Let's say I consider the Banach–Tarski paradox unacceptable, meaning that I would rather do all my mathematics without using the axiom of choice. As my foundation, I would presumably have to use ZF, ZF plus other axioms, or an approach in which sets were not fundamental.

Suppose that all I want is enough analysis to express all existing theories in physics. Is ZF enough? If not, then is there any attractive, utilitarian system of the form ZF+x, where x represents some other axiom(s), that does suffice, without allowing Banach-Tarski?

Wikipedia has a list of statements that are equivalent to choice: http://en.wikipedia.org/wiki/Axiom_of_choice#Equivalents The only one that seems obviously relevant is Blass's result that you need choice to prove that every vector space has a basis. But if all I care about is vector spaces that would actually be used in physics (probably nothing fancier than the space of functions from $\mathbb{R}^m$ to $\mathbb{R}^n$), does this matter? I.e., are the spaces for which you need choice to prove the existence of a basis too pathological to be of interest to a physicist? In cases of physical interest, it seems like it would be trivial to construct a basis explicitly.

Is Solovay's theorem relevant? I'm confused about the role played by the existence of inaccessible cardinals.

I'm a physicist, not a mathematician, so I would appreciate answers pitched at the level of a dilettante, not that of a professional logician.

[EDIT] André Nicolas asks: "[...] why should Banach-Tarski be unacceptable?" Fair enough. Let me try to clarify what I had in mind. The real number system contains stuff that is physically meaningless, but (a) I have a clear idea of which of its features can't mean anything physical (e.g., the distinction between rationals and irrationals), and (b) doing math in $\mathbb{Q}$ would be much less convenient than doing math in $\mathbb{R}$. Similarly, I might prefer to think of my $dy$'s and $dx$'s as infinitesimals, and although those are unphysical, I understand what's unphysical about them, and they're convenient. But when it comes to choice, it's not obvious to me how to distinguish physically meaningful consequences from physically meaningless ones, and it's not obvious that I would lose any convenience by limiting myself to ZF.

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    $\begingroup$ For a fair bit of basic analysis, Countable Choice, or Dependent Choice, is enough. We really want, for example, sequential convergence in the reals to be equivalent to convergence. And a basis in the algebraic sense is not usually what we need for infinite dimensional spaces. But why should Banach-Tarski be unacceptable? It involves really weird sets that presumably would not show up in models from Physics. $\endgroup$ – André Nicolas Jan 29 '12 at 22:36
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    $\begingroup$ @Ben: I think the space of smooth functions $\mathbb{R}^n \to \mathbb{R}^m$ is very fancy, and I'm a mathematician! $\endgroup$ – Zhen Lin Jan 30 '12 at 0:15
  • $\begingroup$ By "basis" I assume you mean a Hamel basis (so every element in the vector space is a finite linear combination of basis elements). I'm not sure how badly you need Hamel bases in mathematical physics, and in much of mathematical analysis and PDE, for that matter. $\endgroup$ – Stefan Smith Mar 11 '13 at 23:16
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    $\begingroup$ It's worth noting that the axiom of dependent choice is far more different from the full axiom of choice than most people think. The axiom of dependent choice follows naturally from considering objects generated by an algorithm that is allowed to instantiate existentially quantified objects at any point. Even without the axiom that gives an infinite set (but there are still infinite classes), this is enough for mathematics that applies to the real world (as is currently known), since algorithms can describe any definable process and the real world doesn't seem to have any infinite object. $\endgroup$ – user21820 Jun 24 '14 at 2:10
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    $\begingroup$ Feferman has, I think, spent quite a bit of intellectual effort on just this question; see, for example, math.stanford.edu/~feferman/papers/psa1992.pdf. $\endgroup$ – LSpice Aug 29 '14 at 23:51

An interesting question, that would take many pages to begin to answer! We make a small disjointed series of comments.

In the last few years, there has been a systematic program, initiated by Friedman and usually called Reverse Mathematics, to discover precisely how much we need to prove various theorems. The rough answer is that for many important things, we need very much less than ZFC. For many things, full ZF is vast overkill. Small fragments of second-order arithmetic, together with very limited versions of AC, are often enough.

