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1 ARE211, Fall2012 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Contents 1. Analysis (cont) Continuous Functions Weierstrass (extreme value) theorem An inverse image formulation of continuity Upper and Lower Hemi continuous correpondences Alternative definitions of hemicontinuity Concepts related to upper hemicontinuity Summary: the three formulations (courtesy of Steve Buck) Some examples Application: Berge s Theorem and the demand correspondence 23 Preliminary draft only: please check for final version 1. Analysis (cont) 1.9. Continuous Functions A function is continuous if it maps nearby points to nearby points. Draw the graph without taking pen off paper. Graph is connected. Formally: 1

2 2 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Definition: Consider f : X R k. Fix x 0 X. The function f is continuous at x 0 if whenever {x m } m=1 is a sequence in X which converges to x 0, then {f(x m )} m=1 converges to f(x 0). The function f is continuous if it is continuous at x, for every x X. While this definition seems obvious and intuitive, all is not what it seems to be. We ve noticed ad nauseum that whether or not a sequence converges depends on the metric and the universe. In this section, we now have two different sequences, one in the domain and one in the range; so we have to worry about what kinds of things converge in the domain and what kinds of things converge in the range. So we have roughly double the number of counter-intuitive possibilities. To see what kinds of things can happen, consider the following question: Question: let f : X R k and let X be endowed with the discrete metric. What can we say about the continuity of f? Answer: f is continuous. In other words, if X is endowed with the discrete metric, then every function is continuous! The reason for this is that the discrete metric makes convergence an extremely stringent requirement: (x n ) converges to x, iff, eventually, the sequence is constant at x. But whenever this stringent condition is satisfied, i.e., we have a constant sequence in the domain, the resulting sequence in the range, {f(x n )} is also, trivially, constant. That is, convergence in the domain is so difficult to accomplish that whenever it is accomplished, convergence in the range is assured. Now consider the reverse question: Question: Let f : X R k, where R k is endowed with the discrete metric. What can we say about the continuity of f. Answer: (first attempt). An obvious first shot at an answer to this question is: f is continuous iff it is constant. The if part is right but the only if part is very wrong. (1) It matters what the metric is on X. If the metric on X is the discrete metric, then, as we just saw, every function is continuous.

3 ARE211, Fall (2) Now let s assume that the metric on X is the Euclidean metric and restrict attention to functions that map X to R 1. In this case, there are non-constant functions which are continuous, provided that the domain has holes in it. For example, (a) if the space X is a discrete space in the Euclidean metric (e.g., if X is the integers), then, obviously, every function f : X R is continuous (b) consider the function f : R\{0} R, defined by 1 if x > 0 f(x) = 1 if x < 0 (1) In this case, there s a hole in the domain at zero and we don t have to worry about sequences that would converge to zero if the domain were R. (c) Finally, let X = {1/n : n N} and consider the function f : X R defined by f(1/n) = ( 1) n. This function would seems to be discontinuous: for example, if you consider the sequence {x n } in X defined by x n = 1/n, the corresponding sequence in the range, {f(1/n)}, bounces up and down between 1 and 1. It looks like this sequence in the range ought to fail the requirement that the definition of continuity imposes, even with the regular metric on the range of the function, and most certainly with the discrete metric on the range. But, since 0 / X, the sequence {1/n} doesn t converge in X indeed, it doesn t even have any subsequences that converge so the definition of continuity imposes absolutely no restriction on any sequence in the range. Answer: (second (and correct) attempt). If f : X R k, where X is endowed with a Euclidean metric and R k is endowed with the discrete metric, then f is continuous iff f is constant on every connected subset of X. I haven t defined connected yet, but, intuitively, a set is connected if it is, well, connected. More precisely, a set C R is connected if for any x,y C, the entire line segment connecting x to y is also in C. In the example given by display (1) above, the domain X has two connected subsets, i.e., R ++ and R Weierstrass (extreme value) theorem. For us, the most important result relating to continuity is the following theorem, sometimes know as the extreme value theorem.

4 4 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Theorem: (Weierstrass) Consider a function f : X R 1, where both X and R are endowed with the Euclidean metric. If X is a compact set and f is continuous on X, then f attains a global maximum and a global minimum on X. Example: I ve emphasized that our definition of compactness (closed and bounded) is specific to Euclidean spaces and Euclidean metrics. To see why it s important to remember this, consider 1/x if x > 0 f : [0,1] [0,1], defined by f(x) =. Now put the discrete metric on [0,1]. With 1 if x = 0 this metric, f is continuous; moreover, the function clearly doesn t have a minimmum. Now since [0, 1] looks like it s compact, this looks like a counter-example to the Weierstrass theorem, but it isn t, because [0,1] isn t compact in the discrete metric. Intuitive sketch of the proof: (1) show that if f is continuous and X is compact, then the image of the function must be bounded. (We ll prove this in a separate lemma.) Fig. 1 illustrates why both conditions are jointly sufficient for this condition 1/x if x (0,1] (a) In the first panel, the function f(x) =. It has a compact domain, 0 if x = 0 [0,1], but is discontinuous. It isn t bounded. (b) In the second panel, the function f(x) = 1/x if x (0,1]. It has a non-compact domain (0,1], but is continuous. It isn t bounded. (c) In the third panel, the function has a compact domain [0,1] and is continuous. It is bounded. It can go up arbitrarily high, but eventually, to maintain continuity, it has to come back down again to hit the vertical axis at some finite point. (2) let f denote the supremum of the image of the function. Since f is bounded we know that the supremum is a real (i.e., finite) number, not infinity. Our task now is to show that this supremum is actually attained, and is hence the maximum value of the function. (3) pick a sequence {x n } such that the sequence {f(x n )} converges to the supremum. We can find such a sequence by Theorem B, proved a few lectures ago.

