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I have a post about this theorem before If $X$ is separable, then the closed unit ball of $X^*$ is weak-star metrizable. Some calculus helps needed!. But what if $X=\{0\}$ (separable), how to show the closed unit ball of $X^*$ is weak-star metrizable?

Thank you!!

Answer Lee
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The dual of $X=\{0\}$ is also $\{0\}$ and if there is something like a unit ball here it’s at most a singleton, hence metrisable. What’s the interest in such a degenerated case?

Henno Brandsma
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  • Thank you for your reply. Because the general proof does not include this case. Yes. $X^={0}$ and I think the closed unit ball of $X^$ is ${0}$ too. But I don't understand why it is metrizable. – Answer Lee Dec 08 '17 at 20:32
  • @AnswerLee there’s only one metric on a singleton. And only one topology too. – Henno Brandsma Dec 08 '17 at 20:40
  • I am so sorry my background is not that strong. Can you be more specific please? Really appreciate!! – Answer Lee Dec 08 '17 at 20:45
  • @AnswerLee any finte discrete space is metrisable, but here it's even more trivial; there is nothing to prove. – Henno Brandsma Dec 08 '17 at 22:18