PARADOXICAL DECOMPOSITIONS OF GROUPS AND THE SETS THEY
ACT UPON
A Thesis
Presented to the
Department of Mathematics
and the
Faculty of the Graduate College
In Partial Fulfillment
of the Requirements for the Degree of
Master’s of Arts in Mathematics
University of Nebraska at Omaha
by
Joseph Lee
May 2010
Supervisory Committee:
Dr. Andrzej Roslanowski
Dr. Griff Elder
Dr. Valentin Matache
Dr. Wai-Ning Mei
PARADOXICAL DECOMPOSITIONS OF GROUPS AND THE SETS THEY
ACT UPON
Joseph Lee, MA
University of Nebraska, 2010
Advisor: Dr. Andrzej Roslanowski
The Banach-Tarski paradox states that a solid ball in R3 can be partitioned in such a
way that by simply rotating and translating those pieces, you end up with two exact
copies of the original ball you started with. We give a detailed decomposition of nine
pieces and nine different isometries that can be used to produce two copies of the
unit ball. We also show this paradox is specifically due to the fact that the group of
isometries in R3 has a free subgroup with two generators. In the second half of this
paper, we introduce the property that prevents this paradox from occuring and show
the group of isometries in R and R2 possess this property.
i
Common Symbols
A,B,X, Y, Z sets
A an algebra of sets
A(X) the algebra induced by a set X
fr(X) the free group generated by a set X
G,H groups
Gn the group of isometries of Rn
g, h elements of a group G
H ≤ G H is a subgroup of G
H ∼= G H is isomorphic to G
µ a measure
N the set of natural numbers, {1, 2, 3, . . .}
P (X) the power set of X
R a relation
R∗ the transitive closure of a relation R
Rn the n-dimensional Euclidean space
R a ring of sets
Tg a translation of a G-set by an element g ∈ G
Tn the group of translations of Rn
SGn the group of orientation-preserving isometries of Rn
SOn the group of rotations about the origin of Rn
[x]R the equivalence class of x with respect to a relation R
X a space
X a class of subsets of X
4 the symmetric difference
ii
Contents
1. Introduction 1
1.1. On the Banach-Tarski Paradox 1
1.2. Timeline 5
1.3. Mathematicians 5
1.4. Contents of the Thesis 8
2. Prelude: Relations, Free Groups, and Group Actions 10
2.1. Relations 10
2.2. Groups 13
2.3. Free Groups 14
2.4. Group Actions 18
3. Paradoxical Groups 22
4. A Free Group of Rotations 28
5. Hausdorff and Banach-Tarski Paradoxes 35
5.1. The Hausdorff Paradox 35
5.2. A Basis for the Paradoxical Decomposition of the Sphere 36
5.3. A Decomposition of the Sphere 38
5.4. The Banach-Tarski Paradox 40
6. Interlude: Measure and Integration 46
7. Amenable Groups 57
8. Examples of Amenable Groups 74
8.1. Isometries in R and R2 74
8.2. No Paradoxes in R or R2 80
9. Conclusion 83
References 84
1
1. Introduction
1.1. On the Banach-Tarski Paradox. In 1638, Galileo Galilei made an interest-
ing observation: there is a one-to-one correspondence between the natural numbers,
{1, 2, 3, 4, 5, . . .}, and the squares of natural numbers, {1, 4, 9, 16, 25, . . .}, i.e., for any
natural number n, there is another natural number n2, and for any natural number
n2, there is a number n that is a natural number [3]. While this perhaps seems ob-
vious, note what this implies. The set of all natural numbers, {1, 2, 3, 4, 5, . . .} has
the same number of elements as the set of all squares, {1, 4, 9, 16, 25, . . .}. Observe
any number that is a square is a natural number, but not every natural number is a
square. Moreover, there is also a one-to-one correspondence between the set of nat-
ural numbers, {1, 2, 3, 4, 5, . . .}, and the set of natural numbers that are not squares,
{2, 3, 5, 6, 7, . . .}, by the following assignments:
1→ 2 4→ 6 7→ 10
2→ 3 5→ 7 8→ 11
3→ 5 6→ 8 etc.
Thus, the set of natural numbers can be split apart into two completely separate sets
that in effect have the same size as the original set.
We may observe that even though we split the natural numbers into two separate
sets, because both of those sets also had an infinite number of elements, we were
able to do the following. The natural numbers were split into two sets, squares and
non-squares. Then each of those sets were “shrunk” back to the original set:
Squares Non-squares
{1, 4, 9, 16, 25, . . .} {2, 3, 5, 6, 7, . . .}
1→ 1 2→ 1
4→ 2 3→ 2
9→ 3 5→ 3
16→ 4 6→ 4
25→ 5 7→ 5
2
In fact, any way you divide up the natural numbers, assuming each division has an
infinite amount of numbers, each division can be “shrunk” back to the original set
of natural numbers. Note we say “shrunk” because some numbers get shifted more
than others. Shifting is an entirely different process. When we “shrink,” it changes
the distances between our numbers. If, however, we were to simply shift these sets of
numbers, the distance between each number would be preserved.
Since ancient times, the ability to shift something has been important in mathe-
matics. Consider the following figures.
Here, we can say figure A and figure B have the same area because we can cut a piece
off figure A and move it over to the side to form figure B, as depicted. This is an
intuitive way to determine the size of many different types of figures. As we move
the pieces by translation and rotation, this should not effect the area.
In 1924, Stefan Banach and Alfred Tarski did something similar to Galileo but
without shrinking. They were able to cut a solid ball into pieces, move those pieces
around, as mentioned above, but they ended up with two whole copies of the original
sphere. Hence, after cutting the pieces and moving them around, the volume doubled.
The theorem they published was appropriately named the Banach-Tarski paradox.
Intuitively, it should not be possible to take something, cut it apart, move it around
without shrinking or stretching, and end up with twice what you started with.
When examining this result, there is an obvious question to consider: is there
some mistake in the proof, or in other words, is there an assumption made that is
actually not correct? For years after its publication, many mathematicians believed
3
that this was in fact the case. To understand this possible flaw, we must introduce
the development of axiomatic set theory. In mathematics, it is not possible to prove
everything—meaning, there are things that are so basic one is forced to simply believe
that they are true. Through the work of two mathematicians, Ernst Zermelo and
Abraham Fraenkel, these facts which must be accepted were reduced to nine axioms.
All nine were easily accepted by mathematicians, that is until the publication of the
Banach-Tarski paradox. Banach and Tarski relied heavily on the use of the ninth
axiom, the axiom of choice. The axiom states: if we are given any number, possibly
infinite, of collections of objects, it is possible to make a selection from each collection.
Simply put, it is always possible to make a choice. This is often stated in the sock
and shoe problem.
If given an infinite number of pairs of shoes, it is always possible to name a selection
of one shoe from each pair, i.e., every right show. However, if you have an infinite
number of pairs of socks, there is no way to name a selection from each pair. The
axiom of choice, in this case, tells us even though it is impossible to name a selection
of one sock from each pair, it is nonetheless possible to choose one sock from each
pair. By accepting this axiom, while we cannot give an explicit definition of the set
of our choices, we assume that such a set exists.
The Banach-Tarski paradox relies heavily on the axiom of choice, so for years,
mathematicians used this proof as the reason this axiom should not be accepted. After
some time passed however, the axiom of choice once again became widely accepted by
mathematicians as a whole. The axiom is necessary for many areas of mathematics
today, and ultimately, it would be even more counter-intuitive to reject the axiom. It
would be hard to accept that, even if you cannot name a choice, it is impossible to
make one.
4
Accepting the axiom, however, means that we are still left with this paradoxical
theorem. Ultimately, we resolve this paradox by returning to our discussion of deter-
mining the area, or in this case volume, of figures. Recall these two figures had the
same area.
We argued this by moving C to D. In fact, we can say the area of this figure is the area
of A plus the area of C (or D). This, however, does not work with the Banach-Tarski
paradox. In this paper, we divide the sphere into nine separate pieces, but unlike
in the picture above, if we move those nine pieces around, we do not get the same
volume. The reason for this paradox, and the ultimate unravelling of our mystery, is
that the nine pieces we divided the ball into are actually not measurable.
This explanation, of course, is also somewhat counter-intuitive, as it suggests that
there are things that we simply cannot measure. In this paper, we will formalize
what is necessary for something to be measurable, but basically it requires two of the
concepts mentioned already:
• when you combine two separate figures that are measurable, the union of the
two is also measurable,
• when a figure that is measurable is moved, it has the same measure as it had
before it was moved.
To be correct mathematically, the latter is for a stricter form of measurement called
an invariant measurement, but this stricter form will be necessary for much of our
work.
Our conclusion is that paradoxical sets, sets that can be divided up and moved
around to form two copies of the original set, are directly linked to the fact that their
paradoxical decompositions are not measurable.
5
1.2. Timeline. The following are some of the contributions relevant to the field of
paradoxical groups.
• In 1638, Galileo Galilei showed that the natural numbers could be partitioned
into two pieces which were both equinumerous with all natural numbers [3].
• In 1905, Giuseppe Vitali showed that there existed a non-measurable subset
of the real number line [16] (compare with Proposition 7.6).
• In 1914, Hausdorff showed that the sphere, minus a countable number of
points, was paradoxical [5].
• Also in 1914, Wac law Sierpinski and Stefan Mazurkiewicz gave an example of
a paradoxical subset of the plane [9].
• In 1924, Stefan Banach and Alfred Tarski showed a ball in R3 was paradoxical
[1].
• In 1991, Janusz Pawlikowski showed that the Hahn-Banach theorem is enough
to construct a paradoxical decomposition of the ball [14].
• In 1994, Randall Dougherty and Matthew Foreman showed the Banach-Tarski
paradox could be done with pieces that possess the Property of Baire [2].
1.3. Mathematicians. In the study of paradoxical groups, there are two results that
are quintessential to this topic: the Hausdorff paradox and the Banach-Tarski para-
dox. Here we introduce three mathematicians who were crucial to the development of
paradoxical groups. The following bibliographies are paraphrased from the respected
University of St. Andrews mathematics history website “The MacTutor History of
Mathematics archive” [11].
Felix Hausdorff was born on November 8, 1868, in Breslau, Germany (present day
Wroc law, Poland). He was born into a rich Jewish family, so Hausdorff did not have
to work to support himself and was encouraged to pursue academics ranging from
literature and music to mathematics. In 1891, Hausdorff graduated with his doctorate
thesis on a mathematical topic in astronomy. He continued to publish papers in this
6
field, but he began to focus more on literary works. Hausdorff went on to publish
multiple literary works until, when in 1904, he began his preeminent work in set
theory. In 1914, Hausdorff published Grundzuge der Mengenlehre which contained
his paradoxical decomposition of almost the entire sphere. Hausdorff is famous for
numerous other contributions in mathematics, including the generalized continuum
hypothesis, Hausdorff spaces, Hausdorff measure, and Hausdorff dimension. In 1942,
Hausdorff took his own life to avoid being sent to a concentration camp by the Nazis.
Stefan Banach was born on March 30, 1892, in Krakow, Poland (then Austria-
Hungary). Banach was raised by his father, as his mother left him right after his birth.
During his secondary education, Banach mathematical prowess became evident and
math was seemingly his only interest. He became friends with a classmate, Witold
Wilkosz, who also went on to become a professor of mathematics. Banach went on
to an engineering university, where he supported himself by tutoring young students.
In 1916, by a chance meeting with Hugo D. Steinhaus, Banach began to work in
mathematical research. He published his first paper in 1918 and shortly thereafter
continued to publish many papers. Banach became part of and helped found the
Mathematic Society of Krakow, which would later become the Polish Mathematical
Society. In 1920, Banach was offered an assistantship at the Jan Kazimierz University
of Lwow. At this time, Banach submitted a doctorate thesis and was awarded his
degree with no other university course work. This thesis has since been called the birth
of functional analysis. In 1924, Banach was promoted to full professor, and this same
year, he published the joint paper with Tarski, Sur la decomposition des ensembles
de points en partiens respectivement congruent [1], on the paradoxical decomposition
of the sphere. Banach would continue produce many works until the start of World
War II. However, due to the Nazi occupation in 1941, Banach was forced to work in
a German lab feeding lice. He died of lung cancer in 1945.
Alfred Tarski was born on January 14, 1902, in Warsaw, Poland (then Rus-
sian Empire). He was born into a Jewish family and his name at birth was Alfred
7
Figure 1. Banach, Tarski. Sur la decomposition des ensem-bles de points en partiens respectivement congruent. Funda-menta Mathematicae 6: 244-277, 1924. Available online athttp://matwbn.icm.edu.pl/ksiazki/fm/fm6/fm6127.pdf.
8
Teitelbaum. In 1915, Russia withdrew from Warsaw and a vibrant math commu-
nity was formed at the re–founded Warsaw University. In 1918, Teitelbaum began
studying at the Warsaw University, where he would come to study under Stanislaw
Lesniewski, Jan Lukasiewicz, Wac law Sierpinski, and Stefan Mazurkiewicz. At age
19, Teitelbaum published his first paper on set theory. In 1923, Teitelbaum converted
to Roman Catholicism and changed his last name to Tarski. These changes reflected
both a strong Polish identity and a desire to receive a university appointment, which
would have been difficult due to anti-Semitic sentiments. In 1924, Tarski published,
along with Stefan Banach, Sur la decomposition des ensembles de points en partiens
respectivement congruent. Still, Tarski had difficulties getting appointments due to
his Jewish heritage. In 1939, Tarski traveled to the United States. When Germany
invaded Poland two weeks later, Tarski was allowed to stay in the United States and
was eventually able to get his family to join him. In 1943, Tarski received an appoint-
ment at University of California, Berkeley and would continue to teach until 1973.
He died on October 26, 1983. Tarski has been called the founder of the American
school of mathematical logic and philosophy of mathematics.
1.4. Contents of the Thesis. Section 2 provides some of the necessary background
needed to develop the theory of paradoxical groups. Subsection 2.1 covers the theory
of relations which is used in Subsection 2.3 to define free groups. Subsection 2.2
introduces groups for Subsection 2.3 and Section 7. Subsection 2.4 introduces group
actions which we use to move around the pieces to form our paradoxical decomposi-
tions. Section 3 introduces the main topic of the thesis, paradoxical groups. Section
4 shows a group of rotations in R3 is isomorphic to the paradoxical group introduced
in Section 3. Section 5 ties everything together to form a paradoxical decomposition
of almost the entire sphere (the Hausdorff paradox) and the ball (the Banach-Tarski
paradox).
The second half of this thesis focuses on showing that this paradox is not possible in
R and R2. Section 6 defines measure and integration. Section 7 introduces amenable
9
groups—groups that are measurable in a strict manner that do not allow for such
paradoxes that were shown in Section 5. Section 8 concludes the second half of the
paper by showing that no bounded subset of R and R2 with non-empty interior is
paradoxical.
10
2. Prelude: Relations, Free Groups, and Group Actions
2.1. Relations. We will review the basic concepts on the theory of relations which
we will use to establish equivalence classes.
Definition 2.1. (1) A relation R on a set X is a set of ordered pairs of elements
of X. By convention, if an ordered pair (a, b) ∈ R, then we write aRb.
(2) A relation R is said to be
• reflexive if aRa for every a ∈ X.
• symmetric if aRb implies bRa for any a, b ∈ X.
• transitive if aRb and bRc implies aRc for any a, b, c ∈ X.
• an equivalence relation if R is reflexive, symmetric and transitive.
Definition 2.2. The transitive closure R∗ of a relation R is the relation introduced
as follows: aR∗b if and only if there exists a finite sequence a1, a2, . . ., an such that
aRa1, a1Ra2, . . ., anRb.
Proposition 2.3. The transitive closure R∗ of a relation R on a set X is transitive.
Proof. Suppose aR∗b and bR∗c for some a, b, c ∈ X. Then there exist finite sequences
a1, a2, . . . , an and b1, b2, . . . , bm such that
aRa1, a1Ra2, . . . , anRb and bRb1, b1Rb2, . . . , bmRc.
When we combine these two finite sequences, we get another finite sequence a1, a2,
. . ., an, b, b1, b2, . . ., bm such that
aRa1, a1Ra2, . . . , anRb, bRb1, b1Rb2, . . . , bmRc,
showing that aR∗c. Thus, R∗ is transitive. �
Proposition 2.4. The transitive closure R∗ of a reflexive and symmetric relation R
on a set X is an equivalence relation.
11
Proof. We have already shown that R∗ is transitive, so we must now show that R∗ is
reflexive and symmetric. Let R be a relation that is reflexive and symmetric, and let
R∗ be its transitive closure. Since for every a ∈ X, aRa, the finite sequence of just
a witnesses the fact that aR∗a. Thus, R∗ is reflexive. Now suppose aR∗b for some
a, b ∈ X. Then there exists a finite sequence such that
aRa1, a1Ra2, . . . , anRb.
But R is symmetric, so we also have
a1Ra, a2Ra1, . . . , bRan,
or if we rearrange our sequence,
bRan, anRan−1, . . . , a1Ra.
Thus, R∗ is symmetric, so R∗ is an equivalence relation. �
Definition 2.5. An operation (or a binary operation) ◦ on a set X is a mapping
from X ×X into X.
Definition 2.6. Let ◦ be an operation on a set X, and let R be a relation on X.
• We say R is pre-normal with respect to the operation ◦ if the following condi-
tion is satisfied:
if aRa′ and b ∈ X, then b ◦ aRb ◦ a′ and a ◦ bRa′ ◦ b.
• We say R is normal with respect to the operation ◦ if the following condition
is satisfied:
if aRa′ and bRb′, then b ◦ aRb′ ◦ a′ and a ◦ bRa′ ◦ b′.
Theorem 2.7. Let ◦ be an operation on a set X, and let R be a relation on X. Then
(1) if R is transitive and pre-normal, then R is normal,
(2) if R is reflexive and normal, then R is pre-normal,
12
(3) if a R is pre-normal, then its transitive closure R∗ is normal.
Proof. (1) Let R be transitive and pre-normal. Suppose aRa′ and bRb′ for some
a, a′, b, b′ ∈ X. As R is pre-normal, we have
b ◦ aRb ◦ a′ and b ◦ a′Rb′ ◦ a′,
and as R is transitive, we have b ◦ aRb′ ◦ a′. Similarly, as R is pre-normal, we
have
a ◦ bRa′ ◦ b and a′ ◦ bRa′ ◦ b′,
and as R is transitive, we have a ◦ bRa′ ◦ b′. Thus R is normal.
(2) Let R be a reflexive and normal. As R is reflexive, if b ∈ X then bRb. Now
suppose aRa′. Then
a ◦ bRa′ ◦ b and b ◦ aRb ◦ a′,
so R is pre-normal.
(3) As R∗ is transitive by Proposition 2.3, we simply must show R∗ is pre-normal.
Suppose aR∗a′ and b ∈ X. Then there exists a finite sequence a1, a2, . . . , an
such that
aRa1, a1Ra2, . . . , anRa′.
Since R is pre-normal, we have
b ◦ aRb ◦ a1, b ◦ a1Rb ◦ a2, . . . , b ◦ anRb ◦ a′
and
a ◦ bRa1 ◦ b, a1 ◦ bRa2 ◦ b, . . . , an ◦ bRa′ ◦ b.
So
b ◦ aR∗b ◦ a′ and a ◦ bR∗a′ ◦ b.
Thus, R∗ is pre-normal, and by (1), R∗ is normal.
13
�
Definition 2.8. Let R be a reflexive and symmetric relation on a set X, and let R∗
be its transitive closure. We define [x]R∗ = {a : aR∗x} and call it the equivalence
class of x. If the equivalence relation R∗ is clear, we may simply write [x].
2.2. Groups. In this subsection, we introduce groups for the following subsection,
as well as Section 7.
Definition 2.9. Let ◦ be an operation on a set G and let g, h, k ∈ G. Then
• ◦ is associative if and only if g ◦ (h ◦ k) = (g ◦ h) ◦ k,
• ◦ has an identity if and only if there exists e ∈ G such that g ◦ e = e ◦ g = g
for all g ∈ G,
• ◦ has inverses if and only if for any g ∈ G, there exists g−1 ∈ G such that
g ◦ g−1 = g−1 ◦ g = e.
If ◦ is associative, has an identity, and has inverses, then the set G with the operation
◦ is called a group. If the operation is clear, then we simply say that G is a group.
Definition 2.10. Let G be a group. If g ◦h = h◦ g for any g, h ∈ G, then G is called
Abelian.
Definition 2.11. Let G be a group and H ⊆ G. If H is also a group with the same
operation as G, then H is called a subgroup of G.
Definition 2.12. Let G be a group, g ∈ G, and H be a subgroup of G.
(1) We define the set gH = {g ◦h : h ∈ H} and call it a left coset of G. Similarly,
we define Hg = {h ◦ g : h ∈ H} and call it a right coset of G.
(2) If gH = Hg for every g ∈ G, H is called a normal subgroup of G.
Proposition 2.13. Let H be a normal subgroup of a group G. Define the set G/H =
{gH : g ∈ G}, and define the operation � : G/H ×G/H −→ G/H by g1H � g2H =
(g1 ◦ g2)H. Then G/H with the operation � is a group and is called a factor group of
G.
14
Proof. First we will note the operation � is well-defined, i.e., for g1H = g2H and
g3H = g4H that g1H � g3H = g2H � g4H. So let g1H = g2H and g3H = g4H.
g1H � g3H = (g1 ◦ g3)H = g1(g3H) = g1(g4H) = (g1 ◦ g4)H
= H(g1 ◦ g4) = (Hg1)g4 = (Hg2)g4 = H(g2 ◦ g4)
= (g2 ◦ g4)H = g2H � g4H.
To show � is associative, observe
g1H � (g2H � g3H) = g1H � (g2 ◦ g3)H
= (g1 ◦ (g2 ◦ g3))H
= ((g1 ◦ g2) ◦ g3)H
= (g1 ◦ g2)H � g3H
= (g1H � g2H)� g3H.
Note H is the identity of G/H as gH �H = H � gH = gH for any gH ∈ G/H. Also
observe (gH)−1 = g−1H as gH � g−1H = g−1H � gH = eH = H. Thus G/H is a
group. �
2.3. Free Groups. In this subsection, we define a free group generated by a set
X. Free groups are key to our paradoxical decomposition of the sphere. Specifically,
in subsequent sections, we will focus on the free group with two generators, i.e.,
X = {a, b}.
Definition 2.14. Let X be a non-empty set where elements of X are characters
in some alphabet. For every x ∈ X, we consider a formal object x−1 /∈ X and let
X−1 = {x−1 : x ∈ X}.
• A word from X is a finite string x1x2x3 . . . xn where xi ∈ X ∪ X−1 for all
i ≤ n. Let W (X) be the set of all possible words of X including the empty
word, e.
• We define lh : W (X) −→ Z by lh(x1x2x3 . . . xn) = n and we call n the length
of x1x2x3 . . . xn.
15
• We define the binary operation · on W (X) to be concatenation, i.e.,
if x1x2x3 . . . xn ∈ W (X) and y1y2y3 . . . ym ∈ W (X) then
(x1x2x3 . . . xn) · (y1y2y3 . . . yn) = x1x2x3 . . . xny1y2y3 . . . ym.
This operation is well-defined, associative, and has an identity (the empty word).
However, we do not have inverses. Note that a−1 is not the inverse of a as aa−1 is not
the empty word. To correct this problem, we will identify some words, in particular,
aa−1 = a−1a = e.
Let u, v ∈ W (X). We define the relation R on W (X) as follows. Let u, v ∈ W (X).
We say uRv if and only if u = v or v can be obtained from u by adding or removing
a word of the form xx−1 or x−1x, where x ∈ X.
Let R∗ be the transitive closure of R (see Definition 2.2).
Lemma 2.15. The transitive closure R∗ forms an equivalence relation on W (X).
Proof. From Proposition 2.4, we only need to show R is reflexive and symmetric. By
definition, R is reflexive, i.e., uRu for all u ∈ W (X). Now suppose uRv, i.e. v can
be obtained from u by adding or removing a word of the form xx−1 or x−1x, where
x ∈ X. Then u can be formed from v by doing the opposite addition or removal of
that same word xx−1 or x−1x. Thus uRv, and R is symmetric. �
We will denote the set of equivalence classes of W (X) with respect to R∗ by fr(X).
We will denote the R∗ equivalence class of x ∈ X by [x].
Definition 2.16. Let u ∈ W (X). We say u is a reduced word if and only if for all
v ∈ W (X) such that [u] = [v], lh(u) ≤ lh(v).
Let [u], [v] ∈ fr(X). Define the binary operation ◦ on fr(X) by [u] ◦ [v] = [u · v].
Lemma 2.17. The relation R∗ is normal with respect to the operation ·. Conse-
quently, the operation ◦ defined above is well-defined.
16
Proof. First, we will show that R∗ is pre-normal with respect to ·. Let u, v, w ∈ W (X)
and suppose uR∗v. Then there exists a finite sequence uRu1, u1Ru2, . . ., unRv. Since
uRu1, u1 can be formed from u by adding or removing a word of the form xx−1 or
x−1x, where x ∈ X. Then u1 · w can be formed by doing the corresponding addition
or removal from u · w. Thus u · wRu1 · w, and by an analogous argument, we have
u · wRu1 · w, u1 · wRu2 · w, . . . , un · wRv · w.
Thus, u ·wR∗v ·w. By the same argument, we may show that w · uR∗w · v. Thus, R∗
is pre-normal with respect to the operation ·. As R∗ is an equivalence relation, R∗ is
transitive, and by Proposition 2.7, R∗ is normal. �
Theorem 2.18. The set fr(X) with the operation ◦ is a group and is called the free
group generated by X.
Proof. First, note that [e], the equivalence class of the empty word, is the identity
element, as for any [u] ∈ fr(X),
[u] ◦ [e] = [u · e] = [u] = [e · u] = [e] ◦ [u].
To show the operation ◦ is associative, recall the operation · was associative, so for
any [u], [v], [w] ∈ fr(X),
[u] ◦ ([v] ◦ [w]) = [u · (v · w)] = [(u · v) · w] = ([u] ◦ [v]) ◦ [w].
Finally, to show any element of fr(X) has an inverse with respect to ◦, let [u] ∈ fr(X).
