If $ \bigcup N_\alpha $ is stationary, then $ \{ \min(N_\alpha) \}$ is stationary

Question

It's the last set-theoretic question for tonight, I promise.

I'm trying to figure out why the following holds true:

Suppose we have a regular, uncountable cardinal $ \kappa $ and a disjoint family of non-stationary subsets $ \{ N_\alpha \subset \kappa, \alpha < \kappa \} $. Then the sum $ \bigcup\limits_{\alpha < \kappa} N_\alpha $ is stationary if and only if the set $ \{\min(N_\alpha), \alpha < \kappa\}$ is stationary.

The $ \Leftarrow $ direction is obvious. For the other one, suppose that $ C \in \kappa $ is a club, and $ C $ intersects some $ N_\alpha $. I can't see how the fact that they are disjoint and non-stationary might mean that $ C $ contains a minimum. I would appreciate a hint

Answer 1

HINT: It seems easier to show the contrapositive of the forward implication. Suppose that $C$ is a club disjoint from $\{\min N_\alpha:\alpha<\kappa\}$.

• Show that for each $\alpha<\kappa$ the set $C\cap\min N_\alpha$ has a maximum element $\gamma_\alpha$.

Define $f:\bigcup_{\alpha<\kappa}N_\alpha\to\kappa$ by

$$\large f(\xi)=\gamma_{\min\{\alpha<\kappa:\xi\in N_\alpha\}}\;,$$

and observe that $f$ is regressive (a pressing-down function). Is there any stationary set on which $f$ is constant?