1873_solutions

The right is and the limit from the left is to see why

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Unformatted text preview: from a set S into R, prove that the function |f | is also continuous from S into R. Hint: Use the preceding exercises and the fact that whenever t and x belong to S we have ||f t | |f x | | |f t f x |. 6. Suppose that a and b are real numbers, that a  b and that f : a, b R. 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 |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 . 7. Given that f is continuous on a closed set H and that x n is a convergent sequence in the set H, prove that the sequence f x n is also convergent. Prove that this assertion is false if we omit the assumption that S 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), the sequence f 1/n fails to converge. Actually, much more can be proved: If H is any set of real numbers 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. If you elect to read the optional section on the distance function of a set that follows this set of exercises then you will find a solution of this stronger assertion there. 8. Prove that if a set S has no limit points then every function f : S R is continuous on S. This exercise follows at once from the fact stated in Theorem 8.7.5 that every function must be continuous at a number that is not a limit point of its domain. 9. Prove that if S is a set of real numbers and if no limit point of S belongs to the set S then every function 193 f : S R is continuous on S. This exercise follows in exactly the same way as Exercise 8. 10. Suppose that f and g are functions from a set S to R and that f is continuous at a given number a at which the function g fails to be continuous. a. What c...
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