six pennies and steals eight more, how many does he have? We first convert this to the mathematical problem: 6 + 8 =? Then by pure mathematics (addition, in this case) we obtain 14 as the solution. Finally, we decide that Johnny has fourteen pennies. The mathematical modeling of a Real World problem is not always this Two 127 pure mathematics Fig. 29. Applied mathematics. straightforward. When we try to interpret our mathematical solution in terms of the original problem, we might discover that it just does not fit. If we start with six blobs of water and add eight more blobs, we may end up with only four or five rather large pUddles. This outcome does not shake our faith in arithmetic at all. It does show that we need to revise the model and try again (perhaps by measuring volume instead of counting blobs). From the vast array of mathematical concepts we must select those (if any!) that accurately model the essential feature of the problem to be solved. |
CHAPTER 6 CARDINAL NUMBERS AND THE AXIOM OF CHOICE EQUINUMEROSITY We want to discuss the size of sets. Given two sets A and B, we want to consider such questions as: (a) Do A and B have the same size? (b) Does A have more elements than B? Now for finite sets, this is not very complicated. We can just count the elements in the two sets, and compare the resulting numbers. And if one of the sets is finite and the other is infinite, it seems conservative enough to say that the infinite set definitely has more elements than does the finite set. But now consider the case of two infinite sets. Our first need is for a definition: What exactly should "A has the same size as B" mean when A and B are infinite? After we select a reasonable definition, we can then ask, for example, whether any two infinite sets have the same size. (We have not yet officially defined "finite" or "infinite," but we will soon be in a position to define these terms in a precise way.) An Analogy In order to find a solution to the above problem, we can first consider an analogous problem, but one on a very simple level. 128 Equinumerosity 129 Fig. 30. Are there exactly as many houses as people? Imagine that your mathematical education is just beginning-that you are on your way to nursery school. You are apprehensive about going, because you have heard that they have mathematics lessons and you cannot count past three. Sure enough, on the very first day they show you Fig. 30 and ask, "Are there exactly as many houses as people?" Your heart sinks. There are too many houses and too many people for you to count. This is just the predicament described earlier, where we had sets A and B that, being infinite, had too many elements to count. But wait! All is not lost. You take your crayon and start pairing people with houses (Fig. 31). You soon discover that there are indeed exactly as many houses as people. And you did not have to count past three. You get a gold star and go home happy. We adopt the same solution. Fig. 31. How to answer the question without counting. Definition A set A is equinumerous to a set B (written A ~ B) iff there is a one-to-one function from A onto B. A one-to-one function from A onto |
B is called a one-to-one correspondence between A and B. For example, in Fig. 30 the set of houses is equinumerous to the set of people. A one-to-one correspondence between the sets is exhibited in Fig. 31. Example The set w x w is equinumerous to w. There is a function J mapping w x w one-to-one onto w, shown in Fig. 32, where the value of 130 6. Cardinal Numbers and the Axiom of Choice 1 - -7 - - 1 6 - I"'" 1 i""'I"'" 1 1""'1""'1""'1 1 - -4 - -9 - - Fig. 32. w x w ~ w. J(m, n) is written at the point with coordinates <m, n). In fact we can give a polynomial expression for J: J(m, n) = t[(m + n)2 + 3m + n], as you are asked to verify in Exercise 2. Example The set of natural numbers is equinumerous to the set 0 of rational numbers, i.e., (}) ~ 0. The method to be used here is like the one used in the preceding example. We arrange 0 in an orderly pattern, then thread a path through the pattern, pairing natural numbers with the rationals as we go. The pattern is shown in Fig. 33. We define f: (}) --+ 0, where f(n) is the rational next to the bracketed numeral for n in Fig. 33. To ensure that f is one-to-one, we skip rationals met for the second (or third or later) time. Thusf(4) = -1, and we skip -2/2, -3/3, and so forth. [0] 0/1 ---- 1/1 [I] [II] [10] 2/1 ---- 3/1 • •• [5] - 2/1 - - - - 1;1 [4] t t -2/2 [3] - 1/2 - - - 0/2 - - - 1/2 [2] • •• [6] -2/3 ---- -1;3 ---- 0/3 ---- 1/3 ---- 2/3 [9] [7] [8] t t 2/2 t t + 3/2 [12] t ~ 3/3 [15] |
