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The Urysohn's metrization theorem states that for any second countable and regular $(X,\tau)$ there exists a topological embedding $f:X\to [0,1]^{\mathbb{N}}$.

What puzzles me is not the theorem itself, but why (and if) $f$ can be chosen so that the image $f(X)$ is a Borel subset of $[0,1]^{\mathbb{N}}$. This is something that is not clear to me from the proof itself and I've been looking for some nice arguments of how to show this, if such exists. As Martin pointed out in the comments, if we know that $X$ is also completely metrizable then $f$ can be chosen so that $f(X)$ is Borel. But is $N_{2}$ and $T_{3}$ enough?

If such $f$ would exist, would it follow nicely from some topological arguments? Or do we have to explicitly define the metric in $[0,1]^{\mathbb{N}}$ and work it out with the function $f$? Also, in case the proof is very lengthy, I would really appreciate if you could include the book where I could find the full proof and its details so I can study it. For that reason I included the reference-request tag as well.

Thanks for all in advance.

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    @Martin By $N_{2}$ I meant second countability and by $T_{3}$ regularity. And thanks a lot, your comments have been a valuable input to this question.2012-12-22

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Borel sets $S$ in complete metric spaces (like $[0,1]^{\mathbb N}$) have lots of nice properties, one of which is that either $S$ is countable or $S$ has a perfect subset. ("Perfect" means nonempty, closed, and having no isolated points.) Notice that this property is preserved by homeomorphisms. There are subsets $X$ of $\mathbb R$ that do not have this property. [The proof uses the axiom of choice. One type of example is a Bernstein set, a set such that both it and its complement meet every uncountable closed subset of $\mathbb R$.] Such an $X$ is second countable and regular, but it cannot be homeomorphic to a Borel set in $[0,1]^{\mathbb N}$.