Milman converse for compact generating sets

The conclusion is immediate if $K=\mathrm{\varnothing }$. Otherwise $A\mathrm{\ne }\mathrm{\varnothing }$, because the closed convex hull of the empty set is empty. Put $B=\bar{A}$. Since $K$ is closed by [F3] and contains $A$, one has $B\subseteq K$; hence $B$ is closed in compact $K$ and compact by [F4].

Assume for contradiction that $w\in \mathrm{ext}K\setminus B$. The open set $X\setminus B$ contains $w$, so translation gives a zero-neighborhood $V$ with $w+V\subseteq X\setminus B$. Continuity of subtraction at $(0,0)$ gives a zero-neighborhood $W$ with $W-W\subseteq V$, and [F2] gives an open convex zero-neighborhood $U\subseteq W$. Thus $(w+U-U)\cap B=\mathrm{\varnothing }$.

The family $\{b+U:b\in B\}$ is an open cover of the nonempty compact set $B$. Compactness supplies a listed subcover $O_1,\dots ,O_n$ with $n\ge 1$. For $i\in n=\{0,\dots ,n-1\}$ put $C_i=\{b\in B:b+U=O_{i+1}\}$. Each $C_i$ is nonempty because $O_{i+1}$ belongs to the displayed cover. Apply [F6] to the function $i\mapsto C_i$ with domain $n$; if $c$ is the resulting choice function on its family of values, set $b_{i+1}=c(C_i)$. Then $b_{i+1}\in B$, $O_{i+1}=b_{i+1}+U$, and hence $B\subseteq {\bigcup }_{j=1}^n(b_j+U)$. Put $B_j=B\cap (b_j+U)$ and $K_j=\stackrel{}{\mathrm{co}}(B_j)$. Each $B_j$ is nonempty because it contains $b_j$.

Each $K_j$ is a nonempty compact convex subset of $K$: the closed convex set $K$ contains $B_j$, hence its closed convex hull, and $K_j$ is closed in compact $K$, so [F4] applies. Moreover $w\mathrm{\notin }{K}_j$. Indeed $\mathrm{co}(B_j)\subseteq b_j+U$ by convexity; if $w$ lay in its closure, the open neighborhood $w+U$ of $w$ would meet $b_j+U$, giving $b_j\in w+U-U$, contrary to $b_j\in B$ and step 2.1.

Let $H=\mathrm{co}(K_1\cup \cdots \cup K_n)$. By [F5], $H$ is compact and therefore closed in the Hausdorff ambient space, and every point of $H$ is ${\sum }_{j=1}^n{t}_j{x}_j$ with $x_j\in K_j$, $t_j\ge 0$, and ${\sum }_j{t}_j=1$. Since $A\subseteq B\subseteq {\bigcup }_j{B}_j\subseteq H$, closedness and convexity of $H$ give $K=\stackrel{}{\mathrm{co}}(A)\subseteq H$; conversely every $K_j\subseteq K$ and $K$ is convex, so $H\subseteq K$. Hence $H=K$.

Apply the representation in step 5.1 to $w$: write $w={\sum }_j{t}_j{x}_j$ with $x_j\in K_j$. If exactly one coefficient is positive, it equals one and gives $w=x_j\in K_j$, contradicting step 4.1. Otherwise, for each $i$ with $t_i>0$ one has $0<t_i<1$ and may write $w=t_i{x}_i+(1-t_i)y_i$, where $y_i=(1-t_i{)}^{-1}{\sum }_{j\mathrm{\ne }i}{t}_j{x}_j\in K$.

Since $w$ is extreme, [F1] applied to the strict representation in step 6.1 gives $x_i=w$ for every positive coefficient $t_i$. At least one coefficient is positive, so $w\in K_i$ for some $i$, again contradicting step 4.1. Therefore no such $w$ exists and $\mathrm{ext}K\subseteq B=\bar{A}$.

If $A$ is compact, then it is closed by [F3], so $\bar{A}=A$ and step 7.1 gives $\mathrm{ext}K\subseteq A$. This proves both assertions, including the empty case from step 1.1.