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 ...

So here's what I myself thought of after a good night's sleep:
For $1\le m<k$ let 
$$\begin{align}U_{m,k}&=\operatorname{span}\{\,c_i-c_m\mid m<i\le k\,\}\\U_m&=\bigcup_{k\ge m}U_{m,k}\\U&=\bigcap_mU_m\end{align}$$
Since $U_{m,m}\subseteq U_{m,m+1}\subseteq U_{m,m+2}\subseteq\ldots$ is eventually stationary, we have $U_m=U_{m,k_m}$ for some $k_m$. Since $U_1\supseteq U_2\supseteq\ldots$ is eventually stationary, we have $U=U_{m_0}=U_{m_0,k_0}$ for some $m_0$ and $k_0$.
Let $$a=\frac{c_{m_0}+c_{m_0+1}+\ldots+c_{k_0}}{k_0-m_0+1}.$$
Then for $r$ small enough $B_r(a)\cap(a+U)$, the $r$-ball around $a$ intersected with $a+U$, is in the convex hull of $c_{m_0}\,\ldots,c_{k_0}$.
For fixed $m$ we have
$$\langle v,w\rangle = \lim_{k\to\infty}\left\langle \frac{c_k}{|c_k|},w\right\rangle = \lim_{k\to\infty}\left\langle \frac{c_k-c_m}{|c_k|},w\right\rangle$$
so that $w\perp U_m$ implies $w\perp v$. We conclude $v\in U_m$ and $v\in U$.
Let $t>0$. For $k\gg 0$ we have  $|c_k-a| > t$ and $\left|\frac{c_k-a}{|c_k-a|}-v\right|<\frac r{2t}$ and hence find for $u=2tv-2t\frac{c_k-a}{|c_k-a|}$ that $u\in B_r(0)\cap U$.
Then $a+tv$ is a convex combination of $a+u$ and $a+\frac{t}{|c_k-a|}(c_k-a)$ and the latter is a convex combination of $a$ and $c_k$. We conclude $a+tv\in C$.
The set $A:=\{\,a'\in C\mid a'+[0,\infty)\cdot v\subseteq C\,\}$ is clearly convex.
Now let $c\in C$ and $a'=xc+ya$ with $x>0,y\ge 0, x+y=1$ be a convex combination of $a\in A$ and $c$. Then $a'+tv$ is a convex combination of $c$ and $a+\frac txv$, hence $a'+tv\in C$ and $a'\in A$. This shows that $A$ is a dense convex subset of $C$. Especially, $A$ contains the relative interior of $C$.