5 ARE211, Fall X = [0,1] i.e., X compact ; f discontinuous X = (0,1] i.e., X not compact; f continuous X = [0,1] i.e., X compact; f continuous Figure 1. Compactness plus continuity implies boundedness (4) while the sequence {x n } needn t converge, it follows from the compactness of X that there must exist a subsequence {y n } of {x n } such that {y n } converges to y X. (5) since f is continuous, {f(y n )} must converge to f(y). But by our choice of the sequence {x n } (step 3) and because {y n } is a subsequence of {x n }, {f(y n )} converges also to f. Hence f(y) = f. This step is tricky, so we ll review what happens here. (a) the sequence {x n } is choosen so that {f(x n )} converges to f. Here we use the definition of the sup plus the fact that f is bounded. (b) the subsequence {y n } of {x n } is choosen so that {y n } converges to y. Here we use the fact that the domain of f is compact. (c) by continuity {f(y n )} must converge to f(y). (d) since {y n } is a subsequence of {x n }, it must converge to the same thing that {x n } converges to (e) hence f(y) = f.

6 6 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) (6) Since f is the supremum of f(x) = {f(x) : x X}, then f(y) = f f(x), x X. A common source of puzzlement is: since f(x n ) already converges to f, why do I need to pick a subsequence {y n } and show that f(y n ) also converges to f? The reason is that {x n } doesn t necessarily converge to some x X, so I can t invoke continuity to establish that f(x n ) converges to f(x), for some x X. For example, let X = (0,1) and consider f : X R, defined by f(x) = x. Clearly f doesn t attain a maximum on X. The theorem doesn t apply because X = (0,1) isn t compact. But it s instructive to see where the proof would break down if we tried to apply it. Using the notation of the proof, f = 1. Pick x n = 1 1/(n + 1) and note that for all n, f(x n ) > f 1/n. This is step (3) of the intuitive sketch above. But with X = (0,1), we can t go to step (4), and thus get ourselves a y. And without a y, we can t go to step (5), and invoke continuity to show that f(y) = f. The point of the example is that having the sequence x n such that the f(x n ) s approach f doesn t do us much good, without further help. Before proceeding to the formal proof, we ll prove a simple Fact about sequences and subsequences. Recall that to show that (t n ) is a subsequence of (x n ), you always have to show the existence of a strictly increasing function τ : N N such that for all n N, t n = x τ(n). Now note: Fact: If τ : N N is a strictly increasing mapping, then for all n, τ(n) n. Note that this fact only holds because of the discrete nature of N. It certainly would not be true if N were replaced by either R + or Q +. For example, consider f : Q + Q + defined by for x 0, f(x) = x/2. f( ) is strictly increasing, but f(x) x only if x = 0. (The reason why this counterexample works for Q + but not for N, is that for x N, x/2 / N.) To prove the Fact, we ll argue by induction. (1) Initial step: τ(1) 1. (2) Inductive step: suppose that τ(n) n; then τ(n + 1) n + 1. Proof of the Inductive step: Let τ(n) = k N, k n. since τ(n+1) > τ(n) and τ(n+1) N, τ(n + 1) k + 1 n + 1.

7 ARE211, Fall Now for the lemma establishing that continuity plus compactness implies boundedness. Lemma: Consider f : X R, where both X and R are endowed with the Euclidean metric. If f is continuous and X is compact, then f is bounded. Proof of the Lemma: We ll just show that the function is bounded above, by proving that if X is compact and f isn t bounded, then f cannot be continuous. (1) Assume that f isn t bounded, i.e., for all n N, x n X such that f(x n ) > n. 1 (2) Since X is compact, the sequence {x n } contains a convergent subsequence. Call this subsequence {y n } and let x X denote its limit. (3) Since f is defined on X, f( x) R, that is, f( x) < N, for some N N. (4) Define τ : N N by y n = x τ(n) for all n. (5) Now pick n N + 1 and note that by the Fact above, τ(n) n N + 1. Moreover, by assumption, f(y n ) = f(x τ(n) ) > τ(n) N + 1. Hence for all n > N + 1, f(y n ) f( x) > 1, so that f is not continuous at x. (6) Similarly, f is bounded below. Proof of the Theorem: (1) Let f denote the supremum of the image of X under f, i.e., the surpremum of the set {f(x) : x X}. (2) By the Lemma above, f R, i.e., f is the least upper bound for the set f(x). (3) By Theorem B from a few lectures ago, for all n, there exists x n such that f(x n ) > f 1/n. (4) Since X is compact, the sequence {x n } contains a convergent subsequence. Call this subsequence {y n } and let x X denote its limit. (5) Since f is continuous, the sequence {f(y n )} converges to f( x) R. To complete the proof, we ll show that {f(y n )} also converges to f. (6) Since {y n } is a subsequence of {x n }, there exists a strictly increasing mapping τ : N N such that for all n, y n = x τ(n). We need to show that ǫ > 0, N N s.t. n > N, f(y n ) B( f,ǫ). (7) Note first that for all n, f(y n ) f, since f is an upper bound for the set {f(x) : x X}. 1 In other words, we can build a sequence {xn} in the following way: for each n N, choose some point in the domain such that f of this point exceeds n; call that point x n.

