Existence of direction for polyhedral set I refer to Lemma 4.42 of this lecture notes on linear programming, about the relationship between the boundedness of a polyhedral set and its directions.

Let $P$ be a non-empty polyhedral set. Then the set of directions of $P$ is
  empty if and only if $P$ is bounded.

Here we define a polyhedral set $P = \{x \in \mathbb{R}^n : Ax \leq b, x \geq 0\}$ where $A$ is a $m \times n$ matrix. A direction $d$ of $P$ is defined to be some $d \in \mathbb{R}^n - \{0\}$ satisfying $\{x_0 + \lambda d : \lambda \geq 0 \} \subseteq P$ for all $x_0 \in P$. Here's the proof for one direction of the lemma:

If $P$ has no directions, then there is some absolute upper bound on the value of $|x|$ for all $x \in P$. Let $r$ be this value. Then trivially, $B_{r+1}(0)$ contains $P$ and so $P$ is bounded.

Is there an easy way to see that "no directions" imply "upper bound on $|x|$"? It does not seem obvious to me; if there was no upper bound, I can't seem to construct a direction for the sake of contradiction.
 A: Suppose that $P$ is not bounded. There is then a sequence $(x_i)_{i\geq0}$ of points in $P$ such that $|x_i|\to\infty$ as $i\to\infty$. We may assume that $x_i\neq x_0$ for all $i\geq1$, and then the sequence of vectors $$d_i=\frac{x_i-x_0}{|x_i-x_0|}$$ takes values in the unit sphere, which is compact, so by replacing our sequence by a subsequence, we may assume that there is a unit vector $d$ such that $d_i\to d$ as $i\to\infty$.
Let now $\lambda>0$ and let us show that the segment $[x_0,\lambda d+x_0]$ is contained in $P$. This will imply that the ray $\{td+x_0:t\geq0\}$ is contained in $P$. 
Let $t\in[0,1]$. Clearly the point $$z_t=(1-t)x_0+t(\lambda x+x_0)=x_0+t\lambda d$$ is the limit as $i$ goes to $\infty$ of the vectors $$y_i=x_0+t\lambda\frac{x_i-x_0}{|x_i-x_0|}=\left(1-\frac{t\lambda}{|x_i-x_0|}\right)x_0+\frac{t\lambda}{|x_i-x_0|}x_i.$$
There is an $N$ such that for $i\geq N$ we have $\frac{t\lambda}{|x_i-x_0|}\in[0,1]$ and for such $i$ the point $y_i$ is in the segment $[x_0,x_i]$, which is contaned in $P$. As $P$ is closed, we see that $z_t\in P$.
As this works for all $t\in[0,1]$, we see that in fact the whole segment $[x_0,\lambda d+x_0]$ is contained in $P$, as we wanted.

The above shows that for each $x\in P$ the set $D_x$ of all unit vectors $d$ such that the ray $\{td+x:t\geq0\}$ is contained in $P$ is not empty. 
Now let $x$ and $y$ be two points of $P$ and let $d\in D_x$. We want to show that $d\in D_y$.
As $d\in D_x$, then $x_i=x+id\in P$ for all $i\in\mathbb N_0$, so the sequence $$y,x_1,x_2,x_3,\dots$$ goes to infinity. It follows, from the above, that $d'=\lim\frac{x_i-y}{|x_i-y|}$ is in $D_y$. Using the triagular inequality, we see at once that $d'$ is equal to $d$ and therefore that $d\in D_y$, as we wanted.
It follows that $D_x$ does not really depend on $x$. Any of its elements are directions of $P$.
