# Approximating a real in the ground model

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).

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.