I want to show that a closed convex set $S \subseteq R^n$ is bounded if and only if $S$ contains no rays.
Where $r \in S$ is a ray of $S$ if $x \in S$ implies that $x+\mu r \in S$ for all $\mu \in R_{+}^1$.
I am looking at the general case of when $S$ is a closed convex set.
Let's say $0\in S$, and let $x_n=r_n \omega_n\in S$, with $r_n\to\infty$ and $\omega_n$ on the unit sphere. By compactness of the sphere, we can assume that $\omega_n\to\omega$. Then the ray $t\omega$, $t\ge 0$, is contained in $S$ because $t\omega_n\in S$ for all $n$ with $r_n\ge t$ by convexity, and these points converge to $t\omega$.
The converse (a bounded [convex] set doesn't contain a ray) is obvious.