Every Ultrafilter Converges implies Every Filter has Limit Point

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

Let every ultrafilter on $S$ be convergent.

Then every filter on $S$ has a limit point.

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

By the Ultrafilter Lemma, there exists an ultrafilter $\FF'$ such that $\FF \subseteq \FF'$.

By hypothesis, $\FF'$ converges to some $x \in S$.

This, by Limit Point iff Superfilter Converges, implies that $x$ is a limit point of $\FF$.

Also see

 * Equivalence of Definitions of Compact Topological Space