Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble with proving this theorem: A metric space is separable iff it is homeomorphic to a totally bounded metric space. There is a link on Wikipedia to book by S. Willard, but it is stated there as a fact leaving it to the reader as an exercise to prove it. Any help would be appreciated.

share|cite|improve this question
Which direction is causing trouble? Or is it both directions? – Brian M. Scott Nov 29 '12 at 20:14
Homeomorphic to totally bounded metric space implies separable is easy: $X$ homeomorphic to $Y$ totally bounded. Any completion of $Y$ is then compact and thus separable. Therefore $X$ is homeomorphic to a subspace of a separable metric space and hence is separable. Done. – kahen Nov 29 '12 at 20:34
To Brian M. Scott: Separable => homeomorphic to a totally bounded space is causing trouble. To kahen: Isn't every totally bounded space separable? I'm not sure, but is it necessary to use the argument with completion? – Josef Ondřej Nov 29 '12 at 20:58
@kahen: U cannot claim that "Any completion of Y is then compact" cos closure of Y may not be equal to completion of Y! – Bear and bunny Nov 27 '13 at 17:23
up vote 7 down vote accepted

First, you’re right that a totally bounded metric space is automatically separable, so that there’s no need to go through the completion: the union of finite $2^{-n}$-nets for $n\in\omega$ is a countable dense subset.

Now assume that $\langle X,d\rangle$ is a separable metric space. Without loss of generality assume that $d(x,y)\le 1$ for all $x,y\in X$, and let $D=\{x_n:n\in\omega\}$ be a dense subset of $X$. Define the map

$$f:X\to[0,1]^\omega:x\mapsto\big\langle d(x,x_n):n\in\omega\big\rangle\;.$$

Now show that $f$ is an embedding of $X$ into the compact metrizable space $[0,1]^\omega$, the Hilbert cube; being compact, the Hilbert cube is totally bounded in any compatible metric, and total boundedness is hereditary, so $f[X]$ is totally bounded in any metric inherited from $[0,1]^\omega$.

share|cite|improve this answer
Can you tell me what's the meaning of $2^-n$-nets? More precisely, what's the meaning of a net? – Bear and bunny Nov 27 '13 at 4:13
@Frank: In this context an $\epsilon$-net in a metric space $\langle X,d\rangle$ is a set $A\subseteq X$ such that $X=\bigcup_{x\in A}B(x,\epsilon)$. Another way to say it is that for each $x\in X$ there is a $y\in A$ such that $d(x,y)<\epsilon$. – Brian M. Scott Nov 27 '13 at 17:38
what does ω mean? And what is $[0,1]^ω$? – Bear and bunny Nov 27 '13 at 17:57
@Frank: $\omega$ is the set of natural numbers, $\{0,1,2,\ldots\}$; you can replace it by $\Bbb N$ if you like. $[0,1]^\omega$ is the product of countably infinitely many copies of the space $[0,1]$. – Brian M. Scott Nov 27 '13 at 17:59
@Frank: Yes, it is. – Brian M. Scott Dec 5 '13 at 18:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.