Then u = (x1)t1(x2)t2 . . . (xn)tn , where xi ∈ X, and
(xi)ti =
x−1i if t = −1
xi if t = 1
17
Let (u)−1 = (xn)−tn(xn−1)−tn−1 . . . (x1)−t1 . Then
[u] ◦ [(u)−1] = [u · (u)−1]
= [(x1)t1(x2)t2 . . . (xn)tn · (xn)−tn(xn−1)−tn−1 . . . (x1)−t1 ]
= [(x1)t1(x2)t2 . . . (xn−1)tn−1 · (xn−1)−tn−1(xn−2)−tn−2 . . . (x1)−t1 ]
= . . . = [e]
= [(x1)−t1 · (x1)t1 ]
= [(x1)−t1(x2)−t2 · (x2)t2(x1)t1 ]
= . . . = [(x1)−t1(x2)−t2 . . . (xn)−tn · (xn)tn(xn−1)tn−1 . . . (x1)t1 ]
= [(u)−1 · u] = [(u)−1] ◦ [u].
Thus, [u]−1 = [(u)−1] ∈ fr(X).
Hence, fr(X) is a group. �
The following theorem is the main property of free groups, which can be used to
define free groups in an abstract way.
Theorem 2.19. If G is a group, X is a set, and f : X −→ G is a function, then
there exists a unique homomorphism f : fr(X) −→ G such that f([x]) = f(x) for all
x ∈ X.
Proof. Let G be a group, X be a set, and f : X −→ G be a function. Now define
f : fr(X) −→ G by
f([x1x2 . . . xn]) = f(x1)f(x2) . . . f(xn) (∗)
where f(x−1) = (f(x))−1 and (f(x))−1 is the inverse of f(x) in G. First, we must
show f is well-defined. To show f is well-defined, recall [x1x2 . . . xn] = [y1y2 . . . ym]
if and only if y1y2 . . . yn can be formed from x1x2 . . . xn by the addition and removal
of words of the form zz−1 or z−1z. However, when we apply f to [x1x2 . . . xn] or
[y1y2 . . . ym], any words of the form zz−1 or z−1z have no effect on the output, as
f(z)f(z−1) = f(z)f(z)−1 = eG = f(z)−1f(z) = f(z−1)f(z).
18
Therefore, if [x1x2 . . . xn] = [y1y2 . . . ym], then f([x1x2 . . . xn]) = f([y1y2 . . . ym]). Thus,
f is well-defined.
To show f is a homomorphism, observe
f([x1x2 . . . xn] ◦ [y1y2 . . . ym]) = f([x1x2 . . . xn · y1y2 . . . ym])
= f(x1)f(x2) . . . f(xn)f(y1)f(y2) . . . f(ym)
= f([x1x2 . . . xn])f([y1y2 . . . ym]).
Thus, f is a homomorphism.
By its definition, f([x]) = f(x). Moreover, since [x1x2 . . . xn] = [x1]◦ [x2]◦ . . .◦ [xn],
it follows that any homomorphism with this property is defined by (∗). �
2.4. Group Actions. In subsequent sections, we will view rotations as group actions
acting on a sphere. To that end, we develop the necessary theory of group actions.
Definition 2.20. Let G be a group and X a set. We say G operates on X (on the
left) when there is a mapping a : G×X −→ X such that for all g, h ∈ G and x ∈ X,
a(gh, x) = a(g, a(hx)) and a(eG, x) = x, where eG is the identity of G. If G operates
on X, we call X a G-set. We abuse the notation by writing gx for a(g, x). Typically,
the group G is a set of permutations of X so this abuse should cause no confusion.
Let X be a G-set, g ∈ G, and x, y ∈ X. Then
• the mapping Tg : X −→ X defined by Tg(x) = gx is called a translation,
• the set orb(x) = {Tg(x) : g ∈ G} is called the orbit of x,
• define the relation RG on X by xRGy if and only if x ∈ orb(y).
For the remainder of this section, assume that G is a group and X is a G-set.
Example 2.21. The group G is a G-set.
Proof. Let g, h, x ∈ G. Then g(hx) = (gh)x as the group is associative and eGx = x
by the definition of the identity of G. Thus, G is a G-set. �
19
As G is a G-set, we will use the group action notation in groups, in particular
Tg : G −→ G such that Tg(h) = gh, and we will call Tg a group translation. Also,
we will freely switch between the notation Tg(x) and gx for a translation by group
element g.
Proposition 2.22. The relation RG defined in Definition 2.20 forms an equivalence
relation on X.
Proof. We must show RG is reflexive, symmetric, and transitive. First, note x ∈
orb(x) so xRGx. Thus RG is reflexive. To show RG is symmetric, assume xRGy. In
other words, x = Te(x) ∈ orb(y), so there exists a g ∈ G such that Tg(y) = gy = x.
Then
y = eGy = (g−1g)y = g−1(gy) = g−1x = Tg−1(x).
Hence, y ∈ orb(x) so yRGx. Thus, RG is symmetric. To show RG is transitive,
we suppose xRGy and yRGz, that is x ∈ orb(y) and y ∈ orb(z). Thus, there exist
g, h ∈ G such that
Tg(y) = gy = x and Th(z) = hz = y.
Then
Tgh(z) = (gh)z = g(hz) = g(y) = x,
and thus, x ∈ orb(z). Hence xRGz andRG is transitive. AsRG is reflexive, symmetric,
and transitive, RG forms an equivalence relation on X. �
Definition 2.23. Let X be a G-set and x ∈ X. Then the set Gstab(x) = {g ∈ G :
gx = x} is called the stabilizer of x.
One easily checks that Gstab(x) is a subgroup of G.
Proposition 2.24. Let g, h ∈ G. If gGstab(x) = hGstab(x) then gx = hx.
20
Proof. Let x ∈ X and g, h ∈ G. Suppose gGstab(x) = hGstab(x). Since, g ∈ gGstab(x),
g ∈ hGstab(x), so there exists a g′ ∈ Gstab(x) such that hg′ = g. Then gx = hg′x =
hx. �
Proposition 2.25. If Gstab(x) = {eG} and gx = hx for g, h ∈ G, then g = h.
Proof. Let gx = hx. Then x = g−1hx so g−1h ∈ Gstab(x). Thus g−1h = eG so
g = h. �
Definition 2.26. Let X be a G-set and x ∈ X. Define the mapping πx : G −→ orb(x)
by πx(g) = gx.
Definition 2.27. Let X be a set and let X1, X2, . . . , Xn be subsets of X such that
• X1 ∪X2 ∪ . . . ∪Xn = X,
• for all i 6= j, Xi ∩Xj = ∅. Then X1, X2, . . . , Xn is called a partition of X.
Proposition 2.28. Let x ∈ X such that Gstab(x) = {eG}. If A1, A2, . . ., An is a
partition of G, then πx[A1], πx[A2], . . ., πx[An] is a partition of orb(x).
Proof. Suppose Gstab(x) = {eG} for some x ∈ X and A1, A2, . . ., An is a partition of
G. First, we will show
πx[A1] ∪ πx[A2] ∪ . . . ∪ πx[An] = orb(x).
Suppose y ∈ orb(x). Then there exists g ∈ G such that gx = y, and as A1 ∪ A2 ∪
. . . ∪ An = G, g ∈ Ai for some i ∈ {1, 2, . . . n}. Therefore,
y = πx(g) ∈ πx[Ai] ⊆ πx[A1] ∪ πx[A2] ∪ . . . ∪ πx[An].
The other inclusion follows immediately. Now we want to show πx[Ai] ∩ πx[Aj] = ∅
for i 6= j. Suppose toward contradiction that y ∈ πx[Ai] ∩ πx[Aj] for i 6= j. Then
there exist g ∈ Ai and h ∈ Aj such that Tg(x) = gx = y and Th(x) = hx = y. But
then gx = hx, so (h−1g)x = x. Thus h−1g = eG so g = h, by our assumption that
21
Gstab(x) = {eG}. Hence, g = h ∈ Ai ∩ Aj 6= ∅, contradicting i 6= j and A1, A2, . . .,
An being a partition.
Thus, πx[A1], πx[A2], . . ., πx[An] is a partition of orb(x). �
22
3. Paradoxical Groups
Consider the free group with two generators, F = fr({a, b}). We want to partition
F into pieces which are in some sense similar. Define
Fa = {[w] ∈ F : w is a reduced word and w begins, on the left, with a},
(see Definition 2.16 for reduced word). Similarly, define Fa−1 , Fb, and Fb−1 . Note that
this way, we have formed a partition of F ,
F = Fa ∪ Fa−1 ∪ Fb ∪ Fb−1 ∪ {[e]}.
We now consider group translations of the parts of this partition. First, we will
examine T[a−1][Fa]. Let [ax1x2 . . . xn] ∈ Fa, where ax1x2 . . . xn is a reduced word, so
in particular x1 6= a−1. Then
T[a−1]([ax1x2 . . . xn]) = [a] ◦ [x1x2 . . . xn] = [x1x2 . . . xn] ∈ F \ Fa−1 .
Conversely, consider [x1x2 . . . xn] ∈ F \ Fa−1 , where x1x2 . . . xn is a reduced word so
in particular x1 6= a−1. Then [ax1x2 . . . xn] ∈ Fa so
[x1x2 . . . xn] = [a−1ax1x2 . . . xn] ∈ T[a−1][Fa].
Thus, we have shown
T[a−1][Fa] = F \ Fa−1 .
Clearly, as T[e] is the identity mapping, we have
T[e][Fa−1 ] = Fa−1 .
Similarly, we have T[b−1][Fb] = F \Fb−1 and T[e][Fb−1 ] = Fb−1 . Thus, sets Fa and Fa−1 ,
after being mapped by T[a−1] and T[e] respectively, create a copy of F . Also, sets Fb
and Fb−1 get mapped by group translation to a copy of F . In other words, we have
divided F into five pieces and after using group translations on four of those pieces,
23
ended up with two copies of F . However, note any translation of the last part of our
partition, {[e]}, is non-empty. It is our goal, however, to get exactly two copies of F ,
so we will add the fifth piece, {[e]}, to one of the other four and make the following
modification:
A1 = Fa \ {[a], [aa], [aaa], . . .}
A2 = Fa−1 ∪ {[e]} ∪ {[a], [aa], [aaa], . . .}
B1 = Fb
B2 = Fb−1 .
Now we examine T[a−1][A1]. Let [ax1x2 . . . xn] ∈ A1 where ax1x2 . . . xn is a reduced
word, so in particular x1 6= a−1 and xi 6= a for some i ∈ {1, 2, . . . , n}. Then
T[a−1]([ax1x2 . . . xn]) = [a−1] ◦ [ax1x2 . . . xn] = [x1x2 . . . xn],
a reduced word such that x1 6= a−1 and xi 6= a for some i ∈ {1, 2, . . . , n}. Also,
T[a−1]([ax1x2 . . . xn]) 6= [e] as [a] /∈ A1.
Thus,
T[a−1]([ax1x2 . . . xn]) ∈ F \ A2.
Conversely, let [x1x2 . . . xn] ∈ F \A2, where x1x2 . . . xn is a non-empty reduced word
such that x1 6= a−1 and xi 6= a for some i ∈ {1, 2, . . . , n}. Then [ax1x2 . . . xn] ∈ A1 so
[x1x2 . . . xn] = [a−1ax1x2 . . . xn] = [a−1] ◦ [ax1x2 . . . xn] ∈ T[a−1][A1].
This completes the argument that
T[a−1][A1] = F \ A2.
24
Also, from our previous arguments, we have all of the following:
T[e][A2] = A2
T[b−1][B1] = F \B2
T[b−1][B2] = B2.
We have now partitioned F into four pieces and moved A1 and A2 by group trans-
lations to get the entire group F , and we have also moved B1 and B2 by group
translations to get the entire group F . Also, these two copies of F do not overlap.
This effect is formalized in the following definition.
Definition 3.1. We say that a group G is n–paradoxical if there exist a partition A1,
A2, . . ., Ai, . . . An of G and translations T1, T2, . . .,Ti, . . ., Tn such that
T1[A1], T2[A2], . . . , Ti[Ai] and Ti+1[Ai+1], Ti+2[Ai+2], . . . , Tn[An]
are each partitions of G.
Corollary 3.2. The free group on two generators is 4–paradoxical.
Using Proposition 2.28, we may transfer the paradoxical decomposition of a group
G onto a G–set X, getting a G–paradoxical decomposition of X. To describe this
effect fully, we need the following definition.
Definition 3.3. Suppose that a group G acts on a space X. We say that the set
X ⊆ X has a G–paradoxical decomposition if there exist
• a partition A1, A2, . . . , Ai, Ai+1, . . . , An of X, and
• elements g1, g2, . . . , gi, gi+1, . . . , gn of G
such that for some i < n,
• g1[A1], g2[A2], . . . , gi[Ai] is a partition of X, and
• gi+1[Ai+1], gi+2[Ai+2], . . . , gn[An] is a partition of X.
Recall gx = Tg(x) so g[A] = Tg[A]. Also, if G is understood to be our group acting
on X, we may simply say X is paradoxical.
25
Theorem 3.4. If G is paradoxical and operates on a set X in such a way that for
all x ∈ X, Gstab(x) = {eG}, then X has a G–paradoxical decomposition.
Proof. Let G be paradoxical. Then we have a partition
B1, B2, . . . , Bi, Bi+1, . . . , Bn
of G and elements g1, g2, . . . , gi, gi+1, . . . , gn of G such that both
g1[B1], g2[B2], . . . , gi[Bi] and gi+1[Bi+1], gi+2[Bi+2], . . . , gn[Bn]
are partitions of G. Also, assume for any x ∈ X that Gstab(x) = {eG}. Then gx = hx
only if g = h by Proposition 2.25.
