If $X$ is a separable Banach space, then do we know that its unit ball has a countably dense subset contained in the unit ball?

This isn't obvious to me.

  • $\begingroup$ Well, doesn't separable mean there is a countably dense subset? $\endgroup$ – Christopher A. Wong Apr 5 '13 at 22:36
  • $\begingroup$ Yes, but I want this countably dense subspace of the unit ball to be contained in the unit ball too. $\endgroup$ – user58514 Apr 5 '13 at 22:39
  • $\begingroup$ If a subset $S \subset X$ is dense, then obviously $S \cap B$ is dense in $B$, where $B$ is an open unit ball. $\endgroup$ – xyzzyz Apr 5 '13 at 22:40
  • $\begingroup$ Sorry, my brain must be fried. Why is it obvious that $S\cap B$ is dense in B? $\endgroup$ – user58514 Apr 5 '13 at 22:43
  • $\begingroup$ Oh, you are assuming B is the open ball. I mean the closed one. $\endgroup$ – user58514 Apr 5 '13 at 22:45

A metric space is separable if and only if it is second countable. Second countability passes to subspaces.

  • $\begingroup$ Nice answer! Better than what I was writing up ^^ $\endgroup$ – Olivier Bégassat Apr 5 '13 at 22:49
  • $\begingroup$ Were you going to write something like:\\ take a point on the boundary. Take an open neighbourhood of it. This open neighbourhood must intersect the open unit ball. Take an element in this intersection. There exists an open set contained in this intersection. Which contains an element of our countably dense subset. Hence the (closed) unit ball is dense? $\endgroup$ – user58514 Apr 5 '13 at 22:54
  • $\begingroup$ But yes, thank you for your elegant answer @yup! $\endgroup$ – user58514 Apr 5 '13 at 22:56
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    $\begingroup$ Second countablility is a "hereditary" property. What is interesting is that separability is not. $\endgroup$ – ncmathsadist Apr 5 '13 at 23:12

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