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.

An ordinal $\alpha$ with $\aleph_\alpha = \alpha$ , built as the supremum of the tower $\aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}, \dots$ , and its cofinality is $\aleph_0$

Example

Work in ZF; no choice principle is used. There is a tower of ordinals with

$T(0) = \aleph_0, \qquad T(n+1) = \aleph_{T(n)} \quad (n \in \omega),$

so that $T(1) = \aleph_{\aleph_0} = \aleph_\omega$ and $T(2) = \aleph_{\aleph_\omega}$ (The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ ). Put

$\alpha \;=\; \sup\{\, T(n) : n \in \omega \,\} \;=\; \bigcup\{\, T(n) : n \in \omega \,\}.$

Then

$\aleph_\alpha \;=\; \alpha, \qquad \operatorname{cf}(\alpha) = \aleph_0 ,$

so $\alpha$ is an infinite cardinal (Cardinal (initial ordinal) and cardinality) fixed by the aleph operation, and it is singular (Cofinality $\operatorname{cf}(\alpha)$ , and regular and singular cardinals).

So the aleph operation has fixed points, even though it is strictly increasing and satisfies $\beta \le \aleph_\beta$ at every ordinal (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 $\alpha \le \aleph_\alpha$ ). The power operation behaves differently: assuming the Axiom of Choice, so that $2^{\kappa}$ is a cardinal at all, $\kappa < 2^{\kappa}$ at every cardinal (Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$ , and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$ ), so there are no fixed points there at all. That comparison is an aside; nothing below uses it, and the example itself stays in ZF.

Facts & Assumptions

Given: ZF, with no choice principle. Write $n + 1 = n \cup \{n\}$ for the successor of $n \in \omega$ (The natural numbers $\mathbb{N}$ (von Neumann)).

[L1]

The operation $\beta \mapsto \aleph_\beta$ is defined at every ordinal, takes infinite cardinal values, is strictly increasing, satisfies $\beta \le \aleph_\beta$ , and is continuous at limits, meaning $\aleph_\lambda = \bigcup\{\aleph_\beta : \beta \in \lambda\}$ for limit $\lambda$ (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 $\alpha \le \aleph_\alpha$ , The successor cardinal $\kappa^{+}$ , the alephs $\aleph_\alpha$ , the beths $\beth_\alpha$ , successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ ).

[L2]

A class rule $G$ on functions whose domain is an ordinal determines exactly one class function $T$ defined at every ordinal with $T(\beta) = G(T \restriction \beta)$ (Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal).

[L3]

Transfinite induction is valid on any well-order, in particular on $(\omega, \in)$ (Transfinite induction, Trichotomy and well-ordering of the ordinals).

[L4]

$\omega$ is the least limit ordinal, is closed under successor, and its elements are exactly the ordinals below it, with $n + 1 = \{0, \dots, n\}$ ( $\omega$ is the least limit ordinal, Successor and limit ordinals, Ordinal (von Neumann), The natural numbers $\mathbb{N}$ (von Neumann)).

[L5]

For a set $D$ of ordinals, $\bigcup D$ is an ordinal and the least upper bound of $D$ ; ordinals satisfy trichotomy; $\alpha \subseteq \beta$ iff $\alpha \in \beta$ or $\alpha = \beta$ ; $\alpha \notin \alpha$ (Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

[L6]

[L7]

An injective map onto its range is a bijection to that range; for a well-orderable set $X$ , $\lvert X\rvert$ is the least ordinal equinumerous with $X$ , and equinumerous sets receive the same cardinal (Injection, surjection, bijection, 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]

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).

Verification

technique · direct

1.1

Apply [L2] to the rule sending a function $h$ whose domain is an ordinal to $\aleph_{\bigcup \operatorname{ran}(h)}$ , which is given by a formula; this yields exactly one class function $T$ , defined at every ordinal, with $T(\beta) = \aleph_{\bigcup \operatorname{ran}(T \restriction \beta)}$ , and in particular $T(0) = \aleph_{\bigcup \varnothing} = \aleph_0$ .

