Limit Point iff Superfilter Converges
Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $\mathcal F$ be a filter on $S$.
Let $x \in S$.
Then $x$ is a limit point of $\mathcal F$ if and only if there exists a superfilter $\mathcal F'$ of $\mathcal F$ on $S$ which converges to $x$.
Proof
Let $x$ be 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 $x$.
$\Box$
Let there be a filter $\mathcal F'$ on $X$ satisfying $\mathcal F \subseteq \mathcal F'$ which converges to $x$.
Let $U \subseteq X$ be a neighborhood of $x$.
Let $F \in \mathcal F$.
Then:
- $U, F \in \mathcal F'$
and therefore:
- $U \cap F \in \mathcal F'$
Because $\varnothing \notin \mathcal F'$ it follows that:
- $U \cap F \ne \varnothing$
This holds for any neighborhood $U$ of $x$.
Hence $x$ is a limit point of $F$.
Therefore:
- $x \in \overline F$
Because this holds for all $F \in \mathcal F$:
- $\displaystyle x \in \bigcap \left\{{\overline F : F \in \mathcal F}\right\}$
and so $x$ is a limit point of $\mathcal F$.
$\blacksquare$