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.
Fixed points of a normal function form a club
Statement
In ZFC, if is normal and is regular uncountable, then is club.
Facts & Assumptions
Normal ordinal functions: Normal functions are strictly increasing and continuous at nonzero limits.
; 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: Countable sets of ordinals are bounded below regular uncountable kappa.
The recursion theorem: A specified self-map can be iterated on omega.
Proof
Given: The objects and hypotheses in the statement.
Transfinite induction gives : zero is automatic, successors use strict increase, and limits use continuity. Given , iterate , . If some adjacent terms agree, that term is a fixed point above . Otherwise the sequence is strictly increasing and its supremum is a nonzero limit.
In the latter case continuity and cofinality of the in give . This proves unboundedness in both cases.
If fixed points are unbounded in a nonzero limit , monotonicity and continuity give . Thus the fixed-point set is closed. Zero may or may not be fixed; no claim depends on it.
Depends on
- Normal ordinal functions
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ 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
- The recursion theorem
Used by
Dependency tree · two levels
25 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
- Welch, Lemma 2.13, p.21; live original reread (standard reference, not scraped)