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Uniformly equivalent metrics on finite set

  1. Jul 5, 2010 #1

    radou

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    Yet another proof I'd like to check.

    Statement. Let X be a finite set. One has to show that every two metric functions d1, d2 on X are uniformly equivalent.

    Proof. If X is finite, then X = {x1, ..., xn}. We have to find constants A and B such that for every x, y in X, we have d1(x, y) <= A d2(x, y) & d2(x, y) <= B d1(x, y). Let S1 = {d1(x, y) / d2(x, y) | x, y from X} and S2 = {d2(x, y) / d1(x, y) | x, y from X}. If we take A = max S1 and B = max S2, the proof is completed. (For example, take some x and y, then d1(x, y)/d2(x,y) <= K1, and so on, for every x and y in X ; if we take A = max{K1, K2, ...}, then d1(x, y)/d2(x,y) <= A holds, for all x, y. Analogous for the other condition we need.)
     
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  3. Jul 5, 2010 #2

    Office_Shredder

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    Just as a technical point, in your definitions of S1 and S2 you have to specify that x and y are not equal
     
  4. Jul 6, 2010 #3

    radou

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    Oh yes, of course - thanks!
     
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