proof of every filter is contained in an ultrafilter (alternate proof)

Let π”˜ be the family of filters over X which are finer than β„±, under the partial orderMathworldPlanetmath of inclusion.

Claim 1. Every chain in U has an upper bound also in U.


Take any chain β„­ in π”˜, and consider the set π’ž=βˆͺβ„­. Then π’ž is also a filter: it cannot contain the empty setMathworldPlanetmath, since no filter in the chain does; the intersectionMathworldPlanetmath of two sets in π’ž must be present in the filters of β„­; and π’ž is closed under supersetsMathworldPlanetmath because every filter in β„­ is. Obviously π’ž is finer than β„±. ∎

So we conclude, by Zorn’s lemma, that π”˜ must have a maximal filter say 𝒰, which must contain β„±. All we need to show is that 𝒰 is an ultrafilterMathworldPlanetmath. Now, for any filter 𝒰, and any set YβŠ†X, we must have:

Claim 2. Either U1={Z∩Y:Z∈U} or U2={Z∩(X\Y):Z∈U} (or both) are a filter subbasis.


We prove by contradictionMathworldPlanetmathPlanetmath that at least one of 𝒰1 or 𝒰2 must have the finite intersection property. If neither has the finite intersection property, then for some Z1,…⁒Zk we must have


But then


and so 𝒰 does not have the finite intersection property either. This cannot be, since 𝒰 is a filter. ∎

Now, by Claim 2, if 𝒰 were not an ultrafilter, i.e., if for some Y subset of X we would have neither Y nor X\Y in 𝒰, then the filter generated 𝒰1 or 𝒰2 would be finer than 𝒰, and then 𝒰 would not be maximal.

So 𝒰 is an ultrafilter containing β„±, as intended.

Title proof of every filter is contained in an ultrafilter (alternate proof)
Canonical name ProofOfEveryFilterIsContainedInAnUltrafilteralternateProof
Date of creation 2013-03-22 17:52:54
Last modified on 2013-03-22 17:52:54
Owner brunoloff (19748)
Last modified by brunoloff (19748)
Numerical id 8
Author brunoloff (19748)
Entry type Proof
Classification msc 54A20