How "much" does (Grigorieff) forcing destroy an ultrafilter? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:17:45Z http://mathoverflow.net/feeds/question/62981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62981/how-much-does-grigorieff-forcing-destroy-an-ultrafilter How "much" does (Grigorieff) forcing destroy an ultrafilter? Peter Krautzberger 2011-04-25T22:56:44Z 2011-05-31T15:43:54Z <p><strong>Introduction.</strong> I recently revisited Shelah's model without P-points and I was wondering how "badly" Grigorieff forcing destroys ultrafilters, i.e., what kind of properties can survive the destruction of the "ultra"ness.</p> <p><strong>An example.</strong> Given a free (ultra)filter $F$ on $\omega$, <strong>Grigorieff forcing</strong> is defined as <code>$$G(F) := \{ f:X \rightarrow 2: \omega \setminus X \in F \},$$</code> partially ordered by reverse inclusion. A simple density argument shows that <strong>"$G(F)$ destroys $F$"</strong>, i.e., the filter generated by $F$ in a generic extension is <strong>not</strong> an ultrafilter (the generic real being the culprit).</p> <p>Of course, there are many forcing notions that specifically destroy ultrafilters (also, Bartoszynski, Judah and Shelah showed that whenever there's a new real in the extension, some ground model ultrafilter was destroyed).</p> <p>My question is: </p> <p><strong>If $F$ is destroyed, how far away is $F$ from being the ultrafilter it once was?</strong> </p> <p>Maybe a more positive version: <strong>Which properties of $F$ can we destroy while preserving others?</strong> </p> <p>This might seem awfully vague, so before you vote to close let me explain what kind of answers I'm hoping for.</p> <ul> <li><strong>Positive answers.</strong> <ul> <li>If the forcing is $\omega^\omega$-bounding and $F$ is rapid, then $F$ will still be rapid. That's a very clean and simple preservation. </li> <li>In Shelah's model without P-points, all ground model Ramsey ultrafilters stop being P-points but "remain" Q-points.</li> </ul></li> <li><strong>"Minimal" answers.</strong> Is it possible that $F$ together with the generic real generates an ultrafilter, i.e., there are only two ultrafilters extending $F$? For Grigorieff forcing, I'd expect this needs at least a Ramsey ultrafilter. But maybe other forcings have this property?</li> <li><strong>Negative answers.</strong> Say $F$ is a P-point; can $F$ still be extended to a P-point? Shelah tells us that forcing with the full product $G(F)^\omega$ denies this. Is it known whether $G(F)$ already denies this? Do other forcing notions allow this?</li> </ul> <p>I know there is a lot of literature on <strong>preserving ultrafilters</strong> (mostly P-points, I think) but I'm more interested in the case where the ultrafilter is actually destroyed. But I'd welcome anything that sheds light on this.</p> <p>PS: community wiki, of course.</p> http://mathoverflow.net/questions/62981/how-much-does-grigorieff-forcing-destroy-an-ultrafilter/64287#64287 Answer by Andreas Blass for How "much" does (Grigorieff) forcing destroy an ultrafilter? Andreas Blass 2011-05-08T11:51:23Z 2011-05-08T20:00:25Z <p>Here's a proof that, if $F$ is an ultrafilter and $g$ is $F$-Grigorieff-generic, then <code>$F\cup\{g\}$</code> does not generate an ultrafilter in the extension. Define a real $x:\omega\to2$ (also viewed as $x\subseteq\omega$ as usual) by letting <code>$x(n)=\sum_{k=0}^ng(k)$</code> modulo 2. (Technically, I should fix the obvious names for $g$ and $x$, but for simplicity let me omit the resulting dots over the letters.) Suppose, toward a contradiction, that $x$ or its complement is in the filter generated by <code>$F\cup\{g\}$</code>. Then there is a condition $p$ and there is a set $B\in F$ such that either (1) $p$ forces $B\cap g\subseteq x$ or (2) $p$ forces $B\cap g\subseteq\omega-x$. Fix two numbers <code>$a&lt;b$</code> such that neither of them is in the domain of $p$ and such that $b\in B$. (This can be done because $B$ is in $F$ while the domain of $p$ isn't.) Now form two extensions $q$ and $q'$ of $p$ as follows. Both of them have the value 1 at $b$ (so they force $b\in g$ and therefore $b\in B\cap g$); they are both defined at $a$ but take opposite values there; and they are both defined and equal at all other numbers smaller than $b$. Then one of them forces $b\in x$ and the other forces $b\notin x$. This is absurd, as both extend $p$, which already decided between (1) (which will require $b\in x$) and (2) (which will require $b\notin x$).</p> http://mathoverflow.net/questions/62981/how-much-does-grigorieff-forcing-destroy-an-ultrafilter/66564#66564 Answer by Peter Krautzberger for How "much" does (Grigorieff) forcing destroy an ultrafilter? Peter Krautzberger 2011-05-31T15:43:54Z 2011-05-31T15:43:54Z <p>I wanted to add two comments that I received in 'meatspace'. I hope this isn't too inappropriate.</p> <ul> <li>If $F$ is a P-filter, and the forcing is proper, than $F$ generates a P-filter in the extension. </li> <li>If $F$ is a Q-filter, i.e., every finite-to-one map becomes injective on a set in $F$, and the forcing is $\omega^\omega$-bounding, then $F$ generates a Q-filter in the extension.</li> </ul> <p>Proofs of these facts can be found, e.g., in <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, Chapter VI, Section 4 and 5 resp.</a></p> <p>One more example from myself.</p> <ul> <li>If $F$ is an <a href="http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000006649/Krautzberger_Idempotent_Filters_and_Ultrafilters.pdf;jsessionid=34AA50646DF451390677F49D912E1212?hosts=" rel="nofollow">idempotent filter</a> then $F$ will remain an idempotent filter in any forcing extension. In particular, if $F$ is an idempotent ultrafilter, it will still extend to an idempotent ultrafilter.</li> </ul>