*This was [asked and bountied at MSE](https://math.stackexchange.com/questions/4036036/are-there-nonstandard-mathsfpa-models-without-delta1-1-cuts) with no response:*


My question is the following:



> 
> Is there a nonstandard model $\mathcal{M}\models\mathsf{PA}$ such that $\mathcal{M}$ has no $\Delta^1\_1$-with-parameters-definable nonempty proper successor-closed initial segments?
> 
> 
> 


Here "$\Delta^1\_1$" is meant in the sense of the **standard semantics** of second-order logic - so a $\Delta^1\_1$ subset of $\mathcal{M}$ doesn't need to be "internal" to $\mathcal{M}$ in any nice sense.


If we replace $\Delta^1\_1$ with $\Pi^1\_1$ the answer is trivially negative since the cut of standard naturals is $\Pi^1\_1$; no parameters are needed here. If we replace $\Delta^1\_1$ with $\Sigma^1\_1$ this answer again becomes negative since for each nonstandard $a\in\mathcal{M}$ the set of elements infinitely below $a$ is $\Sigma^1\_1(a)$ over $\mathcal{M}$. (See [here](https://math.stackexchange.com/a/4036011/28111).) However, I don't see a way to get a $\Delta^1\_1$ cut in a nonstandard model of $\mathsf{PA}$. On the other hand, I don't see how to build a nonstandard model without a $\Delta^1\_1$ cut. In particular, a natural hope might be to look at a nontrivial ultrapower of $\mathbb{N}$, but while $\Sigma^1\_1$ formulas are preserved by taking ultrapowers, $\Delta^1\_1$-ness (= an equivalence between a $\Sigma^1\_1$ formula and a $\Pi^1\_1$ formula) doesn't obviously need to be.



Can a nonstandard model of $\mathsf{PA}$ be "$\Delta^1_1$-well-ordered?"
I believe you don't need this, but assume that there is a strongly inaccessible cardinal $\kappa$. Fix a first-order completion $T$ of $\mathsf{PA}$ and let $\mathcal{M}$ be a saturated model of $T$ of cardinality $\kappa$. I will show that $\mathcal{M}$ has no $\Delta^1\_1$-with-parameters-definable cuts.


*Claim.* For any $\Sigma^1\_1(a)$ formula $\varphi(x,a)$, there is a closed set $F\_\varphi \subseteq S\_{x}(a)$ of types such that $\mathcal{M} \models \varphi(b,a)$ if and only if $\mathrm{tp}(b/a) \in F\_\varphi$ (where $\mathrm{tp}(x/y)$ is the first-order type of $x$ over $y$).


*Proof.* Given $\varphi(x,a)$, by compactness, there is a set of formulas $\Lambda(x,a)$ such that for any $b \in \mathcal{M}$,


* there exists an elementary extension $\mathcal{N} \succeq \mathcal{M}$ for which $\mathcal{N} \models \varphi(b,a)$


if and only if $\mathcal{M} \models \Lambda(b,a)$. Since $\mathcal{M}$ is saturated, it is resplendent, and if there is such an elementary extension for a given $b$, then we actually have that $\mathcal{M} \models \varphi(b,a)$ (since some expansion of $\mathcal{M}$ by a predicate satisfies the part of $\varphi(b,a)$ after the set quantifier). Clearly the other direction holds, so we have that $F\_\varphi$ is the set of types corresponding to the partial type $\Lambda(x,a)$. $\square\_{\text{claim}}$


Assume that there is a $\Delta^1\_1$-with-parameters-definable cut, so in other words, assume that we have two $\Sigma^1\_1$ formulas $\varphi(x,a)$ and $\psi(x,a)$ such that $\varphi(\mathcal{M},a)$ is the cut and $\psi(\mathcal{M},a)$ is the complement of the cut. Let $F\_\varphi$ and $F\_\psi$ be as in the claim.


Since $\mathcal{M}$ is saturated, it is $\omega$-saturated. This implies that $F\_\varphi$ and $F\_\psi$ are disjoint (otherwise a type in their intersection would be realized) and cover $S\_x(a)$ (otherwise a type in the complement of their union would be realized). Therefore, they are actually clopen, and correspond to some first-order formula $\chi(x,a)$ and its negation, but then $\chi(x,a)$ is a definable cut, which cannot happen, as $T$ is an extension of $\mathsf{PA}$. $\square$


As for getting rid of the inaccessible, I believe a special model will be sufficient, since they are resplendent for finite expansions. I actually think a computably saturated model might be sufficient too, since the first-order theory we're trying to expand to is c.e. (finitely axiomatizable, even).


