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If a cone is pointed, does that imply it is convex? It feels like it is true, but I want to be sure, since I can't seem to find it outright stated anywhere. For a cone $K$, if $\forall x \neq 0 \in K$, $-x \notin K$, then the cone must be restricted to half of the orthants in whatever dimension we are in. Based on the structure of cones it feels true, based on how cones look like. This is a far cry from a formal proof, hence why I'm concerned it isn't true.

Certainly the inverse isn't true. $R^n$ is a convex cone, but isn't pointed.

MegaZeroX
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    You seem to be using rather specialised definitions of "cone" and "pointed". At any rate, I don't understand them. – TonyK Sep 22 '19 at 21:15
  • A cone (how I've heard it to be defined) is a set of points such that if a ray passed from the origin to all points in the set, every point that the ray passed over was also in the set. Pointed I heard defined exactly as I posted, where for any point other than the origin in $K$, the reflection across the origin is not also in the set. – MegaZeroX Sep 22 '19 at 21:20
  • Or alternatively, a set where if $x \in K$, then $\theta x \in K$ for all $\theta \in R^+$ – MegaZeroX Sep 22 '19 at 21:31
  • So a cone is the same as a star domain with the distinguished point $x_0$ at the origin? – TonyK Sep 22 '19 at 21:43
  • Yes. At least the definition I saw in my convex and combinatorial optimization class. – MegaZeroX Sep 24 '19 at 22:59

2 Answers2

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Hint: think of a two-scoop ice-cream cone.

Rob Arthan
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  • Oh, right. Alternatively, in 2d, If you have just 2 vectors shooting into the same orthant, but nothing between them, it obviously isn't convex. Thank you! – MegaZeroX Sep 22 '19 at 21:38
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    @MegaZeroX: my pleasure! You are right that I could have said "think of a V-sign", but the ice-cream cone is nicer to think about. – Rob Arthan Sep 22 '19 at 21:41
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No: take a small-enough non-convex planar figure, imbed it in a hyperplane $x+y+z=c$ with $c$ large enough so that the imbedded figure is entirely in the first orthant. Then take all positive scalar multiples of the points in the imbedded image. Pointed, not convex.

paul garrett
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  • But the definition of cone I heard at least requires it to be centered on the origin. If c doesn't equal 0 here, than it can't be a cone – MegaZeroX Sep 22 '19 at 21:30
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    My intention was to have $c$ just specify the plane into which the given figure is imbedded. Then the process of talking all non-negative scalar multiples makes the resulting object include $0$. – paul garrett Sep 22 '19 at 21:38
  • Oh I think I see it. Thanks! – MegaZeroX Sep 22 '19 at 21:40