Interior and closure of a convex set are convex

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 $\Omega\subseteq X$ a convex set. Then:

the interior $\operatorname{int}(\Omega)$ is convex, possibly empty;
the closure $\overline{\Omega}$ is convex.

Remarks

Context. Convexity is stable under two basic topological operations in normed spaces, which is essential for analyzing convex feasible regions.

Proof sketch. For $\overline{\Omega}$ , approximate points by sequences in $\Omega$ and use convexity plus limit arguments. For $\mathrm{int}(\Omega)$ , if $x,y\in\mathrm{int}(\Omega)$ then small balls around $x$ and $y$ lie in $\Omega$ ; convex combinations of these balls give a neighborhood of $\lambda x+(1-\lambda)y$ contained in $\Omega$ , so the combination lies in $\mathrm{int}(\Omega)$ .