We all know and love Cohen reals, and we can (and often do) define the Cohen forcing as partial functions $p\colon\omega\to 2$ with finite domain. The Prikry–Silver forcing is defined as partial functions $p\colon\omega\to 2$ with co-infinite domain.

These two couldn't be any more different. For example, Cohen reals are aggressively non-minimal, whereas Prikry–Silver reals are minimal.

We can look at a similar situation with other forcings that are given by finite conditions. For example $\operatorname{Col}(\omega,\omega_1)$ is a forcing notion whose conditions are finite partial functions $p\colon\omega\to\omega_1$. We can ask what would be the Prikry–Silver analogue of this forcing, then. That is, $\{p\colon\omega\to\omega_1\mid\operatorname{dom} p\text{ is co-infinite}\}$.

Interestingly, assuming CH this is the same as the standard collapsing forcing. This follows from the fact that the cardinality of the partial order is $2^{\aleph_0}$, which under CH is just $\aleph_1$, and we know that any forcing of size $\aleph_1$ which collapses $\omega_1$ is equivalent to $\operatorname{Col}(\omega,\omega_1)$.

Question. Is this so-called "Prikry–Silver collapse" provably equivalent to the standard collapsing forcing?

  • $\begingroup$ Prikry–Silver adds a minimal real, it is proper, and so it most certainly does not collapse $\omega_1$. $\endgroup$
    – Asaf Karagila
    May 20, 2021 at 14:56

1 Answer 1


If $\mathrm{CH}$ fails then $\mathrm{Col}(\omega, \omega_1)$ does not add a generic for the "Prikry-Silver collapse" $\mathbb P$: Let $\mathbb U$ be $(\mathcal{P}(\omega)/I)^+$ where $I$ is the ideal of finite subsets of $\omega$. The map $$\pi:\mathbb P\rightarrow \mathbb U,\ p\mapsto [\omega\setminus\mathrm{dom}(p)]_I$$ is a projection so that $\mathbb P$ adds a $\mathbb U$-generic filter. It is hence enough to show that under $\neg\mathrm{CH}$, $\mathrm{Col}(\omega,\omega_1)$ does not add a $\mathbb U$-generic filter. Suppose it does. In this case, there is a projection $$\mu:\mathrm{Col}(\omega,\omega_1)\rightarrow \mathrm{RO}(\mathbb U)$$ but this implies that there is $p\in\mathbb U$ and a set $D\subseteq\mathrm{RO}(\mathbb U)$ of size $\omega_1$ that is dense below $p$. This contradicts the fact that $\mathbb U$ has an antichain of size continuum below every condition.


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