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

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.
