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Ralf Schindler
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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 thatthis answer in thisthat paper (cf. Theorem 6 of thisthat paper which states a weaker result). Is it true that if $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{{\mathbb S}_\omega}$?

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 that answer in this paper (cf. Theorem 6 of this paper which states a weaker result). Is it true that if $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{{\mathbb S}_\omega}$?

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

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Ralf Schindler
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  • 17

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 that answer in this paper (cf. Theorem 6 of this paper which states a weaker result). Is it true that if $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{{\mathbb S}_\omega}$?