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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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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. –  André Nicolas Jan 29 '12 at 22:36
@Ben: I think the space of smooth functions $\mathbb{R}^n \to \mathbb{R}^m$ is very fancy, and I'm a mathematician! –  Zhen Lin Jan 30 '12 at 0:15
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. –  Stefan Smith Mar 11 '13 at 23:16
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. –  user21820 Jun 24 '14 at 2:10
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. –  Jade NB Aug 29 '14 at 23:51

6 Answers 6

up vote 13 down vote accepted

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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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. –  Pete L. Clark Jan 29 '12 at 23:49
@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). –  Hurkyl Apr 17 '14 at 6:34
@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. –  Hurkyl Apr 17 '14 at 6:50
@Hurkyl: Obviously, I am speaking about admissable idealizations, not physical objects. But as admissible idealizations, there is definitely such a thing as a "randomly chosen number" in any reasonable mathematical universe, you need to speak about it to describe a specific state of an infinite 2d Ising model, a specific instance of a distribution random field, and so on and so on, and it is also completely consistent to speak about such things! It only conflicts with a stupid metaphysical principle you like, namely uncountable choice. Sorry, but your metaphysical principle should go. –  Ron Maimon Apr 17 '14 at 7:12
@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? –  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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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 $|I|>|\mathbb R|$. –  Asaf Karagila Jan 29 '12 at 23:10
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. –  Asaf Karagila Jan 29 '12 at 23:44
@Emilio: Anyone for the Axiom of Determinateness (AD)? It has interesting consequences. One issue (among many) is plausibility. –  André Nicolas Jan 30 '12 at 0:55
@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. –  Asaf Karagila Apr 17 '14 at 18:02
@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. –  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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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. –  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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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. –  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 http://mathoverflow.net/questions/146813/is-sigma-additivity-of-lebesgue-measure-deducible-from-zf

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The system you want is ZF+ dependent choice + "All sets of reals are Lebesgue measurable". This is the comfortable system for physics.

The main theorems which are required for the development of analysis and physical mathematics only require dependent choice, not full continuum choice. The new axiom that all sets of reals are Lebesgue measurable then drastically simplifies measure theory and prevents paradoxes in probability that make intuitive arguments fail. This allows you to avoid the intuition traps that have forced rigorous mathematics and matheamtical physics to diverge.

For an example of such an intuition trap, consider R as a vector space over Q. Can you find an algebraic basis for this vector space?

The answer is obviously no, when you can speak naturally about probability, for the following reason: pick two independent Gaussian random numbers x and y, with unit variance. They have a certain probability distribution P(n), defined on positive integers n, of being made up of a certain number n basis elements.

Now combine them into another Gaussian random number of unit variance by adding them up with rational coefficients, for example z=(3x+4y)/5. Then z has a decomposition of m+n basis elements (where n is the number of basis elements of x and m is the number of basis elements of y). But z is again a Gaussian of unit variance! So the number of basis elements of z must have the same probability distribution as that of x and of y.

But you can easily prove that it is impossible for the sum of two independent positive numbers distributed with distribution P to have the same distribution P. The reason is simply that it is a probability distribution on positive integers, so if there is a nonzero probability at 1, the sum distribution is only nonzero at 2. In general, if it is first nonzero at N, then the sum distribution is first nonzero at 2N, and it cannot be the same distribution.

But the answer in standard ZFC is that yes, you can find an algebraic basis! It just simply happens that the sets involved in the argument are not measurable, so you are simply not allowed to speak about decomposing a randomly chosen Gaussian number into a basis, this concept is meaningless. This means that randomly chosen real numbers in ZFC are second-class citizens, you are not allowed to perform the same operations on such numbers as you would on numbers which are specified by set operations, rather than random picking.

Similar probabilistic disproofs, using randomly picked numbers, can be given to all the statements which are the counterintuitive consequences of choice. The Banach Tarsky theorem is an obvious example: if you pick a random number, the probability that it lands in two spheres is greater than the probability than it lands in one, so no transformation can map the two spheres onto one (except if you remove the ability to talk about randomly picked numbers).

