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It is well-known that in Sacks model there are P-points and even Ramsey ultrafilters, but what the usual (i.e. findable in the literature) proofs for these facts do is proving that ground model P-points (or Ramsey ultrafilters) are preserved by Sacks forcing (i.e. still generate an ultrafilter in the extension). Thus, the only examples of P-points or Ramsey ultrafilters that I know of in Sacks model are those that already existed in the ground model, hence they all have weight $\aleph_1$ (i.e. they are generated by $\aleph_1$ many elements, since the ground model satisfies CH). So my question is:

Is it known whether there are P-points and/or Ramsey ultrafilters of weight $\aleph_2$ in Sacks model? (actually my real question is: if yes, how are they constructed?, I guess)

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I assume that by "the Sacks model" you mean the result of a countable-support iteration of Sacks forcing for $\omega_2$ steps over a ground model satisfying GCH. On that understanding, think the answer to your question is no. If $U$ is a P-point in the final model, then, by a reflection argument, there will be ordinals $\alpha<\omega_2$ such that $U\cap V[G_\alpha]$ is a P-point in $V[G_\alpha]$ (where $G_\alpha$ is the part of the generic ultrafilter that was produced by the first $\alpha$ stages of the iteration). Since $V[G_\alpha]$ satisfies CH, this P-point $U\cap V[G_\alpha]$ is generated by $\aleph_1$ sets. Furthermore, because it's a P-point, it will be preserved through the later stages (from $\alpha$ to $\omega_2$) of the iteration. That means that the $\aleph_1$ generators of $U\cap V[G_\alpha]$ actually generate all of $U$ in the final model.

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  • $\begingroup$ Yes, that's exactly what I meant by "the Sacks model"... and thanks for your answer! Now I'm wondering, since your argument here uses that P-points are preserved, what happens to other kinds of ultrafilters (that are not preserved)? In other words, do you know if Sacks model has any ultrafilters of weight $\aleph_2$ at all? $\endgroup$
    – David Fernandez-Breton
    Commented Jun 3, 2014 at 17:58
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    $\begingroup$ @DavidFernandezBreton Yes, it's a theorem of ZFC that there are always ultrafilters on $\omega$ that can't be generated by fewer than $2^{\aleph_0}$ sets. The proof uses an independent family $F$ of $2^{\aleph_0}$ sets and forms the ultrafilter generated by the sets in $F$ and the complements of all intersections of infinitely many sets from $F$. $\endgroup$
    – Andreas Blass
    Commented Jun 3, 2014 at 18:31

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