Real Analysis with Economic Applications - Chapter B ppt

In particular, we havelittle to say here about cardinality theory and ordinal numbers.1 We shall, however,cover two relatively advanced topics here, namely, the theory of order isomorphi

Chapter B

Countability

This chapter is about Cantor’s countability theory which is a standard prerequisitefor elementary real analysis Our treatment is incomplete in that we cover only thoseresults that are immediately relevant for the present course In particular, we havelittle to say here about cardinality theory and ordinal numbers.1 We shall, however,cover two relatively advanced topics here, namely, the theory of order isomorphismsand the Schröder-Bernstein Theorem on the “equivalence” of infinite sets If you arefamiliar with countable and uncountable sets, you might want to skip Section 1 andjump directly to the discussion of these two topics The former one is put to gooduse in the last section of the chapter which provides an introduction to ordinal utilitytheory, a topic that we shall revisit a few more times later The Schröder-BernsteinTheorem is, in turn, proved via the Tarski Fixed Point Theorem, the first of the manyfixed point theorems that are discussed in this book This theorem should certainly

be included in the tool kit of an economic theorist, for it has recently found a number

of important applications in game theory

In this section we revisit set theory, and begin to sketch a systematic method ofthinking about the “size” of any given set The issue is not problematic in the case offinite sets, for we can simply count the number of members of a given finite set, anduse this number as a measure of its size Thus, quite simply, one finite set is “morecrowded” than another, if it contains more elements than the other But how canone extend this method to the case of infinite sets? Or, how can we decide whether

or not a given infinite set is “more crowded” than another such set? Clearly, thingsget icy with infinite sets One may even think at first that any two infinite sets areequally crowded, or even that the question is meaningless There is, however, anintuitive way to approach the problem of ranking the sizes of infinite sets, and it isthis approach that the current section is about As a first pass, we will talk aboutthe sense in which an infinite set can be thought of as “small.” We will then applyour discussion to compare the sizes of the sets of integers, rational numbers, and realnumbers Various other applications 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), Kaplansky (1977), Enderton (1977), and/or Devlin (1993) My favorite

is, by far, Enderton (1977).

2 We owe our ability 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 provided a precise manner in which infinity can be understood, studied and

Let us begin by defining the two fundamental concepts that will play a tal role in what follows.

fundamen-Dhilqlwlrq A set X is called countably infinite if there exists a bijection f thatmaps X onto the set N of natural numbers X is called countable if it is either finite

or countably infinite X is called uncountable if it is not countable

So, quite intuitively, we can “count” the members of a countable set just like

we could count the natural numbers An infinite set like X = {x1, x2, } is thuscountable, for we can put the members of X into one-to-one correspondence with thenatural numbers:

x1 ←→ 1 x2 ←→ 2 · · · xm ←→ m · ··

Conversely, if X is countably infinite, then it can be enumerated as X = {x1, x2, }

To see this formally, let f be a bijection from X onto N Then f is invertible sition A.2), and the inverse function f−1 is a bijection from N onto X But this meansthat we must have X = {f−1(1), f−1(2), } Thus, if we let xi = f−1(i), i = 1, 2, ,

(Propo-we may write X = {x1, x2, } nice and easy

Remark 1 Let X be a countable subset of R Can we meaningfully talk aboutthe sum of the elements of X? Well, it depends! If X is finite, there is of course

no problem; the sum of the elements of X, denoted x∈Xx is well-defined.3 When

X is countably infinite, we need to be a bit more careful The natural inclination

is to take an enumeration {x1, x2, } of X, and define x∈Xx as ∞xi, providedthat ∞xi ∈ R Unfortunately, this need not well-define x∈Xx— what if we used adifferent enumeration of X? You see, the issue is none other than the rearrangement ofinfinite series that we discussed in Section A.3.5 The right definition is: x∈Xx :=

∞xi where {x1, x2, } is any enumeration of X, provided that ∞xi ∈ R and

∞xi is invariant under rearrangements (If any of the latter two conditions fails,

we say that x∈Xxis undefined.) In particular, thanks to Proposition A.10, x∈Xx

even manipulated It would not be an exaggeration to say that Cantor’s ideas were decades 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 some eminent figures (such as Klein, Kronecker and Poincaré), found Cantor’s theory of infinite sets nonsensical, which has later found to provide a sound foundation for much of mathematics at 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 one can say something like this.

