Core Equals Interior for Convex Sets in Normed Spaces

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 normed vector space and let $\Omega\subset X$ be convex with int(Ω) $\neq\emptyset$ .

Theorem:

\operatorname{core}(\Omega)=\operatorname{int}(\Omega).

Context: The equality identifies the purely algebraic notion core(Ω) with the usual topological interior once the set is convex and has nonempty interior. The proof in the notes uses the geometric lemma segments from interior points stay in the interior.

Proof sketch (idea): The inclusion $\operatorname{int}(\Omega)\subset \operatorname{core}(\Omega)$ is direct. For the reverse direction, translate so that $0\in\operatorname{int}(\Omega)$ and use convexity plus the segment lemma to show any core point must lie in the interior.