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

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.

‡ 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.

$\aleph_0$ 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

Statement

Let $\operatorname{cf}$ , regular and singular be as in Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals, and let $\aleph_\alpha$ be as in The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ . Then:

(a) In ZF. $\operatorname{cf}(\aleph_0) = \aleph_0$ , so $\aleph_0$ is regular.

(b) Assuming the Axiom of Choice (The Axiom of Choice). $\aleph_{\alpha+1}$ is regular for every ordinal $\alpha$ .

(c) In ZF. $\operatorname{cf}(\aleph_\omega) = \aleph_0$ , and $\aleph_0 < \aleph_\omega$ , so $\aleph_\omega$ is singular.

(d) Assuming the Axiom of Choice. Every infinite cardinal below $\aleph_\omega$ is regular, so $\aleph_\omega$ is the least singular infinite cardinal.

Clause (b) is where the Axiom of Choice becomes indispensable, and the hypothesis is not decoration. The proof spends it once, to select an injection $g(\xi) \to \aleph_\alpha$ for each $\xi$ below the cofinality, and there is no canonical such family to fall back on: the sets $g(\xi)$ are ordinals, but the injections are not determined by them. Clauses (a), (c) and the classification half of (d) are choice free.

Facts & Assumptions

Given: ZF, with the Axiom of Choice assumed only in clauses (b) and (d). Throughout, a map is called cofinal when its range is a cofinal subset of the target (Cofinal subset of an ordinal).

[L1]

[L2]

$\operatorname{cf}(\alpha)$ is the least length of a cofinal map into $\alpha$ , and a strictly increasing cofinal map of that length exists (Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals, For every ordinal $\alpha$ there is a least ordinal $\beta$ admitting a map $\beta \to \alpha$ with cofinal range, and that map may always be taken strictly increasing).

[L3]

$\aleph_0 = \omega$ ; $\aleph_{\alpha+1} = \aleph_\alpha^{+}$ is the least cardinal strictly above $\aleph_\alpha$ ; $\aleph_\lambda = \bigcup\{\aleph_\beta : \beta \in \lambda\}$ at limits; every $\aleph_\alpha$ is an infinite cardinal and the operation is strictly increasing (The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ , The clauses at $0$ , at a successor and at a limit determine exactly one operation $\alpha \mapsto \aleph_\alpha$ , in ZF, and — assuming the Axiom of Choice — exactly one operation $\alpha \mapsto \beth_\alpha$ ; each value is an infinite cardinal, each is strictly increasing and continuous at limits, and $\alpha \le \aleph_\alpha$ , For every set $A$ the Hartogs number $\aleph(A)$ is a cardinal, and for every cardinal $\kappa$ it is the least cardinal strictly above $\kappa$ ; this is a theorem of ZF).

[L4]

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, Cardinal (initial ordinal) and cardinality).

[L5]

For an infinite cardinal $\mu$ and a cardinal $\nu$ with $0 \ne \nu \le \mu$ , $\mu \otimes \nu = \mu$ (Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$ , $\kappa \oplus \lambda = \kappa$ , and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$ , Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$ , proved in ZF from the canonical well-order of $\kappa \times \kappa$ , Cardinal sum $\kappa \oplus \lambda$ , product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$ , and why they are written apart from the ordinal operations).

[L6]

For cardinals $\kappa \le \lambda$ iff $\kappa \preceq \lambda$ ; $A \preceq B$ with both well-orderable gives $\lvert A\rvert \le \lvert B\rvert$ ; $\otimes$ is monotone (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 $\lambda$ ).

[L7]

For a well-orderable set $X$ , $\lvert X\rvert$ is the least ordinal equinumerous with $X$ , satisfies $X \approx \lvert X\rvert$ and $\lvert \alpha\rvert \le \alpha$ , and equals $\alpha$ 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 $A \preceq B$ ).

[L8]

Every family of nonempty sets has a choice function (The Axiom of Choice, Choice function).

[L9]

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, every nonempty set of ordinals has an $\in$ -least element, $\omega$ is the least limit ordinal, and every strictly increasing map of ordinals is injective (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, $\omega$ is the least limit ordinal, Injection, surjection, bijection).

[L10]

Every infinite cardinal is $\aleph_\alpha$ for exactly one ordinal $\alpha$ (Every infinite cardinal is $\aleph_\alpha$ for exactly one ordinal $\alpha$ , in ZF; and, assuming the Axiom of Choice, every infinite set is equinumerous with exactly one aleph).

Proof

technique · direct

1.1

Claim (a): $\omega$ is a limit ordinal by [L9], so $\operatorname{cf}(\omega)$ is an infinite cardinal by [L1], hence $\omega \le \operatorname{cf}(\omega)$ by [L4]; and $\operatorname{cf}(\omega) \le \omega$ by [L1], so $\operatorname{cf}(\aleph_0) = \aleph_0$ .

L1L3L4L9

1.2

The set $C = \{\aleph_n : n \in \omega\}$ exists by Replacement, is contained in $\aleph_\omega$ and is cofinal in it, since $\aleph_\omega = \bigcup C$ by [L3] means every $\zeta \in \aleph_\omega$ lies in some $\aleph_n$ and hence satisfies $\zeta \le \aleph_n$ ; moreover $n \mapsto \aleph_n$ is injective by the strict increase in [L3], so $C \approx \omega$ and $\lvert C\rvert = \aleph_0$ by [L7].

