2
$\begingroup$

Possible Duplicate:
Does separability follow from weak-* sequential separability of dual space?
$\omega^*$-separability of $l_\infty^*$.

Recently I read a Theorem stating, Let $X$ be a Banach space which is separable then every weakly compact subset is metrizable. I noticed that the separability is used in order to prove that $X^{*}$ is weakly* separable and in the rest of proof separability of $X$ is no longer used.

Motivated by that my question is, let $X$ be a Banach space if $X^*$ is $\mathrm{weakly}^*$ separable then does this imply that $X$ is separable, if not this would weaken the hypothesis.

  • 0
    Related: http://math.stackexchange.com/questions/43365/does-separability-follow-from-weak-sequential-separability-of-dual-space2012-07-03
  • 0
    This is probably inspired by a proof of the Eberlein–Šmulian theorem, so you might be interested in the references given [here](http://math.stackexchange.com/q/153194) (I recommend having a look at Whitley's paper mentioned in the accepted answer). Seeing that the thread linked to by @Asaf gives a negative answer to a weaker question I'm voting to close as a duplicate.2012-07-03
  • 0
    Actually $(\ell^\infty)^\ast$ is weak$^\ast$-separable, as I argue [here](http://math.stackexchange.com/q/41851), so that's a much more elementary counterexample.2012-07-03
  • 0
    @t.b. thanks that is much better!Also could this be a good counterexample because of the schur property of $l_1$ ?2012-07-03
  • 1
    I'm not exactly sure what you have in mind. It's also a good example because $\ell^\infty$ is a [Grothendieck space](http://math.stackexchange.com/q/128388/5363) :)2012-07-04

0 Answers 0