Banach-Tarski Paradox

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The Banach-Tarski paradox is that statement that a ball can be cut into 5 pieces, which can then be reassembled into two balls of equal volume. It relies on the Axiom of Choice. It is called a paradox because it violates our physical intuition that this is impossible, though this paradox has been proven under Zermelo-Fraenkel set theory. Note that, since the pieces must be non-measurable, this cannot be done with an actual object.

The Mathematics Behind It

This section contains material appropriate for a college undergraduate.

Theorem

All sets discussed here are point sets in $\mathbb {R} ^{3}\$.

A rigid motion of a set is a translation, rotation, or a composition of those two kinds of motion. A rigid motion of a set $A\$ is often written $g(A)\$.

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A, B \ } be sets. We say there is a rigid body equivalence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A \cong B \ } if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A \ } can be partitioned into finitely many sets Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1, ..., A_n \ } and there are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \ } rigid motions $g_{1},...,g_{n}\$ such that $g_{1}(A_{1})\cup ...\cup g_{n}(A_{n})=B\$.

We say there exists a paradoxical decomposition of a set $A\$ if $A=B\cup C\$, and $A\cong B\$ and $A\cong C\$.

The Banach-Tarski theorem (aka the Banach-Tarski paradox) states that the solid ball admits a paradoxical decomposition.

Removing Points with Rigid Body Motions

First, lets examine why a circle is rigid-body equivalent to a circle with a point removed. Pick an irrational number $\alpha \$, and let $R_{\alpha }\$ be the rigid motion which is rotation by $\alpha \$ degrees counterclockwise. Now we pick a point in the circle $P\$ and we form an infinite "necklace" $\left\{{P,R_{\alpha }(P),R_{2\alpha }(P),...}\right\}\$. It is clear that this necklace must be infinite, because if there was any $n\$ such that $R_{n\alpha }(P)=P\$, then $\alpha \$ would not be irrational.

Since our necklace is a countable set and the circle is uncountable, there are points in the circle which are not in the necklace. So we partition the circle into two sets, the necklace, and everything left over. Observe that by performing the rigid motion $R_{\alpha }\$ on the necklace, we simply get $\left\{{R_{\alpha }(P),R_{2\alpha }(P),...}\right\}\$, that is, all the original points remain exactly where they were, except now the first point is gone. Re-combining the rotated necklace with those points not in the necklace gives us a circle with a point removed.

With this procedure, it becomes a simple matter to remove a point from the sphere or the interior of a ball - just pick a circle lying in the sphere or ball which contains the point you wish to remove, and partition the sphere or ball into three pieces - the necklace, the rest of the circle the necklace is on, and the everything in the sphere or ball which isn't in that circle.

Removing countably many points from a circle is a trickier matter. Let $A=\left\{{a_{1},a_{2},...}\right\}\$ be countable set of points in the circle, where it is understood that the $\alpha \$ measure the degree of the angle the point can be found at. We construct a set $W=\left\{{{\frac {a_{m}-a_{k}}{n}}:(m,k,n)\in \mathbb {N} ^{3}}\right\}\$, which is also a countable set, since it is in one-to-one correspondence with the three dimensional lattice.

Make sure you understand the content of $W\$. If the first few values of $A=\left\{{6,2{\sqrt {3}},{\frac {e}{\pi }},...}\right\}\$, then $W\$ contains, among other things, $6-2{\sqrt {3}},(2{\sqrt {3}}-{\frac {e}{\pi }})/653,\$ etc. Now, since $W\$ is countable, and the angles of $\mathbb {S} ^{1}\$ are uncountable, there is an angle in $\mathbb {S} ^{1}\$ which is not in $W\$. Pick one such angle, and call it $\theta \$. This is our magic rotation angle.

Note that if we rotate $A\$ by an angle not in $W\$, it is guaranteed to not carry any point in $A\$ to any other point in $A\$ because of our clever definition. Rotation by $\theta \$, for example, cannot carry $a_{7}\$ to $a_{1}1\$, because then $\theta =a_{11}-a_{7}\in W\$, but we specifically said $\theta \notin W\$. Furthermore, 12 rotations of $\theta \$ could never carry $a_{7}\$ to $a_{11}\$, because then $\theta ={\frac {a_{11}-a_{7}}{12}}\$, but that's in $W\$ too, since we constructed it to contain not just all the differences of angles in $A\$, but all the fractions of those differences as well.

So we construct the infinite super-necklace $T=A\cup R_{\theta }(A)\cup R_{2\theta }(A)...\$ and finally, observe that $R_{\theta }(T)=R_{\theta }(A)\cup R_{2\theta }(A)\cup ...=T-A\$.

We form the partition $\mathbb {S} ^{1}=T\cup (\mathbb {S} ^{1}-T)\$ and then note $R_{\theta }(T)\cup (\mathbb {S} ^{1}-T)=\mathbb {S} ^{1}-A\$ and we have shown that $\mathbb {S} ^{1}\cong \mathbb {S} ^{1}-A\$

The Free Group

Before attempting to understand the free group, it would be very beneficial to the reader to have and understanding of a mathematical group.

Let $a,a',b,b'\$ be "letters." We will form "words" by stringing finitely many of these letters together, subject to just a few simplification rules: $aa'=a'a=bb'=b'b=e\$, where $e\$ is the identity, shorthand for the "word" of zero letters.

Interpretation

Some view the Banach-Tarski paradox as an absurd result, and evidence that the Axiom of Choice is false. This is the reaction of many non-mathematicians upon hearing this result. However, as one grows in mathematical understanding one finds it is impossible to avoid - mathematicians have learned to accept the existence of non-measurable sets, comfortable in the knowledge that they can never exist in the real world.

It can also be viewed as a mathematical affirmation of the biblical story of the loaves and the fishes (as Jesus would not have had to violate the conservation of mass, but merely partition loaves and fishes into non-measurable subsets).

External Links

• Banach-Tarski Paradox at MathWorld