CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)‡ rests on unproved material (inherited)

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‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$ , and $\operatorname{cf}(2^{\kappa}) > \kappa$ ; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $\kappa$ be an infinite cardinal (Cardinal (initial ordinal) and cardinality). Then:

(a) $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations, Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals);

(b) $\operatorname{cf}(2^{\kappa}) > \kappa$ ;

(c) in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ (The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ ).

Clause (c) is a genuine restriction on the continuum, and it is proved in ZFC rather than quoted. It rules out every value of $2^{\aleph_0}$ whose cofinality is $\aleph_0$ , and it selects none.

Facts & Assumptions

Given: The Axiom of Choice and an infinite cardinal $\kappa$ .

[L1]

If $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I}\kappa_i < \prod_{i \in I}\lambda_i$ (König's theorem: assuming the Axiom of Choice, if $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i$ ).

[L2]

For a constant family, $\prod_{i \in I}\kappa = \kappa^{\lvert I\rvert}$ ; and $\sum_{i \in I}\kappa_i = \lvert \bigcup_{i \in I}(\{i\} \times \kappa_i)\rvert$ (The sum $\sum_{i \in I} \kappa_i$ and the product $\prod_{i \in I} \kappa_i$ of an indexed family of cardinals, defined under the Axiom of Choice).

[L3]

[L4]

For cardinals $\kappa \le \lambda$ iff $\kappa \preceq \lambda$ ; $A \preceq B$ with both well-orderable gives $\lvert A\rvert \le \lvert B\rvert$ ; $\mu^{\nu} \le \mu^{\rho}$ for $\nu \le \rho$ and $\mu \ne 0$ ; and $(\mu^{\nu})^{\rho} = \mu^{\nu \otimes \rho}$ (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 $\lambda$ ).

[L5]

$\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ (Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ ).

[L6]

$\kappa < 2^{\kappa}$ , and $2^{\kappa}$ is a cardinal (Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$ , and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ ).

[L7]

Every infinite cardinal is a limit ordinal (claim (b) of Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense, Successor and limit ordinals), and a cardinal $\kappa$ is infinite exactly when $\omega \subseteq \kappa$ , that is $\omega \le \kappa$ (Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations).

[L8]

For a well-orderable set $X$ , $\lvert X\rvert$ is the least ordinal equinumerous with $X$ , $X \approx \lvert X\rvert$ , $\lvert \alpha\rvert \le \alpha$ , and $\lvert \alpha\rvert = \alpha$ exactly when $\alpha$ is a cardinal (A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L9]

Assuming the Axiom of Choice every set is well-orderable, and a product of nonempty sets is nonempty (The well-ordering theorem, The Axiom of Choice, Choice function).

[L10]

Ordinals satisfy trichotomy, $\alpha \subseteq \beta$ iff $\alpha \in \beta$ or $\alpha = \beta$ , the union of a set of ordinals is its least upper bound, and every nonempty set of ordinals has an $\in$ -least element; a function is injective when equality of two values forces equality of the corresponding inputs (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, Well-order and well-ordered set, Injection, surjection, bijection).

Proof

technique · direct

1.1

Put $\beta = \operatorname{cf}(\kappa)$ ; since $\kappa$ is a limit ordinal by [L7], [L3] makes $\beta$ an infinite cardinal with $\beta \le \kappa$ and supplies a strictly increasing cofinal $g : \beta \to \kappa$ ; set $\kappa_\xi = \lvert g(\xi)\rvert$ for $\xi \in \beta$ , so each $\kappa_\xi$ is a cardinal with $\kappa_\xi \le g(\xi) < \kappa$ by [L8], and $\kappa = \bigcup\{g(\xi) : \xi \in \beta\}$ , because $\zeta \in \kappa$ gives $\zeta \cup \{\zeta\} \in \kappa$ and hence $\zeta \cup \{\zeta\} \le g(\xi)$ for some $\xi$ , putting $\zeta \in g(\xi)$ .

L3L7L8L10

2.1

$\kappa \le \sum_{\xi \in \beta} \kappa_\xi$ : for each $\xi$ the set of bijections $g(\xi) \to \kappa_\xi$ is nonempty by [L8], so [L9] supplies such a $b_\xi$ for all $\xi$ at once, and $\zeta \mapsto (\xi_\zeta, b_{\xi_\zeta}(\zeta))$ , with $\xi_\zeta$ the $\in$ -least $\xi$ having $\zeta \in g(\xi)$ , is an injection of $\kappa$ into $\bigcup_{\xi \in \beta}(\{\xi\} \times \kappa_\xi)$ ; [L2] and [L4] then give the inequality.

step 1.1L2L4L8L9L10

3.1

Claim (a): applying [L1] to the families $(\kappa_\xi)_{\xi \in \beta}$ and the constant family $\lambda_\xi = \kappa$ , which satisfy $\kappa_\xi < \kappa$ by step 1.1, gives $\sum_{\xi \in \beta}\kappa_\xi < \prod_{\xi \in \beta}\kappa = \kappa^{\lvert \beta\rvert}$ by [L2], and $\lvert \beta\rvert = \beta = \operatorname{cf}(\kappa)$ by [L8], since $\beta$ is a cardinal; with step 2.1 this is $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ .

step 1.1step 2.1L1L2L8

4.1

Claim (b): put $\mu = 2^{\kappa}$ , an infinite cardinal by [L6] and [L7] since $\omega \le \kappa < \mu$ ; were $\operatorname{cf}(\mu) \le \kappa$ , then step 3.1 applied to $\mu$ together with [L4] and [L5] would give $\mu < \mu^{\operatorname{cf}(\mu)} \le \mu^{\kappa} = (2^{\kappa})^{\kappa} = 2^{\kappa \otimes \kappa} = 2^{\kappa} = \mu$ , which [L10] forbids; so trichotomy leaves $\kappa < \operatorname{cf}(2^{\kappa})$ .

step 3.1L4L5L6L7L10

5.1

Claim (c) is step 4.1 at $\kappa = \aleph_0 = \omega$ , which is an infinite cardinal by [L7].

step 3.1step 4.1L7∎

Remarks

What clause (b) rules out, concretely. If $2^{\aleph_0}$ were $\aleph_\omega$ then its cofinality would be $\aleph_0$ by $\aleph_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph $\aleph_{\alpha+1}$ is regular; $\operatorname{cf}(\aleph_\omega) = \aleph_0$ , so $\aleph_\omega$ is singular, and under choice it is the least singular infinite cardinal, contradicting clause (c); that refutation is carried out in FALSE: $2^{\aleph_0} = \aleph_\omega$ . The same test applies to any proposed value whose cofinality can be computed. Clause (c) is a restriction and not a determination: it excludes values and selects none.

Why the cofinality, and not the size, is the obstruction. Clause (a) says a cardinal is strictly smaller than itself raised to its own cofinality. Read contrapositively, a cardinal $\mu$ that is a power $2^{\kappa}$ cannot have its cofinality drop to or below $\kappa$ , because raising $\mu$ to that exponent would not increase it. The whole argument is the interaction of two facts, König's inequality and the second exponent law, and the second is where Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ enters.

Where the Axiom of Choice is spent here. Three times: in the definitions of $\sum$ , $\prod$ and $2^{\kappa}$ ; in step 2.1, to select a bijection $g(\xi) \to \lvert g(\xi)\rvert$ for every $\xi$ at once; and inside König's theorem: assuming the Axiom of Choice, if $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i$ itself. None of the three is removable by a canonical construction, which is why the whole corollary carries the hypothesis in its statement.

Depends on

Used by

FALSE: 2^ℵ₀ = ℵ_ω False statement
What each result on this page costs in choice, and where the continuum escapes what ZFC can decide Remark

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 128 results over 37 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Kearnes, Cardinal Arithmetic, König’s Corollary (standard reference, not scraped)
König's theorem (set theory) (Wikipedia) (standard reference, not scraped)
Cofinality (Wikipedia) (standard reference, not scraped)
Cardinality of the continuum (Wikipedia) (standard reference, not scraped)

Assuming the Axiom of Choice: κ<κcf⁡(κ)\kappa < \kappa^{\operatorname{cf}(\kappa)}κ<κcf(κ) for every infinite cardinal κ\kappaκ, and cf⁡(2κ)>κ\operatorname{cf}(2^{\kappa}) > \kappacf(2κ)>κ; in particular cf⁡(2ℵ0)>ℵ0\operatorname{cf}(2^{\aleph_0}) > \aleph_0cf(2ℵ0​)>ℵ0​