Real Analysis with Economic Applications - Chapter B ppt

Chapter B Countabilit y This chapter is about Cantor’s coun tability theory whic h i s a standard prerequisite for eleme ntary real analysis. Our tre atment is incom p lete in that we cover only th ose results that are im m ediately r elevant for the p resent course. In particular, we have little t o sa y here a bout ca rdinalit y theory a nd ordinal numbers. 1 We shall, however, cover tw o relativ ely ad vanced topics here, namely, the theory of o rder isomorphisms and the Schröder-Bernstein Theorem on the “equivalence” of infinite sets. If you are familiar with countable and uncountable sets, you might wan t to skip Section 1 and jump directly to the d is cu s sion of these two topics. The former o n e is put to good use in the last section of the chapter which pro v ides an introduction to ordinal utilit y theory, a topic that we s hall revisit a few m o re times later. The Schröder-Bernstein Theorem is, in turn, proved via the Tars ki F ixed Point Theorem, the first of the many fixe d point th eorems that are discussed in this book. Th is theorem should certainly be inclu d ed in the tool kit of an economic th eo rist, for it has recen tly found a number of impor tant application s in game the ory. 1 Countable and Uncountable Sets In this section we revisit set theory, and begin to sketch a syste matic meth od of thinking about the “size” o f any given set. The issue is n ot problematic in th e c ase of finite sets, for w e can simply count the number of members of a giv en finite set, and use this number as a measure o f its size. Thus, quite simply, one finite set is “more crow ded” than another, if it contains more elemen ts than the other. But how can one extend t his method t o the case o f infinite sets? O r, how c an we decide whether or not a given infinite set is “more crowded” than another suc h set? C learly, things get icy with infinite sets. O ne may even t hink at firstthatanytwoinfin ite se ts are equally crowded, or ev en that the question is meaningless. There is, how ever, an in tu itive way to approach the problem of ranking the sizes of infin ite sets, an d it is this approa ch that th e current section is about. As a first pass, we will talk about the sense in which an infin ite set can be thought o f as “small.” We will then a pp ly our discussion to compare the sizes of the sets of in tegers, rational numbers, and real n u mbers. Various other app lica t ion s will be given as we proceed. 2 1 For a rigorous introduction to these topics, and to set theory in general, you should consult on outlets like Halmos (1960), Kaplan sky (1977), Enderton (1977), and/or Devlin (1993). My favorite is, by far, Enderton (1977). 2 We ow e our abilit y to compare the sizes of infinite sets to the great German mathematician Georg Cantor (1845-1818). While the notion of infinity was debated in philosophy for over two thousand years, it was Cantor who pr ovided a precise manner in which infinity can be understood, studied and 63 Let us begin b y defining the two fundamen tal concepts that will play a fundamen- tal role in what follows. Dhilqlwlrq. AsetX is ca lled coun tably infinite if there exists a bijection f that maps X onto the set N o f natural numbers.Xis called coun table if it is either finite or countably infinite.Xis called uncoun table if it is not countable. So, quite intuitively, we can “cou nt” the members of a countable set just like w e could coun t the natural n umbers. An infinite set like X = {x 1 ,x 2 , } is thus countable, for w e can put the members of X in to one-to-one correspondence with the natural numbers: x 1 ←→ 1 x 2 ←→ 2 ··· x m ←→ m ··· Conv ersely, if X is countably infinite, then it can be enumerated as X = {x 1 ,x 2 , }. To see this for mally, let f be a bijection from X on to N. Then f is invertible (Propo- sition A.2), and the inverse fun ct ion f −1 is a bijection f r om N onto X. But this means that we must have X = {f −1 (1),f −1 (2), }. Th us, if we let x i = f −1 (i),i=1, 2, , we may write X = {x 1 ,x 2 , } nice and easy. Remark 1. Let X be a counta ble subset of R. Can we meaningfully talk about the sum of the elements of X? Well, it de pends! If X is finite, ther e is o f course no problem ; the s um of t he elements of X, denoted  x∈X x is w e ll-defined. 