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 $V^{{\mathbb S}_\omega}$?
2 Answers
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 Ppoint. Then I think we need to get to the nittygritty of Laver's proof.

1$\begingroup$ Okay, if $\tau$ is a name for a subset of $\omega$, then there are densely many $p = (T_i : i<\omega) \in {\mathbb S}_\omega$ which come with some $(l_n : n<\omega)$ s.t. for each $n$, if $i<n$, then $T_i$ has $2^n$ nodes of length $l_n$, and if $i \geq n$, then $T_i$ has one node of length $l_n$, and if $(s_i : i<\omega)$ is a sequence s.t. $s_i$ is a node in $T_i$ of length $l_n$, each $i$, then the condition $((T_i)_{s_i} : i<\omega)$ decides ``$n \in \tau$.'' $\endgroup$ Sep 4, 2018 at 8:54

1$\begingroup$ Fix such a $p$, $(l_n : n<\omega)$. You say (if I understand you correctly) that if $A \in U$, then there is an infinite $A' \subset A$ and some $q = (T'_i : i<\omega) \leq p$ s.t. if $n \in A'$ and $(s_i : i<\omega)$ is a sequence s.t. $s_i$ is a node in $T'_i$ of length $l_n$, each $i$, then the condition $((T_i)_{s_i} : i<\omega)$ decides ``$n \in \tau$'' either always positively or always negatively, independently from the choice of $n$, $(s_i : i<\omega)$. How do you prove that? $\endgroup$ Sep 4, 2018 at 8:54

2$\begingroup$ Restricting to levels in $A$, namely let $T^*_i=T_i\restriction \{l_n: n\in A\}$. Color the nodes according how it decides $n\in \tau$. Apply Laver's theorem to $\Pi_{i<\omega} T_i^*$ with the coloring. $\endgroup$ Sep 4, 2018 at 11:07

2$\begingroup$ Yeah I was once convinced that the same statement holds for Ppoints as well but now I find no other way than getting into the Laver proof (even so I don't see things immediately work). On another note, I believe the ultrafilter generated in the forcing extension is still selective: any decreasing sequence of elements in $U$ (the sequence may not be in $V$ but each individual element can be WLOG assumed to be in $V$), then we use the Sacks property of the forcing plus fusion to determine each element with possibily finite error. $\endgroup$ Sep 4, 2018 at 11:40

2$\begingroup$ I think the characterization in terms of selectivity is key: for any $\langle A_n\in U : n\in \omega\rangle$, there exists $A\in U$ such that $A\backslash n \subset A_n$ for all $n\in \omega$. This is a finer pseudointersection which guarantees Ramseyness. $\endgroup$ Sep 4, 2018 at 14:44
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.

$\begingroup$ I agree with what K. P. says concerning "$A \in U$," but I don't see the answer to this more specific question either, which is one of the reasons why I asked my original question. $\endgroup$ Sep 3, 2018 at 13:04