8 8 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) (8) Moreover, it follows from the construction of {x n } that for all n, f(y n ) = f(x τ(n) ) > f 1/τ(n) f 1/n (by the Fact we proved at the beginning of the lecture, i.e., τ(n) n, for all n). (9) It follows therefore that for an arbitrarily chosen ǫ > 0, we can pick N ǫ N, N ǫ > 1/ǫ, so that 1/τ(N ǫ ) 1/N ǫ < ǫ. (10) To complete the proof, observe that for all n > N ǫ, f(y n ) = f(x τ(n) ) ( f 1/τ(n), f] ( f 1/τ(N ǫ ), f] ( f ǫ, f] ( f ǫ, f + ǫ) = B( f,ǫ) So we have, {f(y n )} f( x) R (by continuity) and {f(y n )} f R. (by the previous argument). (11) Since a sequence can have only one limit, 2 f( x) = f. Since f is an upper bound for {f(x) : x X}, we now have f( x) f(x), for all x X An inverse image formulation of continuity In analytic treatments of continuity topics, the concepts are typically introduced with a sequential formulation, i.e., something along the lines of a function f is continuous at x 0 if whenever a sequence in the domain converges to x 0, then the image of the sequence under f converges to f(x 0 ) before turning to other, slightly more abstract formulations, typically involving open sets, sometimes called open set formulations. Our next result establishes for continuity the relationship between the sequential formulation (used above) and the open set formulation. Definition: Given a mapping f : X Y, and O Y, f 1 (O) is the subset of X that f maps into O, i.e., f 1 (O) = {x X : f(x) O}. f 1 (O) is called the inverse image of O under f. The open set definition of continuity is: a function is continuous iff the inverse image of every open subset of the range of the function is an open set in the domain. 2 We have noted this before but haven t proved it.

9 ARE211, Fall Y Y S ( f(x) f(x n ) [ f(x) O ( f(x n ) [ ) x n x f 1 (S) if part of the proof (contra-positive) f continuous, f 1 (S) not open implies S not open X [ x n x W f 1 (O) only if part of the proof (contra-positive) f discontinuous, O open, f 1 (O) not open X Figure 2. f is continuous iff S open = f 1 (S) open Theorem: A function f : X Y is continuous iff for every open set O Y, f 1 (O) is an open subset of X. 3 Intuition for the proof: See Fig. 2. In the left panel of the Fig. 2 we have a continuuous function; in the right, the function is discontinuous. The left panel illustrates the contrapositive proof of the if part of the theorem; the right panel illustrates the only if part. (NOTE: in the left panel, we choose a non-open set in the domain; in the right panel, we choose an open set in the range.) The if part of the theorem (left panel of Fig. 2) (1) pick a set S in the range of f such that its inverse image f 1 (S) is not open. We ll show that if f is continuous, then S can t be open. (2) Since f 1 (S) is not open, it contains a boundary point. Call this boundary point x. (3) In the figure we can pick a sequence (x n ) that approaches x from the left; all of the points in the sequence get mapped to points below the bottom of S 3 It is useful sometimes to formulate theorems in the formal language of logic. This theorem is a good example. First I define three statements (1) given a function f, the statement C(f) is true if f is continuous (2) given a set O, the statement A(O) is true if O is open (3) given a function-set pair (f, O), the statement B(f, O) is true if f 1 (O) is open Our theorem can now be written as f, C(f) ` O, A(O) B(f, O) We ll now make two different kinds of contra-positive arguments. (1) To prove the part the theorem, we ll show: f, given C(f), O, B(f, O) A(O). (2) To prove that part the theorem, we ll show: f, given C(f), O s.t. (A(O) B(f, O)) This is a great exercise for getting the hang of formal logical relationships.

