TITLE: Exam question about stationary sets
QUESTION [5 upvotes]: This is an old exam question that I don't have an answer to. I am posting my solution in hope that if I have errors, someone will point them out, and for future generations of students taking this course.

Let $\{A_\alpha \mid \alpha < \aleph_1\}$, disjoint set of non-stationary sets, such that $\bigcup_{\alpha < \aleph_1}A_\alpha$ is stationary. Prove that $\{\min A_\alpha \mid \alpha < \aleph_1\}$ is stationary.

REPLY [5 votes]: Let $A := \bigcup_{\alpha < \aleph_1}A_\alpha$, $M := \{\min A_\alpha \mid \alpha < \aleph_1\}$.
Assume that $M$ is not stationary, so there is a club set, $C \subset \omega_1 \setminus M$. Let $B := A\cap C$. B is stationary.
Because the sets are disjoint, for every $\beta \in B$, there is a single $\alpha_\beta$ such that $\beta \in A_{\alpha_\beta}$. By the construction of $B$, $\beta \neq \min A_{\alpha_\beta}$.
$h := \beta \mapsto\min A_{\alpha_\beta}$ is regressive, and by Fodor's lemma we get that there is some stationary $B_0 \subset B$ and $\gamma$ such that for every $\beta \in B_0$, $h(\beta) = \gamma$.
$A_\alpha$ are disjoint, hence there is $\alpha_0$ such that for every $\beta \in B_0$, $\beta \in A_{\alpha_0}$ and it follows that $A_{\alpha_0}$ is stationary, which is a contradiction.