Real Analysis with Economic Applications - Chapter E potx

Chapter E Con tinuit y II A function that maps every elemen t of a given set to a nonempty subset of another set is called a correspondence (or a mu ltifunction). Such maps arise quite frequ e ntly in optimization theory and theoretic al econom ics. While this is not really a standard topic in real analysis, this chapter is devoted to the a nalysis of correspondences, because of their importance for economists. Our first task i s to understand in what sense a correspondence can be v iew ed as “contin uous.” After a brief se t-th eoretical overview of co rrespondences, we thu s spend some time examining various con tinuity concepts for correspondences. These concepts are needed to state and pro ve Berge’s Maximum Theorem whic h tells one when the s o lut ion to an o ptimization problem depends on the parameters of the prob- lem in a “ continuo us” way. Along with a few r elatively str aightforwa rd ap plication s that attest to the importance of this result, the cha p ter touches also upon a major topic of optimization theory, namely, the theory of stationary dynamic programming. We discuss, in particular, t he issue of existence, u niqueness and monotonicit y of solu- tions o f a dynam ic programming problem at s om e length, and illustrate o u r find in gs through a st anda rd topic in macroeconom ics, namely, the one-sector op timal grow th model. Delving in to the theory of contin uous correspondences more deeply, the rest of the c hap ter is a bit more advanced than the earlier sections. In th is part, we introduce the partition of unity, and pro v e the famous Michael Selection Theorem which pro vides sufficient conditions to “find” a continuous function within a given correspondence (in the sense that the graph of the function is contained in that of the co rrespondence). In turn, by using this result and the Brouwer Fixed Point Theorem, we derive the celebrated Kakutani Fixed P oint Theorem whic h is used frequently in equilibrium analysis. While it is of a different flav o r, N ad ler’s C ontraction Correspondence T h e- orem also gets some at tention in the same section, for this result ties in closely with our earlier work on con traction s. Som e of these fixed point theorem s are put to good use in the final section o f the chapter where we elaborate on t h e notion of Nash equilibrium. 1 1 As for general references, I should note that Berge (1963) and Border (1989) provide more comprehensive treatments of continuous correspondences than is given here, Berge in topological spaces, Border in Euclidean spaces. (But be warned that Berge includes the property of compact- valuedness in th e definition of upper hemicontinuit y, so, for instance, an upper hemicontinuous correspondence always satisfies the closed graph pro perty in his definition.) The se tw o books a lso provide quite comprehensiv e analyses of topics like parametric con tinuity (in optimization problems), continuous selections, and fixed point theory for correspondences. These issues are also studied in Sundaram ( 1998) at an in troductory level, and in Klein and Thompson (1984) and Aliprantis and Border (1999) at a more advanced level. I will mention more specialized references as we go along. 213 1 Corr espondence s By a correspondence Γ from a nonempty set X into another nonem pty set Y, we mean a m ap from X into 2 Y \{∅}. Thus, for eac h x ∈ X, Γ(x) is a nonempty subset of Y. 2 We write Γ : X ⇒ Y to denote that Γ is a correspondence from X in to Y. Here X is called the domain of Γ, and Y the codomain of Γ. For any S ⊆ X, we let Γ(S):=  {Γ(x):x ∈ S}. (Note. Γ(∅)=∅.)ThesetΓ(X) is called the range of Γ. If Γ(X)=Y ,wesay that Γ is a surjectiv e correspondence,andifΓ(X) ⊆ X, we refer to Γ as a self-correspondence on X. Of course, e very function f ∈ Y X can be viewed as a particular correspondence from X into Y. Indeed, there is n o difference between f and the co rrespondence Γ : X ⇒ Y defined by Γ(x):={f(x)}. 