L3L7L9

1.3

Setting up claim (b): let $\lambda = \aleph_\alpha$ , $\kappa = \aleph_{\alpha+1}$ , and suppose $\beta = \operatorname{cf}(\kappa) < \kappa$ , with $g : \beta \to \kappa$ strictly increasing and cofinal by [L2]; then $\kappa$ is a limit ordinal by [L4], so $\beta$ is an infinite cardinal by [L1], and $\beta \le \lambda$ because $\beta$ is a cardinal below the least cardinal strictly above $\lambda$ ([L3]); moreover $\kappa = \bigcup\{g(\xi) : \xi \in \beta\}$ , since for $\zeta \in \kappa$ the ordinal $\zeta \cup \{\zeta\}$ also lies in $\kappa$ and is $\le g(\xi)$ for some $\xi$ , putting $\zeta \in g(\xi)$ .

L1L2L3L4L9

2.1

Claim (c): step 1.2 and [L1] give $\operatorname{cf}(\aleph_\omega) \le \aleph_0$ ; and $\aleph_\omega$ is an infinite cardinal, hence a limit ordinal by [L4], so $\operatorname{cf}(\aleph_\omega)$ is an infinite cardinal and $\aleph_0 \le \operatorname{cf}(\aleph_\omega)$ by [L1] and [L4]; therefore $\operatorname{cf}(\aleph_\omega) = \aleph_0 < \aleph_\omega$ by the strict increase in [L3], and $\aleph_\omega$ is singular.

step 1.2L1L3L4

2.2

Claim (b): each $g(\xi)$ of step 1.3 lies in $\kappa$ , so $\lvert g(\xi)\rvert$ is a cardinal below $\kappa$ and hence $\lvert g(\xi)\rvert \le \lambda$ by [L3] and [L7], and the set $I_\xi$ of injections $g(\xi) \to \lambda$ is nonempty; a choice function from [L8] on $\{I_\xi : \xi \in \beta\}$ supplies injections $e_\xi : g(\xi) \to \lambda$ for all $\xi$ at once, and $\zeta \mapsto (\xi_\zeta, e_{\xi_\zeta}(\zeta))$ , with $\xi_\zeta$ the $\in$ -least $\xi$ having $\zeta \in g(\xi)$ , is then an injection $\kappa \to \beta \times \lambda$ ; so $\kappa \le \beta \otimes \lambda = \lambda$ by [L6] and [L5], contradicting $\lambda < \kappa$ , and therefore $\operatorname{cf}(\kappa) = \kappa$ .

step 1.3L5L6L7L8L9

3.1

Claim (d) and the conclusion: an infinite cardinal $\kappa < \aleph_\omega$ is $\aleph_\alpha$ for exactly one $\alpha$ by [L10], and $\alpha \in \omega$ , since $\omega \le \alpha$ would give $\aleph_\omega \le \aleph_\alpha$ by the strict increase in [L3]; so $\kappa$ is $\aleph_0$ , regular by step 1.1, or $\aleph_{n+1}$ for some $n \in \omega$ , regular by step 2.2; with step 2.1 this makes $\aleph_\omega$ the least singular infinite cardinal.

step 1.1step 2.1step 2.2L3L9L10∎

Remarks

Why regularity of $\aleph_1$ is not a theorem of ZF. The proof of clause (b) selects one injection for each $\xi$ below the cofinality, and that selection is the entire content of the choice hypothesis: a union of countably many countable sets is not provably countable in ZF, and the same phenomenon is what would otherwise force $\operatorname{cf}(\aleph_1) = \aleph_1$ . The choice ledger at the end of this page records how far this can fail.

Why $\aleph_\omega$ is singular for a completely different reason. Nothing is chosen in clause (c): the map $n \mapsto \aleph_n$ is definable, and it is short and cofinal simply because the index $\omega$ is a limit ordinal reached from below in $\omega$ steps. Singularity of $\aleph_\omega$ is therefore a fact about the index, not about the size, and clause (c) holds in ZF.

What "least singular" means and what it does not. Clause (d) locates $\aleph_\omega$ among the alephs: everything below it is regular, under choice. It says nothing about which cardinals above $\aleph_\omega$ are singular, and it says nothing about $2^{\aleph_0}$ , whose position in the aleph hierarchy is not determined by anything on this page. What is determined is a constraint on that position, and it is Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$ , and $\operatorname{cf}(2^{\kappa}) > \kappa$ ; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ .

Depends on

Used by

cf(ℵ_ω) = ℵ₀, computed from the cofinal map n ↦ ℵₙ Example
FALSE: 2^ℵ₀ = ℵ_ω False statement
FALSE: ℵ_α is regular for every ordinal α 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 38 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

UCL, Axiomatic Set Theory, Ch. 4: Cardinal Arithmetic (standard reference, not scraped)
Regular cardinal (Wikipedia) (standard reference, not scraped)
Cofinality (Wikipedia) (standard reference, not scraped)