1873_solutions

# 0 prove that this function does indeed fail to be

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Unformatted text preview: se a sequence u n in S converging to u and a sequence v n in S converging to v. Using the facts that u n u and v n v and f u n f u and f v n f v as n Ý, choose N such that whenever n N we have 233 d f un , f u  d f vn , f v  3 3  d un, u  3 and d v n , v   . 3 Since d uN, vN d u N , u  d u, v  d v, v N         3 3 3 and therefore d f uN , f vN We conclude that d f u ,f v  d f uN , f vN  3 .  d f vN , f v    . 3 3 3 Second Proof. (less messy but a little too slick, 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  . Now suppose that u and v are any points of S satisfying the inequality d u, v  . Choose a sequence u n in S converging to u and a sequence v n in S converging to v. Since u n u and v n v and f u n f u and f v n f v as n Ý, we see that d un, vn d u, v and d f u n , f v n d f u , f v . Using the fact that d u, v  , choose N such that the inequality d u n , v n   holds whenever n N. Thus, for all n N we have d f u n , f v n  and therefore . d f u , f v  n Ý d f un , f vn lim d f u , f uN 8. a. Given that S is a set of real numbers, that a S S and that f x  x1a for all x S, prove that f is continuous on S but not uniformly continuous. (Use this exercise.) The result follows at once. b. Given that S is a set of real numbers and that S fails to be closed, prove that there exists a continuous function on S that fails to be uniformly continuous on S. The result follows from part a. c. Is it true that if S is an unbounded set of real numbers then there exists a continuous function on S that fails to be uniformly continuous on S? No, it isn’t true. Every function defined on the set Z of integers is uniformly continuous there. 9. Given that f is a function defined on a metric space X, prove that the following conditions are equivalent: This exercise was left in accidentally after being elevated to the status of a theorem. It appears as Theorem 8.15.4. a. The function f fails to be uniformly continuous on the space X. b. There exists a number  0 and two sequences t n and x n in X such that d x n , t n d f xn , f tn 0 as n Ý and for every n. 10. Is it true that the composition of a uniformly continuous function with a uniformly continuous function is uniformly continuous? Yes, it’s true. Suppose that f is a uniformly continuous function from a metric space X to a metric space Y and that g is a uniformly continuous function from Y to a metric space Z. To show that the composition g f is uniformly continuous, suppose that  0. Using the fact that g is uniformly 234 continuous on the space Y, choose   0 such that, whenever y 1 and y 2 belong to Y, and d y 1 , y 2  , we have d g y 1 , g y 2  . Now, using the fact that f is uniformly continuous on the space X, choose   0 such that, whenever x 1 and x 2 belong to X and d x 1 , x 2  , we have d f x 1 , f x 2  . Then, whenever x 1 and x 2 belong to X and d x...
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