Infinite Set of Naturals in Forcing Extension

Just a random doubt that came up while talking to a friend. Assume we have some forcing $$\mathbb P$$ that behaves pretty nicely, so for example we can assume it doesn't collapse $$\omega_1$$, it doesn't collapse $$\mathbb R$$, and if we call $$G$$ a generic for $$\mathbb P$$ then $$\mathcal P(\omega)^V\subseteq \mathcal P(\omega)^{V[G]}$$.

Is this enough to prove the following statement? Given any infinite set $$X\in \mathcal P(\omega)^{V[G]}$$, is there some infinite $$Y\subseteq X$$ such that $$Y\in V$$ (so $$Y\in \mathcal P(\omega)^V$$)? To me, it feels like something that should be pretty obvious, since if every subset that lives in $$V$$ of some set $$X$$ is finite, then it feels like $$X$$ should be finite as well. But neither me nor my friend haven't come up with a convincing proof of why this is true, and probably the answer is "trivially true" or "trivially false" so I am hoping for some enlightenment.

There is a very general observation that for any infinite $$x \subseteq \omega$$, there is an infinite $$x' \subseteq \omega$$, such that any infinite $$y \subseteq x'$$ codes $$x$$ in a very simple way:

Let $$s\colon [\omega]^{<\omega} \to \omega$$ be some bijection. Then let $$x' = \{ s(x \cap n) : n \in \omega \}$$. Whenever $$y \subseteq x'$$ is infinite, then $$\bigcup_{k \in y} s^{-1}(k) = x$$.

So this really has nothing to do with the properties of $$\mathbb{P}$$. As soon as you add a new real, your property will fail.

• great observation! thank you! Commented May 4 at 17:33

This is not true. For a counterexample, consider the "single Cohen" forcing where $$\mathbb{P}$$ is the set of functions from a finite subset of $$\omega$$ to $$\{ 0, 1 \}$$, ordered by reverse inclusion, and let $$\dot S := \{ (p, \check n) \mid n \in \operatorname{dom}(p) \land p(n) = 1 \}$$ be the name for the new subset of $$\omega$$ being added. Then it's straightforward to show $$\dot S^G$$ must be infinite. However, if you have any $$Y \in \mathcal{P}(S) \cap V$$, then in particular there must be some $$p\in G$$ such that $$p \Vdash \check Y \subseteq \dot S$$. But from here, it is straightforward to conclude that for each $$n\in Y$$, $$n\in \operatorname{dom}(p)$$ and $$p(n) = 1$$; since $$\operatorname{dom}(p)$$ is finite, that implies that $$Y$$ is finite.

Also, by standard arguments, $$\mathbb{P}$$ satisfies the countable chain condition, so this forcing doesn't collapse any cardinals.

• That's not random real forcing, that's Cohen forcing. Commented May 4 at 16:46
• Oh, I thought Cohen forcing was restricted to adding enough new subsets to change the cardinality of $P(\omega)$, whereas this is just adding one new subset (and then the ones that follow from combinations with previous sets, of course). Commented May 4 at 16:48
• No. This is the "single Cohen real" forcing. Random real forcing uses positive-measure closed (or Borel) subsets of $\mathcal{P}(\omega)$. Commented May 4 at 16:50
• Thank you Daniel, indeed I hadn’t thought of this easy counterexample. Anyway, now I am curious whether there could be a condition to add about $\mathbb P$ to get the desired property. Surely if it doesn’t change $\mathcal P(\omega)$ then it’s trivially true. But what if we add reals? Cohen forcing doesn’t work for your example, but random? Sacks? Commented May 4 at 17:15
• @alvoi No, as per my more general answer. Commented May 4 at 17:26