3 When X is coun tably infinite, w e need to be a bit m ore careful. The natural inclination is to take an enumeration {x 1 ,x 2 , } of X,anddefine  x∈X x as  ∞ x i , provided that  ∞ x i ∈ R. Unfortunately, this need not w ell-define  x∈X x —whatifweuseda different enumeration of X? You s ee, the issue is non e other th an th e rearran gem ent of infinite series that w e discussed in Section A.3.5. The right definition is:  x∈X x :=  ∞ x i where {x 1 ,x 2 , } is any en u m e ration of X, provided that  ∞ x i ∈ R and  ∞ x i is invariant under rearrangements. (If any of the latter two c o nd ition s fails, we say that  x∈X x is undefined.) In particular, thanks to Proposition A.10,  x∈X x is well-defined when X ⊆ R + .  even m anipula ted. It would not be an exaggeration to say that Cantor’s ideas were deca des ahead of his contemporaries. His strength did not stem from his capability to do hard proofs, but from his ability to think “outside the box.” Most mathematicians of his cohort, including som e eminent figures (such as Klein, Kronecker and Poincaré), found Cantor’s theory of infinite sets nonsensical, which has lat e r found to provide a sound foundation for muc h of mathematics a t large. After Cantor, so says James (2002), p. 214, “mathematics was never to be the same again.” There are not a whole lot of people in history about whom o ne can say something like this. There a re m any references that detail Cantor’s life and work. My favorite is Dauben (1980) from which one learns some mathematics as well. 3 Recall that we handle the exceptional case X = ∅ by convention:  x∈∅ x =0. 64 Letuslookatafewexamplesofcountablesets. N is countable, a s it is evident from the definition of countability. It is also easy to see that any subset of N is countable. For instance, the set of all even natural numbers is countably infinite, because we can put this set in one-to-one correspondence with N: 2 ←→ 14←→ 2 ··· 2m ←→ m ··· Similarly, the set of all prime nu mbers is countable. 4 So, in a sense, there are as man y ev en (or prime) numbers as there are natural numbers. 5 Since the set of all even numbers is a proper subset of N, thi s may appear counter-intuitive at first. Nonetheless, this sort of a thing is really in the natu re of the notion of “infinit y,” and it simply tells us that one has to dev e lop a new kind of “intuition” to deal with infinite sets. The following result is a step to wards this directio n. Proposition 1. Every subset of a co untable set is countable. Pro of. Let X be a coun table set, and tak e any su bset S of X. If S is finite, there isnothingtoprove,sosay|S| = ∞. Then X m ust be countably infinite, and thus we can enumerate it as X = {x 1 ,x 2 , }. No w define the self-map f on N inductively as follows: f(1) = min{i ∈ N : x i ∈ S} f(2) = min{i ∈ N : i>f(1) and x i ∈ S} f(3) = min{i ∈ N : i>f(2) and x i ∈ S} (So x f(1) is the first term of the sequence (x 1 ,x 2 , ) that belongs to S, x f (2) is the second, and so o n.) Now c onsider g ∈ S N defined by g(i):=x f(i) . Since g is a bijection (isn’t it?), we are done.  Exercise 1 . Show that if B is a countable set, and if there exists an injection from aset A into B, then A must be countable. 4 This set is, in fact, countably infinite, fo r according to a celebrated theorem of Euclid, there ar e infinitely many prime numbers. Euclid’s proof of this theorem is almost as famous as the theorem itself: Suppose there are finitely many primes, say, x 1 , , x m , and show that 1+  m x i is prime. (One needs t o use in the a rgument the fact that every integer k>1 is either prime o r a product of primes, but this ca n easily be proved by using the Principle of Mathematical Induction.) 