[14] - 2/4 ___ -1/4 ___ 0/4 ___ 1/4 ___ 2/4 ___ 3/4 [13] Example The open unit interval Fig. 33. w ~ iQI. (0, 1) = {x E IR I 0 < x < I} is equinumerous to the set IR of all real numbers. A geometric construction Equinumerosity 131 of a one-to-one correspondence is shown in Fig. 34. Here (0, 1) has been bent into a semicircle with center P. Each point in (0, 1) is paired with its projection (from P) on the real line. There (0, 1) ~ IR. Let f(x) = tan n(2x - 1 )/2. Then f maps (0, 1) one-to-one (and continuously) onto IR. is also an analytical proof that As the above example shows, it is quite possible for an infinite set, such as IR, to be equinumerous to a proper subset of itself, such as (0, 1). (For finite sets this never happens, as we will prove shortly.) Galileo remarked in 1638 that (}) was equinumerous to the set {O, 1, 4, 9,... } of squares of natural numbers, and found this to be a curious fact. The p Fig. 34. (0, 1) ~ IR. squares are in some sense a small part of the natural numbers, e.g., the fraction of the natural numbers less than n that are squares converges to o as n tends to infinity. But when viewed simply as two abstract sets, the set of natural numbers and the set of squares have the same size. Similarly the set of even integers is equinumerous to the set of all integers. If we focus attention on the way in which even integers are placed among the others, then we are tempted to say that there are only half as many even integers as there are integers altogether. But if we instead view the two sets as two different abstract sets, then they have the same size. Example For any set A, we have q> A ~ A2. To prove this, we define a one-to-one function H from q> A onto A2 as follows: For any subset B of A, H(B) is the characteristic function of B, i.e., the function fa from A into 2 for which if x E B |
, if x E A-B. Then any function 9 E A2 is in ran H, since 9 = H({x E A I g(x) = I}). The following theorem shows that equinumerosity has the property of being reflexive (on the class of all sets), symmetric, and transitive. But it 132 6. Cardinal Numbers and the Axiom of Choice cannot be represented by an equivalence relation, because it concerns all sets. In von Neumann-Bernays set theory, one can form the class E = {<A, B) I A ~ B}. Then E is an "equivalence relation on V," in the sense that it is a class of ordered pairs that is reflexive on V, symmetric, and transitive. But E is not a set, lest its field V be a set. In Zermelo-Fraenkel set theory, we have only the "equivalence concept" of equinumerosity. Theorem 6A For any sets A, B, and C: (a) A ~ A. (b) (c) If A ~ B, then B ~ A. If A ~ Band B ~ C, then A ~ C. Proof See Exercise 5. In light of the examples presented up to now, you might well ask whether any two infinite sets are equinumerous. Such is not the case; some infinite sets are much larger than others. Theorem 6B (Cantor 1873) (a) The set (}) is not equinumerous to the set IR of real numbers.'(b) No set is equinumerous to its power set. Proof We will show that for any function f: (}) --+ IR, there is a real number z not belonging to ranf. Imagine a list of the successive values of f, expressed as infinite decimals: f(O) = 236.001..., f(l) = -7.777..., f(2) = 3.1415..., (In Chapter 5 we did not go into the matter of decimal expansions, but you are surely familiar with them.) We will proceed to construct the real z. The integer part is 0, and the (n + l)st decimal place of z is 7 unless the (n + l)st decimal place of f(n) is 7, in which case the (n + l)st decimal place of z is 6. For example, |
in the case shown, z = 0.767.... Then z is a real number not in ranf, as it differs from f(n) in the (n + 1 )st decimal place. The proof of (b) is similar. Let 9 :A --+ 9 A; we will construct a subset B of A that is not in ran g. Specifically, let B = {x E A I x ¢ g(x)}. Finite Sets 133 Then B c:::; A, but for each x E A, x E B ¢> x ¢ g(x). Hence B #- g(x). The set IR happens to be equinumerous to f!l>0J, as we will soon be able to prove. A larger set is f!l>1R, and f!l>f!I>lR is larger still. Before continuing our consideration of infinite sets, we will study the other alternative: the sets that are "small" at least to the extent of being finite. Exercises 1. Show that the equation f(m, n) = 2m(2n + 1) - 1 defines a one-to-one correspondence between OJ x OJ and OJ. 2. Show that in Fig. 32 we have: l(m, n) = [1 + 2 +... + (m + n)] + m = t[(m + n)2 + 3m + n]. 3. Find a one-to-one correspondence between the open unit interval (0, 1) and IR that takes rationals to rationals and irrationals to irrationals. 4. Construct a one-to-one correspondence between the closed unit interval [0, 1] = {x E IR I 0 ~ x ~ 1} and the open unit interval (0, 1). 5. Prove Theorem 6A. FINITE SETS Although we have long been using the words "finite" and "infinite" in an informal way, we have not yet given them precise definitions. Now is the time. Definition A set is finite iff it is equinumerous to some natural number. Otherwise it is infinite. Here we rely on the fact that in our construction of OJ, each natural number is the set of all smaller natural numbers. For example, any natural number is itself a finite set. We want to check that each finite set S is equinumerous to a unique number n |
