Extreme points of an intersection of convex set with countably many linear spaces Let $V$ be some `nice' vector space and let $T: V\to \mathbb{R}$ be a linear functional over $V$.  
Define
\begin{align}
M=  K \cap  \bigcup_{i \in \mathbb{N} } \{ v \in V: T(v)=c_i \} 
\end{align}
where $K$ is some compact and convex subset of $V$.  Moreover, $K$ has at most $n$ extreme points. 
That is, $M$ is an intersection of $K$ with countably many hyperplanes. 
The question I have is, can we say something about the extreme points of $M$?  
The general answer, I suspect, is that it is impossible to say something without extra assumptions on $T$. So, we would have to make some assumptions on $T$. 
Some motivation: The following result can be shown when the intersection is finite. 
Let $\tilde{M}=K \cap  \bigcup_{i=1}^m \{ v \in V: T(v)=c_i \} $. 
Then, one can show, with little restriction on $T$, that the extreme points of $ \tilde{M}$ can be represented as a convex combination of at most $m$ extreme pints of $K$. 
 A: The last paragraph is wrong.  Consider the case $m=1$.  $\tilde{M}$ is the intersection of $K$ with one hyperplane.  In general this will not contain any extreme point of $K$, so its extreme points will not be convex combinations of $m=1$ extreme points of $K$.
What is true is that every extreme point of $M$ is an 
extreme point of the intersection of $K$ with one hyperplane,
and this is a convex combination of two extreme points of $K$.
Namely, suppose $p = \sum_{i=1}^r t_i p_i$, $t_i \in (0,1)$, $\sum_i t_i = 1$, is a convex combination of $r > 2$ extreme points of $K$. Say $T(p) = c$.  If any $T(p_j) = c$, then 
$p$ is a convex combination of $p_j$ and $(p - t_j p_j)/(1-t_j)$ which are both in $M$, so not an extreme point.  Otherwise some 
$T(p_i) > c$ and some $< c$.  Relabelling, suppose $T(p_1) > c$
and $T(p_2) < c$.  Then 
$$q = \frac{T(p_1) - c}{T(p_1)-T(p_2)} p_2 + \frac{c - T(p_2)}{T(p_1)- T(p_2)} p_1$$ is a nontrivial convex combination of $p_1$ and $p_2$ which is in $M$, and $p$ is a convex combination of this and some other member of $M$, thus not an extreme point.
