Definability properties of box-open subsets of Polish space

Let $$X$$ be a perfect Polish space $$X$$, so that $$X^\omega$$ is also a Polish space under the product topology. Call a subset $$\mathcal{X} \subseteq X^\omega$$ box-open if it is an open subset of $$X^\omega$$ under the box topology. What is known about the descriptive set-theoretic properties (w.r.t. the Polish topology) of box-open subsets of $$X^\omega$$? For instance, are they necessarily Borel? If not, are they necessarily analytic, $$\Sigma_2^1$$ etc? I'm also interested in the particular case where $$X = \mathbb{R}$$ (with the Euclidean metric).

• You changed "Polish" to "uncountable Polish", after Pedro Sánchez Terraf answered with an example of a countable space. What you probably mean is "perfect Polish". Commented Jun 14 at 15:03

(Blowing up Pedro's counterexample to get a perfect Polish set:)

Take the space $$X=(\omega\times \mathbb R)$$. Then for every $$\bar n = (n_k)_k\in \omega^ \omega$$ the set $$X_{\bar n} :=\prod \{n_k\}\times \mathbb R$$ is open in the box topology. For any subset $$S \subseteq \omega^ \omega$$ the set $$X_S:=\bigcup_{\bar n\in S} X_{\bar n}$$ is also open. These are $$2^{\mathfrak c}$$ many sets.

The usual notion of definability (with parameters in $$\mathbb R$$) will give you only $$\mathfrak c$$ many sets.

EDIT: (Since you are interested in $$\mathbb R$$: The space $$\mathbb R$$ contains a homeomorphic copy of the space I described.)

Take any countable set $$X$$ with the discrete topology, hence it is also Polish. Since singletons are open in $$X$$, the base of the box topology on $$X^\omega$$ includes the set of all of its singletons in turn, and then it $$X^\omega$$ is discrete.

Hence there is no definability property the box-open sets have in general.

• Thanks Pedro. I forgot to include the condition that $X$ needs to be uncountable. Commented Jun 14 at 15:01