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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 $0$.

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  • $\begingroup$ $e_n \to 0$ weakly but $\sqrt{n} e_n$ doesn't. Why is $0$ in the weak closure of $\sqrt{n} e_n$ ? $\endgroup$
    – reuns
    Commented Apr 17, 2017 at 22:36
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    $\begingroup$ See math.stackexchange.com/questions/42337/on-the-weak-closure $\endgroup$
    – keej
    Commented Apr 17, 2017 at 22:42
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    $\begingroup$ If $X$ is a separable Banach space the norm-closure of the unit ball of $X^*$ is metrizable (and compact) in the weak$^*$ topology. But in general the weak$^*$ topology on $ X^*$ is not metrizable $\endgroup$ Commented Apr 17, 2017 at 23:23
  • $\begingroup$ @user1952009 A detailed proof: math.stackexchange.com/questions/231176/… More generally, if you have an orthogonal sequence $(x_n)$ s.t. $0\leq \|x_n\|\leq \sqrt{n}$, for every $n$, then $0$ belongs in the weak closure of $\{x_n:n\in \mathbb{N}\}$. One can have the same result even if he substitutes $(\sqrt n)$ by a sequence $(a_n)$ satisfying an appropriate condition (by looking carefully at the proof for the $\sqrt{n}$ case, the condition that $(a_n)$ needs to satisfy becomes obvious). $\endgroup$ Commented Apr 17, 2017 at 23:39

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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.

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