This question already has an answer here:
- Compactness in the weak* topology 2 answers
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)