Call a topological space $X$ standard Borel if $X$ is standard Borel as a measurable space (equipped with its Borel $\sigma$-algebra), i.e. if there is a Borel isomorphism between $X$ and a Polish space.

Clearly all Polish spaces are standard Borel by definition, but the converse is not true (any Borel subset of a Polish space is standard Borel, but it is Polish under the subspace topology if and only if it is $G_\delta$).

Is there a purely topological characterization of which topological spaces are standard Borel?

If an exact characterization is hopeless, I am interested in sufficient or necessary topological conditions that are weaker than Polish. For example, is any standard Borel topological space second-countable? Metrizable? Homeomorphic to a Borel subset of a Polish space?

  • $\begingroup$ The space of Schwartz distributions is another example of a Borel-standard space that is not second countable. However, although the topology doesn't have a countable base, it has a countable network, i.e. a countable family of sets (not required to be open sets) such that each open set is expressible as a union of some of them. $\endgroup$ Dec 13, 2022 at 17:59

2 Answers 2


Here are two examples showing that none of your candidate notions work.

First, we can observe that every Quasi-Polish space (https://doi.org/10.1016/j.apal.2012.11.001) admits a Baire class 1 isomorphism to a Polish space, and thus has a standard Borel $\sigma$-algebra. However, take e.g. the Scott domain $\mathcal{O}(\mathbb{N})$, with underlying set $\mathcal{P}(\mathbb{N})$ and the topology generated by $\{U \subseteq \mathbb{N} \mid n \in U\}$. This space is not Hausdorff, so clearly not metrizable and not isomorphic to any subspace of a Polish space.

For our second example, let us consider the space $\mathbb{R}[X]$ of polynomials of the reals. It is topologized as the limit of the compact Polish space of polynomials of degree up to $n$ and coefficients bounded by $n$. This space is not second-countable, but it is separable, so again, it is not metrizable. As there is a $\Delta^0_2$-bijection between $\mathbb{R}[X]$ and the Polish space $\mathbb{R}^*$, it again has a standard Borel $\sigma$-algebra.

  • $\begingroup$ Thanks for the examples! It seems like standard Borel spaces can be much more general than what I thought. $\endgroup$ Dec 14, 2022 at 20:55

Since every Polish space is Borel isomorphic to a zero-dimensional compact metric space, it suffices to characterize topological spaces, which are Borel isomorphic to a zero-dimensional compact metric space.

Such a characterization can look as follows:

A nonempty topological space $X$ is standard Borel if and only if there exists a sequence $(\mathcal F_n)_{n\in\omega}$ of finite Borel partitions of $X$ such that

  1. $\mathcal F_0=\{X\}$;

  2. for every $n\in\omega$ and $A\in\mathcal F_{n+1}$ there exists $B\in\mathcal F_n$ such that $A\subseteq B$;

  3. for every decreasing sequence $(F_{n})_{n\in\omega}\in\prod_{n\in\omega}\mathcal F_n$ the intersection $\bigcap_{n\in\omega}\mathcal F_n$ is a singleton.

By a Borel partition of a topological space $X$ we understand any cover of $X$ by pairwise disjoint nonempty Borel subsets of $X$.

  • $\begingroup$ Interesting. I was hoping for a more topological condition though (something that doesn't mention the Borel sets). $\endgroup$ Dec 14, 2022 at 20:53
  • 1
    $\begingroup$ @AntoineLabelle Except for the requirements of Borel partitions, this characterization is rather set-theoretical than topological. That is a characterization of standard Borel spaces among sets endowed with a $\sigma$-algebra; such $\sigma$-algebras do not need to be related to any topology. $\endgroup$ Dec 14, 2022 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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