10 10 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) (4) Since f is continuous, these f(x n ) s must accumulate to f(x). (5) Hence by defn, f(x) is a boundary point of S, so S can t be open. The only if part of the theorem (right panel of Fig. 2) (1) we ll show that if f is not continuous, we can find an open set in the range of f whose inverse image (i.e., corresponding set in the domain) is a non-open set. (2) Now let f be discontinuous at x. As you can see from the figure, we have identified an open set O, part of which is sticking into the hole that s created by the discontinuity. (3) But the part of O that s in the hole doesn t correspond to any point in the domain: the edge that s in the middle of O (i.e., f(x)) maps into an edge in the domain. (4) That is, we have an open set in the range whose inverse image is not open in the domain. We ll now make this precise. Proof: (1) to prove the if part of the theorem, we need to show that if f is continuous, then the inverse image of any open sets is open. Fix an arbitrary set S Y such that f 1 (S) isn t open in X. We ll argue that if f is continuous, then S isn t open in Y. This will prove that when f is continuous, S open in Y implies f 1 (S) is open in X. (a) if f 1 (S) isn t open there must exist a point x in f 1 (S) (i.e. such that f(x) S) which is a boundary point of f 1 (S). (b) i.e., (x n ) x such that for all n, f(x n ) / S (c) since f is continuous, f(x n ) must converge to f(x). (d) but this means that f(x) is a boundary point of S. (e) conclude that S isn t open (2) We ll now prove the only if part of the theorem, for the case Y = R. We need to show that if the inverse image of each open set is open, then f is continuous. We ll show that if f is not continuous, then there exists an open set O Y such that f 1 (O) isn t open in X. (a) if f isn t continuous, there exists x X, a sequence {x n } in X, ǫ > 0 and for all n N, f(x n ) f(x) > ǫ.

11 ARE211, Fall (b) Let O = (f(x) ǫ,f(x) + ǫ). Clearly f(x) O so that x f 1 (O). We ll show that x is not an interior point of f 1 (O) and conclude that f 1 (O) is not open. (c) For x to be an interior point of f 1 (O), there must exist some open set W such that x W f 1 (O). (d) pick an arbitrary open set W containing x. Since {x n } converges to x, there exist n sufficiently large that x n W. (e) Since by assumption (step 2a), f(x n ) / O, it follows that x n / f 1 (O). (f) Since W was chosen arbitrarily, we have established that there does not exist an open set which contains x and is itself contained in f 1 (O). (g) Conclude that f 1 (O) is not open in X Upper and Lower Hemi continuous correpondences Consider a correspondence ξ mapping a metric space S to a metric space T. (The standard notation denoting a correspondence is ξ : S T. The (as opposed to ) signifies that it is not necessarily the case that for every s S, {ξ(x)} is a singleton subset of T. {ξ(x)} may be any set at all, including the empty set.) For correspondences, we need to generalize the notion of continuity. There s upper hemi-continuity and lower hemi-continuity. If both properties are satisfied a correspondence is continuous. I m going to present three alternative definitions of hemi-continuity; as always, I ll present the first as a definition, the other two as equivalency theorems. You ve most probably seen the second and third in math camp. The first, I think, is particularly instructive, which is why I m adding it. The first pair of definitions generalizes the notion of an inverse image just discussed. Definition: Given a set O T, the upper inverse image of O, denoted ξ 1 (O), is the set {s S : ξ(s) O}. This is also known as the strong inverse image. Definition: Given a set O T, the lower inverse image of O, denoted ξ 1 (O), is the set {s S : ξ(s) O }. This is also known as the weak inverse image.

12 12 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) ( O ) ( ) ( ) ξ 1 (O) ξ 1 (O) Figure 3. Upper and Lower inverse images of ξ Note that the requirement of a non-empty intersection with O is much weaker than the requirement of containment in O, so that for every subset O of the range of the correspondence, the upper inverse is contained in the lower inverse. Fig. 3 illustrates the two concepts: (1) the red set is the graph of the correspondence ξ. (2) now consider the (open) set in the range, denoted by the blue interval O. (3) the upper inverse image of O, ξ 1 (O), is the small dark green interval in the domain. Verify that for this example it happens to be an open interval. (4) the lower inverse image of O is the larger light green interval in the domain. Verify that for this example it is also an open interval. We can now define upper- and lower-hemi-continuity. For obvious reasons, I ll call this pair the inverse image definitions. Definition: ξ is said to upper hemi continuous if for every open set O T, the upper inverse image of O, ξ 1 (O) S, is an open set.

13 ARE211, Fall ξ ū 1 ((1, 2)) = {0} closed ξ u 1 ((1, 2)) = 1 ξ l (( 1, 1)) = {0} closed ξ 1 l (( 1, 1)) = R ξu is uhs but not lhc (0,0) (1,2) ξl is lhc but not uhc (0,0) ( 1,1) x x Figure 4. ξ u and ξ l Definition: ξ is said to be lower hemi continuous if for every open set O T, the lower inverse image ξ 1 (O) S, is an open set. Definition: ξ is said to be continuous if it is both upper- and lower-hemi-continuous. To see the difference between the two definitions, compare the two mirror image correspondences, ξ u : R R and ξ l : R R, defined by R if s = 0 ξ u = {0} if s 0 ξ l = {0} if s = 0 R if s 0 the graph of ξ u has a monstrous jump at zero in the domain, and the graph of ξ l has a monstrous hole at zero in the domain. More precisely, (1) consider the open interval (1,2), and observe that the lower inverse image, ξ 1 ū (1,2), is the closed set {0}, establishing that ξ u is not lhc; (2) consider the open interval ( 1,1), and observe that the upper inverse image, ξ 1 l ( 1,1), is the closed set {0}, establishing that ξ l is not uhc Alternative definitions of hemicontinuity. Here s the second pair of definitions of upper and lower hemi-continuity. I call these the neighborhood definitions. But of course, as usual, I m presenting the definitions as theorems.

