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

Question

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?

Answer 1

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.

Answer 2

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 $X$.