Neighborhood Sub-Basis Criterion for Filter Convergence

Theorem

Let $\struct {S, \tau}$ be a topological space.

Let $\FF$ be a filter on $S$.

Let $p \in S$.

Let $\FF$ converges to $p$.

Then it contains every neighborhood of $p$.

The set of neighborhoods of $p$ is trivially a neighborhood sub-basis at $p$.

$\Box$

Let $S_p$ be a neighborhood sub-basis at $p$.

Let $S_p \subseteq \FF$.

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

Then by the definition of neighborhood sub-basis, there is a finite $T_N \subseteq S_p$ such that:

$\bigcap T_N \subseteq N$

Since a filter is by definition closed under finite intersections:

$\bigcap T_N \in \FF$

Then:

$\bigcap T_N \in \FF$

and:

$\bigcap T_N \subseteq N$

so by the definition of a filter:

$N \in \FF$

Because $\FF$ contains every neighborhood of $p$, $\FF$ converges to $p$.

$\blacksquare$