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Convex combination in compact convex sets.

Let K be a nonempty, compact, convex set in $R^n$. I am trying to prove that every $x\in K$ is a convex combination of no more than $n+1$ extreme points of K.

This is what I got so far. Use induction on $n$. First $n=0$, that is single-point sets, is trivial. Suppose we have the result for all sets. B, with $dim(B) \le n-1$. Let $K$ have dimension $n$ and $x \in K$ and let $x_0$ be an extreme point of $K$. Take the line segment $[x_0,x]$ and extend it to get $\{t|(1-t)x_0+tx \in K\} = [0,\alpha]$ for some $\alpha$. Let $y=(1-\alpha)x_0+\alpha x$. Since $\alpha \ge 1$, $x=t_0 x_0 + (1-t_0)y$ where $t_0=1-\frac{1}{\alpha} \ge 0$. By construction, let $y \in \partial K$ be the point where the ray from $x_0$ through $x$ exists $K$.

Then, I should use a hyperplane separation theorem to finish this problem, but I don't know how to use it. Also, is this proof is right? Please help me out! Thank you.

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Alice M
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    I am not sure exactly what you are trying to prove here. Are you trying to show that a compact convex set is the convex hull of its extreme points? Reducing the number of required extreme points to $n+1$ is straightforward, it just relies on the fact that a matrix with dimensions $(n+1) \times m$ has a non-trivial null space when $m > (n+1)$. – copper.hat Apr 04 '12 at 05:44
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    See also Robert's answer here. – t.b. Apr 04 '12 at 05:49

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