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?
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.