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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
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Necessary constraints on the regular-cardinal continuum function

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let κ and λ range over infinite regular cardinals (Cofinality cf⁡(α), and regular and singular cardinals). Then the continuum function κ↦2κ satisfies:

(a) κ<2κ;

(b) 2κ≤2λ whenever κ≤λ;

(c) cf⁡(2κ)>κ.

Clause (a) excludes values at most κ, clause (b) requires monotonicity, and clause (c) is König's stronger cofinality bound. Since cf⁡(μ)≤μ (cf⁡(α)≤α; cf⁡(0)=0 and cf⁡(α+1)=1; for a limit ordinal λ the value cf⁡(λ) is an infinite cardinal with cf⁡(cf⁡(λ))=cf⁡(λ), so it is regular; and every cofinal subset of λ has cardinality at least cf⁡(λ), 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 κ≤λ.

[F1]

Under the Axiom of Choice, 2κ=∣P(κ)∣ for every cardinal κ, and κ<2κ. (Assuming the Axiom of Choice, 2κ=∣P(κ)∣, and Cantor's theorem in cardinal form: κ<2κ)

[F3]

Cardinals compare by injections: κ≤λ if and only if there is an injection κ→λ, and if κ≤λ then κμ≤λμ. (Commutativity, associativity, distributivity and monotonicity of ⊕ and ⊗, the unit laws, the two exponent laws, and κ≤λ if and only if κ injects into λ)

[F4]

The Axiom of Choice: every family of nonempty sets has a choice function. (The Axiom of Choice)

Proof technique: direct.

Proof

1.1

Assume [F4]. Let κ be an infinite cardinal. By [F1], 2κ=∣P(κ)∣ is a cardinal and κ<2κ. Restricting to infinite regular κ gives clause (a).

F1F4given
1.2

Let κ≤λ be infinite regular cardinals. Every subset of κ is a subset of λ, so the inclusion P(κ)⊆P(λ) is an injection; by the injection criterion of [F3], ∣P(κ)∣≤∣P(λ)∣. Applying [F1] at κ and at λ turns this into 2κ≤2λ, which is clause (b).

F1F3
1.3

Clause (c) is [F2] at κ: cf⁡(2κ)>κ for every infinite cardinal κ, in particular for every infinite regular one.

F2
2.1

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 2κ with ∣P(κ)∣ and through König's theorem -- and no further selection is made in steps 1.2 and 1.3. ∎

F1F2step 1.1step 1.2step 1.3

Depends on

Used by

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Sources