Banach-Tarski, the Axiom of Choice, and Notions of Size

1 10 2006

At the Australasian Association of Philosophy conference in July, someone raised a question about the Banach-Tarski paradox (that link actually explains and proves it quite nicely) that was supposed to be a problem for someone’s position. It was pointed out that this depended on the Axiom of Choice (and therefore perhaps wasn’t that important), but Jason Grossman said that this wasn’t quite true. In discussion afterwards, he sent me to this paper by Randall Dougherty and Michael Foreman, where the relevant results were published (back in 1994).

It turns out that this is not quite right – full Banach-Tarski-like results do in fact depend on the Axiom of Choice. (This can be seen because the Axiom of Determinacy, which is an alternative to choice that is believed to be consistent, entails that all sets of reals are Lebesgue measurable, which would mean that only sets of the same Lebesgue measure are equidecomposable.) However, Dougherty and Foreman do prove some similarly paradoxical-sounding results without choice, which is what Grossman was referring to.

Perhaps the most surprising result is that for any two closed balls in Rn, one can take a finite collection of disjoint open(!) subsets of the first and rearrange them to form a set whose closure is the other. One can even arrange things so that these sets are dense in each ball. Intuitively, there is a sense in which a dense open subset of a set is “almost equal” to that set (this is in Baire’s sense, rather than the senses of Borel or Lebesgue), and there is also a sense in which open sets are extremely well-behaved. Thus, the result shows that one can take two balls of vastly different sizes, and show that they are “almost the same” as one another, using well-behaved sets and well-behaved tranformations of those sets (rotations and translations). This sounds bad – but it may make weaken the case of Banach-Tarski against the Axiom of Choice, since you can get something almost as bad even without it. However, I think this isn’t quite right. Instead, it seems to suggest both that Baire’s sense of “almost all” just isn’t as good as Lebesgue’s, and perhaps also that open sets aren’t as nice as we thought.

Baire’s idea is that a dense open subset of a nice topological space is almost all of it. (We call the complement of a dense open subset, “nowhere dense”, because its closure contains no open interval – in fact it contains nothing outside of itself.) This was reinforced when he proved the Baire Category Theorem, which shows that the intersection of countably many dense open sets is itself always dense (anything including such a countable intersection is said to be “co-meager”, and its complement, a subset of the countable union of nowhere dense sets, is said to be “meager”). Traditionally, meager sets are thought of as “almost empty” and co-meager sets are thought to contain “almost all” points. In particular, the boundary of any open or closed set is meager. A set is said to have “the property of Baire” if its symmetric difference with some open set is meager – thus, it is “almost open”.

By contrast, Lebesgue introduced a much more precise notion of measure, assigning a real number to every measurable set, and saying that sets of measure 0 are “almost empty” and their complements contain “almost all” points. In the reals, this is done by declaring the measure of any open or closed interval to be the difference between the endpoints, and then saying that this measure is additive for disjoint unions of measurable sets. (In higher dimensions, one starts with rectangular products of open or closed intervals in the reals, saying that the measure of the product is the product of the measures, and continuing as before.) Finally, any subset of a measure 0 set is also declared to be measurable, and to have measure 0. It turns out that a set is Lebesgue measurable iff its symmetric difference with some open set has measure 0, so being measurable is Lebesgue’s counterpart of the property of Baire.

It has long been known that Baire’s and Lebesgue’s notions are different. It turns out that one can find dense open sets of arbitrarily small Lebesgue measure (consider the countable collection of basic open sets with rational endpoints, and take one open subset of each of measure at most x/2n, and their union will be a dense open subset of measure at most 2x), and thus the countable intersection of some of them will be a co-meager set of measure 0, and its complement will be a meager set of full measure.

Now, because Lebesgue measure is countably additive, and open sets are all Lebesgue measurable, we can see what must be going on in the Dougherty-Foreman case mentioned above. The dense open sets in the decomposition must have Lebesgue measure less than that of the small ball, but their boundary must be a meager set whose Lebesgue measure makes up for the difference. Thus, although we can rearrange parts of the small ball to be dense in the large one, we haven’t filled “almost all” of it in Lebesgue’s sense. If we stick with Lebesgue measure rather than Baire-influenced notions, we don’t get the paradox unless we use the Axiom of Choice. And with the Axiom of Choice, the sets needed for the decomposition are non-Lebesgue-measurable. And as a further blow against the Baire notion, the central result of Dougherty and Foreman is a Banach-Tarski-style decomposition where every set involved has the property of Baire. Thus, sets with the property of Baire aren’t really all that nice, since we can use them to decompose small sets into large ones.

At one point in the past, it looked like Baire and Lebesgue had come across two different but equally good notions of “almost equality” for sets of reals (and sets in other topological spaces as well). However, it looks like more recent developments in set theory have shown that Lebesgue’s is in general more useful. The Banach-Tarski paradox doesn’t occur without the Axiom of Choice (or at least, some axiom on the way towards it, since we need to construct non-Lebesgue-measurable sets), but with choice we can do it with sets with the property of Baire. Something seemingly nearly as bad occurs with open sets even without choice, but it turns out that it only looks bad when using Baire’s notion of almost equality. So the Banach-Tarski paradox remains approximately as paradoxical as it always was, though we have more understanding now of it through some of its relatives.

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One response

2 10 2006
Richard

In case you do not have access to JSTOR, “this paper” may also be obtained at

http://www.math.uci.edu/sub2/Foreman/homepage/BT.pdf

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