Let C $\subset$ $\mathbb{R^3}$ be a non-empty, convex, closed set.
Prove that either C = $\partial$C or $\forall$ ${x}$ $\in$ C, $\forall$ n $\in$ $\mathbb{N}$, $\exists$ ${y}$ $\in$ int(C) : $\Vert x-y \Vert$ $\le$ $\frac 1n$ hold.
int (C) being the interior of C, and $\partial$C being the boundary of C
I know that if n $\ge$ 2 and C does not contain any straight line, then C is the convex hull of $\partial$C, and if int(C) is not empty, then $\partial$C cannot be convex, but I don't understand how to use these facts to prove the property
Suppose $C \neq \partial C$. Since $C$ is closed, $\partial C \subset C$. Let $x \in C$. Either $x \in \partial C$ or $x \not\in \partial C$. If $x \in \partial C$, let $z \in C\backslash\partial C$, otherwise let $z \in \partial C$. Since $C$ is convex, the line segment connecting $x$ with $z$ is a subset of int($C$). For any $n \in \mathbb{N}$, you can therefore find a $y \in$ int($C$) close enough to $x$.