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Elementary initial segments form a club
Statement
In ZFC, if is regular uncountable and has universe in a finitary language of size less than , then
is club in .
Facts & Assumptions
Skolem witness closure on a cardinal: Fewer than kappa finite-arity functions suffice for elementary restrictions; the existential witness criterion is proved there.
Closure points form a club: Every self-map of kappa has club many closure points.
; 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: Regularity bounds every fewer-than-kappa collection of ordinals below kappa.
Proof
Given: The objects and hypotheses in the statement.
Use the witness family . For , let . There are fewer than kappa values: the finite-string cardinal count in the witness lemma bounds all tuples, and multiplying by still gives fewer than kappa. Regularity therefore gives .
The nonzero closure points alpha of g form an unbounded set. Since , such alpha are limits. Every finite tuple below alpha is contained in some ; hence every h value on it is below alpha. This includes the empty tuple for constants. The witness lemma gives . Thus is unbounded.
If delta is a nonzero limit point of , every finite tuple below delta lies below some . Language-function closure at alpha makes the restriction to delta a substructure. Any existential formula true in M with such a tuple has a witness below alpha by elementarity there, hence below delta. The witness criterion proves elementarity at delta. Thus itself is closed, completing the club claim.
Depends on
- Skolem witness closure on a cardinal
- Closure points form a club
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ 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
Used by
Dependency tree · two levels
28 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
- Kamensky, Theorem 1.4.7 with complete proof, p.7 (standard reference, not scraped)