Proposition 2

Let x be an extremal point of a compact convex set A in a Hausdorff locally convex space E. Then for every open neighbourhood V of x in E, there exists an open half-space F in E such that $x \in F \cap A \subset V \cap A$ (in other words, the traces on A of the open half-spaces containing x, form a fundamental system of neighbourhoods of x in A).

For every open half-space D of E containing x, the set $A \cap \overline{D}$ is a compact neighbourhood of x in A, and the intersection of all these neighbourhoods is precisely the point x (any two distinct points can be strictly separated by a closed hyperplane (II, p. 38, prop. 4). By prop. 1 of GT, I, § 9.2, it is sufficient to prove that the sets $A \cap \overline{D}$ form a filter base. Now if we write $L_D = A \cap (E - D)$, the set $L_D$ is convex, compact and contained in the convex set $A - \{x\}$; if $D_1, D_2$ are two open half-spaces of E containing x, the convex envelope B of $L_{D_1} \cup L_{D_2}$ is therefore contained in $A - \{x\}$; but B is a compact set (II, p. 14, prop. 15), therefore there exists a closed hyperplane H that separates x strictly from B (II, p. 38, prop. 4) and if the open half-space determined by H and containing x is D, then we have $L_{D_1} \cup L_{D_2} \subset L_D$, therefore $A \cap \overline{D} \subset (A \cap \overline{D_1}) \cap (A \cap \overline{D_2})$.