proof of compact pavings are closed subsets of a compact space


Let (K,𝒦) be a compactPlanetmathPlanetmath paved space (http://planetmath.org/paved space). We use the ultrafilterMathworldPlanetmath lemma (http://planetmath.org/EveryFilterIsContainedInAnUltrafilter) to show that there is a compact paving 𝒦′ containing 𝒦 that is closed under arbitrary intersectionsMathworldPlanetmath and finite unions.

We first show that the paving 𝒦1 consisting of all finite unions of elements of 𝒦 is compact. Let β„±βŠ†π’¦1 satisfy the finite intersection property. It then follows that the collectionMathworldPlanetmath of finite intersections of β„± is a filter (http://planetmath.org/Filter). The ultrafilter lemma says that β„± is contained in an ultrafilter 𝒰.

By definition, the ultrafilter satisfies the finite intersection property. So, the compactness of 𝒦 implies that β„±β€²β‰‘π’°βˆ©π’¦ has nonempty intersection. Also, every element S of β„± is a union of finitely many elements of 𝒦, one of which must be in 𝒰 (see alternative characterization of ultrafilter (http://planetmath.org/AlternativeCharacterizationOfUltrafilter)). In particular, S contains the intersection of β„±β€² and,

β‹‚β„±βŠ‡β‹‚β„±β€²β‰ βˆ….

Consequently, 𝒦1 is compact.

Finally, we let 𝒦′ be the set of arbitrary intersections of 𝒦1. This is closed under all arbitrary intersections and finite unions. Furthermore, if β„±βŠ†π’¦β€² satisfies the finite intersection property then so does

ℱ′≑{Aβˆˆπ’¦1:BβŠ†A⁒ for some ⁒Bβˆˆβ„±}.

The compactness of 𝒦1 gives

β‹‚β„±=β‹‚β„±β€²β‰ βˆ…

as required.

Title proof of compact pavings are closed subsets of a compact space
Canonical name ProofOfCompactPavingsAreClosedSubsetsOfACompactSpace
Date of creation 2013-03-22 18:45:07
Last modified on 2013-03-22 18:45:07
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 28A05