# Have I totally misunderstood the Banach–Tarski paradox?

I was just reading about the Banach–Tarski paradox, and after trying to wrap my head around it for a while, it occurred to me that it is basically saying that for any set A of infinite size, it is possible to divide it into two sets B and C such that there exists some mapping of B onto A and C onto A.

This seems to be such a blatantly obvious, intuitively self-evident fact, that I am sure I must be missing something. It wouldn't be such a big deal if it was really that simple, which means that I don't actually understand it.

Where have I gone wrong? Is this not a correct interpretation of the paradox? Or is there something else I have missed, some assumption I made that I shouldn't have?

• I'm no expert on the topic but my interpretation was that it's not just that there is "some mapping" but that we can do it just with rotation and translation (isometries) in $\mathbb{R}^3$ which is less than obvious. I think the standard proof involves breaking the sphere into four pieces instead of two. – hexomino Jan 20 '17 at 13:14
• Indeed, it's not "infinite sets", it's "infinite sets with extra structure", and it's not "some mapping", it's a very specific kind of mapping. But your intuition is how I think about the "paradox" as well. Infinite sets are weird. People already find it surprising when they first learn $|2\mathbb N| = |\mathbb N|$. Once you wrap your head around that, then I don't think BTP should be all that surprising. It's the same thing but with bijections that preserve more structure. – Jack M Jan 20 '17 at 13:41
• You are thinking about it in terms of cardinality of the sets, but really, it is talking about volume; taking a 3-dimensional sphere $A$ in $\mathbb{R}^{3}$, and the volume of $A$ being equal to both $B$ and $C$. – Morgan Rodgers Jan 20 '17 at 13:48
• No, Banach-Tarski is saying something much more complicated - the "maps" allowed for BT are rigid motions, not just any bijections. – Thomas Andrews Jan 20 '17 at 14:51
• Thinking about it in terms of volume can be misleading - (some of) the pieces involved will necessarily be non-measurable. – Klaus Draeger Jan 20 '17 at 15:08

That's not what the paradox says. It says that you can take the unit ball in $\mathbb{R}^3$, divide it in certain disjoint subsets, then you can rotate and translate these subsets to obtain two unit balls. You need at least $5$ weird subsets if you want to do this 'explicitly'. The weird thing about this construction is that it seems that you somehow doubled the volume of the ball simply by cutting it into several parts.

The simple explanation is that there was absolutely no reason to expect that the volume should be preserved under the construction, as some of the disjoint subsets are not measurable, i.e. have no volume.

A first step to understanding the paradox is showing that it is impossible to define a meaningful measure on all subsets of $\mathbb{R}$ that is translation-invariant and such that the measure of an interval $[a,b]$ is $b-a$ (and a bunch of other desired properties). You can look up Vitali sets as an easy example of non-measurable sets. These certain subsets in the paradox are also going to be very wild, much like the Vitali sets.

Edit: To avoid any confusion. I just want to remark that the Banach-Tarski paradox is in fact not a paradox. Mathematically speaking this construction of "the doubling of the ball" is possible.

• And "have no volume" in this case does not mean "has zero volume." – David K Jan 20 '17 at 13:27
• Small nitpick : the construction to obtain two spheres from one sphere only requires $4$ pieces. You need five pieces to obtain two balls from one ball. And I think there is a variant with less pieces than that. – Arnaud D. Jan 20 '17 at 14:19
• @Benubird The important thing is that the locations of the points and the distances between them matter, because once you split the set into a finite number of subsets, you can only transform the subsets through rigid motions - no stretching, cutting, or rearranging is allowed within each subset. It turns out you can't do this with the unit interval in a way that produces two unit intervals, but you can with the ball in 3-space. – MartianInvader Jan 20 '17 at 20:34
• @Benubird: To stress something that is already included in MartianInvader's comment, what you (Benu) say would apply to a construction where you “disintegrate” a ball into all its (uncountably many) points and then rebuild something else with the same points. But the B-T paradox shows that you can do that construction decomposing the ball into finitely many parts (and very few at that), each keeping its “shape”. – DaG Jan 21 '17 at 18:04
• I feel your edit is not necessary. According to Google, the definition of paradox is "a seemingly absurd or contradictory statement or proposition which when investigated may prove to be well founded or true" which fits the Banach-Tarski paradox perfectly. – user159517 Jan 21 '17 at 18:29

In addition to the other answer, you might be interested to learn that extending the statement of the Banach-Tarski paradox to $\mathbb{R}$ or $\mathbb{R}^2$ doesn't work.

See here for a more detailed discussion.

This shows that the statement is in fact deeper than providing bijections between infinite sets.

