When is a convex cone in $\mathbb{R}^n$ finitely generated by a subset?

Question

A convex cone is a set $C\subseteq\mathbb{R}^n$ closed under adittion and positive scalar multiplication. If $S\subseteq\mathbb{R}^n$ we consider $p(S)$ defined as the set of all positive linear combinations (this is, all linear combinations where the coefficients are positive). Of course $p(S)$ is a convex cone.

If $C$ is a convex cone, is there $S\subseteq C$ finite such that $C=p(S)$?

The answer to this question is no. For instance, take $S$ as any two independent vectors in $\mathbb{R}^2$ and take $p(S)$ without its boundary. The obtained convex cone is not finitely generated.

It's simpler than that. It can be proved that $p(S)$ is always closed in $\mathbb{R}^n$ for any $S\subseteq\mathbb{R}^n$ finite. Then how about this?

If $C$ is a closed convex cone, is there $S\subseteq C$ finite such that $C=p(S)$?

Does anyone know a counterexample or is it true?

Answer 1

Counterexample in $\Bbb R^3$: the full cone with basis a disk (cone, for the layman). A finitely generated convex cone in $\Bbb R^3$ would just be some sort of full infinite pyramid with convex polygonal basis.

Addition: I've been tipped by the person next to me that it is however true for polyhedral cones, id est, cones in the form $\{x\in\Bbb R^n\,:\,Ax\ge 0\}$ for some matrix $A$. Truth to be told, I cannot weight on this, though.

Answer 2

Regarding the Addition part of G.Sassatelli

The claim is true for polyhedral cones.

Proof: (Sketch)

Let $C:=\{x\in\Bbb R^n\,:\,Ax\ge 0\}$, and $B \subset R^n$ be the closed Unit ball w.r.t $\|\|_{\infty}$ norm, i.e., $B$ is Unite Square centered origin. Now $C \cap B$ is compact polyhedral which generates $C$, in another words , $C=P (C \cap B)$. Note that for generating $C$ we don't need whole set $C \cap B$, but its extreme points are enough! Since $C \cap B$ is compact polyhedral it has finite extreme points, Therefore $C = P(ext(C \cap B))$.