3 Conv ersely, if Γ is s ingle-valued,thatis, |Γ(x)| =1for all x ∈ X, then it can be though t of as a function mapping X into Y. We will t hus identify the terms “single-valued correspondence” and “ function” in the follo wing discussion. Before we mo ve on to the formal i nvestigation of continuou s correspondences, w e p resent some b asic exam ples from econ om ics which may serve as a motiva tion for muc h of what follows. Som e of these examples will be further developed in due course. E{dpsoh 1. [1] For any n ∈ N,p∈ R n ++ and ι > 0, define B(p, ι):={x ∈ R n + : n  i=1 p i x i ≤ ι}, which is c a lled, you would surely recall, the budget set of a consumer w ith i ncome ι at prices p (Example D.4.[2]). If we treated p and ι as variables, then it would be necessary to v iew B as a c orrespondence. We have B : R n+1 ++ ⇒ R n + . [2] Let X be a nonempt y set and  apreferencerelationonX. Recall that the upper contou r set of an alternative x ∈ X with respect to  is defined a s U  (x):={y ∈ X : y  x}. 2 Alternatively, one ma y view a correspondence simply as a (binary) relation from X to Y, thereby identifying a correspondence with i ts graph. A moment’s reflection shows tha t the situa tion is analogous to the wa y one may view a function as a particular relation (Section A.1.5). By the way, the math literature is not unified in the way it refers to correspondences. Some mathematicians call them multifunctions, some many-valued maps, and still others refer to them as set-valued maps. In the economics literature, however, the term correspondence se ems widely agreed upon. 3 It is not true that f(x)=Γ(x) here, so from a purely formal point of view we cannot quite say that “f is Γ.” But this is splitting hairs, really. 214 Since  is reflexive, x ∈ U  (x), so U  (x) = ∅ for an y x ∈ X. Th us we ma y view U  as a well-defined self-correspondence on X. This corresponden ce c ontains all the information that  ha s: y  x iff y ∈ U  (x) for any x, y ∈ X. In practice it is simp ly a m atter of convenience to w ork with  or with U  . [3] Let X be a metric space and  an upper semicontin uous preference relation on X.Letc(X) denote the set of all nonempty comp act subsets of X,anddefine C  : c(X) ⇒ X by C  (S):={x ∈ S : y  x fo r no y ∈ S}. By w ha t is p r oved in Exam ple C.7, C  is a (compact-valued) c orre sponden ce — it is called the choice corresponden ce induced b y . [4] Let T beanynonemptyset,∅ = S ⊆ T and ϕ ∈ R T . The canonical optimiza- tion problem is to find the maximum v alue that ϕ attains on S. (We say that ϕ is the objec tive function of t he p roblem, and S is its constraint s e t.) Put more precisely, t o “solve” th is problem means to id entif y all y ∈ S such that ϕ(y) ≥ ϕ(x) for all x ∈ S, that is, to c ompute arg max{ϕ(x):x ∈ S} := {y ∈ S : ϕ(y) ≥ ϕ(x) for all x ∈ S}. Thissetisreferredtoasthesolution set of the prob lem. No w take any nonempty set Θ, call it the parameter space, and suppose that the constrain t set of our canonical optimization problem depends on the parameter θ ∈ Θ. The constraint of the pr ob le m would the n be modeled by means of a c orre sponden ce of the form Γ : Θ ⇒ T, wh ich is ca lled the constrain t correspondence of the problem. This leads to the follow ing formulation, which indeed captures many o ptimization problems that arise in economics: For each θ ∈ Θ, maximize ϕ(x) subjec t t o x ∈ Γ(θ). Clearly, the solution s et arg max{ϕ(x):x ∈ Γ(θ)} o f t he problem depends i n t his case on the value of θ. Assume next that T is a m et ric spa ce, and ϕ ∈ C(T). If Γ : Θ ⇒ T is compact- valued (that is, Γ(θ) is a compact subset o f T for each θ ∈ Θ), we can t hen think of the solution to our p rob lem a s a correspondence from Θ into T. For, in this case, Weierstrass’ T heorem makes sure that σ : Θ ⇒ X is w e ll-d e fined by σ(θ):=argmax{ϕ(x):x ∈ Γ(θ)}. (1) Naturally enough, σ is called the solution correspondence of the problem. Understanding how the contin uit y (and o ther) properties o f σ depend o n those o f ϕ and Γ is one of the main issues studied in optimization theory. We will turn to this matter i n Section 3.  