Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Uniform strict separation of compact and closed convex sets

Statement

Assume HB. Let K be a nonempty compact convex subset and C a nonempty closed convex subset of a locally convex real or complex TVS, with K∩C=∅. There are a nonzero continuous scalar-linear f, α∈R and ε>0 such that Re⁡f(k)≤α−ε<α+ε≤Re⁡f(c)(k∈K, c∈C). Hausdorffness is not required.

Facts & Assumptions

Given: HB, the stated TVS, and K,C with the stated hypotheses.

[F1]

Convex sets and real parts of the continuous dual have their TVS meanings (Local convexity, convex and balanced sets, and the continuous dual).

[F2]

Translations and nonzero dilations are homeomorphisms (Translations, dilations and absorption in a topological vector space).

[F3]

Every zero-neighborhood has an open convex refinement (Open and closed balanced convex zero-neighborhood refinements).

[F4]

Compactness of K means every relative open cover of K has a finite subcover (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[F6]

Choice for a finite indexed list of nonempty sets is a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[F7]

Open convex separation supplies u(a)<β≤u(c) for a nonzero continuous scalar-linear functional with real part u (Continuous separation when one convex set is open).

[A1]

HB is explicitly assumed as an additional principle over ZF (The real dominated-extension principle as an additional hypothesis over ZF).

Proof

1.1F2F3F5

Consider all pairs (k,N) with k∈K and N an open convex zero-neighborhood such that k+N⊆X∖C. There is such a pair above every k: since C is closed, (X∖C)−k is an open zero-neighborhood, and it admits an open convex refinement. Each Gk,N=K∩(k+12N) is open in K and contains k. The family of all such sets covers K; it is defined by a property, without choosing a neighborhood for every k.

2.1F1F2F4F6step 1.1

Compactness gives finitely many cover members G1,…,Gm with m≥1. For each j its set of representing pairs (k,N) is nonempty. Finite choice gives representatives (kj,Nj) for this finite list. Put W=⋂j=1m12Nj. It is an open convex zero-neighborhood.

3.1F1F2step 1.1step 2.1

For k∈K, some j has k=kj+n/2 with n∈Nj. If w∈W, write w=n′/2 with n′∈Nj. Convexity gives n/2+n′/2∈Nj, so k+w∈kj+Nj⊆X∖C. Therefore (K+W)∩C=∅. The set K+W is nonempty and open as a union of translates of W, and convex because the convex combinations of its K and W components stay in those respective sets.

4.1F1F5F7F8A1step 3.1

Apply open separation to K+W and C, using HB once. Obtain a nonzero continuous scalar-linear f and β∈R such that u(k+w)<β≤u(c), where u=Re⁡f. The restriction u∣K is continuous, since real part is continuous and restrictions are continuous. It attains a maximum d=u(k∗) for some k∗∈K. Since 0∈W, the strict inequality at k∗+0 gives d<β. This attainment step turns pointwise strict separation into a uniform gap.

5.1step 4.1algebra∎

Put α=(d+β)/2 and ε=(β−d)/2>0. Then α−ε=d and α+ε=β, so u(k)≤d=α−ε<α+ε=β≤u(c) for all k,c. A singleton K is allowed and simply has its sole value as the maximum. Nonemptiness of K is used for attainment and of C in open separation; no other separation axiom or choice principle is used.

Depends on

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Sources