Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17‡ 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 1 statement 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: every Borel subset of the real line is a countable union of countable intersections of open and closed sets

Statement

Every Borel subset of R belongs to the class obtained by taking countable intersections of open and closed sets and then countable unions of those intersections.

Facts & Assumptions

Given: The Borel hierarchy on R formed by alternating countable unions and countable intersections from the open and closed sets.

‡ [L1]

For every ordinal 1≤α<ω1, the additive and multiplicative Borel classes on R differ, so the Borel hierarchy does not stabilize at any countable stage (The Borel hierarchy on the real line never stabilizes at a countable stage ‡).

Refutation

technique · contradiction
1.1assume-contra

Suppose, for contradiction, that every Borel set belongs to the displayed fixed finite-stage class.

2.1step 1.1algebra

That class would then equal the Borel sigma-algebra. Since the Borel sigma-algebra is already closed under complements, countable unions, and countable intersections, every further stage would add no set, so the hierarchy would stabilize at that countable stage.

3.1step 2.1‡ L1discharge-contradiction∎

The stabilization in step 2.1 contradicts [L1]. Hence the asserted finite description does not contain every Borel subset of R.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · one level

1 result within one dependency step 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