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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
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A compact set and a disjoint closed set have a positive norm-distance gap

Statement

Let X be a real or complex normed space. If KX is nonempty compact, CX is nonempty closed, and KC=, then there is δ>0 with kcδ(kK, cC). No convexity, completeness, HB, or infinite choice principle is required.

Facts & Assumptions

[F1]

Closed means open complement; an open set contains a positive-radius ball about each of its points (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[F2]

A compact subset is compact for its restricted metric, so every intrinsic open cover has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

[F3]

A natural-number-indexed finite family of nonempty sets has a choice function in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[F4]

Every nonempty finite list of real numbers has a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum).

[F5]

The induced metric is d(x,y)=xy over either scalar field, with the norm triangle inequality (Real and complex scalar conventions for normed spaces).

Proof

Given: A normed X, nonempty compact K, nonempty closed C, and KC=.

1.1

Form the set of all admissible pairs T={(a,r)K×(0,):B(a,3r)C=} and the family U={KB(a,r):(a,r)T}. For each fixed kK, the open complement of C contains k, so some s>0 has B(k,s)C=. Then (k,s/3)T and kKB(k,s/3). Thus U covers K without selecting radii for all k simultaneously.

givenF1
2.1

Each V=KB(a,r) in this family is open for the restricted metric on K. Indeed, if kV, then rka>0, and every yK with yk<rka satisfies yayk+ka<r, so is in V. Thus U is an intrinsic open cover of K.

step 1.1F1F5algebra
3.1

Compactness gives a finite subcover V0,,Vn1 with n1, since K. For each index j<n define Wj={(a,r)T:Vj=KB(a,r)}. Each Wj is nonempty by the definition of U. Applying finite choice to the function jWj supplies pairs (aj,rj)Wj for these finitely many indices. Repeated Vj or Wj cause no problem: a choice function on the set of values can be evaluated at each Wj.

step 1.1step 2.1F2F3
4.1

The finite list r0,,rn1 consists of positive reals, so its minimum δ exists and is positive, since it equals one of those reals.

step 3.1F4
5.1

For any kK choose an index j<n with kVj, possible because the finite family covers K. For every cC, admissibility gives caj3rj, whereas kaj<rj. Hence ckcajkaj>2rjδ. This proves the uniform bound for all k,c.

step 3.1step 4.1F5algebra

Source notes

Brezis Theorem 1.7 proof, p.7, closed-minus-compact step expanded; Teschl Corollary 5.4 proof, p.140, finite-cover step specialized to normed spaces.

Depends on

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Sources