If $G$ is an arbitrary countable subset of a Hilbert space $H$ (over complex numbers), does $G$ somehow generate a separable Hilbert subspace? Preferably having $G$ as a countable dense subset? If so, what exactly is the procedure? If the claim is not true, what would be the counterexample?
Any help will be appreciated.
The closure of the span of $G$ would be a separable Hilbert space. It is unlikely that $G$ is in fact a countable dense subset of this space, however. For a simple counterexample, consider the sequence space $\ell_2$. This is generated by the standard basis vectors (which is a countable set), but these are clearly not dense in $\ell_2$. However, the set of linear combinations of elements of $G$ with coefficients in $\mathbb{Q}(i)$ is a countable dense subset.