Exposed points of a compact convex set. Suppose that $K$ is a compact convex set in a topological vectorial space locally convex such that $K$ is the closed convex hull of its exposed points $(x_\mu)$. I would like to know if it is true that there exists a set of functionals $(f_\mu)$, such that $f_\mu$ exposes $x_\mu$, that separes the points of $K$, i.e. for all distinct points x and y of K, it exists $\mu$ such that $f_\mu(x)\neq f_\mu(y)$? 
Someone has a tip to show this or un conterexample?
Thank you for your help 
 A: This answer works with extreme points instead of exposed points and is therefore not an answer to the original question! (I'll leave it here for reference)
The statement is


*

*wrong in the uncountably infinite dimensional case

*true in the finite dimensional case

*I'm not sure in the countably infinite dimensional case


Let $I$ be an uncountably infinite index set.
Consider the cube $C=\prod_{n\in I} [0,1]\subset\mathbb R^I$ in the product topology. By Tychonov $C$ is compact, because all $[0,1]$ are compact. Because $\mathbb R$ is locally convex, the product space is also locally convex.
Assume there is some continuous functional $f:\mathbb R^I\to\mathbb R$ that "exposes" the extreme point $0$. Then we can assume $0\equiv f(0) < f(e_i)$ for all $i\in I$, where $e_i$ is the $i$-th unit vector and also an extreme point of $C$. For all $J\subset I$ let $e_J$ be the vector with elements
$$ (e_J)_j = \begin{cases} 1,& j\in J\\ 0,& j\notin J\end{cases}$$
One can see that
$$ \mathbb R\ni f(e_I) = \sum_{i\in I} f(e_i) \overset{\text{$I$ uncountable, $f(e_i)>0$}} = \infty $$
which is a contradiction.
Proof:
Let $\Lambda$ be the set of all finite subsets of $I$ ordered by inclusion. Then $(e_J)_{J\in\Lambda}$ is a net that converges to $e_I$ (because it converges pointwise). Therefore $\lim_{J\in\Lambda} f(e_J)\to f(e_I)$. We get the contradiction
$$ \infty = \sum_{i\in I} f(e_i) \equiv \lim_{J\in\Lambda} \sum_{j\in J} f(e_j) \overset{\text{$J$ finite}}= \lim_{J\in\Lambda} f(\sum_{j\in J} e_j)
= \lim_{J\in\Lambda} f(e_J) = f(e_I) \in \mathbb R $$
Alternative proof without nets: Let $\varepsilon>0$ be arbitrary and let $U$ be an open neighbourhood of $e_I$ such that $|f(v-e_I)|<\varepsilon$ for all $v\in U$. By definition of the product topology we can choose $U$ such that $U=\prod_{i\in I_0} U_i\times\prod_{i\in I\setminus I_0}\mathbb R$ for a finite subset $I_0\subset I$ and open neighbourhoods $U_i$ of $1\in\mathbb R$.
Let $N\in\mathbb N$ be arbitrary and let $J_N\supset I_0$ be finite with $\sum_{j\in J_N} f(e_j)>N$ (exists because $I$ is uncountable and all $f(e_i)>0$). Then $J_N\supset I_0$ implies $e_{J_N}\in U$, therefore
$$f(e_I)=f(e_{J_N})+f(e_I-e_{J_N}) > N - \varepsilon$$
Since $N$ was arbitrary we get a contradiction to $f(e_I)\in\mathbb R$ being finite.
