I'm trying to understand if the following claim is true: Assume that in $M, \mathbb{P}=\mathrm{Fn}(I,2)$ and $|I|> \omega$. Assume that $G$ is $M$-generic. Let $f \in \omega^{\omega} \cap M[G]$.
Then there is $X \in M$ countable, such that there is an $H ⊆ Fn(X,2)$ $M$-generic s.t. $f \in M[H]$. This claim, I believe, shows up in the answer here.
I'm having a hard time to see why this is true. It`ll be nice if someone could explain this to me or give some hints. is it an ordinary ccc argument?
Thanks.
There are many ways to see this. While not the shortest argument, I believe the generality and elegance of the following proof is well worth the additional time it takes to understand it: (It's essentially taken from Kunen's book on Set Theory, p. 266ff.)
Let us fix some transitive model $M \models \mathrm{ZFC}$. We work in/over $M$.
Definition. Let $\mathbb P$ be a forcing and fix $\tau \in M^{\mathbb P}$. A nice name for a subset of $\tau$ is a name of the form $$ \bigcup \{ \sigma \times A_{\sigma}\mid \sigma \in \operatorname{dom}(\tau) \}, $$ where each $A_{\sigma} \subseteq \mathbb P$ is an antichain in $\mathbb P$.
Fact.(Lemma IV.3.10 of Kunen's book) If $\mathbb P$ is a forcing, $\tau, \mu \in M^{\mathbb P}$, then there is a nice name $\theta \in M^{\mathbb P}$ for a subset of $\tau$ such that $$ 1 \Vdash_{\mathbb P}^{M} \mu \subseteq \tau \implies \mu = \theta. $$
Now let's return to your situtation and fix a $\mathrm{Fn}(I,2)$-name $\mu$ such that $\mu^G = f$. Since $f \subseteq (\omega \times \omega)$ we may assume, without loss of generality, that $$ 1 \Vdash_{\mathrm{Fn}(I,2)}^{M} \mu \subseteq \check{(\omega \times \omega)}. $$
By the fact we may thus fix a nice name $\theta \in M^{\mathrm{Fn}(I,2)}$ for a subset of $\check{(\omega \times \omega)}$ such that $$ 1 \Vdash_{\mathrm{Fn}(I,2)}^{M} \theta = \mu. $$ For future reference let us write $$ \theta = \{ \sigma \times A_{\sigma} \mid \sigma \in \operatorname{dom}(\check{(\omega \times \omega)}) \}. $$ Next note that $$ \operatorname{dom}{\check{(\omega \times \omega)}} = \{ \check{(m,n)} \mid m,n \in \mathbb{N} \} $$ is countable. Furthermore, since $\mathrm{Fn}(I,2)$ satisfies the ccc, every antichain appearing in the nice name $\theta$ is countable. Let $$ X := \bigcup \{ \operatorname{dom}(p) \mid \exists \sigma \in \operatorname{dom}(\check{(\omega \times \omega)}) \colon p \in A_{\sigma} \}. $$
$X$ is countable and $\theta$ is a $\mathrm{Fn}(X,2)$-name as can be easily verified.
The rest is now an easy computation:
Exercise. $H := G \cap \mathrm{Fn}(X,2)$ is $\mathrm{Fn}(X,2)$-generic over $M$ and $\theta^H = \theta^G = f \in M[H] \subseteq M[G]$.
For more details and additional hints see Example III.3.64, Definition IV.3.8, Lemma IV.3.10, Lemma IV.4.2 and Lemma IV.4.10 of Kunen's book on Set Theory.