## 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?

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I believe Mathias forcing never adds Cohen reals (regardless of the kind of ultrafilter) the reason being that it has Laver´s property. When the ultrafilter is Ramsey (I remember reading somewhere that), Mathias forcing is equivalent to Laver forcing (restricted to the ultrafilter), so I would guess that for this case it might be a bit easier to show it has Laver´s property. – Ramiro de la Vega May 14 2012 at 19:06
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 2012 at 19:23
i didn't think about the Laver property, that's a very good suggestion – Justin Palumbo May 14 2012 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 2012 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 2012 at 20:04