Exercise 22, § 4

Let $E$ be a Banach space of countable type and infinite dimension. (a) Now prove that the set $\mathscr{K}$ of convex, balanced and compact subsets A of $E$, such that $E_A$ is of infinite dimension, is infinite and has a cardinal $\leq 2^{\mathrm{Card}(\mathbf{N})}$ (TG, IX, p. 114, exerc. 17). For every $x_0 \in E$ and every $A \in \mathscr{K}$, the set $x_0 + A$ contains a free subset of cardinal $2^{\mathrm{Card}(\mathbf{N})}$ (II, p. 85, exerc. 24, c)). (b) Let $x_0 \neq 0$ in $E$. Now prove that there exists a family $(y_A)_{A \in \mathscr{K}}$ such that $x_0$ and the $y_A$ form a free family and that one has $y_A \in x_0 + A$ for every $A \in \mathscr{K}$ (well order $\mathscr{K}$ and reason by transfinite induction using a)).

(c) Let $f \in E^*$ be a linear form such that $f(x_0) = 1$ and $f(y_A) = 0$ for every $A \in \mathscr{H}$, and let $H = f^{-1}(0)$. Now prove that every subset $M$ of $H$ which is convex, balanced and semi-complete, is necessarily of finite dimension (observe that otherwise, $M$ would contain a convex, balanced and compact set $A$ of infinite dimension, hence $y_A$ would belong to $H \cap (x_0 + M)$).

(d) Now prove that $H$, endowed with the topology induced by that of $E$, is not ultrabornological, although being bornological and barrelled (III, p. 46, exerc. 14). (Using c), prove that if $H$ were ultrabornological, its topology would be the finest locally convex topology, and deduce a contradiction.)