Assume we are given a convex set $A$. A direction of this set is a unit vector $\bar x$ for which $\forall a \in A, \ \forall c>0, \ (a+c \bar x) \in A$. In other words it is a direction you can extend any vector in $A$ indefinitely without leaving $A$.
Assume that $A$ is unbounded, ie. $\forall x \in A, \ \exists \ (z_n) \in A \ \forall n \in \Bbb N, \ \lim_{n\rightarrow\infty} |x-z_n| = \infty$. Ie, for every point in $A$ there is a sequence that gets infinitely far from that point. (Is this definition OK?)
Now, I wish to prove that there exists at least one direction for every unbounded convex set.
Select a point from $A$, call it $x$. There is a sequence $z_n$ as mentioned above for this point $x$. Now calculate the vector $\lim \frac{z_n-x}{|z_n-x|}, \ n \rightarrow \infty$. This is a direction.
It is a direction because if you select any other point in $A$, say $y$. Because the set is convex, the vector $y-z_n$ is also in $A$ for all $n \in \Bbb N$ and also $|y-z_n| \rightarrow \infty, \ n \rightarrow \infty$. (This is intuitively clear for two and three dimensional real spaces. Is this true generally?). Thus, if we go to infinity, the two vectors $\lim \frac{z_n-x}{|z_n-x|}, \ n \rightarrow \infty$ and $\lim\frac{z_n-y}{|z_n-y|},\ n \rightarrow \infty$ are parallel. (Again, geometric intuition. Is this true more generally?)
This was my solution to a homework problem many years ago and it kinda stuck with me. Often wondered if there is anything wrong with it. Any comments?