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# ans1-1 - ECON 831 Solution Homework#1 1 Let(X d be a metric...

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ECON 831 Solution Homework #1 1. Let ( X,d ) be a metric space, and x,y,z X . By the triangle inequality, we have: d ( x,y ) d ( x,z ) + d ( y,z ) d ( x,y ) - d ( y,z ) d ( x,z ) (1) d ( y,z ) d ( x,z ) + d ( x,y ) d ( y,z ) - d ( x,y ) d ( x,z ) (2) From (1) and (2), we conclude that | d ( x,y ) - d ( y,z ) | ≤ d ( x,z ) . 2. Let ( X,d ) and ( X,ρ ) be 2 metric spaces. (i) We show that ( X, max { d,ρ } ) is also a metric space by verifying the three properties that the function max { d,ρ } needs to satisfy in order to be a metric. (1) Show x = y max { d ( x,y ) ( x,y ) } = 0 : x = y d ( x,y ) = 0 = ρ ( x,y ) ( d and ρ are metrics) max { d ( x,y ) ( x,y ) } = 0 max { d ( x,y ) ( x,y ) } = 0 d ( x,y ) = 0 and ρ ( x,y ) = 0 ( d , ρ nonnegative functions) x = y ( d is a metric) (2) Symmetry property holds obviously because of the symmetry of d and ρ as metrics. (3) Triangle inequality: We rst apply the triangle inequality to d and ρ at points x,y,z X to get: ( d ( x,y ) d ( x,z ) + d ( z,y ) ρ ( x,y ) ρ ( x,z ) + ρ ( z,y ) - Suppose that d ( x,y ) ρ ( x,y ) : max { d ( x,y ) ( x,y ) } = d ( x,y ) max { d ( x,y ) ( x,y ) } ≤ d ( x,y ) + d ( z,y ) max { d ( x,y ) ( x,y ) } ≤ max { d ( x,y ) ( x,y ) } + max { d ( z,y ) ( z,y ) } 1

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- Similarly if d ( x,y ) ρ ( x,y ) . ¥ (ii) We show that ( X, min { d,ρ } ) is not a metric space by providing a counterexample where the triangle inequality does not hold for the function
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