Krein Milman Theorem, why the closed convex hull is needed? I have a question about the Krein Milman that I'm having difficulty answering.  Here is the statement of the theorem. 
If ${K} $is a non-empty compact convex subset of a locally convex space $ {X}$, then ${\text{ext}\;K\neq\emptyset}$ and ${K=\overline{\text{co}}(\text{ext}\;K)}.$
My question is, why must it be the closed convex hull?
Surely when working in a locally convex space with a compact space the convex hull is already closed.  I have tried to think of an example in which ${K={\text{co}}(\text{ext}\;K)}$ is not true but haven't been able to think of anything.  
Does this problem only occur when we are dealing with infinite dimensional spaces?
I do know that the set of extreme points of a finite dimensional set need not be closed if this is of any help.
 A: 
Surely when working in a locally convex space with a compact space the convex hull is already closed.

No, that need not be. Consider $\ell^2(\mathbb{N})$, let $(e_n)_{n\in\mathbb{N}}$ be the "standard basis". Define $v_n = \tfrac{1}{n+1}\cdot e_n$ and
$$M := \{ 0\} \cup \left\{v_n : n\in\mathbb{N}\right\}.$$
$M$ is compact, and the convex hull of $M$ is not closed, since every element of the convex hull has only finitely many nonzero coordinates, but
$$x = \sum_{n=0}^\infty 2^{-(n+1)}\cdot v_n \in \overline{\operatorname{co}}(M)$$
has infinitely many nonzero coordinates. Let $K = \overline{\operatorname{co}}(M)$. Then $K$ is a compact convex set, which we can describe as
$$K = \left\{x\in \ell^2(\mathbb{N}) : (\forall k)(x_k \geqslant 0) \land \sum_{n=0}^\infty (n+1)\cdot x_n \leqslant 1\right\}.\tag{1}$$
It is easy to see that the right hand side of $(1)$, let's call that $L$, is a compact convex set, and it is clear that it contains $M$, hence it contains $\overline{\operatorname{co}}(M)$. For $N\in\mathbb{N}$, let $P_N\colon \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ be the orthogonal projection onto the subspace spanned by the $e_n$ with $n\leqslant N$. Then we have $P_N(x) \to x$ for all $x\in \ell^2(\mathbb{N})$, and $P_N(L) \subset \operatorname{co}(M)$, whence $L \subset \overline{\operatorname{co}}(M)$, so indeed $L = K = \overline{\operatorname{co}}(M)$.
Now let's see that $M = \operatorname{ext} K$. The inclusion $M\subset \operatorname{ext} K$ is easy (though if properly done a little tediously) to verify. If $x\in K\setminus M$, then either $x = c\cdot v_m$ for some $m\in\mathbb{N}$ with $0 < c < 1$, then
$$x = c\cdot v_m + (1-c)\cdot 0$$
shows that $x$ is not an extreme point of $K$, or there are (at least) two indices $m < n$ with $x_m > 0$ and $x_n > 0$. Let $w = v_m - v_n$. Then for small enough $\varepsilon > 0$, both $x+\varepsilon w$ and $x-\varepsilon w$ lie in $K$, and $x$ is the midpoint of the segment between these two, hence not an extreme point of $K$.
So we have an example of a compact convex set $K$ in a locally convex space - even a Hilbert space - with
$$\operatorname{co}(\operatorname{ext} K) \subsetneq K = \overline{\operatorname{co}}(\operatorname{ext} K).$$

Does this problem only occur when we are dealing with infinite dimensional spaces?

Yes. In finite-dimensional spaces, Carathéodory's theorem guarantees that the convex hull of a compact set $M$ is again compact, since it puts an upper bound on the number of points that are required in a convex combination. In infinite-dimensional spaces, the number of points of $M$ required to write an $x\in \operatorname{co} (M)$ as a convex combination of points of $M$ can be arbitrarily large, and that allows the existence of points in the closed convex hull that cannot be written as a convex combination of finitely many points of $M$.
