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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.

Asaf Karagila
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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
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    @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 00:15
  • 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
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    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 02:10
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    Feferman has, I think, spent quite a bit of intellectual effort on just this question; see, for example, http://math.stanford.edu/~feferman/papers/psa1992.pdf. – LSpice Aug 29 '14 at 23:51
  • The axiom of choice gets a lot of undeserved negative press. It is actually a theorem of ZF plus the assumption that the universe consists entirely of sets you can construct. So, before you can even consider rejecting choice, you must first assert the existence of an unconstructible set! –  Aug 28 '17 at 16:34
  • Also, insofar as we call real numbers "physical", there is nothing unphysical about infinitesimals -- analysis in the standard and nonstandard models are not just 'physically' indistinguishable from one another, they are theoretically indistinguishable as well. The only way we can even make sense of the idea that the nonstandard model has infinitesimals is if we allow ourselves to work with both the standard and nonstandard models simultaneously, and we see the nonstandard models has positive reals of magnitude smaller than any appearing in the standard model. –  Sep 04 '17 at 06:33
  • In my opinion, the main problem with axiom of choice is that

    Axiom of choice states that one can make a choice of arbitrary elements from infinite number of non-empty sets, without specifying any law or rule.

    To memorize or write down such choice one would need an infinite amount of memory.

    But in our universe no information-processing system can store or communicate infinite amount of information. This makes such choice effectively non-reachable with scientific method, because scientific method requires reproducibility. It also requires ability to communicate the results.

     – Anixx Jun 19 '21 at 14:50
  • @user14972 Axiom of Choice postulates that sets that require infinite memory to construct, are constructible. Thank, very much, don't need this! – Anixx Jun 19 '21 at 14:53

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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?

