Compact imply sequential compactness on sequences?

Asked Mar 19 '20 at 12:32

Active Mar 19 '20 at 12:53

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Let $Q=\{q_1,q_2,\dots,q_n,\dots\}$ be a set included in the dual of a Banach space $X^*$. If we know that $\overline{Q}$ is compact in the weak* topology, is it true that $\overline{Q}$ is sequentially compact in the same topology?

I know that if $X$ is separable then compactness is equivalent to sequential compactness.

Generally, it is well-known that the two concepts are independent. My question is if they coincide on particular sets, like Q. So for countable sets that are relatively compact.

I have tried to use the definition of compact sets with nets. I think that this one is the best for this problem.

edited Mar 19 '20 at 12:53

asked Mar 19 '20 at 12:32

Maxim Diana

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Weak* topology is only defined on a dual space. – Kavi Rama Murthy Mar 19 '20 at 12:33
You are right. I have modified the question. – Maxim Diana Mar 19 '20 at 12:37
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Does this answer your question? Compactness in the weak* topology – Sebastian Cor Mar 19 '20 at 12:43
No. It is well-known that the two concepts are independent. My question is if they coincide on particular sets, like $Q$. So for countable sets... – Maxim Diana Mar 19 '20 at 12:45
@MaximDiana: But an answer to your question follows essentially immediately, because if you have any compact but not sequentially compact $K\subset X^*$, just take $Q$ to be a sequence in $K$ with no convergent subsequence. – Eric Wofsey Mar 19 '20 at 15:28
How can you construct such a sequence? – Maxim Diana Mar 19 '20 at 16:01
By the definition of a not sequentially compact space, $K$ has such a sequence. – Alex Ravsky Apr 07 '20 at 04:49

Compact imply sequential compactness on sequences?

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