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The problem is:

If $(X,d)$ is a separable metric space then there exists a metric $d'$ that is topologically equivalent to $d$ and such that $(X,d')$ is totally bounded.

I know that if $(X,d)$ is separable then there exists an homeomorphism $f \colon(X,d) \to(Y,d_2)$ where $Y$ is a subset of the Hilbert Cube in $L_2$ (and $d_2$ is $L_2$ metric).

$f$ is defined as follows:

Let $\{B_n\}_{n \in\mathbb{N}}$ be a numerable base of open sets of $(X,d)$ and let

$f_n\colon X \to[0,1/n] $ , $f_n(x) = \dfrac{1}{n} \dfrac{ \,d(x,X-B_n)}{ ( 1 + d(x,X-B_n) ) }$

(Note that: $f_n$ is continuous for all $n$, $|f_n(x)| \leq 1/n$, for all $n$ and $x$, and $f_n(x) > 0$ iff $x \in B_n$, as the complement of $B_n$ is closed) )

Then, we define $f$ as:

$f:(X,d)\to (Y,d_2)$

$x \longmapsto f(x)=(f_1(x),f_2(x),..)$

(and $Y = f(X)$)

The hint is to use the following metric $d'$:

$d'(x,y) = d_2(f(x), f(y))$, which I know is top equiv to $d$.

The problem is: How can I prove that $(X ,d' )$ is totally bounded?

Could you help me?

Alex Ravsky
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Quinn
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  • Any help? I've been trying to prove total boundness, but I just don't know how use the hypothesis of separability and the definition of "f". – Quinn Dec 11 '14 at 03:04
  • I would use the fact that rel. sequentially compact implies total boundedness. Just an idea... – user1537366 Dec 11 '14 at 03:07
  • What is rel. sequentially compact? I know that total boundness implies separability, but what do you mean with that concept? – Quinn Dec 11 '14 at 03:14
  • $M$ is relatively sequentially compact in $(X,\rho)\iff$ every sequence in $M$ has a convergent subsequence. Maybe you learnt a different term for this concept? – user1537366 Dec 11 '14 at 03:18
  • Ok, that's the same as compactness... But I don't see how can I use it. I only know that (X,d) is separable. Nothing else. – Quinn Dec 11 '14 at 03:30
  • Sequential compactness is indeed equivalent to compactness for a metric space. But the two concepts are not equivalent for all topological spaces. – Dan Asimov May 28 '23 at 04:06

1 Answers1

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It seems the following.

The closure $\overline{Y}$ of the set $Y$ in the Hilbert Cube is compact, so the space $(\overline{Y},d_2|\overline{Y})$ is totally bounded and its subspace $(Y,d_2|Y)$ is totally bounded too.

Alex Ravsky
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