LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Unboundedly many stationary fibres yield a partition

Statement

In ZFC let $κ$ be regular uncountable and $g : S \to κ$ be regressive, with stationary $S \subseteq κ ∖ {0}$ . If ${α \in S : g (α) \geq β}$ is stationary for every $β < κ$ , then $S$ has a partition into $κ$ stationary sets.

Facts & Assumptions

[F1]

Fodor’s pressing-down lemma: A regressive function on a stationary subset of regular uncountable kappa has a stationary fibre.

[F2]

Basic stationary-set calculus: Supersets of stationary sets are stationary.

Proof

Given: The objects and hypotheses in the statement.

1.1

For each $β < κ$ , apply Fodor to the stated stationary tail domain. Its stationary constant subset lies in a fibre $g^{- 1} ({γ})$ with $γ \geq β$ . Thus $B = {γ < κ : g^{- 1} ({γ}) stationary}$ is unbounded. Regularity gives $∣ B ∣ = κ$ , so its increasing enumeration has domain kappa.

F1F2

2.1

The fibres at values in B are pairwise disjoint stationary sets. Keep them all, and adjoin every remaining point of S to the fibre at the least value in B. That enlarged fibre remains stationary and disjoint from all the others, and the union is now S.

F2step 1.1∎

Depends on

Used by

Dependency tree · two levels

5 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

Lietz, Theorem 5.14 proof, case 1 after Claim 5.17, p.44 (standard reference, not scraped)