Countable support product of Sacks forcings and selective ultrafilters If $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{\mathbb S}$, where ${\mathbb S}$ is Sacks forcing. The same is true with ${\mathbb S}$ being replaced by ${\mathbb S}_n$, the product of $n$ copies of Sacks forcing, $n<\omega$ (Halpern and Pincus, 1981), and I can see a proof of that. How about ${\mathbb S}_\omega$, the full (= ctble.) support product of $\omega$ copies of Sacks forcing? People refer to R. Laver, "Products of infinitely many perfect trees," 1984, for a positive answer, but I don't find this answer in that paper (cf. Theorem 6 of that paper which states a weaker result). Is it true that if $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{{\mathbb S}_\omega}$?  
 A: The collection of possible large sets is analytic, namely: $\{A\subset \omega: \forall i<\omega\ \exists U_i\subset T_i \text{ $U_i$ is perfect and } f\restriction \bigcup_{n\in A}\Pi_{i<\omega} U_i(n) \text{ is constant}\}$ (here we can assume the length of the roots of $T_i$ goes to infinity so the coloring $f$ is coded by a real). The finish by a theorem of Mathias: if $P\subset [\omega]^\omega$ is analytic, and $U$ is Ramsey, then there exists $A\in U$ such that $[A]^\omega\subset P$ or $[A]^\omega\cap P =\emptyset$. The first option must appear by density of the set under $\subseteq$. 
I saw  this argument first from Olga Yiparaki's thesis: On some tree partitions.
To trigger another problem, it may be interesting to think about what happens if the ultrafilter is merely a P-point. Then I think we need to get to the nitty-gritty of Laver's proof.
A: I'm not allowed to comment yet but your question looks equivalent to having the set $A$ in $\mathrm{HL}_\omega$ (see Laver's paper) belong to the ultrafilter. I do not recall how flexible Laver's proof is for this to be possible. I would look at the proof of "selective is Ramsey" for inspiration.
