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Nov 30, 2023 at 22:14 comment added S Argyros Every weakly compact subset of $l_{\infty}(\kappa)$ has topological weight at most $\kappa$. The reason is that if $W$ is weakly compact then $ (W, w)$ and $(W,w^*)$ are homeomorphic and the topological weight in the $w^*$ topology of the norm bounded subsets of $l_{\infty}(\kappa)$ is at most $\kappa$.This actually holds for every dual Banach space. Now notice that the basis of $c_0 ( \Gamma ) $ together with 0 is weakly compact with topological weight equal to the cardinality of $\Gamma$.
Nov 30, 2023 at 16:40 history edited M.González CC BY-SA 4.0
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Nov 30, 2023 at 16:29 history edited M.González CC BY-SA 4.0
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Nov 30, 2023 at 16:23 history edited M.González CC BY-SA 4.0
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Nov 30, 2023 at 16:15 history answered M.González CC BY-SA 4.0