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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (claude-sonnet-5 + deepseek-v4-pro)verified 2026-08-05 (claude-sonnet-5) rests on unproved material (inherited)
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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.

For every set AA the Hartogs number (A)\aleph(A) is a cardinal, and for every cardinal κ\kappa it is the least cardinal strictly above κ\kappa; this is a theorem of ZF

Statement

Work in ZF; no choice principle is used. For a set AA let (A)\aleph(A) be its Hartogs number (Hartogs: an ordinal that does not inject into a given set), the least ordinal (Ordinal (von Neumann)) admitting no injection into AA. Then:

(a) (A)\aleph(A) is a cardinal (Cardinal (initial ordinal) and cardinality), for every set AA, well-orderable or not.

(b) If κ\kappa is a cardinal then κ<(κ)\kappa < \aleph(\kappa), and every cardinal λ\lambda with κ<λ\kappa < \lambda satisfies (κ)λ\aleph(\kappa) \le \lambda. So (κ)\aleph(\kappa) is the least cardinal strictly above κ\kappa.

(c) If κ\kappa is an infinite cardinal then so is (κ)\aleph(\kappa).

What this supplies, and what it does not. It gives a successor operation on cardinals in ZF alone: there is always a next one, and it is definable. It says nothing about how large the next cardinal is compared with 2κ2^{\kappa}; that comparison is not decided by the axioms in use here.

Facts & Assumptions

Given: A set AA and a cardinal κ\kappa, in ZF, with no choice principle.

[L1]

(A)\aleph(A) is an ordinal that does not inject into AA, and it is the least such; consequently every ordinal β(A)\beta \in \aleph(A) does inject into AA (Hartogs: an ordinal that does not inject into a given set).

[L2]

An ordinal κ\kappa is a cardinal when no ακ\alpha \in \kappa satisfies ακ\alpha \approx \kappa (Cardinal (initial ordinal) and cardinality, Equinumerous sets, ABA \approx B and ABA \preceq B).

[L4]

Ordinals satisfy trichotomy, αβ\alpha \subseteq \beta iff αβ\alpha \in \beta or α=β\alpha = \beta, and every element of an ordinal is an ordinal (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals).

[L5]

A composition of an injection with a bijection is an injection, a subset inclusion is an injection, and the identity is a bijection (Injection, surjection, bijection).

[L6]

For a well-orderable XX, X=X\lvert X\rvert = X when XX is a cardinal, and αα\lvert \alpha\rvert \le \alpha (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).

Proof

technique · direct
1.1

Every β(A)\beta \in \aleph(A) injects into AA, and (A)\aleph(A) itself does not.

L1
1.2

If λ\lambda is a cardinal with κ<λ\kappa < \lambda then λ\lambda does not inject into κ\kappa: an injection λκ\lambda \to \kappa would give λκ\lambda \le \kappa by [L3], contradicting κλ\kappa \in \lambda and trichotomy.

L3L4
2.1

Claim (a): if β(A)\beta \in \aleph(A) had β(A)\beta \approx \aleph(A), then composing a bijection (A)β\aleph(A) \to \beta with an injection βA\beta \to A supplied by step 1.1 would inject (A)\aleph(A) into AA, which step 1.1 forbids; so no element of (A)\aleph(A) is equinumerous with it and (A)\aleph(A) is a cardinal by [L2].

step 1.1L2L5
2.2

First half of claim (b): κ\kappa injects into κ\kappa by the identity, so κ(κ)\kappa \ne \aleph(\kappa) by step 1.1; and (κ)κ\aleph(\kappa) \in \kappa is impossible, since then (κ)κ\aleph(\kappa) \subseteq \kappa by [L4] and the inclusion would inject it into κ\kappa; trichotomy leaves κ(κ)\kappa \in \aleph(\kappa), that is κ<(κ)\kappa < \aleph(\kappa).

step 1.1L4L5
3.1

Second half of claim (b) and claim (c): a cardinal λ\lambda with κ<λ\kappa < \lambda does not inject into κ\kappa by step 1.2, so (κ)λ\aleph(\kappa) \le \lambda by the minimality in [L1]; with step 2.2 and step 2.1 this makes (κ)\aleph(\kappa) the least cardinal strictly above κ\kappa; and if ωκ\omega \le \kappa then ωκ<(κ)\omega \le \kappa < \aleph(\kappa) by step 2.2, so (κ)\aleph(\kappa) is infinite.

step 1.2step 2.1step 2.2L1L4L6

Remarks

Why claim (a) is stated for an arbitrary set. For a cardinal κ\kappa only the special case is needed here, but the general case costs the same two lines and is exactly what Tarski: the Axiom of Choice is equivalent to the statement that A×AAA \times A \approx A for every infinite set AA, so extending Hessenberg's theorem from the alephs to arbitrary sets is exactly as strong as choice uses: there, AA is an arbitrary infinite set that is not known to be well-orderable, and the argument needs (A)\aleph(A) to be an infinite cardinal before Hessenberg's theorem can be applied to it.

No power set is involved. Hartogs: an ordinal that does not inject into a given set builds (A)\aleph(A) from the well-ordered subsets of AA, so the successor cardinal is obtained without ever forming P(A)\mathcal{P}(A) as a size. That separation is what makes the aleph hierarchy a ZF construction while the beth hierarchy is not.

The notation collides with the ordinal successor, and this page keeps them apart. α+=α{α}\alpha^{+} = \alpha \cup \{\alpha\} is the ordinal successor (Ordinal (von Neumann)) and is almost never a cardinal: an infinite cardinal is a limit ordinal, so κ{κ}\kappa \cup \{\kappa\} is not one. The cardinal successor is (κ)\aleph(\kappa), which is much larger. Where an item on this page writes the superscript on an ordinal letter, as 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 does with α+\alpha^{+}, it carries the published ordinal meaning α{α}\alpha \cup \{\alpha\}. This item writes the cardinal successor as (κ)\aleph(\kappa) throughout; the abbreviation κ+:=(κ)\kappa^{+} := \aleph(\kappa), on a cardinal letter only, is introduced later on this page, and the superscript on an ordinal letter keeps the ordinal meaning there too.

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