Is it true that a sphere is a convex set?
Because the thing that I know is that the space enclosed by the sphere is the one convex.
thanks for the help.
The word "sphere" is used in two different senses. Sometimes it means a solid ball, and sometimes it means the surface of a ball. In mathematics, it's better to use the word "ball" for the solid ball, and to reserve the word "sphere" for the surface of the ball. In that case, the sphere is not convex. Of course, the ball is convex.
A similar thing happens in two dimensions, where the word "circle" is sometimes used to mean a disk, and sometimes (more accurately) it refers to the perimeter of the disk.
I would like to see how one shows that a sphere $\displaystyle S_r(x_0):=\{x \in E: ||x-x_0||=r\}$ in an arbitrary (not necessarily $\displaystyle \mathbb{R}^n$) normed space $\displaystyle E$ is non-convex.
I would say that this is simply due to the fact that for $\displaystyle x_1,x_2 \in S_r(x_0), t \in [0,1]$ the function $\displaystyle f(t):=||(1-t)(x_1-x_0)+t(x_2-x_0)||$ is non-constant on $\displaystyle (0,1)$. But is it really an adequate argument? There seems to be too much of intuition coming from $\displaystyle \mathbb{R}^n$ in it, don't you find?