{f 1 (U), U F} is an open cover of A. Since A is compact there is a finite subcover of A, {f 1 (U 1 ),...,f 1 (U n )}, {U 1,...


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1 44 CHAPTER 4. CONTINUOUS FUNCTIONS In Calculus we often use arithmetic operations to generate new continuous functions from old ones. In a general metric space we don t have arithmetic, but much of it is recovered if the range space is a normed vector space or the real numbers. The text has some theorems pointing out that the limit of sum is the sum of the limits, and that when they make sense, the limit of a product is the product of the limits and the limit of a quotient is the quotient of the limits. Compactness and connectedness are preserved by continuous functions. Theorem If M 1 is connected and f : M 1 M 2 is continuous, then f(m 1 ) is connected. Proof. Suppose not. Let f(m 1 ) = V 1 V 2. Then M 1 = f 1 (V 1 ) f 1 (V 2 ) is a separation of M 1, which doesn t exist. Theorem Let A be a subset of a metric space (M 1, d 1 ), and suppose f : A (M 2, d 2 ) is a continuous function. If A is compact, then f(a) is compact. Proof. Let F be a family of open sets covering f(a), with U F. Then {f 1 (U), U F} is an open cover of A. Since A is compact there is a finite subcover of A, {f 1 (U 1 ),...,f 1 (U n )}, so is a cover of f(a). {U 1,..., U n } Proposition Let A be a subset of a metric space (S 1, d 1 ), and suppose f : A (S 2, d 2 ) is a continuous function. If A is compact, and f is onetoone, then f 1 is continuous. Proof. Since f : A f(a) is onetoone we have a well defined function g = f 1 : f(a) A, and g 1 = (f 1 ) 1 = f. Suppose K is closed in A. Since A is compact, so is K. Then g 1 (K) = f(k) is compact in f(a). But this means g 1 (K) is closed, and so g 1 (K) is closed for every closed set K A.
2 4.1. MAPPINGS FROM ONE METRIC SPACE TO ANOTHER 45 Here are some theorems generalizing results used frequently in Calculus. Theorem If K M is compact and f : K R is continuous, then f has a maximum and a minimum on K. Proof. We know that f(k) R is compact, so bounded. Moreover the set f(k) is closed, so it contains sup x K f(x) and inf x K f(x), which is the claim. In the introductory remarks about connectednes we noted that a version of the Intermediate Value Theorem would hold for continuous functions f : M R if (M, d) was a path connected metric space. A somewhat more general result is true. Theorem Suppose (M, d) is a metric space, f : A M R is continuous, and A is connected. If x, y A, then for every real number c between f(x) and f(y) there is some z A with f(z) = c. Proof. The argument is by contradiction. If not, then the range of f : A R is separated by the open sets U = (, c) and V = (c, ). The open sets f 1 (U), f 1 (V ) provide a separation of A, which can t exist. An other concept that we encountered in our earlier analysis course is the notion of uniform continuity. In terms of ǫ δ language the idea is that in some situations we can pick δ independent of x. We say f : M 1 M 2 is uniformly continuous if for every ǫ > 0 there is a δ > 0, independent of x M 1, such that d 2 (f(x), f(y)) < ǫ if d 1 (x, y) < δ. In Calculus this property is commonly seen when we have a function with a bounded continuous derivative, since f(y) f(x) = y It s useful to know that compactness is enough. x f (t) dt y x sup f (t). t Theorem Suppose (M 1, d 1 ) and (M 2, d 2 ) are metric spaces, and f : M 1 M 2 is continuous. If M 1 is compact, then f is uniformly continuous.
3 46 CHAPTER 4. CONTINUOUS FUNCTIONS Proof. Given ǫ > 0 and x M 1, find disks D(x, δ x ) such that d 1 (x, y) < δ x implies d 2 (f(x), f(y)) < ǫ/2. Shrink the radii, looking at sets D(x, δ x /2). This open cover of M 1 has a finite subcover by sets D(x n, δ n /2). Take δ = min n δ n /2. Now suppose d 1 (x, y) < δ. Since every point is in a set of the cover, x is within δ < δ n /2 of one of the previously selected points x n. Then d 1 (x n, y) d 1 (x n, x) + d 1 (x, y) < δ n, so d 2 (f(x), f(y)) d 2 (f(x), f(x n )) + d 2 (f(y), f(x n )) < ǫ. Here are a few words about treating subsets of metric spaces as independent metric spaces. Suppose (M, d) is a metric space and A M. Then the metric subspace generated by A is the metric space (A, d A ), where d A is the restriction of d to A A. Define a set U A to be open in A if U A is open as a subset of the metric space (A, d A ). Define a set F A to be closed in A if F A is closed as a subset of the metric space (A, d A ). Theorem a) U A is open in A if and only if there is an open set U M such that U A = U A. b) F A is closed in A if and only if there is an closed set F M such that F A = F A. Proof. a) First observe that for x A, D A (x, ǫ) = D M (x, ǫ) A. Suppose U A is open in A, and let x U A. Then there is some ǫ > 0 such that D A (x, ǫ) U A. We have U A = x UA D A (x, ǫ x ) = A [ x UA D M (x, ǫ x )] = A U. Suppose U A = A U, where U is open in M. For every x U A there is an ǫ > 0 such that D M (x, ǫ x ) U. But then we have D A (x, ǫ x ) = A D M (x, ǫ x ) A U = U A. b) Suppose F A is closed in A. Then A \ F A is open in A, so by part (a) A \ F A = A U
4 4.1. MAPPINGS FROM ONE METRIC SPACE TO ANOTHER 47 for some open U in M. Taking complements in A gives F A = A \ (A U) = A [M \ U], which shows that F A = A F for a closed set F M. Suppose that F A = A F for a closed set F M. Then and F A is closed in A. A \ F A = A [M \ F] = A U,
5 48 CHAPTER 4. CONTINUOUS FUNCTIONS
6 Chapter 5 Uniform Convergence 5.1 Uniform Convergence Power series provide one example of the following situation. We represent a function f(x) by a sequence or series of elementary functions. The elementary functions have good properties, like ease of integration or differentiation, and we want these operations to extend to the limit function. Usually this is handled with the concept of uniform convergence. We distinguish two definitions. Suppose A is a set and (M 1, d) is a metric space. Assume f n : A M is a sequence of functions. The sequence {f n } converges pointwise to f : A M if for every ǫ > 0 and every x A there is an N(x, ǫ) such that d(f(x), f n (x)) < ǫ, n N. The sequence {f n } converges uniformly to f : A M if for every ǫ > 0 and every x S there is an N(ǫ) such that d(f(x), f n (x)) < ǫ, n N, x S. Draw an epsilon tube about f to illustrate the definitions. Very little can be said about the pointwise limits of functions. For instance we can find sliding tents f n : [0, 1] R such that f n 0 pointwise, but 1 f n = 1, 0 49
7 50 CHAPTER 5. UNIFORM CONVERGENCE or 1 f n. 0 Similarly, we can construct a sequence of infinitely differentiable functions g n (x) = tan 1 (nx), which converges pointwise to the function g = π/2, x > 0, π/2, x < 0, 0, x = 0.
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