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I know from this question that it's an open problem whether or not the existence of a dense orthonomral basis for every real or complex Hilbert space $(\text{B}_\text{orth})$ implies the axiom of choice. This tells me that the axiom of dependent choice, $(\text{DC})$, isn't strong enough to imply that every non-seperable Hilbert space has an orthonormal basis (otherwise it wouldn't be an open problem because if $(\text{B}_\text{orth})\Rightarrow(\text{AC})$ then $(\text{DC})\Rightarrow(\text{AC})$ and the Solovay model would be unsatisfiable) so it's not immediately impossible that you can prove the existence of a non-Lebesgue measurable set, $(\text{not-LM})$, from $(\text{B}_\text{orth})$.

I know from the paper The Hahn-Banach theorem implies the existence of a non-Lebesgue measurable set by Foreman and Wehrung that the full Hahn-Banach theorem is enough to prove $(\text{not-LM})$ and I know that a weakened version of the Hahn-Banach theorem for Hilbert spaces is somewhat immediate, but the proof of $(\text{not-LM})$ in that paper relies on the equivalence of Hahn-Banach and the existence of a finitely additive measure on every Boolean algebra and I'm not sure how that statement is weakened for weakened versions of Hahn-Banach.

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James Hanson
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  • These can be difficult questions, worthy of publications in the right circumstances. One cursory search yielded no results; but the rest will have to wait for tomorrow I'm afraid. – Asaf Karagila Aug 02 '15 at 21:26
  • And just to be clear, what do you mean an orthonormal basis? The span is dense or the span is everything? – Asaf Karagila Aug 02 '15 at 21:28
  • I mean that the basis is dense. I'll edit the question. Are there even non-seperable Hilbert spaces with full orthonormal bases, not just dense orthonormal bases? – James Hanson Aug 02 '15 at 21:35
  • Well, I'm not that well-versed in functional analysis, or the intricacies of the terminology. :-) – Asaf Karagila Aug 02 '15 at 21:38
  • @AsafKaragila: I deleted that comment because I realized I was using AC to choose a countable subset! – Nate Eldredge Aug 02 '15 at 22:13
  • @Nate: But it was the principle that counts. I got the gist. That's the important part. And now I am one tiny iota closer to being prepared mentally for a six credits course in [mostly functional] analysis next year! – Asaf Karagila Aug 02 '15 at 22:15
  • @AsafKaragila: But now it makes me wonder about the Hilbert space $\ell^2(A)$ where $A$ is amorphous... – Nate Eldredge Aug 02 '15 at 22:17
  • @Nate: That's actually easy. Brunner has a paper about Hilbert spaces which are like that, $\ell_2(D)$ where $D$ has a Dedekind-finite power set. They have a Hamel basis which is orthonormal or something. I references this paper often on here/MO. "Garnir's dream spaces with Hamel bases." – Asaf Karagila Aug 02 '15 at 22:22
  • @AsafKaragila: I was reading that paper because of this question and I have a perhaps weaker question. Is it genericaly possible to use something like construction 2 (the Hilbert space over Russel's socks) to construct a choice function if you assume the space has a self-adjoint operator? I tried to read their reference [3] for this statement but the paper is far too technical for me and I can't even locate the statement that's supposed to correspond to construction 2 having no self-adjoint operators. – James Hanson Aug 05 '15 at 01:54

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