. The number n can then be used as a count of the elements in S. 134 6. Cardinal Numbers and the Axiom of Choice We first need the following theorem, which implies that if n objects are placed into fewer than n pigeonholes, then some pigeonhole receives more than one object. Recall that a set A is a proper subset of B iff A <;; B and A #- B. Pigeonhole Principle No natural number is equinumerous to a proper subset of itself. Proof Assume thatf is a one-to-one function from the set n into the set n. We will show that ranf is all of the set n (and not a proper subset of n). This suffices to prove the theorem. k+=kU{k] A k k p. I t • k Fig. 35. In! we interchange two values to obtain J. k We use induction on n. Define: T = {n E OJ I any one-to-one function from n into n has range n}. Then 0 E T; the only function from the set 0 into the set 0 is 0 and its range is the set O. Suppose that k E T and that f is a one-to-one function from the set e into the set e. We must show that the range of f is all of the set e; this will imply that e E T. Note that the restriction f ~ k off to the set k maps the set k one-to-one into the set e. Case I Possibly the set k is closed under f. Then f ~ k maps the set k into the set k. Then because k E T we may conclude that ran (f ~ k) is all of the set k. Since f is one-to-one, the only possible value for f (k) is the number k. Hence ranf is k u {k}, which is the set e. Finite Sets 135 Case II Otherwise f(p) = k for some number p less than k. In this case we interchange two values of the function. Define j by J(p) = f(k), J(k) = f(p) = k, J(x) = f(x) for other x E e (see Fig. 35). Then J maps the set e one-to-one into the set e, and the set k is closed under]' So by Case I, ran J = e. |
But ran J = ran f. -l Thus in either case, ran f = k +. So T is inductive and equals w. Corollary 6C No finite set is equinumerous to a proper subset of itself. Proof This is the same as the pigeonhole principle, but for an arbitrary finite set A instead of a natural number. Since A is equinumerous to a natural number n, we can use the one-to-one correspondence 9 between A and n to "transfer" the pigeonhole principle to the set A. Suppose that, contrary to our hopes, there is a one-to-one correspond ence f between A and some proper subset of A. Consider the composition 9 0 f 0 g-1, illustrated in Fig. 36. This composition maps n into n, and it is one-to-one by Exercise 17 of Chapter 3. Furthermore its range C is a ranf; then g(a) En - C.) Thus proper subset of n. (Consider any a in A n is equinumerous to C, in contradiction to the pigeonhole principle. -l The foregoing proof uses an argument that is useful elsewhere as well. We have sets A and n that are "alike" in that A;:::; n, but different in that they have different members. Think of the members of A as being red, the members of n as being blue. Then the function g: A --+ n paints red A n Fig. 36. What f does to A, 9 0 fog - I does to n. 136 6. Cardinal Numbers and the Axiom of Choice things blue, and the function g- 1: n --+ A paints blue things red. The composition 9 0 f 0 g-1 paints things red, applies f, and then restores the blue color. Corollary 6D (a) Any set equinumerous to a proper subset of itself is infinite. (b) The set OJ is infinite. Proof The preceding corollary proves part (a). Part (b) follows at once from part (a), since the function (1 whose value at each number n is n+ maps OJ one-to-one onto OJ -l {O}. Corollary 6E Any finite set is equinumerous to a unique natural number. Proof Assume that A ~ m and A ~ n for natural numbers m and n. Then m ~ n. By trichotomy and Corollary 4M, |
either m = n or one is a proper subset of the other. But the latter alternative is impossible s{nce m ~ n. Hence m = n. -l For a finite set A, the unique n E OJ such that A ~ n is called the cardinal number of A, abbreviated card A. For example, card n = n for n E OJ. And if a, b, c, and d are all distinct objects, then card{a, b, c, d} = 4. This is because {a, b, c, d} ~ 4; selecting a one-to-one correspondence is the process called "counting." Observe that for any finite sets A and B, we have A ~ card A and card A = card B iff A ~ B. What about infinite sets? The number card A measures the size of a finite set A. We want "numbers" that similarly measure the size of infinite sets. Just what sets these "numbers" are is not too crucial, any more than it was crucial just what set the number 2 was. The essential demand is that we define card A for arbitrary A in such a way that card A = card B iff A ~ B. Now it turns out that there is no way of defining card A that is really simple. We therefore postpone until Chapter 7 the actual definition of the set card A. The information we need for the present chapter is embodied in the following promise. Promise For any set A we will define a set card A in such a way that: (a) For any sets A and B, card A = card B iff A ~ B. (b) For a finite set A, card A is the natural number n for which A ~ n. Finite Sets 137 (In making good on this promise, we will use in Chapter 7 additional axioms, namely the replacement axioms and the axiom of choice. If you plan to omit Chapter 7, then regard card A as an additional primitive notion and the promise as an additional axiom.) We define a cardinal number to be something that is card A for some set A. By part (b) of the promise, any natural number n is also a cardinal number, since n = card n. But card OJ is not a natural number (card OJ #- n = card n, since OJ is not equinumerous to n). Just what set card OJ is will not be revealed |
