The Banach-Tarski Paradox & The Axiom of Choice
Jason M. Lutz
November 1, 2011
Abstract
We discuss the Banach-Tarski paradox, often cited as evidence against the use of
the Axiom of Choice. I’ll argue that this paradox is not so implausible, and examine
the role of the Axiom of Choice in the proof.
xkcd # 804: Pumpkin Carving1
The Banach-Tarski theorem was actually first developed by King Solomon,
but his gruesome attempts to apply it set back set theory for centuries.
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Introduction
1.1
Notation
• F2 = ha, bi is the free group on 2 generators; it consists of reduced words in a,
b, a−1 and b−1 .
• S 2 := {~x ∈ R3 : |~x = 1}, the hollow unit sphere in R3 .
• SO2 := the group of rotations of the sphere. It acts on S 2 and the unit ball.
• For an equivalence relation ∼ on a set X, [x] := {y ∈ X : xỹ}.
1
via xkcd.com, used with permission.
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1.2
The Axiom of Choice
Known Aliases: Zorn’s Lemma, Well-Ordering Principle.
A statement of the axiom of choice: Let S be a collection of non-empty sets. Then
there exists a function
[
f :S→
C
C∈S
such that f (C) ∈ C for each C ∈ S.
We need the axiom of choice to tell us how to pick one sock from an infinite
number of pairs, but the same is not true for shoes, as there is a natural
rule: pick the (left or right) shoe each time. The socks, however, are
(presumably) indistinguishable.
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The Banach-Tarski Paradox
2.1
Imprecise Statement of the Paradox
2
A solid ball in R3 may be ‘‘sliced’’ into a finite number of pieces, with
the pieces rotated and translated, then carefully reassembled to form two
balls each identical to the original.
2.2
Adressing Objections
• Using this process, it would be possible to repeatedly decompose and reassemble a pea
enough times to make an object the size of the sun. The problem: mathematical
spheres are arrangements of uncountably many points, and is infinitely
divisible. A pea is not. At some point (perhaps at the atomic level) separating
the parts cannot be undone.
• This means that there is no concept of volume. True, to a degree. Or, rather,
we must accept that there are sets so wild that their size (volume)
cannot be measured in any way that will be consistent with out concept
of volume.
2
via Wikipedia, used with permission.
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2.3
A Nicer Example
We may ‘‘slice’’ N in the same manner to ‘‘get’’ two copies of N:
N
↓ slice
evens
odds
2n
2n − 1
↓
translate
n
∈N
2.4
↓
n
reassemble
∈N
Formalize the “duplication”
Definition 2.1. Let G be a group acting on a set X, with E ⊂ X. E is called G-paradoxical
if there exist pairwise disjoint subsets A1 , . . . , An , B1 , . . . , Bm of E and g1 , . . . , gn , h1 , . . . , hm
such that
[
[
E = gi · (Ai ) = hj · (Bj ).
In this case, we say E has a (G-)paradoxical decomposition. [Wag93]
Loosely speaking, there are disjoint subsets of E which can be split, and
then rearranged via the action of G to cover G twice.
Example 2.2. Consider F2 = ha, bi, the free group of rank 2, whose elements are
(reduced) words in a, a−1 , b, b−1 . Let F2 act on itself by left multiplication. Then F2 is
F2 -paradoxical.
Let S(a) be the set of words in F2 which start (on the left) with a, similarly defined for
S(a−1 ), S(b), and S(b−1 ). Then
F = {e}∪S(a)∪S(a−1 )∪S(b)∪S(b−1 ).
Now, if w ∈ F2 , then either w begins with a, or it does not. If the former, then w ∈ S(a). If
the latter, then a−1 w is reduced, and so w = aa−1 w ∈ a · S(a−1 ).
So, we may also write
F = S(a)∪a · S(a−1 ) = S(b)∪b · S(b−1 ),
so F2 is F2 -paradoxical.
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2.5
More Precise statement of the paradox
Theorem 2.3. Banach-Tarski Paradox: S 2 is SO2 -paradoxical (the hollow sphere is
paradoxical with respect to the group of rotations) Moreover, the unit ball is G3 paradoxical (paradoxical with respect to the group of isometries (reflections,
rotations, and translations) of R3 ). [Wag93]
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A sketch of the proof
(Adapted from Banach-Tarski paradox - Wikipedia)
Step 1: Exhibit a subgroup H of the group of rotations of S 2 which is paradoxical.
We want two generators - rotations of infinite order that interact
trivially; rotations of irrational multiplies of π radians are ideal.
Put H := hA, Bi, where A, B are rotations of arccos(1/3) about the x-axis and z-axis,
respectively. Then H ' F2 , and so H is H-paradoxical as in Example 2.2.
Step 2: Build subsets of S 2 that are compatible with the action of this H, showing that S 2
is H-paradoxical.
S 2 is partitioned into orbits [x] by the action of H. Put S := {[x] : x ∈ S 2 }.
[
By The axiom of choice, there exists a choice function f : S →
[x] such that
[x]∈S
f ([x]) = y ∈ [x]. Put M := f (S) ( S 2 .
Now, almost every point of S 2 can be reached in one way by applying the
proper rotation to the proper element of M . Some of you may recognize this
technique from the construction of a Lebesgue non-measurable set by Vitali.
Partition S 2 into points that are reached by elements in S(A), S(A−1 ), S(B) or S(B −1 ) in
this manner. Then, S 2 is H-paradoxical.
Step 3: Extend to the unit ball (w/o its center) using rays from the origin.
Now that we know how to build a paradoxical decomposition of S 2 , extend
this to the unit ball (w/o the center) by rays from the origin; rays are
preserved by rotations, so all points on the rays move together under the
action of H
Step 4: Account for the rest of the points. We’ve failed to address
1. points that are fixed by some element of H, and
2. the center of the sphere.
Each case is aided by countability.
We deal with them:
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1. Note that there are countably many fixed points, so there is a diameter of the sphere
which touches none of these fixed points (if not, then each point of the sphere
would be on a line with one of the countably many points on it, and so
the sphere would be a countable collection of points, which is absurd,
as the equator of the circle ‘‘acts like’’ [0, 2π)), and there is a rotation
about this diameter which fixes none of the points in question (each point is fixed
by exactly one rotation in [0, 2π), and so such a rotation that fixes no
point must exist, else [0, 2π) is countable, which is absurd).
2. Similarly, but we need to use translations, not just rotations, so we need the group of
isometries of R3 , not just the rotations of the sphere.
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Quotes about the Axiom of Choice
1. “The Axiom of Choice is obviously true, the Well-Ordering Principle is obviously false;
and who can tell about Zorn’s Lemma.” Jerry Bona, University of Illinois at Chicago.
References
[Wag93] S. Wagon. The Banach-Tarski paradox. Encyclopedia of Mathematics and its
Applications. Cambridge University Press, 1993.
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