products of compact pavings are compact

Suppose that (Ki,𝒦i) is a paved space for each i in index setMathworldPlanetmathPlanetmath I. The productPlanetmathPlanetmathPlanetmath ( ∏i∈IKi is the set of all functionsMathworldPlanetmath x:I→⋃iKi such that xi∈Ki for each i, and the product of subsets SiβŠ†Ki is

∏i∈ISi={x∈∏i∈IKi:xi∈Si⁒ for each ⁒i∈I}.

Then, the product paving is defined by

∏i∈I𝒦i={∏i∈ISi:Siβˆˆπ’¦i⁒ for each ⁒i∈I}.
Theorem 1.

Let (Ki,Ki) be compactPlanetmathPlanetmath paved spaces for i∈I. Then, ∏iKi is a compact paving on ∏iKi.

Note that this result is a version of TychonoffPlanetmathPlanetmath’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 equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to Tychonoff’s theorem.

Theorem 1 is simple to prove directly. Suppose that {Aj:j∈J} is a subset of ∏i𝒦i satisfying the finite intersection property. Writing Aj=∏i∈ISi⁒j for Si⁒jβˆˆπ’¦i gives

β‹‚j∈Jβ€²Aj=∏i∈I(β‹‚j∈Jβ€²Si⁒j) (1)

for any Jβ€²βŠ†J. By the finite intersection property, this is nonempty whenever Jβ€² is finite, so β‹‚j∈Jβ€²Si⁒jβ‰ βˆ…. Consequently, {Si⁒j:j∈J}βŠ†π’¦i satisfies the finite intersection property and, by compactness of 𝒦i, the intersectionDlmfMathworldPlanetmath β‹‚j∈JSi⁒j is nonempty. So equation (1) shows that β‹‚j∈JAj is nonempty.

Title products of compact pavings are compact
Canonical name ProductsOfCompactPavingsAreCompact
Date of creation 2013-03-22 18:45:09
Last modified on 2013-03-22 18:45:09
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Theorem
Classification msc 28A05
Related topic SumsOfCompactPavingsAreCompact