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 $V^{{\mathbb S}\_\omega}$? 



Countable support product of Sacks forcings and selective ultrafilters
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.


