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Is $\mathbb R^n\setminus\{\mathbf{0}\}$ a convex set?

I read the convex analysis book (R.T. Rockafellar), in the book he wrote " convex cone may or may not contain the origin point". Then a question occur to me that the whole space $\mathbb{R}^n$ is a convex cone, so it may not contain the origin point too, i.e.$\mathbb R^n\setminus\{\mathbf{0}\}$. But the origin point $0$ is not in the line segment that joins points $(-x,0)$ and $(x,0)$, thus the whole space is not a convex cone, which makes me confused.

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    $\begingroup$ Is the segment $[-x, x]$ entirely contained in $\Bbb R^n \setminus \{0\}$ for non-zero $x$? $\endgroup$ – Arnaud Mortier Feb 20 '18 at 11:07
  • $\begingroup$ What does "convex" mean to you? $\endgroup$ – Umberto P. Feb 20 '18 at 11:07
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    $\begingroup$ It is not, why do you have that intuition? $\endgroup$ – klirk Feb 20 '18 at 11:10
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    $\begingroup$ You should look at the definition of convex pointed and convex blunt cones. In 3d that is exactly what you'd expect. $\endgroup$ – LinAlg Feb 21 '18 at 2:06
  • $\begingroup$ @LinAIg, thanks. The definition of blunt convex cone answers my question. The convex cone $\mathbb {R}^n\setminus \{\mathbf{0}\}$ is blunt, which can be excluded from the definition of convex cone (depending on the definition of the author or in the context). $\endgroup$ – bin Feb 21 '18 at 7:45
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$\mathbb R^n\setminus\{\mathbf 0\}$ is not a convex set for any natural $n$, since there always exist two points (say $(-1,-1,\dots,-1)$ and $(1,1,\dots,1)$) where the line segment between them contains the excluded point $\mathbf 0$.

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  • $\begingroup$ But convex cone (the closed half space), it may or may not contain the origin point. $\endgroup$ – bin Feb 20 '18 at 11:18
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    $\begingroup$ @bin How does that even relate to the original question? A convex cone is by its own wording convex! $\endgroup$ – Parcly Taxel Feb 20 '18 at 11:20
  • $\begingroup$ @José Carlos Santos and @ Parcly Taxel, thanks for the prompt answer. I read the convex analysis book (R.T. Rockafellar), in the book he wrote " convex cone may or may not contain the origin point". Then a question occur to me that the closed half space is a convex cone, so it may or may not contain the origin point too. But from same argument as you said, closed half space is not a convex cone, which make me confused. $\endgroup$ – bin Feb 20 '18 at 11:28
  • $\begingroup$ @bin No, a half-space is a convex cone too. I never said anything about convex cones. You are confused. $\endgroup$ – Parcly Taxel Feb 20 '18 at 11:31
  • $\begingroup$ Sorry, I am not saying you said half space is not convex cone. I mean if we accept that " the convex cone may or may not contain the origin point". Then the closed half space is a convex cone, so it may not contain the origin point too, the two points, e.g. $-x$ and $x$ on the negative and positive axis, the line segment between them contains the excluded origin point $0$. $\endgroup$ – bin Feb 20 '18 at 11:36
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No, it is not. The line segment going from $(1,0,0,\ldots,0)$ to $(-1,0,0,\ldots,0)$ isn't contained in it.

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so my answer is no , you can take any two antipodan point for example i take $n =3$ and i take p=(x,y,z)and the antipodale point q=(-x,-y,-z) the segment [p,q] in not containd in your set

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The set $\Bbb R^n-\{\underbrace{000.....0}_{n\ \ times}\}$ is not convex because of the same arguement of Mr. Parcly Taxel but is connected and your intuition probably leads you to that

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A set is convex if every line between two of its points is contained in the set.

I drew a picture showing that $\mathbb R^2\setminus D$ with $D$ a disc is not convex. The sitution is still the same for $\mathbb R^2\setminus \{0\}$, its just harder to draw.

enter image description here

As you can see, the segment from $A$ to $B$ leaves the highlighted area, i.e. $\mathbb R^2\setminus D$

This immediately proofs the situation for $\mathbb R^n\setminus\{0\}$, as a set is non-convex if it contains a non-convex subset.

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