It is well known that a subset $Y$ of a Polish space $X$ is completely metrisable iff it is a $G_\delta$ subset. This relates a relative topological property of the subspace $Y \subset X$ to an absolute topological property of $Y$.

I wonder are there more general versions of this result where $G_\delta$ is replaced by some other level of the Borel hierarchy and complete metrisability is replaced by some more complex topological property?

The first fact can be used to prove $\mathbb Q^\omega \not \cong \mathbb P^\omega$ where $\mathbb P = \mathbb R - \mathbb Q$ is the set of irrational numbers.

It is well-known $\mathbb P$ is completely metrisable and this extends to the countable power. Next we find a $G_\delta$ subset $G \subset \mathbb R^\omega$ with $\mathbb P^\omega \subset G \subset \mathbb R^\omega - \mathbb Q^\omega$.

$$G = \bigcap_{i \in \omega} \{ x \in \mathbb R^\omega: \text{ all } x_i \ne q_n\}$$

where $\{q_1,q_2,\ldots\}$ is an enumeration of $\mathbb Q$.

Since both spaces are dense in $\mathbb R^\omega$ and any two dense $G_\delta$ sets intersect it follows $\mathbb Q^\omega$ is not $G_\delta$ hence not completely metrisable.

Here is a harder problem: Suppose instead of $\mathbb Q^\omega$ and $\mathbb P^\omega$ we're interested in the spaces $\mathbb Q^{ \oplus \, \omega}$ and $\mathbb P^{ \oplus \, \omega}$ of choice functions $f: \omega \to \mathbb R$ with all but finitely many coordinates equal to zero and the rest in $\mathbb Q$ or $\mathbb P$ respectively.

In this case $G_\delta$ sets are no use. To see $\mathbb P^{ \oplus \, \omega}$ is not $G_\delta$ first observe it is dense in $\mathbb R^{ \omega}$. Then prove $\{x \in \mathbb P^{ \omega}: \text{ all } x_i \ne 0\}$ is $G_\delta$. Then observe the two sets are disjoint so the first cannot be $G_\delta$. In particular $\mathbb P^{ \oplus \, \omega}$ is not $G_\delta$ so cannot be completely metrisable.

**Edit:** Actually the space $\mathbb Q^{\oplus \omega}$ is countable since you can surject $\mathbb Q^{1} \cup \mathbb Q^{2} \cup \ldots$ onto it. So here is a harder question. What if we are interested in the spaces of choice functions with infinitely many coordinates equal to zero and the rest in $\mathbb Q$ or $\mathbb P$ respectively?

Getting back to my original question, suppose we could find some level in the Borel Hierarchy that contains $\mathbb P^{ \oplus \, \omega}$ but not $\mathbb Q^{ \oplus \, \omega}$ or vice-versa. Would this translate to some absolute topological property that distinguishes the spaces?

I'd also appreciate any other ideas to prove the spaces are (not) homeomorphic.