Subset of separable Hilbert space

Question

I have a separable Hilbert space H. How can one prove that the closed unit ball B is separable?

I'm told it's trivial but I can't see it.

My initial idea is to take the set A (dense set of H) and intersect it with B. This new set is clearly constable and it's closure is in B. but how can I show that B is in th closure of this set?

Thanks for any help

Look at what happens when $B$ isn't the closure of $B\cap A$ (you should get a contradiction, so anticipate it). — Aweygan, Jan 08 '17 at 16:42
Your approach will work in this case, but it isn't powerful enough in general. For instance, the unit sphere of $H$ is also separable, but it might not intersect your set $A$ at all. — Nate Eldredge, Jan 08 '17 at 17:19

John Wayland Bales · Answer 1 · 2017-01-08T22:53:32.217

Note that metric spaces are regular.

Show that if $B$ is the closure of an open subset of a separable regular space $H$ and that $A$ is a countable dense subset of $H$ then $A\cap B$ is a countable dense subset of $B$ in the topology inherited from $H$.

Consider two cases:

$P$ is a point in the interior $B\,\backslash\, \beta\{B\}$ of $B$.
$P$ is a point on the boundary $\beta\{B\}\subseteq H$

Case 1: Let $U$ be an open set in $B$ containing $P$ but no point of the boundary of $B$ in $H$. Then since $U$ is also an open set in $H$ it contains a point of $A$.

Case 2: Let $U$ be an open set in $H$ containing $P$. Then $U\cap(B\,\backslash\, \beta\{B\})$ is an open set in $H$ as well as an open set in $B$ so it contains a point of $A\cap B\subset A$.

Thus $B$ is separable.

Subset of separable Hilbert space

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