# Elementary substructure proof that kappa-cc forcing preserves stationary sets

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?

• You are right, you want to show that if $G$ is $\mathbb P$-generic over $M$ then $\delta$ is a limit point of $\dot C^G$ and you are already almost there! So for every $\alpha\in X\cap\kappa$ you want to show that there is some $\alpha\leq\beta\in\dot C^G$ with $\beta\in X\cap\kappa$. Note that you can take $\beta=\dot f^G(\alpha)$... Mar 13 at 15:04

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.