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# sol-04 - MATH 453 SOLUTIONS TO ASSIGNMENT 4 OCTOBER 7 2004...

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M  453 S   A  4 O  7, 2004 Exercise 2 from Section 17, page 100 Using Theorem 17.2, A is closed in Y implies that it is the intersection of Y and a closed subset C of X : A = C Y . Since Y is given to be closed in X , A is the intersection of two closed sets in X , and hence is closed in X . Exercise 6 from Section 17, page 101 (a) ¯ B is a closed set that contains B , so it contains A as well. Hence ¯ A ¯ B . (b) ¯ A ¯ B is closed, being the finite union of closed sets, and contains A B , so it must contain the closure A B as well. The other inclusion is a special case of (c). (c) For each α , A α A α , so ¯ A α A α . Consequently, ¯ A α A α . An example where equality fails is given by A n = [1 / n , 1], n Z + in R . Each A n is closed, so ¯ A n = A n = (0 , 1]. However, A n = (0 , 1] = [0 , 1]. Exercise 7 from Section 17, page 101 The given “proof” fails because, while every neighborhood U of x must intersect some A α , each such neighborhood may intersect a di ff erent A α , leaving x in the closure of none. For

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