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:



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



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> For each $\alpha < \omega\_1$, the set $\mathsf{WO}\_{\alpha}$ of all well orderings of $\mathbb N$ with order type $\alpha$ is Borel.
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**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](https://mathoverflow.net/questions/440926/can-every-borel-set-be-partitioned-into-leq-aleph-1-f-sigma-delta-set) of mine.



Can you fit a $G_\delta$ set between these two sets?
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 $x\notin\mathrm{WO}$.


