# Does the axiom of choice appear to be “true” in the context of physics?

I have been wondering about the axiom of choice and how it relates to physics. In particular, I was wondering how many (if any) experimentally-verified physical theories require axiom of choice (or well-ordering) and if any theories actually require constructability. As a math student, I have always been told the axiom of choice is invoked because of the beautiful results that transpire from its assumption. Do any mainstream physical theories require AoC or constructability, and if so, how do they require AoC or constructability?

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I've never bothered tracing what depends on AC and what doesn't, but I suspect it runs deep enough to touch most of the math underlying physics. For instance, it's good to know that we're talking about something that exists when we use bases for infinite-dimensional vector spaces. – Chris White Nov 10 '12 at 4:41
I think that Banach - Tarski theorem which depends crucially upon choice axiom may have some physical meaning - e.g. in terms of creation of more than one particles out of one when given with enough energy. However, the question of whether this is so or not belongs more to the domain of philosophy than physics. – user10001 Nov 10 '12 at 12:20
@ChrisWhite: right, however physicist very often assume other things that actually don't exist for e.g. general infinite-dimensional vector spaces, neither with or without the axiom of choice. – leftaroundabout Nov 10 '12 at 12:45
I suspect much of physics wouldn't need the full strength AC. Much can be done with countable AC. But, as @ChrisWhite says, measure theory would founder without full AC, although I suspect someone will come up with a measure theory without full AC one day. For me, though, the classic example that catches my eye here is not measure theory, but Tychonov's theorem - the product of compact sets is compact - is equivalent to AC but this is a very applied-maths-sounding theorem: it would be hard to say "throw that one out" - it's going to underpin many mathematical physics ideas. – WetSavannaAnimal aka Rod Vance Nov 3 '13 at 23:06

No, nothing in physics depends on the validity of the axiom of choice because physics deals with the explanation of observable phenomena. Infinite collections of sets – and they're the issue of the axiom of choice – are obviously not observable (we only observe a finite number of objects), so experimental physics may say nothing about the validity of the axiom of choice. If it could say something, it would be very paradoxical because axiom of choice is about pure maths and moreover, maths may prove that both systems with AC or non-AC are equally consistent.

Theoretical physics is no different because it deals with various well-defined, "constructible" objects such as spaces of real or complex functions or functionals.

For a physicist, just like for an open-minded evidence-based mathematician, the axiom of choice is a matter of personal preferences and "beliefs". A physicist could say that any non-contractible object, like a particular selected "set of elements" postulated to exist by the axiom of choice, is "unphysical". In mathematics, the axiom of choice may simplify some proofs but if I were deciding, I would choose a stronger framework in which the axiom of choice is invalid. A particular advantage of this choice is that one can't prove the existence of unmeasurable sets in the Lebesgue theory of measure. Consequently, one may add a very convenient and elegant extra axiom that all subsets of real numbers are measurable – an advantage that physicists are more likely to appreciate because they use measures often, even if they don't speak about them.

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You're really hating on the axiom of choice, and it's not clear why. If you want a new measure theory, you're perfectly free to come up with a new definition of "measure." No need to throw out a huge chunk of math to do it. And all the "open-minded" mathematicians you speak of died a long time ago. – Chris White Nov 10 '12 at 20:17
I am not "hating it", I am mostly indifferent towards it and slightly prefer non-AC over AC. I hope it's not a heresy yet. ;-) No one needs to throw any papers in maths – I just said that the detailed technical parts of those papers that depend on the axiom of choice are irrelevant for physics and irrelevant for any branch of maths that resembles the methods in physics. And that there's no scientific evidence - and can't be any scientific evidence - in favor or against the axiom of choice. – Luboš Motl Nov 10 '12 at 21:12
Whether someone died isn't decisive about statements of validity and consistency of assumptions and theories in maths or science. And the independence of the axiom of choice of the other axioms - i.e. the consistency of the other axioms with AC as well as non-AC (one of them) - was proved by Paul Cohen in the 1960s. Whoever doesn't understand that this means that AC and non(AC) are equally consistent with maths shouldn't call himself or herself a mathematician. Maybe he or she is an activist but not a rationally thinking person. – Luboš Motl Nov 10 '12 at 21:15

Rigorous arguments in functional analisis are made much simpler by employing the axiom of choice. As we are free to model our physics in any set theory we like, and any set theory containing ZF contains a model of ZFC, we are entitled to use this simplification without fear of inconsistency. Discarding the axiom of choice would only make concepts and proofs more tedious, without giving any higher degree of assurance of the results.

For example, the standard proof of the spectral theorem for self-adjoint operators depends on the axiom of choice, I believe, and much in mathematical physics depends on the spectral theorem.

