complete ultrafilter and partitions

If U is an ultrafilterMathworldPlanetmath on a set S, then

U is κ-completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath there is no partitionMathworldPlanetmathPlanetmath of S into κ-many pieces for which each piece Xα of the partition is not in U.

We prove the case of σ-completeness; the case of arbitrary infiniteMathworldPlanetmath cardinality follows closely. For the direction, let P be a partition of S into ω many pieces, all of which do not belong to U, and write S=n=1ωXn to illustrate this partition. Now, =S=n=1ωXn. Since, by our assumptionPlanetmathPlanetmath, each of the Xn do not belong to U, we have XnU for each n<ω as U is an ultrafilter. Thus, (n=1ωXn)U by σ-completeness. This, however, means U, contradicting the definition of a filter.

Note that the converseMathworldPlanetmath states that every partition P of S into ω-many pieces has a (unique) piece X1U. To prove this, let Yn be a collectionMathworldPlanetmath of ω many members of U and let Y=n=1ωYn. Now consider the partition {Pι:ιω} of SY:

for each sSY, put sPι if ι is the least index for which sYι.

It is easy to verify that each sSY belongs to a unique Pι, the collection of Pι’s is indeed a partition of SY.

Along with Y, {Pι:ιω} partitions S into 0=ω many pieces. A (unique) piece of this partition belongs in U: Pι*U or YU. But, PιYι=U by the definition of Pι. This excludes the possibility for the former to belong in U (cf. alternative characterization of filter) and so YU.

Thus, starting from an arbitrary collection {Yn} of ω-many members of U, we have identified a partition of S for which the unique piece which belongs to U is Yn. Therefore, U is σ-complete.

Title complete ultrafilter and partitions
Canonical name CompleteUltrafilterAndPartitions
Date of creation 2013-03-22 18:55:52
Last modified on 2013-03-22 18:55:52
Owner yesitis (13730)
Last modified by yesitis (13730)
Numerical id 4
Author yesitis (13730)
Entry type Definition
Classification msc 03E02