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.


1 Answer 1


I have posted a complete answer to this question on MathOverflow.

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.

  • $\begingroup$ I wonder what's the shortest time interval between answer posting and bounty awarding - for the record, this was 8 seconds by my computer. $\endgroup$ Mar 3, 2021 at 3:27

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