TITLE: Metrizability of weak topology on separable Hilbert space
QUESTION [5 upvotes]: The weak* topology on the dual of a separable space is metrizable.  On a Hilbert space, the weak topology and the weak* topology coincide, and the dual of the Hilbert space is itself.  Thus, on a separable Hilbert space, the weak topology is metrizable.
What's the error in the reasoning here?  There must be an error because of the example $\{\sqrt{n} e_n \}$ being a set whose weak closure includes $0$ but no sequence in the set converges weakly to $0$.

REPLY [3 votes]: It is not true that the weak* topology on the dual of a separable Banach space is metrizable.  What is true that the weak* topology is metrizable on any bounded subset of the dual.  Your set $\{\sqrt{n} e_n\}$ is not bounded, so it does not give a contradiction.