Intro_to_real_Analysis_hmwrk_3

Intro_to_real_Analysis_hmwrk_3 - Brandy M. Hicks MTHSC...

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Unformatted text preview: Brandy M. Hicks MTHSC 653.001 - Homework # 3 1. Let ( X,d ) be a metric space. Let x be any point in X . Prove that { x } c is open. Proof: Let ( X,d ) be a metric space. We wish to show that { x } c is open. { x } c = X \{ x } . Let U = X \{ x } . Let a U . Let r = d ( x,a )- 1 r > 0 and B r ( a ) { x } = B r ( a ) does not include any point of { x } U is open { x } c is open. 2. Let x and y be distinct points in a metric space X . Show r > 0 such that B r ( x ) and B r ( y ) are disjoint. Proof: (Proof by contradiction): Let ( X,d ) be a metric space. Let x,y X such that x 6 = y . Let r = d ( x,y )- 1 2 . Thus we have B r ( x ) X and B r ( y ) X . Assume that B r ( x ) and B r ( y ) are not disjoint. Then z B r ( x ) and z B r ( y ). d ( x,y ) d ( x,z ) + d ( z,y ) d ( x,y ) r + r d ( x,y ) 2 r d ( x,y ) 2 d ( x,y )- 1 2 d ( x,y ) d ( x,y )- 1 But this implies d ( x,y ) < d ( x,y )- 1 which is impossible since d ( x,y ) = | x- y | and an absolute value cannot be negative = =, we have reached a contradiction.=, we have reached a contradiction....
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Intro_to_real_Analysis_hmwrk_3 - Brandy M. Hicks MTHSC...

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