MA, PFA and selective ultrafilters. I have two questions:


*

*Anybody knows where can I find a proof of $\text{MA}\to\exists p\in\mathbb{N}^\ast\text{ selective (Ramsey)}$?

*It is known if $\text{PFA}\to\exists p\in\mathbb{N}^\ast\text{ selective (Ramsey)}$?
Thanks!
 A: For a proof that there is a Ramsey ultrafilter assuming MA, see Theorem 6.11 of these notes. Meanwhile, since PFA implies MA, PFA certainly implies that there is a Ramsey ultrafilter.

The proof that PFA implies MA amounts to showing that c.c.c. forcings are proper; this is a nice argument, so I'll sketch it here. 
EDIT: As Andreas points out below, PFA is often (usually?) stated in the form "If $\mathbb{P}$ is proper and $\mathcal{D}$ is a collection of $\aleph_1$-many dense sets, then there is a $\mathcal{D}$-generic filter." If we adopt this version, then the argument is much harder: we need to show that this form of PFA implies $2^{\aleph_0}=\aleph_2$, which is harder than it sounds. I'm using the seemingly-stronger form of PFA, which says we can meet fewer-than-continuum-many dense sets in a proper forcing and that $2^{\aleph_0}>\aleph_1$, which is how it was taught to me.
Also, I'm using the definition of "properness" in terms of elementary submodels rather than stationarity preserving, since I find it more intuitive.
Recall that $\mathbb{P}$ is proper if for all countable $M\prec H_\theta$ with $\mathcal{P}(\mathbb{P})\in H_\theta$, and all $q\in \mathbb{P}\cap M$, there is some $q'\le q$ such that whenever $r\le q'$ and $D\subseteq \mathbb{P}$ is dense and $D\in M$, there is an $s\in D\cap M$ compatible with $r$. (Such a $q'$ is called $(\mathbb{P}, M)$-generic.)
I claim that the trivial condition $\mathbb{1}$ in a c.c.c. forcing is $(\mathbb{P}, M)$-generic. Why? Well, let $r\in\mathbb{P}$ and $D\subseteq\mathbb{P}$ be dense with $D\in M$. Without loss of generality, $D=\{p: \exists a\in A(p\le a)\}$ for some maximal antichain $A$ with $A\in M$ (why?). 
Since $\mathbb{P}$ is c.c.c., we have that $A$ is countable. But now we use the following neat fact:

If $X\in M\prec H(\theta)$ with $\theta$ infinite and $X$ is countable, then $X\subseteq M$.

Hint for proof: by elementarity $M$ thinks $X$ is countable too, so $M$ thinks $f$ is an injection of $A$ into $\mathbb{N}$ for some $f\in M$...
So $A\subseteq M$. Since $D$ is dense, there is some $s\le r$ with $s\in D$. By definition of $A$, we have $s\le a$ for some $a\in A$, and since $D\supseteq A\subseteq M$ we have $a\in D\cap M$. So we're done.