Exercise 1. H Define Γ : R + ⇒ R + by Γ(θ):=[0, θ], and consider the function f ∈ C(R) defined by f(t):=sint. Now d efine σ as in (1), and give an explicit formula f or it. What is σ(R + )? 215 Exercise 2. H Let (X, d) be a discrete metric space, and let ∅ = S ⊆ X. Define the self-correspondence Γ on X by Γ(x):={y ∈ S : d(x, y)=d(x, S)}. Give an explicit formula for Γ. What is Γ(X)? Exercise 3.LetΓ be a self-correspondence on a nonempty set X. Show that {Γ(x): x ∈ X} is a partition of X iff there exists an equivalence relation ∼ on X such that Γ(x)=[x] ∼ for all x ∈ X (Section A.1.3). Exercise 4. H Let X and Y be two nonempty sets and Γ : X ⇒ Y a c orrespondence. We say that Γ is injective if Γ(x)∩Γ(x  )=∅ for any distinct x, x  ∈ X, and that it is bijective if it is both injective and surjective. Prove that Γ is bijective iff Γ = f −1 for some f ∈ X Y . Exercise 5.LetΓ be a self-correspondence on a nonempty set X. An elemen t x of X is ca lled a fixed poin t of Γ if x ∈ Γ(x), and a nonempty subset S of X is called a fixed set of Γ if S = Γ(S). (a) Give an example of a self-correspondence that does not have a fixed point but that does have a fixed set. (b) P rove that if a self-correspondence has a fixed point, then it has a fixed set. 2 Contin uit y of C orrespondences Since correspondences are generalizations of fu nctions, it seems reasonable to ask if we can ex tend the no tion of “continuity,” w h ich we ha ve originally defined for func- tions, to the realm o f corresponden ces. Of course, w e should be consistent with our original definition in the sense that the notion of “contin uity” that we might define for a correspondence should r educe to ordinary contin uit y when t hat correspondence is single-valued. T here are at least two reasonable continuit y notions for correspon- dences that sa tisfy this requirement. We will tak e ea c h of these in turn. 2.1 Up per Hemico nt inuity Dhilqlwlrq. Fo r an y two metric spaces X and Y , a correspondence Γ : X ⇒ Y is said to be upper hemicontinuous at x ∈ X, if,foreveryopensubsetO of Y with Γ(x) ⊆ O, there exists a δ > 0 such that Γ(N δ,X (x)) ⊆ O. Γ is c a lle d upper hemicontin uous on S ⊆ X if it is upper hemicontinuous at each x ∈ S, and upper hemicontin uous if it is upper hemicon tinu ou s on t h e entire X. Clearly, this definition mimics that of con tinuity of a function, and it reduces to the definition of the l atter wh en Γ i s single-v alued. That i s to sa y, every upper hemi- contin uous single-valued correspondence “ is” a continuous function, and conv ersely. 216 Intuitively speaking, upper hemicontinuity at x sa ys that a small pertu r ba tion of x does not cause the i mage set Γ(x) to “suddenly” get large. What we mean by this is illustrate d in Figure 1 which depicts three correspond enc e s mapp in g [0, 1] into R + . You should not pass t his point in t he lecture before becoming a bsolutely certain that Γ 1 and Γ 3 are upper hemi contin uous everywhere but x 1 and x 2 , respectively, while Γ 2 is upper hemicontinuous. ∗∗∗∗FIGU R E E.1 ABOUT HERE ∗∗∗∗ Exercise 6. Let X and Y be two metric spaces, an d Γ : X ⇒ Y a correspondence. Define Γ −1 (O):={x ∈ X : Γ(x) ⊆ O} for a ll O ⊆ Y. (Γ −1 (O) is called the upper inverse image of O under Γ.)ShowthatΓ is upper hemicontinuous iff Γ −1 (O) is open in X for every open subset O of Y. Wh ile this ex ercise high lights the ana logy between the n otio ns o f continuity and upper hemicontinuity (recall Pro position D.1), there are som e important differences regarding the imp lications of these properties. Most