compact pavings are closed subsets of a compact space

Recall that a paving $𝒦$ is compact if every subcollection satisfying the finite intersection property has nonempty intersection. In particular, a topological space is compact (http://planetmath.org/Compact) if and only if its collection of closed subsets forms a compact paving. Compact paved spaces can therefore be constructed by taking closed subsets of a compact topological space. In fact, all compact pavings arise in this way, as we now show.

Given any compact paving $𝒦$ the following result says that the collection ${𝒦}^{\prime }$ of all intersections of finite unions of sets in $𝒦$ is also compact.

Theorem 1.

Suppose that $\mathrm{(}K\mathrm{,}\mathrm{K}\mathrm{)}$ is a compact paved space. Let ${\mathrm{K}}^{\mathrm{\prime }}$ be the smallest collection of subsets of $X$ such that $\mathrm{K}\mathrm{\subseteq }{\mathrm{K}}^{\mathrm{\prime }}$ and which is closed under arbitrary intersections and finite unions.Then, ${\mathrm{K}}^{\mathrm{\prime }}$ is a compact paving.

In particular,

$$𝒯\equiv \{K\setminus C:C\in {𝒦}^{\prime }\}\cup \{\mathrm{\varnothing },K\}$$

is closed under arbitrary unions and finite intersections, and hence is a topology on $K$.The collection of closed sets defined with respect to this topology is ${𝒦}^{\prime }\cup \{\mathrm{\varnothing },K\}$ which, by Theorem 1, is a compact paving. So, the following is obtained.

Corollary.

A paving $\mathrm{(}K\mathrm{,}\mathrm{K}\mathrm{)}$ is compact if and only if there exists a topology on $K$ with respect to which $\mathrm{K}$ are closed sets and $K$ is compact.