Let $U\in V$ be an ultrafilter on $\omega$. We say $U$ is preserved under forcing with $\mathbb{P}$ if $\Vdash \forall x\subset \omega \ \exists Z\in U \ Z\subset x \vee Z\subset x^c$. In other words, $U$ generates an ultrafilter in $V[G]$. Familiar examples of such ultrafilters:

  • Ramsey ultrafilters, P-points are preserved under Sacks forcing and Miller forcing (also their products)
  • P-points/Ramsey ultrafilters are also preserved under iterations of Sacks forcing with countable support
  • There are also combinatorial characterizations of ultrafilters preserved by Sacks forcing, related to Halpern-Lauchli theorem, see https://www.math.wisc.edu/~miller/res/ultra-s.pdf

My question is: is it known that an ultrafilter preserved by Sacks forcing necessarily needs to be preserved by side-by-side products of Sacks forcing (finite/countable support)? There are probably more ad-hoc examples of forcing $\mathbb{P}$ that preserves an ultrafilter $U$ but not the product $\mathbb{P}\times\mathbb{P}$ (is there any)? Maybe there is something easy that I overlook.

  • $\begingroup$ Yes indeed thanks for pointing out. Edited. $\endgroup$
    – Jing Zhang
    Oct 12, 2016 at 0:48
  • $\begingroup$ It makes sense to ask about the countable support product of Sacks forcing. But finite support? The sequence of all first values of the Sacks reals will be a Cohen real. $\endgroup$
    – Goldstern
    Jul 19, 2017 at 16:32
  • 1
    $\begingroup$ @Goldstern: sorry it was bad phrasing I intended to ask finite products or infinite products with countable support $\endgroup$
    – Jing Zhang
    Jul 19, 2017 at 16:35

2 Answers 2


Regarding your final request, here is an example of a forcing notion $\mathbb{P}$ that preserves all ground-model ultrafilters on $\omega$, but $\mathbb{P}\times\mathbb{P}$ destroys all ground model ultrafilters.

Namely, assume CH and let $\mathbb{P}=T$ be a self-specializing Suslin tree, which is a Suslin tree with the property that forcing to add a branch $g$ through the tree makes $T$ a special $\omega_1$-tree off the generic branch; that is, if $b$ is any node not on the generic branch $g$, then the subtree $T_b$ of conditions extending $b$ is special in $V[g]$. In particular, forcing with $T$ once adds no reals, since it is a Suslin tree, and therefore preserves all ultrafilters on $\omega$. But forcing with the tree twice $T\times T$ amounts to forcing with a special Aronszajn tree in the second step and therefore collapses $\omega_1$ and consequently destroys all ground model ultrafilters on $\omega$.

  • $\begingroup$ Thanks for a nice example. I've been looking at mainly proper forcings whose product remains proper (or rather just Sacks forcing). But this example is great. $\endgroup$
    – Jing Zhang
    Oct 12, 2016 at 1:51
  • $\begingroup$ Also any references on how to construct such trees (assuming maybe only CH but I suspect diamond)? Thanks. $\endgroup$
    – Jing Zhang
    Oct 12, 2016 at 2:02
  • $\begingroup$ If you follow the link, Gunter Fuchs has a construction from $\Diamond$. I'm not sure if he ever published the argument. I recall that it was very nice. $\endgroup$ Oct 12, 2016 at 2:05
  • $\begingroup$ Gunter Fuchs has told me that evidently the construction of a self-specializing Suslin tree was known before his rediscovery, so he didn't publish. But I'm not sure of a reference. Paul Larson has mentioned that it was known earlier. $\endgroup$ Oct 13, 2016 at 23:05
  • $\begingroup$ Thanks! I think it might be a good exercise to figure this out (though I don't know how hard it is). But any sketch or pointer would be appreciated ;-) . $\endgroup$
    – Jing Zhang
    Oct 14, 2016 at 21:46

Let $\mathbb M$ be Miller forcing. Then $\mathbb M$ adds an unbounded real, so $\mathbb M\times \mathbb M\times\mathbb M$ adds a Cohen real (Velickovic; also an unpublished result of Shelah). Hence this forcing destroys every ultrafilter from the ground model.

So for every ultrafilter $U$ in the ground model, one of the following is true:

  • $\mathbb M$ destroys $U$.
  • $\mathbb M$ preserves $U$, but $M\times M$ destroys $U$
  • $P:=\mathbb M\times\mathbb M$ preserves $U$, but $P\times P$ destroys $U$.

The first alternative cannot happen for P-points, but I am not sure which of the other two may hold.

  • 1
    $\begingroup$ For P-points, the second alternative holds, since $\mathbb{M}\times \mathbb{M}$ always adds a splitting real. For non-P-points, the first alternative holds. $\endgroup$
    – Jing Zhang
    Jul 19, 2017 at 17:08

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