How statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Regular cofinality strata are stationary
Statement
In ZFC, if is an infinite regular cardinal with , then is stationary in .
Facts & Assumptions
Cofinality strata, trace, and reflection: The stratum consists of ordinals with exactly the specified cofinality.
; and ; for a limit ordinal the value is an infinite cardinal with , so it is regular; and every cofinal subset of has cardinality at least , a value that is attained: Cofinality bounds the length of cofinal sequences; infinite cofinalities are regular cardinals.
Transfinite recursion: A specified least-next and limit rule recurses along lambda.
Proof
Given: The objects and hypotheses in the statement.
Given a club , define a strictly increasing sequence in : start at , take the least greater C point at successors, and at nonzero limits take the supremum. Every such supremum is below theta since its index is below , and closure places it in C. Likewise belongs to C.
This sequence gives . If a cofinal subset had size , assign to each the least with . These indices would be unbounded in lambda: a bound would bound B below delta. Regularity of lambda rules this out. Therefore , so C meets the stratum. Since C was arbitrary the stratum is stationary.
Depends on
- Cofinality strata, trace, and reflection
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ and $\operatorname{cf}(\alpha + 1) = 1$; for a limit ordinal $\lambda$ the value $\operatorname{cf}(\lambda)$ is an infinite cardinal with $\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda)$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\operatorname{cf}(\lambda)$, a value that is attained
- Transfinite recursion
Used by
Dependency tree · two levels
24 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, Proposition 5.13, p.42 (standard reference, not scraped)
- Vasey, Example 14.13(6), p.82 (standard reference, not scraped)