Prove a closed ball in $\mathbb R^3$ has an infinite number of extreme points. How do I show that a closed ball in $\mathbb R^3$ has an infinite number of extreme points ?
(Closed ball is written as $S = \{(x,y,z) \in \mathbb R^3 | \sqrt {x^2 + y^2 + z^2} \le R \}$)
I know that $z$ is an extreme points of a convex subset $C$ if and only if $z \in \text{conv}(\{x,y\}) \Rightarrow z = x \lor z = y$.
Geometrically, I do verify that this property hold.
Also, $z$ is an extreme point if it's a face of dimension $0$.
Can someone help me proving this rigorously ? 
 A: Rather denot by  $B = \{(x,y,z) \ | \ x^2 + y^2 + z^2 \le R^2\}$  your set, that is $\{ u \ | \ ||u||\le R\}$
Recall the Cauchy inequality 
$$||u||^2 ||v||^2 \ge \langle u, v \rangle ^2$$
with equality if and only if the vectors $u$, $u$ are proportional. 
Assume now $||u|| = R$, then for every $v \in B$ we have
$$|\langle u,v\rangle | \le ||u|| \cdot ||v|| = R \cdot ||v|| \le R^2$$
We have the equality 
$$|\langle u,v\rangle | = R^2 $$
if and only if $v = \pm u$. We conclude that if $||u|| = R$ then for every $v$, $||v|| \le R$ we have:
$$\langle u,v \rangle \le R^2$$
with equality if and only if $v=u$. 
Moreover, if $v \in \mathbb{R}^3$ so that $\langle u, v\rangle \le R^2$ for every $u \in S$  then necessarily $||v|| \le R$. Indeed, apply this with $u = \frac{R}{||v||} \cdot v$. 
Now we see that 
$$B = \cap_{u \in S} \{ v \ | \ \langle u,v\rangle \le R^2 \}$$
an intersection of hyperplanes, hence a convex set ( this of course can be proved directly).
Moreover, every point $u \in S$ is the intersection of the ball $B$ with the hyperplane $H_u \colon = \{ v \ | \ \langle u,v\rangle =R^2 \}$ ( $u$ is an exposed point). It follows that the point $u$ is extreme. Indeed, if $u = \lambda_1 v_1 + \lambda_2 v_2$ then 
$\langle u, v_i \rangle \le R^2$ so we get $\langle u, \sum \lambda_i v_i \rangle \le R^2$. However, we do have equality, so we need to have equality $\langle u, u_i \rangle = R^2$ whenever $\lambda_i \ne 0$, but that implies $u_i = u$. 
$\bf{Added:}$
$B$ has only $0$-dimensional faces and $B$ itself. Why? A face is an extreme convex subset. That is, if a point $u \in F$ has a decomposition $u = \sum \lambda_i u_i$, $\lambda_i >0$, $\sum \lambda_i =1$, then $u_i \in F$. Now, take a face that contains two points $u_1$, $u_2$. The full open segment $(u_1, u_2)$ is in the interior of $B$. Take the point $1/2( u_1 + u_2)$. This is in F. For any other point $u \in B$, there exists a segment with one end at $u$ and containing $1/2( u_1 + u_2)$ inside ( because $1/2( u_1 + u_2)$ is in the interior of $B$). But that implies $u\in F$. 
What we have used is $B$ is strictly convex, that is, whenever $u_1$, $u_2$ are two distinct points on the boundary $\partial B$, the open segment $(u_1, u_2)$ is in the interior of $B$. 
A: Points on the surface (which form an $(n-1)$-sphere) of an $n$-disc, $D^n:=\left\{v \in \mathbb{R}^n\,|\,\|v\|_2\leq R\right\}$, where $R>0$ and $n \in \mathbb{N}$, are all the extreme points of $D^n$.
Hint: See Minkowski's Inequality (http://www.imomath.com/index.php?options=595&lmm=0).
