= Solution
\b[True.] We prove <weak-star topology on an entire infinite-dimensional Banach dual is not metrizable>. Suppose instead that $E'$ has a countable local base $(W_n)$ at zero. Choose a basic <weak-star topology> neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.
For any $x\in E$, the set $\{\phi:|\phi(x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forces
$$
x\in\operatorname{span}S_n.
$$
Indeed a finite-dimensional span is norm closed, and the <Hahn-Banach theorem> supplies a <bounded linear functional> vanishing on it and nonzero at any point outside it.
It follows that $E=\bigcup_n\operatorname{span}S_n$. Each span is a proper finite-dimensional closed <vector subspace> and has empty interior in the infinite-dimensional <Banach space> $E$. This contradicts the <Baire category theorem>. Hence
$$
\boxed{E'\text{ with its weak-star topology is not metrizable}.}
$$
The uniform norm bound that made the metric work on $B'$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore \b[all true].
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