Destroying the P-filter-property - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:15:05Z http://mathoverflow.net/feeds/question/66567 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66567/destroying-the-p-filter-property Destroying the P-filter-property Peter Krautzberger 2011-05-31T16:03:26Z 2011-06-11T21:47:18Z <p>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., <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.pl/1235419814" rel="nofollow">Shelah, Proper and Improper Forcing, VI.5</a>)</p> <p>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.</p> <p>So my question is:</p> <blockquote> <p>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? </p> <p>More spectacularly, is there maybe a forcing notion that could preserve a P-point as an ultrafilter while destroying the P-filter property?</p> </blockquote> <p>EDIT: As Martin Goldstern pointed out, I should add that I'm interested in filters on $\omega$.</p> http://mathoverflow.net/questions/66567/destroying-the-p-filter-property/66885#66885 Answer by Goldstern for Destroying the P-filter-property Goldstern 2011-06-04T12:42:07Z 2011-06-04T12:42:07Z <p>(A very partial answer.) Assuming CH, the answer seems to be "no" (for filters on countable sets).</p> <p>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. </p> http://mathoverflow.net/questions/66567/destroying-the-p-filter-property/67539#67539 Answer by Goldstern for Destroying the P-filter-property Goldstern 2011-06-11T20:30:07Z 2011-06-11T21:47:18Z <p>A second partial (and not very deep, sorry) answer: Assume that at least one of the following holds: </p> <ol> <li><p>zero sharp does not exist, i.e., Jensen's covering lemma holds. </p></li> <li><p>The continuum is below $\aleph_\omega$ (or at least: your filter is generated by less than $\aleph_\omega$ many sets). </p></li> </ol> <p>Then the answer is again "no" (if the forcing is not allowed to collapse cardinals). </p> <p>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.)</p>