until Chapter 7. Meanwhile we give it the name that Cantor gave it: card OJ = ~o. The symbol ~ is aleph, the first letter of the Hebrew alphabet. In general, for a cardinal number K, there will be a great many sets A of cardinality K, i.e., sets with card A = K. (The one exception to this occurs when K = 0.) In fact, for any nonzero cardinal K, the class K" = {X I card X = K} of sets of cardinality K is too large to be a set (Exercise 6). But all of the sets of cardinality K look, from a great distance, very much alike-the elements of two such sets may differ but the number of elements is always K. In particular, if one set X of cardinality K is finite, then all of them are; in this case K is a finite cardinal. And if not, then K is an infinite cardinal. Thus the finite cardinals are exactly the natural numbers. ~o is an infinite cardinal, as are card IR, card f!jJOJ, card f!jJf!jJOJ, etc. Before leaving this section on finite sets, we will verify a fact that, on an informal level, appears inevitable: Any subset of a finite set is finite. (Later we will find another proof of this.) Lemma 6F If C is a proper subset of a natural number n, then C ~ m for some m less than n. Proof We use induction. Let T = {n E OJ I any proper subset of n is equinumerous to a member of n}. Then 0 E T vacuously, 0 having no proper subsets. Assume that k E T and consider a proper subset C of k +. Case I C = k. Then C ~ k E e. Case II C is a proper subset of k. Then since k E T, we have C ~ m for mE k E k+. 138 6. Cardinal Numbers and the Axiom of Choice Case III Otherwise k E C. Then C = (C n k) u {k} and C n k is a proper subset of k. Because k E T, there is mE k with C n k ~ m. Say f is a one-to-one correspondence between C n k and m; then fu {<k, m)} is a one-to-one correspondence between C and m +. Since m |
E k, we have m+ E k+. Hence C ~ m+ E k+ and k+ E T. Thus T is inductive and coincides with OJ. Corollary 6G Any subset of a finite set is finite. Proof Consider A S B and let f be a one-to-one correspondence between B and some n in OJ. Then A ~ f [A] s n and by the lemma f [A] ~ m for some m §. n. Hence A ~ m §. n E OJ. -l Exercises 6. Let K be a nonzero cardinal number. Show that there does not exist a set to which every set of cardinality K belongs. 7. Assume that A is finite and f: A -+ A. Show that f is one-to-one iff ranf= A. 8. Prove that the union of two finite sets is finite (Corollary 6K), without any use of arithmetic. 9. Prove that the Cartesian product of two finite sets is finite (Corollary 6K), without any use of arithmetic. CARDINAL ARITHMETIC The operations of addition, multiplication, and exponentiation are well known to be useful for finite cardinals. The operations can be useful for arbitrary cardinals as well. To extend the concept of addition from the finite to the infinite case, we need a characterization of addition that is correct in the finite case, and is meaningful (and plausible) in the infinite case. In Chapter 4 we obtained addition on OJ by use of the recursion theorem. That approach is unsuitable here, so we seek another approach. The answer to our search lies in the way addition is actually explained in the elementary schools. First-graders are not told about the recursion theorem. Instead, if they want to add 2 and 3, they select two sets K and L with card K = 2 @nd card L = 3. Sets of fingers are handy; sets of apples are preferred by textbooks. Then they look at card(K u L). If they had the good sense to select K and L to be disjoint, then card(K u L) = 5. The same idea is embodied in the following definition of addition. In the Same vein, we can include multiplication and exponentiation. Cardinal Arithmetic 139 Definition Let K and A. be any cardinal numbers. (a) K + A. = card(K u L), where K and L |
are any disjoint sets of cardinality K and A., respectively. (b) K· A. = card(K x L), where K and L are any sets of cardinality K and A., respectively. (c) KA = card LK, where K and L are any sets of cardinality K and A., respectively. In every case it is necessary to prove that the operation is well defined; that is, that the outcome is independent of the particular sets K and L selected to represent the cardinals. Also for the case of finite cardinals, we should check that the above definitions are not in open conflict with operations on OJ defined in Chapter 4. First consider addition. To add two cardinals K and A., the definition demands that we first select disjoint sets K and L with card K = K and card L = A.. This is possible; if our first choices for K and L fail to be disjoint, we can switch to K x {O} and L x {I}. For then, since K ~ K x {O} and L~Lx{I}, we have card(K x {O}) = K and card(L x {I}) = A.. And the sets K x {O} and L x {I} are disjoint. Having selected disjoint representatives K and L for K and A., we form their union K u L. Then by definition K + A. = card(K u L). We must verify that this sum is independent of the particular disjoint sets K and L selected. This verification is accomplished by part (a) of the following theorem. For suppose K 1, Ll and K 2, L2 are two selections of disjoint sets of cardinality K, A.. Since card Kl = card K2 = K, we have K 1 ~ K 2; similarly Ll ~ L2. The following theorem then yields K 1 u Ll ~ K2 U L 2, whence card(Kl u L 1 ) = card(K2 u L 2 ). Thus we have an unambiguous sum K + A.. Parts (b) and (c) of the theorem perform the same service for multiplication and exponentiation. Theorem 6H Assume that Kl ~ K2 and Ll ~ L 2. If Kl n Ll = K2 n L2 = 0, then Kl u Ll ~ K2 U L 2· (a) (b) Kl x Ll ~ K2 |