5 This observation is popularized by means of the following anectode. One night countably infi- nitely man y passengers came to Hilbert’s Hotel, each looking for a room. Now Hilbert’s Hotel did contain c ountably infinitely man y rooms, but that night all of the rooms w ere occupied. This was no problem for Hilbert. He asked ev erybody staying in the hotel to come down to the lobb y, and reallocated e ach of them using only the even-numbered rooms. This wa y all of the newco mers could be accommodated in the odd-numbered rooms. (Here Hilbert refers to David Hilbert, one of the most prominent figures in the history o f mathematics. We will meet him later in the course.) 65 Exercise 2. H Show that every infinite set has a countably infinite subset. Proposition 1 shows how one can obtain new coun t ab le sets by “shrinking” a countable set. It is also possible to “expand” a coun table set in o rder to obtain another countable set. For instance, Z + is coun table, for f : i → i +1 defines a bijection from N ∪ {0} on to N. 6 Similarly, th e set Z of all integers is countab le. Indeed, the map f : N → Z\{0}, defined by f(i):= i+1 2 if i is odd, and f (i):=− i 2 if i is even, is a bijection. M ore generally, we have the follo wing useful resu lt. Proposition 2. A countable union o f countable sets is coun ta ble. Pro of. Let X i be a countable set for each i ∈ N. Amoment’sreflection will co nvince y ou that we would be done if we could sho w that X := X 1 ∪X 2 ∪··· is a co untable set. (But make su re you are really convinced.) It is without loss of gen erality t o assume that X i ∩ X j = ∅ for ve r y distinct i and j. (Right?) Since each X i is coun table, we ma y enumerate it as X i = {x i 1 ,x i 2 , } (but note t hat X i may be finite). Now d efine the mapping f : X → N 2 by f(x i k ):=(i, k). (Is f well-defined? Why?). Clearly, f is injectiv e, a nd h ence, by Exercise 1, we m ay conclude that X is coun table, pro vided that N 2 is countable. The proof is thus complete in view of the next exercise . 7  Exercise 3 . Counting the members of A × B as indicated in Figure 1, show that if A and B are countable sets, t hen so is A × B. Exercise 4.(A nother Proof for Proposition 2 ) Using the notation of the proof of Prop osition 2, define g ∈ N X by g(x k i ):=2 i 3 k . Show that g is injective and invoke Exercise 1 t o conclude that X is countable. ∗∗∗∗FIGURE B.1 ABOUT HERE ∗∗∗∗ An importa nt implication of these results con c erns the c ou ntability of the set Q of all rational n u mbers. Indeed, we have Q =  {X n : n ∈ N} where X n = { m n : m ∈ Z} for a ll n ∈ N. Bu t, o bviously, there is an injection from each X n in to Z, so since Z is countable, so is each X n (Exercise 1 ). Therefore, by Pr oposition 2, we find that Q must be countably infinite. (By Proposition 1, Q ∩ I is a lso countably i nfinite for an y interval I in R.) Corollary 1. (Cantor) Q is countable. 6 What would Hilbert do if a new customer came to h is hotel w hen all rooms were occupied? 7 Hidden in the proof is the Axiom of Choice. Quiz. Where is it? (Hint. This is subtle. Let A i be the set of all enumerations of X i (that is, the set of all bijections from N (or a finite set) on to X i ),i=1, 2, True, I know that each A i is nonempty, because each X i is countable. But in the proof I work with an element of X ∞ A i , don’t I?) 66 So, there are as many rational num bers as there are natural num bers! This is truly a “deep” observation. It tells us that Q can be enumerate d as {q 1 ,q 2 , }, while w e do not at all know ho w to “construct” such an enumeration. Suffices it to note that Cantor himself was fascina ted with this theorem about which he is often quoted as saying “I see it, but I don’t believe it.” 8 The coun ta bility of Q has profound implications. For instance, it implies that any (nondegenera te) interval partition of R must be countab le, an observation