On the other hand, already on the level of theoretical physics, one often replaces scrupulously integral by finite sums, takes limits irrespective of their mathematical existence, and employs lots of other mathematically dubious trickery to get quickly at the results.

So on this level of reasoning, nothing depends on subtilities that make a difference only when one begins to care about precise definitions and arguments in the presence of infinity.

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I taught a graduate course in math physics a couple years ago where I proved the spectral theorem for unbounded operators in separable Hilbert spaces. I did not use the AC but just the (countable) axiom of dependent choice. The two places in the proof that I remember where the AC can show its ugly face are: 1) decomposing the space into cyclic subspaces, 2) things around the Riesz-Markoff theorem for constructing spectral measures of given vectors. 1) is fine in a separable space but 2) gave me more troubles. There was the issue with mass escaping at infinity and all that... – Abdelmalek Abdesselam Aug 5 at 13:58
...which in its simplest form relates to the dual of $l^{\infty}$. When I searched the literature, I was quite shocked to discover that $(l^{\infty})'\neq l^1$ can only be proven using AC. Because of that, I now adopted the point of view that no mathematical physics should use the metaphysics of the AC, somewhat similar to the Luboš' view. – Abdelmalek Abdesselam Aug 5 at 14:01
@AbdelmalekAbdesselam: Why does AC have an ugly face? The problems inherited by and avoided by AC are well understood, and that there are problems only shows that once beyond the countable realm one has multiple notions of infinity, so that one can choose the more tractable one. Since the constructible model within any model of ZF satisfies ZFC, there cannot be anything wrong with the axiom. – Arnold Neumaier Aug 25 at 16:08
as you know De gustibus non est disputandum, so I find the AC ugly, you find it beautiful, but there is no problem with this disagreement. I am perfectly fine with pure mathematicians working on or using the AC. However, in mathematical physics I don't think it should be used, and so far I did not see an instance where one needs to use it. – Abdelmalek Abdesselam Sep 4 at 12:36

The textbook formulation of functional analysis depends on the axiom of choice, eg via Hahn-Banach.

This means that discarding the axoim of choice will break the textbook formulation of quantum mechanics as well. However, as we're dealing with (separable) Hilbert spaces, there exists countable bases and we should be able to replace the axiom of choice with a less 'paradox' alternative like the Solovay model and still get the right physics.

The full Hahn-Banach theorem cannot be recovered, though, as it implies the existence of an unmeasurable set.

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This is just wrong, Christoph. A textbook presentation of a math problem may decide to believe the axiom of choice but one may do all the things at least equally well in systems, like Solovay models, that assume the AC is false. Nothing in quantum physics would break down if one used non-AC in all textbooks. Your suggestion that one uses the AC with infinite bases in QM is wrong, too. All the structures that matter in QM, like the Hilbert space of L^2 integrable functions (well, some equivalence classes), are continuous and well-behaved, incompatible with the discrete AC-like selection. – Luboš Motl Nov 10 '12 at 7:21
@LubošMotl: please re-read my answer - I do not disagree – Christoph Nov 10 '12 at 7:29
@LubošMotl: clarified my answer a bit, but imo it was fine as it was... – Christoph Nov 10 '12 at 7:43
All the functional analysis that physicists use it is restricted to cases where the Hahn-Banach theorem is only used with at worst countable dependent choice, and you really don't need it for physics, as Lubos Motl explains clearly. This is pro-choice FUD, the "full Hahn-Banach theorem" is going on about vector spaces of basis size aleph_continuum, and nonsense like that. – Ron Maimon Nov 11 '12 at 4:54

The following paper may be of interest:

Norbert Brunner, Karl Svozil, Matthias Baaz, "The Axiom of Choice in Quantum Theory". Mathematical Logic Quarterly, vol. 42 (1) pp. 319-340 (1996).

The abstract is as follows:

We construct peculiar Hilbert spaces from counterexamples to the axiom of choice. We identify the intrinsically effective Hamiltonians with those observables of quantum theory which may coexist with such spaces. Here a self adjoint operator is intrinsically effective if and only if the Schrödinger equation of its generated semigroup is soluble by means of eigenfunction series expansions.

Also relevant is the fact that classical analysis doesn't require much more than dependent choice, which is consistent with "All sets of reals are Lebesgue measurable". However the combination of the two statements requires a stronger assumption as a theory (inaccessible cardinals).

What does baffle me, however, with physicists that have strong objections to the Banach-Tarski paradox, that it makes much less sense that a set can be partitioned into strictly more [non-empty] parts than elements. And that is a consequence of having all sets Lebesgue measurable.

So while you may sleep quietly knowing that you cannot partition an orange into five parts and combining the parts into two oranges (thus solving world hunger), you have an equally disturbing problem. You can cut out a line [read: the real numbers] into more parts than points.

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