Let C be a selector from the orbits of X, that is C ⊆ X and |C ∩ orb(x)| = 1 for
all x ∈ X. For any k ∈ {1, 2, . . . , n} and x ∈ C, define
Axk = {πx(g) : g ∈ Bk}
(see 2.26 for the definition of πx). Also define
Ak =⋃{Axk : x ∈ C}.
Claim 3.4.1. For each x ∈ C, Ax1 , Ax2 , . . . , A
xn is a partition of orb(x).
Proof of the Claim. Immediate by Proposition 2.28. �
Claim 3.4.2. A1, A2, . . . , An is a partition of X.
Proof of the Claim. First, we will show A1 ∪ A2 ∪ . . . ∪ An = X. Let y ∈ X. Then
y ∈ orb(x) for some x ∈ C, as {orb(x) : x ∈ C} is a partition of X (as C was a selector
of all the orbits). But Ax1 , Ax2 , . . . , A
xn is a partition of orb(x) (by Claim 3.4.1), so
y ∈ Ax1∪Ax2∪. . .∪Axn. Then for some i ∈ {1, 2, . . . , n}, y ∈ Axi ⊆ Ai ⊆ A1∪A2∪. . .∪An.
The other inclusion follows immediately.
26
Now we want to show Aj ∩ Al = ∅ for j 6= l. Suppose towards contradiction
that y ∈ Aj ∩ Al and j 6= l. Then for some x1, x2 ∈ C, y ∈ Ax1j ∩ Ax2l . But then
y ∈ orb(x1) ∩ orb(x2), so orb(x1) = orb(x2). Thus x1 = x2 (by the definition of
C). Then y ∈ Ax1j ∩ Ax1l , so for some g1 ∈ Bj and g2 ∈ Bl, y = g1x1 = g2x2. By
Proposition 2.25, g1 = g2, so Bj ∩Bl 6= ∅, a contradiction.
Thus, A1, A2, . . . , An is a partition of X. �
Claim 3.4.3. Let x ∈ C. Then g1[Ax1 ], g2[Ax2 ], . . . , gi[Axi ] is a partition of orb(x).
Proof of the Claim. First note, for any k ∈ {1, 2, . . . , i},
gk[Axk] = {gkgx : g ∈ Bk} = {g′x : g′ ∈ gk[Bk]}.
So
g1[Ax1 ] ∪ g2[Ax2 ] ∪ . . . ∪ gi[Axi ]
= {g′x : g′ ∈ g1[B1]} ∪ {g′x : g′ ∈ g2[B2]} ∪ . . . ∪ {g′x : g′ ∈ gi[Bi]}
= {g′x : g′ ∈ g1[B1] ∪ g2[B2] ∪ . . . ∪ gi[Bi]}
= {g′x : g′ ∈ G} = orb(x).
Also, if y ∈ gk[Axk] ∩ gl[Axl ], then y = g1x = g2x for some g1 ∈ gk[Bk] and g2 ∈ gl[Bl].
By Proposition 2.25, g1 = g2, so gk[Ak] ∩ gl[Al] 6= ∅. Hence, k = l. �
Claim 3.4.4. Let x ∈ C. Then gi+1[Axi+1], gi+2[Axi+2], . . . , gn[Axn] is a partition of
orb(x).
Proof of the Claim. Similar to Claim 3.4.3. �
Claim 3.4.5. g1[A1], g2[A2], . . . , gi[Ai] is a partition of X.
Proof of the Claim. First, we will show
g1[A1] ∪ g2[A2] ∪ . . . ∪ gi[Ai] = X.
27
Let y ∈ X. Then y ∈ orb(x) for some x ∈ C. As g1[Ax1 ], g2[Ax2 ], . . . , gi[Axi ] is a
partition of orb(x), y ∈ gk[Axk] for some k ∈ {1, 2, . . . , i}. But
gk[Axk] ⊆ gk[Ak] ⊆ g1[A1] ∪ g2[A2] ∪ . . . ∪ gi[Ai],
so y ∈ g1[A1] ∪ g2[A2] ∪ . . . ∪ gi[Ai]. The other inclusion follows immediately.
Now we want to show gj[Aj] ∩ gl[Al] = ∅ for j, l ∈ {1, 2, . . . , i} and j 6= l. Suppose
towards contradiction that y ∈ gj[Aj] ∩ gl[Al], for j 6= l. Also, y ∈ orb(x) for some
x ∈ C. Thus,
y ∈ gj[Aj] ∩ orb(x) ∩ gl[Al].
Note for any k,
gk[Ak] ∩ orb(x) = gk[Ak] ∩ gk[orb(x)] = gk[Ak ∩ orb(x)] ⊆ gk[Axk].
Thus
y ∈ gj[Axj ] ∩ gl[Axl ],
so gj[Axj ] ∩ gl[Axl ] 6= ∅, contradicting Claim 3.4.3.
Thus, g1[A1], g2[A2], . . . , gi[Ai] is a partition of X. �
Claim 3.4.6. gi+1[Ai+1], gi+2[Ai+2], . . . , gn[An] is a partition of X.
Proof of the Claim. Similar to Claim 3.4.5. �
By Claims 3.4.2, 3.4.5, and 3.4.6, X is G-paradoxical.
�
28
4. A Free Group of Rotations
We now will construct a free group of rotations in R3. Let θ and ρ be rotations by
an angle of arccos 13
about the z and x axises respectively. Then the linear transfor-
mations θ, θ−1, ρ, ρ−1 have the following representation in the standard basis.
θ =
13−2√
23
0
2√
23
13
0
0 0 1
θ−1 =
13
2√
23
0
−2√
23
13
0
0 0 1
ρ =
1 0 0
0 13−2√
23
0 2√
23
13
ρ−1 =
1 0 0
0 13
2√
23
0 −2√
23
13
We should also note that while we have chosen these angles to work with, any
angles chosen that are irrational with respect to the circle would form a free group
of rotations. Every composition of transformations θ, θ−1, ρ, ρ−1 can be thought of as
a word in W ({θ, ρ}). We will identify the word in W ({θ, ρ}) by the corresponding
rotations.
Lemma 4.1. Let w be a non-empty reduced word that ends with the letter θ or θ−1.
Then w(1, 0, 0) 6= (1, 0, 0).
Proof. We will argue by induction on the length, n, of w that
w(1, 0, 0) =1
3n(a, b√
2, c)
for some a, b, c ∈ Z and that 3 does not divide b. Once that is shown, this will imply
w(1, 0, 0) 6= (1, 0, 0) for every non-empty word, as 0 is divisible by 3.
For n = 1, observe the following
θ(1, 0, 0) = 13(1, 2√
2, 0)
θ−1(1, 0, 0) = 13(1,−2
√2, 0).
29
For n = 2, observe the following:
θθ(1, 0, 0) = 19(−7, 4
√2, 0)
ρθ(1, 0, 0) = 19(3, 2√
2, 8)
ρ−1θ(1, 0, 0) = 19(3, 2√
2,−8)
θ−1θ−1(1, 0, 0) = 19(−7,−4
√2, 0)
ρθ−1(1, 0, 0) = 19(3,−2
√2,−8)
ρ−1θ−1(1, 0, 0) = 19(3,−2
√2, 8).
Now suppose
w(1, 0, 0) =1
3k(a, b√
2, c)
for every reduced word w of length k ≤ n− 1 that ends with the letter θ or θ−1. We
will keep the convention that for k = n− 1,
w′(1, 0, 0) =1
3n−1(a′, b′
√2, c′),
for k = n− 2,
w′′(1, 0, 0) =1
3n−2(a′′, b′′
√2, c′′),
etc.
We want to show that for word w of length n, w(1, 0, 0) is of the form
1
3n(a, b√
2, c)
where a, b, c ∈ Z. To show this, we consider 4 cases:
(A1) w = θw′, where w′ is a reduced word of length n− 1 ending with the letter θ
or θ−1. By the inductive hypothesis,
w′(1, 0, 0) =1
3n−1(a′, b′
√2, c′)
where a′, b′, c′ ∈ Z. Then
w(1, 0, 0) =1
3n(a′ − 4b′, (b′ + 2a′)
√2, 3c′).
30
Then as a′, b′, c′ ∈ Z, so are (a′ − 4b′), (b′ + 2a′), and (3c′).
(A2) w = θ−1w′. Then
w(1, 0, 0) =1
3n(a′ + 4b′, (b′ − 2a′)
√2, 3c′).
Then as a′, b′, c′ ∈ Z, so are (a′ + 4b′), (b′ − 2a′), and (3c′).
(A3) w = ρw′. Then
w(1, 0, 0) =1
3n(3a′, (b′ − 2a′)
√2, c′ + 4b′).
Then as a′, b′, c′ ∈ Z, so are (3a′), (b′ − 2a′), and (c′ + 4b′).
(A4) w = ρ−1w′. Then
w(1, 0, 0) =1
3n(3a′, (b′ + 2a′)
√2, c′ − 4b′).
Then as a′, b′, c′ ∈ Z, so are (3a′), (b′ + 2a′), and (c′ − 4b′).
In any case now, w(1, 0, 0) is of the form 13n
(a, b√
2, c) where a, b, c ∈ Z.
Now to argue that 3 does not divide b, we must consider 12 cases:
(B1) w = θρw′′ where w′′ is a reduced word of length n− 2 ending with the letter
θ or θ−1. By the inductive hypothesis,
w′′(1, 0, 0) =1
3n−2(a′′, b′′
√2, c′′)
where a′′, b′′, c′′ ∈ Z. Then
w(1, 0, 0) = θρw′′(1, 0, 0) = θ 13n−1 (3a′′, (b′′ − 2c′′)
√2, c′′ + 4b′′)
= 13n
(x1, [(b′′ − 2c′′) + 6a′′]
√2, z1).
Note that ρw′′ is a reduced word of length n − 1, so by the inductive hypothesis,
b′′ − 2c′′ = b′, and 3 does not divide b′. So
b = (b′′ − 2c′′) + 6a′′ = b′ + 6a′′,
and as 3 divides 6a′′ but does not divide b′, 3 does not divide b = b′ + 6a′′.
31
(B2) w = ρρw′′. Then
w(1, 0, 0) =1
3n(x2, (−7b′′ − 4c′′)
√2, z2) =
1
3n(x2, (2b
′ − 9b′′)√
2, z2).
So b = 2b′ − 9b′′, and as 3 divides −9b′′ but does not divide 2b′, 3 does not divide
b = 2b′ − 9b′′.
(B3) w = θ−1ρw′′. Then
w(1, 0, 0) =1
3n(x3, (b
′ − 6a′′)√
2, z3).
So b = b′ − 6a′′, and as 3 divides 6a′′ but does not divide b′, 3 does not divide
b = b′ − 6a′′.
(B4) w = θρ−1w′′. Then
w(1, 0, 0) =1
3n(x4, (b
′ + 6a′′)√
2, z4).
So b = b′ + 6a′′, and as 3 divides 6a′′ but does not divide b′, 3 does not divide
b = b′ + 6a′′.
(B5) w = θ−1ρ−1w′′. Then
w(1, 0, 0) =1
3n(x5, (b
′ − 6a′′)√
2, z5).
So b = b′ − 6a′′, and as 3 divides 6a′′ but does not divide b′, 3 does not divide
b = b′ − 6a′′.
(B6) w = ρ−1ρ−1w′′. Then
w(1, 0, 0) =1
3n(x6, (−7b′′ − 4c′′)
√2, z6) =
1
3n(x6, (2b
′ − 9b′′)√
2, z6).
So b = 2b′ − 9b′′, and as 3 divides −9b′′ but does not divide 2b′, 3 does not divide
b = 2b′ − 9b′′.
(B7) w = θθw′′. Then
w(1, 0, 0) =1
3n(x7, (−7b′′ + 4a′′)
√2, z7) =
1
3n(x7, (2b
′ − 9b′′)√
2, z7).
32
So b = 2b′ − 9b′′, and as 3 divides −9b′′ but does not divide 2b′, 3 does not divide
b = 2b′ − 9b′′.
(B8) w = ρθw′′. Then
w(1, 0, 0) =1
3n(x8, (b
′ − 6c′′)√
2, z8).
So b = b′ − 6c′′, and as 3 divides 6c′′ but does not divide b′, 3 does not divide
b = b′ − 6c′′.
(B9) w = ρ−1θ−1w′′. Then
w(1, 0, 0) =1
3n(x9, (b
′ + 6c′′)√
2, z9).
So b = b′+6c′′, and as 3 divides 6c′′ but does not divide b′, 3 does not divide b = b′+6c′′.
(B10) w = ρθ−1w′′. Then
w(1, 0, 0) =1
3n(x10, (b
′ − 6c′′)√
2, z10).
So b = b′ − 6c′′, and as 3 divides 6c′′ but does not divide b′, 3 does not divide
b = b′ − 6c′′.
(B11) w = θ−1θ−1w′′. Then
w(1, 0, 0) =1
3n(x11, (−7b′′ − 4a′′)
√2, z11) =
1
3n(x11, (2b
′ − 9b′′)√
2, z11).
So b = 2b′ − 9b′′, and as 3 divides −9b′′ but does not divide 2b′, 3 does not divide
b = 2b′ − 9b′′.
