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Let $X$ be a Banach space and $K$ be a weakly compact subset of $X$, i.e. compact set with respect to the $\sigma(X, X^{*})$-topology. Is it true that $K$ is weakly separable? Since weak topology is never metrizable for infinite dimensional space, it is impossible to construct a countable dense subset using a metric. Also, I tried to use Eberlein-Smulian theorem, which states that weakly compactness is equivalent to weakly sequentially compactness. But this isn't helpful to me.

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  • $\begingroup$ If this is not true in general, it will be helpful to show for $l_\infty$. $\endgroup$ – Seewoo Lee Jun 3 '18 at 9:57
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Let $H$ be a non-separable Hilbert space, say one with an orthonormal basis $\{e_t\}$. Let $K$ be the closed unit ball of $H$. By Reflexivity and Banach Alaoglu Theorem $K$ is weakly compact. Suppose there is a countable dense set $\{f_n\}$ in $K$. Since the weak closure of $M \equiv span \{f_n:n \geq 1\}$ is equal to the norm closure we see that each $e_t$ is in the separable Hilbert space spanned by the $f_n$'s. This is a contradiction.

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