Suppose $U$ is a non-principal ultrafilter on $\omega$, and let us define $\tau(U)$ to be the minimum cardinality of a family $\mathcal{X}\subseteq U$ such that $\mathcal{X}$ does not have an infinite pseudo-intersection, that is, there is no infinite $A$ such that $A\setminus B$ is finite for all $B\in \mathcal{X}$.

Claim: $\tau(U)\leq\mathfrak{s}$ for any $U$.

The point is that $U$ will contain a splitting family of cardinality $\mathfrak{s}$, and this splitting family has no infinite pseudo-intersection. (To see the first statement, note that if $\mathcal{X}$ is a splitting family and we replace some of the elements of $\mathcal{X}$ by their complements, then the resulting collection is still a splitting family. Since an ultrafilter will contain one of $X$ and $\omega\setminus X$ for each $X\in\mathcal{X}$, we may as well assume $\mathcal{X}\subseteq U$.)

Question: Is there (in ZFC) an ultrafilter $U$ on $\omega$ for which $\tau(U)=\mathfrak{s}$?

I conjecture that the answer is "no", and that this negative answer will be witnessed in the original Blass-Shelah model of NCF from the paper below.

Blass, Andreas; Shelah, Saharon, There may be simple $P_{\aleph _ 1}$- and $P_{\aleph _ 2}$-points and the Rudin-Keisler ordering may be downward directed, Ann. Pure Appl. Logic 33, 213-243 (1987). ZBL0634.03047.

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    $\begingroup$ Is it true that $\tau(U)$ is a regular cardinal, for every non-principal ultrafilter $U$ on $\omega$? $\endgroup$ – Santi Spadaro Feb 16 '20 at 23:22
  • $\begingroup$ I don't know the answer to that one. A priori, I don't see why this would need to occur, but I have not thought deeply about it. $\endgroup$ – Todd Eisworth Feb 17 '20 at 1:11
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    $\begingroup$ I think another good candidate for a counterexample could be the Mathias model. $\endgroup$ – Will Brian Feb 17 '20 at 17:58
  • $\begingroup$ I guess it’s really about reflection dealing with questions like “If every small subfamily has a pseudointersection, does the whole family?” $\endgroup$ – Todd Eisworth Feb 17 '20 at 19:19

The answer is no -- it is consistent that every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$.

I had an idea for proving this earlier today, using the Mathias model. I couldn't quite make things work, and I ended up talking about the problem with Alan Dow for a good part of the afternoon. (1) We still think the Mathias model could work, but it seems tricky. (2) There's a fix: if you interleave Laver forcings with Mathias forcings in a certain way, then the resulting iteration does work. (I'll sketch this below.) (3) I learned from Alan that your question has been studied already. The consistency of "every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$" is already known (via a different argument than the Mathias-Laver iteration sketched below), and the characteristic $\tau(U)$ has been studied quite a bit.

A rich source of information on $\tau(U)$ is the following paper by Brendle and Shelah:

Jörg Brendle and Saharon Shelah, ``Ultrafilters on $\omega$ -- their ideals and their characteristics,'' Transactions of the AMS 351 (1999), pp. 2643-2674. (available here)

What you call $\tau(U)$ is in this paper called $\pi \mathfrak{p}(U)$. They prove, among other things, that

$\bullet$ If $U$ is not a $P$-point, then $\tau(U) \leq \mathfrak{b}.$

$\bullet$ If $U$ is not a $P$-point, then the cofinality of $\tau(U) \leq \mathfrak{b}$ is uncountable. (This provides a partial answer to Santi's question in the comments.)

$\bullet$ A characterization of $\tau(U)$ is given in terms of an ideal defined from Ramsey-null sets.

$\bullet$ It is consistent that $\tau(U) < \mathfrak{s}$ for all $U \in \omega^*$.

The first two results are in Section 2, the next in Section 3, and the last in Section 7. The last result answers your question, of course, but I should also mention another relevant paper:

Alan Dow and Saharon Shelah, ``Pseudo P-points and splitting number,'' Archive for Mathematical Logic 58 (2019), pp. 1005-10027. (available here)

In the Brendle-Shelah paper, they prove $\sup_{U \in \omega^*}\tau(U) < \mathfrak{s}$ is consistent, but the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ is only one. In the Dow-Shelah paper, they use a more complicated matrix iteration to make the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ arbitrarily large.

Finally, let me sketch the idea I mentioned above. The idea is to do a countable support iteration that uses Laver forcing at limit steps of cofinality $\omega_1$, and uses Mathias forcing everywhere else. The iteration is of length $\omega_2$ and CH holds in the ground model. Let $V[G]$ denote the result of such an iteration, and let $U \in \omega^*$ in $V[G]$. By reflection, there is some intermediate model $V[G_\alpha]$ with $\alpha < \omega_2$ where $U \cap V[G_\alpha]$ is an ultrafilter in $V[G_\alpha]$, and where $\alpha$ has cofinality $\omega_1$. At this stage, we force with Laver forcing, and this adds a length-$\omega_1$ tower to $U \cap V[G_\alpha]$. All the subsequent Laver forcings and Mathias forcings preserve the fact that this tower has no pseudo-intersection, and so this tower is a witness to the fact that $\tau(U) = \aleph_1$ in the final extension. (See Theorem 7.11 from this paper of Alan's for more detail on the last two sentences.) Finally, the Mathias forcings enable us to get $\mathfrak{s} = \mathfrak{c}$. This is well-known to be true in the Mathias model (although not in the Laver model), and it's true in this model for essentially the same reasons.

  • $\begingroup$ Perfect! And thank you for the references! $\endgroup$ – Todd Eisworth Feb 17 '20 at 22:36
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    $\begingroup$ This can be modified to push up $\mathfrak{h}$ as well by sprinkling in more Mathias forcing on a stationary/co-stationary set of ordinals of cofinality $\omega_1$. Can the Dow-Shelah construction accomplish the same? $\endgroup$ – Todd Eisworth Feb 18 '20 at 3:04

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