4
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ The existence of Ramsey ultrafilters is independent of ZFC. Do you know what happens if there are no Ramsey ultrafilters? (If the OP had forgotten to say "non-principal" in the question, this would be trivial: The bottom element of the RK order (the class of principal ultrafilters) would not be the lower part of a gap. $\endgroup$
    – Andreas Blass
    Commented Aug 8, 2023 at 21:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .