# sol3 - Problem Set 3 18.100B/C Fall 2011 Michael Andrews...

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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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sol3 - Problem Set 3 18.100B/C Fall 2011 Michael Andrews...

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