Forcing problem. Find a subset of a set in the extension which is uncountable according to ground model.

I want to apply forcing method in the following context. I do not want any solution, just a tiny hint how to begin.

I have a countable transitive model for ZFC which I call $$M$$. Also I know that for some set $$A$$ and the so called forcing poset $$\mathbb{P}$$ are elements of $$\mathcal{M}$$. Let $$G$$ be an $$\mathcal{M},\mathbb{P}$$-generic filter over $$\mathcal{M}$$.

According to the model $$\mathcal{M}$$, $$\mathbb{P}$$ is countable but $$A$$ has $$\aleph_1$$- many elements. According to the extended model $$\mathcal{M}[G]$$, there is some subset of $$A$$ which is uncountable, B.

What I want is to find a subset of $$B$$, $$C$$ so that $$\mathcal{M}[G]$$ thinks that

1. $$C$$ is a subset of $$B$$
2. $$C$$ is uncountable.

My attempt in finding such set $$C$$ sadly does not get me anywhere. I only have rough ideas. The main problem is: What do people (who know what they are doing) do to start? Perhaps begin with finding a fitting name for $$A$$ and $$B$$? Do I try to find dense sets which $$G$$ should intersect?

In the case of 2.: Maybe I can use that $$\mathbb{P}$$ has the countable chain condition? I believe that in this case the cofinalities and cardinalities are preserved by our forcing poset $$\mathbb{P}$$. But how to continue then?

This is my first ever forcing try.

• $\mathcal M$ has a hard time thinking that $C$ is a subset of $B$, because $\mathcal M$ doesn't see $B$. I think what you wanted (and what Noah's answer achieved) is that $\mathcal M[G]$ thinks that $C\subseteq B$. Commented Nov 10, 2022 at 18:37
• @AndreasBlass Oh hey, I misread that! Commented Nov 10, 2022 at 18:59

For simplicity, let's just assume $$A=\omega_1$$. We have a name $$\nu$$ (with $$\nu[G]=B$$) such that $$\mathcal{M}[G]\models\vert\nu[G]\vert=\aleph_1$$ and in the ground model $$\mathcal{M}$$ we're trying to "predict" infinitely many countable ordinals that wind up in $$\nu[G]$$.
My first hint is to think dynamically: a name $$\nu$$ functions as a "bucket," and conditions in $$\mathbb{P}$$ "put things into" the bucket. More concretely, one thing we can do in $$\mathcal{M}$$ is tell - for each specific $$p\in\mathbb{P}$$ - what elements $$p$$ puts into the bucket $$\nu$$: $$X_p=\{\alpha\in\omega_1: p\Vdash\check\alpha\in\nu\}\in\mathcal{M}.$$ My second hint is really a refinement on this first hint: note that $$G$$ is countable in $$\mathcal{M}[G]$$ (since it's countable in $$\mathcal{M}$$) and $$\nu[G]=\bigcup_{p\in G}X_p.$$ Now, do you see a cute way to apply the pigeonhole principle here?
If $$\nu[G]$$ is uncountable in $$\mathcal{M}[G]$$, then some $$X_p$$ (with $$p\in G$$) must be uncountable in $$\mathcal{M}[G]$$. But such an $$X_p$$ is a subset of $$\nu[G]$$ which is in $$\mathcal{M}$$ ...
Note that we're crucially using not just c.c.c.-ness of $$\mathbb{P}$$, but countability of $$\mathbb{P}$$ (in the ground model that is). It's a good follow-up exercise to show that this is necessary:
Think about adding a Cohen-generic subset of $$\omega_1$$ (that is, forcing with finite partial functions $$\omega_1\rightarrow 2$$) ...