Consider $X = \{(x_1,x_2 \dots x_n) | x_i \geq 0\}$, a subset of $R^n$. Is $A(X)$ a closed set when $A$ is a linear map from X to $R^m$? While I was trying to prove Farkas' Lemma in Linear Programming (Pg. 19 of these slides), I came across the question posted in the title. If $A$ is invertible, then the result is straightforward. But what happens if $A$ is not? I'm also aware that there are linear transformations that are not closed (a non-example). However, the set that I have here - $X$, is much "nicer" than the non-examples that I have seen.
 A: There are two interpretations of the set $X$ in your question.
1)) The index $i$ in the definition of $X$ is fixed. A set $A(\Bbb R^n)$ is a linear subspace of $\Bbb R^m$, and is thus closed. Let  $e_i\in\Bbb R^n$ be the vector whose $i$-th coordinate equals $1$ and the other coordinates equals zero. Let $Y=\{(x_1,x_2 \dots x_n)\in\Bbb R^n| x_i=0\}$. If $A(e_i)\in A(Y)$ then $A(\Bbb R^n)=A(Y)=A(X)$, and so the latter set is closed. Otherwise pick any linear map $f:A(\Bbb R^n)\to \Bbb R$ such that $f(A(Y))=0$ and $f(A(e_i))=1$. Clearly, $f$ is continuous and $A(X)=f^{-1}([0,\infty))$ is a closed subspace of $A(\Bbb R^n)$.
2)) $X = \{(x_1,x_2 \dots x_n) | x_i \geq 0\mbox{ for all }i\in\{1,\dots,n\}\}$. Then $A(X)$ a cone generated by a finite set $\{A(e_1),\dots, A(e_n)\}$. By Weyl’s Theorem, $A(X)$ is polyhedral and therefore closed (see, for instance, [Paf, Theorem 1.8] and Definition 1.3 of a polyhedral cone).
References
[Paf] Andreas Paffenholz, Polyhedral Geometry and Linear Optimization. Summer Semester 2010.
