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Or another way to put it: Could the axiom of choice, or any other set-theoretic axiom/formulation which we normally think of as undecidable, be somehow empirically testable? If you have a particular scheme for testing it, that's great, but even the existence or non-existence of a proof regarding potential testability is wonderful.

How about something a little simpler: can we even test the Peano axioms? Are there experiments that can empirically verify theorems not provable by them?

This is a slightly fuzzy question, so to clarify what I mean, consider this: the parallel postulate produces good, useful geometry, yet beyond its inapplicability to the sphere, there's evidence to suggest that the universe is actually hyperbolic - this can be considered an experimental evidence "against" the parallel postulate in our universe.

Edit: Thanks to all the people who answered - I understand the concerns of those who don't like this questions, and I appreciate all those who answered a more modest interpretation that I should, in retrospect, have stated. That is, "Is the axiom of choice agreeable with testable theories of mathematical physics, is it completely and forever irrelevant, or is it conceivably relevant but in a way not yet known," to which I got several compelling answers indicating the former.

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    $\begingroup$ I remember an American Mathematical Monthly article where the authors showed that you can use the axiom of choice to predict the future (well, in a rather non-constructive way)... :-) $\endgroup$
    – Andrea Ferretti
    Commented Jun 8, 2010 at 10:45
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    $\begingroup$ Well, for sure quantum mechanics assumes those Hilbert spaces where wavefunctions live have a basis. I don't know if assuming that those particular Hilbert spaces have a basis depends on the axiom of choice. Certainly the statement that every vector space has a basis does. $\endgroup$
    – Marc Alcobé García
    Commented Jun 8, 2010 at 20:11
  • $\begingroup$ Does Banach-Tarski count as evidence that AC is "false"? Does the existence of non-measurable sets, which implies the existence of coins with the property that even if you toss them a gazillion times the ratio (number of heads)/(total tosses) does not show any signs of converging to anything, count as evidence that it's "false"? Is Goodstein's theorem an experiment which empirically verifies something not provable in PA? These are really just vague comments on what seems to me to be a vague question. $\endgroup$
    – Kevin Buzzard
    Commented Jun 8, 2010 at 21:13
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    $\begingroup$ @Kevin: Since Banach-Tarski does not apply to matter made up of atoms, why would it say anything about falsity of some mathematics? Littlewood has an essay saying that trying to make probability theory fit anything in the real world is problematic. $\endgroup$
    – Gerald Edgar
    Commented Jun 9, 2010 at 1:00
  • $\begingroup$ @GeraldEdgar, what is the name of this essay? $\endgroup$
    – goblin GONE
    Commented Mar 6, 2016 at 22:25

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Before answering your question about the axiom of choice, let me take another set-theoretic axiom: "There exists an inaccessible cardinal." This axiom implies that ZFC is consistent. One could argue that this has the following "real-world consequence": If you set a computer running to look for a contradiction in ZFC, it will never find one. More generally, large cardinal axioms imply that smaller cardinal axioms are consistent with ZFC. Thus running your computer to search for contradictions in large cardinal axioms is a way to "test" larger cardinal axioms.

If you buy that, then one way to "test" an axiom is to look at what $\Pi^0_1$ sentences (i.e., sentences of the form "for all integers $n$, $P(n)$ holds" where $P(n)$ is some statement about the number $n$ that can be checked by a terminating computer program) the axiom implies. Coming back now to the axiom of choice, there is unfortunately no way to test it in this manner, because it is a theorem that any $ \Pi^0_1$ sentence (in fact, any first-order sentence of arithmetic) is a theorem of ZFC if and only if it is a theorem of ZF. The same goes for statements like the continuum hypothesis: any arithmetical theorem of ZFC + CH is already a theorem of ZF.

