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

Question

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.

Answer 1

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 $\omega_1\rightarrow 2$) ...