# When does compactness imply sequential compactness? [duplicate]

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

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

• No: see this answer to an earlier question. Specifically, the unit ball in $\ell^{\infty*}$ is not sequentially compact. – Brian M. Scott Apr 26 '13 at 15:44
• 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. – David Mitra Apr 26 '13 at 16:22