# $M$ is compact iff $M$ is homeomorphic to a closed subset of $H^{\infty}$

(a) Let $M$ be a metric space. If there exists a countable subset $X$ of $M$ such that $\overline{X}=M$, $M$ is said to be separable. Prove that a compact metric space is separable.

(b) Let $M$ be a separable metric space. Prove that there is a 1-1, continuous function $f$ from $M$ into $H^{\infty}$.

(c) Prove the following theorem. A metric space $M$ is compact if and only if $M$ is homeomorphic to a closed subset of $H^{\infty}$.

($H^{\infty}$ denote the set of all real sequences $\{a_n\}$ such that $|a_n|\le 1$ for every positive integer $n$. The metric on $H^{\infty}$ is given by $$d(\{a_n\},\{b_n\})=\sum_{i=1}^{\infty}\frac{|a_n-b_n|}{2^n}$$)

I already did parts (a) and (b). For the forward direction of (c), suppose $M$ is compact. By (a), $M$ is separable. By (b), there exists a 1-1, continuous function $f$ from $M$ into $H^{\infty}$. Since $M$ is compact and $f$ is continuous, $f(M)$ is compact and thus closed. Since $f$ is 1-1, $f^{-1}$ exists, and since $f$ is continuous, $f^{-1}$ is continuous on $f(M)$. This function $f$ shows the homeomorphism between $M$ and a closed subset of $H^{\infty}$.

Now for the backward direction, suppose that $M$ is homeomorphic to a closed subset of $H^{\infty}$. I don't know where to go from here.

-
What is $H$? Please edit your question to define it. –  dfeuer Jun 19 '13 at 16:52
It was defined in the OP what is $\,H^\infty\,$ but the editor thought it was superfluous for some reason... –  DonAntonio Jun 19 '13 at 16:56
That is very strange. The history indicates that I deleted it, which I know is not the case. –  dfeuer Jun 19 '13 at 16:57
@dfeuer That's because you suggested an retag edit. It's okay now. –  PJ Miller Jun 19 '13 at 16:58
Do you know that $H^\infty$ is compact? –  Martin Jun 19 '13 at 16:59

We first show that $H^{\infty}$ is compact, i.e. every sequence in $H^{\infty}$ has a convergent subsequence. First, suppose the sequence is $\{\{a_{ij}\}_j\}_i$. We find a subsequence $\{n_i^1\}_i$ of $\{i\}_i$ such that the sequence $\{a_{n_i^11}\}_i$ converges. (This is possible by Bolzano-Weierstrass.) We then find a subsequence $\{n_i^2\}_i$ of $\{n_i^1\}_i$ such that the sequence $\{a_{n_i^22}\}_i$ converges. Note also that $\{a_{n_i^21}\}_i$ converges because $\{n_i^2\}_i$ is a subsequence of $\{n_i^1\}_i$. We proceed like this. And to get a convergent subsequence of $\{\{a_{ij}\}_j\}_i$, we simply take the indices to be the first term of $\{n_i^1\}_i$, the second term of $\{n_i^2\}_i$, and so forth. It can be verified that this subsequence indeed converges in $H^{\infty}$.
Now suppose that $M$ is homeomorphic to a closed subset of $H^{\infty}$. Since $H^{\infty}$ is compact, the closed subset of it must be compact too. Since there exists a continuous function from that compact set onto $M$, $M$ is also compact, and we are done.