Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

up vote 3 down vote accepted

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 $p_i \Vdash \check{\alpha_i} \in \check{S} \cap \dot{C}$.

share|improve this answer
    
Thank you, jolubo. –  Kuhndog Jun 10 '12 at 20:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.