notably, while a c ontinu ous func- tion maps co mpact s ets to com p act sets (Proposition D.3), this is not the case with upper hemicon tinuous correspondences. For instance, wh ile Γ :[0, 1] ⇒ R + defined by Γ(x):=R + is obviously u pper hemicontinuous, i t doesn’t map e ven a s ingleton set to a compact set. (Suppose y ou wanted to imitate the proof of Proposition D.3 to show that an upper hemicontinuous image of a compact set is compact. Where w ould the argumen t fail?) Ho wev er, in most applications, the correspondences that one deals with have some additional structure that might circumv ent this p roblem. In particular, if every value o f t he co rrespondence was a compact set, t hen we would be okay here. Dhilqlwlrq. For any two metr ic spaces X and Y , a co rrespondence Γ : X ⇒ Y is said to be compact-valued,ifΓ(x) is a compact sub set of Y for each x ∈ X. Similarly, Γ is said to be closed -valued if the image of every x under Γ is a closed subset of Y. Finally, if Y is a subset o f a Euclidean space, and Γ(x) is convex for each x ∈ X, then we sa y that Γ is convex-valued. Under the hypothesis of compact-v a luedness, we can prove a result for upper hemicontin uous correspondences t hat parallels P roposition D.3. Proposition 1. Let X and Y be two met ric spaces. If Γ : X ⇒ Y is a compact- valued and upper hemicontin uous correspondence, then Γ(S) is comp act in Y for a ny compact subset S of X. 217 Proof. Take any c ompact-valued and upper hemicon tin uous correspondence Γ : X ⇒ Y. Let O be an open c over o f Γ(S), where S ⊆ X is compact. We wish to find a finite subset of O th at would a lso c o v er Γ(S). No te that, for eac h x ∈ S, O is also an open cover of Γ(x), so, since Γ(x) is com pact, there exist finitely many open sets O 1 (x), ,O m x (x) in O such that Γ(x) ⊆  m x O i (x)=:O(x). By Exercise 6, Γ −1 (O(x)) is open in X for each x ∈ X. Moreov er, Γ(S) ⊆  {O(x):x ∈ S} so that S ⊆  {Γ −1 (O(x)) : x ∈ S}, that is, {Γ −1 (O(x)) : x ∈ S} is an open co ver of S. By compactness of S, there f ore, th ere exist finitely ma ny points x 1 , , x m ∈ S such that {Γ −1 (O(x i )) : i =1, , m} covers S. But then {O(x 1 ), , O(x m )} must cover Γ(S). (Why?) Theref ore, {O j (x i ):j =1, , m x i ,i=1, , m} is a finite subset of O that cover s Γ(S).  Recallin g how useful the sequential ch aracterization of contin u ity of a function is, we now ask if it is pos sible to give such a charac te rization f or u p per hemicont inu- ity o f a correspondence. The answer i s yes, at least i n t he case o f compact-valued correspondences. (Try the following characterization in Figure 1.) Proposition 2. Let X an d Y be two metric spaces, and Γ : X ⇒ Y a correspon- dence. Γ is upper hemicontinuous at x ∈ X if, for a ny (x m ) ∈ X ∞ and (y m ) ∈ Y ∞ with x m → x and y m ∈ Γ(x m ) for each m, there exists a subsequen c e of (y m ) that conver g es to a poin t in Γ(x). If Γ is compact-valued , then the converse is also true. Proof. If Γ is not u pper hem icon tinuous a t x, then there e xists an open subset O of Y with Γ(x) ⊆ O and Γ(N 1 m ,X (x))\O = ∅ for any m ∈ N. But then there ex ist (x m ) ∈ X ∞ and (y m ) ∈ Y ∞ such t hat x m → x an d y m ∈ Γ(x m )\O for e ac h m. (Why?) Sin ce Γ(x) ⊆ O and eac h y m belongs to the closed set Y \O, no subsequence of (y m ) can converge t o a point in Γ(x). (Recall Proposition C.1.) Co nversely, assum e that Γ is a compact-valued correspondence which is upper hemicontinuous at x ∈ X.If(x m ) is a sequence in X with x m → x,thenS := {x, x 1 ,x 2 , } is sequentially com pact in X. (W hy?) By T heorem C.2 and Proposition 1, then, Γ(S) is sequentially compact in Y .So,ify m ∈ Γ(x m ) for each m, then (y m ) —beingasequenceinΓ(S) — m ust possess a subsequence t hat con verges to some y ∈ Γ(S). With an i nnocen t abuse o f notation, we denote this subsequence a lso by (y m ), and write y m → y. We wish to show that y ∈ Γ(x). Let y/∈ Γ(x), and observe that this implies ε := d Y (y, Γ(x)) > 0 because Γ(x) is a compact (hence cl osed) set (E xercise D .2). Now d e fine T :=  z ∈ Y : d Y (z,Γ(x)) ≤ ε 2  . Since d Y (·, Γ(x)) is a con tinuou s map (Example D.1.