Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26‡ rests on unproved material
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 2 statements not proved in this library. Every dependency marked ‡ below is recorded with a citation but 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.

FALSE, relative to an inaccessible cardinal: ZF + DC proves that a nonmeasurable subset of R exists

Statement

Assume ZFC together with the existence of an inaccessible cardinal is consistent. FALSE. ZF + DC proves that a nonmeasurable subset of R exists. Equivalently, relative to this consistency hypothesis, ZF + DC alone cannot guarantee a construction of such a set.

Facts & Assumptions

Given: The consistency of ZFC together with the existence of an inaccessible cardinal, and the external consistency-strength results recorded on the published choice pages.

‡ [L1]

If ZFC together with the existence of an inaccessible cardinal is consistent, then so is ZF + DC + "every set of reals is Lebesgue measurable" (Solovay's model: ZF + DC with every set of reals measurable ‡).

‡ [L2]

If ZF + DC + “every set of reals is Lebesgue measurable” is consistent, then so is ZFC + “there exists an inaccessible cardinal”; in contrast, Con(ZF) implies the consistency of ZF + DC + “every set of reals has the Baire property” (Shelah 1984: the inaccessible is needed for measurability, not for the Baire property ‡).

Refutation

technique · direct
1.1

By [L1], the stated consistency hypothesis supplies a model of ZF + DC in which every set of reals is Lebesgue measurable.

‡ L1
2.1

If ZF + DC proved that a nonmeasurable subset of R exists, every model of ZF + DC would contain one. The model from step 1.1 contains none, so the asserted theorem of ZF + DC is false relative to the stated consistency hypothesis. This is precisely the consistency-strength obstruction recorded in [L2].

step 1.1‡ L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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