X L 2· (c) ~(L2)K2. (LIlK l • 140 6. Cardinal Numbers and the Axiom of Choice Proof Since K 1 ~ K 2' there is a one-to-one function f from K 1 onto K 2; since Ll ~ L 2, there is a one-to-one function 9 from Ll onto L 2. Then for (a), the function h, defined by h(x) = {f(X) g(x) if xEKl' if x E Ll' maps Kl u Ll one-to-one onto K2 u L 2· (We need Kl n Ll = 0 to be sure h is a function; we need K2 n L2 = 0 to be sure it is one-to-one.) For (b), the function h, defined by h«x, y») = <f(x), g(y) (for x E Kl and YELl)' maps Kl x Ll one-to-one onto K2 x L 2· Fig. 37. H(j) = f 0 j 0 g- '. Finally for (c), the function H, defined by HU}=fojog-l, maps (LdK l one-to-one onto (L2)K 2 (see Fig. 37). For clearly HU} is a function from L2 to K 2' To see that H is one-to-one, suppose that j and l' are different functions from Ll into K l. Then j(t) #- 1'(t) for some tELl' Then compute HU) and HV) at g(t): HU)(g(t)) = fU(t)) #- fV(t)) = HV)(g(t)) since f is one-to-one. Hence the function HU) is different from the function HU'), i.e., H is one-to-one. And to see that ran H is all of (L2)K 2, into K 2' Then d = HU), where consider any function d from L2 j=f-lodog. -l Remark The proof of part (c) is our first proof dealing with cardinal exponentiation. Observe that it involves constructing a function H whose arguments and values are themselves functions. In such cases it is imperative to use a rationally chosen notation, |
e.g., an uppercase letter for Hand lowercase letters for the arguments of H, as in the expression" HU)." Here Cardinal Arithmetic 141 the function H must be one-to-one. But the function j is in general not one-to-one, nor is HU). It may help to think of H as a "super function" that assigns certain lower-level functions to other lower-level functions. The super function in this proof is one-to-one, but the lower-level functions are not necessarily one-to-one. Examples 1. 2 + 2 = 4. Proving this amounts to finding disjoint sets K and L with K ~ 2 and L ~ 2, and then checking that K u L ~ 4. (This is not the same as the verification of 2 + 2 = 4 in Chapter 4, because now we are using the addition operation for cardinal numbers. But Theorem 6J assures us that the answer is unchanged.) 2. For m and n in OJ, m· n = card m x nand mn = card nm. 3. For any natural number n, n + ~o = ~o and n. ~o = ~o (unless n = 0). Also ~o + ~o = ~o and ~o· ~o = ~o· The last equation follows from an earlier example showing that OJ x OJ ~ OJ. To prove that 2 + ~o = ~o' it suffices to show that {a, b} u OJ ~ OJ (where a and b are not in OJ). The function f(a) = 0, f(b) = 1, f(n)=n++ establishes this. The other equations are left for you to check. Observe that cancellation laws fail for infinite cardinals, e.g., 2 + ~o = 3 + ~o'but 2 #- 3. K. ° = 0, 4. For any cardinal number K, K + ° = K, K. 1 = K. 5. Recall that 12) K = {0} for any set K and that K0 = 0 for nonempty K. In terms of cardinal numbers, these facts become for any K, for any nonzero K. In particular, 00 = 1. 6. For any set A, the cardinality of its power set is 2card A. This is because by the definition of exponentiation, 2card A = |
card ( A2). And we have shown that A2 ~ & A, and hence 2card A = card( A2) = card & A. In particular, card &OJ = 2No • (The term "power set" is rooted in the fact that card & A equals 2 raised to the power card A.) 142 6. Cardinal Numbers and the Axiom of Choice 7. By Cantor's theorem and the preceding example, K #- 2" for any K. In particular ~o #- 2No • 8. For any cardinal K, You are invited to explain how this fact is obtained. K + K = 2· K. The following theorem lists some of the elementary facts of cardinal arithmetic. (Parts of the theorem will later be seen to be rather trivial.) Theorem 61 For any cardinal numbers K, A, and /1: 1. K + A = A + K and K. A = A. K. 2. K + (A + /1) =(K + A) + /1 and K' (A' /1) = (K' A)' /1. 3. K' (A + /1) = K. A + K' /1. 4. KA+I' = KA. KI'. 5. 6. (K' At = KI'. AI'. (KA)!t = KA·I'. Proof Take sets K, L, and M with card K = K, card L = A, and card M = /1; for convenience choose them in such a way that any two are disjoint. Then each of the equations reduces to a corresponding statement about equinumerous sets. For example, K' A = card K x L and A' K = card(L x K); consequently showing that K. A = A. K reduces to showing that K x L ~ L x K. Listed in full, the statements to be verified are: 1. K u L ~ L u K and K x L ~ L x K. 2. K u (L u M) ~ (K u L) u M and K x (L x M) ~ (K x L) x M. 3. K x (L u M) ~ (K xL) u (K x M). 4. (L uM)K ~ LK x MK. 5. M(KxL)~MKxML. 6. M(L |
K) ~ (LXM)K. Most of the verifications are left as exercises. In the case of item 6 we want a one-to-one function H from M(LK) onto (LXM)K. ForfE M(LK), let H(f) be the function whose value at <I, m) equals the value of the function f(m) at I. To see that H is one-to-one, observe that if f #- 9 (both belonging to M(LK)), then for some m, the functionsf(m) and g(m) differ. This in turn implies that for some 1,f(m)(/) #- g(m)(/). Hence H(f)(l, m) = f(m)(/) #- g(m)(/) = H(g)(/, m) so that H(J) #- H(g). Cardinal Arithmetic 143 Finally, to see that ran H is all of (L x M) K, consider any function j E (LXM)K. Then j = H(f), where (for mE M) f(m) is the function whose value at IE L is j(l, m). -l The next theorem reassures us that for finite cardinals (i.e., for natural numbers), the present arithmetic operations agree with those of Chapter 4. In Chapter 4, exponentiation was defined only in passing; for completeness we include it here. Theorem 6J Let m and n be finite cardinals. Then m+n=m+w n, m· n = m·w n, mn = mn, where on the right side we use the operations of Chapter 4 (defined by recursion) and on the left side we use the operations of cardinal arithmetic. Proof We use induction on n. First we claim that for any cardinal numbers K and A. the following identities are correct. (al) K+O=K. (a2) K+(A.+l)=(K+A.)+1. (ml) K·O=O. (m2) K·(A.+l)=K·A.+K. KO = 1. (el) (e2) KHI = K". K. In each case, the equation is either trivial or is an immediate consequence of Theorem 61 (or both). The second piece of information we need is that |
