Mathias forcing with Ramsey ultrafilters, and Cohen reals [closed]

Edit/update: One reason this question never received an answer is because it was founded on a faulty premise! The Blaszczyk-Shelah paper I mentioned below does not prove that Mathias forcing with a nowhere-dense ultrafilter does not add Cohen reals. That statement is false. (I am not quite sure where I got the misconception). They do construct a sigma-centered forcing which doesn't add Cohen reals using a nowhere dense ultrafilter and topological methods, but the forcing is something else.

In fact, one can show that Mathias forcing $\mathbb{M} _ {\mathcal{U}}$ relative to an ultrafilter $\mathcal{U}$ adds Cohen reals exactly when $\mathcal{U}$ is not selective. For selective ultrafilters $\mathbb{M} _ {\mathcal{U}}$ has the Laver property, which implies it doesn't add Cohen reals. And if $\mathcal{U}$ isn't selective, that means one can partition $\omega=\bigcup_{k<\omega}A_k$ where each $A_k\not\in\mathcal{U}$, and so that no $A\in\mathcal{U}$ has the cardinality of its intersections with the $A_k$ bounded by some $M<\omega$ independent of $k$. Let $g$ be the generic real. Let $x\in V[g]\cap\omega^\omega$ be the real defined by having $x(n)=k$ where $g(n)\in A_k$. Then, let $y(i)$ equal the number of times the $i$th distinct digit of $x$ repeats. It is not hard to see that $y\in V[g]\cap\omega^\omega$ is a Cohen real.

I'll leave the question as I wrote it below, since I'm not bashful about being mistaken. Although it is probably appropriate to close this as 'not a real question'.

Given an ultrafilter $\mathcal{U}$ on $\omega$ the corresponding Mathias forcing $\mathbb{M}_\mathcal{U}$ is the forcing consisting of conditions $\langle s,A\rangle$ where $s$ is a finite subset of $\omega$ and $A\in\mathcal{U}$. The ordering is given by $\langle t,B\rangle\leq\langle s,A\rangle$ if $B\subseteq A$ and $t$ is an end extension of $s$, with $t\setminus s\subseteq A$. (So $\mathbb{M} _\mathcal{U}$ is just like Prikry forcing, but the relevant objects are defined on $\omega$.)

An ultrafilter $\mathcal{U}$ is nowhere dense if whenever $F:\omega\rightarrow\mathbb{R}$ there is some $A\in\mathcal{U}$ whose image under $A$ is nowhere dense. Blaszczyk and Shelah have shown that when $\mathcal{U}$ is nowhere dense, then $\mathbb{M} _\mathcal{U}$ does not add a Cohen real. In fact, they proved that the existence of a $\sigma$-centered forcing adding no Cohen real is equivalent to the existence of a nowhere dense ultrafilter.

Their proof that $\mathbb{M} _\mathcal{U}$ adds no Cohen real mostly uses topological methods rather than forcing theoretic ones, and I find it a bit opaque. I was wondering if a simpler proof was known if we add some stronger hypotheses on the ultrafilter. My question is:

Is there a (relatively) simple proof that when $\mathcal{U}$ is a Ramsey ultrafilter $\mathbb{M}_\mathcal{U}$ adds no Cohen real? What about for p-points?

closed as not a real question by Justin Palumbo, François G. Dorais♦May 28 '12 at 2:44

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

• vanilla Mathias forcing (without an ultrafilter) doesn't add Cohen reals, but Mathias forcing with an ultrafilter very well might; in Shelah's model with no nowhere dense ultrafilter every Mathias forcing with an ultrafilter and indeed every sigma-centered forcing will add a Cohen real – Justin Palumbo May 14 '12 at 19:23
• i didn't think about the Laver property, that's a very good suggestion – Justin Palumbo May 14 '12 at 19:24
• Ramiro, your suggestion seems to be right; I think the same diagonal arguments that show vanilla Mathias forcing has the Laver property shows that Mathias forcing relative to a Ramsey ultrafilter does.. if you wanted to add your suggestion as an answer I would certainly upvote it... – Justin Palumbo May 14 '12 at 20:04
• in the meantime I've 'strengthened' the question by also asking about p-points, where it isn't clear that the forcing has the Laver property (and I would guess, perhaps, it does not) – Justin Palumbo May 14 '12 at 20:04
• (Closed per author's request.) – François G. Dorais May 28 '12 at 2:45