Is every extreme point in a compact convex set contained in a defining supporting hyperplane?

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Let $K \subseteq X$ be a compact convex subset of a locally convex space $X$. Let $k \in K$ be an extreme point.

Question 1: Does there exist a supporting hyperplane of $X$ containing $k$?

I think the answer is “yes” via some Hahn-Banach argument, although I’m a little confused about this at the moment. But what I really want to know is the following:

Question 2: Suppose that $K = \cap_i H_i$ where each $H_i$ is a closed half-space. EDIT: Suppose also that for each $i$ the face $K \cap \partial(H_i)$ is not empty. end EDIT Then is $k$ contained in the boundary of some $H_i$?

That is, assuming the answer to Question 1 is “yes”, I want to know whether I can guarantee that the supporting hyperplane can be chosen from a list of hyperplanes I already have.

Notes:

  • I’m aware that the extreme point $k$ doesn’t have to be exposed — i.e. it need not be the case that $\{k\} = K \cap Y$ for some supporting hyperplane $Y$. But I want to know whether we have $\{k\} \subseteq K \cap Y$ for some supporting hyperplane $Y$.

  • If $K$ is the intersection of a finite number of half-spaces, I’m pretty sure the answer to both questions is yes. Even in finite dimensions, I’m not sure about the answer if $K$ is the intersection of infinitely many half-spaces.

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The answer to question 2 is: "no" even in $X = \mathbb R$. Indeed, $$ [0,\infty) = \bigcap_{n \in \mathbb N} [-1/n, \infty) $$ and the extreme point $0$ is not a boundary point of any $[-1/n, \infty)$.

After the edit, we need at least two dimensions, I guess: Take $X = \mathbb R^2$ and $$ X = \bigcap_{q \in \mathbb Q \cap [0,2 \pi]} \{ x \in \mathbb R^2 \mid x_1 \cos(q) + x_2 \sin(q) \le 1\}. $$ That is, we write the unit circle as a countable intersection of half spaces. Note that the boundary of each half space intersects the unit circle.