Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniformly equivalent metrics on finite set

  1. Jul 5, 2010 #1


    User Avatar
    Homework Helper

    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.)
  2. jcsd
  3. Jul 5, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Homework Helper

    Oh yes, of course - thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook