Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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.

An algebra closed under countable disjoint unions is a sigma-algebra

Statement

Let A be an algebra of subsets of X. If the union of every pairwise disjoint sequence in A belongs to A, then A is a sigma-algebra on X.

Facts & Assumptions

Given: An algebra A on X with the stated closure under countable disjoint unions, and a sequence (An)nN in A.

[L1]

An algebra is closed under complements and finite unions (Algebras of subsets).

[L2]

A sigma-algebra is an algebra closed under countable unions (Sigma-algebras).

Proof

technique · direct
1.1

For nN put Bn:=Ank<nAk. The union with k<n is finite and is empty when n=0, so [L1] gives BnA. The sets Bn are pairwise disjoint and nBn=nAn.

L1algebra
2.1

The assumed disjoint-union closure applied to (Bn) gives nAnA. Thus A has the countable-union axiom in [L2] and is a sigma-algebra.

step 1.1givenL2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 2 results over 2 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