Let $M$ be a model of $I\Delta_0$. Recall that a definable cut is a definable (possibly with parameters) subset $I$ of $M$ that is non-empty, downwards closed, and closed under successor.

If we consider the set $C=\{x \in M : (\forall I\text{ definable cut}) x \in I\}$, then standard arguments show that $C$ satisfies $B\Sigma_1$ and is closed under addition, multiplication, and exponentiation. Visser's paper The small-is-very-small principle goes into some more detail about the theory that $C$ satisfies.

Aside from the obvious case where $M \models \mathsf{PA}$, I'm curious about the extent to which $C$ can resemble $M$.

Question 1. If $M \not \models \mathsf{PA}$, can $M$ and $C$ have the same first-order theory?

Question 2. If $M \not \models \mathsf{PA}$, can $M$ and $C$ be isomorphic?

Note that the second question doesn't a priori follow from the first because we can't really capture the notion of the intersection of all definable cuts of $M$ in a first-order way.

If these situations are possible, I'm curious if it entails a certain amount of induction on $M$, although I don't expect this to be the case.

Question 3. If $M \not \models \mathsf{PA}$ and $M\equiv C$ or $M \cong C$, does it follow that $M \models I\Sigma_n$ for some $n > 0$? Does it follow that $M \models \mathrm{SuperExp}$?

$\mathrm{SuperExp}$ is, of course, the axiom that states that the superexponential function (also known as tetration) is total.

  • 1
    $\begingroup$ If $M\equiv C$, then (on top of the already mentioned $I\Delta_0+\mathrm{Exp}+B\Sigma_1$) $M$ satisfies the local reflection principle for restricted/cut-free provability. That is, it is a model of Peano Basso. $\endgroup$ Apr 20, 2021 at 6:45


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.