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If $d$ is a metric for the Polish space $X$ and if $\left \{ x_n \right \}_n$ is a countable dense subset of $X,$ then it is easy to see that the map $f:x\mapsto \left \{ \min(1,d(x,x_n)) \right \}_n$ is a homeomorphism onto a subset $Y$ of $[0,1]^{\mathbb N}$.

My problem is showing that $Y$ is a $G_{\delta }.$ $Y$ cannot be closed because this would imply that $X$ is compact, but if I could show that the $Y=\overline Y\cap \left \{ y\in [0,1]^{\mathbb N}:osc _{f^{-1}}(y)=0 \right \},$ that would do it.

On the other hand, the result follows immediately from Lavrentiev's theorem because appealing to it, we can find a $G_{\delta }$ set $G$ such that $Y\subset G$ and $f(X)=G, $ which is only possible if $G=Y.$

This seems like cheating, though, especially because the aforementioned theorem is not trivial. Does anyone have another argument?

Matematleta
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    I think that showing $Y=\overline Y$ cannot work, because it would imply that $X$ is compact, which need not be true. A related question is the following: https://math.stackexchange.com/questions/19344/polish-spaces-and-the-hilbert-cube – s.harp Jul 11 '17 at 08:46
  • How about Mazurkiewicz's theorem? Since $f$ is an embedding, $f(X)$ is Polish, hence a $G_{\delta}$. – Daniel Fischer Jul 11 '17 at 15:19
  • @DanielFischer: Yes that works. I used Lavrentiev, which I think also works. I want to do this without the big guns, though, especially because in this case I was able to construct a specific homeomorphism. – Matematleta Jul 12 '17 at 00:28

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