Uniformly equivalent metrics on finite set

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SUMMARY

The discussion focuses on proving that any two metric functions, d1 and d2, defined on a finite set X are uniformly equivalent. The proof involves identifying constants A and B such that for all x, y in X, the inequalities d1(x, y) ≤ A d2(x, y) and d2(x, y) ≤ B d1(x, y) hold. The sets S1 and S2 are defined as the ratios of the metrics, and the maximum values from these sets provide the necessary constants A and B. It is crucial to note that x and y must be distinct in the definitions of S1 and S2.

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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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Just as a technical point, in your definitions of S1 and S2 you have to specify that x and y are not equal
 
Office_Shredder said:
Just as a technical point, in your definitions of S1 and S2 you have to specify that x and y are not equal

Oh yes, of course - thanks!
 

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