[3]), T is a c losed set. Moreover, by definition, we have Γ(x) ⊆ int Y (T ) and y/∈ T (Figure 2). But, since Γ is upper hem icontinuous at x, there ex ists a δ > 0 such that Γ(N δ,X (x)) ⊆ int Y (T ).Thus 218 there is an M ∈ N such that y m ∈ Γ(x m ) ⊆ int Y (T ) for all m ≥ M. But t hen, since T is a close d set and y m → y, we must have y ∈ T, a contradiction.  ∗∗∗∗FIGU R E E.2 ABOUT HERE ∗∗∗∗ Here is a nice application of t h is result. E{dpsoh 2. (The Budget Correspondence)Fixanyn ∈ N, and define B : R n+1 ++ ⇒ R n + by B(p, ι):={x ∈ R n + : px ≤ ι}, (2) where px stands for the inner product of the n-vecto rs p an d x, th at is, px :=  n p i x i . We claim that B, which i s aptly called the budget c orr espondence, is upper hemi- contin uous. The proof is by means of Pro position 2. Take a n arbitrary (p, ι) ∈ R n+1 ++ , a nd sequences (p m , ι m ) and (x m ) (in R n+1 ++ and R n + , res pectively) such that lim(p m , ι m )=(p, ι) and x m ∈ B(p m , ι m ) for eac h m. Si nce (p m i ) ∈ R ∞ ++ con verges to a strictly positive real number, we have p ∗ i := inf{p m i : m ∈ N} > 0 for each i =1, , n. Sim ilarly, ι ∗ := sup{ι m : m ∈ N} < ∞. Butitisplainthatx m ∈ B(p ∗ , ι ∗ ) for each m, while B(p ∗ , ι ∗ ) is obviou sly a closed and bounded subset of R n + .Bythe Hein e-Borel Theorem and Theore m C.2, therefore, there exists a subsequence (x m k ) whic h converges to some x ∈ R n + . But then , by a s traightforward continuity argu men t, px = lim p m k x m k ≤ lim ι m k = ι, that is, x ∈ B(p, ι). Since (p, ι) w as arbitrarily chosen in R n+1 ++ we may in voke P roposition 2 to conclude tha t B is upper hemicon tin uous.  Exercise 7. Consider the s elf-correspondence Γ on [0, 1] defined as: Γ(0) := (0, 1] and Γ(t):=(0,t) for all 0 <t≤ 1. Is Γ upper hemicontinuous? Does Γ satisfy the sequential property considered in Proposition 2? Exercise 8. H Give an e xample of a compact-valued and u pper hemicon tin uous corre- spondence Γ :[0, 1] ⇒ R such that Υ :[0, 1] ⇒ R, defined by Υ(t):=bd R (Γ(t)), is not upper hemicontinuous. Exercise 9.Foranygiven n ∈ N,define Γ : R n ⇒ S n−1 by Γ(x):={y ∈ S : d(x, y)=d(x, S n−1 )}. (Recall that S n−1 is the n − 1 dimensional unit sphere (Section D.8.2).) Is Γ well- defined? Is it compact-valued? Is i t upper hemicontinuous? Exercise 10. H Define B : R n+1 + ⇒ R n + by (2). Is B upper hemicontinuous? Exercise 11. H Give an alternative proof for Proposition 1 by using Theorem C.2 and Proposition 2. 219 Exercise 12. H Let X and Y be t wo metric spaces and f ∈ Y X .Wesaythatf is a closed map if f(S) is closed in Y for every closed subset S of X. (a)Showthatf is a closed surjection iff f −1 is an upper hemicontinuous correspon- dence. (b)If f is a c on tinuous surjection, does f −1 need to be up per hemicon tinuous? Exercise 13.Let X and Y be two metric spaces, and Γ : X ⇒ Y a correspondence. Define Γ : X ⇒ Y by Γ(x):=cl Y (Γ(x)). Sho w tha t if Γ is u p per hemicontinu ous, then so is Γ. Is the converse necessarily true? Exercise 14 . Let X and Y be tw o metric spaces, and Γ 1 and Γ 2 correspondences from X into Y. (a)Define Φ : X ⇒ Y by Φ(x):=Γ 1 (x) ∪ Γ 2 (x), and show that if Γ 1 and Γ 2 are upper hem icontin uous, t hen so is Φ. (b) Assume that Γ 1 (x) ∩ Γ 2 (x) = ∅ for any x ∈ X, and define Ψ : X ⇒ Y by Ψ(x):=Γ 1 (x) ∩ Γ 2 (x). Show that if Γ 1 and Γ 2 are compact-valued and upper hemicontinuous, t hen so is Ψ. Exercise 15.Let Γ 1 and Γ 2 be an y two correspondences t hat map a metric space X in to R n . Define Φ : X ⇒ R 2n and Ψ : X ⇒ R n by Φ(x):=(Γ 1 (x), Γ 2 (x)) and Ψ(x):={y 1 + y 2 :(y 1 ,y 2 ) ∈ Γ 1 (x) × Γ 2 (x)}, respectiv ely. Show that if Γ 1 