There are many 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.

Let us look at a few examples of countable sets N is countable, as it is evidentfrom the definition of countability It is also easy to see that any subset of N iscountable 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:

Similarly, the set of all prime numbers is countable.4 So, in a sense, there are asmany even (or prime) numbers as there are natural numbers.5 Since the set of alleven numbers is a proper subset of N, this may appear counter-intuitive at first.Nonetheless, this sort of a thing is really in the nature of the notion of “infinity,”and it simply tells us that one has to develop a new kind of “intuition” to deal withinfinite sets The following result is a step towards this direction

Proposition 1.Every subset of a countable set is countable

Proof Let X be a countable set, and take any subset S of X If S is finite, there

is nothing to prove, so say |S| = ∞ Then X must be countably infinite, and thus wecan enumerate it as X = {x1, x2, } Now define the self-map f on N inductively asfollows:

Exercise 1 Show that ifB is a countable set, and if there exists an injection from

a setA intoB,then Amust be countable.

4 This set is, in fact, countably infinite, for according to a celebrated theorem of Euclid, there are 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 + mx i is prime (One needs to use in the argument the fact that every integer k > 1 is either prime or a product of primes, but this can easily be proved by using the Principle of Mathematical Induction.)

5 This observation is popularized by means of the following anectode One night countably nitely many passengers came to Hilbert’s Hotel, each looking for a room Now Hilbert’s Hotel did contain countably infinitely many rooms, but that night all of the rooms were occupied This was

infi-no problem for Hilbert He asked everybody staying in the hotel to come down to the lobby, and reallocated each of them using only the even-numbered rooms This way all of the newcomers could

be accommodated in the odd-numbered rooms (Here Hilbert refers to David Hilbert, one of the most prominent figures in the history of mathematics We will meet him later in the course.)

Exercise 2.H Show that every infinite set has a countably infinite subset.

Proposition 1 shows how one can obtain new countable sets by “shrinking” acountable set It is also possible to “expand” a countable set in order to obtainanother countable set For instance, Z+ is countable, for f : i → i + 1 defines abijection from N ∪ {0} onto N.6

Similarly, the set Z of all integers is countable.Indeed, the map f : N → Z\{0}, defined by f(i) := i+12 if i is odd, and f (i) := −2i if

i is even, is a bijection More generally, we have the following useful result

Proposition 2.A countable union of countable sets is countable

Proof Let Xibe a countable set for each i ∈ N A moment’s reflection will convinceyou that we would be done if we could show that X := X1∪X2∪··· is a countable set.(But make sure you are really convinced.) It is without loss of generality to assumethat Xi ∩ Xj = ∅ for very distinct i and j (Right?) Since each Xi is countable, wemay enumerate it as Xi ={xi

Exercise 3 Counting the members ofA× B as indicated in Figure 1, show that if

Aand B are countable sets, then so is A× B

Exercise 4 (Another Proof for Proposition 2 ) Using the notation of the proof of Proposition 2, define g ∈ NX byg(xk

i) := 2i3k Show that g is injective and invoke Exercise 1 to conclude that X is countable.

∗ ∗ ∗ ∗ FIGURE B.1 ABOUT HERE ∗ ∗ ∗ ∗

An important implication of these results concerns the countability of the set Q ofall rational numbers Indeed, we have Q = {Xn: n ∈ N} where Xn ={mn : m∈ Z}for all n ∈ N But, obviously, there is an injection from each Xn into Z, so since Z

is countable, so is each Xn (Exercise 1) Therefore, by Proposition 2, we find that

Q must be countably infinite (By Proposition 1, Q ∩ I is also countably infinite forany interval I in R.)

Corollary 1 (Cantor) Q is countable

6 What would Hilbert do if a new customer came to his hotel when 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) onto

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?)

