Is a metrizable and compact subset of $E^*$ is a metrizable complete subset ? for the weak* topology

Question

If we have a Banach space $E$. and we consider it's dual $E^*$, it is a Banach space. so we consider the weak* topology ($\sigma(E^*,E)$) on $E^*$.

So My question is : If we have a set $B\subset E^*$, that is compact and Metrizable (weak* topology).

Why is $B$ is Metrizable complete ?

Answer 1

All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.

Answer 2

Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.