proof of compact pavings are closed subsets of a compact space
Let (K,π¦) 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 π¦β² containing π¦ that is closed under arbitrary intersections
and finite unions.
We first show that the paving π¦1 consisting of all finite unions of elements of π¦ is compact.
Let β±βπ¦1 satisfy the finite intersection property. It then follows that the collection of finite intersections of β± is a filter (http://planetmath.org/Filter). The ultrafilter lemma says that β± is contained in an ultrafilter π°.
By definition, the ultrafilter satisfies the finite intersection property. So, the compactness of π¦ implies that β±β²β‘π°β©π¦ has nonempty intersection. Also, every element S of β± is a union of finitely many elements of π¦, one of which must be in π° (see alternative characterization of ultrafilter (http://planetmath.org/AlternativeCharacterizationOfUltrafilter)). In particular, S contains the intersection of β±β² and,
ββ±βββ±β²β β . |
Consequently, π¦1 is compact.
Finally, we let π¦β² be the set of arbitrary intersections of π¦1. This is closed under all arbitrary intersections and finite unions. Furthermore, if β±βπ¦β² satisfies the finite intersection property then so does
β±β²β‘{Aβπ¦1:BβA for some Bββ±}. |
The compactness of π¦1 gives
ββ±=ββ±β²β β |
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 |