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 $$\mathcal{B} := \{ F \cap U | F \in \mathcal{F} \text{ and } U \text{ is a neighborhood of } x \}$$. Then $$\mathcal{B}$$ is filter basis. 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 $$\mathcal{F}$$. $$\square$$