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Unformatted text preview: Math 104  Homework 5 Solutions Lectures 2 and 4, Fall 2011 1. Let ( X,d ) be a metric space, p a point of X , and r > 0 a positive number. Show that the closed ball of radius r around p : S := { q X  d ( q,p ) r } is closed in X . Hint: Show that the complement of S is open in X . (Draw a picture of the situation in R 2 to get some intuition.) Proof 1. Note that the complement of S in X can be described via X \ S = { x X  d ( x,p ) > r } . Let x X \ S and set s = d ( x,p ) r . Since x X \ S , d ( x,p ) > r so s > 0. We claim that B s ( x ) X \ S , which will show that X \ S is open. Indeed, let y B s ( x ) so that d ( x,y ) < s . Then d ( y,p ) d ( x,p ) d ( x,y ) > d ( x,p ) s = r where the first inequality follows from the triangle inequality d ( x,p ) d ( x,y ) + d ( y,p ). Hence y X \ S , so B s ( x ) X \ S as claimed. We conclude that X \ S is open in X and thus that S is closed in X . Proof 2. We show directly that S is closed in X . To this end, suppose that ( x n ) is a sequence in S converging to x X . By the triangle inequality, we have d ( x,p ) d ( p,x n ) + d ( x n ,x ) r + d ( x n ,x ) for any n. Let > 0 and choose N large enough so that d ( x N ,x ) < , which exists since ( x n ) x . Then the above inequality gives d ( x,p ) r + d ( x N ,x ) < r + . Since this holds for all > 0, we conclude that d ( x,p ) r so x S and S is closed in X as claimed. 2. For each of the following subsets of R , determine its interior, closure, and boundary in R , and whether or not the given subset is open, closed, or neither in R . ( a ) 1 n n N ( b ) ( 1 , 0) (0 , 1) ( c ) ( R \ Q ) [ e, ) Make sure to justify your answers! Solution. (a) Call this set A . First, A has empty interior in R since any interval around one of the 1 n will contain some irrational, and so no such interval will be contained in A . The closure of....
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 Fall '08
 RIEMAN
 Math

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