Let $X$ be a separable Banach space, $M \subset X$ a linear subspace. Must $M$ be a Borel set in $X$?


I believe the answer is "no," since I have seen authors who are careful to talk about "Borel subspaces". But I have not been able to find a counterexample.


If the answer is indeed "no", does every infinite-dimensional separable Banach space contain a non-Borel dense linear subspace?



Non-Borel subspace of Banach space
This is an answer, but not the "right" answer". 


Presumably you mean that $X$ is infinite dimensional and hence has Hamel dimension the continuum $c$. For every subset of a given Hamel basis you get the linear subspace spanned by the subset, and these subspaces are different for different subsets of the basis. Thus $X$ has $2^c$ linear subspaces but only $c$ Borel sets (since $X$ is separable).


EDIT: I just noticed the second question. Consider subsets of the basis that all contain one fixed countable subset whose span is dense.


