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.
2026-04-21 13:29:17.1776778157
Is the unit ball of a separable Banach space itself separable?
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A metric space is separable if and only if it is second countable. Second countability passes to subspaces.