# Are separable Hilbert spaces also separable in the topological sense?

A Hilbert space is called separable if it has a dense, countable subset, which occurs iff it admits a countable orthogonal basis. Separability in the topological sense is the same as saying the topological space $(X,\mathcal{T})$ has a countable base.

Since all Hilbert spaces are equipped with an inner product, which can be used to define a metric via $d(f,g) = (f-g,f-g)^{1/2}$, then one call always define a metric topology on any Hilbert space. Let us call this the nominal metric topology on a Hilbert space $X$.

Now suppose $X$ is a separable Hilbert space. Is the topological space $(X,\mathcal{T})$ necessarily separable in the topological sense when $\mathcal{T}$ is the nominal metric topology on $X$? What about if $\mathcal{T}$ is another topology? (some easy examples clearly fail--the discrete topology, for instance)

• Separability in the topological sense means having a countable dense subset. A topological space with a countable base is called second countable. This is stronger than separability in general, but the two conditions are equivalent for metric spaces. – carmichael561 Sep 7 '17 at 3:11
• @carmichael561 I see. The book I got this out of defined separability the way I did, but probably assumes metrizability tacitly. – Mortified Through Math Sep 7 '17 at 3:13
• separability for topologies is never defined as having a countable base. – Henno Brandsma Sep 7 '17 at 5:16