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?

  • $\begingroup$ 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. $\endgroup$ – Mark Meckes Oct 6 '10 at 18:52
  • $\begingroup$ What about the kernel of a non-continuous linear functional? $\endgroup$ – Robin Chapman Oct 6 '10 at 19:04
  • $\begingroup$ @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. $\endgroup$ – Nate Eldredge Oct 6 '10 at 19:17
  • $\begingroup$ Sorry, I meant to say cofinite. $\endgroup$ – Mark Meckes Oct 6 '10 at 23:58
  • $\begingroup$ But from Bill's comments on Gerald's answer, it looks like even what I meant to write may not be right. $\endgroup$ – Mark Meckes Oct 7 '10 at 13:24

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.


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?

  • $\begingroup$ Yes; that gives the "right" answer to Nate's question. $\endgroup$ – Bill Johnson Oct 7 '10 at 16:14
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 11 '10 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.

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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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 6 '10 at 20:37
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    $\begingroup$ Gerry, $\ell_2$ is $F_\sigma$ in $\ell_4$ but not closed, which contradicts part of what you wrote. $\endgroup$ – Bill Johnson Oct 6 '10 at 20:43
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    $\begingroup$ 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.) $\endgroup$ – Bill Johnson Oct 7 '10 at 2:41
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    $\begingroup$ 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. $\endgroup$ – Gerald Edgar Oct 7 '10 at 15:10

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