ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck 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.

Assuming countable choice, the countable-cocountable family is a sigma-algebra

Example

Assume ${A C}_{ω}$ . For a set $X$ , define

$C : = {A \subseteq X : A is at most countable or X ∖ A is at most countable} .$

Then $C$ is the countable-cocountable sigma-algebra on $X$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a set $X$ .

[L1]

Under countable choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

At most countable means finite or countably infinite (Finite, countably infinite, countable, uncountable).

[L3]

Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L4]

A sigma-algebra contains the empty set and is closed under complements and countable unions (Sigma-algebras).

Verification

technique · direct

1.1L2algebra

The empty set is at most countable. Complementation exchanges the two alternatives in the definition of $C$ , so $C$ contains $\emptyset$ and is complement-closed.

1.2L1L2L3

Let $(A_{n})$ lie in $C$ . If every $A_{n}$ is at most countable, [L1] makes $⋃_{n} A_{n}$ at most countable. If some $A_{j}$ is cocountable, then $X ∖ ⋃_{n} A_{n} \subseteq X ∖ A_{j}$ is at most countable by [L3]. In either case the union lies in $C$ .

2.1step 1.1step 1.2L4∎

Steps 1.1 and 1.2 verify all axioms in [L4], so $C$ is a sigma-algebra on $X$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

R. F. Bass, Real Analysis for Graduate Students, version 5.0, Example 2.3 (standard reference, not scraped)