1873_solutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online