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Proof: $X^\ast$ separable $\implies X$ separable

Suppose $X$ is a normed vector space. Does $X^\ast$ separable imply $X$ separable?

If $X$ is complete, the answer is yes and this is well known. What about if we drop the completeness hypotesis? Is the statement still true? I'm a little puzzled... Thanks.

P.S. I find this question but it's not really the same thing, so I decided to open this new one.

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Romeo
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    Why isn't it the same thing? – Davide Giraudo Nov 10 '12 at 23:08
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    In a separable metric space, any subset is separable. This is because a metric space is separable if and only if it possesses a countable basis of open sets. Given a subset you only need to intersect this basis with it to get a countable basis for the subset. In particular, if you can prove that the completion of $X$ is separable you'll get separability of $X$ for free. – Giuseppe Negro Nov 10 '12 at 23:44
  • @DavideGiraudo : the questions are very similar, I know. But I wasn't looking for a proof of this fact; if I'm not wrong, in the linked question a proof of this fact - when $X$ is a Banach space - is provided. I was simply looking for an answer to the question: "Is completeness necessary"? Indeed, Giuseppe Negro's comment answers completely to my question. Thank you very much. – Romeo Nov 11 '12 at 09:20
  • I don't see where is the space is assumed complete in the link. – Davide Giraudo Nov 11 '12 at 11:03
  • In t.b. answer, he wrote: "Always state what you assume! For instance, your choice of $Y$ later on indicates that you work with a real Banach space (who should know that this is assumed if you don't state it?)" – Romeo Nov 11 '12 at 11:08

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