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Unformatted text preview: MATH 255 ASSIGNMENT 2, short solutions 1. [10 points] Let X be the collection of all sequences of positive integers. If x = ( n j ) ∞ j =1 and y = ( m j ) ∞ j =1 are two elements of X , set k ( x,y ) = inf { j : n j 6 = m j } and d ( x,y ) = ( if x = y 1 k ( x,y ) if x 6 = y. Prove that d is a metric on X . Solution. The only nontrivial property to check is the triangle inequality. Let x,y,z ∈ X . If x = y or y = z or z = x , then d ( x,z ) ≤ d ( x,y ) + d ( y,z ) obviously holds. Otherwise, if x = ( n j ), y = ( m j ), z = ( k j ), then n j = m j for j < k ( x,y ) and m j = k j for j < k ( y,z ). This implies that n j = k j for j < min { k ( x,y ) ,k ( y,z ) } . So, k ( x,z ) ≥ min { k ( x,y ) ,k ( y,z ) } and 1 k ( x,z ) ≤ max 1 k ( x,y ) , 1 k ( y,z ) ≤ 1 k ( x,y ) + 1 k ( y,z ) . Hence, d ( x,z ) ≤ d ( x,y ) + d ( y,z ). 2. [10 points] For x,y ∈ R , set d ( x,y ) = arctan  x y  ....
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This document was uploaded on 02/22/2011.
 Spring '09
 Integers

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