for a finite cardinal n, n + 1 = n+ (with cardinal addition). This holds because nand {n} are disjoint sets of cardinality nand 1, respectively, and hence n + 1 = card(n u {n}) = card n+ = n+. It remains only to go through the motions of the induction. Consider any m E OJ and let T = {n E OJ I m + n = m + w n}. 144 6. Cardinal Numbers and the Axiom of Choice Then ° E A since n + 0= n = n + ro 0, by (al) and (AI), where (AI) is given in Theorem 41. Suppose that k E T. Then m + e = m + (k + 1) =(m+k)+1 =(m+rok)+1 = (m +rok)+ by (a2) since kET by (A2). Hence e E T, T is inductive, and T = w. For multiplication and exponentiation the argument is identical. -l Corollary 6K If A and B are finite, then A u B, A x B, and B A are also finite. Proof Let m = card A and n = card B. Then we calculate: card(A x B) = card A. card B = m. n = m 'ro nEW. A similar argument applies to B A and mn. For union we must use disjoint sets: A u B = A u (B - A). B - A is a subset of a finite set and hence (by Corollary 6G) is finite. Let k = card(B - A). Then card(A u B) = m + k = m +rok E W. -l The above corollary can also be proved without using arithmetic (Exercises 8 and 9). Exercises 10. Prove part 4 of Theorem 61. 11. Prove part 5 of Theorem 61. 12. The proof to Theorem 61 involves eight instances of showing two sets to be equinumerous. (The eight are listed in the proof of the theorem as statements numbered 1-6.) In which of these eight cases does equality actually hold? 13. Show that a finite union of finite sets is finite. That is, show that if B is a finite set whose members are themselves finite sets, then U B is finite. 14. Define a |
permutation of K to be any one-to-one function from K onto K. We can then define the factorial operation on cardinal numbers by the equation K! = cardUlfis a permutation of K}, where K is any set of cardinality K. Show that K! is well defined, i.e., the value of K! is independent of just which set K is chosen. Ordering Cardinal Numbers 145 ORDERING CARDINAL NUMBERS We can use the concept of equinumerosity to tell us when two sets A and B are of the same size. But when should we say that B is larger than A? Definition A set A is dominated by a set B (written A ~ B) iff there is a one-to-one function from A into B. For example, any set dominates itself. If A c:::; B, then A is dominated by B, since the identity function on A maps A one-to-one into B. More generally we have: A ~ B iff A is equinumerous to some subset of B. This is just a restatement of the definition, since f is a function from A into B iff it is a function from A onto a subset of B (see Fig. 38). F A Fig. 38. F shows that A ~ B. We define the ordering of cardinal numbers by utilizing the concept of dominance: card A ~ card B iff A ~ B. As with the operations of cardinal arithmetic, it is necessary to check that the ordering relation is well defined. For suppose we start with two cardinal numbers, say K and A. In order to determine whether or not K ~ A, our definition demands that we employ selected representatives K and L for which K = card K and A = card L. Then K~A iff K~L. But the truth or falsity of" K ~ A" must be independent of which selected representatives are chosen. Suppose also that K = card K' and A = card £. To avoid embarrassment, we must be certain that K ~ L iff K' ~ £. To prove this, note first that K ;::= K' and L ;::= £ (because card K = card K' and card L = card £). If K ~ L, then we have one-to-one maps (i) from K' onto K, (ii) from K into L, and (iii) from L onto £. By composing the |
three functions, we can map K' one-to-one into £, and hence K' ~ £. 146 6. Cardinal Numbers and the Axiom of Choice We further define K < A. iff K ~ A. and K #- A.. Thus in terms of sets we have card K < card L iff K ~ Land K * L. Notice that this condition is stronger than just saying that K is equinumerous to a proper subset of L. After all, OJ is equinumerous to a proper subset of itself, but we certainly do not want to have card OJ < card OJ. The definition of" <" has the expected consequence that K ~ A. iff either K < A. or K = A.. Examples 1. If A <:::; B, then card A ~ card B. Conversely, whenever K ~ A., then there exist sets K <:::; L with card K = K and card L = A.. To prove this, start with any sets C and L of cardinality K and A., respectively. Then C ~ L, so there is a one-to-one function f from C into L. Let K = ranf; then C ~ K <:::; L. 2. For any cardinal K, we have 0 ~ K. 3. For any finite cardinal n, we have n < ~o. (Why?) For any two finite cardinals m and n, we have m~n => m<:::;n => m~n. Furthermore the converse implications hold. For if m ~ n, then m ~ nand there is a one-to-one function f: m --+ n. By the pigeonhole principle, it is impossible to have n less than m, so by trichotomy m ~ n. Thus our new ordering on finite cardinals agrees with the epsilon ordering of Chapter 4. 4. K < 2" for any cardinal K. For if A is any set of cardinality K, then 9 A has cardinality 2". Then A ~ 9 A (map x E A to {x} E 9 A), but by Cantor's theorem (Theorem 6B) A * 9A. Hence K ~ 2" but K #- 2", i.e., K < 2". In particular, there is no largest cardinal number. The first thing to prove about the ordering we have defined for cardinals is that it actually behaves like something we would be |
