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)

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    $\begingroup$ 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. $\endgroup$ – carmichael561 Sep 7 '17 at 3:11
  • $\begingroup$ @carmichael561 I see. The book I got this out of defined separability the way I did, but probably assumes metrizability tacitly. $\endgroup$ – Mortified Through Math Sep 7 '17 at 3:13
  • $\begingroup$ separability for topologies is never defined as having a countable base. $\endgroup$ – Henno Brandsma Sep 7 '17 at 5:16

By definition "separable" means having a countable dense subset (for topological, metric or Hilbert spaces). For Hilbert spaces we could define it as having a countable orthogonal base as well (this only makes sense in an inner product space, of course). This is the equivalent to saying its induced metric topology is separable. For metric spaces this then further implies that its metric topology has a countable base. (but e.g. the Sorgenfrey line is separable but does not have a countable base, so we need to assume something on the space.)


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