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

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 $$F \in \mathcal F$$, $$x \in \bigcap \{ \overline F : F \in \mathcal F \}$$ and thus $$x$$ is a limit point of $$\mathcal f$$.