Every Ultrafilter Converges implies Every Filter has Limit Point

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let every ultrafilter on $S$ be convergent.


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


Proof

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

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

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

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

$\blacksquare$


Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI), by way of Ultrafilter Lemma.

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.


Also see