# products of compact pavings are compact

Suppose that $(K_{i},\mathcal{K}_{i})$ is a paved space for each $i$ in index set $I$. The product (http://planetmath.org/GeneralizedCartesianProduct) $\prod_{i\in I}K_{i}$ is the set of all functions $x\colon I\rightarrow\bigcup_{i}K_{i}$ such that $x_{i}\in K_{i}$ for each $i$, and the product of subsets $S_{i}\subseteq K_{i}$ is

 $\prod_{i\in I}S_{i}=\left\{x\in\prod_{i\in I}K_{i}\colon x_{i}\in S_{i}\text{ % for each }i\in I\right\}.$

Then, the product paving is defined by

 $\prod_{i\in I}\mathcal{K}_{i}=\left\{\prod_{i\in I}S_{i}\colon S_{i}\in% \mathcal{K}_{i}\text{ for each }i\in I\right\}.$
###### Theorem 1.

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

Note that this result is a version of Tychonoff’s theorem applying to paved spaces and, together with the fact that all compact pavings are closed subsets of a compact space, is easily seen to be equivalent to Tychonoff’s theorem.

Theorem 1 is simple to prove directly. Suppose that $\{A_{j}\colon j\in J\}$ is a subset of $\prod_{i}\mathcal{K}_{i}$ satisfying the finite intersection property. Writing $A_{j}=\prod_{i\in I}S_{ij}$ for $S_{ij}\in\mathcal{K}_{i}$ gives

 $\bigcap_{j\in J^{\prime}}A_{j}=\prod_{i\in I}\left(\bigcap_{j\in J^{\prime}}S_% {ij}\right)$ (1)

for any $J^{\prime}\subseteq J$. By the finite intersection property, this is nonempty whenever $J^{\prime}$ is finite, so $\bigcap_{j\in J^{\prime}}S_{ij}\not=\emptyset$. Consequently, $\{S_{ij}\colon j\in J\}\subseteq\mathcal{K}_{i}$ satisfies the finite intersection property and, by compactness of $\mathcal{K}_{i}$, the intersection $\bigcap_{j\in J}S_{ij}$ is nonempty. So equation (1) shows that $\bigcap_{j\in J}A_{j}$ is nonempty.

Title products of compact pavings are compact ProductsOfCompactPavingsAreCompact 2013-03-22 18:45:09 2013-03-22 18:45:09 gel (22282) gel (22282) 5 gel (22282) Theorem msc 28A05 SumsOfCompactPavingsAreCompact