which is certainly worth keeping in min d. T h is is because in any giv en nondegenerate in terval, there exists at least one rational number (in fact, there are infinitely many of them). Th us, if there exi sted uncoun tably many nono verlapping intervals in R, we could deduce that Q contains an uncoun table subset, which co n tradicts Proposition 1. In fact, we can say som ething a little stronger in this regard. Proposition 3. Let I be a set of nondegenerate in tervals in R suc h that |I ∩ J| ≤ 1 for any I, J ∈ I.ThenI is countable. Pro of. Given the nondegeneracy hy pothesis, for each I ∈ I, there exist real num- bers a I <b I with (a I ,b I ) ⊆ I, s o b y Proposition A.6, we can find a q I ∈ Q with q I ∈ I. Define f : I → Q by f(I):=q I . Since any two mem bers of I overlap at most at one point, f is injective. (Yes?) T hus the claim follows from the countability of Q and Exercis e 1.  So much for counta ble sets, what of uncountable ones? As you might guess, there are plenty of them. Foremost, our belov ed R is uncoun table. Indeed, one cannot enumerate R as {x 1 ,x 2 , }, or what is the same thing, one cannot exhaust all real numbers by counting them lik e one coun ts the natural numbers. Put differently, R is “more crowded” than any coun table set. Proposition 4. (Can tor) R is unco un table. There are various w ays of proving this result. The proof we presen t here has the a dvantage of making transparen t the reliance of t he result o n t he Completeness Axiom. It is ba sed on the f ollow ing useful fact. Can tor’s Nested In terval Lemma. Let I m := [a m ,b m ] be a closed interval for each m ∈ N.IfI 1 ⊇ I 2 ⊇ ···,then  ∞ I i = ∅. If, in addition, b m − a m → 0, then  ∞ I i is asingleton. 8 To be more precise, however, I should note that Cantor made this remark about the possibility of constructing a bijection between a given line and a plane. While the spirit of this is simila r to the argument that yields the countability of Q, it is considerably deeper. See D auben (1980) for a detailed historical account of the matter. 67 Pro of. I 1 ⊇ I 2 ⊇ ··· im plies that (a m ) is a bounded and increasing sequence while (b m ) is a bounded and decreasing sequence. By Proposition A.8, both of t hese sequences conv erge (and this we o we to the Com p leteness Axiom), so lim a m = a and lim b m = b for some real n u mbers a and b. It is easy to see that a m ≤ a ≤ b ≤ b m for all m, and that [a, b]=  ∞ I i . (Check!) The second claim follows from the fact that b − a = lim b m − lim a m = lim(b m − a m ).  We can now prov e the uncountability of the reals by mea n s of a method that w e will later use on a few oth er occasions as well. (Note. This method is sometimes referred to as “butterfly h unting.” You’ll s ee wh y in a s econd.) ProofofProposition4. To deriv e a c ontradiction, suppose th at [0, 1] = {a 1 ,a 2 , } =: I 0 . We wish to find a real n umber (a “butterfly”) w hich is not e qual to a i for any i ∈ N. Divid e I 0 in to an y three closed nondegenerate intervals (say, [0, 1/3], [1/3, 2/3] and [2/3, 1]). Clearly, a 1 does not belong to at l east one of these thre e in tervals, call an y on e such interval I 1 . (There ha s to be a “butterfly” in I 1 .) Now play the same game with I 1 , tha t is, divide I 1 in to three closed nondegenera te intervals, and observe that a 2 does not belong to at least one of these subin tervals, say I 2 . (There has to be a “butterfly” in I 2 .) Continuing this w ay, w e obtain a nested sequence (I m ) of closed in tervals, so Can tor’s Nested Interval Lemm a y ields  ∞ I i = ∅. (Aha! We caught our “ bu tterfly” in  ∞ I i .) But, by construction, a i /∈ I i for e ach i, so we must have  ∞ I i = ∅, a contradiction.  