1873_solutions

Intervals that do not intersect with each other and

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Unformatted text preview: s 0, Ý . 7. Suppose that f is a uniformly continuous function from a subset S of a metric space X into a metric space Y. a. Prove that if x n is a Cauchy sequence in the set S then the sequence f x n is a Cauchy sequence in the space Y. Suppose that x n is a Cauchy sequence in the set S, and, to show that f x n is a Cauchy sequence in the space Y, suppose that  0. Using the fact that f is uniformly continuous on S, choose   0 such that whenever t and x belong to S and d t, x  , we have d f t , f x  . Using the fact that x n is a Cauchy sequence, choose an integer N such that, whenever m N and n N, we have d x m , x n  . Then, whenever m N and n N, we have d f xm , f xn  . b. Prove that if the space Y is complete then whenever a sequence x n in the set S converges to a point x X the sequence f x n converges in the space Y. We assume that Y is complete and that x n is a sequence in S that converges to a point x of X. Note that we cannot simply claim that f x n f x because we have not said that the point x must belong to the set S. However, the sequence x n must be a Cauchy sequence and therefore, by part a, the sequence f x n must be a Cauchy sequence in Y, and must therefore converge in Y because Y is complete. c. Prove that if x S and if x n and t n are two sequences in the set S both of which converge to x and if the space Y is complete then lim lim f x n  n Ý f t n . nÝ We assume that x n and t n are sequences in the set S and that both of these sequences converge to x. We now consider the sequence x 1 , t 1 , x 2 , t 2 , . More precisely, we define un  t n/2 if n is even x if n is odd. n1 /2 Since the sequence u n also converges to x, it is also a Cauchy sequence. Therefore the sequence f u n is convergent in the space Y. Since both lim n Ý f x n and lim n Ý f t n are partial limits of the sequence f u n we deduce that lim lim lim f x n  n Ý f t n  n Ý f u n . nÝ d. Given that Y is complete and that x S S, explain how we can use Part c to extend the definition of the function f to the point x in such a way that the extension is continuous on the set S Þ x . The continuity of f at the point x is really very simple. If y n is any sequence in the set S Þ x converging to x then it is clear that f y n f x . This part of the exercise also follows at once from part e that we shall prove below. e. Prove that if Y is complete then there exists a uniformly continuous function g from the set S into Y such that g x  f x for every point x S. For each point x S S we define f x by choosing a sequence x n in S that converges to x and defining f x  n Ý f xn . lim We shall now show that this extension of the function f to S is uniformly continuous on S. Suppose that  0. First Proof (a bit messy, perhaps). Using the fact that f is uniformly continuous on S, choose   0 such that whenever u and v belong to S and d u, v   we have d f u , f v  /3. Now suppose that u and v are any points of S satisfying the inequality d u, v  /3. Choo...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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