# sums of compact pavings are compact

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  (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 $\mathcal{K}_{i}$ is defined as

 $\sum_{i\in I}\mathcal{K}_{i}=\left\{\sum_{i\in I}S_{i}\colon S_{i}\in\mathcal{% K}_{i}\cup\{\emptyset\}\text{ is empty for all but finitely many }i\right\}.$
###### Theorem.

Let $(K_{i},\mathcal{K}_{i})$ be compact  paved spaces for $i\in I$. Then, $\sum_{i}\mathcal{K}_{i}$ is a compact paving on $\sum_{i}K_{i}$.

The paving $\mathcal{K}^{\prime}$ consisting of subsets of $\sum_{i}\mathcal{K}_{i}$ of the form $\sum_{i}S_{i}$ 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).

Title sums of compact pavings are compact SumsOfCompactPavingsAreCompact 2013-03-22 18:45:15 2013-03-22 18:45:15 gel (22282) gel (22282) 5 gel (22282) Theorem msc 28A05 disjoint unions of compact pavings are compact ProductsOfCompactPavingsAreCompact direct sum of pavings disjoint union of pavings