Is the dual space of $\ell^{\infty}$ contains a countable total subset ? I'm trying to understand the proof of Mr. Robert Whitley "Projecting $m$ onto $c_0$" and he claims in his proof that this is true . but I think that if this is true than $\ell^{\infty}$ is separable and this is not true . thanks for any help .
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3@LucasHenrique No, it is not. The dual of a non-separable space is non-separable (https://math.stackexchange.com/questions/2388541/proving-that-a-banach-space-is-separable-if-its-dual-is-separable). This also answers the question. – MaoWao Apr 02 '21 at 12:48
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1You might want to check whether or not the text you are reading assumes different set theory axioms. When the Axiom of Choice is ommitted, strange things may happen! – Ruy Apr 02 '21 at 14:25
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Proving that $(\ell^\infty)'$ is larger than $\ell^1$ needs the axiom of choice? But omitting choice we still won't get that $(\ell^\infty)'$ is separable, unless ZFC is inconsistent. – reuns Apr 02 '21 at 14:41
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@reuns: see https://www.sciencedirect.com/science/article/pii/S0019357798800396 – Abdelmalek Abdesselam Apr 02 '21 at 22:48