# compact pavings are closed subsets of a compact space

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

###### Theorem 1.

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

In particular,

 $\mathcal{T}\equiv\left\{K\setminus C\colon C\in\mathcal{K}^{\prime}\right\}% \cup\left\{\emptyset,K\right\}$

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 $\mathcal{K}^{\prime}\cup\{\emptyset,K\}$ which, by Theorem 1, is a compact paving. So, the following is obtained.

###### Corollary.

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

Title compact pavings are closed subsets of a compact space CompactPavingsAreClosedSubsetsOfACompactSpace 2013-03-22 18:45:01 2013-03-22 18:45:01 gel (22282) gel (22282) 6 gel (22282) Theorem msc 28A05