About the Axiom of Choice, for a fair bit of basic analysis, it is pleasant to have Countable Choice, or Dependent Choice, at least for some kinds of sets. We really want, for example, sequential continuity in the reals to be equivalent to continuity. One could do this without full DC, but DC sounds not unreasonable to many people who have some discomfort with the full AC. This was amusingly illustrated in the early $20$-th century. A number of mathematicians who had publicly objected to AC turned out to have unwittingly used some form of AC in their published work.

Next, bases. For finite dimensional vector spaces, there is no problem, we do not need any form of AC (though amusingly we do to prove that the Dedekind definition of finiteness is equivalent to the usual definition.)

For some infinite dimensional vector spaces, we cannot prove the existence of a basis in ZF (I guess I have to add the usual caveat "if ZF is consistent"). However, an algebraic basis is not usually what we need in analysis. For example, we often express nice functions as $\sum_0^\infty a_nx^n$. This is an infinite "sum." The same remark can be made about Fourier series. True, we would use an algebraic basis for $\mathbb{R}$ over $\mathbb{Q}$ to show that there are strange solutions to the functional equation $f(x+y)=f(x)+f(y)$. But are these strange solutions of any conceivable use in Physics?

Finally, why should the Banach-Tarski result be unacceptable to a physicist as physicist? It is easy to show that the sets in the decomposition cannot be all measurable. In mathematical models of physical situations, do non-measurable sets of points in $\mathbb{R}^3$ ever appear?

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    $\begingroup$ I agree especially with the final paragraph. To be honest, I suspect that trying to divide mathematics into "physical" and "unphysical" is wrong-headed: physics does not accept or reject parts of mathematics in any meaningful sense. Rather, there are some parts of mathematics that have proved useful (often, indispensably so) in modelling and analyzing various physical phenomena and theories and other parts of mathematics that have not (yet) been so applied. The idea that physical reality is itself somehow a mathematical object has not been taken very seriously since Kant's time. $\endgroup$ – Pete L. Clark Jan 29 '12 at 23:49
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    $\begingroup$ @Ron: While I certainly understand "the coefficient on the basis vector $v$" is a non-measurable function of the reals and so you really shouldn't be talking about the coefficients of a random variable for the same reason you shouldn't be trying to measure a non-measurable set, I can't figure out what exactly you think is contradictory in your argument (and why). $\endgroup$ – user14972 Apr 17 '14 at 6:34
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    $\begingroup$ @Hurkyl: There are operations that you can apply to real numbers in ZFC, like "decompose into a basis over Q" which cannot be applied to randomly selected numbers, to random variables. This makes for a completely mentally retarded distinction between "random variables" and "real numbers" which is useless and counterproductive, and makes developing analysis and physics a pain in the ass. Physicists use random numbers all the time, and they can't be forced to remember which operations are allowed. It is also ridiculous and anachronistic, because it is possible to just use a Solovay model. $\endgroup$ – Ron Maimon Apr 17 '14 at 6:41
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    $\begingroup$ @Ron: There is no such thing as a "randomly chosen number". And quite frankly, the heavy use of probability theory in modern physics tells me that we should really stop pretending the universe has the definiteness property. And if you really want to push the physical arguments, you can't measure precisely enough to meaningfully speak of a "randomly chosen number" anyways: the best you can do is to pin it down to some open (and thus measurable) set. $\endgroup$ – user14972 Apr 17 '14 at 6:50
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    $\begingroup$ @RonMaimon For example, there is no basis of R considered as a vector space over Q. Proof? Pick two Gaussian random reals x and y of unit variance and note that (3x+4y)/5 is a Gaussian random real of unit variance, which is a rational linear combination of x and y, so that it is impossible to decompose x and y and the sum into a basis consistently. I know what Gaussian random variables are (not Gaussian random reals, which do not exist) but I fail to grasp your argument. Why is it impossible to decompose, for each ω in the sample space, x(ω) and y(ω) and (3x(ω)+4y(ω))/5 into a basis? $\endgroup$ – Did Apr 17 '14 at 8:32

Blass's theorem is a very strong one indeed. If the axiom of choice does not hold then there is a vector space without a basis. It is unusual to be able and tell which vector space it is (unless assuming more, or constructing the model directly).

