We define a subset $A$ of an ordinal $\alpha$ as stationary iff we have $A$ intersecting every closed and unbounded subset of $\alpha$. Equivalently, one can define this as
$$\forall f:\alpha\mapsto\alpha,\exists\beta\in A,\forall\gamma\in\beta(f(\gamma)\in\beta)$$
I'm curious about $\omega_1$, and whether or not the following is satisfied:
For every set $A$ of stationary subsets of $\omega_1$, define $B$ as the set of ordinals $\beta$ such that $\beta\cap A_\star$ is stationary in $\beta$ for all $A_\star\in A$. Whenever $B$ is uncountable, is $B$ a stationary subset of $\omega_1$?
By having $\min A_\star$ be unbounded, we can make $B$ empty, and by similar arguments we can make it contain countably many ordinals. However, I'm not sure if we can make $B$ uncountable without trivially being stationary.
It may also be helpful to note that in order for $\beta\cap A_\star$ to be stationary in countable $\beta$, we must have $\sup(\beta\setminus A_\star)\in\beta$.
My curiosity on whether or not the above is true or not stems from trying to see if the above is equivalent to a large cardinal, namely something like weakly compact cardinals. Hence why my intuition says the above is false.
The set of countable ordinals which aren't successors of limit ordinals forms a stationary set, since it contains every countable limit ordinal. This reflects at every countable limit ordinal which isn't a limit of limit ordinals. However, the set of limits of limit ordinals is club, so this set isn't stationary. In general, replacing $\omega_1$ with any ordinal fails to have this property by the same construction as above.