14 14 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Theorem: ξ : S T is upper hemi continuous iff for every s S and every open set U such that ξ( s) U there exists an open neighborhood V of s such that ξ(s) U for every s V. Theorem: ξ : S T is lower hemi continuous iff for every s S, and every open set U such that U ξ( s), there exists an open neighborhood V of s such that U ξ(s), for every s V. Proof: Sketch of the proof of the u.h.c. theorem (1) if or = (nbd condition above implies that upper inverse images of open sets are open). We ll show that if this inverse image property is not satisfied, then the neighborhood condition of the theorem is not satisfied either. (a) suppose that ξ 1 (U) is not an open set, for some open set U T. (b) this means that ξ 1 (U) contains a boundary point, call it s. (c) now consider the open neighborhood U of ξ(s). (i) since s is a boundary point of ξ 1 (U), for any nbd V of s, there exists s V such that ξ(s) U. (ii) thus, we ve established that there does not exist a neighborhood V of s such that ξ(s) U for every s V. (2) only if or = (relevant property of upper inverse image implies above nbd condition): Do it as an exercise. To illustrate the u.h.c. theorem, look again at Fig. 4. (1) in the left panel, there are two cases to consider, to verify that ξ u satisfies the neighborhood condition (a) s = 0: in this case the only nbd U of ξ u ( s) is R; necessarily, the image of every nbd of s maps into R. (b) s 0: let V be any neighborhood of s that doesn t contain zero and pick ǫ > 0. Clearly B(0,ǫ) contains the image under ξ u of every s V. (2) in the right panel, let s = 0 and let U be the green nbd, ( 1, 1): clearly, there is no nbd V of 0 such that for all s V, ξ(s) U.

15 To illustrate the l.h.c. theorem, look yet again at Fig. 4. ARE211, Fall (1) in the left panel, let U be the green set on the vertical axis and let V be any nbd of zero. Clearly, U ξ(s) =, for every 0 s V, establishing that ξ u is not l.h.c. (2) in the right panel, there are two cases to consider. (a) s = 0: let U denote the green set (-1,1); Clearly, for any nbd V of zero, U ξ(s) for any s V. (b) s 0: in this case, ξ( ) is constant on some neighborhood V of s, so this is again easy. The final pair of definitions are called the sequential definitions. The sequential definition of upper hemi-continuity is not equivalent to the first two. It is, however, equivalent to the other definitions provided that the correspondence is compact-valued. (A correspondence ξ : S T is compactvalued if for every s S, ξ(s) is a compact set.) The sequential definition of lower hemi-continuity is equivalent to the first two. Theorem: A compact-valued correspondence ξ is u.h.c iff for every s S, every sequence (s n ) converging to s and every sequence (t n ) with t n ξ(s n ), there is a convergent subsequence of (t n ) such that lim n t n ξ( s). Theorem: The correspondence ξ is l.h.c iff for every s S, any t ξ( s), and any sequence (s n ) converging to s, there exists a sequence (t n ) such that t n ξ(s n ) and lim n t n = t. To illustrate these sequential definition, we ll redefine our example above, but use the same figure [ 1,1] if s = 0 {0} if s = 0 ψ u = ψ l = {0} if s 0 [ 1,1] if s 0 In the right panel, consider the sequence s n = 1/n converging to 0, and the sequence t n = 1/2, for all n. The limit of (t n ) = 1/2, but 1/2 / ψ l (0). Now in the left panel, consider again the sequence s n = 1/n 0 and 1/2 ψ u (0). There does not exist a sequence (t n ) such that t n ξ(1/n) and t n 1/2.

16 16 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) To see why the compact-valued caveat in the sequential defn of u.h.c. is needed, we need to show that the sequential definition fails for a u.h.c. correspondence for which for some s, ξ(s) is (a) open but bounded and (b) closed but unbounded (a) open but bounded: consider the constant correspondence ξ : R = R defined by ξ(x) = (0, 1), for all x. This correspondence is continuous: in particular, it satisfies the other two defns of u.h.c. But it fails to satisfy the sequential property above. To see this, let s n = 1/n, s = 0 and t n = 1/n. Clearly, t n 0 but 0 / ξ(0). (b) closed but unbounded: consider the constant correspondence ξ : R = R defined by ξ(x) = R, for all x. This correspondence is again continuous but again fails the sequential property. To see this, let s n = 1/n, s = 0 and t n = n. In this case, {t n } does not contain a convergent subsequence Concepts related to upper hemicontinuity. Definition: A correspondence ξ : S T is said to be closed-valued if for every s S, ξ(s) is a closed set. Given ξ : S T, the graph of ξ is defined as {(s,t) S T : t = ξ(s)}. This is a set, like any other set, and it may be open, closed or neither. Definition: A correspondence ξ : S T is said to have a closed graph if its graph is closed. Note the relationship between the concepts of being closed-valued and having a closed graph: closed graph correspondences are necessarily closed-valued but the converse is not necessarily true. It s helpful to think of it this way: (1) a correspondence is closed-valued if every vertical line segment in the graph contains all of its accumulation points (2) a correspondence has closed-graph if every line segment in the graph contains all of its accumulation points