Each of these intuition conflicts amounts to the intuition that it is possible to choose a number uniformly randomly from [0,1], for example by flipping coins for the binary digits. This is the key in Solovay's construction. When this is possible (it is always convergent), and when this generated number can be assigned membership to arbitrary sets, then it follows that every set is Lebesgue measurable. Conversely, if every set is Lebesgue measurable, it makes sense to speak about random uniform picks from [0,1] without contradiction, because every question you ask about this number has a well-defined probability of being true.

So in this kind of universe, you can talk coherently about random real numbers, about random Gaussian picks, and so on, without fear of hitting a contradiction. This is required for many constructions in statistical and quantum physics.

For example, this allows you to define path integrals and statistical limits, intuitively. To define the free scalar field, simply choose a Gaussian random real with appropriate variance, and do a Fourier transform. To define a stochastic differential equation, pick a noise, and solve the equation, and show that the limit exists. These measure theory arguments become super-annoying when there are non-measurable sets lurking about, because you need to prove that the non-measurable sets never appear when you are transforming your noise to your solution.

Using such a system does not discard any mathematics, you can still talk about classical choice constructions, because every model of ZF has an inner model (Godel's L) which obeys the axiom of choice. So within your Solovay measurable model, you still have a normal usual choice model of set theory. The theorems about Banach Tarsky then you simply reinterpret as theorems about the L submodel. The Banach Tarsky decomposition is then simply a decomposition of all the L-points of a sphere so that they are disjointly mapped to all the L-points of two spheres, something which is neither counterintuitive or surprising, because the L subuniverse is a small measure zero dust within the full universe.

The issue of which model best approximates something like "mathematical reality" is not one which has a logical positivist answer, but if you ask my own opinion, the measurable universe is always better thought of as reality. The constructions of objects which lead to non-measurable sets of reals are always in conflict with intuition, and are incompatible with a simple procedure which denies their existence. But these ridiculous constructions are self-consistent within an inner model, so that you do not lose the arguments made using them, you simply reinterpret those arguments as applying to an impoverished sub-universe within your true universe.

This has been advocated many times within mathematics, but mathematicians are conservative and stupid, and take forever to get with the ball. This should be changing now, but it shouldn't have taken 40 years.

As for the inaccessible cardinals required in the argument, this is a ridiculous propaganda made against these models. When you are proving relative consistency, you often need to go to models where you can explicitly describe a model of the previous axioms as a set. Within ZFC, the strongly inaccessible cardinal is simply the first normal set which can be used as a model for ZFC, and this is what it is used for by Solovay. The essential argument for why measurability is consistent, the fact that you can consistently pick random real numbers, is not itself at all sensitive to large cardinals, it's just that translating this construction into a full model of ZF you need to start in a universe where the ZF operations can be applied to a set which already makes a model for ZFC.

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To date, I have never seen an argument against choice of this sort that cannot be countered with "you are thinking about measurable sets, not sets" (or similar). Because of this, I have never found these persuasive, because they simply boil down to "I don't want to think about non-measurable sets, therefore nobody else should either." –  Hurkyl Apr 17 '14 at 6:47
Please see Shelah's celebrated paper "Can you take Solovay's inaccessible away" from 1984. In that paper Shelah proves mathematically (not based on intuition and "it just feels right" arguments) that if there is a model of $\sf ZF+DC+LM$ (where $\sf LM$ means all sets are Lebesgue measurable) then there is a model of $\sf ZFC+I$ (where $\sf I$ denotes an inaccessible cardinal). Maybe you know set theory much better than me, but I have high doubts that you know set theory that much better than Shelah. Your last paragraph is pure nonsense. Also, Tarski and not Tasrky. –  Asaf Karagila Apr 17 '14 at 8:18
@RonMaimon: You best keep the personal insults to a minimum, which would be none. –  Arthur Fischer Apr 17 '14 at 15:59
@Ron: You keep claiming that the inaccessible is merely used to produce a model of $\sf ZFC$. This is FALSE. And since this has been a well-established mathematical fact, and you claim otherwise, then the burden is on YOU to produce a correct proof. Please let me know when you do, I'll be very interested in reading it. –  Asaf Karagila Apr 17 '14 at 18:05
@RonMaimon Sorry but I must be a little harsh now: you have simply no idea of the domain you pretend to describe (probability theory since the mid 1930s, say)). To say that one "end(s) up introducing filtrations", as if this was a curse, is mere nonsense. –  Did Apr 17 '14 at 18:06

protected by Asaf Karagila Nov 17 '14 at 5:02

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