Hamel bases without (too much) axiomatic set theory

Hamel bases have cropped up on the periphery of my mathematical interests a few times over my mathematical career, but I have never found the time or had a real need to look into them at any depth. Most of what I know comes from the 1966 book "A First Course in Functional Analysis" by M. Davies, in which he uses them to prove the existence of discontinuous solutions of the functional equation $f(x+y) = f(x) + f(y)$.

My questions/queries:

• Is it possible to talk (meaningfully) about Hamel bases without invoking the axiom of choice?

• am I correct in my primitive, intuition-led understanding: "we can't explicitly exhibit a Hamel basis because that would be "equivalent" (in some obscure way that I cannot define precisely) to explicitly exhibiting a "choice function"?

• can anyone give me a nice reference where an old-fashioned analyst (well, old, at least) could read up on such matters without getting too heavily involved in axiomatic set theory or foundations of maths texts?

• Are you considering the wolfram definition or the general definition? $\;$ – user57159 Sep 9 '13 at 1:46
• I had not seen the Wolfram definition. I was working from the definition which goes: "a set $\Gamma$ is a Hamel basis if $1\in\Gamma$ and for each real $x$ there are uniquely determined $x_1, x_2, \dots \in \Gamma$ and non-zero rationals $r_1, r_2, \dots$ such that $x = r_1x_1 + r_2x_2 + \dots$. – Old John Sep 9 '13 at 1:48
• That's almost the Wolfram definition. $\;$ – user57159 Sep 9 '13 at 1:55
• You have created the tag mornington-crescent for this question. It would be nice, if you edited tag-excerpt and tag-wiki, so that other users know what this tag is intended for. (I do not know what this tag is about just by the name of the tag.) – Martin Sleziak Sep 9 '13 at 7:17
• Yes, @MartinSleziak, I had seen the proof, and don't find ZL daunting. It was the second point in my question that was bugging me and I thought might get me into deeper territory. – Old John Sep 9 '13 at 8:45

In general, you can't even posit the existence of a Hamel basis without the axiom of choice; and even with the axiom of choice, you can't get your hands on them. If you were to say "pick a basis of $\mathbb{R}$ over $\mathbb{Q}$" then you'd already have invoked the axiom of choice; and then you wouldn't be able to write down what a general vector in the basis looks like. This is characteristic of any construction in mathematics that depends on the axiom of choice, since it's the only axiom which says "this thing exists" without also saying what it looks like.

The sense in which the existence of Hamel bases is equivalent to the axiom of choice is as follows: if every vector space has a (Hamel) basis, then the axiom of choice holds; and if the axiom of choice holds, then every vector space has a basis. This equivalence means that, if the axiom of choice fails, then there is a vector space without a basis; likewise, if there exists a vector space without a basis, then the axiom of choice fails. But the question of whether the axiom of choice holds or fails (and hence of whether Hamel bases exist or don't exist) is unprovable, in a very concrete sense... the response of most mathematicians to this fact is "we'll just admit the axiom of choice".

It's not true that the existence of a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ implies the full axiom of choice. However, it does imply a weaker form of the axiom of choice which is [still] unprovable.

Beyond this, I don't know what to say: it's tough to answer an axiomatic set theoretic question without talking about axiomatic set theory.

• Thanks, I think that pretty much agrees with what I have believed for many years to be true, but it is nice to have it confirmed that I am not too far off-beam. – Old John Sep 9 '13 at 1:55
• @IttayWeiss: I said that in my answer! :P – Clive Newstead Sep 9 '13 at 2:00
• oh, I could have sworn I read it as "I'm not sure if the existence of a Hamel basis..... implies the full axiom of choice". Oh well, I'm either reading things that are not there, or you edited your answer while I was typing the comment. I hope it's the latter ;) – Ittay Weiss Sep 9 '13 at 2:02
• @PeteL.Clark Perhaps your last comment could be made into a separate question. (In my opinion, an interesting one.) IIRC the implication "Every vector space has a basis" $\Rightarrow$ AC is due to A. Blass; as he frequents both MSE and MO, there is a hope of getting an expert answer. – Martin Sleziak Sep 9 '13 at 9:01
• @PeteL.Clark Today I accidentally (I was looking for something else) stumbled upon Theorem 4.47 in Herrlich's Axiom of Choice. It says that AC is equivalent to "In every vector space over $\mathbb Q$ each generating set contains a basis." K. Keremedis. Extending independent sets to bases and the axiom of choice. is given there as a reference. I have not looked into the proof, yet. – Martin Sleziak Sep 16 '13 at 17:41

Hamel basis implies discontinuous real-valued solutions to the functional equation $f(x+y)=f(x)+f(y)$ and such a function is known to be non-measurable. Solovay proved that ZF + countable Dependent Choice is consistent with "every set of reals is Lebesgue measurable" so that

a Hamel basis cannot be proved to exist in ZF + DC. Informally, it cannot be done in classical analysis without uncountable, set-theoretic forms of choice

• OK. I rather like the existence of Hamel bases, so I guess then I have to go along with AC, and not just DC. – Old John Sep 9 '13 at 2:12
• I thought "Solovay proved that" if ZF+IC is consistent then "ZF + countable Dependant Choice is ...". $\hspace{.45 in}$ – user57159 Sep 9 '13 at 2:29
• @RickyDemer, yes, I think it is known that "you can't take Solovay's inaccessible away" for proving his result. – zyx Sep 9 '13 at 2:31
• The inaccessible is irrelevant here. All solutions to Cauchy's functional equation are continuous if all sets of reals have the property of Baire. Now, (over $\mathsf{ZF}$) we have that $\mathsf{DC}$ plus all sets of reals have the property of Baire is equiconsistent with $\mathsf{ZF}$. – Andrés E. Caicedo Sep 9 '13 at 3:16
• Thanks (@AndresCaicedo ). I assumed it is also irrelevant in the sense that if ZF+IC were inconsistent, so much of set theory would be reconsidered that is more basic than Solovay's work, that the whole question would get a fresh analysis. – zyx Sep 9 '13 at 3:27

It is shown here that the existence of a Hamel basis for every vector space implies the Axiom of Choice in ZF.

• Thanks - that paper looks like it could well be understandable to an old analyst! – Old John Sep 9 '13 at 2:01

To anyone finding this question now, Beriashvili-Schindler-Wu-Yu have recently found a model in which choice fails but which has a Hamel basis: http://wwwmath.uni-muenster.de/u/rds/hamel_basis.pdf.

• To be clear, in case anybody wonders: The term Hamel basis is used in the narrow sense of a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space in that paper. – Daniel Fischer Jan 17 '17 at 22:02