LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Strictly separate a point from a nonempty closed convex Euclidean set

Statement

Let $n \geq 1$ , let $E \subseteq R^{n}$ be nonempty, closed, and convex, and let $z \notin E$ . Then there are a nonzero $u \in R^{n}$ and a real $c$ such that

$⟨ u, e ⟩ \leq c < ⟨ u, z ⟩ (e \in E) .$

Thus an affine hyperplane strictly separates the point from the entire set.

Facts & Assumptions

Given: The Euclidean inner product and norm, a nonempty closed convex $E \subseteq R^{n}$ , and $z \notin E$ .

[F1]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ x ∥_{2}^{2} = ⟨ x, x ⟩$ and the squared norm expands bilinearly.

[F2]

A convex subset of $R^{m}$ contains every line segment between two of its points: for $e_{0}, e_{1} \in E$ and $t \in [0, 1]$ , $(1 - t) e_{0} + t e_{1} \in E$ .

[L1]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm satisfies the triangle inequality, so $e \mapsto ∥ z - e ∥_{2}$ is continuous by the reverse triangle inequality.

[L2]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: a closed bounded subset of $R^{n}$ , $n \geq 1$ , is compact.

[L3]

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

Proof

technique · direct

1.1givenL1L2L3choose

Choose $e_{0} \in E$ and set $R = ∥ z - e_{0} ∥_{2} + 1$ . The set $K = E \cap {e : ∥ z - e ∥_{2} \leq R}$ is nonempty and bounded; it is closed because $E$ is closed and the distance map is continuous by [L1]. Hence [L2] makes it compact, and the continuous function $e \mapsto ∥ z - e ∥_{2}^{2}$ attains a minimum at some $e_{*} \in K$ by [L3].

2.1step 1.1givenalgebra

This $e_{*}$ is nearest to $z$ in all of $E$ : every $e \in E ∖ K$ satisfies $∥ z - e ∥_{2} > R > ∥ z - e_{0} ∥_{2} \geq ∥ z - e_{*} ∥_{2}$ . Put $u = z - e_{*}$ ; since $z \notin E$ , $u \neq 0$ .

3.1F1F2step 2.1algebra

Fix $e \in E$ and put $v = e - e_{*}$ . For $0 < t \leq 1$ , [F2] gives $e_{*} + t v \in E$ , so minimality of $e_{*}$ and [F1] imply $2 t ⟨ u, v ⟩ \leq t^{2} ∥ v ∥_{2}^{2}$ . If $⟨ u, v ⟩ > 0$ , then $v \neq 0$ ; setting $t = \min {1, ⟨ u, v ⟩ / ∥ v ∥_{2}^{2}}$ contradicts this inequality whether the minimum is $1$ or the quotient. Hence $⟨ u, v ⟩ \leq 0$ .

4.1F1step 3.1algebra∎

Therefore $⟨ u, e ⟩ \leq ⟨ u, e_{*} ⟩$ for every $e \in E$ , while $⟨ u, z ⟩ = ⟨ u, e_{*} ⟩ + ∥ u ∥_{2}^{2} > ⟨ u, e_{*} ⟩$ . Taking $c = ⟨ u, e_{*} ⟩$ proves strict separation.

Strictly separate a point from a nonempty closed convex Euclidean set

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources