$S$ is bounded if and only if $D=\{\mathbf 0\}$ The set of all recession directions of the polyhedron $S=\{\mathbf x:\mathbf A\mathbf x \le \mathbf b ,\mathbf x \ge \mathbf 0\}$ is equal to  the set $D=\{\mathbf d:\mathbf A\mathbf d \le \mathbf 0 ,\mathbf d \ge \mathbf 0\}$, Use this to show that the set $S$ is bounded if and only if $D=\{\mathbf 0\}$

Assume $S$ is bounded by $M$ and let $\mathbf d$ be a recession directions of the polyhedron, then $\forall \mathbf x \in S,\forall \lambda \ge 0: \mathbf x +\lambda \mathbf d \in S$, since $S$ is bounded by $M$ it follows that $$\forall \mathbf x \in S,\forall \lambda \ge 0:|| \mathbf x +\lambda \mathbf d || \le M$$
I somehow was going to show that $||\mathbf d|| \le 0$ and since from the definition of norms $||\mathbf d|| \ge 0$ so $||\mathbf d||=0$ which shows that $\mathbf d =\mathbf 0$, but I could not prove the result.
 A: On one hand if $\mathbf{d} \in D$ is nonzero and $\mathbf{x} \in S$ is any point then $A (\mathbf{x} + \lambda \mathbf{d}) = A \mathbf{x} + \lambda A \mathbf{d} \leq A \mathbf{x} \leq \mathbf{b}$ for all $\lambda > 0$, which shows that $S$ is unbounded (because $\mathbf{x} + \lambda \mathbf{b} \in S$ for arbitrarily large $\lambda$). This proves the contrapositive of the argument which you couldn't complete.
On the other hand the other direction is harder and a question equivalent to this direction has been answered before on this site: If $P$ is an unbounded polyhedron, there exists a point $c \in P$ and a vector $d \neq 0 $ such that $ \forall \lambda \geq 0$, $c+ \lambda d \in P$
But the argument is just that if $(\mathbf{x}_n)$ is a sequence in $S$ converging to infinity and $\mathbf{x} \in S$ is a fixed point, then the differences $\mathbf{x}_n - \mathbf{x}$ are eventually nonzero, in which case the new sequence of unit vectors $\mathbf{y}_n := \frac{\mathbf{x}_n - \mathbf{x}}{\lVert \mathbf{x}_n - \mathbf{x} \rVert}$ is eventually well-defined. By compactness of the finite dimensional unit ball there exists a subsequence $(\mathbf{y}_{n_k})$ converging to some $\mathbf{y}$. Of course $\mathbf{y}$ is a unit vector, so cannot be zero. The claim is then that $\mathbf{y} \in D$: to check this just observe (as in the linked question, since $A \mathbf{x_n} \leq \mathbf{b}$ for all $n$) that
$$
A \mathbf{y}
= \lim_{k \to \infty} A \mathbf{y}_{n_k}
= \lim_{k \to \infty} \frac{A \mathbf{x}_{n_k} - A \mathbf{x}}{\lVert \mathbf{x}_{n_k} - \mathbf{x} \rVert}
\leq \lim_{k \to \infty} \frac{\mathbf{b} - A \mathbf{x}}{\lVert \mathbf{x}_{n_k} - \mathbf{x} \rVert} = 0.
$$
In particular the last limit is zero because $\lVert \mathbf{x}_{n_k} \rVert \to \infty$ and thus $\lVert \mathbf{x}_{n_k} - \mathbf{x} \rVert \to \infty$ as $k \to \infty$. Because $\mathbf{y}$ is nonzero $D$ cannot be $\{\mathbf{0}\}$, so this completes the proof.
