Destroying the P-filter-property - MathOverflow

Destroying the P-filter-property

2011-05-31T16:03:26Z

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 $\omega$.

Answer by Goldstern for Destroying the P-filter-property

2011-06-04T12:42:07Z

(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.

Answer by Goldstern for Destroying the P-filter-property

2011-06-11T20:30:07Z

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

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