Banach–Tarski paradox = Contradiction?

[Jcrew]

Why don't mathematicians consider the Banach–Tarski paradox a contradiction that disproves the validity of the axiom of choice?

[blah_blah]

because they don't want to get rid of the positive consequences of assuming the axiom of choice. a huge portion of modern analysis falls apart without it.

in the framework of measure theory, we can simply view the sets produced in the banach tarski paradox as nonmeasurable. in fact, it is a result of solovay that nonmeasurable sets do not exist in ZF without the assumption of AC.

another interesting consequence of AC: it is trivial using the axiom of choice to construct a discontinuous additive function using a hamel basis for R over Q. we can also prove that measurable additive functions are linear (taking the convolution of exp(if(x)) with the standard mollifier when f is measurable + additive). again without AC, all discontinuous additive functions no longer exist, which is interesting because their existence doesn't seem to be related to measure theory in the first place.

[borisp]

The key idea worth noting is that the AoC asserts the existence of something, namely a choice function, without actually asserting its computability. (An example of something that exists but is not computable is the smallest integer that a human being will never compute, assuming the human race exists for finite time.)

To say that a huge portion of modern analysis "falls apart" without it is somewhat narrow in scope...the existence of bases for arbitrary vector spaces and the existence of maximal ideals in arbitrary commutative rings both depend on the AoC.

What really falls apart is the convenience of stating various theorems that hold for "all" of something. Mathematicians are content to overlook this theoretical speedbump since an abundance of concrete examples exist where we can compute bases, maximal ideals, and so on, explicitly. In other words, if you are ever actually using mathematics to answer a concrete question, the AoC will never come up, since you must actually present whatever data or function you need.

One way to think about it is that whenever you use the AoC to prove a theorem, you have informally added "proof of concept" to your burden of proof. By this, I mean that any theorem or result that uses the AoC must be accompanied by an abundance of examples; otherwise, it is meaningless for "real life" use. Hence, we are content to ignore Banach Tarski and similar paradoxes, since they do not involve things that are explicitly computable.

P.S. Another consequence of the AoC is that we can consistently assign a meaningful limit to all bounded sequences...this is the Hahn-Banach theorem applied to the limit functional on the space of convergent sequences as a subspace of bounded sequences. So I guess (1, -1, 1, -1, ...) has a limit somewhere out there...

[Siegmund]

Short answer: don't lose sleep over it.

Not too much to add to this thread, except to comment that if one carefully inspects all the assumptions behind banach-tarski, it's just a highly non-intuitive result (to illustrate that nonmeasurable sets are highly non-intuitive objects)... you'll hear people describe Banach-Tarski as "how to cut an orange up and assemble it into two oranges" - but no piece of physical matter meets the conditions. In that light, B-T isn't all that much more paradoxical than things like putting the even numbers in 1-1 correspondence with the whole numbers - you might even call it the uncountable-infinity equivalent of that.

I would venture to say that neither nonmeasurable sets nor the axiom of choice put in many appearances outside the graduate mathematics classroom - I, for instance, could likely spend the rest of my life as a practicing statistician without having a client confront me with either one (or even do original research in various areas of mathematics other than the dark corners of analysis!)

I will confess that when I, as a graduate student, took real analysis, I a) thought there was a lot of excessive hype over the axiom of choice -- I might have felt differently if I had a desire to pursue the foundations of analysis further but I didnt -- and b) had B-T explained to me a couple times, sort of shrugged, didn't find it all that paradoxical, just another odd counterexample to file away for homework.

[gaming_mouse]

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In that light, B-T isn't all that much more paradoxical than things like putting the even numbers in 1-1 correspondence with the whole numbers - you might even call it the uncountable-infinity equivalent of that.

[/ QUOTE ]

Indeed. Mathematicians often enjoy playing up the drama of paradoxes like this.... "You would have never guessed it! There are actually the same number of integers and rationals! But its true, and here's a proof!"