17 ARE211, Fall The concept of upper-hemi-continuity is similar to, but stronger than, the property of having a closed graph. We have the following relationship. Theorem: Let ξ : S T be a closed-valued correspondence such that T is compact. Then ξ is u.h.c iff its graph is closed. Note that the condition on T is much stronger than the condition that ξ is compact-valued. The latter property requires only that for every s S, ξ(s) is a compact set. To see the role of compactness of T, consider the correspondence ξ : R R defined by {0} if s = 0 ξ(s) = {0,1 + 1/s} if s 0 (2) Look at the graph of this function: clearly it s closed in R 2, i.e., it contains all its boundary points. (Indeed, in this example, every element of the graph is a boundary point.) But it fails the definition of u.h.c, i.e., the upper inverse image of the open set U = ( 1,1), ξ 1 (( 1,1)) = {0}. Why does the relationship between closed-graphness and u.h.c. break down in the correspondence defined by display (2)? It s the same idea that we ve seen before (e.g., R is closed since it contains all of its accumulation points. Infinity isn t an accumulation point, so we don t have to worry that it isn t contained in R.) Similarly, in the case of ξ, sequences in the graph go off into outer space, but these sequences don t accumulate to anything, so we don t have to worry about them. But when the graph is contained in a compact set, every sequence has to contain a subsequence that converges to something. So the closed-graph property does not imply u.h.c. Nor is it the case that u.h.c. implies a closed graph. To see this, consider the correspondence ξ : R R, defined by, for s R, ξ(s) = (0,1). This function is u.h.c. but it is does not have a closed graph.

18 18 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Summary: the three formulations (courtesy of Steve Buck). Remember: any one of the three formulations of each concept can be used as the definition of that concept However, once you pick one of them as your definition then the other two statements follow as if and only if theorems. Note also that the sequential definition of upper hemi-continuity has a caveat that the other two definitions of u.h.c. do not. 1(a) Inverse image formulation of continuity. A function f : X Y is called continuous if for every open set O Y, f 1 (O) is an open set of X. 1(b) Neighborhood formulation of continuity. A function f : X Y is called continuous if for every x X and every neighborhood U of f(x) there is a neighborhood V of x such that f(x) U for every x V. 1(c) Sequential formulation of continuity. A function f : X R n is called continuous at x 0 X if whenever {x m } m=1 converges x 0 then {f(x m )} converges to f(x 0 ); the function f is continuous if it is continuous at x for every x X. 2(a) Inverse image formulation of upper hemicontinuity. A correspondence ξ : S T is called upper hemicontinuous if for every open set O T, the upper inverse image of O, ξ 1 (O) S, is an open set. 2(b) Neighborhood formulation of upper hemicontinuity. A correspondence ξ : S T is called upper hemicontinuous if for every s S and every neighborhood U of ξ(s) there is a neighborhood V of s such that ξ(s) U for every s V. 2(c) Sequential formulation of upper hemicontinuity. A compact-valued correspondence ξ : S T is called upper hemicontinuous if for every s S, every sequence {s n } converging to s and every sequence {t n } with t n ξ(s n ), there is a convergent subsequence {t nk } of {t n } such that lim t n n k = t ξ(s).

19 ARE211, Fall (a) Inverse image formulation of lower continuity. A correspondence ξ : S T is called lower hemicontinuous if for every open set O T, the lower inverse image of O, ξ 1 (O) S, is an open set. 3(b) Neighborhood formulation of lower hemicontinuity. A correspondence ξ : S T is called lower hemicontinuous if for every s S, and every open set U T with U ξ(s), there exists a neighborhood V of s such that U ξ(z), for every z V. 3(c) Sequential formulation of lower hemicontinuity. A correspondence ξ : S T is called lower hemicontinuous if for every s S, any t ξ(s), and any sequence {s n } converging to s, there exists a sequence {t n } such that t n ξ(s n ) and lim n t n = t.

