4
$\begingroup$

Two metric spaces X and Y are called equivalent if: $d_X (x,x_n) \to 0 \Leftrightarrow d_Y (x,x_n) \to 0 $ with $ n \to \infty $

I wonder whether, if you took a certain set (for example a finite set, the natural numbers, or any other, compact, non compact, complete, not complete set...) whether it is possible to construct infinitely many different metric spaces, so that no two of them are equivalent.

$\endgroup$
1
  • $\begingroup$ The notions of "compact" and "complete" don't apply to plain old sets. Being compact is a property of topological spaces, and being complete is a property of metric spaces. $\endgroup$ Jun 10, 2015 at 21:44

3 Answers 3

7
$\begingroup$

The answer depends only on the cardinality of the set in question - if two sets have the same cardinality then we can always transfer metrics back and forth between them using a bijection. Thus the two other answers essentially answer every case: if your set is finite, then there is only one equivalence class; while if the set is infinite, there are infinitely many.

$\endgroup$
5
$\begingroup$

The set $\Bbb{Q}$ admits infinitely many non equivalent metrics: consider for example all $p$-adic metrics, where $p$ runs among all prime numbers.

$\endgroup$
4
  • $\begingroup$ OK but the question is about equivalent metrics so your answer is not at all valid. $\endgroup$
    – Piquito
    Jun 10, 2015 at 14:23
  • 1
    $\begingroup$ @LuisGomezSanchez Read better the question: "I wonder now, if you took a certain set, [...] if it were possible to construct infinitely many different metric spaces, so that, no two of them are equivalent." $\endgroup$
    – Crostul
    Jun 10, 2015 at 14:26
  • $\begingroup$ The p-adic numbers really do seem interesting and have some nice properties. I will read up on them, though the material seems to be possible to become difficult quickly, if you approach it from the wrong path. $\endgroup$
    – Imago
    Jun 10, 2015 at 17:20
  • $\begingroup$ @Crostul: you are right, I have read quite wrongly the statement (I even believed the question was just for finite sets; I feel I have "surmenage" and I leave StackExchange for a while). Who does not know the p-adic my God! $\endgroup$
    – Piquito
    Jun 11, 2015 at 17:19
4
$\begingroup$

On a finite set, $d_X(x, x_n)$ can converge to $0$ if and only if there exists $N\in\mathbb N$ such that $x_n=x$ for all $n>N$.

Therefore, by your definition of equivalence, all metrics on a finite set are equivalent.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .