Lemma 2

Suppose $K$ contains at least two points. For every $\varepsilon > 0$, there exists a partition of $K$ into two non-empty subsets $K_1$ and $K_2$, having the following properties :

a) $K_1$ is convex and compact ; b) we have $\| x_1 - x_2 \| < \varepsilon$ for every $x_1$ and $x_2$ in $K_2$.

Let $L$ be the closure of the set of all extremal points of $K$. By the Krein-Milman th. (II, p. 55, th. 1), $K$ is the closed convex envelope of $L$. Since $K$ contains at least two points, so does $L$. For every $x \in L$, let $A_x$ be the set of all $y \in L$ such that $\| x - y \| \leq \varepsilon / 4$. By the hypothesis, on $K$, there exists a countable subset $D$ of $L$ such that $L = \bigcup_{x \in D} A_x$. Since the norm is lower semi-continuous, each of the sets $A_x$ is closed. Applying Baire’s th. (GT, IX, § 5, No. 3, th. 1) to the compact space $L$, we see that there exists a point $a$ in $D$ and an open subset $U$ in $E$ such that $L \cap U$ is non-empty and is contained in $A_a$. Since $L$ contains at least two points, and since $E$ is Hausdorff, we can choose $U$ in such a way that $L \notin U$.

Let $M$ be the closed convex envelope of $L \cap \complement U$. For every real number $t$ such that $0 < t < 1$, let $M_t$ be the set of all vectors of the form $tx_1 + (1 - t) x_2$ with $x_1 \in M$ and $x_2 \in K$; this is a non-empty, compact convex subset of $K$. We shall prove that $M_t \neq K$ by reductio ad absurdum. Suppose that $M_t = K$; then every extremal point $x$ in $K$ belongs to $M_t$, hence can be written in the form $x = tx_1 + (1 - t) x_2$ with $x_1 \in M$ and $x_2 \in K$. This implies that $x = x_1 = x_2$, and so $x \in M$. By Krein-Milman th. (II, p. 55, th. 1), we have $K = M$, and $K$ is the closed convex envelope of $L \cap \complement U$. By II, p. 56, corollary, this implies that $L \subset L \cap \complement U$, which contradicts the relation $L \cap U \neq \varnothing$.

Put $d = \sup_{x \in K, y \in K} \| x - y \|$ and choose a real number $t$ such that $0 < t < 1$ and $t < \varepsilon / 4d$. Put $K_1 = M_t$ and $K_2 = K - M_t$. By the preceding argument, the sets $K_1$ and $K_2$ are non-empty, and $K_1$ is convex and compact. Let $M'$ be the closed convex envelope of $L \cap U$. Since $K$ is the closed convex envelope of the set $L = (L \cap \complement U) \cup (L \cap U)$, it is also the closed convex envelope of $M \cup M'$. Let $x_1$ and $x_2$ be two points in $K_2$; for $i = 1, 2$, there exist $y_i \in M$, $z_i \in M'$ and a real number $\alpha_i$ such that $0 \leq \alpha_i \leq 1$ and $x_i = \alpha_i y_i + (1 - \alpha_i) z_i$. If $\alpha_i \geq t$, then

; this contradicts the assumption that $x_i \notin M_i$. Hence $\alpha_i < t$, for $i = 1, 2$, and so For every point $z$ in $M'$, we have $\| z - a \| \leq \varepsilon / 4$, since $L \cap U \subset A_a$, and so, in particular $\| z_i - a \| \leq \varepsilon / 4$. Thus This completes the proof.