TITLE: The dual of $\ell_\infty$ is weak* separable
QUESTION [5 upvotes]: Where can I find the proof that $\ell_\infty^*$ is weeak$^*$-separable?
I want to re-examine the proof of that fact.

REPLY [7 votes]: I'm not entirely sure what you're asking. I suppose you're asking why $l_{\infty}^{\ast}$ is weak$^{\ast}$-separable. To see this, simply note that $l_1$ is dense in the weak$^{\ast}$-topology of $l_{\infty}^{\ast}$ (by Goldstine's theorem). Obviously, $l_1$ contains a countable norm-dense subset $S$, which is also weak$^{\ast}$-dense because the norm-closure is always smaller than the weak$^\ast$-closure. You'll find a proof of the (rather tautological) Golstine theorem on the Wikipedia page I linked to.
Added later: Of course, this does by no means imply that $l_{\infty}^{\ast}$ is second countable in the weak$^{\ast}$-topology. In fact, for an infinite-dimensional Banach space the weak$^{\ast}$-topology is not even first countable (even if its restriction to the unit ball is metrizable for duals of separable Banach spaces).
Added even later: The only thing used in the above argument is of course the separability of $l_1$. Verbatim the same argument shows

The bidual of a separable Banach space is weak$^{\ast}$-separable.

However, I don't know whether the converse holds.
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