For $\mathcal{M}$ a (countable) nonstandard model of $\mathsf{PA}$, let $\mathsf{SS}(\mathcal{M})$ be the set of sets of natural numbers coded by elements of $\mathcal{M}$. There are various ways to define this, for example $$\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;(k\in X\iff p\_k\vert a)\}$$ (where $p\_k$ is the $k$th prime and we conflate $\mathbb{N}$ with its canonical image in $\mathcal{M}$).


I'm curious about how this could differ from the following analogue: let $$\mathsf{SS}^-(\mathcal{M})=\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;[k\in X\iff \exists n\in\mathbb{N}\;(p\_k^n\not\vert a)]\}.$$


Intuitively, elements of $\mathbb{N}$ are **prevented** from entering $X$ by corresponding primes dividing $a$ "too much." (This version has come up in a separate problem I'm playing with, and I'd like to understand it better.)


It's easy to show that $\mathsf{SS}(\mathcal{M})$ is the set of elements of $\mathsf{SS}^-(\mathcal{M})$ whose complements are also in $\mathsf{SS}^-(\mathcal{M})$, and it's not hard to show that every $\mathcal{M}$ has an elementary extension $\mathcal{N}$ such that $\mathsf{SS}^-(\mathcal{N})=\mathsf{SS}(\mathcal{N})$. However, this still leaves a lot open. In particular:



> 
> Is there an $\mathcal{M}$ with $\mathsf{SS}(\mathcal{M})\not=\mathsf{SS}^-(\mathcal{M})$?
> 
> 
> 


Equivalently per the above, is there an $\mathcal{M}$ whose $\mathsf{SS}^-$ is not closed under complementation?



A "negative" standard system
If $\mathcal{M}$ is a nonstandard model of PA then for any set $X \in \mathsf{SS}(\mathcal{M})$, $X' \in \mathsf{SS}^-(\mathcal{M})$. Thus if $\mathsf{SS}(\mathcal{M}) = \mathsf{SS}^{-}(\mathcal{M})$ then $\mathsf{SS}(\mathcal{M})$ is closed under the Turing jump. Since not every Scott set is closed under the jump, $\mathsf{SS}(\mathcal{M})$ is not always equal to $\mathsf{SS}^-(\mathcal{M})$.


**Claim.** If $X \in \mathsf{SS}(\mathcal{M})$ then $X' \in \mathsf{SS}^-(\mathcal{M})$.


*Proof.* Suppose $X$ is in $\mathsf{SS}(\mathcal{M})$. Thus there is some $a$ such that
$$
\forall k \in \mathbb{N}\, (k \in X \iff p\_k \mid a).
$$
Now let $b$ be a fixed nonstandard number in $\mathcal{M}$ and let $c$ be a nonstandard number such that
$$
\forall n, k < b\, (p\_k^n \mid c \iff \varphi^a\_k(k) \text{ does not converge in $\leq n$ steps}).
$$
Here, using $a$ as an oracle for $\varphi\_k(k)$, means that when $\varphi\_k(k)$ asks a question about $m$ to the oracle, we check if $p\_m \mid a$ to determine the answer. Note that the existence of such a $c$ can be proved in PA.


Now let $Y = \{k \in \mathbb{N} \mid \exists n \in \mathbb{N} \, (p\_k^n \nmid c)\}$. I claim that $Y = X'$. To show this it is enough to show that for all $n, k \in \mathbb{N}$, $p\_k^n \mid c$ if and only if $\varphi^X\_k(k)$ does not converge in $\leq n$ steps.


Note that for $n$ and $k$ standard natural numbers, the convergence or nonconvergence of $\varphi^X\_k(k)$ within $n$ steps is witnessed by a standard natural number (encoding the transcript of the computation). And since running $\varphi^X\_k(k)$ for $n$ steps never requires asking questions of the oracle at nonstandard numbers, there is no difference between using $X$ and $a$. Thus $\mathcal{M}$ can check that the witness to convergence or divergence of $\varphi^X\_k(k)$ within $n$ steps is also a witness to the convergence or divergence of $\varphi^a\_k(k)$ within $n$ steps and therefore by definition of $c$, $p\_k^n \mid c$ if and only if $\varphi^X\_k(k)$ does not converge in $\leq n$ steps.


