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solutions-assignment2

# solutions-assignment2 - MATH 255 ASSIGNMENT 2 short...

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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 ) = ( 0 if x = y 1 k ( x,y ) if x 6 = y. Prove that d is a metric on X . Solution. The only non-trivial 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 | . Show that d is a metric on R . Solution. Again, the only non-trivial property to check is the triangle inequality. Let s 0 be fixed and consider the function h ( t ) = arctan( s + t ) - arctan t - arctan s. h (0) = 0 and h 0 ( t ) = 1 1 + ( s + t ) 2 - 1 1 + t 2 .

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