Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Strong separation of a closed and a compact convex set

Statement

Assume the Axiom of Choice. Let C,K⊆X 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: The Axiom of Choice and 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).

[A1]

The Axiom of Choice supplies the Hahn–Banach premise in [F3].

Proof

technique · direct
1.1F1F2givenchoose

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

2.1step 1.1F3A1

The thickening W=C+B(0,r) is open and convex and is disjoint from K. Under [A1], apply [F3] to W,K to obtain nonzero f with Re⁡f(w)<Re⁡f(k) for w∈W,k∈K.

3.1step 2.1algebra∎

For c∈C, take the supremum over b∈B(0,r) in the inequalities from step 2.1. Since sup⁡∥b∥<rRe⁡f(b)=r∥f∥, this gives Re⁡f(c)+r∥f∥≤Re⁡f(k) for every k∈K. Hence sup⁡CRe⁡f+r∥f∥≤inf⁡KRe⁡f, a positive gap.

Depends on

Used by

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Sources