Suppose that $(K_i,\mathcal{K}_i)$ is a paved space for each $i$ in an index set $I$ The direct sum, or disjoint union, $\sum_{i\in I}K_i$ is the union of the disjoint sets $K_i\times\{i\}$ The direct sum of the paving $\mathcal{K}_i$ is defined as \begin{equation*} \sum_{i\in I}\mathcal{K}_i=\left\{\sum_{i\in I}S_i\colon S_i\in\mathcal{K}_i\cup\{\emptyset\}\text{ is empty for all but finitely many }i\right\}. \end{equation*}

The paving $\mathcal{K}^\prime$ consisting of subsets of $\sum_i\mathcal{K}_i$ of the form $\sum_iS_i$ where $S_i=\emptyset$ for all but a single $i\in I$ is easily shown to be compact. Indeed, if $\mathcal{K}^{\prime\prime}\subseteq \mathcal{K}^\prime$ satisfies the finite intersection property then there is an $i\in I$ such that $S\subseteq K_i\times\{i\}$ for every $S\in\mathcal{K}^{\prime\prime}$ Compactness of $\mathcal{K}_i$ gives $\bigcap\mathcal{K}^{\prime\prime}\not=\emptyset$

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