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I recently came across the following problem:

Let $ K \subseteq \mathbb{R}^2 $ be a closed convex cone (meaning K is closed under non negative linear combinations) and I am asked to show it is polyhedral meaning it is the intersection of finitely many half spaces, or alternatively that it is finitely generated $ K = cone(A) $ where $ A $ is a finite set and cone(A) denotes all possible non-negative linear combinations of elements of A, or alternatively that K has a finite number of extreme vectors meaning vectors $x\in K$ such that if $ x=y+z $ where $ y,z\in K $ are non-negative multiples of x.

I have given three possible definitions of polyhedral cone but for some reason I cannot seem to see why a convex cone which is closed in $\mathbb{R}^2$ satisfies either one of the three possible definitions. I certainly need and appreciate all the help I can get on this.

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    $\begingroup$ This is false: consider a round cone (like the things pictured at en.wikipedia.org/wiki/Cone) $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '16 at 20:26
  • $\begingroup$ @MarianoSuárez-Álvarez: Thanks but those pictures appear to be three dimensional whereas my problem is in $ \mathbb{R}^2 $ and the cone is closed, did I miss something or can you counter this on the plane $ \mathbb{R}^2 $? $\endgroup$ – Don John Prep Oct 13 '16 at 20:31
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    $\begingroup$ Ah! I missed that :-| $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '16 at 20:37
  • $\begingroup$ There is never need in this site to add comments of the type «can you help me?», for the unique and only purpose of the site is to answer questions and therefore help askers. $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '16 at 20:42
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Let $K$ be a closed cone in the plane. If $K$ is not contained in a closed halfspace, then $K$ coincides with the plane. Suppose then that it is contained in a closed halfspace, whose boundary is a line $L$. Let now $L'$ be a line parallel to $L$ contained in the interior of that halfspace. The intersection $K\cap L'$ is a convex closed subset of $L'$, so it is a closed interval. Now consider cases: the interval may be the whole of $L'$, a closed halfline, or a bounded closed interval.

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    $\begingroup$ Thanks for the answer, ould you please explain what you mean by this approach of intersection with the line L being an interval? $\endgroup$ – Don John Prep Oct 13 '16 at 21:13
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    $\begingroup$ Well, I mean exactly that! That the intersection of K with my line L' is a closed interval. I don't understand what you don't understand. $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '16 at 23:26
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    $\begingroup$ I mean I don't understand how you used this to solve the original problem? $\endgroup$ – Don John Prep Oct 14 '16 at 2:11

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