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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)



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

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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