One might still wonder whether there is some other way to test mathematical statements using physical experiments. It seems unlikely to me, mainly because as finite creatures we can make only finitely many physical observations, so I think that the only mathematical statements that we will be able to reject definitively on the basis of physical experiments will be finitary ones, and first-order arithmetic should be able to express any finitary mathematical statement. It's true that, as some others have mentioned, some physicists have used the axiom of choice to construct physical theories, but if one of these theories were to be contradicted by experiment, we would probably just say that this disproves the physical theory, not the axiom of choice itself.

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  • $\begingroup$ @goblin : I don't understand your comment. Are you saying that we might look at $\Pi^0_1$ consequences of ZF + X where X is some random axiom? How would this provide evidence for or against AC? $\endgroup$
    – Timothy Chow
    Commented Mar 3, 2016 at 15:53
  • $\begingroup$ Sorry, my previous comment was typo-ridden. What I meant was: we can potentially "test" AC by considering the $\Pi_1$ consequences of X+AC where X is a collection of axioms distinct from ZF. $\endgroup$
    – goblin GONE
    Commented Mar 4, 2016 at 2:40
  • $\begingroup$ For a dumb example: X could be {ZF + "If AC is true, then ZFC is consistent."} $\endgroup$
    – goblin GONE
    Commented Mar 4, 2016 at 3:09
  • $\begingroup$ "Coming back now to the axiom of choice, there is unfortunately no way to test it in this manner, because it is a theorem that any Π01Π10 sentence (in fact, any first-order sentence of arithmetic) is a theorem of ZFC if and only if it is a theorem of ZF. The same goes for statements like the continuum hypothesis: any arithmetical theorem of ZFC + CH is already a theorem of ZF." Do you have a source for this? $\endgroup$
    – Christopher King
    Commented Oct 18, 2016 at 18:49
  • $\begingroup$ @PyRulez : It's a consequence of Shoenfield absoluteness. en.wikipedia.org/wiki/… $\endgroup$
    – Timothy Chow
    Commented Oct 18, 2016 at 20:06
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The 1996 paper The Axiom of Choice in Quantum Theory by Brunner, Svozil and Baaz contains the following provocative statement in the first paragraph: "Hence the very notion of a self-adjoint operator as an observable of quantum theory may become meaningless without the axiom of choice." The authors arrive at this conclusion by "constructing peculiar Hilbert spaces from counterexamples to the axiom of choice."

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    $\begingroup$ I don’t see how this is a “real-world consequence” at all: it seems just to be evidence that in the absence of AC, the mathematical formalism of quantum theory needs to be formulated slightly differently. This is a nice point, but a familiar phenomenon, nothing to do with physics — when a definition has many forms, equivalent under AC, the standard one is not “right” in absence of AC. I can’t access the full paywalled paper, but from the abstract, it sounds like the authors may also suggesting such a conclusion: “one needs to restrict to certain well-behaved spaces and/or operators”. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 15, 2019 at 10:27
  • $\begingroup$ @Peter doesn't Bishop+Bridges treat the basics of spectral theory of operators? $\endgroup$
    – David Roberts ♦
    Commented Feb 15, 2019 at 12:44
  • $\begingroup$ @DavidRoberts: Yes. Having read the linked paper (now that I’m at work and can access it), I’m now not convinced even by its arguments that the formalism needs to be changed in the absence of AC. Rather, it takes some fun “pathological” Hilbert spaces — the first one, for instance, is based on Russell’s socks — and points out that certain principles related to quantum theory, e.g. existence of self-adjoint operators (“observables”), fail there. One can respond to this either by changing the notion of observables (as they suggest), or by simply accepting it, under the expectation (cont’d) $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 15, 2019 at 13:32
  • $\begingroup$ such spaces don’t occur in nature, any more than Russell-like sets of socks do. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 15, 2019 at 13:33
  • $\begingroup$ @Peter What does it mean for a Hilbert space to "occur in nature"? $\endgroup$
    – dezign
    Commented Nov 27 at 23:27
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I think this question is the wrong way around (but this is also a philosophical issue).

The question is not if a theory has real world consequences but if the theory fits the parts of reality that is in the focus of the researcher.

Mathematics is about deducing lemmas and theorems from axioms so there are no consequences for the real world. All the lemmas and theorems are already there when you choose the axioms (sometimes you don't find them within some hundred years - this is also about computability). The question is if these theorems describe the real world and give correct forecasts when you insert real world variables into them.

If you choose too few axioms you won't get a very rich theory, if you choose too many you will get something with which you could prove anything and its counterpart. So it is really a tightrope walk to find the "right" axioms - also trial and error over the centuries.

After that you model reality with your lemmas/theorems and fit it to reality. If it works - fine, if not, try something else. I think it all boils down to that.

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    $\begingroup$ Isn't this basically what I just said? This is not rhetorical: I'm a little sick and feverish so my answer may not make as much sense as I think it does ;). $\endgroup$
    – jeremy
    Commented Jun 8, 2010 at 8:05
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    $\begingroup$ Well, to be quite honest: Me too - perhaps the same virus? Could be we are both delusional right now and when we'll read our answers a few days from now won't understand them any more ;-) But back to the point: I think the difference to your answer is that mine is more general, yours is more concrete and you say that there are no better or worse choices but I think there are more appropriate ones (they fit better and are better building blocks). But generally I agree with your answer (and just upvoted it :-) $\endgroup$
    – vonjd
    Commented Jun 8, 2010 at 8:14
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This is a slightly fuzzy question, so to clarify what I mean, consider this: the parallel postulate produces good, useful geometry, yet beyond its inapplicability to the sphere, there's evidence to suggest that the universe is actually hyperbolic - this can be considered an experimental evidence "against" the parallel postulate in our universe.

But that doesn't mean the same thing as your question is asking. The generalization of Euclidean geometry is not just hyperbolic or spherical geometry, but differential geometry. And a lot of the power of general relativity (or any Yang-Mills theory) comes from its general differential geometric structure (in other words, that it's a principal bundle with certain gauge group, etc), not the specific "geometry." This is analogous to thinking of differential equations v.s. initial conditions and specific solutions.

From the point of view of theoretical physicists, in a sense the answer is the same. If it is sensible to have math with and without the axiom of choice, one could reasonably expect that there are physical situations that can be described with and without the axiom of choice.

In other words, it may not be reasonable to say that AC is "empirically testable," as some systems may be described by "X+AC" and some others may be described by "X". Analogously, some systems are hyperbolic, e.g., special relativity's geometry, and others are not, e.g., generic Yang-Mills, string theory, etc--this does not mean that "geometry" is testable, it simply means that specific systems have specific descriptions. Not that any one formulation is in any way "better" than others from an experimental standpoint.

So I do not believe it makes sense to ask if the "universe" satisfies AC (or any other property) in this way.

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Unfortunately, I can't remember either the details or an appropriate reference, but I once read of an amusing proposal made by a physicist to exploit ideas in relativity, which, if the geometry of the universe was as it might possibly be, would allow you to do infinitely many computational steps in finite time (though presumably it wasn't finite time from the point of view of the computer). If it worked, then one could test number theoretic statements by simply running through all the integers. (This is in non-serious answer to your question about the Peano axioms.)

Well, I wrote that, but I now see that there's a problem because the input size would tend to infinity, so the computer would need infinite memory as well. I imagine the physicist concerned had thought about that but can't remember enough to be sure.

