# What (and how many) pieces does the Banach-Tarski Paradox break a sphere into?

Wikipedia says

There exists a decomposition of the ball into a finite number of non-overlapping pieces which can then be put back together in a different way to yield two identical copies of the original ball.

So my question is, how many pieces is the ball broken into, and what are their shapes like?

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–  user17762 May 4 '12 at 2:31
Stan Wagon's book works.bepress.com/stan_wagon/13 or amazon.com/… is excellent in describing this. –  Ross Millikan May 4 '12 at 2:36
Downvoted for not noticing that the Wikipedia article, linked in the question, says it's five. –  MJD May 4 '12 at 2:37
@MarkDominus: Indeed, I deserve it. I didn't happen to get that far in the article since it was kind of over my head >_< thanks. –  Mehrdad May 4 '12 at 2:52
You may also be interested on this question on the number of pieces needed for more than two copies. –  Andres Caicedo Jan 12 at 7:40

From the article to which you linked:

In fact, there is a sharp result in this case, due to Robinson: doubling the ball can be accomplished with five pieces, and fewer than five pieces will not suffice.

A reference is given to R. M. Robinson, On the Decomposition of Spheres, Fund. Math. 34:246–260.

It's impossible to describe the shapes of all of the pieces. Another quotation from the article:

Unlike most theorems in geometry, this result depends in a critical way on the axiom of choice in set theory. This axiom allows for the construction of nonmeasurable sets, collections of points that do not have a volume in the ordinary sense and for their construction would require performing an uncountably infinite number of choices.

Very roughly speaking, the construction of such a decomposition requires the ability to make too many arbitrary choices for them to be describable in any concrete terms. There are models of set theory in which such choices cannot be made, and in which such decompositions do not exist. This applies to four of the pieces in the $5$-piece decomposition mentioned above; the fifth is a single point.

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I thought one of the five pieces consisted of the single point at the center. –  MJD May 4 '12 at 2:35
@Mark: That's definitely true of the decomposition that I've seen. If didn't actually mean that every piece had to be non-measurable; I hope that it's clearer now. –  Brian M. Scott May 4 '12 at 2:47
But at most one piece of five can be measurable ... regardless of the decomposition. –  Mark Bennet May 4 '12 at 8:24