*Fix some computable bijection $b$ between $\omega$ and $2^{<\omega}$. For $r\in 2^\omega$, let $$[r]=\{f\in 2^\omega: \forall\sigma\prec f(b^{-1}(\sigma)\in r)\}$$ be the closed subset of Cantor space coded by $r$. For $M\models PA$ nonstandard, let $\mathcal{S}(M)$ be the standard system of $M$ thought of as a topological space (namely, as a subspace of Cantor space).*




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Say that a **closed set pattern** on a topological space $(X,\tau)$ is an assignment $c$ of $\tau$-closed sets to points in $X$. 


*EDIT: while it doesn't impact this question or the [followup question](https://mathoverflow.net/questions/353801/detecting-comprehension-topologically), it seems natural in retrospect to also add the condition that the relation "$x\in c(y)$" be closed in the product topology; that is, the pattern itself should also be closed.*


Every nonstandard $M\models PA$ has a corresponding closed set pattern on $\mathcal{S}(M)$ given by $$c\_M: r\mapsto [r]\cap SS(M).$$ If $M$ is countable the space $\mathcal{S}(M)$ is homeomorphic to the rationals, so any interesting behavior is concentrated on $c\_M$.


I would like to understand how closed set patterns of the form $c\_M$ behave, and the following seems a good starting point. Say that closed set patterns $c\_1,c\_2$ on $\mathcal{X},\mathcal{Y}$ respectively are **equivalent** (and write $c\_1\sim c\_2$) if they differ by a homeomorphism - that is, if there is an $H:\mathcal{X}\cong\mathcal{Y}$ satisfying $$x\in c\_1(y)\leftrightarrow H(x)\in c\_2(H(y))$$ for all $x,y\in\mathcal{X}$. My question is:



> 
> Are there countable nonstandard $M,N\models PA$ such that $c\_M\not\sim c\_N$?
> 
> 
> 


The knee-jerk approach to a positive answer would be a back-and-forth argument, but since the assignment of closed sets to reals isn't continuous in any good sense that doesn't seem to work here. On the other hand, I don't even see how to start approaching a negative answer.



The "higher topology" of countable Scott sets
Given any topological space $X$ and subset $F\subseteq X$, define the Cantor-Bendixson sequence of $F$ in $X$ as:


* $F^{(0)} = F$
* $F^{(\alpha +1)} = F^{(\alpha)} \setminus \{x \in F^{(\alpha)} : x \text{ is isolated in }F^{(\alpha)}\}$
* $F^{(\beta)} = \bigcap\_{\alpha < \beta} F^{(\alpha)}$, $\beta$ a limit ordinal.


Now we'll define the CB-rank of $F$, written $CB(F)$, to be the least ordinal $\alpha$ such that $F^{(\alpha)} = F^{(\alpha +1)}$ (I think this is a slightly non-standard definition). Note that this doesn't actually depend on the ambient space $X$. The typical argument gives us that for second countable $F$, $CB(F) < \omega\_1$ (specifically, take a countable base for the topology on $F$, each set in this base can only be removed at most once in the sequence), and crucially $CB(F)$ only depends on the topological properties of $F$.


Fix a non-standard model $M$ of $PA$. Now, since $\mathcal{S}(M)$ is countable, we have that $\gamma = \sup \_{r \in \mathcal{S}(M)}CB(c\_M(r))$ is also a countable ordinal.


Now fix a countable closed subset $F \subseteq 2^{\omega}$ with $CB(F) > \gamma$ (such a set always exists, since $\alpha$ is countable). Pick a real $r$ such that $[r] = F$ and now take a countable model $N$ of $PA$ such that $r \in \mathcal{S}(N)$ and $F \subseteq \mathcal{S}(N)$. This is always possible by your comment that every Scott set is the standard system of some countable model of $PA$. (EDIT: But also just compactness and the downward Löwenheim–Skolem theorem, since we don't really care about the particular Scott set in question.)


So now clearly we have $c\_N(r) = F$, so $CB(c\_N(r)) = CB(F) >CB(c\_M(s))$ for every $s \in \mathcal{S}(M)$, and thus we have $c\_M\not\sim c\_N$.