In particular, finite dimensional vector spaces always have a basis, since such basis is finite and thus completely definable in the universe.

Most of the basic analysis would require the axiom of countable choice, or the axiom of dependent choice. Both would be enough for almost every theorem you learn in basic calculus class - but neither is enough for Banach-Tarski. You may wish to add something like the ultrafilter lemma, however once there is a free ultrafilter over $\mathbb N$ there are unmeasurable sets - if that would bother you.

In general to prove that a space has a basis may require some choice, for example $\mathbb R$ as a vector space over $\mathbb Q$ requires choice. If however your interest is in finitely dimensional vector spaces then you can relax, since those would be fine regardless to the axiom of choice. There are infinitely dimensional spaces which have explicit basis as well, for example all the infinite sequences which are eventually zero.

Once you go beyond that it becomes harder and harder to produce a basis without the axiom of choice, but your needs might not go that far.

By Solovay's theorem I suppose you mean his model in which every set is measurable, and such. This is irrelevant, and in fact it holds a horrible secret:

In Solovay's model we can cut $\mathbb R$ into more parts then it has elements. Namely, we can cut $\mathbb R$ into non-empty parts and have more parts than real numbers. This sort of partition might sound very bizarre and pathological, much like the Banach-Tarski paradox. However such partitions can be a handful in some parts of set theory.

You may want to think that you're screwed basing yourself on ZF either way, but the problem is that mathematics almost always have this way to sting you in the back, no matter how you put it. You can simply put "new limitations" (e.g. limit yourself to measurable sets, which still gives you a rich and fulfilling world) and just use mathematics as you first wanted.

One of the least known facts about choice is that the definitions of continuity: $\epsilon-\delta$ and sequential continuity are not equivalent without some choice. If you have used this before, you have used the axiom of choice.

The point above is that the axiom of choice simply allows us to control many infinitary processes in a very simple way. While physics itself does not really talk about infinite processes (at least not as far as I know) you should be able to get away from that if you ditch the axiom of choice. However you may want to keep enough of it to ensure that what you have approximated with finite parts is continuously carried over to the limit point. This is in its essence the principle of dependent choice (and to a lesser extent the axiom of countable choice).

Something to Read:

  1. Finite dimensional subspaces of a linear space
  2. Finite choice without AC
  3. Why worry about the axiom of choice?
  4. Does the Axiom of Choice (or any other “optional” set theory axiom) have real-world consequences? [closed]
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    $\begingroup$ Ben, what I am saying is that one can write $\mathbb R=\bigcup\{A_i\mid i\in I\}$, where $A_i$'s are disjoint, nonempty and $|I|>|\mathbb R|$. $\endgroup$ – Asaf Karagila Jan 29 '12 at 23:10
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    $\begingroup$ Ben, of course that if the axiom of choice does not hold then there are counterexamples to it. The point in my remark on Solovay's model is that if Banach-Tarski serves as a pitching point against the axiom of choice since the decomposition is paradoxical, I cannot imagine how you can calmly accept slicing $\mathbb R$ into more parts than elements! Furthermore, in this model every part is measurable!! When I first heard this, I shook my head in disbelief. $\endgroup$ – Asaf Karagila Jan 29 '12 at 23:44
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    $\begingroup$ @Emilio: Anyone for the Axiom of Determinateness (AD)? It has interesting consequences. One issue (among many) is plausibility. $\endgroup$ – André Nicolas Jan 30 '12 at 0:55
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    $\begingroup$ @Ron: You can PROVABLY map $\Bbb R$ onto $\omega_1$. In the absence of choice not every surjection has an inverse injection, so you can't map $\omega_1$ into $\Bbb R$ again. But you can map $\Bbb R$ onto $\omega_1$. Using this map, you can create this decomposition. If you want to downvote me, go ahead, but try to do it for mathematically correct reasons. $\endgroup$ – Asaf Karagila Apr 17 '14 at 18:02
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    $\begingroup$ @Ron: Also, the partition has $2^{\aleph_0}+\aleph_1$ parts. Not $\aleph_1$ parts. In Solovay's model, by the way, the real numbers are not "that much large" than $\aleph_1$. They can't even be mapped onto $\aleph_2$ (making them quite standardly small), let me repeat that ONTO. NOT BIJECTIVELY. Finally, I haven't used the term "randomly chosen real number". So I have no idea what your first comment is talking about. $\endgroup$ – Asaf Karagila Apr 17 '14 at 18:36

This answer consists basically of two remarks.

