Here is how I understand the Banach–Tarski paradox, based on the Wikipedia article : with a clever partitioning, one can decompose a solid ball into two solid balls, each identical to the first one.

I hear this paradox cited here and there a lot, but I don't really get what makes it so interesting.

For instance, it is easily accepted that $[0,1]$ and $[0,2]$ are isomorphic.

We can by the exact same reasoning show that a cube on $\mathbb{R}^3$ is isomorphic to twice itself.

To me, it seems that the 'Banach-Tarski paradox' immediately follows.

Have I missed something ?

PS : I am talking about the raw version of the paradox, not the numerous extensions that have been made of it, like showing that the decomposition can be chosen in such a way that the parts can be moved continuously into place without running into one another, etc.

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    You can decompose the ball into finitely many pieces and reassamble them **using only rigid motions** (rotations + translations) to get two balls. In the "isomorphy" of $[0,1]$ and $[0,2]$, the bijection between the two intervals is **not length preserving**, whereas the transformations in the Banach Tarski paradox **are**. – PhoemueX Sep 28 '14 at 21:10
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    The motivation for the Banach-Tarski paradox (and the Hausdorff paradox which preceded it) is that it establishes a limitation on what you can do with measure theory. Namely, it shows that there is no possible way to extend the definition of measure so that all sets are measurable, if certain basic axioms are to be satisfied. – bof Sep 29 '14 at 07:42

3 Answers3


The reason it is interesting is in the exact nature of the isomorphism involved in the Banach Tarski paradox, namely, isomorphism up to rigid motion of $\mathbb{R}^3$.

What the paradox says is that a solid ball in $\mathbb{R}^3$ can be partitioned into a finite number of sets $A_1,…,A_K$, and those sets $A_1,…,A_K$ can be moved around by rigid motions of $\mathbb{R}^3$, so that they consitute two solid balls each of the same size as the original. That is not what is going on with $[0,1]$ and $[0,2]$.

Added to address some comments: The Banach-Tarski paradox does not occur in $\mathbb{R}$ or in $\mathbb{R}^2$. The underlying reason for this difference with $\mathbb{R}^3$ is that the group of rigid motions of $\mathbb{R}$ or of $\mathbb{R}^2$ is amenable, whereas the group of rigid motions of $\mathbb{R}^3$ contains a free group of rank $>1$ and therefore is not amenable.

Lee Mosher
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    Is there no way to transform $[0,1]$ into $[0,1]\cup[1,2]$ using rigid motions ? – Hippalectryon Sep 28 '14 at 21:15
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    @TheGame No because rigid motions preserve things such as length. – Benjamin Sep 28 '14 at 21:18
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    @Benjamin: That is not really a valid argument here. Rigid motions also preserve "volume", but obviously this is violated in the Banach Tarski paradox. So we have to find another argument for why this is not possible. Somewhere I read that the lebesgue measure on (the Borel sets of) $\Bbb{R}$ can be extended to a finitely additive measure on all subsets of $\Bbb{R}$, invariant under rigid motions. **This** would make the argument valid. Banach Tarski then implies that this is not possible for Lebesgue measure on $\Bbb{R}^3$. – PhoemueX Sep 28 '14 at 21:21
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    @PhoemueX yes, that's why BT is always done in three dimensions. – Kevin Arlin Sep 28 '14 at 21:23
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    See also this question http://math.stackexchange.com/questions/380184/proof-that-there-is-no-banach-tarski-paradox-in-bbb-r2-using-finitely-additi. – PhoemueX Sep 28 '14 at 21:24
  • @PhoemueX I interpreted The Game's question as not involving a decomposition of $[0,1]$, but rather if there was a rigid motion of $[0,1]$ itself that had $[0,2]$ as its image. If decompositions were permitted then I admit you are right. Either way, considering the nature of the main question, I probably should have read it in the second. – Benjamin Sep 28 '14 at 21:29

The map from $[0,1]$ to $[0,2]$ does not preserve measure. The map which "recomposes" the unit ball does. Which reinforces the point that "outlaw sets" can disobey many "laws" (in this case, outlaws means non-measurable, and disobey a law means to have a strange behavior under measure preserving maps).

Asaf Karagila
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  • If you mapped only the irrationals $x\in[0,1]\to x+1$, under certain measures you'd have a mapping from $[0,1] \to [0,2)$, at least, but that's definitely not a finite number of sets. – Mark Hurd Sep 30 '14 at 07:08
  • @Mark: Yes, under *certain measures*. Under other measures you have a measure preserving bijection between any two sets of cardinality continuum which include $0$. – Asaf Karagila Sep 30 '14 at 07:10

Banach-Tarkis paradox shows that it is impossible to have a Banach measure in 3 dimensions. This is a bit disappointing, because it would allow us to "measure" all sets, in some sense, if it existed. Also, this is surprising, because Banach measures exist in dimension $1$ and $2$.

A Banach measure is a much like a "normal" measure, except:

  1. It is only required to be finitely additive.
  2. It is supposed to be defined for all sets and be preserved under isometries.

Such measures are less useful for the purpose of developing integration theory, and so on, but it is not clear a priori (at least to me) that they are less interesting than the "normal" measures.

As for the "isomorphism" between $[0,1]$ and $[0,2]$, this shows that we can't have a measure that is invariant under scaling, which is not surprising at all (it would be rather surprising if you had such invariance!).

Jakub Konieczny
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