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 the 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? If so, how do they require AoC or constructability?
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asked Nov 10, 2012 at 4:23
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2$\begingroup$ 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. $\endgroup$– user10851Nov 10, 2012 at 4:41
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$\begingroup$ Related: physics.stackexchange.com/q/14939/2451 , physics.stackexchange.com/q/20370/2451 and links therein. $\endgroup$– Qmechanic ♦Nov 10, 2012 at 4:56
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$\begingroup$ 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. $\endgroup$– user10001Nov 10, 2012 at 12:20
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$\begingroup$ @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. $\endgroup$– leftaroundaboutNov 10, 2012 at 12:45
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1$\begingroup$ I have edited the title of this question to bring it closer to the core query as presented in the text (and to bring it closer to this site's core topics). $\endgroup$– Emilio PisantyMay 9, 2018 at 10:06
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5 Answers
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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-constructible 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.
answered Nov 10, 2012 at 5:38
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3$\begingroup$ 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. $\endgroup$– user10851Nov 10, 2012 at 20:17
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5$\begingroup$ 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. $\endgroup$ Nov 10, 2012 at 21:12
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2$\begingroup$ 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. $\endgroup$ Nov 10, 2012 at 21:15
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$\begingroup$ This argument form comes up a lot, but it's really worth noting that the existence of nonconstructible sets is a prerequisite for the failure of the axiom of choice, and is not a consequence of its truth! (e.g. the axiom of constructibility implies the axiom of choice) $\endgroup$– user5174Sep 11, 2016 at 19:07
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1$\begingroup$ I am not sure what you meant by "constructible sets" here. I was talking about measurable sets. Physicists want sets - subsets of real numbers - to be measurable. In mathematics, non-measurable sets may arise and be "constructed". Those are considered pathological by physicists (they allow one to be divided in 2 equal parts whose total measure is greater than the original one...) - and most mathematicians - and these pathologies are only inevitable if one assumes the axiom of choice. Without the axiom of choice, it's consistent to assume that all subsets of the interval (0,1) are measurable. $\endgroup$ Sep 12, 2016 at 10:27
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Rigorous arguments in functional analysis 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 integrals 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 subtleties that make a difference only when one begins to care about precise definitions and arguments in the presence of infinity.
answered Nov 12, 2012 at 17:45
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2$\begingroup$ 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... $\endgroup$ Aug 5, 2015 at 13:58
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1$\begingroup$ ...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. $\endgroup$ Aug 5, 2015 at 14:01
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$\begingroup$ @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. $\endgroup$ Aug 25, 2015 at 16:08
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1$\begingroup$ 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. $\endgroup$ Sep 4, 2015 at 12:36
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1$\begingroup$ The point is you don't have to adopt anything having to do with the chimeras where ZFC vs Solovay makes a difference. Time spent thinking about this is time not spent on real problems like, e.g., investigating ferromagnetic inequalities for non-Abelian systems. $\endgroup$ Oct 27, 2021 at 12:14
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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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$\begingroup$ One reason I am okay with the paradox you mention about partitioning a set into more parts than elements is because I don't believe in the well-definedness of cardinals larger than $\aleph_0$. Countless results in set theory show that you can make $2^{\aleph_0}$ look like almost any aleph you like, and alephs can collapse together in different theories. So to me the whole business is "unphysical", and I can believe in the existence of the real line without believing that we can "measure its size" in the cardinal sense in any nice linear order along with everything else in the (math) universe. $\endgroup$ Sep 3, 2019 at 1:32
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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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$\begingroup$ 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. $\endgroup$ Nov 10, 2012 at 7:21
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$\begingroup$ @LubošMotl: please re-read my answer - I do not disagree $\endgroup$ Nov 10, 2012 at 7:29
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$\begingroup$ @LubošMotl: clarified my answer a bit, but imo it was fine as it was... $\endgroup$ Nov 10, 2012 at 7:43
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4$\begingroup$ 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. $\endgroup$ Nov 11, 2012 at 4:54
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The axiom of choice can be applied in all mathematical problems concerning physics. However there it is not required because there are at most potentially infinite sets and every element can be chosen without an axiom.