willing to call an ordering. After all, just using the symbol "~" does not confer any special properties, but it does indicate the expectation that special properties will be forthcoming. For a start, we ask if the following are valid for all cardinals K, A., and /1: 1. K ~ K. K ~ A. ~ /1 => K ~ /1. 2. 3. K ~ A. & A. ~ K => K = A.. 4. Either K ~ A. or A. ~ K. Ordering Cardinal Numbers 147 The first is obvious, since A ~ A holds for any set A. The second item follows at once from the fact that A~B&B~C = A~C. (We prove this by taking. the composition of two functions.) The third item is nontrivial. But the assertion is correct, and is called the Schroder Bernstein theorem. We will also prove the fourth item, but that proof will require the axiom of choice. First we will prove the Schroder-Bernstein theorem, which will be a basic tool in calculating the cardinalities of sets. Typically when we want to calculate card S for a given set S, we try to squeeze card S between upper and lower bounds. If possible, we try to get these bounds to coincide, K ~ card S ~ K, whereupon the Schroder-Bernstein theorem asserts that card S = K. We will see examples of this technique after proving the theorem. SchrOder-Bernstein Theorem (b) For cardinal numbers K and A, if K ~ A and A ~ K, then K = A. If A ~ Band B ~ A, then A ~ B. (a) Proof It is done with mirrors (see Fig. 39). We are given one-to-une functionsf: A --+ Band g: B --+ A. Define Cn bX recursion, using the formulas and Co = A - Cn+ = g[f[CnU ran 9 Thus Co is the troublesome part that keeps 9 from being a one-to-one correspondence between Band A. We bounce it back and forth, obtaining C l' C2, •••• The function showing that A ~ B is the function h: A --+ B defined by h(x) = {f(X) g-1(X) if x E Cn otherwise. for some n, Note that in the second case (x E A |
bqt x ¢ Cn for any n) it follows that x ¢ Co and hence x E ran g. So g-1(XJ makes sense in this case. A: B: Fig. 39. The Schroder-Bernstein theorem. 148 6. Cardinal Numbers and the Axiom of Choice Does it work? We must verify that h is one-to-one and has range B. Define Dn = f[C n], so that Cn+ = g[Dnl To show that h is one-to-one, consider distinct x and x' in A. Since both f and g-1 are one-to-one, the only possible problem arises when, say, x E Cm and x' ¢ Unero Cn' In this case, whereas h(x)=f(x)EDm' h(x') = g-1(X') ¢ Dm, lest x' E Cm+. So h(x) #- h(x'). Finally we must check that ran h exhausts B. Certainly each Dn c:::; ran h, because Dn = h[Cnl Consider then a point y in B - UneroDn' Where is g(y)? Certainly g(y) ¢ Co. Also g(y) ¢ Cn+, because Cn+ = g[Dn], y ¢ Dn, and 9 is one-to-one. So g(y) ¢ C n for any n. Therefore h(g(y)) = 9 - 1 (g(y)) = y. This shows that y E ran h, thereby proving part (a). Part (b) is a restatement of part (a) in terms of cardinal numbers. -l The Schroder-Bernstein theorem is sometimes called the "Cantor Bernstein theorem." Cantor proved the theorem in his 1897 paper, but his proof utilized a principle that is equivalent to the axiom of choice. Ernst Schroder announced the theorem in an 1896 abstract. His proof, published in 1898, was imperfect, and he published a correction in 1911. The first fully satisfactory proof was given by Felix Bernstein and was published in an 1898 book by Borel. Examples The usefulness of the Schroder-Bernstein theorem in calculat ing cardinalities is indicated by the following examples. 1. If A c:::; B c:::; C and A ~ C, then all |
three sets are equinumerous. To prove this, let K = card A = card C and let Ie = card B. Then by hypothesis K ~ Ie ~ K, so by the Schroder-Bernstein theorem K = Ie. 2. The set!R of real numbers is equinumerous to the closed unit interval [0, 1]. For we have (0,1) c:::; [0, 1] c:::;!R, and (as noted previously)!R ~ (0, 1). Thus by the preceding example, all three sets are equinumerous. (For a more direct construction of a one-to-one correspondence between!R and [0, 1], we suggest trying Exercise 4.) 3. If K ~ Ie ~ /1, then, as we observed before, K ~ /1. We can now give an improved version: K ~ Ie < /1 K < Ie ~ /1 => => K < /1, K < /1. Ordering Cardinal Numbers 149 For by the earlier observation we obtain K ~ /1; if equality held, then (as in the first example) all three cardinal numbers would coincide. 4. IR ~ "'2, and hence IR ~ f!J>OJ. Thus the set of real numbers is equi numerous to the power set of OJ. To prove this it suffices, by the Schroder-Bernstein theorem, to show that IR ~ "'2 and "'2 ~ IR. To show that IR ~ "'2, we construct a one-to-one function from the open unit interval (0, 1) into "'2. The existence of such a function, together with the fact that IR ~ (0, 1), gives us IR ~ (0, 1) ~ "'2. The function is defined by use of binary expansions of real numbers; map the real whose binary expansion is 0.1100010... to the function in "'2 whose successive values are 1, 1, 0, 0, 0, 1, 0,.... In general, for a real number z in (0,1), let H(z) be the function H(z): OJ -+ 2 whose value at n equals the (n + l)st bit (binary digit) in the binary expansion of z. Clearly H is one-to one. (But H does not have all of "'2 for its range. Note that 0.1000:.. |
