Prikry-Silver forcing $\mathbb{V}$ (sometimes just Silver forcing) is the forcing notion consisting of all partial functions $p:\omega\rightarrow 2$ with co-infinite domain. In "Combinatorics on ideals and forcing with trees" Marcia Groszek mentions (without proof) that a Prikry-Silver real has minimal real degree, but not minimal degree.

That the Prikry-Silver real $r$ has minimal real degree means that whenever $s$ is a real in $V[r]$ that doesn't belong to $V$ we have $V[r]=V[s]$. A proof of this can be extracted from some more general results in the seminally named "Combinatorics on ideals and forcing" by Serge Grigorieff.

That $r$ doesn't have minimal degree means that there is some object $A\in V[r]$ for which $V[A]$ is different from both $V$ and $V[r]$. Can anyone point me to a proof of this fact?


I don't know a reference for this, but the $A$ that you are looking for is the collection of all domains of conditions in the Silver generic filter.
Equivalently, you can consider the set of all complements of domains of conditions in the filter. This is a non-principal ultrafilter on $\omega$.

Silver forcing can be considered as an iteration of the following two forcing notions: First you force with $\mathcal P(\omega)/fin$, which gives you a nonprincipal ultrafilter on $\omega$, and then you force with Grigorieff forcing with respect to that ultrafilter. (Grigorieff forcing is like Silver forcing, only that the complements of the domains are in the filter, not just infinite.) The first forcing, $\mathcal P(\omega)/fin$, is $\sigma$-closed and therefore doesn't add any reals at all. The second forcing does all the adding of reals.

The extension is not minimal since you force with an iteration. I am leaving out the details, but it shouldn't be hard to show that the iteration that I am talking about is equvalent to Silver forcing. (Consider instead of $\mathcal P(\omega)/fin$ the equivalent p.o. of infinite subsets of $\omega$. Map each Silver condition $p$ to the pair $(\omega\setminus\mbox{dom}(p),p)$. This should be a dense embedding of Silver forcing into the iteration.)

I don't have access to the Grigorieff paper or to MathSciNet right now, but possibly it says something about this iteration.

  • 1
    $\begingroup$ This is very nice. For some reason I wasn't expecting something this natural. Thanks! $\endgroup$ Apr 16 '12 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.