Proposition 14.4.1.label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, and $S = \bracsn{y \in F|\ \norm{y}_F \le 1}$ be the closed unit ball of $F$, then
- (1)
$S$ is separable with respect to the $\sigma(F, E)$-topology.
- (2)
$S$ is metrisable with respect to the $\sigma(F, E)$-topology.
- (3)
For any $A \subset F$, $A$ is separable with respect to the $\sigma(F, E)$-topology.
- (4)
If the duality is norming, then there exists $\seq{y_n}\subset F$ such that for each $x \in E$, $\norm{x}_{E} = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$.
Proof. (1), (2): Let $D \subset E$ be a countable dense subset. By the Arzelà-Ascoli Theorem, $S$ is embedded as a subspace of $K^{D}$. By Theorem 6.3.10, $\real^{D}$ is metrisable. By Proposition 5.5.7, $K^{D}$ is separable. Thus $S$ is also metrisable and separable by Proposition 8.1.2.
(3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By Proposition 8.1.2, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable.$\square$
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