If $X$ is a set of ordinals, define $f(X, \alpha)$ such that $f(X, 0) = X$, $f(X, \beta + 1) = f(X, \beta) \cap Lim(f(X, \beta))$, where $Lim(Y) = \{$limit points of $Y\}$, and $f(X, \gamma) = \bigcap \{f(X, \beta) : \beta < \gamma\}$ for limit $\gamma$.


Now let $\kappa$ be uncountable regular, and let $X \subset \kappa$ such that for all $\alpha < \kappa$, $f(X, \alpha)$ is unbounded in $\kappa$. Can we then conclude that $X$ is stationary in $\kappa$? I feel the answer is no, but I could not come up with a counter example.



Is there such a sufficient condition for "$X$ is a stationary subset of uncountable regular $\kappa$" involving limit points?
The answer is no. Let $\langle \gamma\_\alpha \rangle\_{\alpha\lt\kappa}$ be a continuous sequence such that $\gamma\_{\alpha+1} \geq \gamma\_\alpha + \alpha$ for every $\alpha\lt\kappa$. Then $X = \kappa\setminus\{\gamma\_\alpha:\alpha\lt\kappa\}$ is nonstationary but $f(X,\alpha)$ is always unbounded in $\kappa$ since $X$ contains arbitrarily long intervals.


