Limit Point iff Superfilter Converges

Theorem
Let $\mathcal F$ be a filter on a topological space $X$ and let $x \in X$.

Then $x$ is a limit point of $\mathcal F$ iff there is a superfilter $\mathcal{F}'$ of $\mathcal F$ on $X$ which converges to $x$.

Proof

 * Assume first that $x$ is a limit point of $\mathcal F$.

Define:
 * $\mathcal B := \left\{{F \cap U : F \in \mathcal F \text{ and } U \text{ is a neighborhood of } x}\right\}$.

Then $\mathcal B$ is filter basis by definition.

Let $\mathcal F'$ be the corresponding generated filter.

By construction we have $\mathcal F \subseteq \mathcal F'$ and $U \in \mathcal F'$ for every neighborhood $U$ of $x$.

Thus $\mathcal F'$ converges to $x$.


 * Assume now that there is a filter $\mathcal F'$ on $X$ satisfying $\mathcal F \subseteq \mathcal F'$ which converges to $x$.

Let $U \subseteq X$ be a neighborhood of $x$ and $F \in \mathcal F$.

Then $U, F \in \mathcal F'$ and therefore $U \cap F \in \mathcal F'$.

Because $\varnothing \not \in \mathcal F'$ it follows that $U \cap F \ne \varnothing$.

Since this holds for any neighborhood $U$ of $x$ we know that $x$ is a limit point of $F$ and therefore $x \in \overline{F}$.

Because this holds for all $F \in \mathcal F$, $\displaystyle x \in \bigcap \{ \overline F : F \in \mathcal F \}$ and thus $x$ is a limit point of $\mathcal F$.