## Non-Borel subspace of Banach space

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?

-
 Wild guess: suppose $X$ is infinite dimensional, pick a Hamel basis of $X$, and let $M$ be the span of a proper infinite subset of that basis. Then I'm guessing $M$ will not be Borel. – Mark Meckes Oct 6 2010 at 18:52 What about the kernel of a non-continuous linear functional? – Robin Chapman Oct 6 2010 at 19:04 @Mark: any infinite-dimensional subspace $M$ of $X$ is the span of an infinite subspace of a Hamel basis for $X$. (Take a Hamel basis for $M$ and use Zorn to extend it to a Hamel basis for $X$.) So any subspace can be produced by your construction, including, say, the closed ones, so it need not produce a non-Borel subspace. – Nate Eldredge Oct 6 2010 at 19:17 Sorry, I meant to say cofinite. – Mark Meckes Oct 6 2010 at 23:58 But from Bill's comments on Gerald's answer, it looks like even what I meant to write may not be right. – Mark Meckes Oct 7 2010 at 13:24

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.

-
 Very nice! Thanks! – Nate Eldredge Oct 6 2010 at 20:18

Similar question was discussed before and answered (in particular to Bill's question in the comments)

http://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager

-
 Yes; that gives the "right" answer to Nate's question. – Bill Johnson Oct 7 2010 at 16:14 Thanks. I saw that question, but failed to unwind it far enough to see that it addressed my question as well. Now I see it. – Nate Eldredge Oct 11 2010 at 5:48

Let $X$ be a separable Banach space. Any linear subspace with the property of Baire (in particular, any linear subspace that is a Borel set) is closed. According to the Axiom of Choice, if $X$ is also infinite-dimensional, then there are discontinuous linear functionals, and their kernels are non-closed (and hence non-Borel) linear subspaces---of course these are dense.

EDIT
The above is clearly wrong, as the comments show. I could try to salvage something maybe saying $G_\delta$-set instead of property of Baire, but that would not answer the original question here.

-
Could you elaborate? Perhaps I myself am dense (pun intended), but I don't see why a Borel subspace must have the Baire property, nor why a subspace with the Baire property must be closed. – Nate Eldredge Oct 6 2010 at 20:37
Gerry, $\ell_2$ is $F_\sigma$ in $\ell_4$ but not closed, which contradicts part of what you wrote. – Bill Johnson Oct 6 2010 at 20:43
So the question remains for the infinite dimensional space $X$: 1. Is every dense codimension one subspace of $X$ Borel? 2. Does there exist a dense codimension one subspace of $X$ that is Borel? (Exercise: Show, for a fixed $X$, that this really is just one question.) – Bill Johnson Oct 7 2010 at 2:41
And the answer should be: A non-closed hyperplane (in a separable Banach space) must be non-Borel. Intuitively, non-closed hyperplanes live way out in Axiom-Of-Choice land, but Borel sets don't... But that's not a proof. – Gerald Edgar Oct 7 2010 at 15:10