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Unformatted text preview: Problem Set 3, 18.100B/C, Fall 2011 Michael Andrews Department of Mathematics MIT October 1, 2011 1 Let X be a metric space and E X . Let cl( E ) denote the closure of E and let int( E ) denote the interior of E . a We proved on the last sheet (problem 2) that for any E , int( E ) is open and that if E is open then int( E ) = E . Thus int(int( E )) = int( E ). Similarly, it is proved in Rudin (2 . 27) that for any E , cl( E ) is closed and that if E is closed then cl( E ) = E . Thus cl(cl( E )) = cl( E ). b Lemma: E F X = int( E ) int( F ) and cl( E ) cl( F ). Proof: Suppose E F ; then E cl( F ) and by 2 . 27 ( a ), ( c ) of Rudin, cl( E ) cl( F ). Also, E c F c gives cl( E c ) cl( F c ). By what we proved in problem 2 of the last sheet this gives (int( E )) c (int( F )) c so that int( E ) int( F ). Lemma: int(cl(int(cl( E )))) = int(cl( E )). Proof: int(cl( E )) cl( E ) = cl(int(cl( E ))) cl(cl( E )) = cl( E ) = int(cl(int(cl( E )))) int(cl( E )) and int(cl( E )) cl(int(cl( E ))) = int(cl( E )) = int(int(cl( E ))) int(cl(int(cl( E )))) . Corollary: cl(int(cl(int( E )))) = cl(int( E )). Proof: We proved last time (problem 2) that cl( E c ) = (int( E )) c and we also have int( E c ) = ((int( E c )) c ) c = (cl(( E c ) c )) c = (cl( E )) c ....
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 Fall '10
 Prof.KatrinWehrheim
 Math

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