Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-26
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.

For every positive ε there is a dense open subset of (0,1) of Lebesgue measure below ε

Example

Assume the Axiom of Countable Choice. For every real ε>0 there is an open set U⊆(0,1) such that U is dense in (0,1) and λ1(U)<ε.

Facts & Assumptions

Given: The Axiom of Countable Choice and a real ε>0.

[L1]

Assuming countable choice, a box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai) (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included).

[F1]

Let μ be a measure and let (Ek)k∈N be measurable. Then μ(⋃kEk)≤∑kμ(Ek) (Finite and countable subadditivity of measures).

[F2]

The rationals are countably infinite (Q is countably infinite).

Verification

technique · direct
1.1F2choose

Since the rationals in (0,1) are countably infinite, fix an enumeration (qk)k∈N of Q∩(0,1).

2.1step 1.1L1F1F3algebra

Put U:=⋃k∈N((qk−ε2−k−3, qk+ε2−k−3)∩(0,1)). This set is open, and [L1], [F1] and [F3] give λ1(U)≤∑k=0∞ε2−k−2=ε/2<ε.

3.1step 1.1step 2.1∎

Every rational point of (0,1) lies in U, so U is dense in (0,1).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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