If $X \in V[G]$, then $X \cap V \in V[G]$?

Question

I hope this is not a silly question, but I can't think of either a proof or counterexample of this seemingly simple problem.

Let $V[G]$ be a generic extension of $V$. Is it true that for all $X \in V[G]$, we have that $X \cap V \in V[G]$ as well?

Answer 1

Yes. Because $X$ is a set, when we write $X\cap V$, what we really mean is $X\cap V_\alpha$, for a sufficiently large $\alpha$.

Since $V$ is a subclass of $V[G]$, it means that $V_\alpha\in V[G]$, so the intersection is there as well.

Note that we didn't even need the genericity of $G$ here.

Answer 2

This is true. In general, we can introduce a predicate on members of $V[G]$ which indicates whether they are in $V$, and this predicate satisfies all the usual forcing properties. For example, it's definable in $V$ whether a name for an element of $V[G]$ is in $V$. In the partial order formulation, $p \Vdash \sigma \in \dot V$ if for every $q\le p$ there exists $r \le q$ and $x$ (in $V$) such that $r \Vdash (\sigma = \check x)$. Similarly, in the boolean algebra valued model formulation, $\lVert \sigma \in \dot V \rVert = \sum_x \lVert \sigma = \check x \rVert = \sup \{ p \in B \mid \exists x, p = \lVert \sigma = \check x \rVert \}$. (Here, ${\cdot} \in \dot V$ is syntactic sugar for a predicate (or unary relation) symbol being introduced into the first-order forcing language for $V[G]$.)

Therefore, repeating the usual proof of the selection axiom, we can also define a name for $X\cap V = \{ x\in X \mid x\in V \}$.