Why is the weak* topology not in general metrizable?

A Banach space is a topological group under addition. The dual is a topological group under the weak$^*$ topology. The weak$^*$ topology is weaker than the operator norm topology, so is it first-countable? It is also Hausdorff. (Seems intuitively obvious, might require the Hahn-Banach Theorem to make rigorous). So by the Birkhoff-Kakutani Theorem it should be metrizable. But I believe this is only the case for the finite-dimensional case. Where is my reasoning wrong?

• "The weak* topology is weaker than the operator norm topology, so is first-countable" is, except in the finite-dimensional case, wrong. – Daniel Fischer Jan 11 '14 at 16:44
• First countability doesn't transfer to stronger or weaker topologies. Discrete topology and trivial topology are both always first countable and any topology is between the two. – tomasz Jan 11 '14 at 16:52

It turns out that, if $X$ has uncountable base (as a linear space), then the weak$^*$ topology of $X^*$ is not first countable. See
The weak$^*$ topology on $X^*$ is not first countable if $X$ has uncountable dimension.
But, even if the dimension of a normed space is $\aleph_0$, then its dual shall coincide with the dual of the completion of $X$, which will have dimension $2^{\aleph_0}$, and hence the weak$^*$ topology of $X^*$ will not be first countable, as well.
• Is the weak* topology on $X^*$ induced by the completion of $X$ the same as the weak* topology on $X^*$ induced by $X$? – David Mitra May 29 '14 at 22:20
• If and only if $X$ is complete. I'm sure that was a rhetorical question by @DavidMitra, but just in case somebody doesn't know: if $Y$ is a (linear) subspace of the completion of $X$ such that $\lambda\lvert_Y = \mu\lvert_Y \implies \lambda = \mu$ for $\lambda,\mu\in X^\ast$, then the dual space of $X^\ast$ with the weak* topology induced by $Y$ is $Y$, so no two such subspaces induce the same weak* topology on $X^\ast$. – Daniel Fischer May 30 '14 at 23:44