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I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert spaces. I also don't see how to generalize Blass's proof to my question.

To be clear, what I am asking is:

Question Over ZF, does "every Hilbert space have a basis" imply AC, where a Hilbert space is a complete inner product space and a basis for a Hilbert space is a set of orthonormal elements whose span is dense?

[1] A. Blass, "Existence of bases implies the axiom of choice", Contempory Mathematics, Vol. 31, (1984).

Kameryn Williams
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    Considering the fact that this doesn't seem to appear in the literature, I'd suggest that this is still open. My guess is that it's in fact equivalent, and that some variant of Blass' original proof will do the trick. – Asaf Karagila Oct 01 '14 at 04:27
  • Thanks for the comment. That was my guess as well. – Kameryn Williams Oct 02 '14 at 01:33
  • @Jonas: If I'd have had a proof, I'd probably submit it to a journal before posting it here (or at least post it on arXiv). – Asaf Karagila Oct 21 '14 at 17:24
  • @Asaf: Right, but if someone has a proof that has been posted or submitted elsewhere, someone may post a link or citation to the other reference here. Or perhaps a more convincing case that this is an open problem would be a good answer here. – Jonas Meyer Oct 21 '14 at 17:28
  • @Jonas: As I wrote in the comment, I couldn't find it in the literature. I consider my search-fu to be relatively good. So I'm not saying it doesn't exist out there, it's just not likely that it does. (And of course, after posting it to arXiv, I'd have given a summary here...) – Asaf Karagila Oct 21 '14 at 17:33
  • @AsafKaragila: Thanks for your search-fu. As for the implied point about the bounty being likely futile for getting a resolution to the problem posted here, I have no expectation but choose to remain optimistic that some more light may be shed. At the least the bounty leads to greater sharing of an interesting problem. – Jonas Meyer Oct 21 '14 at 17:37
  • @Jonas: No problem, really, it's an interesting question. If I didn't have a whole lot of things on my plate right now, I'd have taken a crack at it. I expect it not to be very difficult, considering the fact that Blass' proof is already out there. I do expect it to be similar in nature somehow. – Asaf Karagila Oct 21 '14 at 17:39
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    @AsafKaragila: I doubt that Blass' proof can be adapted. The reason is that the proof constructs a certain field of homogeneous polynomials from a given family of non-empty sets, a vector space over this field and then constructs a multiple-choice function from a vector space basis. Blass himself remarks that he doesn't know if the Theorem is true if we fix a base field. But this was long ago, what is the current status? Does the existence of vector space bases over $\mathbb{C}$ imply the Axiom of Choice? Such a proof has a better chance to be transferred to Hilbert spaces. – Martin Brandenburg Oct 28 '14 at 06:12
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    The paper "On vector spaces over specific fields without choice" by P. Howard and E. Tachtsis seems to be relevant (I don't have access to it). – Martin Brandenburg Oct 28 '14 at 06:19
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    Related: The paper "All operators on a Hilbert space are bounded" (sic!) by JDM Wright shows that there is a model of $\mathrm{ZF}+\mathrm{DC}$ in which every operator on a Hilbert space is bounded. – Martin Brandenburg Oct 28 '14 at 06:35
  • @Martin: If I remember correctly, if every vector space over $\Bbb Q$ has a basis the axiom of choice holds. I don't know the proof, though, so I can't judge whether or not we can replace $\Bbb Q$ by $\Bbb C$. As for the Wright paper, that is just Solovay's model, and later we learned that we can improve upon that to Shelah's model removing the inaccessible from the game. – Asaf Karagila Oct 28 '14 at 06:39
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    Just a point about Blass' proof. The use of Axiom of Foundation in this proof seems to be essential so it cannot be done in $ZF^-$ alone. –  Oct 30 '14 at 08:56

1 Answers1

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According to the book Albrecht Pietsch: History of Banach Spaces and Linear Operators (Birkhäuser, 2007, DOI: 10.1007/978-0-8176-4596-0) this is an open problem. (Or at least it was at the time of publishing the book.)

I quote from p.586:

We list two more consequences of the axiom of choice, which have already been discussed in 1.2.2 and 1.5.10, respectively:
$(\mathsf{B}_{\mathsf{alg}})$ Every real or complex linear space has a Hamel basis.
$(\mathsf{B}_{\mathsf{orth}})$ Every real or complex Hilbert space has an orthonormal basis.

It is unknown whether $(\mathsf{B}_{\mathsf{alg}}) \overset{?}\Rightarrow (\mathsf{AC})$ or $(\mathsf{B}_{\mathsf{orth}}) \overset{?}\Rightarrow (\mathsf{AC})$. Some partial results were obtained by Bleicher [1964] and Halpern [1966]. Blass [1984, p. 31] showed that the axiom of choice follows if we assume that every linear space over an arbitrary field has a basis.

  • J.D. Halpern [1966] Bases in vector spaces and the axiom of choice, Proc. Amer. Math. Soc. 17, 670–673. DOI: 10.1090/S0002-9939-1966-0194340-1
  • M.N. Bleicher [1964] Some theorems on vector spaces and the axiom of choice, Fund. Math. 54, 95–107. eudml, matwbn
  • A. Blass [1984] Existence of basis implies the axiom of choice, Contemp. Math. 31, 31–33. author's website
Najib Idrissi
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Martin Sleziak
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