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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?

Old John
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  • Are you considering the wolfram definition or the general definition? $;$ –  Sep 09 '13 at 01:46
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    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 09 '13 at 01:48
  • That's almost the Wolfram definition. $;$ –  Sep 09 '13 at 01:55
  • You have created the tag [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 09 '13 at 07:17
  • @MartinSleziak If you look at the edit-history, you will see that I didn't add the tag. It was a sort of private joke. I will see if I can delete the tag. – Old John Sep 09 '13 at 08:33
  • Let me comment on this: without getting too heavily involved in axiomatic set theory. The usual proof of the existence of Hamel basis is a typical application of Zorn's Lemma. So anyone who is well versed in using ZL should be able to follow the proof or perhaps even devise such a proof by himself. The only question is whether you consider Zorn's lemma as something heavily involved in axiomatic set theory. – Martin Sleziak Sep 09 '13 at 08:40
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    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 09 '13 at 08:45
  • @MartinSleziak,I made the tag. http://en.wikipedia.org/wiki/Mornington_Crescent_%28game%29 and a sample game at http://www.youtube.com/watch?v=PVszcAgsBxs – Will Jagy Sep 09 '13 at 17:10
  • Related. – Andrés E. Caicedo Sep 10 '13 at 20:22
  • Just remembered this http://readr.ru/clifford-simak-why-call-them-back-from-heaven.html?page=7 – Will Jagy Sep 11 '13 at 00:48

4 Answers4

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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.

Clive Newstead
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  • 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 09 '13 at 01:55
  • As for a Hamel basis of $\mathbb R$ over $\mathbb Q$, it does not imply the full axiom of choice. The full axiom of choice is equivalent to the claim that every vector space has a basis. – Ittay Weiss Sep 09 '13 at 02:00
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    @IttayWeiss: I said that in my answer! :P – Clive Newstead Sep 09 '13 at 02:00
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    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 09 '13 at 02:02
  • @IttayWeiss: It might have been, I added it a few minutes after posting (but a few minutes before your comment) :) – Clive Newstead Sep 09 '13 at 02:07
  • When you say "Every vector space has a basis", do you mean "Every vector space over every field"? (Probably you do, but could you weaken it, e.g. to vector spaces over $\mathbb{Q}$?) – Pete L. Clark Sep 09 '13 at 02:39
  • Every field, yes. Choice certainly implies this; maybe you don't need every field to imply choice... interesting question. – Clive Newstead Sep 09 '13 at 02:51
  • @Clive: sure, I know that every vector space has a basis! I am asking whether you can restrict the scalar field and still get AC as a consequence. – Pete L. Clark Sep 09 '13 at 04:47
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    @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 09 '13 at 09:01
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    @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
  • @Martin: Thanks. That's (a priori, at least) slightly weaker than what I was asking about, but I consider it to be a pretty satisfying answer nonetheless. – Pete L. Clark Sep 16 '13 at 17:45
  • @PeteL.Clark On the page 68 (the above link) H. Herrlich claims: We may ask whether, like in Theorem 4.45, we can restrict attention in Theorem 4.44 to vector spaces over $\mathbb R$ resp. $\mathbb Q$. The answer to these questions is unknown. However, there is a slightly weaker result in the case of $\mathbb Q$. – Martin Sleziak Sep 16 '13 at 18:21
  • @PeteL.Clark See also: http://math.stackexchange.com/questions/951846/over-zf-does-every-hilbert-space-have-a-basis-imply-ac/1039841#1039841 – Martin Sleziak Feb 05 '15 at 13:42
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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

zyx
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  • 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 09 '13 at 02:12
  • I thought "Solovay proved that" if ZF+IC is consistent then "ZF + countable Dependant Choice is ...". $\hspace{.45 in}$ –  Sep 09 '13 at 02:29
  • @RickyDemer, yes, I think it is known that "you can't take Solovay's inaccessible away" for proving his result. – zyx Sep 09 '13 at 02:31
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    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 09 '13 at 03:16
  • 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 09 '13 at 03:27
  • @AndresCaicedo : $;;;$ Is there a citation for your second sentence that is clear on not using too much Choice? $:$ I had been looking, and would have included that in my previous comment here if I'd found one. $\hspace{.69 in}$ In any case, if that sort of argument can go through, then it will also work for $:$ZF+DC($\omega_1$). $\hspace{.95 in}$ –  Sep 09 '13 at 04:43
  • @RickyDemer I couldn't locate a primary source, but this is in several places. See for example "Strange Functions in Real Analysis", Second Edition, by A.B. Kharazishvili, page 202. The basic (standard) arguments establishing the Steinhaus property easily formalize in $\mathsf{DC}$, and this is all the proof requires. – Andrés E. Caicedo Sep 10 '13 at 20:20
  • @RickyDemer By the way, Kharazishvili's book is particularly careful about keeping track of uses of choice beyond $\mathsf{DC}$. – Andrés E. Caicedo Sep 10 '13 at 20:25
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It is shown here that the existence of a Hamel basis for every vector space implies the Axiom of Choice in ZF.

Mauricio Tec
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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.

Dan Saattrup Nielsen
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    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