# sol-09 - MATH 453 SOLUTIONS TO ASSIGNMENT 9 NOVEMBER 11,...

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M  453 S   A  9 N  11, 2004 Exercise 6 from Section 28, page 181 X is a metric space, so it is Hausdor . Thus X is compact Hausdor and we need only check that f is a continuous bijection. Continuity and injectivity is immediate since f is an isometry. Suppose that f is not onto, and let a X - f ( X ). f ( X ) is compact, being the continuous image of a compact set, so it is closed. Thus X - f ( X ) is open and we can ﬁnd an ε > 0 such that B ( a , ε ) is disjoint from f ( X ). Set x 0 = a and for n > 0, x n = f ( x n - 1 ) = f n ( a ). If n > m , we have d ( x n , x m ) = d ( f n ( a ) , f m ( a )) = d ( f n - m ( a ) , a ) > ε since f n - m ( a ) is in f ( X ) and a isn’t. Thus { x n } has no convergent subsequence and this contradicts the compactness of X . ± Exercise 7 from Section 28, page 181 (a) First note that since X is compact, { d ( x , y ) | x , y X } is a compact subset of R . Thus diam X , deﬁned to be the supremum of the above set, is ﬁnite.

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## This note was uploaded on 02/10/2010 for the course MATH 205 taught by Professor Smith during the Spring '05 term at Adrian College.

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sol-09 - MATH 453 SOLUTIONS TO ASSIGNMENT 9 NOVEMBER 11,...

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