Let $\mathbb{P}$ be a proper notion of forcing, having the Sacks property. Suppose that $\dot{D}$ is a $\mathbb{P}$-name for an infinite subset of $\omega$. I'm looking for a set which approximates $\dot{D}$ both from above and below, that is:

Is there a set $A\subseteq\omega$ (in the ground model) and a $p\in\mathbb{P}$ such that (1) $p\Vdash\dot{D}\subseteq\check{A}$, and (2) for any finitely many elements $a_1,\ldots,a_n\in A$, there is a $q\leq p$ such that $q\Vdash a_1,\ldots,a_n\in\dot{D}$?

It might be useful to note that if $q_n$ is a decreasing sequence in $\mathbb{P}$ such that each $q_n$ decides whether $n\in\dot{D}$, then $A=\{n:q_n\Vdash n\in\dot{D}\}$ satisfies (2), though I don't think it satisfies (1) if $\Vdash\dot{D}\notin\mathbf{V}$. Meanwhile, the Sacks property ensures that there is a set in the ground model satisfying (1), but at least for the set you get by naively applying the Sacks property, it need not satisfy (2).


1 Answer 1


The answer is no. Here is a counterexample: For definiteness, let's work with $\mathbb P$ equal to Sacks forcing, though the proof works verbatim for any reasonable forcing whose generic can be understood as a real. Let $s$ be Sacks generic over $V$ and let $\dot{D}$ be the name for the set in the extension where for each $n \in \omega$ we let $2n \in D$ iff $s(n) = 1$ and $2n + 1 \in D$ iff $s(n) = 0$. So, for instance, if the first four bits of $s$ are, say, 0110 then $1, 2, 4, 7 \in D$ and $0, 3, 5, 6 \notin D$. Note that for each $n$ exactly one of $\{2n, 2n+1\}$ is in $D$. Also clearly we can read off $s$ from $D$ so it's not in the ground model.

Now suppose towards a contradiction that there is an $A \subseteq \omega$ satisfying your conditions for this $\dot{D}$. It can't be the case that there is an $n \in \omega$ so that $2n, 2n+1 \in A$ since that gives us two elements of $A$ for which no $q$ can force both of them to be in $\dot{D}$, contradicting (2) in your question. Also, for no $n$ can it be the case that neither $2n$ nor $2n+1$ is in $A$ since one of them is in $\dot{D}$ and hence in this case $A$ wouldn't cover $\dot{D}$. Therefore exactly one of $2n$, $2n+1$ is in $A$ for each $n$, but this means that $A$ must be equal to $D$, which is a contradiction.

  • $\begingroup$ Thanks Corey! That's a very clear counterexample. $\endgroup$ Commented Dec 5, 2020 at 17:59

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