Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Is this really true? If you take $X$ to be any subset of $[0,1]^{\mathbb N}$, then $X$ is regular and second countable, but you cannot expect it to be Borel... yet, the inclusion map $f: X\hookrightarrow [0,1]^{\mathbb N}$ is certainly an embedding. Maybe $f(X)$ being Borel really comes from the way one usually constructs this embedding $f$? It doesn't seem to be true for any old embedding. – Sam Dec 21 '12 at 19:51
@SamL. Very nice point that you're saying. It also answers one part of my question. Namely, that it does not follow from the topological properties as they stand, but explicitly from the choice of $f$. And the proof that I am looking at uses Urysohn's lemma countably many times appropriately with second countability of $X$, and does not give the explicit function. However, I am looking at some other results in measure theory that use this property quite freely. E.g. when embedding a Polish space to the Hilbert cube and assuming that $f(X)$ is a measurable set respect to a strict Borel measure. – T. Eskin Dec 22 '12 at 14:39
In response to the last sentence of your comment: A subspace of a Polish space is completely metrizable if and only if it is a $G_\delta$. So if you have an embedding of a Polish space into the Hilbert cube, its image must be a $G_\delta$. This is a classic consequence of theorems of Kuratowski, Lavrentiev and Marczewski. Here's a thread discussing this. In some books this is treated under the heading "absolute $G_\delta$'s" (which was the terminology used by the Polish school). – Martin Dec 22 '12 at 14:56
@Martin. Thanks. I knew the connection of $G_{\delta}$ sets and completely metrizable sets, but it didn't occur to me to use it in this context. This is in fact a very nice a way of proving this. Do you think that $N_{2}$ and $T_{3}$ are alone enough for $f(X)$ to be Borel? I think I will slightly rephrase my question in this thread. – T. Eskin Dec 22 '12 at 15:09
I'm not sure what you mean by $N_2$ and $T_3$. A Borel subset of a Polish space is in particular analytic, which forces your space to be a Suslin space. Continuous images of Suslin spaces are analytic (essentially by definition). So, if you take a non-analytic subset of the Hilbert cube (there are such sets for cardinality reasons), then there's no hope to embed it as a Borel set. – Martin Dec 22 '12 at 15:34

1 Answer 1

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 $[0,1]^{\mathbb N}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.