So, there are as many rational numbers as there are natural numbers! This istruly a “deep” observation It tells us that Q can be enumerated as {q1, q2, }, while

we do not at all know how to “construct” such an enumeration Suffices it to notethat Cantor himself was fascinated with this theorem about which he is often quoted

as saying “I see it, but I don’t believe it.”8

The countability of Q has profound implications For instance, it implies that any(nondegenerate) interval partition of R must be countable, an observation which iscertainly worth keeping in mind This is because in any given nondegenerate interval,there exists at least one rational number (in fact, there are infinitely many of them).Thus, if there existed uncountably many nonoverlapping intervals in R, we coulddeduce that Q contains an uncountable subset, which contradicts Proposition 1 Infact, we can say something a little stronger in this regard

Proposition 3.Let I be a set of nondegenerate intervals in R such that |I ∩ J| ≤ 1for any I, J ∈ I Then I is countable

Proof Given the nondegeneracy hypothesis, for each I ∈ I, there exist real bers aI < bI with (aI, bI) ⊆ I, so by Proposition A.6, we can find a qI ∈ Q with

num-qI ∈ I Define f : I → Q by f(I) := qI.Since any two members of I overlap at most

at one point, f is injective (Yes?) Thus the claim follows from the countability of Q

So much for countable sets, what of uncountable ones? As you might guess, thereare plenty of them Foremost, our beloved R is uncountable Indeed, one cannotenumerate R as {x1, x2, }, or what is the same thing, one cannot exhaust all realnumbers by counting them like one counts the natural numbers Put differently, R is

“more crowded” than any countable set

Proposition 4 (Cantor) R is uncountable

There are various ways of proving this result The proof we present here hasthe advantage of making transparent the reliance of the result on the CompletenessAxiom It is based on the following useful fact

Cantor’s Nested Interval Lemma.Let Im := [am, bm]be a closed interval for each

m∈ N If I1 ⊇ I2 ⊇ · · ·, then ∞Ii =∅ If, in addition, bm− am → 0, then ∞Ii is

a singleton

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 similar to the argument that yields the countability of Q, it is considerably deeper See Dauben (1980) for a detailed historical account of the matter.

Proof I1 ⊇ I2 ⊇ · · · implies that (am) is a bounded and increasing sequencewhile (bm) is a bounded and decreasing sequence By Proposition A.8, both of thesesequences converge (and this we owe to the Completeness Axiom), so lim am = aandlim bm = b for some real numbers a and b It is easy to see that am ≤ a ≤ b ≤ bm forall m, and that [a, b] = ∞Ii (Check!) The second claim follows from the fact that

We can now prove the uncountability of the reals by means of a method that wewill later use on a few other occasions as well (Note This method is sometimesreferred to as “butterfly hunting.” You’ll see why in a second.)

Proof of Proposition 4 To derive a contradiction, suppose that [0, 1] = {a1, a2, }

=: I0 We wish to find a real number (a “butterfly”) which is not equal to ai for any

i∈ N

Divide I0 into any three closed nondegenerate intervals (say, [0, 1/3], [1/3, 2/3]and [2/3, 1]) Clearly, a1 does not belong to at least one of these three intervals, callany one such interval I1 (There has to be a “butterfly” in I1.) Now play the samegame with I1, that is, divide I1 into three closed nondegenerate intervals, and observethat a2 does not belong to at least one of these subintervals, say I2.(There has to be

a “butterfly” in I2.) Continuing this way, we obtain a nested sequence (Im)of closedintervals, so Cantor’s Nested Interval Lemma yields ∞Ii = ∅ (Aha! We caughtour “butterfly” in ∞Ii.) But, by construction, ai ∈ I/ i for each i, so we must have

While the theory of countability is full of surprising results, it is extremely usefuland its basics are not very difficult to master However, to develop a good intuitionabout the theory, one needs to play around with a good number of countable anduncountable sets The following exercises provide an opportunity to do precisely this

Exercise 5.H (a) Show thatNm is countable for any m∈ N.

(b) Prove or disprove: N∞ is countable.

Exercise 6 LetAand B be any sets such thatAis 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 LetAandB be any sets such thatAis countable andBis uncountable Show thatA∪ B is uncountable In fact, show that there is a bijection fromB onto

A∪ B

Exercise 8.H For any self-mapf onR, we say that x∈ R is a point of nuity off iff is not continuous atx.Show that iff is monotonic, then it can have

disconti-at most countably many points of discontinuity Give an example of a real function

on Rwhich has uncountably many points of discontinuity.

