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Let $E$ be a Banach space. Suppose that $E'$ is weakly-* sequentially separable, that is, that there exists a countable $D \subset E'$ s.t. every $x' \in E'$ is a limit point of a sequence in $D$. Does it follow that $E$ is separable?

This question arises from a conversation with a friend of mine. He thinks this is true and plans to use it to prove separability of $C(K)$ for a metrizable and compact Hausdorff $K$. I'm not so sure this can work, though. Of course, if $E'$ is norm separable then $E$ is separable, but we are talking about a much weaker topology here.

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  • $\begingroup$ What a coincidence, I asked myself the same thing a few days ago (I didn't want to prove separability of $C(K)$ from that, though). $\endgroup$ – t.b. Jun 5 '11 at 10:34
  • $\begingroup$ Well, not quite the same thing, but still similar though. I admit that I have no idea. $\endgroup$ – t.b. Jun 5 '11 at 10:47
  • $\begingroup$ @Theo: In that link you proved that the non-separable $\ell^\infty$ has got a weak-$\star$ separable dual. This does not disprove the claim of this question because this dual needs not be sequentially separable, right? $\endgroup$ – Giuseppe Negro Jun 5 '11 at 11:09
  • $\begingroup$ No, as far as I can see it doesn't tell usanything about your question. I know for sure that the unit ball of the dual of $\ell^\infty$ is not sequentially separable (e.g. because it would be metrizable then, which it isn't), but I suspect this is true for the whole $\ell^{\infty}$. I'm not so used to thinking about separability in a non-metric setting, though. $\endgroup$ – t.b. Jun 5 '11 at 11:16
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    $\begingroup$ Of course, there are much easier ways to prove that $C(K)$ is separable. $\endgroup$ – Nate Eldredge Jun 5 '11 at 14:49
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Let $JT$ denote the James Tree space, $Q: JT \longrightarrow JT\,^{\prime\prime}$ the canonical embedding, $D_\ast$ a countable norm-dense subset of $JT$, let $D= Q(D_\ast)$ and let $E = JT\,^\prime$. Note that $E$ is nonseparable in the norm topology and that the $w^\ast$-sequential closure of $D$ in $E^{\prime}$ is $E^{\prime}$ since $Q(JT)$ is $w^\ast$-sequentially dense in $JT\,^{\prime\prime}$ (for the last claim, see in particular Corollary 2 of Lindenstrauss and Stegall Examples of separable spaces which do not contain $\ell_1$ and whose duals are non-separable, Studia Math. 54 (1975), p.81--105).

I should mention that until the appearance of the Lindenstrauss-Stegall result cited above, it seems to have been open since the time of Banach whether there could exist a separable Banach space that has nonseparable dual and is $w^\ast$-sequentially dense in its bidual.

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    $\begingroup$ Click here for the Lindenstrauss-Stegall article. $\endgroup$ – t.b. Jun 6 '11 at 7:52
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    $\begingroup$ I'd like to point out that $JT$ is a modification of James's famous example of a non-reflexive Banach space that is isometrically isomorphic to its bidual. $\endgroup$ – t.b. Jun 6 '11 at 7:59
  • $\begingroup$ Thanks for adding the link to the article Theo; in the future I'll try to remember to include such links in my answers for the benefit of readers. $\endgroup$ – Philip Brooker Jun 7 '11 at 23:42
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    $\begingroup$ You're welcome. As I was unaware of this work, this wasn't really much effort, but people tend to be lazy... Give them a clickable way to beautiful maths and they go there, force them to copy-paste the reference into Google and look around a little bit and they'll miss it for sure... And these Polish collections are a bit difficult to browse through if you're not used to them. It's a very nice answer! Thanks a lot, Philip. $\endgroup$ – t.b. Jun 7 '11 at 23:50
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    $\begingroup$ Further to Theo's second comment to my answer, I'll just add that there are many modifications to James' constructions out there and it is worth looking at them when wanting to make a conjecture about Banach spaces - they are important test cases. A good source of background and references on these spaces is the book The James Forest by Helga Fetter and Berta Gamboa de Buen, published in 1997 by Cambridge University Press. $\endgroup$ – Philip Brooker Jun 8 '11 at 0:03

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