Let $E$ - be a complex (or real) vector space. Is it always possible to find a Hausdorff locally convex topology $\mathcal{T}$ on $E$ such that $(E, \mathcal{T})$ is separable?
The motivation goes from this question De Rham cohomology $H_{dR}^1(-):\mathcal{Top}\to\mathcal{Vec}$ is suryective. While answering on that I had to note that not all vector spaces possess a metric in which they are separable. Now I wonder if the same thing works without metrizability.
It seems to me that separability makes some constraint to the cardinality of $E$ or its algebraic basis (also called Hamel basis). I can prove that if $E$ possesses a metrizable topology such that $E$ is separable then $E$ is not larger than continuum. But for arbitrary topology there is an issue with the cardinality of local base of $E$. Also I can't imagine a general construction that gives a topology $\mathcal{T}$ if it is possible for all vector spaces.
Assume that such topology exists. If we consider a local base $U_\alpha$ where $\alpha \in I$ then there is an injective map from $E$ to $2^{S \times I}$ where $S$ is a dense set. This function maps $x \in E$ to a subset $\{(s,\alpha): x \in s + U_\alpha\} \subset S \times I\}$. It is injective since $S$ is dense and $E$ is Hausdorff. Then cardinality of $S \times I$ equals cardinality of $I$ since $S$ is supposed to be countable. Therefore $E$ is not greater then $2^I$. For simple case of metrizable spaces (for which I is countable) it gives that $E$ is not greater then continuum. But in general I don't know how to estimate cardinality of $I$.
