Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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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.

Strong separation of a closed and a compact convex set

Statement

Let C,KX be disjoint nonempty convex sets, where C is closed and K is compact. Then they are strongly separated by a nonzero functional in X.

Facts & Assumptions

Given: Disjoint nonempty convex C,K, with C closed and K compact.

[F2]

A continuous real-valued function on a compact metric space attains its minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F3]

Disjoint convex sets, one open, are strictly separated by a nonzero continuous functional (Separation of disjoint convex sets when one is open).

Proof

technique · direct
1.1

By [F1]--[F2], d(k,C) has a minimum δ on K. It is positive: a zero minimum would put some kK in the closed set C. Choose 0<r<δ.

F1F2givenchoose
2.1

The thickening W=C+B(0,r) is open and convex and is disjoint from K. Apply [F3] to W,K to obtain nonzero f with Ref(w)<Ref(k) for wW,kK.

step 1.1F3
3.1

For cC, take the supremum over bB(0,r) in the inequalities from step 2.1. Since supb<rRef(b)=rf, this gives Ref(c)+rfRef(k) for every kK. Hence supCRef+rfinfKRef, a positive gap.

step 2.1algebra

Depends on

Used by

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Dependency tree · two levels

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Sources