In fact, it is usually just a matter of playing around with definitions, and choosing convenient ones. So you define "equality of sets" but saying if there exists a 1-1 mapping. Well, if you had defined it by saying "every onto mapping must also be 1-1" -- a definition that works fine for equality of finite sets, so why not use that one? -- then the whole thing breaks down. Even numbers and integers are no longer equal, and our "intuition" still works.

[PairTheBoard]

[ QUOTE ]
Well, if you had defined it by saying "every onto mapping must also be 1-1" -- a definition that works fine for equality of finite sets, so why not use that one? -- then the whole thing breaks down. Even numbers and integers are no longer equal, and our "intuition" still works.


[/ QUOTE ]

Our intuition doesn't work so well under that definition when we realize that the even numbers and the even numbers are no longer equal either.

PairTheBoard

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
Well, if you had defined it by saying "every onto mapping must also be 1-1" -- a definition that works fine for equality of finite sets, so why not use that one? -- then the whole thing breaks down. Even numbers and integers are no longer equal, and our "intuition" still works.


[/ QUOTE ]

Our intuition doesn't work so well under that definition when we realize that the even numbers and the even numbers are no longer equal either.

PairTheBoard

[/ QUOTE ]

You are missing the point, or actually making my point.

I am not arguing for one definition over the other, but pointing out that we should not be surprised when definitions developed for finite sets break down (or result in "paradoxes") when applied to infinite sets.

In the case of the diagonal proof, etc, mathematicians have arbitrarily (in the philosophical sense) chosen one specific definition of equality borrowed from finite sets, and taken off from there.

But people are not "wrong" to say that there are more integers than even numbers. They just mean it in a different way. They mean that the integers contain all the even number, but not vice-versa, which is a perfectly reasonable way to talk about "more."

Now the mathematician's choice is of course not arbitrary in the sense that it has led to many useful results and an entire theory of sets.

[pzhon]

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Why don't mathematicians consider the Banach–Tarski paradox a contradiction that disproves the validity of the axiom of choice?

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We don't because it is not a contradiction. A contradiction is when you can prove both A and ~A. The result mentioned in the Banach-Tarski paradox is a surprise, not a contradiction.

Incidentally, you don't need to assume the axiom of choice to use nonexistence results proven using the axiom of choice. These rely on the consistency of the axiom of choice (proven by Godel and Cohen), not the axiom of choice itself; the existence of something that can't exist given the axiom of choice would show that the axiom of choice is false.

For example, you can use the axiom of choice to construct some functions to solve Hilbert's third problem, to determine whether the regular tetrahedron can be decomposed into polyhedra and reassembled into a cube. You can construct an invariant using the axiom of choice which is different for the regular tetrahedron and the cube. Because this is a nonexistence result, the usage of the axiom of choice is removable, and the result does not actually depend on the axiom of choice.

[gaming_mouse]

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Because this is a nonexistence result, the usage of the axiom of choice is removable, and the result does not actually depend on the axiom of choice.

[/ QUOTE ]

hey pzhon,

I don't follow this last part. If you had to use the AC in your proof of the result, how does it not depend on it?

Thanks,
gm

[Siegmund]

I think what pzhon means (he can jump in and correct me if I misunderstood him) is...

If you accept the axiom of choice, then the tetrahedron cannot be dissected into polyhedra and turned into a cube;

The contrapositive is, if a tetrahedron CAN be turned into a cube, you must reject the axiom of choice.

However.... a) we're talking about whacking the tetrahedron into other polyhedra, simple knife cuts, NOT into complicated Banach-Tarski-esque pieces; either this geometric construction is possible or it isn't, regardless of how you feel about the axiom of choice

and.... b) Mathematics with the axiom of choice is consistent; and mathematics without the axiom of choice is also consistent. That is, it's really an axiom - you either take it on faith or you don't, its truth or falsity cannot be proven from the other axioms.

...but IF you could dissect a tetrahedron, you WOULD be able to prove the axiom of choice was false. This is a contradiction of the Godel result pzhon cited.

ON the other hand, if the dissection is impossible, we continue to have no basis for whether or not accepting AoC is a good idea.