Stationary sets

Question

Currently learning about stationary sets, came across this problem:

Let $\{S_\alpha : \alpha<\omega_1\}$ be disjoint pairwise sets with $S_\alpha \subseteq \omega_1$ non-stationary for each $\alpha$. Prove that there exists $A\subseteq \omega_1$ such that $\mid A \mid = \aleph_1$ and $\cup_{\alpha \in A} S_\alpha$ is also not stationary.

So we know that for each $S_\alpha$ there is a closed-unbounded set which has empty intersection with $S_\alpha$. How can we use that info though? Maybe some property of closed-unbounded sets under intersection?

Answer 1

Recursively construct a strictly increasing $\omega_1$-sequence $\langle\beta_\xi:\xi<\omega_1\rangle$ in $\omega_1$ such that $\min S_{\beta_\eta}>\sup_{\xi<\eta}\beta_\xi$ for each $\eta<\omega_1$; the fact that the $S_\alpha$ are pairwise disjoint ensures that this is possible. Let $A=\{\beta_{\xi+1}:\xi<\omega_1\}$, and let $S=\bigcup_{\alpha\in A}S_\alpha$. If $\gamma\in S$, there is a unique $\xi<\omega_1$ such that $\gamma\in S_{\beta_{\xi+1}}$; let $\varphi(\gamma)=\beta_\xi<\min S_{\beta_{\xi+1}}\le\gamma$. Clearly $\varphi$ is a pressing-down function, and for each $\gamma<\omega_1$ the set $\varphi^{-1}\big[\{\gamma\}\big]$ is non-stationary: it is either empty or $S_\alpha$ for some $\alpha\in A$. The pressing-down lemma now implies that $S$ is non-stationary.

Answer 2

I think I have a direction: by induction on $\alpha < \omega_1$, we can choose a set $S_\beta$ out of the given sets such that $minS_\beta > \alpha$ (because each set has a different minimum due to disjointness and we have $\omega_1$ given sets) and that wasn't picked yet. $A$ will be all the $\beta$'s.

Then assuming by negation that $\cup_{\beta \in A}S_\beta$ is stationary, we build a regressive function where for an element $\alpha$ of this union we take the (only) $S_\beta$ to which it is a member of, and $f(\alpha)$ will be defined as $\beta$. It's regressive by the construction of the $S_\beta$'s. Now by Fodor's lemma there is a stationary $S_0 \subseteq$ the union such that all its elements are assigned the same values by $f$, so $S_0$ has to be contained in one of the $S_\beta$'s, which are not stationary. Then we have a contradiction because a stationary set cannot be a subset of a non-stationary set.

Does this sound right?