# How "much" does (Grigorieff) forcing destroy an ultrafilter?

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.

• 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.
• "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?