# Do the higher levels of the Borel hierarchy correspond to absolute topological properties?

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.

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

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.
• 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. – Pedro Sánchez Terraf Jul 10 '19 at 23:35
• 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 $\xi\leq 2$. – Pedro Sánchez Terraf Jul 11 '19 at 13:54