A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is called selective (or Ramsey) if for every partition $\mathcal{P}$ of $\omega$ disjoint with $\mathcal{U}$ there is $A\in\mathcal{U}$ such that $|A\cap B|\le1$ for every $B\in\mathcal{P}$. There are several weakenings of this definition, but still expressing some kind of selectivity; most notably P-points and Q-points.

My question is as follows. Do you know any notion of highly non-selective ultrafilters? A class of ultrafilters about which we can say that they are very far from being selective? Or maybe some way of measuring how ultrafilters are selective?

Selective ultrafilters are exactly the minimal points of the Rudin-Keisler ordering of non-principal ultrafilters on $\omega$, so any ultrafilter which is not a minimal one is not selective. But how can we see this ordering in the context of selectivity?

I know that the question is vague, but maybe you have some association with it. I will appreciate any answer.

  • $\begingroup$ Surely that cannot be the correct definition. $\{\omega\}$ is a partition of $\omega$. $\endgroup$ – Asaf Karagila Dec 5 '16 at 16:52
  • $\begingroup$ Yeah, no element of the partition should be in the ultrafilter. I corrected it. $\endgroup$ – Damian Sobota Dec 5 '16 at 16:59
  • 2
    $\begingroup$ Personally, I always thought of idempotent ultrafilters on $(\mathbb{N},+)$ as particularly good examples of "non-selective" ultrafilters. The recent literature on the Tukey order might also be helpful. $\endgroup$ – Peter Krautzberger Dec 6 '16 at 8:01

Given any set $X$ of $2^{\aleph_0}$ ultrafilters on $\omega$, there exists an ultrafilter on $\omega$ that is Rudin-Keisler above all the members of $X$. Is that the sort of strong non-minimality that you wanted?

Edit, to answer the OP's comment: Here's a result, based on the same idea as in my original answer, but perhaps closer to what you want because it's phrased in terms of partitions rather than the RK ordering. There exist an ultrafilter $\scr U$ and a family $\scr F$ of $2^{\aleph_0}$ partitions of $\omega$ such that, on no set $X\in\scr U$ is there a "functional dependence" between any finitely many of the partitions. By a functional dependence between partitions $P_0,P_1,\dots,P_n$, I mean that, if I tell you that I'm thinking of a point $x\in X$ and I tell you which blocks of $P_1,\dots,P_n$ contain $x$, you cannot infer which block of $P_0$ contains $x$. (In particular, $\scr U$ cannot contain any block of any of the partitions in $\scr F$, nor can it contain a selector for any of those partitions.)

  • $\begingroup$ Yes, I am aware of this fact. But what can we say about the "(non)selectivity" of this upper bound? How can we understand the RK ordering in terms of "selectivity"? $\endgroup$ – Damian Sobota Dec 5 '16 at 17:05
  • $\begingroup$ Andreas, thank you. Where can I read details concerning those two facts you provided in your answer? $\endgroup$ – Damian Sobota Dec 6 '16 at 17:49
  • 1
    $\begingroup$ I'm away from home and can't easily check references, but these facts should be in the book "Theory of Ultrafilters" by Comfort and Negrepontis. The key ingredient in the proof is the existence of a family of $2^{\aleph_0}$ subsets of $\omega$ that are independent, meaning that the intersection of any finitely many of them and the complements of any finitely many others is infinite. The existence of such a family is surely in Jech's "Set Theory". Comfort and Negrepontis have it also but they say "family of large oscillation" instead of "independent family". $\endgroup$ – Andreas Blass Dec 7 '16 at 3:41

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.