1873_solutions

X 0 1 such that f x x this assertion is the one

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Unformatted text preview: of this exercise. Well, of course, such a choice of x n would make f x n Ý which we know now to be impossible. 7. a. Given that S is a set of real numbers, that a for all x S S and that f x  x1a S, prove that f is continuous on S but not uniformly continuous. Hint: Use the preceding exercise. Choose a sequence f xn Ý as n x n in S a that converges to x. Note that Ý. 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. Choose a number a S S and use 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, as we remarked after Exercise 1, every function on the set Z of integers must be uniformly continuous. 8. Is it true that the composition of a uniformly continuous function with a uniformly continuous function is uniformly continuous? Yes, this assertion is true. Suppose that f is a uniformly continuous function on a set S and that g is a uniformly continuous function on a set T that includes the range f S of f. To show that the function g f is uniformly continuous on S, suppose that  0. Using the uniform continuity of g on the set T we choose a number   0 such that the inequality |g u g v |  holds whenever u and v belong to T and |u v |  . Now, using the uniform continuity of f on the set S we choose a positive number  such that the inequality |f s f t |   will hold whenever s and t belong to S and |s t |  . Then whenever s and t belong to S and gft | . |s t |   we have |g f s 9. a. Suppose that f is uniformly continuous on a set S and that x n is a convergent sequence in S. Prove that the sequence f x n cannot have more than one partial limit. We know from the result proved in Exercise 6 d that the sequence f x n is bounded. We choose a partial limit y of f x n and we want to prove that y is the only partial limit of f x n . Suppose that z is any number unequal to y. We define  1 |y z |. 3 Choose a number   0 such that the inequality 203 |f s f t |  holds whenever s and t belong to S and |s t |  /2. We write the limit of x n as x and choose an integer N such that the inequality |x n holds whenever n N. Using the fact that y is a partial limit of the sequence f x n an integer m N such that |f x m y |  . Now given any integer n N, since |x n x m | |x n x |  |x x m |       2 2 we must have f xm |  |f x n and consequently f xm |  2 |y f x n | |y f x m |  |f x n y f xm z f xn Thus if n N, the number f x n cannot lie in the neighborhood z that z fails to be a partial limit of the sequence f x n . x|   2 we choose ,z  of z and we conclude b. In Part a, did you assume that the limit of the sequence x n belongs to S? If so, go back and do the problem again. c. Prove that if f is uniformly continuous on a set S and x n is a convergent sequence in S then the sequence f x n is also convergent. Do not assum...
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