Compact subset of banach space is dentable

Question

Let $X$ be a real Banach Space and $K\subset X$ be compact. We've to show that $K$ is dentable.

A bounded set $B$ of $X$ is said to be dentable if $\forall \epsilon>0$, there exists $x_{\epsilon}\in B$ such that $x_{\epsilon}\not\in \overline{co}(B\setminus B(x_{\epsilon},\epsilon))$

I know two results-

1. For compact $K$, $\overline{co}(K)$ is compact.
2. If $\overline{co}(B)$ is dentable implies $B$ is dentable.

Using these two we can WLOG assume $K$ is compact and CONVEX.

Then by Krein Milman Theorem, $\text{Ext}(K)\neq\emptyset$ and $K=\overline{co}(\text{Ext}(K))$. (Here $\text{Ext}(K)$ denotes the set of all extreme points of $K$)

So I pick $x_0\in\text{Ext}(K)$. My intuition is this $x_0$ will serve our purpose for all $\epsilon>0$ i.e. $x_0\not\in \overline{co}(K\setminus B(x_0,\epsilon))$.

But I cannot prove that. Can anyone help me finish the argument? Thanks for your help in advance.

Answer 1

This is straight forward with Milman's converse to the Krein-Milman theorem:

Suppose $C = \operatorname{\overline{co}} A$ is compact and convex. Then $\operatorname{ext} C \subseteq \overline{A}$.

In this case, if we take $A = K \setminus B(x_0, \varepsilon)$, then $C$ is a compact, convex subset of $K$. Note that $x_0 \notin \overline{A}$, as we have removed all the points $\varepsilon$-close to $x_0$. This means that $x_0 \notin \operatorname{ext} C$. If we had $x_0 \in C$, this would make $x_0$ a non-trivial convex combination of points in $C$, and hence in $K$, which is impossible by assumption. Thus, $x_0 \notin C$, as required.