Wh ile the the ory of countability is full of surprisin g results, it is extremely useful and its basics are not v ery difficult to m aster. H owever, to develop a good intuition about the theory, one needs to play around with a good num ber of coun table and uncountable sets. The follow ing exercises provide an opportunity to do precisely this. Exercise 5. H (a)ShowthatN m is countable for any m ∈ N. (b) Prov e or dispro ve: N ∞ is countable. Exercise 6.Let A and B be any sets such that A is countable, B is uncountable and A ⊆ B. Can B\A = {x ∈ B : x/∈ A} be countable? Is the set of all irrational numbers countable? Exercise 7. H Let A and B be any sets such that A is countable and B is uncountable. Show that A ∪ B is uncountable. In fact, show that there is a bijection from B onto A ∪ B. Exercise 8. H For any self-map f on R,wesaythatx ∈ R is a point of disc onti- nuity of f if f is not continuous at x. Show that if f is monotonic, then it can have at most countably many points of discontinuit y. Give an example of a real function on R which has uncountably many points of discontinuity. 68 ∗ Exercise 9. H Let I be an open interv al and f ∈ R I .LetD f stand for the set of all x ∈ I such that f is differentiable at x.Provethatiff is concave, then I\D f is countable. Exercise 10. H For any sets A and B, we write A  card B, if there exists an injection from B into A but there does not exist an injection from A into B. (a)ShowthatA  card B need not hold even if B is a proper subset of A. (b) Show that R  card Q. (c)Showthat 2 A  card A for any nonempty set A. (d)Showthat2 N is uncountable. (e)(Cantor’s Par adox )Let X = {x : x is a s et}. Use part (c) and the fact that 2 X ⊆ X to establish that X cannot be considered as a “set.” 2LosetsandQ The set of rational numbers is not only countable, but it is also ordered in a natur al wa y (Proposition A.4). These two properties of Q combine nicely in various applica- tions of r eal a na lysis. It may th us be a g ood idea t o see what kinds o f s ets we can in general not only count like Q but also order like Q. In fac t, we will see in Sect ion 4 that this issue is closely related t o a fundamen tal problem in decision theory. In this section, there fore, w e will make its statement precise, and then outline Cantor’s solutio n for it. Let u s first s e e in what w ay we can relate a giv e n linearly ordered coun tab le set to the set of rational n umbers. Proposition 5. (Can to r) Let X be a countable set and  alinearorderonX. There exists a function f : X → Q such that f(a) ≥ f(b) if and only if a  b for any a, b ∈ X. Pro of. The claim is trivial when X is finite,soweassumethatX is countably infinite. Owing to their countability, we may enumerate X and Q as X = {x 1 ,x 2 , } an d Q = {q 1 ,q 2 , }. We co nstruct the function f ∈ Q X as follows. First let f (x 1 ):=q 1 . If x 1  x 2 (x 2  x 1 ), then set f(x 2 ) as the first elemen t (with respect to the subscript s) in {q 2 , } such that q 1 ≥ f(x 2 ) (f(x 2 ) ≥ q 1 , respectively). Proceeding inductively, for an y m =2, 3, , we set f(x m ) as the first elemen t of {q 1 , }\{f(x 1 ), , f(x m−1 )} which has the same or der relation (with respect to ≥)tof(x 1 ), ,f(x m−1 ) as x m has to x 1 , , x m−1 (w it h res pect to ). (Why is f well-defined?) I t follows readily from this construction that, for any a, b ∈ X, we have f(a) ≥ f(b) iff a  b.  69 So is it then true that the order-theoret ic properties of any coun tab le loset is identical to that of the rational numbers? No! While, according to Proposition 5, we can embed the loset (N, ≥) in (Q, ≥) in an order-preserving manner, it is not true that (N, ≥) and (Q, ≥) ar e identical with respect to all order-theoretic properties. For instance, wh ile N has a ≥-minim um, Q does not have a ≥-minim um . The p roblem is that we cannot em bed (Q, ≥) back in (N, ≥). Formally speaking, these two losets are not order-isomorphic. Dhilqlwlrq. Let (X,  X ) and (Y, Y ) be two posets. A map f ∈ Y X is s a id to be order-preserving (or isotonic), provided that f(x)  Y f(y) if and only if x  X y for any x, y ∈ X. If f ∈ Y X is an order-pr eser vin g injection, then it called an order- embedd in g from X into Y. If such an f exists, then we say th at (X,  X ) can be order-embedded i n (Y, Y ). Finally, if f is an order-preservin g bije ction , then it is called an order-isomorphism.Ifsuchanf exists , (X,  X ) and (Y, Y ) are said to be order-isomorphic. If t wo posets are order-isomorphic, th ey are i n distinguisha b le from each other insofar as their order-theoretic properties are concerned; on e can simp ly be thought of as the relabeling of the other in this regard. In con cert w ith this view , “being order- isomorphic” acts as an equivalence relation on any given class of posets. (Proof?) In this sense, the order structures of any two finite losets of the same size are identical, but those of two infinite losets ma y be different from each other. Consider, for instance, (Z + , ≥) an d (Z − , ≥); th e ≥-m in imum of the former is t h e ≥-maximum of the latter. Similarly, the posets ({1, 3, 7, }, ≥) and ({2, 4, 5, }, ≥) are order- theoretically iden tical, but (N, ≥) and (Q, ≥) are n ot. In fact, Proposition 5 sa ys that every c ou ntable loset can be order-embedded in Q, butofcourse,noteveryloset is order-isomorphic t o Q. Here are some more exam p les. Exercise 11.LetA :=  1 2 , 1 3 , 1 4 ,  and B :=  3 2 , 5 3 , 7 4 ,  . Prove: (a) (A, ≥) and (N, ≥) are not order-isomorphic, but (A, ≥) and (A ∪ {1}, ≥) are. (b) (A, ≥) and (A ∪ B, ≥) are not order-isomorphic. Exercise 12.Let (X,  X ) and (Y,  Y ) be two order-isomorphic posets. Show that X has a  X -maximum iff Y has a  Y -maximu m. (The same also applies to the minimum elements, of course.) Exercise 13. H Prove that any poset (S, ) is order-isomorphic to (X,⊇) for some nonempty set X. It is now time to iden tify those countable losets whose order structures are indis- tinguishab le from that of Q. The following property turns o u t to be crucial for th is purpose. 70 Dhilqlwlrq. Let (X,) be a preordered set and S ⊆ X. If, for any x, y ∈ X such that x  y, there exists an element s of S suc h that x  s  y, we say that S is -dense (or, order-dense) in X. If S = X here, we simply say that X is -dense (or, order-dense ). Proposition 6 . (Can tor) Let (X,  X ) and (Y, Y ) be two countab le losets with neither maximum nor minimum elements (with respect to t h eir respective linear orders).Ifboth(X,  X ) and (Y,  Y ) are order-dense, then they are order-isomorphic. Exercise 14. H Pro ve Proposition 6. A special case of Proposition 6 solves the m otivating p rob lem of this section by c haracterizing the losets th at are order-isomorp hic to Q. This findin g is important enough to deserve separate mention, so we state it as a co rollary below. In Section 4, we will see how important this observation is for utility theory. Corollary 2. Any countable and order-den s e loset with neith e r maximum nor mini- mumelementsisorder-isomorphictoQ (and, therefore, to Q ∩ (0, 1)). 3 SomeMoreAdvancedSetTheory 3.1 The Cardinality Ordering We now turn to the problem of com paring the “size” of tw o infinite sets. We hav e already had a head start on this front in Section 1. That section taugh t us that N and Q are “ equ ally crow d ed” (Corollary 1), wh ereas R is “more crowded” than N (Proposition 4). This is because we can put all members of N in a one-to -one cor r espondence w it h all members of Q while mapping ea ch natural number to a real n umber can never exhaust the en tire R. If we wish to generalize this reasoning to compare t he “size” of a ny tw o sets, then we arrive at the f ollowing notion. Dhilqlwlrq. Let A an d B be an y t wo sets. We say that A is cardinally larger than B, denoted A : card B, if t here e xists an injection f from B into A. If, on t he other han d, we can find a bijection from A onto B, then we say t ha t A an d B are cardin ally equiva lent, and deno te this by A ∼ card B. So, a set is countably infinite iff it is cardinally equivalent to N. Moreover, we have learnedinSection1thatN ∼ card N 2 ∼ card Q while [0, 1] : card N but not N : card [0, 1]. Similarly, 2 S : card S but not convers ely, for