products of compact pavings are compact

Suppose that $(K_i,{𝒦}_i)$ 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 $x:I\to {\bigcup }_i{K}_i$ such that $x_i\in K_i$ for each $i$, and the product of subsets $S_i\subseteq K_i$ is

$$\prod _{i\in I}{S}_i=\{x\in \prod _{i\in I}{K}_i:x_i\in S_i\text{ for each }i\in I\}.$$

Then, the product paving is defined by

$$\prod _{i\in I}{𝒦}_i=\{\prod _{i\in I}{S}_i:S_i\in {𝒦}_i\text{ for each }i\in I\}.$$

Theorem 1.

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

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 $\{A_j:j\in J\}$ is a subset of ${\prod }_i{𝒦}_i$ satisfying the finite intersection property. Writing $A_j={\prod }_{i\in I}{S}_{ij}$ for $S_{ij}\in {𝒦}_i$ 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 ${\bigcap }_{j\in J^{\prime }}{S}_{ij}\ne \mathrm{\varnothing }$. Consequently, $\{S_{ij}:j\in J\}\subseteq {𝒦}_i$ satisfies the finite intersection property and, by compactness of ${𝒦}_i$, the intersection ${\bigcap }_{j\in J}{S}_{ij}$ is nonempty. So equation (1) shows that ${\bigcap }_{j\in J}{A}_j$ is nonempty.