# 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 $$K$$.

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.