If points in a convex set $C$ escape to infinity roughly in direction $v$ then an infinite ray in that direction exists Let  $C\subseteq \mathbb R^n$ a convex set.
Assume there is a sequence $\{c_k\}_{k\in\mathbb N}$ with $c_k\in C$, $|c_k|\to\infty$ such that  $v:=\lim \frac 1{|c_k|}c_k$ exists. Does this imply that there exists $a\in C$ with $a+[0,\infty)\cdot v\subseteq C$? Is the set of such $a$ dense in $C$? 
This looks intuitively clear, e.g., in the case $n=2$. I was thinking of projecting the $c_k$ on a hyperplane $\perp v$ through some inner point of $C$ and use these to find a suitable $a$. But these projections don't seem to be well-behaved ...

You probably need that $C$ is closed.
Assume $0\in C$ without loss of generality. Otherwise, if $a\in C$, then $|c_k-a|\to \infty$ and
$$ \frac{c_k-a}{|c_k-a|} = \frac{c_k}{|c_k|}\frac{|c_k|}{|c_k-a|} - \frac{a}{|c_k-a|} \to v.$$
Let $r\in[0,\infty)$. As $|c_k|\to\infty$, for $k$ sufficiently large, we have $r < |c_k|$ and
$\frac r{|c_k|} c_k\in C$ by convexity. If $C$ is closed than, we have $\frac r{|c_k|} c_k \to r v \in C$.
Note:
If $C$ is closed, we have in fact $C+[0,\infty)v \subseteq C$.
