proof of compact pavings are closed subsets of a compact space

Let $(K,𝒦)$ be a compact paved space (http://planetmath.org/paved space). We use the ultrafilter lemma (http://planetmath.org/EveryFilterIsContainedInAnUltrafilter) to show that there is a compact paving ${𝒦}^{\prime }$ containing $𝒦$ that is closed under arbitrary intersections and finite unions.

We first show that the paving ${𝒦}_1$ consisting of all finite unions of elements of $𝒦$ is compact.Let $ℱ\subseteq {𝒦}_1$ satisfy the finite intersection property. It then follows that the collection of finite intersections of $ℱ$ is a filter (http://planetmath.org/Filter). The ultrafilter lemma says that $ℱ$ is contained in an ultrafilter $𝒰$.

By definition, the ultrafilter satisfies the finite intersection property. So, the compactness of $𝒦$ implies that ${ℱ}^{\prime }\equiv 𝒰\cap 𝒦$ has nonempty intersection.Also, every element $S$ of $ℱ$ is a union of finitely many elements of $𝒦$, one of which must be in $𝒰$ (see alternative characterization of ultrafilter (http://planetmath.org/AlternativeCharacterizationOfUltrafilter)). In particular, $S$ contains the intersection of ${ℱ}^{\prime }$ and,

$$\bigcap ℱ\supseteq \bigcap {ℱ}^{\prime }\ne \mathrm{\varnothing }.$$

Consequently, ${𝒦}_1$ is compact.

Finally, we let ${𝒦}^{\prime }$ be the set of arbitrary intersections of ${𝒦}_1$. This is closed under all arbitrary intersections and finite unions. Furthermore, if $ℱ\subseteq {𝒦}^{\prime }$ satisfies the finite intersection property then so does

$${ℱ}^{\prime }\equiv \{A\in {𝒦}_1:B\subseteq A\text{ for some }B\in ℱ\}.$$

The compactness of ${𝒦}_1$ gives

$$\bigcap ℱ=\bigcap {ℱ}^{\prime }\ne \mathrm{\varnothing }$$

as required.