= 0.0111... = 1. For definiteness, always select the nonterminating binary expansion.) To show that "'2 ~ IR we use decimal expansions. The function in "'2 whose successive values are 1, 1, 0, 0, 0, 1, 0,... is mapped to the real number with decimal expansion 0.1100010.... This maps "'2 one-to-one into the closed interval [0, n 5. By virtue of the above example, Consequently the plane IR x IR has cardinality card IR = 2No • Thus the line IR is equinumerous to the plane IR x IR. This will not come as a surprise if you have heard of "space-filling" curves. The next theorem shows that the operations of cardinal arithmetic have the expected order-preserving properties. Theorem 6L Let K, A, and /1 be cardinal numbers. (a) K ~ A = K + /1 ~ A + /1. (b) K ~ A = K. /1 ~ A. /1. K ~ A = KIl ~ All. (c) (d) K ~ A"; /1" ~ /1'-; if not both K and /1 equal zero. Proof Let K, L, and M be sets of cardinality K, A, and /1, respectively. Then MK has cardinality KIl, etc. We assume that K ~ A; hence we may 150 6. Cardinal Numbers and the Axiom of Choice select K and L such that K <;; L. And we may select M so that L,nM=0. Parts (a), (b), and (c) are now immediate, since K u M <;; L u M, K x M <;; Lx M, MK<;;ML. For part (d), first consider the case in which 11 = O. Then K #- 0 and 11" = 0 ~ 11'-· There remains the case in which 11 #- 0, i.e., M #- 0. Take some fixed a EM. We need a one-to-one function G from KM into LM. For any IE KM, define G(f) to be the function with domain L for which G(f)(x) = {~(x) if x E K, if x E L - K. In one line, G(f) = I u ((L - K) x {a}). Then G |
: KM --+ LM and G is clearly one-to-one. -l Example We can calculate the product ~o. 2No by the method of upper and lower bounds: whence equality holds throughout. We would like to show that ~o is the least infinite cardinal; that is, that ~o ~ K for any infinite cardinal K. This amounts (by the definition of ~) to showing that OJ ~ A for any infinite A. We might attempt to define a one-to-one function g: OJ --+ A by recursion: g(O) = some member of A, g(n+) = some member of A - g[n+]. Here A - g[ n +] is nonempty, lest A be finite. A minor difficulty here is that g(n+) is being defined not from g(n) but from g[n+] = {g(O), g(1),..., g(n)}. This difficulty is easily circumvented, and will be circumvented in the proof of Theorem 6N. A major difficulty is the phrase "some member." Unless we say which member, the above cannot possibly be construed as defining g. What is needed here is the axiom of choice, which will enable us to convert "some member" into a more acceptable phrase. Exercises 15. Show that there is no set d with the property that for every set there is some member of d that dominates it. Axiom of Choice 151 16. Show that for any set S we have S ~ s2, but S * s2. (This should be done directly, without use of f?}JS or cardinal numbers. If F: S --+ s2, then define g(x) = 1 - F(x)(x).) 17. Give counterexamples to show that we cannot strengthen Theorem 6L by replacing "~" by "<" throughout. AXIOM OF CHOICE At several points in this book we have already encountered the need for a principle asserting the possibility of selecting members from nonempty sets. We can no longer postpone a systematic discussion of such a principle. There are numerous equivalent formulations of the axiom of choice. The following theorem lists six of them. Others will be found in the exercises. Theorem 6M The following statements are equivalent. (1) Axiom of choice, I. For any relation R, there is a function F c:::; R with dom F = dom |
R. (2) Axiom of choice, II; multiplicative axiom. The Cartesian product of nonempty sets is always nonempty. That is, if H is a function with domain I and if (Vi E 1) H(i) #- 0, then there is a functionfwith domain I such that (Vi E 1)f(i) E H(i). (3) Axiom of choice, III. For any set A there is a function F (a "choice function" for A) such that the domain of F is the set of non empty subsets of A, and such that F(B) E B for every nonempty B c:::; A. (4) Axiom of choice, IV. Let.s4 be a set such that (a) each member of.s4 is a nonempty set, and (b) any two distinct members of.s4 are disjoint. Then there exists a set C containing exactly one element from each member of.s4 (i.e., for each BE.s4 the set en B is a singleton {x} for some x). (5) Cardinal comparability. For any sets C and D, either C ~ D or D ~ C. For any two cardinal numbers K and A, either K ~ A or A ~ K. (6) Zorn's lemma. Let.s4 be a set such that for every chain /11 c:::;.s4, we have U/1l E.s4. (/11 is called a chain iff for any C and D in /11, either C c:::; D or Dc:::; C.) Then.s4 contains an element M (a "maximal" element) such that M is not a subset of any other set in.s4. Statements (1)-(4) are synoptic ways of saying that there exist uniform methods for selecting elements from sets. On the other hand, statements (5) and (6) appea.r to be rather different. Proof in part First we will prove that (1 )-(4) are equivalent. 152 6. Cardinal Numbers and the Axiom of Choice (1) = (2) To prove (2), we assume that H is a function with domain I and that H(i) #- 0 for each i E I. In order to utilize (1), define the relation R = |
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