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.
Necessary constraints on the regular-cardinal continuum function
Statement
Assume the Axiom of Choice (The Axiom of Choice). Let and range over infinite regular cardinals (Cofinality , and regular and singular cardinals). Then the continuum function satisfies:
(a) ;
(b) whenever ;
(c) .
Clause (a) excludes values at most , clause (b) requires monotonicity, and clause (c) is König's stronger cofinality bound. Since (; 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), clause (c) already implies clause (a); an Easton function is specified by monotonicity and this cofinality bound.
Facts & Assumptions
Given: The Axiom of Choice, so that every set has a cardinality, and infinite regular cardinals .
Under the Axiom of Choice, for every cardinal , and . (Assuming the Axiom of Choice, , and Cantor's theorem in cardinal form: )
Under the Axiom of Choice, for every infinite cardinal . (Assuming the Axiom of Choice: for every infinite cardinal , and ; in particular )
Cardinals compare by injections: if and only if there is an injection , and if then . (Commutativity, associativity, distributivity and monotonicity of and , the unit laws, the two exponent laws, and if and only if injects into )
The Axiom of Choice: every family of nonempty sets has a choice function. (The Axiom of Choice)
Proof technique: direct.
Proof
Assume [F4]. Let be an infinite cardinal. By [F1], is a cardinal and . Restricting to infinite regular gives clause (a).
Let be infinite regular cardinals. Every subset of is a subset of , so the inclusion is an injection; by the injection criterion of [F3], . Applying [F1] at and at turns this into , which is clause (b).
Clause (c) is [F2] at : for every infinite cardinal , in particular for every infinite regular one.
Clauses (a), (b) and (c) hold for all infinite regular cardinals, so in ZFC the continuum function on infinite regular cardinals satisfies exactly the displayed constraints. The Axiom of Choice enters only through [F1] and [F2] -- through the identification of with and through König's theorem -- and no further selection is made in steps 1.2 and 1.3. ∎
Depends on
- Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$
- Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ 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
- Cofinality $\operatorname{cf}(\alpha)$, and regular and singular cardinals
- Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$, the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$
- The Axiom of Choice
Used by
Dependency tree · two levels
32 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
- Thomas Jech, Set Theory, Chapter 15: Applications of Forcing, condition (15.7), printed p.232 (standard reference, not scraped)
- Kameryn J. Williams, Math 655 Lecture Notes 2.2, Definition 52, PDF p.11 (standard reference, not scraped)