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Does there exist such a map $\phi: [0, 1] \rightarrow S^2$ which determines some sort of a Banach-Tarski decomposition for the unit interval $[0, 1]$? I did read through Stan Wagon's The Banach-Tarski Paradox but the language is somewhat terse. I did read somewhere that Felix Hausdorff proved that one can chop up the unit interval into countably many pieces, slide the pieces around, and fit them together into the interval $[0, 2]$. Not sure of any journals that describe the exact proof. Is it possible to find $\phi$ explicitly perhaps in terms of the middle-third Cantor set?

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I actually thought this was not possible in dim 1 because in the BT proof you need a free group of isometry, which doesn't exist for R – Glougloubarbaki Nov 25 '11 at 20:05
There is a bijection between $[0,1]$ and $S^2$, so the decomposition on $S^2$ can be mapped to $[0,1]$. But it is not very surprising there, since the rigid motions in $S^2$ are mapped to things that are no longer rigid motions in $[0,1]$. We already know there are bijections between intervals of different lengths. – GEdgar Nov 25 '11 at 20:06
@GEdgar of course I assume the "piecing together" means isometrically, otherwise the problem isn't very interesting – Glougloubarbaki Nov 25 '11 at 20:11
up vote 4 down vote accepted

To get such a paradox using finitely many pieces, as in the Banach-Tarski case, one must move beyond isometries. But if one uses measure-preserving transformations, such a paradox does exist on the line. Reviewing my book on the BTP I find Theorem 7.9 which shows that there is a paradoxical decomposition of the interval using transformations preserve Lebesgue measure. The main point is that there is a Lebesgue-measure preserving function from the interval to the sphere, and that can be used to move the Banach-Tarski Paradox down to the line.

And there is von Neumann's paradox (Thm 7.12) which provides a finitary paradoxical decomposition of the line using epsilon contractions, for any positive epsilon. Since such a contraction is a linear fractional transformation it shrinks Lebesgue measure, so this can be considered paradoxical. Both of these use the Axiom of Choice.

Stan Wagon

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One cannot expect an explicitly constructible example. By an old result of Solovay, if ZF (Zermelo-Fraenkel Set Theory without Choice) plus "there exists an inaccessible cardinal" is consistent (which is generally believed to be the case), then so is ZF plus "every set of reals is Lebesgue measurable."

The sets involved in the paradoxical decomposition cannot all be Lebesgue measurable. Indeed thinking of them as measurable is what makes the decomposition seem paradoxical. Thus some version of the Axiom of Choice is needed, and an explicit set-theoretic construction from just ZF is not possible.

As for "one dimensional" paradoxical decompositions into a finite number of sets, these are not possible. For it can be shown that there is a finitely additive translation-invariant "measure" that extends Lebesgue measure and is defined on all sets of reals. Using such a "measure" we can translate the intuitive notion of paradoxicalness into a proof that there is no paradoxical decomposition. The Banach-Tarski Paradox can be thought of as a proof that there is no translation invariant finitely additive "measure" defined on all subsets of $\mathbb{R}^3$.

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I am sorry to be nitpicky, but you start with a model of ZFC+Inaccessible (although this is the same as without choice since you can always go to $L$) and generate a model of ZF+DC+Every set of reals is Lebesgue measurable. In particular you need somewhat more than just countable choice. I believe that DC($\aleph_1$) would suffice, or perhaps for the continuum... :-) – Asaf Karagila Nov 25 '11 at 20:14
This answer is vast overkill. Even with the Axiom of Choice, there is still no paradoxical decomposition in one dimension, which is what this question asks about. – GEdgar Nov 25 '11 at 20:17
@Asaf Karagila: Thank you, I should have said Dependent Choice. Nitpicky one should be. – André Nicolas Nov 25 '11 at 20:19
@GEdgar: Lebesque measure can be extended to a finitely addive "measure" defined on all sets of reals, which rules out only paradoxical decompositions into a finite number of sets. The OP was asking about decompositions into countably many sets. There are plenty of those, like the one in the standard proof that there are non-measurable sets of reals. – André Nicolas Nov 25 '11 at 20:27

Actually, let's work with the circle ${\mathbb T} = {\mathbb R}/{\mathbb Z}$, and $\mathbb Q_T = {\mathbb Q}/{\mathbb Z}$. A Vitali set $V$ is a subset of $\mathbb T$ such that $\mathbb T$ is the disjoint union of the translates $V + r$ for $r \in \mathbb Q_T$. Of course this can't be constructed explicitly, you need the Axiom of Choice. Now for $r \in [0,1/2)/\mathbb Z$ translate $V + r$ to $V + 2r$, and the disjoint union of these is $\mathbb T$; for $r \in [1/2, 1)/\mathbb Z$ translate $V + r$ to $V + 2r - 1$, and the disjoint union of these is also $\mathbb T$. So we've decomposed $\mathbb T$ into countably many pieces, translated some of them to make $\mathbb T$, and translated the rest to make another copy of $\mathbb T$.

If you take $[0,1)$ instead of $\mathbb T$, you can make this work with a slight adjustment: instead of translating all of $V + r$ by $r$ to get $V+2r$, you have to break $V+r$ into two pieces, translate $(V + r) \cap [0,1-r)$ by $r$ but translate $(V + r) \cap [1-r, 1)$ by $r-1$.

If you want to use $[0,1]$ instead of $[0,1)$ there's a further complication: what do you do with $1$?. But that's not hard: noting that all the translations above were by rational numbers, you just do what I described to the irrationals in $[0,1]$, and separately map the rationals in $[0,1]$ to two copies of itself.

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