solutionshw2-555-2011

solutionshw2-555-2011 - Solutions homework 2. Page 32...

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Solutions homework 2. Page 32 Problem 6 : Clearly d ( x,y ) 0 and d ( x,y ) = 0 if and only if x = y . Hence property (i) holds. It is also clear that d ( x,y ) = d ( y,x ), so it remains to show that the triangle inequality holds. Let x,y,z X . If x = y , then d ( x,y ) = 0 d ( x,z )+ d ( z,y ), since both terms on the right hand side are 0. If x 6 = y , then at least one of x 6 = z or z 6 = y is true. This implies that either d ( x,z ) 1 and/or d ( z,y ) 1. Hence d ( x,z )+ d ( z,y ) 1 = d ( x,y ). Page 32 Problem 7. From the triangle inequality we get d ( x,y ) d ( x,x 0 ) + d ( x 0 ,y ) . Another application of the triangle inequality gives d ( x 0 ,y ) d ( x 0 ,y 0 ) + d ( y 0 ,y ) . Combining these two inequalities we get d ( x,y ) d ( x,x 0 ) + d ( x 0 ,y 0 ) + d ( y 0 ,y ) , which gives d ( x,y ) - d ( x 0 ,y 0 ) d ( x,x 0 ) + d ( y 0 ,y ) . Now interchanging x and x 0 , and y and y 0 in this last inequality gives (using the symmetry property of the metric) d ( x 0 ,y 0 ) -
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