Alphabeta Math
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.

cf(α)α\operatorname{cf}(\alpha) \le \alpha; cf(0)=0\operatorname{cf}(0) = 0 and cf(α+1)=1\operatorname{cf}(\alpha + 1) = 1; for a limit ordinal λ\lambda the value cf(λ)\operatorname{cf}(\lambda) is an infinite cardinal with cf(cf(λ))=cf(λ)\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda), so it is regular; and every cofinal subset of λ\lambda has cardinality at least cf(λ)\operatorname{cf}(\lambda), a value that is attained

Statement

Work in ZF; no choice principle is used. Let cf\operatorname{cf} be the cofinality of Cofinality cf(α)\operatorname{cf}(\alpha), and regular and singular cardinals. Then:

(a) cf(α)α\operatorname{cf}(\alpha) \le \alpha for every ordinal α\alpha (Ordinal (von Neumann));

(b) cf(0)=0\operatorname{cf}(0) = 0, and cf(α+1)=1\operatorname{cf}(\alpha + 1) = 1 for every ordinal α\alpha, where α+1=α{α}\alpha + 1 = \alpha \cup \{\alpha\} (Ordinal addition α+β\alpha + \beta);

(c) for a limit ordinal λ\lambda (Successor and limit ordinals), cf(λ)\operatorname{cf}(\lambda) is an infinite cardinal (Cardinal (initial ordinal) and cardinality) and cf(cf(λ))=cf(λ)\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda), so cf(λ)\operatorname{cf}(\lambda) is a regular cardinal;

(d) for a limit ordinal λ\lambda, every cofinal CλC \subseteq \lambda (Cofinal subset of an ordinal) satisfies cf(λ)C\operatorname{cf}(\lambda) \le \lvert C \rvert, and some cofinal subset of λ\lambda has cardinality exactly cf(λ)\operatorname{cf}(\lambda).

Clause (c) is what discharges the naming obligation of Cofinality cf(α)\operatorname{cf}(\alpha), and regular and singular cardinals: "regular" is defined through cf\operatorname{cf}, and it is a theorem, not a convention, that cf\operatorname{cf} of a limit ordinal is a cardinal at which the definition can be tested.

Facts & Assumptions

Given: ZF, with no choice principle. Throughout, a map f:βαf : \beta \to \alpha is called cofinal when f[β]f[\beta] is cofinal in α\alpha.

[L1]

cf(α)\operatorname{cf}(\alpha) is the least ordinal β\beta admitting a cofinal f:βαf : \beta \to \alpha; for that β\beta a strictly increasing cofinal g:βαg : \beta \to \alpha exists (Cofinality cf(α)\operatorname{cf}(\alpha), and regular and singular cardinals, For every ordinal α\alpha there is a least ordinal β\beta admitting a map βα\beta \to \alpha with cofinal range, and that map may always be taken strictly increasing).

[L2]

CαC \subseteq \alpha is cofinal when every ζα\zeta \in \alpha has some ηC\eta \in C with ζη\zeta \le \eta (Cofinal subset of an ordinal).

[L3]

Ordinals: trichotomy; αβ\alpha \subseteq \beta iff αβ\alpha \in \beta or α=β\alpha = \beta; αα\alpha \notin \alpha; every element of an ordinal is an ordinal; every set of ordinals is well ordered by \in (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, Well-order and well-ordered set).

[L4]

Every ordinal is exactly one of 00, a successor, or a limit, and ω\omega is the least limit ordinal (Successor and limit ordinals, ω\omega is the least limit ordinal).

[L5]

For a well-orderable XX: XXX \approx \lvert X\rvert, the value is a cardinal, equinumerous sets receive the same one, and α=α\lvert \alpha\rvert = \alpha exactly when α\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).

[L6]

For cardinals κλ\kappa \le \lambda iff κλ\kappa \preceq \lambda, and ABA \preceq B with both well-orderable gives AB\lvert A\rvert \le \lvert B\rvert (claim (a) of Commutativity, associativity, distributivity and monotonicity of \oplus and \otimes, the unit laws, the two exponent laws, and κλ\kappa \le \lambda if and only if κ\kappa injects into λ\lambda).

[L7]
[L8]

Precomposing a function with a bijection onto its domain leaves its range unchanged, since ghg \circ h has image g[h[μ]]=g[β]g[h[\mu]] = g[\beta] when h:μβh : \mu \to \beta is onto; a strictly increasing map of ordinals is injective, and satisfies ξηg(ξ)g(η)\xi \le \eta \Rightarrow g(\xi) \le g(\eta), both by trichotomy (Injection, surjection, bijection, Trichotomy and well-ordering of the ordinals).

Proof

technique · direct
1.1

Claim (a): the identity αα\alpha \to \alpha is cofinal by [L2], so the least length in [L1] is at most α\alpha.

