Does a second category set in a topological vector space always contain a basis for the space?

Question

Suppose I have a topological vector space $X$ and a second category set $B$ in that space. Is it the case that $B$ contains a basis for $X$? For instance, in normed linear spaces, an open ball (which is second category) surely contains a Hamel basis for the space because all possible 'directions' are contained in it. I was wondering whether this fact can be extended to both topological vector spaces from NLS as well as to second category sets from open balls. In case we cannot obtain a Hamel basis, can we get a Schauder basis?

Answer 1

No. For instance, let $X$ be any separable infinite-dimensional Banach space. In particular, $X$ is second-countable so it has only $2^{\aleph_0}$ different closed sets and thus only $2^{\aleph_0}$ different countable unions of closed sets with empty interior. Let $(S_\alpha)_{\alpha\in\mathfrak{c}}$ be an enumeration of all such countable unions, so every first category subset of $X$ is contained in some $S_\alpha$. Fix some nonzero vector $x\in X$; we can then by transfinite recursion construct a sequence $(b_\alpha)_{\alpha<\mathfrak{c}}$ such that $\{x\}\cup\{b_\alpha\}_{\alpha<\mathfrak{c}}$ is linearly independent and $b_\alpha\not\in S_\alpha$ for each $\alpha$. (At each step, the set of choices of $b_\alpha$ which would maintain the linear independence is the complement of a proper linear subspace of $X$, which cannot have first category since $X$ is covered by two of its translates, and thus is not contained in $S_\alpha$.) The set $B=\{b_\alpha\}_{\alpha<\mathfrak{c}}$ then has second category in $X$ by construction, but does not contain a Hamel basis for $X$ since its span does not contain $x$. (And if $X$ itself has no Schauder basis, then $B$ cannot contain one either.)