∗ Exercise 9.H LetI be an open interval andf ∈ RI LetDf stand for the set of all

x ∈ I such that f is differentiable at x Prove that if f is concave, then I\Df is countable.

Exercise 10.H For any setsAandB,we writeA card B,if there exists an injection fromB intoA but there does not exist an injection from AintoB

(a) Show thatA card B need not hold even if B is a proper subset ofA

(b) Show that R card Q.

(c) Show that2A card Afor any nonempty set A

(d ) Show that2N is uncountable.

(e) (Cantor’s Paradox ) Let X = {x : x is a set} Use part (c) and the fact that

2X ⊆ X to establish that X cannot be considered as a “set.”

The set of rational numbers is not only countable, but it is also ordered in a naturalway (Proposition A.4) These two properties of Q combine nicely in various applica-tions of real analysis It may thus be a good idea to see what kinds of sets we can

in general not only count like Q but also order like Q In fact, we will see in Section

4 that this issue is closely related to a fundamental problem in decision theory Inthis section, therefore, we will make its statement precise, and then outline Cantor’ssolution for it

Let us first see in what way we can relate a given linearly ordered countable set

to the set of rational numbers

Proposition 5.(Cantor) Let X be a countable set and a linear order on X Thereexists a function f : X → Q such that

f (a)≥ f(b) if and only if a  bfor any a, b ∈ X

Proof The claim is trivial when X is finite, so we assume that X is countablyinfinite Owing to their countability, we may enumerate X and Q as

X ={x1, x2, } and Q = {q1, q2, }

We construct the function f ∈ QX as follows First let f (x1) := q1 If x1  x2(x2  x1), then set f (x2) as the first element (with respect to the subscripts) in{q2, } such that q1 ≥ f(x2) (f (x2) ≥ q1, respectively) Proceeding inductively, forany m = 2, 3, , we set f (xm) as the first element of {q1, }\{f(x1), , f (xm−1)}which has the same order relation (with respect to ≥) to f(x1), , f (xm−1)as xm has

to x1, , xm−1 (with respect to ) (Why is f well-defined?) It follows readily fromthis construction that, for any a, b ∈ X, we have f(a) ≥ f(b) iff a  b 

So is it then true that the order-theoretic properties of any countable loset isidentical 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 truethat (N, ≥) and (Q, ≥) are identical with respect to all order-theoretic properties Forinstance, while N has a ≥-minimum, Q does not have a ≥-minimum The problem

is that we cannot embed (Q, ≥) back in (N, ≥) Formally speaking, these two losetsare not order-isomorphic

Dhilqlwlrq Let (X,X) and (Y,Y) be two posets A map f ∈ YX is said to beorder-preserving (or isotonic), provided that

f (x)Y f (y) if and only if xX yfor any x, y ∈ X If f ∈ YX is an order-preserving injection, then it called an order-embedding from X into Y If such an f exists, then we say that (X,X) can beorder-embedded in(Y,Y).Finally, if f is an order-preserving bijection, then it iscalled an order-isomorphism If such an f exists, (X,X)and (Y,Y) are said to

of the latter Similarly, the posets ({1, 3, 7, }, ≥) and ({2, 4, 5, }, ≥) are theoretically identical, but (N, ≥) and (Q, ≥) are not In fact, Proposition 5 saysthat every countable loset can be order-embedded in Q, but of course, not every loset

order-is order-order-isomorphic to Q Here are some more examples

Exercise 11 LetA := 12,13,14, and B := 32,53,74, 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 -maximum (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 identify those countable losets whose order structures are tinguishable from that of Q The following property turns out to be crucial for thispurpose

indis-Dhilqlwlrq Let (X,) be a preordered set and S ⊆ X If, for any x, y ∈ X suchthat x y, there exists an element s of S such 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 (Cantor) Let (X,X) and (Y,Y) be two countable losets withneither maximum nor minimum elements (with respect to their respective linearorders) If both (X,X)and (Y,Y)are order-dense, then they are order-isomorphic

Exercise 14.H Prove Proposition 6.

