Walker Ray Econ 204 Problem Set 2 Suggested Solutions July 22, 2017

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1 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 1. Show that any set in a metric space (X, d) can be written as the intersection of open sets. Take any subset A X and define C = x A c {x} c First, because in a metric space any singleton sets are closed, the complement {x} c is open. So C is the intersection of open sets. Now let s show C = A. Take any element c C. Then c c {x} c x A c x A c {x} c c x x A c c A c c A 1

2 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 2. Suppose the metric space (X, d) is infinite. (a) Show that there exists an open set U such that U and U c are both infinite. (b) Show that there exists an infinite subset Y X such that (Y, d) is a discrete metric space. 1 (a) Note that every neighborhood of a limit point is infinite. We consider two cases: Case 1: X has at least two distinct limit points p q. Let ε = 1 2 d(p, q). Then B ε(p) and B ε (q) are disjoint, and both sets are infinite. So let U = B ε (p) (which is open), and then since B ε (q) U c, U c is infinite as well. Case 2: X has at most one limit point. Then take a sequence {x n } of distinct isolated points. Then U = {x 2n } is an open, infinite set. Further, U is disjoint from the infinite set V = {x 2n+1 }. Then since V U c, U c is also infinte. (b) First let s prove the following useful characterization of discrete metric spaces: a metric space is discrete if and only if every point is isolated. ( = ): Take any x X. Since the metric space is discrete, the singleton set {x} is open. So we can find a neighborhood B(x) {x}, and in fact B(x) = {x}. So x is not a limit point of X, and hence is isolated. ( = ): since every point x is isolated, every singleton set {x} is open. Because any set can be written as the union of singleton sets, every set is open. So the metric space is discrete. For the proof, we construct an infinite set Y of isolated points. The problem is trivial if X is discrete, so suppose not. Since X is not discrete, we can find a limit point p X. Choose a distinct sequence y n p where y n p. Define the set Y = {y n }. Recall a convergent sequence only has one limit point (namely its limit). So the only limit point of Y in X is p. Since p Y and Y X, the metric space (Y, d) has no limit points. In other words, (Y, d) only contains isolated points and hence is discrete. 1 A metric space (X, d) discrete if every subset A X is open. Notice that any set equipped with the discrete metric forms a discrete metric space, but not every discrete metric space necessarily has the discrete metric. 2

3 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 3. Construct a real function that is continuous at exactly one point. One example: f(x) = { x x if x Q if x Q It s clear that the function is discontinuous at any point x 0. What about at x = 0? Note that for any x R: f(x) f(0) = f(x) 0 = f(x) = x So for any ε > 0, simply let δ = ε. If x 0 = x < δ, then f(x) f(0) = x < δ = ε. So f is continuous at x = 0. 3

4 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 4. Take any mapping f from a metric space X into a metric space Y. Prove that f is continuous if and only if f ( A ) f(a) for every set A. (Hint: use the closed set characterization of continuity). I make use of the following properties of images and pre-images of functions. For any sets A X, B Y : 2 A f 1 (f(a)), f(f 1 (B)) B Now for the proof: ( = ): Suppose f is continuous. Take any A X. Since A f 1 (f(a)) and f(a) f(a), we have ) ) A f (f(a) 1 = A f 1 (f(a)) = f (f(a) 1 ) where the final equality follows because we know that f (f(a) 1 is closed from the continuity of f. Then take the image of both sides to get f ( A ) ( )) f (f 1 f(a) f (A) where the final set inclusion follows from the properties above. ( = ): Suppose a function f satisfies f ( A ) f(a) for every set A X. Take C Y closed. We want to show D = f 1 (C) is closed. First note f(d) = f (f 1 (C)) C = C Hence we have f ( D ) f(d) C. Then taking the pre-image of both sides gives f 1 ( f ( D )) f 1 (C) = D From the properties above, D f 1 ( f ( D )) and combining we reach D D. Thus D is closed, and f is continuous. 2 The set inclusions may be proper; try to come up with examples. 4

5 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 5. Prove that a metric space (X, d) is discrete if and only if every function on X into any other metric space (Y, ρ), where Y has at least two distinct elements, is continuous. ( = ): every set in a discrete metric space is open. So for any function f and any set A Y, f 1 (A) is open. In particular this holds for all open subsets of Y and hence f is continuous. ( = ): Let A X. Fix y, y Y with y y. Let f : X Y be given by { y if x A f(x) = y if x A c Then f is continuous, since every function from X to Y is continuous. Since {y} is a closed set, A = f 1 ({y}) is closed. Similarly, {y } is closed, so A c = f 1 ({y }) is closed, which implies that A is also open. Since A X was arbitrary, X is discrete. 5

6 Walker Ray Econ 204 Problem Set 2 Suggested s July 22, 2017 Problem 6. For any continuous function f : [0, 1] R, define the functions T (f) : [0, 1] R and S(f) : [0, 1] R by x T (f)(x) = 1 + f(s)ds 0 { f ( ) x + 1 if x < S(f)(x) = f (1) if x 1 2 Let W (f) = αt (f) + βs(f) for some α, β R. Show that if α + β < 1, then there exists a continuous function f : [0, 1] R such that W (f) = f. Note that T (f), S(f), and W (f) are continuous functions from [0, 1] into R. So T, S, and W are operators on the Banach space C ([0, 1]) endowed with the sup-norm. We wish to show that when α + β < 1, W is a contraction mapping. The result then follows from the contraction mapping theorem. Take any functions f, g C ([0, 1]) and any x [0, 1]. First, note that x T (f)(x) T (g)(x) = (f(s) g(s))ds Next, we have Thus, 0 x f g f g = T (f) T (g) = sup T (f)(x) T (g)(x) f g x [0,1] S(f)(x) S(g)(x) = { ( ) ( f x g x + 1 2) if x < 1 2 f(1) g(1) if x 1 2 f g = S(f) S(g) = sup S(f)(x) S(g)(x) f g x [0,1] W (f) W (g) = (αt (f) + βs(f)) (αt (g) + βs(g)) = α(t (f) T (g)) + β(s(f) S(g)) a T (f) T (g) ) + β (S(f) S(g)) ( α + β ) f g Therefore, W is a contraction mapping with modulus α + β < 1. 6