Suppose that $\mathcal{M}$ is a countable, arithmetically saturated model of $\mathsf{PA}$ with universe $M$, the gap about some $a\in\mathcal{M}$ is defined as the set of all $b \in M$ such that $\mathcal{M}\models b\le a \le t(b) \vee a\le b\le t(a)$ for some term $t(v)$.
Note that the relation "$x$ is in $\mathrm{gap}(y)$" is an equivalence relation, resulting in the fact that each two gaps of $\mathcal{M}$ are either identical or disjoint.
I have been wondering if for any two gaps $\mathrm{gap}(a)<\mathrm{gap}(b)$ there is some $c\in M$ such that $\mathrm{gap}(a)<\mathrm{gap}(c)<\mathrm{gap}(b)$?
Any help (or reference) would be much appreciated, Thanks!
This is in fact true (even for recursively saturated models of $\mathsf{PA}$): there is a gap between any two gaps $\mathrm{gap}(a) < \mathrm{gap}(b)$ in $\mathcal{M}$.
As mentioned in "The Structure of Models of Peano Arithmetic" (by Kossak and Schmerl), page 17, in the case where $\mathcal{M}$ is recursively saturated the elementary cut $\inf (\mathrm{gap}(b))$ is tall, i.e it has no last gap. Therefore $\sup(\mathrm{gap}(a)) < \inf (\mathrm{gap}(b))$ and there is some $\mathrm{gap}(a) < c < \mathrm{gap}(b)$ which must satisfy $\mathrm{gap}(a) < \mathrm{gap}(c) < \mathrm{gap}(b)$.