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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 ?

BrianTag
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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.

SmileyCraft
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  • It makes sense. Thank you. – BrianTag Jan 14 '19 at 11:31
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    Not every compact space is sequentially compact (see this thread: https://math.stackexchange.com/questions/220422/compact-space-which-is-not-sequentially-compact). Of course, both properties are equivalent for metric spaces. – MaoWao Jan 14 '19 at 12:47
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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.

Myunghyun Song
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