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    $\begingroup$ Some of my quantum computation friends have mentioned these things, and they all rely on closed timelike curves so that any computation that can be done in finite time can be done in bounded time--including zero time or a negative amount of time! If you're clever, you can get around some of the infinite memory problems by having the computer communicate with past states of itself in a clever manner, IIRC. Then, after explaining this, they laugh and tell me how they're paid to study this... $\endgroup$
    – jeremy
    Commented Jun 8, 2010 at 7:42
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    $\begingroup$ Philip Welch has an excellent mathematical summary of these ideas at maths.bris.ac.uk/~mapdw/chapter-fin.pdf (and several other papers on his web page). The idea of building these strange spacetimes to physically realize infinitary computation stretches back at least to Hogarth 1992, with a lot of work since then, and there is currently resurgent active work on the purely mathematical aspects of infinitary computability, such as that arising in Blum-Shub-Smale machines and infinite time Turing machines, with which I have been involved. $\endgroup$
    – Joel David Hamkins
    Commented Jun 8, 2010 at 12:54
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    $\begingroup$ You can find many such proposals by googling for "hypercomputation." But even if we take such proposals seriously, there is a problem. Say we construct a physical theory that tells us how to build a hypercomputer to solve the halting problem. We build the hypercomputer and ask it, "Would a Turing machine programmed to find a contradiction in ZFC halt?" Say the hypercomputer replies, "No." Have we "proved" that ZFC is consistent? I don't think so. We can't rule out the possibility that ZFC is inconsistent but that there is something wrong with our physical theory about the hypercomputer. $\endgroup$
    – Timothy Chow
    Commented Jun 9, 2010 at 2:16
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    $\begingroup$ Here is a completely serious paper about computation and closed timelike curves, by two top quantum computer scientists: arxiv.org/PS_cache/arxiv/pdf/0808/0808.2669v1.pdf The gist of it is that if closed timelike curves exist, then quantum computers are no more powerful than classical computers (although both become superpowerful, they become equally superpowerful). $\endgroup$
    – Ryan O'Donnell
    Commented Jun 9, 2010 at 2:22
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Asking whether a mathematical axiom can be empirically tested makes little or no sense, as it would be asking the same about the rules of chess. Excuse me for trivializing a little. If we use natural numbers with Peano's axioms to count grains of rice, we may be satisfied, for they give a quite satisfying model. One could also object that after all there are finitely many grains of rice in the world, though. But this say nothing about the consistency of Peano axioms; it only shows the range of applicability of a certain physical interpretation of them.

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    $\begingroup$ I think this is going too far--one can definitely test if a particular system is well-described by a particular set of axioms. One could even imagine there could be a system where you can "exactly" test them. This certainly counts as "empirically testing" but my contention is that there is not an well-defined/objective (i.e., 'experiment independent') notion of this. $\endgroup$
    – jeremy
    Commented Jun 8, 2010 at 8:37
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Stan Wagon mentions the following paper in his book: B. W. Augenstein, Hadron physics and transfinite set theory, International Journal of Theoretical Physics, 23(1984), 1572-9575.

Peter Komjath

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I vaguely recall some (humorous?) exchange about whether it is possible that a bridge would fall down because the calculations in its design had used the Lebesgue integral instead of the Riemann integral... Does anyone know where this was?

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    $\begingroup$ According to Rota's Discrete Thoughts, page 2, F.P. Ramsey asked Wittgenstein: "Suppose a contradiction were to be found in the axioms of set theory. Do you seriously believe that a bridge would fall down?" I think I've heard the Riemann v. Lebesgue variant too, but I can't find the source. $\endgroup$
    – John Stillwell
    Commented Jun 9, 2010 at 2:33
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    $\begingroup$ The Riemann vs. Lebesgue issue is a story about Hamming. He said that if the structural integrity of a particular airplane turned on the distinction between Riemann and Lebesgue integrals, he wouldn't fly in it. For an exact quote and citation, see the list of quotes on the wikipedia page en.wikipedia.org/wiki/Richard_Hamming. (Now no longer there, but see en.wikipedia.org/w/… One source seems to be N. Rose, Mathematical Maxims and Minims (Raleigh NC 1988).) $\endgroup$
    – KConrad
    Commented Jun 9, 2010 at 3:27
  • $\begingroup$ The Hamming quote may be in R. W. Hamming, Mathematics on a Distant Planet, The American Mathematical Monthly, Vol. 105, No. 7 (Aug. - Sep., 1998), pp. 640-650. $\endgroup$
    – Gerry Myerson
    Commented Feb 15, 2019 at 12:14
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I'm not sure how to add comments to other responses (maybe I don't have enough reputation, or maybe I'm just inept). I think the paper that gowers is referring to is on Malament-Hogarth spacetime.

