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Unformatted text preview: p 6 = 2. 1 5. We have max i n  x i  p n X  x i  p , hence k x k k x k p . Also, n X  x i  p n max i n  x i  p , hence k x k n1 p k x k p . 6. Using the naming convention for constants in the previous problem, k x k q 1 c q k x k d p c q k x k p . Similarly, k x k q 1 d q k x k c p d q k x k p . 7. For z A d ( x , z ) d ( x , y ) + d ( y , z ) , hence inf z A d ( x , z ) d ( x , y ) + inf z A d ( y , z ) , which gives, by denition, d ( x , A ) d ( x , y ) + d ( y , A ) . Hence, d ( x , A )d ( y , A ) d ( x , y ) . Similarly d ( y , A )d ( x , A ) d ( x , y ) , hence  d ( x , A )d ( y , A )  d ( x , y ) . 2...
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 Fall '08
 Stopple,J

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