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 } = φ ⇒ 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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This note was uploaded on 10/23/2009 for the course MTHSC 101 taught by Professor Staff during the Fall '08 term at Clemson.

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Intro_to_real_Analysis_hmwrk_3 - Brandy M Hicks MTHSC...

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