I'm trying to solve exercise (H1) of chapter VII on Kunen's Introduction to Independence Proofs and I would like some hint. I would prefer a hint instead of the full solution :)
Assume in M that $\kappa>\omega$, $\kappa$ is regular and $\mathbb P$ has the $\kappa$-c.c. In $M[G]$, let $C \subset \kappa$ and $C$ c.u.b. Show that there is a $C' \in M$ subse of $C$ c.u.b. in $\kappa$. Hint: in $M[G]$, let $f: \kappa \rightarrow \kappa$ be such that $\forall \alpha < \kappa(\alpha < f(\alpha) \in C)$ and apply lemma 6.8
Lemma 6.8 states that assuming that $\mathbb P \in M$, $A, B \in M$ and in $M$, $\theta$ is a cardinal, $\mathbb P$ has the $\theta$-c.c. and $f \in M[G]$ with $f:A\rightarrow B$, then there is a map $f:A\rightarrow \mathscr P(B)$ with $F \in M$, $\forall a \in A(f(a)\in F(a))$ and $\forall a \in A(|F(a)|<\theta)^M$.
I have noticed that there is already a question like this one, right here If P is k-c.c. and C is club in k in M[G] then C contains a club in M, but in this question, the Kunen's hint was not used. I would like to use it. Any hints?
This is essentially the same idea from Joel's proof: Let $f$ and $F$ be as above. For every $\alpha<\kappa$ define $G(\alpha)=\sup F(\alpha)$. It's easy to see that for every $\alpha$, $\alpha<G(\alpha)$. Now let $G^{\omega}(\alpha)=\lim_{n<\omega}G^{(n)}(\alpha)$. Let $D=Range(G^{\omega})$. $D$ is clearly unbounded, so it's enough to show that $D\subseteq C$. In order to see that, note that for every $\alpha$ and every $n$, $G^n(\alpha)<f(G^n(\alpha))\leq G^{n+1}(\alpha)$ and $f(G^n(\alpha)) \in C$, therefore $G^{\omega}(\alpha) \in C$.