Cloudscape
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André Nicolas
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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
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    @PeteL.Clark: I agree. I don't believe that the foundations of mathematics have any implications for physics, and I don't think physics provides any grounds for accepting or rejecting parts of mathematics. However, I think most mathematicians would agree that as you make a foundational axiomatic system like set theory stronger and stronger, you reach a point of diminishing returns and diminishing plausibility. Most people want to stop before they start assuming things like large cardinals. To me, the full axiom of choice feels like it's already past that point. –  Jan 30 '12 at 01:38
  • The reason Banach Tarski is unacceptable (as is Vitali) is because physicists are used to making intuitive probabilistic arguments which fail when there are nonmeasurable sets. 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. Being able to speak about random picks is so important, that AC is a non-starter. – Ron Maimon Apr 16 '14 at 19:32
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    @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). –  Apr 17 '14 at 06:34
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    @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. – Ron Maimon Apr 17 '14 at 06:41
  • @Hurkyl: It's not just the coefficient that is not measurable, it's the number of basis elements used in the decomposition. The decomposition itself is not measurable. You claim "every real number can be decomposed", and then "but no random variable can be decomposed". This means that you simply decided to never speak about the decomposition of randomly chosen numbers! As if there is no such thing. This is a ridiculous stance, you are simply declaring that random numbers don't exist, and blinding yourself like that is stupid, it is not acceptable to me, or to a lot of other people. – Ron Maimon Apr 17 '14 at 06:45
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    @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. –  Apr 17 '14 at 06:50
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    @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 07:12
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    @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 08:32
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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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Asaf Karagila
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  • Nice answer. But I'm confused by the final remark about Solovay. If I'm understanding correctly, then we're writing $\mathbb{R}=\cup A_\alpha$, where the A are disjoint, and the cardinality of ${A_\alpha}$ is less than the cardinality of $\mathbb{R}$. If you had choice, you could use it to pick a member of each A and map it to that A. This function would be one-to-one, contradicting your statement. So is you statement that the cardinality of ${A}$ is less than the cardinality of the reals, or merely that we can't prove in ZF that it's not less? –  Jan 29 '12 at 23:09
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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
  • OK, so is one of the following true? (1) My reasoning above about the one-to-one function is incorrect. (2) The example you refer to is a counterexample to the axiom of choice within this model -- in which case all I think the example demonstrates is that if you don't assume choice, choice doesn't necessarily hold. –  Jan 29 '12 at 23:23
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    In a manner very fitting to the sense of irony this example carries with it, neither the options you gave is true. The point I was trying to make is that you have a problem with the Banach-Tarski paradox, but this is much much worse. You can actually cut something into more parts than elements. This only demonstrates that if you let go of the axiom of choice simply because Banach-Tarski is hold you back then there is going to be a different paradoxical snake to bite you. – Asaf Karagila Jan 29 '12 at 23:29
  • Thanks for the links to the other questions. But, sorry, you've lost me completely re Solovay. If choice is independent of ZF, then it doesn't surprise me if there are models of ZF in which there are counterexamples to choice. If there is more to what you're saying than this, then I don't understand what it is. –  Jan 29 '12 at 23:41
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    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
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    @Asaf This is very interesting. However, is it possible to avoid both the Banach-Tarski paradox and the paradox in the Solovay model you just mentioned, perhaps allowing countable choice, so that one can still do basic calculus? This seems like a nice compromise (if it is possible). – Emilio Ferrucci Jan 30 '12 at 00:38
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    @Emilio: Anyone for the Axiom of Determinateness (AD)? It has interesting consequences. One issue (among many) is plausibility. – André Nicolas Jan 30 '12 at 00:55
  • @André they all seem very plausible to start with, and then contradict each other and imply horrible paradoxes! – Emilio Ferrucci Jan 30 '12 at 01:29
  • @Asaf: I see. The thing is, just because a particular theory T has a particular model M with properties we don't believe, that doesn't mean that T is a bad theory. If T is "all poodles are dogs," and M is a model of T in which George W. Bush was a good president, that isn't evidence that T is a bad theory. Similarly, the existence of nonstandard models of Peano arithmetic doesn't tell me that Peano arithmetic is a poor way of formalizing our notions about the natural numbers. To the extent that AC sounds reasonable when you first hear it, any theory without AC admits strange models. –  Jan 30 '12 at 01:47
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    @AndréNicolas: I do not understand your concern on plausibility. Care to expand? The reason why I don't know what you mean is that the axiom of determinacy is simply true (not just consistent) in several inner models, under assumptions of large cardinals in the universe. Now, there are absolutely no mathematical arguments against large cardinals (see http://mathoverflow.net/questions/44095/arguments-against-large-cardinals/44185#44185). Instead, they are routinely adopted by most modern set theorists as part of the basic assumptions we start with. So plausibility as a concern is strange. – Andrés E. Caicedo Jan 30 '12 at 02:33
  • @Andres Calcedo: I have an interest in AD, at least in the sense that I published a minor paper that uses it. But, perhaps only because of familiarity, I tend to view full AC as true. – André Nicolas Jan 30 '12 at 04:18
  • @Emilio: My point is that there will always be some "paradox". Be it Banach-Tarski; or this horrid decomposition; or worse. If you want to accept one "paradox" over the other, then it doesn't really matter and you can just use the one which is the most convenient (in this case, just ZFC). – Asaf Karagila Jan 30 '12 at 08:03
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    @Ben: See my previous comment to Emilio. – Asaf Karagila Jan 30 '12 at 08:03
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    @Asaf If I understand correctly though, the Banach-Tarski paradox is a Theorem in ZFC, but the paradoxical decomposition you are referring to is only consistent with ZF. So if one's goal is to use a theory in which no paradoxes are actually theorems, one could renounce the full AC; maybe allowing countable choice actually rules out "Solovay's paradox" altogether. Is this reasoning right? – Emilio Ferrucci Jan 30 '12 at 11:12
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    @Emilio: The Banach-Tarski is indeed a theorem of ZFC, the "paradoxes" I refer to are theorems of ZF+[various non-AC assertions]. The point is that either you will have no choice and thus some paradox will rise - or that you'll have enough choice for some paradox to rise. Either way, you're stuck with paradoxes. Regardless to that, none of them are actually important for doing mathematics (they just guide you that you must walk carefully between the available definitions), so there's no point in "renouncing choice" for the sake of no Banach-Tarski. – Asaf Karagila Jan 30 '12 at 11:23
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    @AsafKaragila: You are forgetting the obvious paradox that choice is incompatible with the statement "consider this randomly chosen real number". This is a paradox which makes it forbidden to talk about random numbers with specific values! Instead, you have to talk about randomly chosen things as second-class objects in the universe, where certain operations cannot be defined. – Ron Maimon Apr 17 '14 at 15:21
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    @AsafKaragila: Downvoted, because of the statement that "in Solovay's model you can decompose R into more parts than it has elements". This is using the ridiculous idea that just because you can't inject aleph-1 into R that aleph-1 somehow has "more elements" than R. Not true, it doesn't, it is simply incomparable. The ordinal tower is separated conceptually from the powersets in Solovay model, and you can't compare the two in size at all. The reals are enormously large, and the ordinals are much smaller, but to demonstrate this by embedding you have to first take the ordinals to be countable. – Ron Maimon Apr 17 '14 at 16:01
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    @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
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    @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 is pretty silly in any case. If you have any partition of R that doesn't inject into R, then you can create this by just putting [0,1] in singleton partitions and the rest into that other partition. It's really meaningless to call that more, since $\mathbb{R} \leq \mathbb{R}$ – Jason Carr Apr 14 '18 at 18:45
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    @Jason: And if the partition is strictly larger? How can adding more elements to a set make it smaller? – Asaf Karagila Apr 14 '18 at 19:10
  • @AsafKaragila No, I'm showing how to construct a strictly larger partition easily. If you have one that is incomparable over injective functions, then you have one which is "larger" over injective functions by just adding $\mathbb{R}$-many elements to the partition. – Jason Carr Apr 14 '18 at 19:32
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    @Jason: Then I really have no idea what is the point you're trying to make with your comment. – Asaf Karagila Apr 14 '18 at 19:33
  • @AsafKaragila The point is it's a rather trivial and intuitive consequence, not a very strong argument. – Jason Carr Apr 15 '18 at 00:31
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    @Jason: Can we skip the pronoun game? What are you talking about? – Asaf Karagila Apr 15 '18 at 07:29
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    The existence of partitions with "more" elements via injections is a rather trivial and intuitive consequence of partitions which are incomparable to $\mathbb{R}$ under injections. For this reason, using such a partition's existence as an argument against Solovay's model is not very strong.

