# Are there nonstandard $\mathsf{PA}$ models without $\Delta^1_1$ cuts?

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 ("cuts")?

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.) 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.

The general idea of the solution is that sufficiently saturated models will always fail to have $$\Delta^1_1$$ definable cuts, even with parameters. Computably saturated countable models should also be sufficient, which avoids set theoretic assumptions.