Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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.

Relative interiors of finite-dimensional convex sets

Statement

For a nonempty set C, define aff⁡C={∑j=1Najcj:N≥1, cj∈C, aj∈R, ∑jaj=1}. With a fixed c0∈C, this is c0+span⁡{c−c0:c∈C}, where span consists of finite linear combinations. An affine Euclidean space is such a translate of a finite-dimensional linear subspace with its induced Euclidean distance. Relative balls use this distance; interior means containing a relative ball, and closure means every relative ball meets the set. Let C be a nonempty convex subset of a finite-dimensional affine Euclidean space H. Its relative interior ri⁡HC is the interior in H=aff⁡C. Then ri⁡HC is nonempty. If w∈ri⁡HC, y∈C‾ H, and 0≤t<1, then (1−t)w+ty∈ri⁡HC.

Facts & Assumptions

Given: A nonempty convex C⊆H in finite dimension. Affine dimension is the dimension of its translation space.

[F1]

A convex subset of Rm contains every line segment between two of its points: every finite convex combination of points of C belongs to C.

Proof

1.1givenF1construct

Let k=dim⁡(aff⁡C). If k=0, then C is a singleton and is open in its affine hull. If k>0, the definition of affine hull supplies k+1 affinely independent points of C spanning it. Their simplex lies in C by [F1]. Its barycenter has all k+1 barycentric coordinates positive; sufficiently small coordinate perturbations preserve positivity and their sum one. Thus a relative neighborhood of that barycenter lies in the simplex, proving ri⁡HC≠∅.

2.1givenF1choosealgebra∎

Fix w,y,t as in the Statement. The case t=0 is w∈ri⁡HC. Suppose 0<t<1 and put q=(1−t)w+ty. Choose r>0 so the relative ball BH(w,r) lies in C. Since y∈C‾ H, choose one y∗∈C with ∥y∗−y∥<(1−t)r/(4t). For every q′∈BH(q,(1−t)r/4), put w∗=(q′−ty∗)/(1−t). Then ∥w∗−w∥<r/2, so w∗∈C and q′=(1−t)w∗+ty∗∈C by [F1]. A relative neighborhood of q lies in C, so q∈ri⁡HC.

Depends on

Dependency tree · 2 levels

Sources