L1L2
1.2

Claim (b): for α=0\alpha = 0 the empty map 000 \to 0 is cofinal vacuously, so cf(0)=0\operatorname{cf}(0) = 0; for α+1\alpha + 1 the map 0α0 \mapsto \alpha is cofinal, since every ζα+1\zeta \in \alpha + 1 satisfies ζα\zeta \le \alpha by [L3], while the empty map into the nonempty α+1\alpha + 1 is not, so cf(α+1)=1\operatorname{cf}(\alpha + 1) = 1.

L1L2L3
1.3

Let λ\lambda be a limit ordinal, β=cf(λ)\beta = \operatorname{cf}(\lambda) and g:βλg : \beta \to \lambda strictly increasing and cofinal by [L1]; then β\beta is a limit ordinal, so ωβ\omega \le \beta by [L4]: β0\beta \ne 0 because 0λ0 \in \lambda and the empty range is not cofinal, and β=γ+1\beta = \gamma + 1 is impossible, since then g(η)g(γ)g(\eta) \le g(\gamma) for all ηβ\eta \in \beta by [L8], so cofinality would give λg(γ)+1\lambda \subseteq g(\gamma) + 1 while g(γ)λg(\gamma) \in \lambda gives g(γ)+1λg(\gamma) + 1 \subseteq \lambda, making λ=g(γ)+1\lambda = g(\gamma) + 1 a successor.

L1L2L3L4L8
1.4

With λ\lambda, β\beta, gg as above, β\beta is a cardinal: if μ=ββ\mu = \lvert \beta\rvert \in \beta then a bijection h:μβh : \mu \to \beta makes gh:μλg \circ h : \mu \to \lambda a map with the same range as gg, hence cofinal, so the least length would be at most μβ\mu \in \beta, contradicting β=cf(λ)\beta = \operatorname{cf}(\lambda); so β=β\lvert \beta\rvert = \beta and [L5] applies.

L1L2L5L8
2.1

Claim (c): β\beta is an infinite cardinal by steps 1.3, 1.4 and [L9]; and writing γ=cf(β)\gamma = \operatorname{cf}(\beta), step 1.1 gives γβ\gamma \le \beta, while a cofinal k:γβk : \gamma \to \beta makes gk:γλg \circ k : \gamma \to \lambda cofinal — given ζλ\zeta \in \lambda pick ξβ\xi \in \beta with ζg(ξ)\zeta \le g(\xi), then ργ\rho \in \gamma with ξk(ρ)\xi \le k(\rho), and g(ξ)g(k(ρ))g(\xi) \le g(k(\rho)) by [L8] — so β=cf(λ)γ\beta = \operatorname{cf}(\lambda) \le \gamma and therefore cf(cf(λ))=cf(λ)\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda).

step 1.1step 1.3step 1.4L1L2L8L9
3.1

Claim (d): a cofinal CλC \subseteq \lambda is a set of ordinals, well ordered by \in by [L3], with order type δ\delta and an order isomorphism e:δCe : \delta \to C by [L7]; then ee is a cofinal map δλ\delta \to \lambda, so βδ\beta \le \delta by [L1], and applying [L5] and [L6] gives β=βδ=C\beta = \lvert \beta\rvert \le \lvert \delta\rvert = \lvert C\rvert using step 2.1; conversely g[β]g[\beta] is cofinal with g[β]=β\lvert g[\beta]\rvert = \beta, since gg is injective by [L8].

step 2.1L1L2L3L5L6L7L8
4.1

Claims (a), (b), (c) and (d) are established, in ZF.

step 1.1step 1.2step 2.1step 3.1

Remarks

Why (c) is restricted to limit ordinals. At 00 and at a successor the cofinality is 00 or 11, neither of which is an infinite cardinal, and the regular/singular vocabulary is not applied there. Since every infinite cardinal is a limit ordinal, the restriction costs nothing where the notion is used.

What clause (d) is for. It converts a cofinality question into a counting question: to show cf(λ)κ\operatorname{cf}(\lambda) \le \kappa it suffices to exhibit any cofinal subset of size κ\kappa, with no attention to its order type. That is how every cofinality on the companion page is computed, and the attainment half is what makes the bound sharp.

Where the strictly increasing witness is spent. Three times, and each time essentially: in step 1.3, to know that a witness of successor length would have a largest value; in step 2.1, to know that gg preserves \le, without which the composite gkg \circ k need not be cofinal; and in step 3.1, to know that gg is injective, without which g[β]g[\beta] need not have cardinality β\beta. That is why For every ordinal α\alpha there is a least ordinal β\beta admitting a map βα\beta \to \alpha with cofinal range, and that map may always be taken strictly increasing proves claim (b) rather than stopping at the existence of a least length.

Depends on

Used by

Cited to discharge well-definedness by Cofinality cf(α), and regular and singular cardinals.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 88 results over 35 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