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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.

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  • $\begingroup$ Set of irrational numbers, thanks! I've edited the question now. $\endgroup$ – Daron Jul 9 '19 at 12:02
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    $\begingroup$ A (very) partial answer: I think that a subset of $\mathbb R$ is locally compact if and only if it is $\Delta^0_2$. $\endgroup$ – Will Brian Jul 9 '19 at 13:41
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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.

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  • $\begingroup$ Can you explain why this is a contradiction please? $\endgroup$ – Daron Jul 10 '19 at 21:11
  • $\begingroup$ Essentially, $B=(\bar f^{-1})^{-1}[A]$ and thus $B$ must be in $\boldsymbol\Pi^0_\zeta\subseteq\boldsymbol\Pi^0_\xi$. But then $B$ can't be $\boldsymbol\Sigma^0_\xi$-complete. $\endgroup$ – Pedro Sánchez Terraf Jul 10 '19 at 23:35
  • $\begingroup$ There is a further technical detail, that in this argument I'm using that $B$ is still complete in the smaller space $H$. I believe this might hold $\xi>2$ but not so certain for $\xi\leq 2$. $\endgroup$ – Pedro Sánchez Terraf Jul 11 '19 at 13:54
  • $\begingroup$ Thanks! I was wondering why the functions seemed to be the wrong way around. $\endgroup$ – Daron Jul 13 '19 at 17:02
  • $\begingroup$ @Daron You're welcome! $\endgroup$ – Pedro Sánchez Terraf Jul 14 '19 at 2:25

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