A special case of Proposition 6 solves the motivating problem of this section bycharacterizing the losets that are order-isomorphic to Q This finding is importantenough to deserve separate mention, so we state it as a corollary below In Section

4, we will see how important this observation is for utility theory

Corollary 2 Any countable and order-dense loset with neither maximum nor mum elements is order-isomorphic to Q (and, therefore, to Q ∩ (0, 1))

3.1 The Cardinality Ordering

We now turn to the problem of comparing the “size” of two infinite sets We havealready had a head start on this front in Section 1 That section taught us that

N and Q are “equally crowded” (Corollary 1), whereas R is “more crowded” than

N (Proposition 4) This is because we can put all members of N in a one-to-onecorrespondence with all members of Q while mapping each natural number to a realnumber can never exhaust the entire R If we wish to generalize this reasoning tocompare the “size” of any two sets, then we arrive at the following notion

Dhilqlwlrq Let A and B be any two sets We say that A is cardinally largerthan B, denoted A:card B, if there exists an injection f from B into A If, on theother hand, we can find a bijection from A onto B, then we say that A and B arecardinally equivalent, and denote this by A ∼card B

So, a set is countably infinite iff it is cardinally equivalent to N Moreover, we havelearned in Section 1 that N ∼cardN2

∼card Q while [0, 1] :card N but not N :card [0, 1].Similarly, 2S :card S but not conversely, for any nonempty set S (Exercise 10) Foranother example, we note that 2N ∼card{0, 1}∞,that is, the class of all subsets of thenatural numbers is cardinally equivalent to the set of all 0-1 sequences Indeed, the

map f : 2N → {0, 1}∞ defined by f (S) := (xSm), where xSm = 1 if m ∈ S and xSm = 0otherwise, is a bijection Here are some other examples.

Exercise 15.H Prove the following facts:

(a) R ∼card R\N; (b) [0, 1]∼card2N ; (c)R ∼card R2 ; (d ) R ∼card R∞

Exercise 16.H Prove: A set S is infinite iff there is a setT ⊂ S withT ∼card S

It is readily verified that A :card B and B :card C imply A:card C for any sets

A, B and C, that is,:card is a preorder on any given class of sets But there is still apotential problem with taking the relation:card as a basis for comparing the “sizes”

of two sets This is because at the moment we do not know what to make of thecase A :card B and B :card A Intuitively, we would like to say in this case that Aand B are equally crowded, or formally, we would like to declare that A is cardinallyequivalent to B But, given our definitions, this is not at all an obvious conclusion.What is more, if it didn’t hold, we would be in serious trouble Assume for a momentthat we can find a set S such that S :card N and N :card S but not S ∼card N.According to our interpretation, we would like to say in this case that the “size” of

S is neither larger nor smaller than N, but S :card N and not S ∼card N entail that S

is an uncountable set! Obviously, this would be a problematic situation which wouldforbid thinking uncountable sets as being much more “crowded” than countable ones.But, of course, such a problem never arises

The Schröder-Bernstein Theorem.9 For any two sets A and B such that A:card

B and B :card A, we have A ∼card B

In the rest of this section we will provide a proof of this important theorem Ourproof is based on the following result which is of interest in and of itself It is thefirst of many fixed point theorems that you will encounter in this text

Tarski’s Fixed Point Theorem Let (X,) be a poset such that

(i) if S ∈ 2X

\{∅} has an -upper bound in X, then sup(S)∈ X;

(ii) X has a -maximum and a -minimum

If f is a self-map on X such that x y implies f(x)  f(y) for any x, y ∈ X, then

it has a fixed point, that is, f (x) = x for some 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 was popularized instead by an 1898 book of Emile Borel In the same year Friedrich Schröder published an independent proof of the fact (but his argument contained a (relatively minor) error which he corrected in 1911).

10 A special case of this result is the famous Knaster-Tarski Fixed Point Theorem: Every preserving self-map on a complete lattice has a fixed point (See Exercise A.16 to recall the definition

order-of a complete lattice.)

