# How similar can a model of $I\Delta_0$ be to the intersection of all of its definable cuts?

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.

• 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. Apr 20, 2021 at 6:45