# Compact subset of banach space is dentable

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.

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.