Suppose $C$ is a convex set in $\mathbb{R}^n$ whose closure contains the open ball $B(x,r)$. Is it true that $C$ contains $B(x,r)$?
Motivation: I am asking this because something like this seems to be implicitly assumed in the proof of the separating hyperplane theorem I am working through (or I might just be generally confused). Thus if there is a simple direct argument that would be nice.
This is undoubtedly true. I'll give a proof in $\mathbb{R}^3$.
Let $a \in B(x,r)$. Then we have $B(a,r_1) \subseteq \overline{C}$ for some small $r_1$. Now let $b_1 b_2 b_3 b_4$ be some tetrahedron in $B(a,r_1)$ that contains $a$ in its interior. For $i = 1, 2, 3, 4$, let $x_{i,n} \in C$ be a sequence of points tending towards $b_i$. For large enough $n$, the point $a$ must be in the interior of $x_{1,n}x_{2,n}x_{3,n}x_{4,n}$.