(B12) w = ρ−1ρ−1w′′. Then
w(1, 0, 0) =1
3n(x12, (−7b′′ + 4c′′)
√2, z12) =
1
3n(x12, (2b
′ − 9b′′)√
2, z12)
So b = b′ + 6c′′, and as 3 divides −9b′′ but does not divide 2b′, 3 does not divide
b = 2b′ − 9b′′.
In any case now, 3 does not divide b. �
33
Lemma 4.2. Let w be a non-empty reduced word that ends with the letter ρ or ρ−1.
Then w(0, 0, 1) 6= (0, 0, 1).
Proof. A very similar argument will work for all non-empty reduced words w that
end with the letter ρ or ρ−1. We will argue by induction on the length, n, of w that
w(0, 0, 1) =1
3n(a, b√
2, c)
for some a, b, c ∈ Z and that 3 does not divide b. Once that is shown, this will imply
w(0, 0, 1) 6= (0, 0, 1) for every non-empty word, as 0 is divisible by 3.
For n = 1, observe the following
ρ(0, 0, 1) = 13(0,−2
√2, 1)
ρ−1(0, 0, 1) = 13(0, 2√
2, 1).
For n = 2, observe the following:
θρ(0, 0, 1) = 19(8,−2
√2, 3)
ρρ(0, 0, 1) = 19(0,−4
√2,−7)
θ−1ρ(0, 0, 1) = 19(−8,−2
√2, 3)
θρ−1(0, 0, 1) = 19(−8, 2
√2, 3)
ρ−1ρ−1(0, 0, 1) = 19(0, 4√
2,−7)
θ−1ρ−1(0, 0, 1) = 19(8, 2√
2, 3).
The inductive step of this proof is the same as Lemma 4.1, replacing w′′(1, 0, 0)
with w′′(0, 0, 1). Thus, suppose
w′′(0, 0, 1) =1
3n−2(a′′, b′′
√2, c′′)
for some a′′, b′′, c′′ ∈ Z and that 3 does not divide b′′ for all non-empty reduced words
w′′ of length n− 2 that end with the letter ρ or ρ−1, and
w′(0, 0, 1) =1
3n−1(a′, b′
√2, c′)
34
for some a′, b′, c′ ∈ Z and that 3 does not divide b′ for all non-empty reduced words
w′ of length n− 1 that end with the letter ρ or ρ−1.
We want to show that w(0, 0, 1) is of the form
1
3n(a, b√
2, c)
where a, b, c ∈ Z. Note this is shown in cases (A1 - A4) in 4.1.
Also, 3 does not divide b by cases (B1 - B12) in 4.1. �
Definition 4.3. Let G∗ be the subgroup of the group of rotations of the sphere S1 =
{(x, y, z) ∈ R3 : x2 + y2 + z2 = 1} generated by {θ, ρ}.
Theorem 4.4. G∗ is isomorphic to fr({a, b}).
Proof. Define f : {a, a−1, b, b−1} −→ G∗ by f(a) = θ, f(a−1) = θ−1, f(b) = ρ, and
f(b−1) = ρ−1. By Theorem 2.19, there exists a homomorphism f : fr({a, b}) −→ G∗
such that
f([x1x2 . . . xn]) = f(x1)f(x2) . . . f(xn).
We claim f is an isomorphism. We will first show f is onto G∗. Note that θ, ρ ∈
rng(f) = {f([w]) : [w] ∈ fr({a, b})}. As rng(f) is a subgroup of G∗ (as f is a
homomorphism), rng(f) = G∗ (as G∗ is the group generated by {θ, ρ}). Next we
show f is injective. We will examine the kernel of f ,
ker(f) = {[w] : f([w]) = eG∗}.
By Lemma 4.1, if w is a reduced word that ends with a or a−1, w(1, 0, 0) 6= (1, 0, 0)
so [w] is not the identity and hence, [w] /∈ ker(f). By Lemma 4.2, if w is a reduced
word that ends with b or b−1, w(0, 0, 1) 6= (0, 0, 1) so again [w] is not the identity and
therefore, [w] /∈ ker(f). Hence, ker(f) = {eG∗}. As the kernel of the homomorphism
f is trivial, f is injective. �
35
5. Hausdorff and Banach-Tarski Paradoxes
5.1. The Hausdorff Paradox. Let S1 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1} and
let G∗ be our free group of rotations defined in Definition 4.4. Our goal is to choose
X1 ⊆ S1 to be a G∗-set under the natural action in such a way that
G∗stab(x) = {eG∗} for all x ∈ X1. (⊕)
Note that θ((0, 0, 1)) = (0, 0, 1), so θ ∈ G∗stab((0, 0, 1)) 6= {eG∗} and thus we cannot
set X1 = S1. In fact, for any rotation g ∈ G∗, both of the poles, pg,1 and pg,2, will
have non-trivial stabilizers. Then let P1 = {pg,1, pg,2 : g ∈ G∗}. Thus, our demand ⊕
requires X1 ⊆ S1 \ P1.
As we have taken points out of S1, we must be concerned that G∗ operating on X1
is well-defined. To ensure gx ∈ X1 we may simply set P1 = {q ∈ orb(p) : p ∈ P1} and
then define X1 = S1 \ P1. Let us argue that for any x ∈ X1 and g ∈ G∗, gx ∈ X1.
Plainly for any s ∈ S1, gs ∈ S1. Therefore as x ∈ S1, gx ∈ S1. As x /∈ orb(p) for any
p ∈ P1, orb(x) ∩ orb(p) = ∅, and thus, gx /∈ orb(p) for any p ∈ P1. Thus gx ∈ X1.
This proves that the action of G∗ on X1 is well-defined.
Let us show that X1 is not something trivial. First note that S1 is uncountable.
Now, we will argue that P1 is countable. Define pi to be the i-th prime number, so
p1 = 2, p2 = 3, etc. Define α : {θ, θ−1, ρ, ρ−1} −→ N by
α(a) =
1 if a = θ
2 if a = θ−1
3 if a = ρ
4 if a = ρ−1.
36
Now define f : G∗ −→ N by
f(g) =
1 if g = eG∗
pα(a1)1 p
α(a2)2 p
α(a3)3 . . . p
α(an)n if g = a1a2a3 . . . an is a reduced word and
ai ∈ {θ, θ−1, ρ, ρ−1}.
This defines an injective function f , so G∗ is equinumerous with a subset of the
natural numbers, so G∗ is countable.
As P1 = {pg,1, pg,2 : g ∈ G∗} = {pg,1 : g ∈ G∗} ∪ {pg,2 : g ∈ G∗}, P1 is a union of
two countable sets, P1 is countable.
Finally,
P1 = {q ∈ orb(p) : p ∈ P1} =⋃p∈P1
orb(p) =⋃p∈P1
{gp : g ∈ G∗}
is a countable union of countable sets, so P1 is countable. Therefore, X1 = S1 \ P1
is uncountable, as we have simply removed a countable number of points from an
uncountable set.
To summarize, we have formed X1 which is almost the entire sphere such that G∗
acts on X1 and each x ∈ X1 has a trivial stabilizer. Therefore, any non-trivial rotation
moves every point of X1 and any partition of our group G∗ creates a partition of X1.
Corollary 5.1 (Felix Hausdorff, [5]). X1 has a G∗ paradoxical decomposition into 4
pieces.
Proof. By Corollary 3.2, Theorem 4.4, and Theorem 3.4. �
5.2. A Basis for the Paradoxical Decomposition of the Sphere. While it is
remarkable in itself that the surface of the sphere, minus a countable number of points,
X1 = S1 \ P1, is paradoxical, we really want to show that the entire surface of the
sphere S1 is paradoxical. Recall how X1 was paradoxically decomposed. We identify
the group of rotations generated by {θ, ρ} with F = fr{a, b}. Then A1, A2, B1, B2 ⊆
37
G∗ were defined on page 23. This creates a partition X1 into TA11 , TA2
1 , TB11 , TB2
1 where
TA11 = {gx : x ∈ C1, g ∈ A1},
TA21 = {gx : x ∈ C1, g ∈ A2},
TB11 = {gx : x ∈ C1, g ∈ B1},
TB21 = {gx : x ∈ C1, g ∈ B2},
and C1 is a selector from all of the orbits in X1.
Thus, we have a partition of S1 into TA11 , TA2
1 , TB11 , TB2
1 , P1. Then
θ−1[TA11 ], TA2
1 , P1
is a partition of S1 and
ρ−1[TB11 ], TB2
1
is a partition of S1 \ P1. Now, we only must fill in P1 of the second copy of S1.
As P1 is countable and S1 \ P1 is uncountable, we may find an axis of rotation
through the origin such that its poles on the sphere, q1, q2, are elements of S1 \ P1.
We want to choose a rotation σ about this axis so that no p ∈ P1 gets moved back to
itself or to another point p′ ∈ P1 no matter how many times σ is applied to it. As P1
is countable, there is only a countable number of points σ must avoid, and there are
only a countable number of ways to reach each point, σn(p) = p′ where n ∈ N. As
the number of rotations is uncountable, we may choose a desired rotation σ. Now let
P ∗1 = {x ∈ X1 : (∃n ∈ N, p ∈ P1)(σn(p) = x)} ⊆ S1 \ P1.
Then
σ−1[P ∗1 ] = P ∗1 ∪ P1def= P1.
Finally, ρ−1[TB11 ], TB2
1 is a partition of S1 \ P1 = (S1 \ P1) ∪ P ∗1 . Then
(S1 \ P1) ∪ σ−1[P ∗1 ] = (S1 \ P1) ∪ P1 = S1.
38
This gives us our second copy of S1 and will give us the basis for our paradoxical
decomposition the entire surface of the sphere.
5.3. A Decomposition of the Sphere. To specify the six pieces used for this
decomposition of the sphere, we will use the following notation. For a positive number
r, let
Sr = {(x, y, z) :√x2 + y2 + z2 = r}
Pr = {p : p is a pole of Sr for some rotation g ∈ G∗}
Pr = {q ∈ orb(p) : p ∈ Pr}
Xr = Sr \ Pr
P ∗r = {x ∈ Xr : (∃n ∈ N, p ∈ Pr)(σn(p) = x)}
Pr = P ∗r ∪ Pr
Cr is a selector from the orbits of Xr
TA1r = {gx : x ∈ Cr, g ∈ A1}
TA2r = {gx : x ∈ Cr, g ∈ A2}
TB1r = {gx : x ∈ Cr, g ∈ B1}
TB2r = {gx : x ∈ Cr, g ∈ B2}
σ is a rotation with poles in Xr such that (∀n ∈ N, p ∈ Pr)(σn(p) /∈ Pr).
For now we consider r to be 1.
Our decomposition of S1, the surface of the sphere of radius 1, is
A1 = TA11
A2 = TA21 ∪ P1
A3 = ρ[(TA11 ∪ TA2
1 ∪ TB11 ) ∩ P ∗1 ]
A4 = TB11 \ ρ[(TA1
1 ∪ TA21 ∪ TB1
1 ) ∩ P ∗1 ]
A5 = TB21 ∩ P ∗1
A6 = TB21 \ P ∗1 .
39
Note that ρ[TA11 ∪TA2
1 ∪TB11 ] ⊆ TB1
1 so ρ[(TA11 ∪TA2
1 ∪TB11 )∩P ∗1 ] ⊆ TB1
1 . The following
rotations will witness that A1,A2,A3,A4,A5,A6 is a paradoxical decomposition of
S1:
g1 = θ−1
g2 = eG∗
g3 = σ−1ρ−1
g4 = ρ−1
g5 = σ−1
g6 = eG∗ .
As we have mentioned before,
g1[A1] = θ−1[TA11 ] = X1 \ TA2
1 and g2[A2] = eG∗ [TA21 ∪ P1] = TA2
1 ∪ P1
is a partition of S1.
Next observe for A3,
g3[A3] = σ−1ρ−1[ρ[(TA11 ∪ TA2
1 ∪ TB11 ) ∩ P ∗1 ]
= σ−1[(TA11 ∪ TA2
1 ∪ TB11 ) ∩ P ∗1 ].
For A4,
g4[A4] = ρ−1[TB11 \ ρ[(TA1
1 ∪ TA21 ∪ TB1
1 ) ∩ P ∗1 ]]
= ρ−1[TB11 ] \ ρ−1[ρ[(TA1
1 ∪ TA21 ∪ TB1
1 ) ∩ P ∗1 ]]
= (TA11 ∪ TA2
1 ∪ TB11 ) \ ((TA1
1 ∪ TA21 ∪ TB1
1 ) ∩ P ∗1 )
= (TA11 ∪ TA2
1 ∪ TB11 ) \ P ∗1 .
For A5,
g5[A5] = σ−1[TB21 ∩ P ∗1 ].
And simply
g6[A6] = TB21 \ P ∗1 .
40
Observe
g3[A3] ∪ g5[A5] = σ−1[(TA11 ∪ TA2
1 ∪ TB11 ) ∩ P ∗1 ] ∪ σ−1[TB2
1 ∩ P ∗1 ]
= σ−1[(TA11 ∪ TA2
1 ∪ TB11 ∪ TB2
1 ) ∩ P ∗1 ]
= σ−1[P ∗1 ] = P1.
Clearly, g3[A3] ∩ g5[A5] = ∅.
Also observe
g4[A4] ∪ g6[A6] = ((TA11 ∪ TA2
1 ∪ TB11 ) \ P ∗1 ) ∪ (TB2
1 \ P ∗1 )
= (TA11 ∪ TA2
1 ∪ TB11 ∪ TB2
1 ) \ P ∗1= X1 \ P ∗1 .
