Suppose $\kappa$ is a regular uncountable cardinal in the ground model $M$, and let $\mathbb P\in M$ be a forcing notion that has the $\kappa$ chain condition in $M$. Further suppose that $S\subseteq \kappa$ is stationary (in $M$). I would like to show that $S$ remains stationary in $M[G]$.
Toward that end, suppose $C\in M[G]$ is club and fix $f:\kappa\to\kappa$ to be its (continuous) increasing enumeration. Also fix names $\dot C,\dot f\in M$ and a condition $p\in G$ such that
$$p\Vdash \text{$\dot C$ is club in $\check \kappa$ and $\dot f$ is an increasing enumeration of $\dot C$}.$$
For $\alpha<\kappa$, let $S_\alpha = \{\beta<\kappa \mid \exists q\leq p, \,q\Vdash \dot f(\check \alpha)=\check \beta\}$. Informally, $S_\alpha$ is the set of "possible values" of $f(\alpha)$ in a generic extension. The $\kappa$-cc tells us that $|S_\alpha|<\kappa$ for all $\alpha$ (any two values in $S_\alpha$ have corresponding incompatible conditions that realize $f(\alpha)$ taking that value), and thus each $S_\alpha$ is bounded in $\kappa$ by regularity. It's also not too hard to show that the sequence of $\min(S_\alpha)$ is strictly increasing.
So now, working in $M$, let $\theta$ be sufficiently large that $H(\theta)$ has all of the information about $\mathbb P,\dot C,\kappa$ that it needs. Let $X\prec H(\theta)$ such that $|X|<\kappa$ and $\mathbb P,\dot C,\kappa,S\in X$. Through an elementary chain argument we can further require that $X\cap \kappa$ is transitive and thus some ordinal $\delta$, and moreover that $\delta\in S$ (this takes some work, but I understand how to do it).
To finish the proof, it would suffice to show that $p\Vdash \delta \in \dot C$. I have in my notes that $\delta$ is a closure point of the map $\alpha\mapsto \sup(S_\alpha)$, which I believe follows from elementarity, but I don't see how to finish off the proof from there (although seemingly I did understand it in the moment because I didn't write anything else). I want to say it has something to do with this implying that $\delta$ is a limit point of $C$, but I don't quite see how to achieve it. Could I get some help?
Best Answer
Since you know that $\delta$ is a closure point of $\alpha\mapsto \sup(S_\alpha)$, you may notice that $\delta$ is also a closure point of $f$. Indeed for all $\alpha<\delta$, since $f(\alpha)\in S_\alpha$, we have $f(\alpha)\le\sup(S_\alpha)<\delta$. So since $f$ is increasing continuous, $f(\delta)\le\delta$.
However $f$ is strictly increasing, so we must have $f(\delta)\ge\delta$, which means that $f(\delta)=\delta$.
Now for all $\alpha<\kappa$ we have $f(\alpha)\in C$, so this means $\delta=f(\delta)\in C$ in particular.