Every Filter has Limit Point implies Every Ultrafilter Converges

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

Let each filter on $S$ have a limit point in $S$.

Then each ultrafilter on $S$ converges to a point in $S$.

Proof
Let $T = \struct {S, \tau}$ be such that each filter on $S$ has a limit point in $S$.

Let $\FF$ be an ultrafilter on $S$.

By hypothesis, $\FF$ has a limit point $x \in S$.

By Limit Point iff Superfilter Converges, there exists a filter $\FF'$ on $S$ which converges to $x$ satisfying $\FF \subseteq \FF'$.

By :
 * $\FF$ is an ultrafilter whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$

Therefore:
 * $\FF = \FF'$.

Thus $\FF$ converges to $x$.

Also see

 * Equivalence of Definitions of Compact Topological Space