Clearly, g4[A4] ∩ g6[A6] = ∅. Moreover,
(g3[A3] ∪ g5[A5]) ∩ (g4[A4] ∪ g6[A6]) = ∅,
and consequently, g3[A3], g4[A4], g5[A5], g6[A6] are all pairwise disjoint.
Finally, we have
g3[A3] ∪ g4[A4] ∪ g5[A5] ∪ g6[A6] = P1 ∪ (X1 \ P ∗1 ) = S1.
Thus g3[A3], g4[A4], g5[A5], g6[A6] is a partition of S1, and since g1[A1], g2[A2] is
a partition of S1, we have paradoxically decomposed S1.
5.4. The Banach-Tarski Paradox. We can now paradoxically decompose the unit
ball
B = {(x, y, z) :√x2 + y2 + z2 ≤ 1}.
Define
X =⋃r∈(0,1]Xr
P =⋃r∈(0,1] Pr
P =⋃r∈(0,1] Pr
P ∗ =⋃r∈(0,1] P
∗r
41
P =⋃r∈(0,1] Pr
C =⋃r∈(0,1]Cr
Tα =⋃r∈(0,1] T
αr (for α ∈ {A1, A2, B1, B2}).
Choose x0 ∈ TA11 ⊆ X1 = S1 \ P1. In a similar manner that we found a rotation σ,
we will argue we may find a rotation γ such that for all n ∈ N
(1) γn(x0) ∈ X,
(2) γn(x0) 6= x0.
Fix two poles in S1 \ {x0}. As there are only countably many ways to reach a point
in P1 and only countably many points in P1 that we must avoid, there are only a
countable number of rotations that do not work for demand (1). For any n ∈ N,
there is at most a finite number of rotations such that γn(x0) = x0. Therefore, there
are only countably many rotations such that γn(x0) = x0 for some n ∈ N, so there is
only a countable number of rotations that do not work for demand (2). But as the
number of rotations about the axis through the fixed poles is uncountable, we may
choose a desired rotation γ.
Define the set Y = {γn(x0) : n ∈ N}, and let the translation τ be such that
τ(x0) = (0, 0, 0).
Our decomposition of B is
B1 = θ[(TA1 ∪ TB1 ∪ TB2) ∩ Y ]
B2 = TA1 \ θ[[(TA1 ∪ TB1 ∪ TB2) ∩ Y ] ∪ {x0}]
B3 = TA2 ∩ Y
B4 = (TA2 \ Y ) ∪ P ∪ {(0, 0, 0)}
B5 = θ[{x0}]
B6 = ρ[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]
B7 = TB1 \ ρ[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]
B8 = TB2 ∩ P ∗
B9 = TB2 \ P ∗.
42
Here we note that θ[TA1 ∪ TB1 ∪ TB2 ] ⊆ TA1 . Easily, B1,B2,B5 is a partition of
TA1 ; B3,B4 is a partition of TA2 ∪ P ∪ {(0, 0, 0)}; B6,B7 is a partition of TB1 ; and
B8,B9 is a partition of TB2 . On the other hand, TA1 , TA2 ∪ P ∪ {(0, 0, 0}, TB1 , TB2
is a partition of B. Therefore, B1,B2, . . . ,B9 is a partition of B as well.
Using the following rotations and translation, we will show that our partition into
9 pieces is a paradoxical decomposition of B:
h1 = γ−1θ−1
h2 = θ−1
h3 = γ−1
h4 = eG∗
h5 = τθ−1
h6 = σ−1ρ−1
h7 = ρ−1
h8 = σ−1
h9 = eG∗ .
Observe for B1,
h1[B1] = γ−1θ−1[θ[(TA1 ∪ TB1 ∪ TB2) ∩ Y ]]
= γ−1[(TA1 ∪ TB1 ∪ TB2) ∩ Y ].
For B2,
h2[B2] = θ−1[TA1 \ θ[[(TA1 ∪ TB1 ∪ TB2) ∩ Y ] ∪ {x0}]]
= θ−1[TA1 ] \ θ−1[θ[[(TA1 ∪ TB1 ∪ TB2) ∩ Y ] ∪ {x0}]]]
= (TA1 ∪ TB1 ∪ TB2) \ [(TA1 ∪ TB1 ∪ TB2) ∩ Y ] ∪ {x0}]
= (TA1 ∪ TB1 ∪ TB2) \ (Y ∪ {x0}).
For B3,
h3[B3] = γ−1[TA2 ∩ Y ].
43
For B4, simply
h4[B4] = (TA2 \ Y ) ∪ P ∪ {(0, 0, 0)}.
Observe
h1[B1] ∪ h3[B3] = γ−1[(TA1 ∪ TB1 ∪ TB2) ∩ Y ] ∪ γ−1[TA2 ∩ Y ]
= γ−1[(TA1 ∪ TA2 ∪ TB1 ∪ TB2) ∩ Y ]
= γ−1[X ∩ Y ] = γ−1[Y ] = Y ∪ {x0}.
Clearly, h1[B1] ∩ h3[B3] = ∅.
Also observe
h2[B2] ∪ h4[B4] =[(TA1 ∪ TB1 ∪ TB2) \ (Y ∪ {x0})
]∪[(TA2 \ Y ) ∪ P ∪ {(0, 0, 0)}
]=[(TA1 ∪ TA2 ∪ TB1 ∪ TB2)
]\[(Y ∪ {x0}) ∪ P ∪ {(0, 0, 0)}
]= [X \ (Y ∪ {x0})] ∪ P ∪ {(0, 0, 0)}.
Clearly, h2[B2] ∩ h4[B4] = ∅. Moreover,
(h1[B1] ∪ h3[B3]) ∩ (h2[B2] ∪ h4[B4]) = ∅,
and consequently, h1[B1], h2[B2], h3[B3], h4[B4] are all pairwise disjoint.
Finally, we have
h1[B1] ∪ h2[B2] ∪ h3[B3] ∪ h4[B4]
= (Y ∪ {x0}) ∪ [X \ (Y ∪ {x0}) ∪ P ∪ {(0, 0, 0)}]
= X ∪ P ∪ {(0, 0, 0)}
= B.
Thus h1[B1], h2[B2], h3[B3], h4[B4] is a partition of B.
Now observe for B5,
h5[B5] = τθ−1[θ[{x0}]] = τ [{x0}] = {(0, 0, 0)}.
44
For B6,
h6[B6] = σ−1[ρ−1[ρ[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]]]
= σ−1[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗].
For B7,
h7[B7] = ρ−1[TB1 \ ρ[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]]
= ρ−1[TB1 ] \ ρ−1[ρ[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]]
= (TA1 ∪ TA2 ∪ TB1) \ [(TA1 ∪ TA2 ∪ TB1) ∩ P ∗]
= (TA1 ∪ TA2 ∪ TB1) \ P ∗.
For B8,
h8[B8] = σ−1[TB2 ∩ P ∗].
And simply
h9[B9] = TB2 \ P ∗.
Observe
h6[B6] ∪ h8[B8] = σ−1[(TA1 ∪ TA2 ∪ TB1) ∩ P ∗] ∪ σ−1[TB2 ∩ P ∗]
= σ−1[(TA1 ∪ TA2 ∪ TB1 ∪ TB2) ∩ P ∗]
= σ−1[X ∩ P ∗] = σ−1[P ∗] = P .
Clearly, h6[B6] ∩ h8[B8] = ∅.
Also observe
h7[B7] ∪ h9[B9] = [(TA1 ∪ TA2 ∪ TB1) \ P ∗] ∪ (TB2 \ P ∗)
= X \ P ∗.
Clearly, h7[B7] ∩ h9[B9] = ∅. Moreover,
(h6[B6] ∪ h8[B8]) ∩ (h7[B7] ∪ h9[B9]) = (P ∗ ∪ P ) ∩ [B \ (P ∗ ∪ P )] = ∅,
and therefore, as h5[B5] = {(0, 0, 0)}, h5[B5], h6[B6], h7[B7], h8[B8], and h9[B9] are
pairwise disjoint.
45
Finally, we have
h5[B5] ∪ h6[B6] ∪ h7[B7] ∪ h8[B8] ∪ h9[B9] = {(0, 0, 0)} ∪ P ∪ (X \ P ∗) = B.
Thus h5[B5], h6[B6], h7[B7], h8[B8], h9[B9] is a partition of B, and since h1[B1],
h2[B2], h3[B3], h4[B4] is a partition of B, we have paradoxically decomposed B.
46
6. Interlude: Measure and Integration
In order to explain why there is no parallel of the Banach-Tarski paradox on the
plane, we need to study measures invariant under isometries. To introduce and de-
velop the concept, we need some background knowledge of measures and integration.
We will develop the theory of integration for finitely additive measures. For a more
standard approach, see [4] or [15].
Definition 6.1. (1) A ring of sets R is a class of subsets of a space X such that
∅ ∈ R and for any X, Y ∈ R,
X ∪ Y ∈ R and X \ Y ∈ R.
(2) An algebra of sets A is a ring of subsets of a space X such that X ∈ A.
(3) A σ-ring (or σ-algebra) S is a ring (respectively, algebra) of sets such that if
Xi ∈ S for all i ∈ N, then
∞⋃i=1
Xi ∈ S.
Example 6.2. Let R be the family of all bounded subsets of R2. Then R is a ring
but not an algebra. Also, R is not a σ-ring.
Definition 6.3. Let A be an algebra of subsets of X and let C ∈ A. Then A(C)
defined by
A(C) = {X ∈ A : X ⊆ C}.
One notices that A(C) is an algebra of subsets of C and is called the algebra induced
by C.
Definition 6.4. A finitely additive measure is a function µ : R −→ [0,∞] such that
• R is a ring
• µ(∅) = 0
47
• if Xi ∈ R for all i ∈ {1, 2, . . . , n} and Xi ∩Xj = ∅ for j 6= k, then
µ(n⋃i=1
Xi) =n∑i=1
µ(Xi)
A σ-additive (or countably additive) measure is a finitely additive measure µ :
R −→ [0,∞] such that R is a σ-ring and if Xi ∈ A for all i ∈ N and Xi ∩ Xj = ∅
for j 6= k, then
µ(∞⋃i=1
Xi) =∞∑i=1
µ(Xi).
Definition 6.5. A finitely subadditive outer measure is a function µ∗ : P (X) −→
[0,∞] defined on all subsets of a set X such that
• µ∗(∅) = 0
• for any X, Y ∈ P (X) where X ⊆ Y , µ∗(X) ≤ µ∗(Y )
• for any Xi ∈ P (X),⋃ni=1 µ
∗(Xi) ≤∑n
i=1 µ∗(Xi)
A σ-subadditive outer measure is a finitely subadditive µ∗ : P (X) −→ [0,∞] such
that for any Xi ∈ P (X),∞⋃i=1
µ(Xi) ≤∞∑i=1
µ∗(Xi).
For our purposes, we are more concerned with finitely additive measures (see Propo-
sition 7.6), so when we say µ is a measure, we mean µ is a finitely additive measure.
However, unless otherwise specified, the same definitions could be stated with a σ-
additive measure as well.
Definition 6.6. (1) A measure space (X,R, µ) is a tuple such that X is a set, R
is a ring of subsets of X, and µ is a measure on R.
(2) A subset X ⊆ X is called measurable if X ∈ R.
(3) Define B(X) to be the class of all real-valued bounded functions f : X −→ R.
(4) A function f ∈ B(X) is called measurable if f−1[I] is measurable for any open
interval I ⊆ R.
48
Typically, integration theory is developed for σ-additive measures. When dealing
with finitely additive measures, one of the difficulties one encounters is dealing with
the measurability of some sets. We will ameliorate this difficulty by working with
measures that measure any set. Having found no good source covering integration
with finitely additive measures, we will present the details fully.
From now on in this section, we will assume (X, P (X), µ) is our mea-
sure space, where µ is a finitely additive measure of total measure 1.
In this setting, every function f : X −→ R will be measurable.
Definition 6.7. (1) For any subset X ⊆ X, the function χX : X −→ {0, 1}
defined by
χX(x) =
0 if x /∈ X
1 if x ∈ X
is called the characteristic function of X.
(2) A simple function is a function f : X −→ R defined by
f(x) =n∑i=1
αiχXi(x),
where αi ∈ R and X1, X2, . . ., Xn is a partition of X.
(3) The integral of a simple function f =∑n
i=1 αiχXiis defined by∫
f(x) dµ(x) =n∑i=1
αiµ(Xi).
Note if f is a simple function, then f ∈ B(X).
We now prove some basic facts about integrals of simple functions, which we will
later show are also true for all bounded functions in general.
Lemma 6.8. Let f, g ∈ B(X) be simple functions and α, β ∈ R. Then αf + βg is a
simple function and∫(αf + βg)(x) dµ(x) = α
∫f(x) dµ(x) + β
∫g(x) dµ(x).
49
Also, |f | is a simple function and |∫f(x) dµ(x)| ≤
∫|f(x)| dµ(x). Hence, if f(x) ≥ 0,
then also∫f(x) dµ(x) ≥ 0. Finally, if f(x) ≤ g(x) for all x ∈ X, then
∫f(x) dµ(x) ≤∫
g(x) dµ(x).
Proof. Let f, g ∈ B(X) be simple functions, i.e.
f(x) =n∑i=1
αiχXi(x) and g(x) =
m∑i=1
βiχYi(x),
where X1, X2, . . ., Xn and Y1, Y2, . . ., Ym are partitions of X. But then there exists
a partition Z1, Z2, . . ., ZN of X, such that
f(x) =N∑i=1
α′iχZi(x) and g(x) =
N∑i=1
β′iχZi(x).
Now let α, β ∈ R and observe
(αf + βg)(x) = α∑N
i=1 α′iχZi
(x) + β∑N
i=1 β′iχZi
(x)
=∑N
i=1[(αα′i + ββ′i)χZi(x)].
Consequently, αf + βg is a simple function and
∫(αf + βg)(x) dµ(x) =
∑Ni=1(αα′i + ββ′i)µ(Zi)
= α∑N
i=1 α′iµ(Zi) + β
∑Ni=1 β
′iµ(Zi)
= α∫f(x) dµ(x) + β
∫g(x) dµ(x).
Concerning |f |, note that, as X1, X2, . . . , Xn are pairwise disjoint,
|f(x)| = |n∑i=1
αiχXi(x)| =
n∑i=1
|αi|χXi(x),
so |f | is a simple function. Furthermore,
|∫f(x) dµ(x)| = |
N∑i=1
αiµ(Xi)| ≤N∑i=1
|αi|µ(Xi) =
∫|f(x)| dµ(x).
50
Finally, for f(x) ≤ g(x) for all x ∈ X,∫g(x) dµ(x)−
∫f(x) dµ(x) =
∫g(x)− f(x) dµ(x) ≥ 0.
�
Definition 6.9. (1) Let f and g be simple functions. We define the function ρ
by
ρ(f, g) =
∫|f − g|dµ.
(2) A sequence {fn} of simple functions is said to be fundamental in the mean if
ρ(fn, fm) −→ 0 as n,m −→∞.
(3) A sequence {fn} of functions fn ∈ B(X) is said to converge in measure to a
function f ∈ B(X) if for every ε > 0,
limn→∞
µ({x : |fn(x)− f(x)| ≥ ε}) = 0.
(4) A sequence {fn} of functions is said to be uniformly bounded if for some
M > 0, we have for every n ∈ N and every x ∈ X, −M < fn(x) < M .
Lemma 6.10. Let f ∈ B(X). Then there exists a uniformly bounded sequence {fn}
of simple functions such that
• {fn} is fundamental in the mean.
• {fn} converges in measure to f .
Proof. As f is bounded, −M < f(x) < M for some M ∈ R. For n ∈ N and for all
k ∈ {0, 1, 2, . . . , 2n − 1}, define
Xn,k = f−1
[[2M
2nk −M,
2M
2n(k + 1)−M
)],
51
and put αn,k = 2M2nk −M . Then, for all n ∈ N, define
fn(x) =n∑k=1
αn,kχXn,k(x).
Observe that |fn(x)| < M for all n ∈ N and x ∈ X. Note that if n < m, then each in-
terval[
2M2mk −M, 2M
2m(k + 1)−M
)is included in exactly one
[2M2nl −M, 2M
2n(l + 1)−M
).
Consequently, each Xm,k is included in one of the Xn,l’s and each Xn,l is divided into
2m−n many Xm,k’s. Moreover, if k, l are such that the inclusion holds, then∣∣∣∣(2M
2mk −M
)−(
2M
2nl −M
)∣∣∣∣ ≤ 2
2n,
so |αm,k − αn,l| ≤ 22n
. Therefore, for each x ∈ X, |fn(x)− fm(x)| ≤ 22n
. Hence,
ρ(fn, fm) =
∫|fn(x)− fm(x)| dµ(x) ≤
∫2
2ndµ(x) =
2
2n.
Thus, as n,m −→∞, ρ(fn, fm) −→ 0, so {fn} is fundamental in the mean.
To show {fn} converges in measure, we note that |fn(x)−f(x)| ≤ 22n
. So let ε0 > 0.
For n > log2
(2ε0
), we have ε0 >
22n
, so
{x : |fn(x)− f(x)| ≥ ε0} = ∅.
Therefore,
µ({x : |fn(x)− f(x)| ≥ ε0}) = 0,
so {fn} converges in measure to f . �
Lemma 6.11. Let {fn}, {gn} be uniformly bounded sequences of simple functions
such that
(1) {fn} and {gn} are fundamental in the mean,
(2) {fn} and {gn} converge in measure to a function f ∈ B(X).
Then
limn→∞
∫fn(x) dµ(x) = lim
n→∞
∫gn(x) dµ(x).
52
Proof. Let {fn}, {gn} be uniformly bounded sequences of simple functions fundamen-
tal in the mean convergent in measure to f . Choose M > 0 such that for all n ∈ N,
−M < fn(x) < M, −M < gn(x) < M, and −M < f(x) < M.
First, we will argue that limn→∞
∫fn(x) dµ(x) exists. Note, by Lemma 6.8, we have
|∫fn(x) dµ(x)−
∫fm(x) dµ(x)| = |
∫fn(x)− fm(x) dµ(x)|
≤∫|fn(x)− fm(x)| dµ(x) = ρ(fn, fm).
But ρ(fn, fm) → 0 when n → ∞ (as {fn} is fundamental in the mean), thus
{∫fn(x) dµ(x)} is a Cauchy sequence, so it converges.
Now, let
L = limn→∞
∫fn(x) dµ(x)− lim
n→∞
∫gn(x) dµ(x)
= limn→∞
(∫fn(x) dµ(x)−
∫gn(x) dµ(x))
= limn→∞
∫(fn(x)− gn(x)) dµ(x). (∗)
Suppose towards contradiction that L 6= 0. Set
ε0 =|L|
8(M + 1)> 0.
Let
Xn = {x : fn(x)− f(x) ≥ ε0}, Yn = {x : gn(x)− f(x) ≥ ε0}.
By assumption (1), we know
limn→∞
µ(Xn) = 0 and limn→∞
µ(Yn) = 0. (∗∗)
By (*) and (**), we may choose n ∈ N such that
µ(Xn) < ε0, µ(Yn) < ε0, and |L−∫
(fn(x)− gn(x)) dµ(x)| < ε0.
For x ∈ X \ (Xn ∪ Yn), we have
|fn(x)− f(x)| < ε0 and |gn(x)− f(x)| < ε0,
53
so also
|fn(x)− gn(x)| < 2ε0.
For x ∈ Xn ∪ Yn, we have only
|fn(x)− gn(x)| < 2M,
but µ(Xn ∪ Yn) < 2ε0. Hence, we may easily conclude
|∫
(fn(x)− gn(x)) dµ(x)| ≤∫|fn(x)− gn(x)| dµ(x)
≤ 2ε0 + 2M · 2ε0
= |L|4(M+1)
+ MM+1· |L|
2
≤ 14|L|+ 1
2|L| = 3
4|L|.
Then
|L−∫
(fn(x)− gn(x)) dµ(x)|
≥∣∣|L| − | ∫ (fn(x)− gn(x)) dµ(x)|
∣∣≥ |L| − 3
4|L| = 1
4|L|.
However,
|L−∫
(fn(x)− gn(x)) dµ(x)| < ε0 = |L|8(M+1)
< 14|L|,
a contradiction. �
Definition 6.12. Let f ∈ B(X). The integral of f is defined as∫f(x) dµ(x) = lim
n→∞
∫fn(x) dµ(x),
for some (equivalently, all) uniformly bounded sequence {fn} of simple functions such
that
• {fn} is fundamental in the mean.
• {fn} converges in measure to f .
Lemmas 6.10 and 6.11 verify that the above definition is correct and indeed does
not depend on the choice of {fn}.
54
Theorem 6.13. Let f, g ∈ B(X). Then αf + βg ∈ B(X) and∫(αf + βg)(x) dµ(x) = α
∫f(x) dµ(x) + β
∫g(x) dµ(x).
Proof. Let f, g ∈ B(X) and let α, β ∈ R. Then there exist uniformly bounded se-
quences {fn}, {gn} fundamental in the mean which converge in measure to f, g re-
spectively such that each fn, gn is simple a function. Choose M > 0 such that for all
n ∈ N and x ∈ X,
−M < fn(x) < M and −M < gn(x) < M.
Note that for all n ∈ N and x ∈ X,
|αfn(x) + βgn(x)| ≤ (|α|+ |β|)M,
so {αfn + βgn} is uniformly bounded.
Now, we will show {αfn + βgn} is fundamental in the mean. Given ε0 > 0. As
{fn}, {gn} are both fundamental in the mean, there exists N ∈ N such that for any
n,m > N ,
ρ(fn, fm) <ε0
2(|α|+ 1)and ρ(gn, gm) <
ε02(|β|+ 1)
.
Then for any n,m > N ,
ρ(αfn + βgn, αfm + βgm) =∫|(αfn + βgn)(x)− (αfm + βgm)(x)| dµ(x)
=∫|αfn(x)− αfm(x) + βgn(x)− βgm(x)| dµ(x)
≤∫|αfn(x)− αfm(x)| dµ(x) +
∫|βgn(x)− βgm(x)| dµ(x)
= |α|∫|fn(x)− fm(x)| dµ(x) + |β|
∫|gn(x)− gm(x)| dµ(x)
= |α|ρ(fn, fm) + |β|ρ(gn, gm)
< |α| ε02(|α|+1)
+ |β| ε02(|β|+1)
≤ ε02
+ ε02
= ε0.
Thus, {αfn + βgn} is fundamental in the mean.
55
Next we will show {αfn + βgn} converges in measure to αf + βg. Since both {fn},
{gn} converge in measure to f and g respectively, we know
limn→∞
µ({x : |fn(x)− f(x)| ≥ ε02(|α|+1)
}) = 0,
limn→∞
µ({x : |gn(x)− g(x)| ≥ ε02(|β|+1)
}) = 0. (◦)
Note that if
|fn(x)− f(x)| < ε02(|α|+ 1)
and |gn(x)− g(x)| < ε02(|β|+ 1)
then
|(αfn(x) + βgn(x))− (αf(x) + βg(x))|
≤ |α||fn(x)− f(x)|+ |β||gn(x)− g(x)|
< |α| ε02(|α|+1)
+ |β| ε02(|β|+1)
≤ ε02
+ ε02
= ε0.
Therefore,
{x : |(αfn(x) + βgn(x))− (αf(x) + βg(x))| ≥ ε0}
⊆ {x : |fn(x)− f(x)| ≥ ε02(|α|+1)
} ∪ {x : |gn(x)− g(x)| ≥ ε02(|β|+1)
}.
Therefore,
0 ≤ µ({x : |(αfn(x) + βgn(x))− (αf(x) + βg(x))| ≥ ε0})
≤ µ({x : |fn(x)− f(x)| ≥ ε02(|α|+1)
}) + µ({x : |gn(x)− g(x)| ≥ ε02(|β|+1)
}).
Hence, remembering (◦),
limn→∞
µ({x : |(αfn(x) + βgn(x))− (αf(x) + βg(x))| ≥ ε0}) = 0,
so {αfn + βgn} converges in measure to αf + βg. Hence,
∫(αf + βg)(x) dµ(x) = lim
n→∞
∫(αfn + βgn)(x) dµ(x)
6.8= α lim
n→∞
∫fn(x) dµ(x) + β lim
n→∞
∫gn(x) dµ(x)
= α∫f(x) dµ(x) + β
∫g(x) dµ(x).
�
56
Lemma 6.14. Let f : X −→ [0,∞) be a bounded function. Then∫f(x) dµ(x) ≥ 0.
Proof. Refer to the proof of Lemma 6.10. If f(x) ≥ 0, then the sets Xn,k determined
there will be empty whenever αn,k < 0. Therefore, fn(x) ≥ 0 for all x ∈ X. Thus,∫f(x) dµ(x) ≥ 0.
�
57
7. Amenable Groups
We like to believe that we have invariant measures that can measure any set.
However, we have shown this belief is not true in R3. In this section, we develop
the theory of groups that do possess an invariant measure that can measure any set,
amenable groups.
Definition 7.1. Let G be a group. If there exists a finitely additive measure µ :
P (G) −→ [0, 1] such that µ(G) = 1 and for any A ⊆ G and g ∈ G, µ(g[A]) = µ(A),
then G is called amenable.
Proposition 7.2. If G is amenable, G cannot be paradoxical.
Proof. Let G be amenable, i.e. there exists a finitely additive measure µ : P (G) −→
[0, 1] such that µ(G) = 1 and for any A ⊆ G and g ∈ G, µ(gA) = µ(A). Suppose
now, arguing by contradiction, that G is paradoxical, i.e. there exists a partition A1,
A2, . . ., An of G and group translations T1, T2, . . ., Ti, . . ., Tn such that
T1[A1] ∪ T2[A2] ∪ . . . ∪ Ti[Ai] = G
and
Ti+1[Ai+1] ∪ Ti+2[Ai+2] ∪ . . . ∪ Tn[An] = G.
Then
µ(G) = µ(A1) + µ(A2) + . . .+ µ(An)
= µ(T1[A1]) + µ(T2[A2]) + . . .+ Ti[Ai] + Ti+1[Ai+1] + . . .+ µ(Tn[An])
= µ(G) + µ(G),
a contradiction. Thus, G cannot be paradoxical. �
Definition 7.3. Let G act on X, and let (X, µ,R) be a measure space.
