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 $G\_\delta$).



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



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