L1L2L5

2.1

By induction along [L3] on $n \in \omega$ , the statement " $T(k) \in T(k+1)$ for every $k \le n$ , and $T(n+1) = \aleph_{T(n)}$ " holds for every $n$ . At $n = 0$ : $\operatorname{ran}(T \restriction 1) = \{T(0)\}$ by [L4], so $T(1) = \aleph_{T(0)} = \aleph_{\aleph_0}$ , and $0 \in \aleph_0$ gives $\aleph_0 \in \aleph_{\aleph_0}$ by the strict increase in [L1], that is $T(0) \in T(1)$ . At $n+1$ : the statement at $n$ makes $T(0) \subseteq \cdots \subseteq T(n+1)$ by [L5], so $\bigcup \operatorname{ran}(T \restriction (n+2)) = T(n+1)$ and $T(n+2) = \aleph_{T(n+1)}$ ; and $T(n) \in T(n+1)$ together with strict increase gives $\aleph_{T(n)} \in \aleph_{T(n+1)}$ , that is $T(n+1) \in T(n+2)$ .

step 1.1L1L3L4L5

3.1

Replacement makes $C = \{T(n) : n \in \omega\}$ a set of ordinals, so $\alpha = \bigcup C$ is an ordinal and the least upper bound of $C$ by [L5]; and $\alpha$ is a limit ordinal, since $\alpha \ne 0$ because $\aleph_0 = T(0) \subseteq \alpha$ , and $\alpha$ is not a successor because step 2.1 gives $T(n) \in T(n+1) \subseteq \alpha$ for every $n$ , so no member of $C$ is largest and no ordinal below $\alpha$ is an upper bound of $C$ .

step 2.1L1L4L5

4.1

$\aleph_\alpha = \alpha$ : continuity in [L1] at the limit ordinal $\alpha$ gives $\aleph_\alpha = \bigcup\{\aleph_\beta : \beta \in \alpha\}$ ; each $\beta \in \alpha$ lies in some $T(n)$ by [L5], so strict increase gives $\aleph_\beta \in \aleph_{T(n)} = T(n+1) \subseteq \alpha$ using step 2.1, whence $\aleph_\alpha \subseteq \alpha$ ; and $\alpha \subseteq \aleph_\alpha$ is the inequality $\beta \le \aleph_\beta$ of [L1] at $\beta = \alpha$ .

step 2.1step 3.1L1L5

4.2

$\operatorname{cf}(\alpha) = \aleph_0$ : the set $C$ is cofinal in $\alpha$ , since $\alpha = \bigcup C$ makes every $\zeta \in \alpha$ a member of some $T(n)$ and hence $\zeta \le T(n)$ ; and $\lvert C \rvert = \aleph_0$ , because $n \mapsto T(n)$ is injective by step 2.1 and [L5], so $C \approx \omega$ and [L7] applies; therefore $\operatorname{cf}(\alpha) \le \aleph_0$ by [L6], while $\operatorname{cf}(\alpha)$ is an infinite cardinal by [L6] and step 3.1, hence $\aleph_0 \le \operatorname{cf}(\alpha)$ by [L8].

step 2.1step 3.1L5L6L7L8

5.1

So $\alpha = \aleph_\alpha$ is an infinite cardinal by [L1] and step 4.1, it is fixed by the aleph operation, and $\operatorname{cf}(\alpha) = \aleph_0 < \alpha$ by step 4.2 and [L5], so it is singular.

step 4.1step 4.2L1L5∎

Remarks

Why a fixed point is not a contradiction. $\beta \le \aleph_\beta$ holds everywhere and the operation is strictly increasing, but neither forces $\beta < \aleph_\beta$ : strict increase compares the values at two different indices and says nothing about the value at one index. The tower converges exactly because at a limit the value is the supremum of the earlier ones, and the supremum of the tower is what the tower was climbing towards.

Nothing is chosen, and nothing beyond Replacement is used. The tower is a definable $\omega$ -indexed family, Replacement makes its range a set, and every step of the verification is a computation. So the whole example is a theorem of ZF, like the aleph hierarchy it is built from.

Its cofinality is the smallest an infinite cardinal can have. $\operatorname{cf}(\alpha) = \aleph_0$ says $\alpha$ is reached by an $\omega$ -indexed family, which is exactly how it was built. So this fixed point is singular, and, assuming the Axiom of Choice, by Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$ , and $\operatorname{cf}(2^{\kappa}) > \kappa$ ; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ it is therefore not a possible value of $2^{\aleph_0}$ . Nothing above claims it is the least fixed point.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 113 results over 36 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

MATH 5001, Fixed Points of the Aleph Sequence (standard reference, not scraped)
Aleph number — fixed points (Wikipedia) (standard reference, not scraped)
Cofinality (Wikipedia) (standard reference, not scraped)

An ordinal α\alphaα with ℵα=α\aleph_\alpha = \alphaℵα​=α, built as the supremum of the tower ℵ0,ℵℵ0,ℵℵℵ0,…\aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}, \dotsℵ0​,ℵℵ0​​,ℵℵℵ0​​​,…, and its cofinality is ℵ0\aleph_0ℵ0​