20 20 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) Some examples Question: We ve noted that unlike a function, a correspondence may be empty valued at some point in its domain. So consider the correspondence ξ : R R defined by if s = 0 ξ(s) = {1/s} if s 0 Question: Is this correspondence upper-hemi-continuous? Answer: No. As an exercise, we ll show how each of the definitions fails. (1) The inverse image definition of u.h.c.: Consider the set R. The upper inverse image of, ξ 1 ( ), is {0} which is a closed set (2) The neighborhood definition of u.h.c.: To satisfy this definition, it must be the case that for every open neighborhood U of ξ(0) there is a neighborhood V of 0 such that ξ(s) U for every s V. This condition is violated: take an arbitrary bounded subset U in the range. Certainly ξ(0) = U. For s sufficiently close to zero ξ(s) lies outside of U. (3) The sequential definition of u.h.c.: (Note that this definition is applicable, because the correspondence is compact valued, even though the range of the correspondence isn t compact.) Consider a sequence {s n } converging to 0 and a sequence {t n } with t n ξ(s n ). The definition requires that this sequence (t n ) converges to t ξ(0). But there is no convergent subsequence of {t n } so this condition is violated. Question: Is this correspondence lower-hemi-continuous? Answer: Yes: we ll just check at s = 0, since that s the only potential problem. (1) The inverse image definition of l.h.c.: Consider the set R. The lower inverse image of is {s S : ξ(s) }. Pretty hard to satisfy this condition! Conclude that the lower inverse image of is which is open (2) The neighborhood definition of l.h.c.: Consider the point s = 0. We need to show that for every open set U T with U ξ(0), there exists a neighborhood V of 0 such that U ξ(z), for every z V. Well, that s easy since once again there are no open set U T with U ξ(0). The condition is satisfied vacuously.

21 ARE211, Fall (3) The sequential definition of l.h.c.: To satisfy this definition, the following has to be true: for (s n ) 0, and t ξ(0), must exist (t n ) such that for all n, t n ξ(s n ) and t n t. Again, the condition is satisfied vacuously, because there is no t ξ(0). This is a good example of the intuitive comment that implosions are consistent with l.h.c. but not with u.h.c. When the image of s is empty, then one could think of this as being the ultimate implosion. Question: Consider the correspondence ξ : [0,1] R whose graph is drawn in Fig. 5 below. Is it u.h.c?, l.h.c? We ll look at some upper and lower inverse images, as an exercise. (1) first look at its graph. What can we say about it? What does this imply? Ans: graph is clearly closed, and contained in a compact set. So we know right away that it s u.h.c. (2) just for practice, look at the set U in the range of the correspondence: (a) what is its upper inverse image? ξ 1 (U) = ( x,1]. Is this set open? (b) what about its lower inverse image? ξ 1 (U) includes points to the left of x, i.e., (z, x) as well as points to the right of x. That is, ξ 1 (U) = (z,1]. (3) for more practice, consider the inverse images of U O (a) what is its upper inverse image? ξ 1 (U O) = (y,1]. (b) what is its lower inverse image? ξ 1 (U O) = (z,1] (4) finally consider the inverse images of O (a) what is its upper inverse image? ξ 1 (O) is empty. (b) what is its lower inverse image? ξ 1 (O) is the not-open interval (w, x]. So, point (1) establishes that ξ is u.h.c., while (4b) establishes that it isn t lower-hemi-continuous. This example illustrates that explosions are consistent with u.h.c. but not l.h.c. In this case, the explosions occur at the two critical points where the correspondence is vertical: as you approach x from the right, the image is single valued, then it explodes and becomes double, and then immediately triple-valued.

22 22 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) ( U ) ( O ) z y 0 w x 1 Figure 5. ξ : [0,1] R ( U ) ( O ) vz y 0 w x 1 Figure 6. ξ : [0,1] R What would happen if you modified the graph, slighly, but removing the points in the graph where it is vertical, i.e., making the correspondence single-valued at x and empty valued at v (see Fig. 6). Now the correspondence is l.h.c. not u.h.c. We re transformed the explosions into implosions.

23 ARE211, Fall P = x 1 x 2 y 2 y ω Γ(0.5) * * * y y 1 1 y (1) y 1 y 2 y (2) y (0.5) Γ(1) Γ(2) Figure 7. The demand correspondence is u.h.c Application: Berge s Theorem and the demand correspondence For economists, by far the most familiar application of correspondences is Berge s maximum theorem, which is used to established that demand is an upper-hemi-continuous correspondence. Below is the version of the theorem given in the ECON 201A notes. For concreteness, think of Γ( ) as a budget set, y ( ) as a demand correspondence, φ( ) as a utility function and φ ( ) as the indirect utility function corresponding to φ( ), i.e., mapping each price vector to the maximized value of φ( ) on the budget set defined by x. Berge s Theorem of the Maximum: If Γ : X Y is a continuous correspondence with nonempty and compact values and φ : Y R is a continuous function, then y : X Y defined by y (x) = argmax y Γ(x) φ(y) is u.h.c. and φ : X R defined by φ (x) = max y Γ(x) φ(y) is a continuous function. Question: : Note that no restrictions are imposed on X or Y. Berge gives us a maximum to the demand function. How do we reconcile this with Weierstrass, which requires that the domain of the demand function is compact? Answer: For each x, Γ(x) is required to be compact, and we apply Weierstrass to Γ(x).

