A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is a p-point (or weakly selective) iff for every partition $\omega = \bigsqcup _{n < \omega} Z_n$ into null sets, i.e each $Z_n \not \in \mathcal{U}$, there exists a measure one set $S \in \mathcal{U}$ such that $S \cap Z_n$ is finite for each $n$.

A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is Ramsey (or selective) iff for every partition as above, there exists a measure one set $S$ such that $|S \cap Z_n| = 1$ for each $n$.

Clearly, every Ramsey ultrafilter is a p-point. What is known about the converse?

I couldn't find anything, not even a consistency result, in any searches I've done or sources I've checked. Is very little known/published about the converse?

up vote 16 down vote accepted


The converse is not true, not even under MA. This is a result of Kunen, and the paper you want to look at is "Some points in $\beta{\mathbb N}$", Math. Proc. Cambridge Philos. Soc. 80 (1976), no. 3, 385–398.

There is a related notion, called $q$-point. These are ultrafilters such that any finite-to-one $f:\omega\to\omega$ is injective on a set in the ultrafilter. A Ramsey ultrafilter is one that is simultaneously a $p$-point, and a $q$-point.

Miller proved ("There are no $Q$-points in Laver's model for the Borel conjecture", Proc. Amer. Math. Soc. 78 (1980), no. 1, 103–106) that it is consistent that there are no $q$-points. The consistency of the non-existence of $p$-points is significantly harder, and due to Shelah (see for example Chapter VI of his "Proper and improper forcing").

There is a fairly extensive literature on related results. You may want to start by looking at Blass' article in the Handbook of Set Theory, "Combinatorial Cardinal Characteristics of the Continuum".

Another small addendum to Andres's and Andreas's answers.

It is also consistent that the answer to your question is yes.

Shelah has constructed a model of ZFC in which there exists (up to isomorphism) exactly one p-point -- and that p-point is, in fact, selective. This construction is Section XVIII.4 in Shelah, Proper and Improper Forcing .

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    Oh, yes. If there is a unique $p$-point, it follows that it must be Ramsey. By the way, the consistency of this uniqueness result is significantly more elaborate than the consistency of "there is a unique Ramsey ultrafilter" (also due to Shelah). – Andrés E. Caicedo Apr 20 '11 at 1:53
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    Andres, is there an easy argument why a unique p-point must be selective? Also, it could be interesting to the OP to point out that you can pick any Ramsey ultrafilter in a CH model and have it survive to become the unique one in an extension. But then again, selective ultrafilters all look the same (in the sense of what Andreas calls 'complete combinatorics'). I would agree that the model is more involved though I have heard good arguments that this is mostly due to the presentation. – Peter Krautzberger Apr 20 '11 at 3:59
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    Peter, isn't it the case that any non-principal ultrafilter Rudin-Keisler below a p-point must itself be a p-point? In which case, if you have only one p-point, it must be minimal under the Rudin-Keisler ordering, and hence Ramsey. – tci Apr 20 '11 at 13:36
  • Doh! Thanks Tanmay! I shouldn't ask questions after midnight... – Peter Krautzberger Apr 20 '11 at 18:20
  • I just looked at Shelah's construction and I think it's worth pointing out that Shelah really needs (or at least thinks he needs) a Ramsey ultrafilter to start with, i.e., the preservation lemma requires the "surviving" P-point to be selective. – Peter Krautzberger Apr 26 '11 at 21:38

A few addenda to Andres Caicedo's answer: It was proved around 1970 by several people (Adrian Mathias was one of them) that the continuum hypothesis (CH) implies the existence of P-points that are not selective. (CH also implies the existence of selective ultrafilters and the existence of Q-points that are not selective. ZFC alone suffices to prove the existence of ultrafilters that are neither P-points nor Q-points.) The more difficult task of producing a model of set theory in which P-points exist but selective ultrafilters don't was achieved in Kunen's paper cited by Andres. It is a famous open problem whether there are models of set theory in which neither P-points nor Q-points exist; it is known that in such a model the cardinal of the continuum must be at least $\aleph_3$. (In contrast, there are models with $2^{\aleph_0}=\aleph_2$ with no P-points and others with no Q-points.)

  • Is it difficult to obtain in ZFC a non-principal ultrafilter which is neither P- nor Q-point? Where can I read about this construction? Thanks in advance. – Damian Sobota Oct 31 '16 at 21:03
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    @DamianSobota No, given any nonprincipal ultrafilter U on the set N of natural numbers, you get an ultrafilter on N^2 which is neither a P-point nor a Q-point by taking the set of those subsets X of N^2 for which almost all (w.r.t. U) vertical sections are big (w.r.t. U), i.e., $\{X\subseteq N^2:\{x:\{y:(x,y)\in X\}\in U\}\in U\}$. – Andreas Blass Oct 31 '16 at 22:57
  • Thank you, Andreas. 1) I can see why it is a non-principal ultrafilter which is not a P-point, but I can't see immediately why it is not a Q-point. Can you give me some further hint? 2) How to construct under CH a Q-point which is not a P-point? (And vice versa -- a P-point which is not a Q-point). Thanks! – Damian Sobota Nov 2 '16 at 16:35
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    @DamianSobota 1. The projection to the second factor, $(x,y)\mapsto y$ is finite-to-one on the set $\{(x,y):x<y\}$, which is in the ultrafilter I suggested, but this projection is not one-to-one on any set in that ultrafilter. 2. If $V_n$ for $n\in N$ and U are non-isomorphic selective ultrafilters on N, then $\{X\subseteq N^2:\{x:\{y:(x,y)\in X\}\in V_x\}\in U\}$ is a Q-point but not a P-point. For a P-point that is not a Q-point, under CH, build the filter inductively, using F_\sigma filters at each stage of the induction. Proofs are way too long to put here (or even into an answer). – Andreas Blass Nov 2 '16 at 17:19
  • Andreas, many, many thanks! – Damian Sobota Nov 2 '16 at 18:27

Another small and slightly trivial addendum:
If there are no p-points, then every p-point is a Ramsey ultrafilter. (Duh!) As Andreas Blass remarked above, this situation is consistent, which is easier to prove than the consistency of a unique p-point. ("It is usually significantly harder to prove there is a unique object than to prove there is none". See Shelah's Proper and improper forcing VI.5)

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