and Γ 2 are compact-valued and up per hemicontinuous, then so are Φ and Ψ. Exercise 16. A metric space is said to have the fixed set p roperty if for e very upper hemicontinuous self-correspondence Γ on X there is a nonempty closed subset S of X such that Γ(S)=S. (a)Isthefixed set property a top ological property? (b)Isittruethatthefixed set property of a metric space is inherited by all of its retracts? Exercise 17.Let T be a metric space, and F anonemptysubsetofC(T ). Define Γ : T ⇒ R by Γ(t):=  {f(t):f ∈ F}. Sho w that Γ is compact-valued and upper hemicontinuous if (a) F is finite, o r more generally, (b) F is compact in C(T ). 2.2 The Closed G raph Propert y By analogy with the gr aph of a function , the graph o f a c orrespondence Γ : X ⇒ Y , denote d by Gr(Γ), is defined a s Gr(Γ):={(x, y) ∈ X × Y : y ∈ Γ(x)}. In turn, we say that Γ has a closed graph if Gr(Γ) is closed in the product metric space X × Y . The follo wing definition express es this in s lightly different words. 220 Dhilqlwlrq. Let X and Y be two m etric spaces. A correspondence Γ : X ⇒ Y is said to be closed at x ∈ X, if for any convergen t sequences (x m ) ∈ X ∞ and (y m ) ∈ Y ∞ with x m → x and y m → y, we have y ∈ Γ(x) whenev er y m ∈ Γ(x m ) for each m =1, 2, Γ is said to hav e a closed graph (or to sa tisfy the close d g raph property)ifitisclosedateveryx ∈ X. Exercise 18.LetX and Y be two metric spaces. Prove that a correspondence Γ : X ⇒ Y has a closed graph i ff Gr(Γ) is closed in X × Y . Wh ile it is som ewh at stand ard , the choice of termin ology here is rather unfortu- nate. In particular, if Γ(x) is a closed set, this does not mean that Γ is closed at x. Indeed, even if Γ : X ⇒ Y is closed-valued, Γ need not have a closed graph. After all, any se lf-map f on R is closed-valued, but of course, a discontinuous self-map on R does not have a closed graph. But the converse is true, that is, if Γ is closed at x, then Γ(x) must be a closed subset of Y. (Why?) In sh ort, the closed gr ap h p ro perty is (much!) more demanding than being c losed-valued. It is worth n oting that having a close d graph can no t really be cons ide re d as a contin uit y pro perty. Indeed, this propert y d oes not reduce to our ordinary notion of continuity in the c as e of s in gle -valued corres pondenc es. For instance, th e graph of the function f : R + → R + where f (0) := 0 and f(x):= 1 x for all x>0, is closed in R 2 , but this function exhibits a serious discon tin uity at the origin. Thus, for single- valued correspondences, having a closed graph is in general a weaker propert y than contin uit y — the latter implies the former but not conversely. (As w e shall see shortly, howev e r, the c o nverse wou ld hold i f the codom ain was compact.) Neverth eless, t he closed graph property i s s till an interesting property to im pose on a correspondence. After all, closedness at x simply says th at “if some points in the images of points nea rby x concentra t e ar ound a particular point in the codo main, that point must be contain ed in the im age of x.” T his statement, at least in tuitiv ely, brings to mind the notion of continuity. What is more, there is in fact a tight connec tion between upper hemicontin uity and t he closed graph property. Proposition 3. Let X an d Y be two metric spaces, and Γ : X ⇒ Y a correspon- dence. (a) If Γ ha s a closed graph, then it need not be upper hemicontin uous. But if Γ has a closed graph and Y is compact, then it i s upper hem icont inuous. (b) If Γ is upper hem icon tinuous, t hen it need not hav e a closed gra ph. But i f Γ is upper hemicontin uous and closed-v alued, then it has a c losed graph. Proof. (a) I t is observed above that the closed graph property does not imply upper hemicontin uit y ev en for single-valued correspondences. To see the second claim, assume that Γ is closed at some x ∈ X and Y is compact. Take any (x m ) ∈ X ∞ and (y m ) ∈ Y ∞ with x m → x and y m ∈ Γ(x m ) for each m. By T heorem C.2, th ere e x ists a strictly increasing seque nce (m k ) ∈ N ∞ and a point y ∈ Y such that y m k → y (as 221 k →∞). Thus , in view of Proposition 2, it is enou gh to show that y ∈ Γ(x). But since x m k → x and y m k ∈ Γ(x m k ) for each k, this is an obvious consequence of the close dness of Γ at x. (b) The se lf-correspondence Γ on R + , defined by Γ(x):=(0, 1), is upper hemicon - tinuous, but it is closed no w here. The proof of the second claim is, on the other hand, ident ical to t he argument give n in the third par agraph of the proof o f Proposition 2. (Use, again , the idea depicte d in Figu re 2.)  The closed graph propert y is often easier to v erify than upper hemicon tin uit y. For this reason , Proposition 3.(a) is used i n practice quite frequently. In f act, even when w e don’t ha v e the compactness of its codomain, w e ma y be able to make use of the closed graph propert y to v erify that a giv en correspondence is upper hemicon tin uous. A case in point is illustrated in the following e xe r cise . Exercise 19 .LetX and Y be two metric spaces, and Γ 1 and Γ 2 two correspondences from X into Y with Γ 1 (x) ∩ Γ 2 (x) = ∅ for any x ∈ X. Define Ψ : X ⇒ Y by Ψ(x):=Γ 1 (x) ∩ Γ 2 (x). Prov e: If Γ 1 is compact-valued and upper hemicon tinuous at x ∈ X, and Γ 2 is closed at x, then Ψ is upper hemicon tinuous at x. Exercise 20 .LetX be a compact metric space, and Γ a self-correspondence on X. Prove: If Γ has a closed graph, then Γ(S) is a closed set whenever S is closed. Give an example to show that the converse claim is false. Exercise 21. H Let X be any metric space, Y a compact metric space, and f ∈ Y X a contin uous f unction. For any fixed ε > 0, define Γ : X ⇒ Y by Γ(x):={y ∈ Y : d Y (y, f(x)) ≤ ε}. Show that Γ is upper hemicontinuous. 4 Exercise 22. H (AFixedSetTheorem)LetX be a compact metric space, and Γ an upper hemicontinuous self-correspondence on X. Prov e that there exists a nonempty compact subset S of X with S = Γ(S). 5 4 Here is an example (which was suggested to me by Kim Border) that shows that compactness of Y is essential for this result. Pick any (q,x) ∈ Q×R\Q with q>x,and let ε := q − x. The self- correspondence Γ on R\Q,defined b y Γ(t):={t ∈ R\Q : |t − x| ≤ ε}, is not upper hemicontinuous at x. To see this, note first that Γ(x):=[2x−q, q)∩R\Q. (Why?) Now let O := (2x−q −1,q)∩R\Q, and observe that O is an open subset of R\Q with Γ(x) ⊆ O. But it is clear that Γ(N δ (x))\O = ∅ for any δ > 0. 5 Digression. There is more to the story. For any metric space X, a nonempty s ubset S of X is said to be an almos t-fixed set of Γ : X ⇒ X, if Γ(S) ⊆ S ⊆ cl X (Γ(S)). The following identifies exactly when Γ has a compact almost-fixed set. Theorem. Γ has a compact almost-fixed set iff X is compact. To prove the “if” part, we let A := {A ∈ 2 X \{∅} : A is closed and A ⊇ cl X (Γ(A))}, and apply Zorn’s Lemma to the poset (A, ⊇). (The “only if” part is a bit trickier.) I hav e actually written on this topic elsewhere; see Ok (2004) for a detailed treatment of fixed set theo ry. 222 [...]... D.1.[3] and Weierstrass’ Theorem.) You are asked below to verify that (c(Y ), dH ) is indeed a metric space We also provide several other exercises here to help you get better acquainted with the Hausdorff metric (For a more detailed analysis, see the first chapter of Nadler (1978).) ∗ ∗ ∗ ∗ FIGURE E. 3 ABOUT HERE ∗ ∗ ∗ ∗ Exercise 31 Let Y be a metric space (a) True or false: If Y is bounded, (2Y {∅},... applied in economic analysis However, our approach is slightly indirect in this regard, for we wish to do some sight-seeing on the way, and talk about a few interesting properties of continuous correspondences that are of independent interest Towards the end of the section, we will also talk about how one may carry the basic idea behind the Banach Fixed Point