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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?

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@user1551 I think a cone is a set of directions, but the word unbounded sounds better than unlimited. –  Valtteri Dec 13 '12 at 0:27
    
@user1551 Well, a half-line and so are the result of this infinite extension, but direction is the name I was taught (in Finnish, so that's a translation) to the actual unit vector of this half-line. –  Valtteri Dec 13 '12 at 0:40
    
I think that the question is not about the convex set containing a half-line, but something more: indeed, here we require that any point of the convex set is the starting point of a half-line contained in the set, and moreover, all of those half-lines are parallel. I think that the word "direction" is better than this wording! :-) –  Giuseppe Negro Dec 13 '12 at 1:24
    
@GiuseppeNegro Right. You've got a good point. –  user1551 Dec 13 '12 at 1:44
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up vote 1 down vote accepted

In my humble opinion, your proof is incomplete (note: I am not saying that it is wrong):

  1. You haven't explained why $d=\lim_{n\rightarrow\infty}\frac{z_n-x}{\|z_n-x\|}$ exists. While $\frac{z_n-x}{\|z_n-x\|}$ has a limit point on the unit sphere is clear, the sequence per se may not have a limit.
  2. Even if $d=\lim_{n\rightarrow\infty}\frac{z_n-x}{\|z_n-x\|}$ exists, you still haven't explained why $\{x+cd: c\ge0\}\subset A$.
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