Neighborhood Sub-Basis Criterion for Filter Convergence

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

Let $\mathcal F$ be a filter on $X$.

Let $p \in X$.

Let $S_p \subseteq \tau$.

Suppose that for each neighborhood $N$ of $p$, there is a finite $T_N \subseteq S_p$ such that:
 * $p \in \bigcap T_N \subseteq N$.

Then if $S_p \subseteq \mathcal F$, $\mathcal F$ converges to $p$.

Proof
Let $N$ be a neighborhood of $p$.

Then by the premise, there is a finite $T_N \subseteq S_p$ such that:
 * $p \in \bigcap T_N \subseteq N$.

Since a filter is closed under finite intersections, $\bigcap T_N \in \mathcal F$.

Then $\bigcap T_N in \mathcal F$ and $\bigcap T_N \subseteq N$, so by the definition of a filter, $N \in \mathcal F$.