In the wikipedia page: https://en.wikipedia.org/wiki/Totally_bounded_space I found the following theorem:

A metric space is separable if and only if it is homeomorphic to a totally bounded metric space.

I am struggling to prove the "$\Rightarrow$" direction. I found here in the forum these two attempts to build the desired homeomorphism:

  1. Separability, total boundness and topological equivalence of metrics.
  2. separable iff homeomorphic to totally bounded.

But I believe that in both cases these functions are not open, hence not homeomorphism.

Please help me find a proof.


1 Answer 1


In the second link the inverse map is continuous: If $d(x^{j},x_n) \to d(x,x_n)$ for every $n$ then then $d(x^{j},x)\to 0$.

Indeed, $d(x^{j},x)\leq d(x^{j},x_n)+d(x_n,x)$. First choose $n$ such that the second term is less than $\epsilon /2$ and then choose $m$ such that the first term is less than $\epsilon /2$ for $j \geq m$.

  • $\begingroup$ Can you please elaborate why does such $m$ exist? $\endgroup$
    – User271828
    May 28 at 5:05
  • $\begingroup$ $d(x^{j},x_n) \to d(x,x_n)$ for the $n$ we have already chosen. SInce the limit here is less than $\epsilon/2$ all terms of the sequence $(d(x^{j},x_n))$ would also be less than $\epsilon/2$ for $j$ sufficiently large. @User31415 $\endgroup$ May 28 at 5:20

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