I am reading Chapter 3: Convexity of Rudin's "Functional Analysis".
Here is the problem I'm having trouble solving (number 18):
Let $K$ be the smallest convex set in $\mathbb{ R}^3$ that contains the points $(1, 0, 1), (1, 0, -1)$, and $(\cos \phi, \sin \phi, 0)$, for $0 \le \phi \le 2 \pi $. Show that $K$ is compact but that the set of all extreme points of $K$ is not compact. Does such an example exist in $\mathbb{R}^2$?
So I need to show that this set is closed and bounded.
Could you help me with that?
Thanks.
Best Answer
You can explicitly write down a parametrization of $K$, showing that it's a continuous image of a compact set. But here is a general argument.
Claim
If $A\subset \mathbb R^n$ is compact and $C$ is the smallest convex set containing $A$, then $C$ is compact.
Proof. The set $C$, also known as the convex hull of $C$ consists of all finite convex combinations of the points of $A$. But Carathéodory's theorem on convex hull, we only need the combinations of $(n+1)$ points of $A$. Let $$\Delta_n=\left\{x\in \mathbb R^{n+1}: x_i\ge 0, \ \sum x_i=1\right\}$$ be the standard $n$-dimensional simplex (a compact set). By the above, $C$ is the image of $A^{n+1}\times \Delta_{n}$ under the map $$\Phi((p_1,\dots, p_{n+1}), (x_1,\dots,x_{n+1}))=\sum_{i=1}^{n+1} x_i p_i$$ which is evidently continuous. Thus $C$ is compact.
Some further hints:
The set of extreme points of $K$ contains almost all of the unit circle in the $xy$-plane, except for one point. This is why it's not closed.
In two dimensions, the set of non-extreme point is open in the induced topology of $\partial K$, because open line segments are open relative to $\partial K$. This is why the set of extreme points is closed in this case.