Segments from interior points stay in the interior

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 ∥ .

Lemma. Let $X$ be a normed vector space and let $\Omega\subset X$ be a convex set with nonempty interior. If $a\in \mathrm{int}(\Omega)$ and $b\in\Omega$ , then

[a,b)\subset \mathrm{int}(\Omega),

where $[a,b)$ is the half-open segment from $a$ to $b$ .

Remarks

Context. This is a key geometric fact for convex sets: interior points "see" the whole set through interior segments. It underlies closure/interior relations for convex sets.

Proof idea. Starting from a ball around $a$ contained in $\Omega$ , use convexity and scaling properties of balls to build a ball around each point $\lambda a+(1-\lambda)b$ (with $\lambda\in(0,1]$ ) that still lies in $\Omega$ .