Proof 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 := supS exists Observethat, 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 the former, we find f(x)  y for any y ∈ S.Thus f (x) is an -upper bound for S in X, which implies f(x)  x In turn, thisyields f (f (x))  f(x) which means that f(x) ∈ S But then, since x = supS, wehave x f(x) Since  is antisymmetric, we must then have f(x) = x 

This theorem is surprisingly general It guarantees the existence of a fixed point

by using the order structure at hand and nothing more For instance, it implies, as

an immediate corollary, the following beautiful result (which is, of course, not validfor decreasing functions; see Figure 2)

∗ ∗ ∗ ∗ FIGURE 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 tion, 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 those obtainedfrom the Brouwer Fixed Point Theorem (which you have probably seen somewherebefore — if not, don’t worry, we’ll talk about it quite a bit in Section D.8) should giveyou an idea about how useful Tarski’s Fixed Point Theorem is Indeed, when onelacks the continuity of the involved individuals’ objective functions, but has ratherthe monotonicity of them, this theorem does the work of the Brouwer Fixed PointTheorem to ensure the existence of equilibrium in a variety of economic models.11

func-Exercise 17 Let(X,)and f satisfy the hypotheses of Tarski’s Fixed Point rem Let Fix(f ) stand for the set of all fixed points of f on X Prove or disprove:

Theo-(Fix(f ),)is a poset with a-maximum and-minimum element.

Let us now turn back to our primary goal of proving the Schröder-BernsteinTheorem It turns out that a clever application of Tarski’s Fixed Point Theoremdelivers this result in a silver platter.12 We will first derive the following intermediatestep which makes the logic of the proof transparent

11 If you don’t know what I am talking about here, that’s just fine The importance of fixed point theory will become crystal clear as you learn more economics This text will help too; pretty much every chapter henceforth contains a bit of fixed point theory and its applications.

12 I learned this method of proof from Gleason (1998), which is also used by Carothers (2000) There are many alternative proofs See, for instance, Halmos (1960), Cox (1968) or Kaplansky (1979).

Banach’s Decomposition Theorem Let X and Y be any two nonempty sets andtake any f ∈ YX

and g ∈ XY Then X can be written as the union of two disjointsets X1 and X2, and Y can be written as the union of two disjoint sets Y1 and Y2such that f (X1) = Y1 and g(Y2) = X2

Proof Clearly, (2X,⊇) is a poset that has a ⊇-maximum and a ⊇-minimum.Moreover, for any (nonempty) class A ∈ 2X,we have sup⊇A = A Now define theself-map Φ on 2X by

Φ(S) := X\g(Y \f(S)),and observe that B ⊇ A implies Φ(B) ⊇ Φ(A) for any A, B ⊆ X We may thusapply Tarski’s Fixed Point Theorem to conclude that there exists a set S ⊆ X suchthat Φ(S) = S Then g(Y \f(S)) = X\S, and hence defining X1 := S, X2 := X\S,

Now the stage is set for settling the score, for the Schröder-Bernstein Theorem is

an almost immediate corollary of Banach’s Decomposition Theorem

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 DecompositionTheorem, there exist a doubleton partition {A1, A2} of A and a doubleton partition{B1, B2} of B such that F := f|A 1 is a bijection from A1 onto B1, and G := g|B 2 is abijection from B2 onto A2 But then h : A → B, defined by

h(x) := F (x), if x ∈ A1

G−1(x), if x ∈ A2

,

This proof is a beautiful illustration of the power of fixed point theorems Likemost proofs that are based on a fixed point argument, however, it is not constructive.That is, it does not tell us how to map A onto B in a one-to-one manner, it justsays that such a thing is possible to do If this worries you at all, then you will bepleased to hear that there are other ways of proving 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 f (X1) from Y, you end up with a set the image of which under

g is X\X 1 , that is, g(Y \f(X 1 )) = X\X 1 But the converse is also true, that is, if we can get our hands on an X1 ⊆ X with g(Y \f(X 1 )) = X\X 1 , then we are done That is, the assertion is true

iff we can find a (possibly empty) subset S of X with g(Y \f(S)) = X\S, that is, a fixed point of φ! And how do you find such a fixed point? Well, with this little structure at hand, Alfred Tarski should be the first person we should ask help from.