# Polish Spaces and the Hilbert Cube

I've been trying to prove that every Polish Space is homeomorphic to a $G_\delta$ subspace of the Hilbert Cube. There is a hint saying that given a countable dense subset of the Polish space $\{x_n : n\in\mathbb{N}\}$ define the function $f(x)=(d(x,x_n))_{n\in\mathbb{N}}$ (where $d$ is a metric on the Polish space). I think I've shown that $f$ is continuous and $1-1$ but I don't know how to prove that the converse of the function is continuous and that the image is a $G_\delta$ set. It's been a really long time since I've dealt with topology so I'm having a hard time coming up with any idea and I'm afraid I've forgotten some well known topological fact (so maybe it's something obvious here I'm not seeing). Any help on how to proceed would be kindly appreciated.

-

I take it that you've already normalized $d$ such that $0 \leq d \leq 1$ (otherwise replace $d$ by $\frac{d}{1+d}$).

As you've said, the function $f: x \mapsto f(x) = (d(x,x_{n}))_{n \in \mathbb{N}}$ is continuous and injective. Let $f(y_{m}) \to f(y)$ be a convergent sequence in $f(X)$. We want to show that $y_{m} \to y$.

By definition of the product topology, we have $d(y_{m},x_{n}) \xrightarrow{m \to \infty} d(y,x_{n})$ for all $n$. Let $\varepsilon > 0$ and pick a point $x_{n}$ such that $d(y,x_{n}) < \varepsilon/3$ by density. Since $d(y_{m},x_{n}) \to d(y,x_{n})$, there is $M$ such that $|d(y_{m},x_{n}) - d(y,x_{n})| < \varepsilon /3$ for all $m \geq M$, so $d(y_{m},x_{n}) < 2 \varepsilon /3$. But then $d(y_{m},y) \leq d(y_{m},x_{n}) + d(x_{n},y)< \varepsilon$ and hence $y_{m} \to y$.

Why is the image a $G_{\delta}$-set? This seems to be much more difficult. I don't see any easier way than to essentially re-prove two classical results on metric spaces which are much more interesting, so I prefer to explain this:

Theorem (Kuratowski) Let $A \subset X$ be a subset of a metrizable space and let $g: A \to Y$ be a continuous map to a completely metrizable space $Y$. Then $g$ can be continuously extended to a $G_{\delta}$-set containing $A$.

Fix a bounded and complete metric on $Y$. For the proof we need the notion of oscillation of $g$ at a point $x \in \overline{A}$ (the closure of $A$ in $X$) defined by $$\displaystyle \operatorname{osc}_{g}(x) = \inf\{\operatorname{diam}g(U \cap A)\,:\, x \in U, \;U\; \text{open}\}.$$ The set $B = \{x \in \overline{A}\,:\,\operatorname{osc}_{g}(x) = 0\}$ is a $G_{\delta}$-set. To see this, note that $B_{n} = \{x \in \overline{A} \,:\, \operatorname{osc}_{g}(x) < \frac{1}{n}\}$ is an open subset of the closed set $\overline{A}$ and $B = \bigcap_{n \in \mathbb{N}} B_{n}$. The continuity of $f$ implies that $A \subset B$. Now define $f: B \to Z$ by $f(x) = \lim g(x_{n})$, where $x_{n} \to x$. It is not hard to show that $f$ is well-defined (because $\operatorname{osc}_{g}(x) = 0$ implies that $g(x_{n})$ is a Cauchy-sequence) and clearly $f$ extends $g$ and is continuous.

The second ingredient we need is:

Theorem (Lavrentiev) Let $X$ and $Y$ be completely metrizable spaces and let $g: A \to B$ be a homeomorphism from $A \subset X$ onto $B \subset Y$. Then there exist $G_{\delta}$-sets $G \supset A$ and $H \supset B$ and a homeomorphism $f: G \to H$ extending $g$.

Let $h = g^{-1}$. Choose $G_{\delta}$-sets $G' \supset A$ and $H' \supset B$ and continuous extensions $g': G' \to Y$ and $h': H' \to X$ by Kuratowski's theorem. Let $Z = \operatorname{graph}(g') \cap \widetilde{\operatorname{graph}}(h') \subset X \times Y$ be the intersection of the graphs (the tilde indicates the 'switch' $\widetilde{(y,x)} = (x,y)$ of coordinates) and let $G = \operatorname{pr}_{X} (Z)$ and $H = \operatorname{pr}_{Y}(Z)$. Obviously, $f = g'|_{G}$ is a homeomorphism of $G$ onto $H$. One can check that $H$ (and thus also $G$ by symmetry) is a $G_{\delta}$-set as follows: The graph of $g'$ is closed in $G' \times Y$ and thus it is a $G_{\delta}$-set and $H$ is its preimage under the continuous map $y \mapsto (h'(y),y)$.

Corollary. If $Y$ is a completely metrizable space and $X \subset Y$ a completely metrizable subspace then $X$ is a $G_{\delta}$-set.

By Lavrentiev's theorem, the inclusion $X \subset Y$ extends to a homeomorphism onto its image.

A further corollary of these ideas is that a subset of a Polish space is Polish if and only if it is a $G_{\delta}$.

More detailed information can be found in any decent book on descriptive set theory, for instance Kechris, Classical descriptive set theory, or Srivastava, A course on Borel sets, both appeared in the Springer Graduate Texts in Mathematics series.

-
Theo, the clever argument in Kechris using the Kuratowski lemma is nice, but there's also a fairly straightforward argument (I think due to Alexandrov) that a Polish subspace $Y$ of a Polish space $X$ is $G_\delta$. The idea is to build a sequence $\mathcal{U}_n$ of open (in $Y$'s closure) covers of $Y$ consisting of sets which are small wrt a compatible metrics for $X$ and whose trace on $Y$ is small wrt a metric for $Y$. One also wants closures of elements of later covers to be contained in elements of prior covers. Letting the diameters go to $0$ gives a witness that $Y$ is $G_\delta$. – user83827 Sep 8 '11 at 22:26
@ccc: Thanks for that. Yes, that looks like the more natural way to do it. Anyway, I like the argument above very much (it's an amalgam of arguments I learned from Kechris and Srivastava). – t.b. Sep 8 '11 at 23:14
I'll take shocking, sneaky argument over a natural one any day (as an added bonus, surprising arguments tend to be easier to remember)! – user83827 Sep 9 '11 at 0:34