any nonem pty set S (Exercise 10). Fo r another e xample, w e n ote that 2 N ∼ card {0, 1} ∞ , that is, the class of all subsets of the natural numbers is cardinally equ ivalen t to the set of all 0-1 seq uences. Indeed, the 71 map f :2 N → {0, 1} ∞ define d by f(S):=(x S m ), where x S m =1if m ∈ S and x S m =0 otherwise , is a bijection. He re are some other examples. Exercise 15. H Prove the following facts: (a) R ∼ card R\N;(b) [0, 1] ∼ card 2 N ;(c) R ∼ card R 2 ;(d) R ∼ card R ∞ . Exercise 16. H Prove: A set S is infinite iff there is a set T ⊂ S with T ∼ card S. It is readily verified that A : card B and B : card C imp ly A : card C for a n y sets A, B an d C, that is, : card is a preorder on an y given class of sets. But there is still a potential problem with taking the relation : card as a basis for comparing the “sizes” of two sets. This is because at the momen t we do not kno w what to mak e of the case A : card B and B : card A. Intuitive ly, we would like t o sa y in this cas e that A and B are equally cro w d ed, or formally, w e w ould like to declar e that A is cardinally equ ivalent t o B. But, given our defin itions, th is is not at all an ob v ious conclusion. What is more, if it didn’t hold, w e would be in serious trouble. Assume for a momen t that we can find a set S suc h that S : card N and N : card S but not S ∼ card N. According t o our interpretation , we would like to s ay in this case that th e “size” of S isneitherlargernorsmallerthanN, but S : card N and not S ∼ card N en tail that S is an uncountable set! Obviously, this would be a problematic situation which wou ld forbid thinking uncoun table sets as being much more “crowded” than countable ones. But, of cour s e , such a problem never arise s. The Sc h röder-Bern stein Theorem . 9 For any two sets A and B such that A : card B and B : card A, we have A ∼ card B. Intherestofthissectionwewillprovideaproofofthisimportanttheorem.Our proof is base d on the following resu lt wh ich is of interest in and of itself. It is the first of man y fixed point theorems that you will encounter in this text. Tarski’s Fixed Point T he o re m. Let (X, ) be a poset such that (i) if S ∈ 2 X \{∅} has an -upper bound in X, then sup  (S) ∈ X; (ii) X has a -maximum and a -min imum. If f is a self-map on X such that x  y implie s f(x)  f(y) for an y x, y ∈ X, then it has a fixed point, tha t is, f(x)=x for som e x ∈ X. 10 9 This result is sometimes referred to as Bernstein’s Theorem, or the Cantor-Bernstein Theorem. This is because Cantor has actually conjectured the result publicly in 1897, but he was unable to prove it. Cantor’s conjecture was proved that year by (then 19 years old) Felix Bernstein. This proof was never published; it w as popularized instead by an 18 98 book of Emile Borel. In the same year Friedric h Schröder published an independent proof of the fact (but his argument con tained a (relatively minor) e rror which he corrected in 1911). 10 A special case of this result is the famous Knaster-Tarski Fixed Point Theorem: Every order- preserving self-map on a complete lattice has a fixed point. (See Exercise A .16 to recall the definition of a c omplete lattice.) 72 [...]... Proof of the Schröder-Bernstein Theorem Let A and B be two nonempty sets, and take any two injections f : A → B and g : B → A By Banach’s Decomposition Theorem, there exist a doubleton partition {A1 , A2 } of A and a doubleton partition {B1 , B2 } of B such that F := f |A1 is a bijection from A1 onto B1 , and G := g |B2 is a bijection from B2 onto A2 But then h : A → B, defined by h(x) := F (x), if... Proposition 8 Obviously, if = ∅, then there is nothing to prove So let = ∅, and assume that X contains a countable -dense set — let’s call this set Y Clearly, there must a} is a then exist a, b ∈ Y such that b a Thus {(a, b) : a, b ∈ Y and b countably infinite set, where (a, b) := {x ∈ X : b x a} We enumerate this set as {(a1 , b1 ) , (a2 , b2 ) , } Each (ai , bi ) ∩ Y is