proof of compact pavings are closed subsets of a compact space
Let be a compact paved space (http://planetmath.org/paved space). We use the ultrafilter lemma (http://planetmath.org/EveryFilterIsContainedInAnUltrafilter) to show that there is a compact paving containing that is closed under arbitrary intersections and finite unions.
We first show that the paving consisting of all finite unions of elements of is compact. Let satisfy the finite intersection property. It then follows that the collection of finite intersections of is a filter (http://planetmath.org/Filter). The ultrafilter lemma says that is contained in an ultrafilter .
By definition, the ultrafilter satisfies the finite intersection property. So, the compactness of implies that has nonempty intersection. Also, every element of is a union of finitely many elements of , one of which must be in (see alternative characterization of ultrafilter (http://planetmath.org/AlternativeCharacterizationOfUltrafilter)). In particular, contains the intersection of and,
Consequently, is compact.
Finally, we let be the set of arbitrary intersections of . This is closed under all arbitrary intersections and finite unions. Furthermore, if satisfies the finite intersection property then so does
The compactness of gives
|Title||proof of compact pavings are closed subsets of a compact space|
|Date of creation||2013-03-22 18:45:07|
|Last modified on||2013-03-22 18:45:07|
|Last modified by||gel (22282)|