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

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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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$\mathbb R^n\setminus\{\mathbf 0\}$ 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 $\mathbf 0$.

This does not contradict the statement that "a convex cone may or may not contain the origin point" because according to the author's definition, a convex cone cannot contain any lines, which means $\mathbb R^n$ is not a convex cone.

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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.