If $(P,\leq)$ is a poset and $p\in P$, then we say that $p$ *is the lower part of a gap* there is $q \in P$, $q>p$ such that $[p,q] = \{p,q\}$. (This is equivalent to the statement that $(\uparrow p) \setminus \{p\}$ contains a minimal element.)


Let $\text{NPU}(\omega)$ be the set of non-principal ultrafilters on $\omega$. The *Rudin-Keisler preorder* on $\text{NPU}(\omega)$ is defined by
$${\cal U} \leq\_{RK} {\cal V} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall U\in{\cal U}) f^{-1}(U)\in {\cal V} .$$


It is easy to see that $\leq\_{RK}$ is reflexive and transitive, but not anti-symmetric. Set ${\cal U}\simeq\_{RK} {\cal V}$ if ${\cal U}\leq\_{RK}{\cal V}$ and ${\cal V}\leq\_{RK}{\cal U}$. So $\text{NPU}(\omega)/\simeq\_{RK}$ is a poset with the Rudin-Keisler order applied to equivalence classes.


Does $\text{NPU}(\omega)/\simeq\_{RK}$ contain elements that are the lower part of a gap?



"Gaps" in the Rudin-Keisler ordering
Yes. Take U, V non-isomorphic Ramsey ultrafilters. Then there is no W that is RK-strictly in between $U$ and $U\cdot V$. The reason is the same as that given in [Infima in the Rudin-Keisler ordering](https://mathoverflow.net/questions/283045/infima-in-the-rudin-keisler-ordering?rq=1)


