# Forcing exercise

This exercise is from Kunen, Set Theory.

Let $M$ be a ctble transitive model of ZFC,$\kappa > \omega$, $\kappa$ regular, $P$ be a notion of forcing that is $\kappa$-closed (i.e. whenever $\gamma < \kappa$ and $\{ p_\eta : \eta < \gamma \}$ is a decreasing sequence of elements of $P$, then $\exists q \in P, \forall \eta < \gamma$, $q \le p_{\eta}$).

Then stationary sets of $\kappa^M$ are preserved under forcing. Could someone help me out with this?

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Any restriction on $\kappa$ or on the stationary sets (i.e. are those stationary sets of $\omega_1^M$? or what?) – Asaf Karagila Jun 8 '12 at 22:08
Yeah, $\kappa > \omega$, $\kappa$ regular. And stationary sets of $(\kappa)^M$. – Kuhndog Jun 8 '12 at 22:13

Let us work in the ground model $M$. Suppose that the claim was not true. This would mean that there are a stationary set $S \subseteq \kappa$, a condition $p \in P$, and a $P$-name $\dot{C}$ such that $p$ forces $\dot{C}$ to be a closed unbounded subset of $\kappa$, but disjoint from $\check{S}$, where $\check{S}$ is the canonical name for $S$.

Now, recall that any true statement in the extension $M[G]$ will be forced by some condition $q \in G$. Thus, for example, the first element of $\dot{C}$ will be decided by some condition $p_0 \leq p$. By deciding, I mean that there is an ordinal $\alpha$ such that $p_0 \Vdash \check{\alpha} = \min{\dot{C}}$.

As $P$ is $\kappa$-closed, we can continue and construct sequences $\langle p_i \mid i < \kappa \rangle$ and $\langle \alpha_i \mid i < \kappa \rangle$ with the properties that $$p_i \Vdash \check{\alpha_i} = \min{(\dot{C} \setminus \{ \check{\alpha_j} \mid j < i \})}.$$

When defining these sequences, the successor step is just as the first step, and at limit stages, say, $i$, we only need to find a condition $p_i$ stronger than the conditions $p_j$ defined before, and this is exactly what the $\kappa$-closedness of $P$ guarantees.

So, in the end we have defined a sequence $\langle \alpha_i \mid i < \kappa \rangle$, which certainly is closed, as $p$ forces $\dot{C}$ to be closed, and also unbounded simply because it is strictly increasing, again something that $p$ quarantees.

Hence there must be some $\alpha_i$ in $S$, and then $p_i \Vdash \check{\alpha_i} \in \check{S} \cap \dot{C}$.

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Thank you, jolubo. – Kuhndog Jun 10 '12 at 20:15