= Solution
\b[True.] The <Banach-Alaoglu theorem> makes $B'$ compact for the <weak-star topology>. Its <compactness> can be seen by embedding it in the product $\prod_{x\in E}[-\|x\|,\|x\|]$: the closed conditions expressing linearity identify a closed subset with $B'$, and the product is compact by the <Tychonoff theorem>. The product topology is exactly <pointwise convergence> on $E$.
Part (i) supplies a compatible metric. Every compact <metric space> has a finite $1/n$-net for each positive integer $n$. The countable union of these finite nets is dense. Therefore
$$
\boxed{B'\text{ is weak-star separable}.}
$$
The argument concerns <topological separability>, with the relative <weak-star topology> specified in the question.
Back to article page