     – Jason Carr Apr 15 '18 at 09:10
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    @Jason: Then your intuition is quite unique, I must say. I'm still not used to this "intuitively", having spent the last seven years working a lot in choiceless settings. Kudos. In any case, my point is that the Banach–Tarski theorem cannot be used as an argument against choice, as a whole, since in its absence, you get a whole other kind of weird partitions. But again, if you find this intuitive, good for you, then – Asaf Karagila Apr 15 '18 at 09:38
  • @AsafKaragila The point I was trying to make is that it's a very simple consequence of an incomparable partition (which ought to exist). It's just that injections are a very poor means of talking about more, since we let $\mathbb{R}$ inject by just adding a copy of $\mathbb{R}$ into the partition. It's just putting A into A + B, where there's no relation between A and B. – Jason Carr Apr 16 '18 at 02:20
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    @Jason: The "ought to exist" and the "intuitive" (from previous comments) are the reasons I find your blasé attitude so odd. Yes, these things exist in Solovay's model, and yes if one has done enough of these proofs it's even easy to see why. But from here to intuitive there is a *long* way. – Asaf Karagila Apr 16 '18 at 06:10
  • @AsafKaragila The ought here is in regard to models which don't have non-measurable sets. We don't expect partitions like R/Q to be a subset of R (in a nice way), nor R a subset of them, so if that were true for some partition, we'd get a 'larger' partition by simple adding R many elements. – Jason Carr Apr 17 '18 at 05:23
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    @Jason: This will be my last comment, because we are really going at it like two broken records. You claim that it's very intuitive that a set can be partitioned into more non-empty sets than it has elements. Kudos for you. You are one in a billion. All that I am claiming is that *in my opinion* arguing that Banach–Tarski is a reason to abandon choice is silly, since you get a weird partition of the reals instead of the BT partition of $\Bbb R^3$. Clearly you disagree. I hope you move to choiceless set theory soon, since your intuition seems to be impeccable. See you at some conference... – Asaf Karagila Apr 17 '18 at 07:12
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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.

Michael Greinecker
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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
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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. :-)

hmakholm left over Monica
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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
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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

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Mikhail Katz
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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.

Timothy
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