24 24 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) The following example, depicted in Fig. 7, illustrates why the result is needed and how it works. The domain of the budget and domain correspondences is the interval at the top of the figure: the space P R ++ denotes the price of good 2 relative to good 1. (It s important that the bottom end of the interval P is a positive number. P hits zero the budget set ceases to be compact valued.) The consumer has an endowment ω 0. The budget set is Γ(p) = {y R 2 + : x y x ω}. Three budget sets for relative prices 0.5, 1 and 2 are drawn at the bottom of the figure. In each of the three bottom panels, the two parallel (red) lines denote the consumer s indifference curves, while the asterisk denotes the endowment vector ω (which is necessarily always in the budget set). Except when x 1 x 2 is unity, the budget correspondence is single valued, but it jumps from the northwest corner of the budget set to the south-east corner as relative prices pass through unity, at which point the consumer is indifferent between all goods on the outer boundary of the budget set, i.e., y (1) is multiple valued. Note that indeed y is compact valued. You should verify that it is upperbut not lower- hemi-continuous. Also, note that φ ( ) is continuous at unity, even though y is not. Proof of Berge s Theorem: We ll begin by proving that the value function φ (x) is continuous. Assume that Γ satisfies the assumptions of the theorem and fix x X. Fix an open set O such that φ (x) O, i.e., y Γ(x) such that φ(y) O and y maximizes φ( ) on Γ(x). We ll use the neighborhood definition of continuity. That is, we need to show that there is an open neighborhood V of x such that for all x V, φ (x ) O. The way we ll do it is to find an open nbd V of x such that for all x V, inf(o) < φ (x ) < sup(o) and hence φ (x ) O.

25 ARE211, Fall V l = Γ 1 (U) φ(y) = φ (x) U = φ 1 (O) ( (} O (nbd of φ (x)) inf(o) x y (x) Γ(x) X Y R Figure 8. Lower hemi-continuity of Γ implies that φ (x ) > inf(o). Legend (1) x is a point in the domain of the correspondence Γ (e.g. price) (2) Γ(x) is a image of x (e.g., budget set) (3) φ(x) function being maximized (e.g., utility); (4) y(x) is the set of maximizers (e.g., demand correspondence); (5) y (x) is the maximum value on Γ(x) (e.g., indirect utility). Fig. 8 finds an open set V l such that for all x V l, inf(o) < φ (x ) (1) O is an open set around φ (x). (2) since φ( ) is continuous, U = φ 1 (O) is open. (3) since U is open the lower inverse image of U, Γ 1 (U) is open in X, (4) by definition of a lower inverse image x Γ 1 (U) iff Γ(x ) U. (5) note that x Γ 1 (U) since φ(y) U by construction, so Γ 1 (U) is an open nbd of x. (6) now for x Γ 1 (U) and y Γ(x ) U, φ(y ) O. (7) since φ (y ) is the maximum of φ( ) on Γ(x ), φ (x ) φ(y ) > inf(o).

26 26 ANALYSIS6: TUE, SEP 11, 2012 PRINTED: AUGUST 22, 2012 (LEC# 6) x V h = Γ 1 (U) φ(y) = φ (x) y (x) sup(o) } ( ( ( U = φ 1 (O ) Γ(x) O (nbd of φ (x)) O (nbd of φ(γ(x)) O = O O X Y ( R Figure 9. Upper hemi-continuity of Γ implies that φ (x ) < sup(o). Fig. 9 finds an open set V h such that φ (x ) < sup(o) if x V h, (1) Pick an open nbd O of φ(γ(x)) such that sup(o ) < sup(o). (2) Let O = O O. O is the union of two open sets and is hence open. (3) since φ( ) is continuous, U = φ 1 (O ) is open. (4) moreover, since φ(γ(x)) O, U contains Γ(x). (5) since Γ is u.h.c., Γ 1 (U) is open in X; by definition of an upper inverse image, x Γ 1 (U) iff Γ(x ) U. (6) note that x Γ 1 (U) since Γ(x) U by construction, so that Γ 1 (U) is an open neighborhood of x. (7) now for x Γ 1 (U) and every y Γ(x ), φ(y ) O. (8) hence φ (y ) < sup(o). Finally, let V = V l V h and note that for all x V φ (x ) O. That is, V is an open neighborhood of x which maps into O, as required.

27 ARE211, Fall We ve now proved that the value function φ ( ) is continuous. To complete the proof, we need to show that y is u.h.c. Since Γ is compact valued, it suffices to show that y satisfies the sequential formulation of u.h.c. (1) Let (x n ) x, let (y n ) y with y n y (x n ), for all n. We need to show that y y (x). (2) Necessarily, y n Γ(x n ), for all n. (3) Since Γ is u.h.c and compact valued, the sequential formulation implies that y Γ(x). (4) If y / y (x), then z Γ(x) and ǫ > 0 such that φ(z) > φ(y) + ǫ. (5) since Γ is l.h.c., there exists (z n ) z with z n y (x n ), for all n. such that z n z. (6) since φ is continuous, it follows that for n sufficiently large, φ(z n ) > φ(y n ), contradicting the assumption that y n y (x n ).