Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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The club filter is never an ultrafilter

Statement

In ZFC, on every regular uncountable κ there is a stationary costationary subset; consequently its club filter is not an ultrafilter (it does not decide every subset by membership or complement membership).

Facts & Assumptions

[F1]

Solovay’s stationary partition theorem: Every stationary subset of kappa partitions into kappa stationary pieces.

[F2]

The club filter and nonstationary ideal: The club filter contains exactly the supersets of clubs; stationary sets meet every club.

Proof

Given: The objects and hypotheses in the statement.

1.1

The whole cardinal is stationary because every club is nonempty. Split it into (Sξ)ξ<κ by Solovay. Then S0 is stationary and its complement contains the stationary S1, so the complement is stationary as well.

F1F2
2.1

Neither S0 nor its complement contains a club, since such a club would be disjoint from the stationary set on the other side. Thus the club filter contains neither side of this partition and does not decide every subset.

F2step 1.1∎

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources