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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

$f(S)$ 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.


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 $\mathbb{R}^n$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.