Let $X$ be a perfect Polish space $X$, so that $X^\omega$ is also a Polish space under the product topology. Call a subset $\mathcal{X} \subseteq X^\omega$ boxopen if it is an open subset of $X^\omega$ under the box topology. What is known about the descriptive settheoretic properties (w.r.t. the Polish topology) of boxopen 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).
2 Answers
(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 boxopen sets have in general.

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