# Over ZF, does "every Hilbert space have a basis" imply AC?

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).

• 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. Oct 1, 2014 at 4:27
• @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. Oct 28, 2014 at 6:12
• 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). Oct 28, 2014 at 6:19
• 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. Oct 28, 2014 at 6:35
• 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.
– user180918
Oct 30, 2014 at 8:56

## 1 Answer

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
• The question about Hamel basis over $\mathbb R$ and $\mathbb Q$ being an open problem is also mentioned in Herrlich's book about AC. See this comment in a related discussion. Feb 5, 2015 at 13:43