http://en.wikipedia.org/wiki/Malament%E2%80%93Hogarth_spacetime

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    $\begingroup$ At the risk of making further feverish, potentially wrong, comments, I believe that when I discussed this particular situation with my QC friend, we decided that this does not actually solve the halting problem because (at the least) long-time signals will be arbitrarily red shifted, so you'd need to detect arbitrarily low energy photons, which you can't because eventually their wavelength will be longer than the universe. I believe there were other problems as well but I do not remember them offhand... $\endgroup$
    – jeremy
    Commented Jun 8, 2010 at 11:44
  • $\begingroup$ You need 50 reputation to add comments. $\endgroup$
    – Ian Morris
    Commented Jun 8, 2010 at 12:04
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The following paper "On Non-measurable sets and Invariant Tori' uses the axiom choice to solve a problem in classical mechanics and discusses the application of the axiom of choice to physics.

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Answering your question specifically concerning real-world consequence of AC, it is worth noting that the answer is strongly dependent on whether or not the universe is discrete or continuous. Although quantum mechanics and high energy physics hint at a fully discrete universe, this is not at all settled. For example it is not known whether or not space-time is discrete. If the universe is discrete, and therefore either finitely or infinitely countable, depending on whether or not the size of the universe is finite (also not known), the full AC is no longer applicable (a choice function for finite sets can be proven within ZF). In this case AC would seem exceedingly unlikely to have real-world consequences.

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    $\begingroup$ Quantum mechanics and high energy physics absolutely do not hint at a "fully discrete universe." There is absolutely nothing 'discrete' about quantum mechanics. This is a frequent misunderstanding and is discussed in many introductory undergraduate quantum mechanics books. No serious physicists believe that spacetime is in any way literally "discrete." $\endgroup$
    – jeremy
    Commented Jun 23, 2010 at 5:40
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    $\begingroup$ I have heard of these ideas; they aren't new, they've been around since at least the '60s. Things like discrete causal networks need to have uncountably many networks to hope to make sense. Most of the actual implementations of it either COMPLETELY fail to reproduce relativity in any limit, or just end up being perverse rewritings of SR or GR's geometry in terms of, e.g., its topology plus mysterious relations. This line of research was pretty much entirely abandoned by mainstream physics by the early to mid '80s. $\endgroup$
    – jeremy
    Commented Jun 23, 2010 at 6:10
  • $\begingroup$ Also, those ideas had absolutely nothing to do with any "discreteness" of quantum mechanics. They have to do with the fact that any individual observer actually sees a countable number of events, and spacetime is modeled on the structure of events (more-or-less lightcones in SR, more subtle in GR). So, the actual history of an observer is modeled on a finite lattice of events. But it turns out that it's not important what you DO observer, but the space of possibilities of what you COULD observe, which is uncountable. So you're just left with describing the points on a spacetime manifold. $\endgroup$
    – jeremy
    Commented Jun 23, 2010 at 6:13
  • $\begingroup$ Eh? Spacetime being discrete would absolutely slaughter the structure of string theory, not to mention GR and QFTs. For specific examples, see what happened with the development of lattice GR (particularly early on) and the special things needed to do in order to make lattice field theory work. (Mainly, what has to be done is account for the fake-discreteness by really knowing everything is discrete.) $\endgroup$
    – jeremy
    Commented Jun 23, 2010 at 6:29
  • $\begingroup$ (cont) It's well-known that spacetime (specifically) can't be discrete, and that other things (generically) are not expected to be discrete without very good reasons. There appears to be no "fundamental" discreteness in physics. $\endgroup$
    – jeremy
    Commented Jun 23, 2010 at 6:29

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