Can you fit a $G_\delta$ set between these two sets?

Question

Every subset of $\mathbb N \times \mathbb N$ can be viewed as a relation on $\mathbb N$. The set $\mathcal P(\mathbb N \times \mathbb N)$ of all relations on $\mathbb N$ has a natural topology with which it is homeomorphic to the Cantor space. A pretty well-known fact from descriptive set theory is:

The subset $\mathsf{WO}$ of well orderings of $\mathbb N$ is a coanalytic subset of $\mathcal P(\mathbb N \times \mathbb N)$.

This is one of the most natural examples out there of a subset of a Polish space that is coanalytic but not Borel. This coanalytic set stratifies nicely into an increasing union of $\omega_1$ Borel sets:

For each $\alpha < \omega_1$, the set $\mathsf{WO}_{\alpha}$ of all well orderings of $\mathbb N$ with order type $\alpha$ is Borel.

Question: Given $\alpha < \omega_1$, is there a $G_\delta$ set $A \subseteq \mathcal P(\mathbb N \times \mathbb N)$ such that $\mathsf{WO}_{\alpha} \subseteq A \subseteq \mathsf{WO}$?

Motivation: If the answer is positive, then we would be able to write this coanalytic set as a union $\bigcup_{\alpha < \omega_1} A_\alpha$ of $G_\delta$ sets, which would imply that there is a partition $\{ A_\alpha \setminus \bigcup_{\xi < \alpha} A_\xi :\, \alpha < \omega_1 \}$ of $\mathsf{WO}$ into $F_{\sigma \delta}$ sets. Because $\mathsf{WO}$ is an example of a complete coanalytic set, this would then imply that every coanalytic set can be partitioned into $\aleph_1$ $F_{\sigma \delta}$ sets, answering another recent question of mine.

Answer 1

No, not for $\alpha\geq\omega$. For let $A$ be $G_\delta$ and suppose that WO$_\alpha\subseteq G_\delta$. Let's show that $A\not\subseteq$ WO. Fix a sequence $\left<A_n\right>_{n<\omega}$ of open sets such that $A=\bigcap_{n<\omega}A_n$. Instead of directly discussing elements of $\mathcal{P}(\mathbb{N}\times\mathbb{N})$, I will discuss functions $x:\mathbb{N}\times\mathbb{N}\to 2$, and functions $\sigma:n\times n\to 2$ for $n<\omega$ as their finite approximations. Say that such a $\sigma$ is \emph{good} if it is (the characteristic function of) a linear order on $n$ (where $\mathrm{dom}(\sigma)=n\times n$). Note that since $\alpha\geq\omega$, every good $\sigma$ extends to some $x\in\mathrm{WO}_\alpha$. We can assume that for each $n$ we can fix a set $B_n$ of good tuples such that $A_n$ is just the set of all $x$ which extend some $\sigma\in B_n$, and $B_n$ is closed under extension, i.e. if $\sigma\in B_n$ and $\sigma'$ extends $\sigma$ and is also good, then $\sigma'\in B_n$.

Now we will construct a sequence $\left<\sigma_n,k_n\right>_{n<\omega}$ consisting of good $\sigma_n\in B_n$ with $\sigma_n\subsetneq\sigma_{n+1}$, and integers $k_n$, as follows. (Letting $x=\bigcup_{n<\omega}\sigma_n$, we will then have $x\in A$. But the plan is to arrange that $x(k_{n+1},k_n)=1$ for all $n$, so $x\notin\mathrm{WO}$.)

So, fix some $\sigma_0\in A_0$ with domain $m\times m$ for some $m>0$, and let $k_0=0$. Given $\sigma_n\in A_n$, and given $k_0<\ldots<k_n$ with $\mathrm{dom}(\sigma_n)=m'\times m'$ for some $m'>k_n$, let $k_{n+1}=m'$, and let $\sigma'_n$ be good and with domain $(m'+1)\times(m'+1)$ and with ``bottom'' element $m'$, i.e. $\sigma'_n(m',i)=1$ for all $i<m'$. Now since $\sigma'_n$ is good, we can find $y\in\mathrm{WO}_\alpha$ such that $\sigma'_n\subseteq y$, and therefore $y\in A_{n+1}$. So let $\sigma_{n+1}\in B_{n+1}$ with $\sigma'_n\subseteq\sigma_{n+1}$. This completes the construction.

Letting $x=\bigcup_{n<\omega}\sigma_n$, note that $x\in A$, but $x(k_{n+1},k_n)=1$ for all $n$, so $x\notin\mathrm{WO}$.

Answer 2

A negative answer follows from $(*)$ Proposition 8.2 of Jacques Stern's paper “Effective partitions of the real line into Borel sets of bounded rank”. Link to paper:

https://www.sciencedirect.com/science/article/pii/0003484380900030

Assume $\alpha<\omega_1$ is a countable ordinal, $(\omega+\omega)\cdot\alpha<\beta<\omega_1$ is an arbitrary countable ordinal sufficiently large, and $A\subseteq\textrm{WO}$ is a $\mathbf{\Sigma}^0_\alpha$ Borel set. We'll show that $\textrm{WO}_{\omega^\beta}\not\subseteq A$.

Assume towards a contradiction that $\textrm{WO}_{\omega^\beta}\subseteq A$. Recall that $S_\infty$ is the Polish group of all permutations on $\omega$, and it admits a logic action on $\mathscr{P}(\omega\times\omega)$ in which each $\textrm{WO}_\gamma$ is an $S_\infty$-orbit. Then the Vaught transform $A^\Delta=A^{\Delta S_\infty}$ of $A$ is still a $\mathbf{\Sigma}^0_\alpha$ Borel set such that $\textrm{WO}_{\omega^\beta}\subseteq A^\Delta$, since $\textrm{WO}_{\omega^\beta}$ is $S_\infty$-invariant. By the cited proposition $(*)$, if $(\omega+\omega)\cdot \alpha<\zeta<\omega_1$ is arbitrary then we also have $\textrm{WO}_{\omega^\zeta}\subseteq A^\Delta$. In particular, there is an unbounded set of $\omega^\zeta=\gamma<\omega_1$ for which $\textrm{WO}_\gamma\subseteq A^\Delta$. On the other hand since $\textrm{WO}$ is $S_\infty$-invariant, $A^\Delta\subseteq\textrm{WO}$ is a Borel subset, and the boundedness theorem for analytic subsets of $\textrm{WO}$ implies $\{\gamma:\textrm{WO}_\gamma\textrm{ intersects }A^\Delta\}$ is bounded below $\omega_1$. This is a contradiction.

Letting $\alpha=3$ gives for any $A\subseteq\textrm{WO}$ a $\mathbf{\Pi}^0_2$ set, $\textrm{WO}_{\omega^{\omega\cdot 7}}\not\subseteq A$ which is weaker than Farmer S's answer.