Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedaudited 2026-09-22
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.

Failure of inaccessibility in L produces a real with correct omega-one

Statement

Work in ZF+Countable Choice. If the ambient ω1 is not an inaccessible cardinal of L, then there is a real x such that ω1L[x] equals the ambient ω1.

Facts & Assumptions

Given: Countable Choice and the hypothesis that the ambient ω1 is not inaccessible in L.

[F1]

The Axiom of Countable Choice (ACω) with Countable choice makes omega-one regular: Countable Choice makes the ambient ω1 regular, and every countable subset of ω1 is bounded.

[F2]

The generalized continuum hypothesis holds in L: L satisfies GCH, so inside L the power-set operation is the cardinal successor and 20=1.

[F3]

Absoluteness, idempotence and minimality of L: constructibility is absolute between the relevant transitive models with the same ordinals, and LL[x]V. No preservation of L-cardinals in V is asserted: in fact, every ordinal below the ambient ω1 is countable in V.

[F4]

Inaccessible and Mahlo cardinals: a cardinal is inaccessible when it is uncountable, regular and a strong limit.

Proof

1.1

The ambient ω1 is regular by Countable Choice, and it is a cardinal in L: an L-definable surjection from a smaller ordinal onto ω1 would still be a surjection in the universe. It is also regular in L, since an L-cofinal map from a smaller ordinal would remain cofinal in the universe.

F1F3
2.1

Hence, if ω1 is not inaccessible in L, it fails one of the three clauses of [F4] there. It is uncountable in L (it is uncountable in the universe and L has the same ordinals) and regular in L by step 1.1, so it is not a strong limit cardinal of L: there is a cardinal μ<ω1 of L with (2μ)Lω1. By GCH in L, (2μ)L=μ+, so the ordinal ω1 equals (μ+)L for some L-cardinal μ<ω1. This μ is infinite, since the L-successor of a finite cardinal is finite whereas the ambient ω1 is uncountable.

F2F4step 1.1
3.1

The ordinal μ is countable in the ambient universe because μ<ω1, so there exists a real x coding a bijection b:ωμ. This is one existential choice from a nonempty set of codes and needs no family-choice principle.

F1step 2.1
4.1

Let β<(μ+)L=ω1. If β=0, it is countable in L[x] trivially. If β>0, then Choice in L and the fact that μ is an infinite L-cardinal imply that L contains a surjection from μ onto β; this map also belongs to L[x] by [F3]. Composing it with the bijection b coded by x makes β countable in L[x]. Thus every ordinal below the ambient ω1 is countable in L[x], so ω1L[x]ω1. Conversely the ambient ω1 is uncountable in L[x], since any bijection with ω in L[x]V would contradict its ambient definition; hence ω1L[x]ω1. Therefore equality holds.

F2F3step 2.1step 3.1
5.1

The steps above produce a real x with ω1L[x]=ω1 from the failure of inaccessibility in L; only this existential real is claimed, not the statement for every real.

step 4.1

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources