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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.

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Which direction is causing trouble? Or is it both directions? –  Brian M. Scott Nov 29 '12 at 20:14
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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! –  Frank_W Nov 27 '13 at 17:23

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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 $[0,1]^\omega$.

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Can you tell me what's the meaning of $2^-n$-nets? More precisely, what's the meaning of a net? –  Frank_W 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 $[0,1]^ω$? –  Frank_W 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
    
U mean the amount of elements in D is uncertain and actually determined by the separpable set that u choose in X. So u use power N over [0,1] to build the hilbert cube. Besides, if the amount of elements in D is finite and supposed by m, then the power N should be fixed to m, is that right? –  Frank_W Nov 28 '13 at 1:04

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