How do you prove that $\{ Ax \mid x \geq 0 \}$ is closed? Let $A$ be a real $m \times n$ matrix.  

How do you prove that $\{ Ax \mid x \geq 0, x \in \mathbb R^n \}$ is closed (as in, contains all its limit points)?  

The inequality $x \geq 0$ is interpreted component-wise.

This fact is used in some proofs of Farkas's lemma.  It seems like it should be easy, but the proof I've seen seems to be unexpectedly complicated.  Is there a very clear / easy / obvious proof of this fact? 

(Note that linear transformations do not always map closed sets to closed sets, as discussed in this question.  For example, let $S = \{ (x,y) \in \mathbb R^2 \mid y \geq e^x \}$ and let $T:\mathbb R^2 \to \mathbb R^2$ such that $T(x,y) = (0,y)$.  Then $S$ is closed, but $T(S)$ is not closed.)

Edit: Here is a simple proof in the case where $A$ has full column rank.  (A very similar proof is given in Nocedal and Wright, in the Notes and References at the end of chapter 12.)
Let $y^*$ be a limit point of $\Omega = \{ Ax \mid x \geq 0, x \in \mathbb R^n \}$.  There exists a sequence $(x_i)_{i=1}^\infty$ of points in $\mathbb R^n$ such that $x_i \geq 0$ for all $i$ and $A x_i \to y^*$ as $i \to \infty$.  Let $B$ be a left inverse for $A$. Then $B A x_i \to B y^*$ as $i \to \infty$.  In other words, $x_i \to x^*$ as $i \to \infty$, where we have defined $x^* = B y^*$.  Clearly, $x^* \geq 0$ and $A x^* = y^*$. This shows that $y^* \in \Omega$.

(Alternatively, you could just note that if $A$ has full column rank then the mapping $x \mapsto Ax$ is a homeomorphism between $\mathbb R^n$ and $R(A)$, so it maps closed sets to closed sets.  This shows that $\Omega$ is a closed subset of $R(A)$, and it follows that $\Omega$ is a closed subset of $\mathbb R^m$.)
We denote by $a_i \in \mathbb R^m$, $i = 1, \ldots, n$ the columns of $A$. By a conic variant of Carathéodory's theorem, each conic combination of $\{a_i\}$ can be written as a conic combination of a linearly independent subset of $\{a_i\}$. Since there are only finitely many linearly independent subsets of $\{a_i\}$, it is sufficient to prove the claim for matrices $A$ which have full column rank (i.e., all columns are linearly independent). But in this case, the claim is easy to establish.
Let $e_1,...,e_n$ be a basis of $R^n$, you can suppose that $A(e_1),...,A(e_l)$ generate the image of $A$, and $(e_{l+1},...,e_n)$ is a basis of $ker(A)$.

if $y$  is an element of the adherence of $C=\{A(x),x\geq 0\}$, $y$ is in the vector subspace generated by $(A(e_1),...,A(e_l)$. You can write $y=y_1A(e_1)+...+y_lA(e_l)$. There exists $(x_m)=A(x_1^me_1+...x^m_ne_n)=x_1^mA(e_1)+...+x_m^lA(e_l)$ which converges towards $y$. This implies that $x_i^m$ converges towards $y_i, i=1,...,l$ (use $\|\|_{\infty}$ for that) this implies that $A(y_1e_1+...+y_le_l)=y$.