I would like to prove that, given a filter $\mathcal{F}$, the collection of all limit points of $\mathcal{F}$ is closed. If I call such a set as $C$, then I need to show $\overline{C} \subseteq C$. If $x \in \overline{C}$, then, for every neighborhood $N_x$ containing $x$, $N_x \cap C \neq \emptyset$, which implies that there exists a filter $\mathcal{F}_2$ such that $\mathcal{F}_2 \to x$. I am sure I can use this result to show that $\mathcal{F} \to x$ as well, but struggling to come up with an idea. Any help will be appreciated.
Condition for two filters to have a common limit
general-topology
Best Answer
Suppose $x$ is in the closure of $C$. We need to show that every open neighborhood of $x$ is in $\mathcal{F}$. So let $x\in U$ be open. Then there is some $y\in U\cap C$, so $U$ is also an open neighborhood of $y$. Since $y\in C$, $U\in \mathcal{F}$, and we're done.