**Introduction.** 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.


**An example.** Given a free (ultra)filter $F$ on $\omega$, **Grigorieff forcing** is defined as
$$ G(F) := \{ f:X \rightarrow 2: \omega \setminus X \in F \},$$
partially ordered by reverse inclusion. A simple density argument shows that **"$G(F)$ destroys $F$"**, i.e., the filter generated by $F$ in a generic extension is **not** an ultrafilter (the generic real being the culprit).


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).


My question is: 


**If $F$ is destroyed, how far away is $F$ from being the ultrafilter it once was?** 


Maybe a more positive version: **Which properties of $F$ can we destroy while preserving others?** 


This might seem awfully vague, so before you vote to close let me explain what kind of answers I'm hoping for.


* **Positive answers.**
	+ 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.
	+ In Shelah's model without P-points, all ground model Ramsey ultrafilters stop being P-points but "remain" Q-points.
* **"Minimal" answers.** 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?
* **Negative answers.** 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?


I know there is a lot of literature on **preserving ultrafilters** (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.


PS: community wiki, of course.



How "much" does (Grigorieff) forcing destroy an ultrafilter?
Here's a proof that, if $F$ is an ultrafilter and $g$ is $F$-Grigorieff-generic, then $F\cup\{g\}$ does not generate an ultrafilter in the extension. Define a real $x:\omega\to2$ (also viewed as $x\subseteq\omega$ as usual) by letting $x(n)=\sum\_{k=0}^ng(k)$ 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 $F\cup\{g\}$. 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 $a<b$ 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 $b\notin x$).