Theorem to the realm of correspondences 5.1... the sequence 1 1 (x1 − m ) obviously converges to x1 , the sequence of image sets (Γ2 (x1 − m )) does not get “close” to Γ2 (x1 ) While this statement is ambiguous in the sense that we do not know at present how to measure the “distance” between two sets, it is intuitive that if 1 (Γ2 (x1 − m )) is to be viewed as getting “close” to Γ2 (x1 ), then there must be at least 1 one sequence (ym ) with ym... 1] See, a pattern is emerging 8 4 here; this is how I got my “guess.” (Iterate one more time if you like) 17 We confine our attention here to Euclidean spaces, because we are about to postulate properties that require us to think about linear combinations of the elements of X It is meaningless to talk about such things in an arbitrary metric space — what we need is a linear space for this purpose Thus... property on this function Of course, this approach necessitates that we have a sensible metric on the associated codomain of sets, and the formulation of this is not really a trivial matter However, in the case of compact-valued correspondences (where the range of the correspondence is contained in the set of all compact subsets of its codomain), this approach becomes quite useful We explore this issue... ), m = 1, 2, (8) Here x0 is any element of a given metric space X While X is called the state space of the problem, x0 is thought of as the initial state of the dynamical system at hand Γ is any self-correspondence on X that lets one know which states are possible “tomorrow” given the state of the system “today.” It is called the transition correspondence of the problem A sequence (xm ) ∈ X ∞ such... Then the issue is to decide how to define the “inverse image” 223 of a set under Γ Of course, we should make sure that this definition reduces to the usual one when Γ is single-valued Well, since f −1 (O) = {x ∈ X : {f (x)} ⊆ O} and f −1 (O) = {x ∈ X : {f(x)} ∩ O = ∅}, there are at least two ways of doing this The first one leads us to the notion of upper inverse image of O under Γ, and hence to upper... for each θ, and Γ(θ) is compact, the compactvaluedness of σ would follow from its closed-valuedness (Proposition C.4) The latter 9 This result was first proved by Claude Berge in 1959 in the case of non-parametric objective functions The formulation we give here - due to Gerard Debreu - is a bit more general Walker (1979) and Leininger (1984) provide more general formulations 229 property is, on the other... do this, we need to first metrize X ∞ in such a way that the requirements of the Maximum Theorem are met While something like this can actually be done, such a direct approach turns out to be a poor method for “solving” (10) Instead, there is an alternative, recursive method that would enable us to bring the power of the Maximum Theorem to the fore in a more striking manner 4.2 The Principle of Optimality... of the prices of the commodities to be zero But it would then cease being compact-valued 225 Exercise 26 Define the correspondence Γ : [0, 1] ⇒ [−1, 1] as Γ(0) := [−1, 1] and 1 Γ(x) := sin x for any 0 < x ≤ 1 Is Γ continuous? Exercise 27.H Let X be a metric space and ϕ ∈ C(X) Show that the correspondence Γ : X ⇒ R, defined by Γ(x) := [0, ϕ(x)], is continuous Exercise 28.H Define the correspondence Γ : . Exam p le D.1.[3] and Weierstrass’ Theorem.) Youareaskedbelowtoverifythat(c(Y ),d H ) is indeed a metric space. We also provide several o ther exercises here to help you get better acquain ted. multifunctions, some many-valued maps, and still others refer to them as set-valued maps. In the economics literature, however, the term correspondence se ems widely agreed upon. 3 It is not true that f(x)=Γ(x). Som e of these fixed point theorem s are put to good use in the final section o f the chapter where we elaborate on t h e notion of Nash equilibrium. 1 1 As for general references, I should note