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 filter $$\mathcal{F}'$$ on $$X$$ satisfying $$\mathcal{F} \subseteq \mathcal{F}'$$ 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 $$\mathcal{F}$$.