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This question already has an answer here:

It is well known that if $ X $ is a metric space then sequential compactness and compactness are equivalent.

Now we consider a normed vector space $ E $ and its dual $ E^\ast $. From Banach Alauglou theorem we know that the closed unit ball in $ E^\ast $ is compact in the weak $ \ast $ topology. Moreover this topology is not metrizable in general. So it is natural to ask:

Is this ball also sequentially compact in the weak $ \ast $ topology? How can I prove it?

In general, when does compactness imply sequential compactness? (I'm loooking for some results about it)

Thanks

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marked as duplicate by 23rd, vonbrand, Micah, Amzoti, Jim Apr 26 '13 at 16:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ No: see this answer to an earlier question. Specifically, the unit ball in $\ell^{\infty*}$ is not sequentially compact. $\endgroup$ – Brian M. Scott Apr 26 '13 at 15:44
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    $\begingroup$ For a Banach space $X$: see the first answer to Brian's link for a sufficient condition ($X$ is separable). Another sufficient condition is that $X^*$ does not contain a copy of $\ell_1$. Also of interest is the following result of Edward Odell and Haskel Rosenthal: If $X$ is separable, then the unit ball of the second dual of $X$ is weak$*$ sequentially compact if and only if $X$ contains no isomorph of $\ell_1$. This result is contained in Joseph Diestel's Sequences and Series in Banach Spaces. $\endgroup$ – David Mitra Apr 26 '13 at 16:22