Have I totally misunderstood the Banach–Tarski paradox?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\Bbb R^3$.