3
$\begingroup$

Let $\text{NPU}(\omega)$ be the set of non-principal ultafilters 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.

It is known that if ${\cal R}$ is a minimal element of Ramsey ultrafilter, then $[{\cal R}]_{\simeq_{RK}}$ is a minimal element of $\text{NPU}(\omega)/\simeq_{RK}$.

Question. Suppose ${\cal U, V}\in \text{NPU}(\omega)$ are non-Ramsey , do $[{\cal U}]_{\simeq_{RK}}, [{\cal V}]_{\simeq_{RK}}$ have an infimum in $\text{NPU}(\omega)/\simeq_{RK}$?

$\endgroup$
2
$\begingroup$

This part only shows the existence of lower bounds, which is not the point. See edit. This is consistently true for example when the near coherence principle of ultrafilters holds. It says, for any two non-principal ultrafilters $U, V$ there exists a finite-to-one $f: \omega\to \omega$ such that $f(U)=f(V)$. Since $f$ is finite-to-one, it makes sure $f(U)=f(V)$ is not principal.

Edit: the answer is consistently no. Suppose there exist two non-isomorphic Ramsey ultrafilters $U, V$, then we claim $U\cdot V$ and $V\cdot U$ do not have an infimum. This follows from the following facts

  1. $U\cdot V, V\cdot U$ are neither P-point nor Q-point
  2. $U\cdot V \simeq V\cdot U$ implies $U\simeq V$
  3. The only elements that are potentially RK below $U\cdot V$ are $V\cdot U$, $U$, $V$ (or the principal ultrafilters)
  4. So the common non-principal lower bounds for $U\cdot V, V\cdot U$ are $U, V$. But they are incomparable.

All of these can be found in Blass' thesis.

$\endgroup$
  • 1
    $\begingroup$ Does NCF actually give you an infimum, or just the existence of lower bounds? This isn't immediately clear to me. $\endgroup$ – Todd Eisworth Oct 9 '17 at 16:05
  • $\begingroup$ I don't think it does. that answer is not right to the point as I didn't finish reading the question. I think the answer is no if there are two non-isomorphic selective ultrafilters. $\endgroup$ – Jing Zhang Oct 10 '17 at 0:11
  • $\begingroup$ Thanks and +1Jing Zhang! Can you elaborate on the non-isomorphic selective ultrafilters? $\endgroup$ – Dominic van der Zypen Oct 10 '17 at 6:42
  • $\begingroup$ @DominicvanderZypen: I think you can try the following to classify the ultrafilters below $U\cdot V$ (similarly $V\cdot U$): suppose $W\leq_{RK} U\cdot V$, let $f: \omega\times \omega\to \omega$ be the witness. Then you can cook up two functions $g: \omega\to \omega$ such that $g(n)=k$ iff $\{m: f(n,m)=k\}\in V$. Similary $h$ reversing the role of $U$ and $V$. By Ramsey, there are 4 cases to consider: g is constant on a $U$-large set, or 1-1 and similarly for $h$. This might help. $\endgroup$ – Jing Zhang Oct 11 '17 at 1:46
  • $\begingroup$ Thanks @JingZhang, it makes sense. Maybe you can put this comment into your answer, I am tipping towards accepting it now because I think I can work out the details $\endgroup$ – Dominic van der Zypen Oct 11 '17 at 6:47

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.