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.



Do the higher levels of the Borel hierarchy correspond to absolute topological properties?
You can use complete Borel sets for this.


Recall that for Polish spaces $X,Y$ and subsets $A\subseteq X$ and $B\subseteq Y$, $A$ is *Wadge-reducible* to $B$ if there is a continuous $f:X\to Y$ such that $f^{-1}(B) = A$. It can be proved (for instance, see Theorem 22.10, and Exercises 22.11 and 24.20 in Kechris' *Classical Descriptive Set Theory*) that $B\in \boldsymbol\Sigma^0\_\xi\setminus\boldsymbol\Pi^0\_\xi$ if and only if $B$ is *complete*, in that every $A\in\boldsymbol\Sigma^0\_\xi$ is Wadge-reducible to it (for $\xi\geq 1$).


This immediately gives that such a $B$ is never homeomorphic to an $A\in \boldsymbol\Pi^0\_\zeta$, with $\zeta\leq\xi$, in some other Polish space $X$. Otherwise, take $f:A\to B$ a homeomorphism; by Lavrentiev's theorem, $f$ extends to homeomorphism $\bar f: G \to H$ where $G,H$ are $G\_\delta$ subsets such that $A\subseteq G \subseteq X$ and $B\subseteq H \subseteq Y$. But then $B$ reduces to $A$ through $\bar f^{-1}$, a contradiction.


