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My notes say that the theorem of Banach-Alaoglu states the following: If $X$ is a normed separable space, then every bounded sequence in $X'$ has a weak-* convergent subsequence.

How is this equivalent to the usual formulation from Wikipedia, etc - i.e. the closed unit ball being weak-*-compact, for a (not necessarily separable) normed space $X$? Or is it a special case?

Davide Giraudo
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amaflix
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2 Answers2

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If $X$ is separable, then the closed unit ball in $X'$ is metrizable with respect to the weak-$\ast$ topology. Hence compactness, given by Banach-Alaoglu, implies sequential compactness.

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It's a special case. If $X$ is separable then it's not hard to show that the closed unit ball of $X'$ is metrizable in the weak-* topology (note $X'$ itself is not weak-* metrizable), so compactness is equivalent to sequential compactness in that case.