partially ordered by so that, by the... owns x to begin with < /b> Then it is quite conceivable that she may exchange x with < /b> z, barring any status quo bias But she likes y better than z, so it shouldn’t be too difficult to convince her to trade z with < /b> y What is more, she would certainly be willing to pay at least a small amount of money to exchange y with < /b> x So, after all this trade, the agent ends up where she started, but she is now a bit poorer... Alfred Tarski should be the first person we should ask help from 74 3.2 The Well-Ordering Principle We now turn to the concept of well-ordering Since this notion will play a limited role in this book, however, our exposition will be brief D Let (X, ) be a poset The relation is said to be a well-ordering if every nonempty subset of X has a -minimum In this case X is said to be well-ordered by , and (X, )... the Schröder-Bernstein Theorem, and some of these actually “construct” a tangible bijection between the involved sets Two such proofs are reported in Halmos (1960) and Kaplansky (1977) 13 Whoa! Talk about a rabbit-out-of-the-hat proof! What is going on here? Let me explain Suppose the assertion is true, so there exists such a decomposition of X and Y Then X is such a set that when you subtract the image... this, however, be warned that transitivity of ∼ is a somewhat problematic postulate The problem is that an individual may be indifferent between two alternatives just because she fails to perceive a difference between them The now-classical argument goes as follows: “I’m indifferent between a cup of coffee with < /b> no sugar, and the same cup of coffee with < /b> one grain of sugar added I’m also indifferent between the... Let f be a self-map on X such that f (x) f (y) whenever x y Define S := {z ∈ X : f (z) z}, and observe that S is nonempty and bounded from above by (ii) (Why nonempty?) By (i), on the other hand, x := sup S exists Observe that, for any y ∈ S, we have f (y) y and x y The latter expression implies that f (x) f(y), so combining this with < /b> the former, we find f (x) y for any y ∈ S Thus f (x) is an -upper bound... countable -dense subset of R+ , but there exists an U ⊆ RR+ with < /b> |U| = 2 that represents All this is nice, but it is clear that we need stronger utility representation results for applications < /b> For instance, at present we have no way of even speaking about representing a preference relation by a continuous utility function (or a set of such functions) In fact, a lot can be said about this issue, but... relation on P(X) such that A ⊂ B implies B A.18 Prove: (a) If X = [0, 1], then is not representable by a utility function (b) If X is uncountable, then is not representable by a utility function 4.3 Utility Representation of Incomplete Preference Relations As noted earlier, there is a conceptual advantage in taking a (possibly incomplete) preorder as the primitive of analysis < /b> in the theory of rational... B. 2 ABOUT HERE ∗ ∗ ∗ ∗ Corollary 3 Every increasing self-map on [0, 1] has a fixed point Or how about the following? If n ∈ N, ϕi : [0, 1]n → [0, 1] is an increasing function, i = 1, , n, and Φ is a self-map on [0, 1]n defined by Φ(t) := (ϕ1 (t), , ϕn (t)), then Φ has a fixed point Noticing the similarity of these results with < /b> those obtained from the Brouwer Fixed Point Theorem (which you have probably . countable for any m ∈ N. (b) Prov e or dispro ve: N ∞ is countable. Exercise 6.Let A and B be any sets such that A is countable, B is uncountable and A ⊆ B. Can B A = {x ∈ B : x/∈ A} be countable?. irrational numbers countable? Exercise 7. H Let A and B be any sets such that A is countable and B is uncountable. Show that A ∪ B is uncountable. In fact, show that there is a bijection from B onto A ∪ B. Exercise. f : A → B and g : B → A. By Banach’s Decomposition Theorem, there exist a doubleton p artition {A 1 ,A 2 } of A and a doubleton partition {B 1 ,B 2 } of B such that F := f| A 1 is a bijection