ExampleConstruction: AI-adaptedVerification: 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.

Assuming countable choice, $\operatorname{cf}(\aleph_{\omega_1}) = \aleph_1$ , so singular does not mean of countable cofinality

Example

Assume the Axiom of Countable Choice $\mathrm{AC}_\omega$ (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )). Then

$\operatorname{cf}(\aleph_{\omega_1}) \;=\; \aleph_1 \;<\; \aleph_{\omega_1},$

so $\aleph_{\omega_1}$ is singular (Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals) and its cofinality is uncountable (Finite, countably infinite, countable, uncountable).

This separates two conditions that the first singular example runs together. A singular cardinal is one reachable from below by a strictly shorter family; it need not be reachable by a countable one. Here the reaching family has length $\omega_1$ and no shorter one will do, and what rules out a shorter one is the boundedness theorem for $\omega_1$ (Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable), which is where $\mathrm{AC}_\omega$ is spent.

Facts & Assumptions

Given: The Axiom of Countable Choice. Write $\omega_1$ for the first uncountable ordinal (The first uncountable ordinal $\omega_1 := \aleph(\omega)$ ).

[L1]

$\aleph_\lambda = \bigcup\{\aleph_\beta : \beta \in \lambda\}$ at a limit ordinal $\lambda$ ; every $\aleph_\alpha$ is an infinite cardinal; the operation is strictly increasing; $\aleph_1 = \omega_1$ (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 $\alpha \le \aleph_\alpha$ ).

[L2]

$\omega_1$ is uncountable, is a cardinal, is a limit ordinal, and every ordinal in it is at most countable ( $\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF, Successor and limit ordinals).

[L3]

$\aleph_1 = \aleph_0^{+}$ is the least cardinal strictly above $\aleph_0$ (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, The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ ).

[L4]

Assuming $\mathrm{AC}_\omega$ , every at most countable $A \subseteq \omega_1$ is bounded below $\omega_1$ : $\sup A = \bigcup A$ lies in $\omega_1$ and dominates every member of $A$ (Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$ , so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).

[L5]

A nonempty set is at most countable if and only if it is a surjective image of $\mathbb{N}$ (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$ , Finite, countably infinite, countable, uncountable).

[L6]

[L7]

For a well-orderable set $X$ , $\lvert X\rvert$ is the least ordinal equinumerous with $X$ , equinumerous sets receive the same one, 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, Cardinal (initial ordinal) and cardinality, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L8]

Every infinite cardinal is a limit ordinal, and a cardinal is infinite exactly when $\omega \le \kappa$ (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, $\omega$ is the least limit ordinal).

[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; and every strictly increasing map of ordinals is injective (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, Injection, surjection, bijection).

Verification

technique · direct

1.1

The set $C = \{\aleph_\alpha : \alpha \in \omega_1\}$ exists by Replacement and is cofinal in $\aleph_{\omega_1}$ : $\omega_1$ is a limit ordinal by [L2], so $\aleph_{\omega_1} = \bigcup C$ by [L1] and every $\zeta \in \aleph_{\omega_1}$ lies in some $\aleph_\alpha \in C$ ; moreover $\alpha \mapsto \aleph_\alpha$ is injective by the strict increase in [L1], so $C \approx \omega_1$ and $\lvert C\rvert = \lvert \omega_1\rvert = \omega_1 = \aleph_1$ by [L7], [L2] and [L1].

L1L2L7L9

1.2

The cofinality is not smaller than $\aleph_1$ . Put $\beta = \operatorname{cf}(\aleph_{\omega_1})$ ; it is an infinite cardinal by [L6] and [L8], since $\aleph_{\omega_1}$ is an infinite cardinal and hence a limit ordinal. If $\beta < \aleph_1$ then $\beta \le \aleph_0$ by [L3], so $\beta = \aleph_0 = \omega$ by [L8], and [L6] supplies a cofinal $f : \omega \to \aleph_{\omega_1}$ . For each $n \in \omega$ let $\alpha_n$ be the $\in$ -least $\alpha \in \omega_1$ with $f(n) \in \aleph_\alpha$ , which exists by [L1] and [L9] and is determined rather than chosen. Then $A = \{\alpha_n : n \in \omega\}$ is a nonempty at most countable subset of $\omega_1$ by [L5], so $\gamma = \sup A \in \omega_1$ by [L4], and every $f(n)$ lies in $\aleph_{\alpha_n} \subseteq \aleph_\gamma$ by [L1] and [L9]. But $\aleph_\gamma \in \aleph_{\omega_1}$ , and cofinality of $f$ would give some $n$ with $\aleph_\gamma \le f(n) \in \aleph_\gamma$ , which [L9] forbids. So $\aleph_1 \le \beta$ .

L1L2L3L4L5L6L8L9

2.1

By [L6] applied to the cofinal set of step 1.1, $\operatorname{cf}(\aleph_{\omega_1}) \le \lvert C\rvert = \aleph_1$ .

step 1.1L6

3.1

Steps 2.1 and 1.2 give $\operatorname{cf}(\aleph_{\omega_1}) = \aleph_1$ by [L9]; and $\aleph_1 < \aleph_{\omega_1}$ by the strict increase in [L1], since $1 \in \omega_1$ by [L2], so $\aleph_{\omega_1}$ is singular with uncountable cofinality by [L2].

step 1.2step 2.1L1L2L9∎

Remarks

Why countable choice appears, and where exactly. It is used once, at [L4]: without it, $\omega_1$ can consistently be the supremum of an $\omega$ -indexed family of countable ordinals, and then the argument of step 1.2 collapses. That dependence is inherited, not introduced here — the published boundedness theorem carries the same hypothesis, and states so in its own title.

What "singular" does and does not mean. Singular says only $\operatorname{cf}(\kappa) \ne \kappa$ . The singular cardinal computed in $\operatorname{cf}(\aleph_\omega) = \aleph_0$ , computed from the cofinal map $n \mapsto \aleph_n$ has countable cofinality, and a reader who meets only that example may take the two conditions to be the same. They are not: here the cofinality is $\aleph_1$ , uncountable, while the cardinal is still singular because $\aleph_1$ is far below $\aleph_{\omega_1}$ .

The pattern behind both computations. For a limit ordinal $\lambda$ the family $\alpha \mapsto \aleph_\alpha$ restricted to $\lambda$ is cofinal in $\aleph_\lambda$ , so $\operatorname{cf}(\aleph_\lambda) \le \lvert \lambda\rvert$ always. The work is entirely in the lower bound, and it is a statement about $\lambda$ rather than about $\aleph_\lambda$ : it asks how short a family can be and still reach $\lambda$ .

Depends on

Used by

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Sources

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

Assuming countable choice, cf⁡(ℵω1)=ℵ1\operatorname{cf}(\aleph_{\omega_1}) = \aleph_1cf(ℵω1​​)=ℵ1​, so singular does not mean of countable cofinality