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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?

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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 '10 at 18:52
    
What about the kernel of a non-continuous linear functional? –  Robin Chapman Oct 6 '10 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 '10 at 19:17
    
Sorry, I meant to say cofinite. –  Mark Meckes Oct 6 '10 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 '10 at 13:24
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3 Answers 3

up vote 9 down vote accepted

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.

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Very nice! Thanks! –  Nate Eldredge Oct 6 '10 at 20:18
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Similar question was discussed before and answered (in particular to Bill's question in the comments)

Are proper linear subspaces of Banach spaces always meager?

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Yes; that gives the "right" answer to Nate's question. –  Bill Johnson Oct 7 '10 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 '10 at 5:48
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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.

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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 '10 at 20:37
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Gerry, $\ell_2$ is $F_\sigma$ in $\ell_4$ but not closed, which contradicts part of what you wrote. –  Bill Johnson Oct 6 '10 at 20:43
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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 '10 at 2:41
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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 '10 at 15:10
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