Uniformly equivalent metrics on finite set

[radou]

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.)

 

[Office_Shredder]

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

 

[radou]

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!