Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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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.

Positive open-set and metric-ball volume

Statement

Let r be positive smooth. Every nonempty open OM has μr(O)>0. If d induces the manifold topology, then for every pM and R>0 the ball Bd(p,R) is Borel and has positive measure. If its closure in M is compact, it also has finite measure.

Facts & Assumptions

Given: Assume ACω. Manifolds are Hausdorff, second countable and smooth, with boundary allowed; n=0 is allowed unless excluded. Densities are pointwise Borel, 0=0, and λ0(R0)=1. Nonempty open set; topology-compatible positive-radius ball.

[F1]

Positive smooth densities give Radon volume: Positive smooth densities give compact-finite Borel measures.

[F2]

Intrinsic density measure and its chart restriction: The measure in any chart is its coefficient integral.

[F4]

Open ball, closed ball and sphere in a metric space: Bd(p,R)={q:d(p,q)<R} for R>0.

Proof

1.1

Choose pO and a chart around it restricted inside O. For n1, continuity and rx(x(p))>0 give a relative coordinate neighborhood where rxc=rx(x(p))/2>0. This neighborhood contains a closed nondegenerate Euclidean box Q in the interior of the half-space: even if p is on the face, move its last coordinate a sufficiently small positive distance and choose a still smaller box. Hence μr(O)Qrxcvol(Q)>0.

F2F3given
2.1

For n=0 the singleton {p} is open with measure r(p)>0, so again μr(O)>0. The empty manifold has no nonempty open subset, making this clause vacuous.

F2step 1.1
3.1

If qBd(p,R), then ϵ=Rd(p,q)>0 and the triangle inequality gives Bd(q,ϵ)Bd(p,R). Thus the ball is open in the metric topology, hence in the manifold topology and Borel. It contains p since d(p,p)=0<R, so the preceding positivity applies. If Bd(p,R) is compact, monotonicity and compact-finiteness give μr(Bd(p,R))μr(Bd(p,R))<.

F1F4step 1.1step 2.1

Depends on

Used by

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