1873_solutions

to see why the f x as x 3 suppose that w is any real

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Unformatted text preview: me up with an example in which the space X is connected. In fact, if X  0, 1 then X is connected and, if we define f x  x, 0 for all x X, then f fails to send open sets to open subsets of R 2 . 7. Suppose that X is a metric space, that k is a positive integer and that f j : X 220 R for each j  1, 2, , k. Suppose that f x  f 1 x , f 2 x , , f k x X. Prove that the function f is continuous on X if and only if each of the functions f j is for every point x continuous on X. This exercise follows almost at once from the corresponding fact about limits. 8. Suppose that a and b are real numbers, that a  b and that f is a function from the metric space a, b to a metric space Y. Prove that the following conditions are equivalent: a. The function f is continuous at the number a. b. For every number  0 there exists a number   0 such that for every number x in the interval a, b that satisfies the inequality x a   we have d f x , f a  . This exercise is obvious because of the fact that |x a |  x a whenever x a, b . You can convert this exercise into one that is a shade more interesting by removing the words “in the interval” from condition b and replace them by the condition x  a. This change would require a slightly more careful choice of  to ensure that it does not exceed b a so that any number x  a that lies within a distance  of a would automatically belong to the interval a, b . 9. Prove that if a metric space X has no limit point and f is a function from X to a metric space Y then f must be continuous on X. This result follows at once from the fact that a function is always continuous at a point in its domain if that point is not a limit point of the domain. 10. Suppose that f is a continuous function from a closed subset H of a metric space X into a metric space Y and suppose that x n is a convergent sequence in the set H. Prove that the sequence f x n is convergent in the space Y. Is this conclusion still valid if we don’t assume that H is closed? Solution: Suppose that f is continuous on a closed set H and that x n is a convergent sequence in H. We define x  n Ý xn. lim Since H is closed we must have x H and therefore f is continuous at the number x. Therefore the fact f x as n Ý. that x n x as n Ý guarantees that f x n To see why the assertion does not remain true without the assumption that H be closed we look at the example in which H 1 n Z n and we define f1 n  0 if n is even 1 if n is odd Since no member of the set H is a limit point of H we know that f is continuous at every member of H. However, in spite of the fact that the sequence 1/n converges (to 0) in the metric space R, the sequence f 1/n fails to converge. Actually, much more can be proved: If H is any subset of a metric space X and H fails to be closed then there exists a convergent sequence x n in H and a continuous function f on H such that the sequence f x n fails to converge. This stronger assertion is harder to prove and we omit the proof at this point. Thi...
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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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