sums of compact pavings are compact

Suppose that $(K_i,{𝒦}_i)$ is a paved space for each $i$ in an index set $I$. The direct sum, or disjoint union (http://planetmath.org/DisjointUnion), ${\sum }_{i\in I}{K}_i$ is the union of the disjoint sets $K_i\times \{i\}$. The direct sum of the paving ${𝒦}_i$ is defined as

$$\sum _{i\in I}{𝒦}_i=\{\sum _{i\in I}{S}_i:S_i\in {𝒦}_i\cup \{\mathrm{\varnothing }\}\text{ is empty for all but finitely many }i\}.$$

Theorem.

Let $\mathrm{(}{K}_i\mathrm{,}{\mathrm{K}}_i\mathrm{)}$ be compact paved spaces for $i\mathrm{\in }I$. Then, ${\mathrm{\sum }}_i{\mathrm{K}}_i$ is a compact paving on ${\mathrm{\sum }}_i{K}_i$.

The paving ${𝒦}^{\prime }$ consisting of subsets of ${\sum }_i{𝒦}_i$ of the form ${\sum }_i{S}_i$ where $S_i=\mathrm{\varnothing }$ for all but a single $i\in I$ is easily shown to be compact.Indeed, if ${𝒦}^{\prime \prime }\subseteq {𝒦}^{\prime }$ satisfies the finite intersection property then there is an $i\in I$ such that $S\subseteq K_i\times \{i\}$ for every $S\in {𝒦}^{\prime \prime }$. Compactness of ${𝒦}_i$ gives $\bigcap {𝒦}^{\prime \prime }\ne \mathrm{\varnothing }$.

Then, as ${\sum }_i{𝒦}_i$ consists of finite unions of sets in ${𝒦}^{\prime }$, it is a compact paving (see compact pavings are closed subsets of a compact space).