First: If you do find Banach-Tarski unacceptable and want to be sure that you live in a world without such decompositions, dropping the axiom of choice is not enough. Since the axiom of choice is consistent, you cannot prove that there is no Banach-Tarski decomposition lurking somewhere. What you need is a theory in which you assume something that blatantly contradicts the axiom of choice.

There are axiom systems that do that and give you a number of "pleasant" consequences. For example assuming ZF+Dependent Choice+Every set of real numbers has the Baire property, leads to several convenient results. For example, you can then show that every two complete norms on the same vector space give you the same topology. You can find more such results in this wonderful book.

The downside of making such an assumption is that it is ot so clear how to view the corresponding set theoretic universe. The axiom of choice seems to be a natural consequence of the notion of an "arbitrary set". The problem might be how you embed your physical theory in set theory, not with set theory itself. Which leads to my second remark.

Second: Being really careful that the mathematical objects you are working with have physical meaning, should lead you not to meddling in the foundations of mathematics but toembracing the theory of measurement, where you have explicit theory of what is meaningful. You can find a readable first introduction here and here.

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    $\begingroup$ The "pleasant property" you want is ZF+dependent choice+every subset of R is measurable. This pleasant property allows you to make probability arguments without worrying about inconsistencies, and Solovay proved it is consistent. It is eminently clear how to view the corresponding set theoretic universe, it is the best model of "reality", in that it rejects the fantastical constructions of choice, without rejecting the ones that are uncontroversial. Solovay's razor makes it easier to cut away the useless parts of measure theory, and some path integrals then become trivial to define. $\endgroup$ – Ron Maimon Apr 16 '14 at 19:36

One possible answer (though there are strong arguments that it sidesteps the question rather than answer it) would be that it suffices to work in ZF + "ZF is consistent".

This would depend on the viewpoint that a physical theory is just a black-box mathematical mechanism into which you can stick a description of an experiment and get a prediction of its results out of. Now consider any physical theory which says "do such-and-such, working in ZFC".

Now, since in ZF+Con(ZF) we can prove that ZFC is consistent and therefore has a model, we can instead change our physical theory to "do such-and-such inside a model of ZFC". This would produce the same result as before. The only snag is that there are several different models of ZFC, which might not all give the same result, but one might always specify one particular model to do the computations in, such as a suitably specified term model.

If you want to keep any intuitive insights into how the real world works that can be extracted from the mathematical structure of the physical theory, it might be a different matter. :-)

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    $\begingroup$ This is not sufficient, as you need "ZF + ZF is consistent" is consistent, and so on, iterated over computable ordinals, and this is equivalent to the consistency of arbitrarily strong theories of the large cardinal type. The main issue is with the metaphysics of the continuum. For a physics application, you need that every subset is Lebesgue measurable, and one must never renounce this, as there is no gain from assuming a non-measurable set, only headaches. $\endgroup$ – Ron Maimon Apr 16 '14 at 19:37

On a rather basic level, ZF is not enough because a standard fact used in many applications, namely the countable additivity of the Lebesgue measure, fails in ZF due to the Feferman-Levy model; see https://mathoverflow.net/questions/146813/is-sigma-additivity-of-lebesgue-measure-deducible-from-zf


In Zermelo-Fraenkel set theory, the axiom of choice is not disprovable so the Banach-Tarski paradox is not disprovable. Adding the negation of the axiom of choice is still not enough to disprove the Banach-Tarski paradox. To prove that all subsets of R^3 have a well defined volume such that the intuitive properties of volume hold, you need to add a stronger axiom like the axiom of determacy or the axiom that all subsets of R^3 are lebesgue measurable. With one of those axioms added, the axiom of choice can be disproven. I think it has probably been proven that adding one of those axioms does not make the theory contradictory which means you can't prove the axiom of choice with one of them added.


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