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Possible Duplicate:
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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    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?)"2012-11-11

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