Interior and closure relations for convex sets with nonempty interior

Theorem. Let $X$ be a normed vector space and let $\Omega\subset X$ be convex with $\mathrm{int}(\Omega)\neq\emptyset$ . Then:

$\overline{\mathrm{int}(\Omega)}=\overline{\Omega}$ .
$\mathrm{int}(\overline{\Omega})=\mathrm{int}(\Omega)$ .

Remarks

Context. For convex sets with nonempty interior, "taking closure" and "taking interior" are tightly compatible. This is special to convexity and can fail for arbitrary sets.

Proof sketch.

The inclusion $\overline{\mathrm{int}(\Omega)}\subset\overline{\Omega}$ is immediate. For the reverse, it suffices to show $\Omega\subset\overline{\mathrm{int}(\Omega)}$ : fix $b\in\Omega$ and $a\in\mathrm{int}(\Omega)$ ; by the interior-segment lemma, points $(1/k)a+(1-1/k)b$ lie in $\mathrm{int}(\Omega)$ and converge to $b$ .
The inclusion $\mathrm{int}(\Omega)\subset\mathrm{int}(\overline{\Omega})$ is obvious. For the converse, use a "push-out" argument: given $b\in \mathrm{int}(\overline{\Omega})$ and $a\in\mathrm{int}(\Omega)$ , move slightly past $b$ along the ray from $a$ through $b$ to find a point $c\in\overline{\Omega}$ , then apply the interior-segment lemma to conclude $b\in\mathrm{int}(\Omega)$ .