Compactness via ultrafilters

Remarks

Suppose $X$ is compact and an ultrafilter $\mathcal U$ converges to no point. Then each $x$ has an open neighbourhood $V_x \notin \mathcal U$, so $X \setminus V_x \in \mathcal U$. Finitely many $V_x$ cover $X$, and the intersection of their complements is empty but lies in $\mathcal U$, a contradiction. Conversely, if an open cover has no finite subcover, the complements of its members generate a proper filter; an ultrafilter that contains it cannot converge to any point. This direction uses the ultrafilter lemma, a weak form of the axiom of choice.

The theorem gives a short proof of Tychonoff's theorem: an ultrafilter on a product converges as soon as all its images under the projections converge. If $X$ is also Hausdorff, each ultrafilter has exactly one limit. For example, on $X = [0, 1]$, the ultrafilter of all sets that contain a fixed point $p$ converges to $p$.

Going deeper

For a student

For general spaces, sequences are not enough to describe compactness, but ultrafilters are. The criterion turns proofs about open covers into proofs about limits. Its best-known use is the proof of Tychonoff's theorem, where convergence in a product is checked coordinate by coordinate. It also leads to the Stone–Čech compactification of a discrete space as the space of ultrafilters on it.