The "higher topology" of countable Scott sets

Question

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, 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.

Answer 1

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 $c_M\not\sim c_N$.

Answer 2

Here's another way to apply Cantor-Bendixson derivatives (following James Hanson): only some $M$s have the property that $ran(c_M)$ is closed under (single) Cantor-Bendixson derivatives. Specifically, let $T$ be the downward closure of the set of strings of the form $0^n1^k0^s$ such that $s=0$ or $\Phi_n(n)$ has halted by stage $k$ - so that the non-isolated paths of $T$ (besides the all-$0$s path) are those of the $0^n1^\infty$-form for $n$ in the halting problem. Any tree $S$ with $[S]=CB([T])$ would enumerate the complement of the halting problem: $n$ is not in the halting problem iff the part of $S$ above $0^n1$ eventually dies out. In particular, if $\mathcal{S}(M)$ doesn't contain the halting problem then $ran(c_M)$ won't contain $[CB(T)]\cap \mathcal{S}(M)$.

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We can also use prunings. For $r\in \mathcal{S}(M)$, let $B_r=\{s\in\mathcal{S}(M): c_M(s)\supseteq c_M(r)\}$. Then we have that $B_r\in ran(c_M)$ for all $r$ iff in $\mathcal{S}(M)$ every tree has a pruning (= subtree with no dead ends and the same paths), which is of course equivalent to being arithmetically closed.

The right-to-left direction is essentially immediate: if $P$ is pruned then $[T]\not\supseteq [P]$ iff for some $\sigma\in P$ we have $\sigma\not\in T$, which is an open condition. In the left-to-right direction, note that a code for $B_r$ lets us enumerate the extendible nodes of the tree coded by $r$ ($\sigma$ is extendible in the tree coded by $r$ iff the real coding the tree of strings $\not\succcurlyeq\sigma$ is not in $B_r$), and the non-extendible nodes of the tree are a priori (relatively) computably enumerable.

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Two final remarks:

• Note that when we shift attention to the $\omega$-models of $WKL_0$ given by the standard systems, the two arguments above are pointing at $ACA_0$: for $M\models PA$ nonstandard, $ran(c_M)$ is closed under (single) Cantor-Bendixson derivatives iff $B_r\in ran(c_M)$ for all $r\in \mathcal{S}(M)$ iff $\mathcal{S}(M)$ is arithmetically closed. I've followed up on this line of thought here.

• All these arguments so far leave open the problem of whether we can have $c_M\not\sim c_N$ for "finer" reasons. Specifically, for $c_1,c_2$ closed set patterns on $\mathcal{X},\mathcal{Y}$, write $c_1\approx c_2$ if there is some $H:\mathcal{X}\cong\mathcal{Y}$ such that $ran(c_2)=\{H[A]: A\in ran(c_1)\}$; then we can ask whether there countable nonstandard $M,N\models PA$ with $c_M\not \sim c_N$ but $c_M\approx c_N$, and I don't see how to attack this at the moment.