I am trying to show that $\operatorname{Ext}B(X)$ does not have to be closed where $B(X)$ is closed unit ball of normed space $X$ and $\operatorname{Ext}B(X)$ is the set of extreme points of unit ball.
My definition for extreme point : $x \in S(X)$ is an extreme point of $B(X)$ if $x$ cannot be written as arithmetic mean of distinct elements of $B(X)$. ($S(X)$ is unit sphere)
I thought that $\Bbb R^n$ can be an example for this and I tried to use $\Bbb R^3$ but I could not find the proper sequence to show that $\operatorname{Ext}B(\Bbb R^3)$ is not closed. I really stuck.
I appreciate any help.
Thank you
Best Answer
Let $C=\operatorname{co} ((\pm 1, 0, \pm 1) \cup \{ (0,y,z)| y^2+z^2 = 1\})$.
The points $\{ (0,y,z)| y^2+z^2 = 1, |y| \neq 1\}$ are extreme, but the points $(0,0,\pm 1)$ are not.
$C$ is convex, $0 \in C^\circ$ and $C=-C$. Then the Minkowski functional is a norm.