3
$\begingroup$

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?

  • 0
    I wanted to add another assumption to your question, your convex set should be closed too. consider the set A={3>y≥0 /\ x≥0} $\cup$ {y=3,x=0} in $R^2$ A is convex and unbounded but it doesn't have any rescission direction because the (1, 0) is the only direction but if we put X=(0,3) the X+1.d= X+(1,0)=(1,3) is not in the set A.2018-11-08

1 Answers 1

2

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 $\{x+cd: c\ge0\}\subset A$.