I want to prove the following:
Let $K$ be a convex polytope. Show that $K$ has a finite number of extreme points.
I have seen the bound for the cardinality of the set of extreme points:
$|E| \leq 2$ $ n\choose 2$
I know that the $n$ comes from the number of half-spaces that form a polytope, the 2 in the binomial coefficient comes from that the extreme points are vertices, because they are in the intersection of two hyperplanes, but I dont know how to write and complete this better :)
but I dont know how to prove it, because if I prove that result I answer the above problem.
Can someone help me to prove this?
A vertex of a $d$ dimensional polytope must be the intersection of at least $d$ of the bounding hyperplanes. Linear algebra tells us that the intersection of fewer than that many hyperplanes must have dimension at least 1, if a point $p$ in our polytope is in such an intersection and not contained in any other hyperplanes, then we can choose a direction inside this intersection and find two points on either side of $p$ still in the polytope, hence $p$ is not extremal.
This gives the bound of $|E| \le {n \choose d}$ and we get the finiteness property you wanted.
The other bound you claim $|E| \le 2{n \choose 2}$ is false, try an $8$ dimensional hypercube.