Proof of Straszewicz's theorem

Question

I'm interested in the following result, known as Straszewicz's theorem:

(1) For $C$ a compact convex set, the set of exposed points $\text{exp}\ C$ is dense in the set of extreme points $\text{ext}\ C$.

After attempting to prove this for about three hours, I gave up and decided to Google. However, I wasn't able to find any proof, save for that in the original paper. My German isn't excellent, but I believe the author first proves

(2) Every open half-space that intersects $C$ contains an exposed point of it.

and then immediately declares

(3) The closed convex hull of all exposed points is the same as that of all extreme points, which is $C$.

as an immediate consequence of (2) and the fact a closed convex set is the intersection of all its supporting hyperplanes.

I understand the proof of (2), and I also understand all of the other auxiliary results that are mentioned, but I fail to see how (3) follows. Further, even though (1) $\Rightarrow$ (3), I fail to see how (3) $\Rightarrow$ (1).

I'm looking for either a completion of the previous argument, or any other complete argument proving (1).

Answer 1

To show $(2) \implies (3)$, we show that $C$ is the intersection of all closed half-spaces containing $\exp C$.

Suppose not; then there is a closed half-space $H$ containing $\exp C$ which does not contain $C$. But then the complement of $H$ is an open half-space intersecting $C$ which does not contain any exposed point of $C$, which contradicts $(2)$.

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To show $(3) \implies (1)$, we need to say that $C$ is the convex hull of the closure $\overline{\exp C}$. This is not true in general; in general, what we can say is that $C$ is the closure of the convex hull, and these are not the same. However, it is true for bounded sets, which is what we have here. I am not sure of what the details are, but Carathéodory's theorem is required.

But suppose we do know that $C$ is the convex hull of the closure $\overline{\exp C}$. Then in particular, every extreme point of $C$ lies in this convex hull.

But if $\mathbf x$ an extreme point of $C$, it is (by definition) not even in the convex hull of $C - \mathbf x$. The only way to get $\mathbf x$ to be in the convex hull of a subset of $C$ is to include $\mathbf x$ in that subset. Therefore $\overline{\exp C}$ must already include every extreme point.

Answer 2

$\neg(3)\Rightarrow\neg(2)$: By $\neg(3)$, there exists $x\in C\setminus\overline{\operatorname{conv}{\exp{C}}}$. Both $\{x\}$ and $\overline{\operatorname{conv}{\exp{C}}}$ are closed subsets of compact $C$, and so compact in their own right. By Hahn-Banach, there exists a hyperplane strictly separating the two; call that hyperplane $\{y:f(y)=0\}$.

W/oLoG $f(x)>0$, so that $$\exp{C}\subseteq\overline{\operatorname{conv}{\exp{C}}}\subseteq\{y:f(y)<0\}$$ But the open half-space $\{y:f(y)>0\}$ contains $x\in C$, but does not contain an exposed point of $C$. Thus $(2)$ fails.

$(3)\Rightarrow(1)$: For any set $X$, we have $$\operatorname{extr}{\overline{\operatorname{conv}{X}}}=\operatorname{extr}{\overline{X}}\tag{a}$$ Moreover, for any $Y\subseteq\operatorname{extr}{X}$, we have $$\operatorname{extr}{Y}=Y\tag{b}$$ Also, since any exposed point is extremal, and the set of extremal points is closed, we have $$\exp{C}\subseteq\overline{\exp{C}}\subseteq\operatorname{extr}{C}$$ Thus \begin{align*} \operatorname{extr}{C}&=\operatorname{extr}{\operatorname{conv}{\operatorname{extr}{C}}} \\ &=\operatorname{extr}{\overline{\operatorname{conv}{\exp{C}}}} \tag{by (3)} \\ &=\operatorname{extr}{\overline{\exp{C}}} \tag{by (a)} \\ &=\overline{\exp{C}} \tag{by (b)} \end{align*}