It is known that if a forcing notion is proper, then every P-filter will generate a P-filter in the generic extension (see, e.g., Shelah, Proper and Improper Forcing, VI.5)

On the other hand, if we start collapsing cardinals, we can destroy the P-filter property. For example, making a base of a P-point countable will add a sequence of elements from the filter (namely, a complete enumeration of that base) that serves as a counterexample for the P-filter property in the extension.

So my question is:

Are there examples of "nicer" (e.g., not collapsing cardinals) forcing notions that destroy P-filters in this way, i.e., add a sequence in the filter that the filter cannot decide?

More spectacularly, is there maybe a forcing notion that could preserve a P-point as an ultrafilter while destroying the P-filter property?

EDIT: As Martin Goldstern pointed out, I should add that I'm interested in filters on $\omega$.

  • $\begingroup$ You mean filters on $\omega$, right? If not, and if you allow me to call a sigma-complete ultrafilter a "p-point", then the answer is easy: Prikry forcing on a measurable cardinal will destroy the p-ness of the measure. Also, namba forcing destroys the p-ness of the club filter on $\omega_2$. (It collapses $\omega_2$, but is nice in the sense that it may preserve $\omega_1$.) $\endgroup$
    – Goldstern
    Jun 4, 2011 at 12:57
  • $\begingroup$ Yes, I mean filters on $\omega$. I'll add that to my question. $\endgroup$ Jun 4, 2011 at 19:10
  • 1
    $\begingroup$ My previous comment suggested that "not collapsing $\omega_1$" is a more suitable niceness property, but that was misguided. If you have a filter generated by a tower of length $\omega_2$, then this will be a p-filter, and Namba forcing will destroy the p-property trivially. (Namba introduces a cofinal $\omega$-sequence to $\omega_2$ without collapsing $\omega_1$) So a better property to demand (to avoid trivial counterexamples) should be "preserves uncountable cofinality". $\endgroup$
    – Goldstern
    Jun 4, 2011 at 22:47

2 Answers 2


(A very partial answer.) Assuming CH, the answer seems to be "no" (for filters on countable sets).

CH implies that every P-filter on $\omega$ is generated by a tower (i.e., an almost decreasing sequence) of length $\omega_1$. In any $\omega_1$-preserving forcing extension any countable sequence of filter sets can be refined to a countable sequence of sets from the tower, so there is a lower bound in the tower.

  • $\begingroup$ Martin, thanks! I hope you don't mind that I'll wait a bit to see if more (partial) answers turn up. $\endgroup$ Jun 4, 2011 at 19:28

A second partial (and not very deep, sorry) answer: Assume that at least one of the following holds:

  1. zero sharp does not exist, i.e., Jensen's covering lemma holds.

  2. The continuum is below $\aleph_\omega$ (or at least: your filter is generated by less than $\aleph_\omega$ many sets).

Then the answer is again "no" (if the forcing is not allowed to collapse cardinals).

Proof: Every new countable subset of the filter base is contained in an old countable subset of the filter base. (This is well known, but for the sake of completeness I give a sketch of the proof. Fix a bijection between the filter base and some ordinal $\alpha$. Every new countable set $A \subseteq \alpha$ is contained in an old set $B$ of size $\aleph_n$, for some $n$. [Under assumption 1, the covering lemma gives $n\le 1$; assumption 2 just says outright that there is some $n$.] Now fix $A$ and choose $n$ as small as possible. Using a bijection from $B$ to $\aleph_n$ in $V$, we may wlog assume that the original set $A$ was a subset of $\omega_n$. If $n>0$, then the countable set $A$ is bounded in $\omega_n$, so we can cover $A$ by an ordinal of cardinality $\aleph_{n-1}$, contradiction.)

  • $\begingroup$ Martin, thanks again. I was really hoping for a positive answer, but hank you for sharing these two partial ones. $\endgroup$ Jun 14, 2011 at 16:57

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