HW05 - f ( M ) is actually closed. To see this, let ( f ( a...

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Modern Analysis 1 Homework 05 1. Let M be a compact metric space. Let f : M M be isometric in the sense a,b M d ( f ( a ) ,f ( b )) = d ( a,b ) . Prove that f is surjective. Suggestion : Let z M . By considering the sequence ( z n ) n =0 given by z n = f ◦ ··· ◦ f ( z ) (with n factors f ) show that z is the limit of a sequence in the image f ( M ). Solution As in the suggestion, let z M and for n N let z n = f n ( z ). Because M is compact, the sequence ( z n : n N ) has a convergent (hence Cauchy) subsequence: say ( z n k : k N ). If ε > 0 then there exists N N such that q > p > N d ( z n q ,z n p ) < ε ; in particular, for any such q > p > N it follows by the isometric nature of f that d ( f n q - n p ( z ) ,z ) = d ( z n q ,z n p ) < ε. This proves that the arbitrarily-chosen z lies in the closure f ( M ) of the image f ( M ). The proof is completed by noting that
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Unformatted text preview: f ( M ) is actually closed. To see this, let ( f ( a n ) : n N ) be a sequence in f ( M ) converging to b in M . From d ( a q ,a p ) = d ( f ( a q ) ,f ( a p )) it follows that a n : n N ) is Cauchy, hence (as compact implies complete) convergent, say a n a ; from d ( f ( a ) ,f ( a n )) = d ( a,a n ) it follows that f ( a n ) f ( a ). Finally, uniqueness of limits forces b = f ( a ) and places b in f ( M ). Remark : The notion of continuity simplies some of this argument: f ( M ) is the image of a compact space under a continuous map, thus compact and so closed. 1...
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This note was uploaded on 11/09/2011 for the course MAA 5228 taught by Professor Robinson during the Fall '09 term at University of Florida.

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