Linear image of a closed cone. Is it closed? Today in my optimization class we needed to use that a certain set, which was the linear image of the closed cone $ S = \{ y \ge 0 \} \subset R^m$ (that is, evey component of $y$ is greater than $0$)$ , was a closed set.
To fix notation, we will call the linear function $f: R^m \to R^n$.
However, we could not figure out how to properly show that $f(S)$ is closed. Is it right? How do we show it?
 A: $f(S)$ is the convex cone generated by $m$ vectors in $\mathbb R^n$.  Namely, if $e_1, \dots, e_m$ is the standard basis of $\mathbb R^m$,  then $S = \{\sum_i \alpha_i e_i :  \alpha_i \ge 0 \text{ for all } i\}$ and
$f(S) = \{\sum_i \alpha_i f(e_i) :  \alpha_i \ge 0 \text{ for all } i\}$.
So what is needed is the following statement:

Proposition.   Any convex cone generated by $m$ vectors in $\mathbb R^n$ is closed.

There is a proof of this fact here.
The brief sketch of the proof is as follows:  
$\def\co{\text{co}}$
For any finite set $B$ in $\mathbb R^n$ let $\co(B)$ denote the convex cone generated by $B$, namely the set $\{\sum_{j = 1}^l \beta_j b_j : l \ge 0, b_j \in B, \beta_j \ge 0\}$.
Let $C = \co(A)$ for a finite set $A$  in $\mathbb R^n$.  For any non-zero $x \in C$, $x$ can be written as $x = \sum_{i =1}^m \alpha_i a_i$, where $\alpha_i \ge 0$, $a_i \in A$, and $m$ is minimal among all such representations.  Then necessarily $\alpha_i >0$ for all $i$, by minimality.  

Lemma.
   $\{a_i, \dots, a_m\}$ is linearly independent.  

This is proved in the source linked above.

Lemma.  If  $\{a_i, \dots, a_m\}$ is linearly independent, then $\co(\{a_i, \dots, a_m\})$ is closed.

It follows from the two lemmas that $C$ is the finite union of of closed cones $\co(\{a_i, \dots, a_m\})$, where $\{a_i, \dots, a_m\}$ is linearly independent.  Hence $C$ is closed.
A: Yes, it is true, assuming that you are talking about a linear map $f$ from $\mathbb{R}^2$ into $\mathbb{R}^n$. In that case, the image of you semiplane can be a point, a line a ray, or a simplane of $\operatorname{Im}f$. In each case, it's a closed subset of $\operatorname{Im}f$ and therefore a closed subset of $\mathbb{R}^n$.
