# Which topological spaces have a standard Borel $\sigma$-algebra?

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?

• 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. Dec 13, 2022 at 17:59

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.

• Thanks for the examples! It seems like standard Borel spaces can be much more general than what I thought. 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 $$X$$.

• Interesting. I was hoping for a more topological condition though (something that doesn't mention the Borel sets). Dec 14, 2022 at 20:53
• @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. Dec 14, 2022 at 21:24