Separability of Hilbert/vector space and its countable basis

Question

The separable space is a one having the countable everywhere dense subset where the "dense" implies a topology. Consider a vector and Hilbert space $V$ of finite/nonfinite dimension. Please enlighten me on some details about relation between this formal definition of separability and availability of countable basis for this vector space. Seems the former should follow from the latter but I was not able to make sure this at the moment. Put differently, if vector space has a countable basis whether it should be separable as a topological space? Need proof just in a nutshell. Can I do without norm here? Certainly, for such a topological question I need the topology itself. But then the question above is meaningless for a `naked' vector space without additional structures like norm, metric, or scalar product. So is the question meaningless even in the case of a finite dimensional $V$-space?

Answer 1

Two types of bases are considered for (topological) vector spaces $X$: a Hamel basis (a set of vectors $B$ such that every $x \in X$ can be written uniquely as some finite sum $x=\sum_{i=1}^n c_i b_i$ where $b_1,\ldots,b_n \in B$, and a Schauder basis where we have a sequence $(b_1, b_2, \ldots)$ of vectors such that for every $x \in X$ there is a unique sequence of scalars $(c_n)_n$ such that $\sum_{n=1}^\infty c_i b_i := \lim_{N \to \infty } \sum_{i=1}^N c_i b_i = x$, where the limit taken is w.r.t. a topology on $X$ that makes the vector space operations continuous (i.e. a topological vector space), most often taken to be some norm-topology.

If a TVS $X$ has a Schauder basis or a countable Hamel basis, it is topologically separable as well: the countable dense subset in either case can be taken to be $$D = \{\sum_{i=1}^n q_i b_i: n \in \mathbb{N}, q_1,\ldots,q_n \in \mathbb{Q}\}$$

which is countable as $\cup_{n=1}^\infty \mathbb{Q}^n$ is a countable set by standard set theory, plus the fact that we can approximate vectors like $c b_i$($ c \in \mathbb{R}$) by vectors of the form $q b_i$, where $q \in \mathbb{Q}$, as the rationals are dense in the reals and scalar multiplication is continuous. (This argument also works over other fields when these fields are separable, like $\mathbb{C}$).

The Wikipedia entry I linked to mentions a famous example by Enflo of a separable (topologically) Banach space without a Schauder basis. Most separable Banach spaces in practice do have a Schauder basis, though. No complete metric TVS can have a countable Hamel basis by Baire's theorem. So in neither definition of basis need a topologically separable TVS have a countable basis.