There is an even stronger version of the paradox:
In other words, a marble can be cut up into finitely many pieces and reassembled into a planet. Or you could disassemble a telephone and reassemble it to make a water lily. OK, you can't do this with a real marble or a real water lily, which are made up of atoms, but you can do it with their shapes.
It is interesting to note that this theorem depends on three dimensions; while intuitively the twodimensional case seems to be easier, it is in fact not true that all bounded subsets of the plane with nonempty interior are equidecomposable. Still, there are some paradoxical decompositions in the plane: a circle can be cut into finitely many pieces and reassembled to form a square of equal area; see Tarski's circlesquaring problem.
Note that in the decomposition, the pieces won't be measurable, and so they will not have "reasonable" boundaries nor a "volume" in the ordinary sense. It is impossible to carry out such a disassembly physically because disassembly "with a knife" can create only measurable sets. This pure existence statement in mathematics points out that there are many more sets than just the measurable sets familiar to most people.
In 1924, Stefan Banach and Alfred Tarski described this paradox, building on earlier work by Felix Hausdorff who managed to "chop up" the unit interval into countably many pieces which (by translation only) can be reassembled into the interval of length 2. He did this in order to show that there can be no nontrivial translation invariant measure on the real line which assigns a size to all sets of real numbers.
Logicians most often use the term "paradox" for a statement in logic which creates problems because it causes contradictions, such as the Liar paradox or Russell's paradox. The BanachTarski paradox is not a paradox in this sense but rather a proven theorem; it is a paradox only in the sense of being counterintuitive. Because its proof prominently uses the axiom of choice, this counterintuitive conclusion has been presented as an argument against adoption of that axiom.
A sketch of the proof follows. Essentially, the paradoxical decomposition of the ball is achieved in four steps:
The free group with two generators a and b consists of all finite strings that can be formed from the four symbols a, a^{1}, b and b^{1} such that no a appears directly next to an a^{1} and no b appears directly next to a b^{1}. Two such strings can be concatenated and converted into a string of this type by repeatedly replacing the "forbidden" substrings with the empty string. For instance: abab^{1}a^{1} concatenated with abab^{1}a yields abab^{1}a^{1}abab^{1}a, which gets reduced to abaab^{1}a. One can check that the set of those strings with this operation forms a group with neutral element the empty string, here denoted ε. We will call this group G.
The group G can be "paradoxically decomposed" as follows: let S(a) be the set of all strings that start with a and define S(a^{1}), S(b) and S(b^{1}) similarly. Clearly,
In order to find a group of rotations of 3d space that behaves just like (or "isomorphic to") the group G, we take two orthogonal axes and let A be a rotation of arccos(1/3) about the first and B be a rotation of arccos(1/3) about the second. (This step cannot be performed in two dimensions.) It is somewhat messy but not too difficult to show that these two rotations behave just like the elements a and b in our group G. We'll skip it. The new group of rotations generated by A and B will be called H. Of course, we now also have a paradoxical decomposition of H.
Step number 3: The unit sphere S^{2} is partitioned into "orbits" by the action of our group H: two points belong to the same orbit if and only if there's a rotation in H which moves the first point into the second. We can use the axiom of choice to pick exactly one point from every orbit; collect these points into a set M. Now (almost) every point in S^{2} can be reached in exactly one way by applying the proper rotation from H to the proper element from M, and because of this, the paradoxical decomposition of H then yields a paradoxical decomposition of S^{2}.
(This sketch glosses over some details. One has to be careful about the set of points on the sphere which happen to lie on an axis of rotation of some matrix in H. On the one hand, there are countably many such points so they "do not matter," and on the other hand it is possible to patch up even those points.)
Finally, connect every point on S^{2} with a ray to the origin; the paradoxical decomposition of S^{2} then yields a paradoxical decomposition of the solid unit ball (minus the origin, but that can be dealt with easily).
References:
Search Encyclopedia

Featured Article
