TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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Disjoint nonempty Euclidean convex sets have a separating hyperplane

Statement

Assume the Axiom of Choice (The Axiom of Choice) and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $C, D \subseteq R^{n}$ be nonempty, disjoint, and convex. Then there is $a \neq 0$ such that

$⟨ a, c ⟩ \leq ⟨ a, d ⟩ (c \in C, d \in D) .$

Thus $C$ and $D$ are separated by a hyperplane in the sense of Supporting and strictly separating hyperplanes in Euclidean space. The inequality need not be strict when the two sets have distance zero.

Facts & Assumptions

Given: The sets and choice principles in the Statement, convexity as in A convex subset of $R^{m}$ contains every line segment between two of its points, and the boundary convention Interior, closure, boundary, limit point, isolated point and dense subset of a metric space.

[A1]

AC and ${A C}_{ω}$ supply the choice functions asserted in The Axiom of Choice and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[L1]

A point outside a nonempty closed convex set can be strictly separated from it (A point outside a nonempty closed convex set is strictly separated from it).

[L2]

Every boundary point of a nonempty convex set has a supporting hyperplane (Every boundary point belonging to a nonempty Euclidean convex set has a supporting hyperplane).

[L3]

The closure of a nonempty convex subset of $R^{n}$ is convex (A convex set and its closure have the same interior and boundary).

Proof

technique · direct

1.1

Put $E = C - D = {c - d : c \in C, d \in D}$ . It is nonempty and convex and omits zero because $C \cap D = \emptyset$ . By [L3], $\overline{E}$ is convex. Either $0 \notin \overline{E}$ , or $0 \in \overline{E} ∖ E$ and therefore $0 \in \partial E$ , since an interior point of $E$ would belong to $E$ .

L3givenalgebra

2.1

In the first case, apply [L1] to $0$ and $\overline{E}$ ; in the second, [A1] licenses the hypotheses of [L2], which applies to $E$ at zero. Each branch gives a nonzero $a$ with $⟨ a, e ⟩ \leq 0$ for every $e \in E$ . Substituting $e = c - d$ gives $⟨ a, c ⟩ \leq ⟨ a, d ⟩$ for all $c \in C, d \in D$ .

step 1.1A1L1L2algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

S. Boyd and L. Vandenberghe, Convex Optimization, §2.5.1 (standard reference, not scraped)
D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 7 (standard reference, not scraped)