# Is the unit ball of a separable Banach space itself separable?

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.

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