Strict Separation of Compact and Closed Convex Sets

Positive definiteness: ∥ x ∥ = 0 \|x\|=0 ∥ x ∥ = 0 iff x = 0 x=0 x = 0 .
Absolute homogeneity: ∥ α x ∥ = ∣ α ∣ ∥ x ∥ \|\alpha x\|=|\alpha|\,\|x\| ∥ α x ∥ = ∣ α ∣ ∥ x ∥ .
Triangle inequality: ∥ x + y ∥ ≤ ∥ x ∥ + ∥ y ∥ \|x+y\|\le \|x\|+\|y\| ∥ x + y ∥ ≤ ∥ x ∥ + ∥ y ∥ .

Let $X$ be a real normed space. Let $K,F\subset X$ be nonempty convex sets and assume:

$K$ is compact,
$F$ is closed, and
$K\cap F=\emptyset$ .

Theorem: The sets $K$ and $F$ can be strictly separated by a closed hyperplane. Equivalently, there exists $x^\ast \in X^\ast$ (see dual space) such that

\sup_{x\in K}\langle x^\ast ,x\rangle < \inf_{y\in F}\langle x^\ast ,y\rangle.

Context: This is a strong separation result in normed spaces. The proof in the notes applies closed hyperplane separation to a ball around $0$ and a translated difference set $F-K$ , using compactness/closedness to guarantee closedness and a positive gap.