How to prove that in a closed, convex, and not bounded set from every point there is a closed half-line that is in the set?

Question

Let $K \subseteq \mathbb{E}^n$ a closed, convex, and not bounded set. How to show that then from every point of $K$ there is a closed half-line that is in $K$.

Answer 1

Wlog the point in $K$ we want to find a half-line from is the origin. For $r>0$, let $U_r=\frac1r\cdot K^\complement$, which is open.

Assume $\bigcup_{r>0}U_r$ covers the unit sphere $S^{n-1}$. Then by compactness a finite cover suffices, $S^{n-1}\subseteq U_{r_1}\cup\ldots\cup U_{r_n}$. Let $R=\max\{r_1,\ldots,r_n\}$. As $K$ is unbounded, pick $x\in K$ with $|x|>R$. Then $\frac1{|x|}x\in S^{n-1}$ and therefore $\in U_{r_i}$ for some $1\le i\le n$. Then $y:=\frac{r_i}{|x|}x\in K^\complement$. But $0<\frac {r_i}{|x|}$ so that $y$ is on the line segment between $0$ and $x$, and so $0\in K$, $x\in K$, $y\notin K$ contradicts the convexity of $K$.

We conclude that $\bigcup_{r>0}U_r$ does not cover the unit sphere. If $v\in S^{n-1}$ is left out, this means that $rv\in K$ for all $r>0$.