The fullest version of the paradox (that is known to me) says that if you have two sets $A, B \subseteq \Bbb R^3$, both of which have non-empty interior, then you can divide $A$ up into some finite number $n$ of disjoint subsets, then move those subsets around isometrically so that they remain disjoint and their union is now $B$.

More formally, if $A, B$ have non-empty interior, there exist sets $A_k, k = 1, ..., n$ such that $A = \bigcup_{k=1}^n A_k$ and $A_j\cap A_k = \emptyset$ when $j \ne k$. And there exist isometries $f_k, k = 1, ..., n$ of $\Bbb R^3$ such that $B = \bigcup_{k=1}^n f_k(A_k)$ and $f_j(A_j) \cap f_k(A_k) = \emptyset$ when $j \ne k$.

Because of the paradox, we know that if $V : \scr P(\Bbb R^3) \to \Bbb R$ is a set function, then one of these 3 conditions must hold:

• there are disjoint sets $A, B \subseteq \Bbb R^3$ such that $V(A\cup B) \ne V(A) + V(B)$, or
• there are isometries $f$ of $\Bbb R^3$ and sets $A \subseteq \Bbb R^3$ such that $V(A) \ne V(f(A))$, or
• $V$ is constant on all sets with interior.

Edit Adding explanation why at least one of the three conditions must hold:

The first two conditions are just "$V$ is not additive" and "$V$ is not preserved under isometries". So if neither of those hold, then $V$ is additive and is preserved by isometries. In this case, let $A$ and $B$ be two sets with interior. Then per the BTP result I gave, we can write $A = \bigcup_{k=1}^n A_k$ and $B = \bigcup_{k=1}^n f_k(A_k)$ for disjoint $A_k$ and $f_k(A_k)$. Hence $$V(A) = V\left(\bigcup_{k=1}^n A_k\right) = \sum_{k=1}^n V(A_k) = \sum_{k=1}^n V(f_k(A_k)) = V\left(\bigcup_{k=1}^n f_k(A_k)\right) = V(B)$$ Thus if $V$ is additive and preserved under isometries, it must have the same value for all sets with interior (and in fact, that value must be $0$, since any set with interior is the disjoint union of two other sets with interior).

The problem is that all three are violations of properties that any concept of "volume" should have:

• Volume is additive: if two sets are disjoint, the volume of their union should be the sum of their volumes.
• Volume is unchanged by isometries. Volume is supposed to depend only on size and shape, not location or orientation. So moving a set around rigidly should not change its volume.
• Volume is obviously not constant on sets with interior. The only way it could be both additive and constant is if it was always $0$. But the volume of the unit cube is $1$.

The conclusion is therefore that it is impossible to define a concept of volume that works for all subsets of $\Bbb R^3$.

• I like this statement of it. Somehow it sparked an idea that no other discussion of BTP has, for me: What if BTP shows how volume actually does work, and we just haven't learnt to adequately manipulate the particles which comprise mass? :) Interesting food for thought (and sci-fi). – Wildcard Jan 20 '17 at 18:38
• But what is special about R3? Why is a set of an infinite points occupying 3 dimensions different from a set of infinite points occupying 2 dimensions, or 1, or n? Couldn't you substitute "line of length X" for "ball of voume X"? – Benubird Jan 20 '17 at 19:10
• Can you explain a little bit more why one of these 3 conditions must hold? I'm not quite seeing it – WorldSEnder Jan 20 '17 at 19:43
• @WorldSEnder - I added an explanation to the post. – Paul Sinclair Jan 20 '17 at 23:56
• @Wildcard - There are no such physical implications. The very concept of exact measure does not exist for physical objects. There is no well-defined exact boundary between what is part of an object and what is not part of the object. The concept breaks down when you look closely enough. When we apply such mathematical concepts as length, area, volume, etc to the physical world, one should always keep in mind that they are idealizations, not exact representations. – Paul Sinclair Jan 21 '17 at 0:41

The Banach-Tarski paradox is the theorem in ZFC that a sphere can be partitioned into finitely many subets, then those subsets can be rearranged into 2 copies of the original sphere using only translation and rotation. Actually according to ZFC, it can be partitioned into only 5 parts.

Some people don't think you can just take on blind faith the axiom of choice. Indeed, some mathematicians work in the formal system of ZF and the axiom of choice is not a theorem of ZF. ZFC is ZF with the assumption of axiom of choice. It turns out that the Banach-Tarski paradox cannot be derived in ZF. According to https://www.emis.de/journals/RSMT/61-4/393.pdf, the axiom of determinacy implies Lebesgue measurability and according to https://brilliant.org/wiki/axiom-of-determinacy/, the axiom of determinacy is consistent with ZF.

• None of the other answers got accepted so I thought that maybe mine would more clearly address the OP's confusion. – Timothy Feb 27 at 0:14
• Axiom of Determinacy. – Andrés E. Caicedo Feb 27 at 4:05