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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?

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  • $\begingroup$ Its not necessarily true , for example consider the linear function $ f: \mathbb{R^2} \to \mathbb{R}$ with $ f(x,y) = x $ . Then $F =\{(x,\frac{1}{x}) : x >0 \} $ is closed but $ f(F) = (0 ,\infty)$ . $\endgroup$ – dem0nakos Apr 9 '18 at 14:07
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    $\begingroup$ @dem0nakos That set is not a semiplane. $\endgroup$ – José Carlos Santos Apr 9 '18 at 14:07
  • $\begingroup$ Oh okay , i actually thought that the question was in general for any closed , anyway ! $\endgroup$ – dem0nakos Apr 9 '18 at 14:54
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$f(S)$ 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.

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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 $\mathbb{R}^n$.

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  • $\begingroup$ Clarification, the map goes from $R^m$ to $R^n$. The notation $y\ge 0$ was meant to signify that every component of $y$ is non negative. $\endgroup$ – Jsevillamol Apr 9 '18 at 14:28

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