The axiom of choice has gotten its bad reputation because it leads to contradictions with uncountable sets. But it is a very natural axiom and is frequently applied without notice in mathematics as Zermelo has correctly pointed out when defending his invention against objections of Borel, Peano, Poincaré, and others. [E. Zermelo: "Neuer Beweis für die Möglichkeit einer Wohlordnung", Math. Ann. 65 (1908) pp. 107-128]
The problem is only, as mentioned above, the application of the axiom to uncountable sets. In the history of mathematics, it was usual to fix factual conventions by an axiom, like "it is possible to draw a straight line from any point to any point" or "if n is a natural number, then n+1 is a natural number". The application of the axiom of choice claims for the first time a counterfactual convention, namely to choose an element without knowing what is chosen.
At least in 1904 it was clear that there are only countably many finite strings of letters, including strings defining mathematical objects. Cantor knew this theorem, as he wrote in a letter to Hilbert in 1906, although he did not believe that it is true. "If König's theorem was true, according to which all 'finitely definable' real numbers form a set of cardinality aleph_0, this would imply that the whole continuum was countable, and that is certainly false." [G. Cantor, letter to D. Hilbert (8 Aug 1906)]
Today there is no doubt that König's theorem is true. In order to maintain transfinite set theory, it is necessary to have the (in this realm) counterfactual axiom of choice for proving the basic theorem of set theory: Every set can be well-ordered. Otherwise a lot of ordinal theory would be unprovable. Therefore set theorists have agreed that the axiom is "not constructive", i.e., we can prove that we can choose every element, but we cannot choose every element. Although Zermelo used the axiom to prove that every set can be well-ordered, i.e., he thought it could be done and not only be proven that it could be done, knowing that in fact it cannot be done. [E. Zermelo: "Beweis, daß jede Menge wohlgeordnet werden kann", Math. Ann. 59 (1904)]
But what is the value of a counterfactual axiom? We could state many other axioms of same value like:
Axiom of three points on a line: Every triple of points belongs to a straight line. (But in most cases provably no geometrical construction can be given.)
Axiom of ten even primes: There are 10 even prime numbers. (But provably no arithmetical method to find them is available.)
Axiom of prime number triples: There is a second triple of prime numbers, besides (3, 5, 7). (But provably this second triple is not arithmetically definable.)
Axiom of meagre sum: There is a set of n different positive natural numbers with sum n*n/2. (This axiom is not constructive. Provably no such set can be constructed.)
All theories based upon such axioms would have the same value as transfinite set theory, namely none.
Keeping this in mind and ignoring absurd attempts to apply uncountable alphabets or infinite definitions to define uncountably many elements, we can be sure that the axiom of choice is true in every world with correct mathematics and therefore without uncountable sets.
Footnote.
Countable and uncountable sets require completed, finished, or actual infinity like the complete set of all natural numbers to quantify over. But what means quantifying over all natural numbers? Does it mean to take only those natural numbers which have the characteristic property of every natural number, i.e., to be followed by infinitely many natural numbers? Then you don't get all of them because always infinitely many are remaining. Or do you take all natural numbers with no exception? Then you include some which are not followed by infinitely many and hence are not natural numbers because they are lacking the characteristic property of every natural number. That means you get more than all natural numbers.
To make a long story short: The set of all natural numbers is a set that cannot exist because its elements cannot be in a set together. Therefore we have at most potential infinity. Actual infinity with countable and uncountable sets is a contradictio in adjecto.
Although it was Cantor's aim to apply set theory to physical, chemical, and even psychological and political problems
"The third part contains the applications of set theory to the natural sciences: physics, chemistry, mineralogy, botany, zoology, anthropology, biology, physiology, medicine etc. It is what Englishmen call 'natural philosophy'. In addition we have the so-called 'humanities', which, in my opinion, have to be called natural sciences too, because also the 'mind' belongs to nature." [G. Cantor, letter to D. Hilbert (20 Sep 1912)]
his ideas have turned out to be inapplicable everywhere and in particular to physics.
