Why is the weak* topology not in general metrizable?

Question

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?

Answer 1

If a topological space is metrizable, then it has to be first countable.

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.