(1) We say that µ is left-invariant with respect to G if
• g[A] ∈ R for each A ∈ R and g ∈ G, and
58
• µ(g[A]) = µ(A).
(2) We say that R is left-invariant with respect to G if
(g ∈ G & A ∈ R) =⇒ g[A] ∈ R.
Lemma 7.4. Let θ be a left-invariant measure on G. If f, h ∈ B(G), g ∈ G, and
h(x) = f(gx) then ∫f(x) dθ(x) =
∫h(x) dθ(x).
Proof. First note if f is a simple function, then h is as well, and easily,∫f(x) dθ(x) =∫
h(x) dθ(x). In the general case, pick a uniformly bounded sequence {fn} fundamen-
tal in the mean and convergent in measure to f . Let hn(x) = fn(gx). Then {hn}
is uniformly bounded, fundamental in the mean, converges in measure to h, and∫fn(x) dθ(x) =
∫hn(x) dθ(x). �
Theorem 7.5. Let G be an amenable group that acts on a space X. If there exists a
finitely-additive measure µ on P (X) that is left-invariant with respect to G, then for
any X ⊆ X such that µ(X) ∈ (0,∞), X is not G-paradoxical.
Proof. Let G be an amenable group and X be a G-set. Then there exists a left-
invariant finitely additive measure µ : P (G) −→ [0, 1] of total measure 1. Also, let µ
be a finitely additive measure µ : P (X) −→ [0,∞] that is left-invariant with respect
to G.
Suppose now towards contradiction that X ⊆ X such that µ(X) ∈ (0,∞) is G-
paradoxical, i.e. there exist
• a partition A1, A2, . . . , Ai, Ai+1, . . . , An of X, and
• elements g1, g2, . . . , gi, gi+1, . . . , gn of G
such that for some i < n,
• g1[A1], g2[A2], . . . , gi[Ai] is a partition of X, and
• gi+1[Ai+2], gi+1[Ai+2], . . . , gn[An] is a partition of X.
59
Then
µ(X) = µ(A1 ∪ A2 ∪ . . . ∪ An)
= µ(A1) + µ(A2) + . . .+ µ(An)
= µ(g1[A1]) + µ(g2[A2]) + . . . µ(gi[Ai]) + µ(gi+1[Ai+1]) + . . .+ µ(gn[An])
= µ(X) + µ(X)
= 2µ(X),
a contradiction, as µ(X) ∈ (0,∞). Thus, X ⊆ X such that µ(X) ∈ (0,∞) is not
G-paradoxical. �
The following proposition explains why we focus on finitely additive measures and
not σ-additive measures.
Proposition 7.6 (Giuseppe Vitali, [16]). Let G be an infinite group. Then there
exists no left-invariant σ-additive measure µ : P (G) −→ [0, 1] of total measure 1.
Proof. Let G be an infinite group, and let H ≤ G be a countably infinite subgroup,
so H = {h1, h2, . . .}. For any g′, h′ ∈ G, define the relation g′Rh′ if and only if
g′H = h′H. Now choose a selector S ⊆ G such that |S ∩ gH| = 1 for all gH ∈ G/H.
Note for any g′ ∈ G, g′ ∈ gH for some g ∈ S. Then
HS = {hS : h ∈ H} = G.
Now suppose, by way of contradiction, that a left-invariant σ-additive measure µ
exists on P (G) of total measure 1. Then
µ(G) =∞∑i=1
µ(hiS).
If µ(S) = 0, then µ(G) = 0. If µ(S) > 0, then µ(G) =∞. In either case, we arrive at
a contradiction, so there exists no left-invariant σ-additive measure on G with total
measure 1. �
60
Theorem 7.7. Let A0 be a subring of an algebra A and G be an amenable group
acting on X. Suppose also that A and A0 are left-invariant with respect to G. If
there exists a left-invariant finitely additive measure µ on A0, then there exists a
left-invariant finitely additive measure µ on A that extends µ.
Proof. Let µ be a finitely additive measure on A0 that is left-invariant with respect
to G. First, we will show there exists a finitely additive measure on A that extends
µ.
To start, we will consider the case if A is finite and proceed by induction on the
number of atoms of A. If A has one atom, then A = {∅,X}, so A0 = A and µ = µ.
Now let A have n atoms and assume for any algebra with less than n atoms that we
can extend its measure. Let B ∈ A0 be an atom of A0. Define C = B′ = X \B ∈ A.
Choose A0 ⊆ B such that A0 is an atom of A. Then A0 /∈ A(C) (see Definition 6.3
and recall A(C) is an algebra of subsets of C), so A(C) has less than n atoms, and
we can find a finitely additive measure ν : A(C) −→ [0, 1] that extends µ|A(C)∩A0 by
our inductive hypothesis (note A(C) ∩ A0 is a subring of A(C)). Note that every
atom A ∈ A, A ⊆ C or A ⊆ B. So for any atom A ∈ A, define
µatom(A) =
ν(A) if A ∈ A(C)
µ(B) if A = A0
0 if A ⊆ B but A 6= A0
Finally, for any X ∈ A, define
µ(X) =∑{µatom(A) : A ⊆ X, and A is an atom of A}
To show µ is finitely additive, let A1, A2, . . . , Ak ∈ A be pairwise disjoint. Then
µ(A1) + µ(A2) =∑{µatom(X) : X ⊆ A1, and X is an atom of A}
+∑{µatom(X) : X ⊆ A2, and X is an atom of A}
=∑{µatom(X) : X ⊆ A1 ∪ A2, and X is an atom of A} = µ(A1 ∪ A2).
61
To show µ extends µ, let D ∈ A0. Then either D∩B = ∅ or B ⊆ D. If D∩B = ∅,
then D ∈ A(C), so µ(D) = ν(D) = µ(D). If instead B ⊆ D, then
µ(D) = µ(B) + µ(D \B) = µ(A0) + µ(B \ A0) + ν(D \B)
= µ(B) + µ(D \B) = µ(D).
Thus µ is a finitely additive measure on A that extends µ.
Consider the case when A is infinite. Let [0,∞]A be the set of all functions f :
A −→ [0,∞] and let it be equipped with the product space topology so that the basic
open sets are of the form
{f ∈ [0,∞]A : f(A1) ∈ U1, f(A2) ∈ U2, . . . , f(An) ∈ Un}
where A1, A2, . . . , An ∈ A and U1, U2, . . . , Un ⊆ [0,∞] are open.
By Tychonoff’s Theorem [10, p.234], we know that [0,∞]A is compact, and thus
[0,∞]A has the finite intersection property, i.e. if F is a family of closed subsets of
[0,∞]A and for any F1, F2, . . . , Fn ∈ F , F1 ∩ F2 ∩ . . . ∩ Fn 6= ∅, then⋂F 6= ∅.
For all finite subalgebras C of A, define
M(C) = {ν ∈ [0,∞]A : ν|C is a measure and ν extends µ|C∩A0}
We will show that each M(C) is a closed subset of [0,∞]A and that the family FM of
all of these sets satisfies the finite intersection property.
To show M(C) is closed, let ν ∈ [0,∞]A \M(C). Then either ν does not extend
µ|C∩A0 or ν|C is not a measure.
If ν does not extend µ|C∩A0 , then for some A ∈ C ∩A0, ν(A) 6= µ(A). Then choose
an open set U ⊆ [0,∞] such that ν(A) ∈ U but µ(A) /∈ U . Then {f ∈ [0,∞]A :
f(A) ∈ U} is an open set disjoint from M(C), and ν ∈ {f ∈ [0,∞]A : f(A) ∈ U}.
If ν|C is not a measure, then there exist A1, A2 ∈ C such that A1 ∩ A2 = ∅ and
ν(A1) + ν(A2) 6= ν(A1 ∪ A2). We will define open sets U1, U2, U2 ⊆ [0,∞] depending
on one of the following three sub-cases.
62
If ν(A1), ν(A2), ν(A1 ∪ A2) ∈ [0,∞), let
ε = |ν(A1 ∪ A2)− [ν(A1) + ν(A2)]|
and let
U1 = (ν(A1)− ε/4, ν(A1) + ε/4)
U2 = (ν(A2)− ε/4, ν(A2) + ε/4)
U3 = (ν(A1 ∪ A2)− ε/4, ν(A1 ∪ A2) + ε/4).
If ν(A1), ν(A2) ∈ [0,∞), but ν(A1 ∪ A2) =∞, let
U1 = (ν(A1)− 1/4, ν(A1) + 1/4)
U2 = (ν(A2)− 1/4, ν(A2) + 1/4)
U3 = (ν(A1) + ν(A2) + 1,∞].
If ν(A1 ∪ A2) ∈ [0,∞), but ν(Ai) =∞ where i ∈ {1, 2}, let
Ui = (ν(A1 ∪ A2) + 1,∞]
U3−i = [0,∞]
U3 = [0, ν(A1 ∪ A2) + 1/2).
In any case, {f ∈ [0,∞]A : f(A1) ∈ U1, f(A2) ∈ U2, f(A1∪A2) ∈ U3} is an open set
disjoint from M(C), and ν ∈ {f ∈ [0,∞]A : f(A1) ∈ U1, f(A2) ∈ U2, f(A1∪A2) ∈ U3}.
Thus, [0,∞]A \M(C) is open, and M(C) is closed.
Now observe that M(C) is non-empty as we have already shown that since C is
finite, µ|C∩A0 can be extended to C.
Also note that if C1, C2, . . . , Cn are finite subalgebras of A, then C1 ∪ C2 ∪ . . . ∪ Cn
generate a finite subalgebra C∗ of A. Then
∅ 6= M(C∗) ⊆M(C1) ∩M(C2) ∩ . . . ∩M(Cn).
Thus, the family FM satisfies the finite intersection property, so⋂FM 6= ∅. Hence,
there exists a finitely additive measure µ on A that extends µ.
63
Now that we have shown there exists a finitely additive measure on A that extends
µ, we need to show there is a measure that is also left-invariant with respect to the
amenable group G. Let ν be a finitely additive measure that extends µ and let θ be
a finitely additive measure on G of total measure one. By our assumption, we know
µ is left-invariant with respect to G, i.e. for any A ∈ A0 and g ∈ G, µ(g[A]) = µ(A).
Now for any set B ∈ A, define fB : G −→ [0,∞] by
fB(g) = ν(g−1B).
Then define
µ(B) =
∫fBdθ.
Note fB ≥ 0 so µ(B) ≥ 0. Also, for any A1, A2 ∈ A such that A1 ∩ A2 = ∅,
fA1(g) + fA2(g) = ν(g−1[A1]) + ν(g−1[A2]) = ν(g−1[A1] ∪ g−1[A2])
= ν(g−1[A1 ∪ A2]) = fA1∪A2(g)
so
µ(A1) + µ(A2) =∫fA1 dθ +
∫fA2 dθ =
∫fA1 + fA2 dθ
=∫fA1∪A2 dθ = µ(A1 ∪ A2).
Thus, µ is a measure.
To show µ is left-invariant, let A ∈ A and let h ∈ G. Then
fh[A](g) = ν(g−1h[A]) = ν((h−1g)−1[A]) = fA(h−1g)
so, remember Lemma 7.4,
µ(h[A]) =
∫fh[A](g) dθ(g) =
∫fA(h−1g) dθ(g)
=
∫fA(g) dθ(g) = µ(A).
Thus, µ is left-invariant.
64
Finally, to show µ extends µ, let A ∈ A0. Then
fA(g) = ν(g−1A) = µ(g−1A) = µ(A)
so
µ(A) =
∫fAdθ =
∫µ(A)dθ = µ(A).
Thus, µ is a left-invariant, finitely additive measure on A that extends µ. �
Theorem 7.8. Let G be a group. If G is finite, then G is amenable.
Proof. Let G be a group and |G| = n. For A ⊆ G, define µ : G −→ [0, 1] by
µ(A) =|A|n.
First note µ(G) = nn
= 1. Now to show that µ is finitely additive, let A1, A2 ⊆ G
such that A1 ∩ A2 = ∅. Then
µ(A1) + µ(A2) =|A1|n
+|A2|n
=|A1 ∪ A2|
n= µ(A1 ∪ A2)
Finally, to show µ is left-invariant, let A ⊆ G and g ∈ G. Then
µ(gA) =|gA|n
=|A|n
= µ(A)
So µ is a left-invariant finitely additive measure on G of total measure 1, and thus,
G is amenable. �
Theorem 7.9. Let H be a subgroup of a group G. If G is amenable, then H is
amenable.
Proof. Let H be a subgroup of a group G and let G be amenable. Then there exists
a left-invariant finitely additive measure µ on G of total measure 1. Let M be a
selector from the right cosets of H so that for all g ∈ G, |M ∩Hg| = 1. Then define
65
ν : P (H) −→ [0, 1] by
ν(A) = µ
(⋃g∈M
Ag
).
Observe
ν(H) = µ
(⋃g∈M
Hg
)= µ(G) = 1.
To show that ν is finitely additive, let A1, A2 ⊆ H such that A1 ∩ A2 = ∅. Then the
sets⋃g∈M A1g and
⋃g∈M A2g are disjoint, and
ν(A1) + ν(A2) = µ
(⋃g∈M
A1g
)+ µ
(⋃g∈M
A2g
)= µ
(⋃g∈M
A1g +⋃g∈M
A2g
)
= µ
(⋃g∈M
(A1g ∪ A2g)
)= µ
(⋃g∈M
(A1 ∪ A2)g
)= ν(A1 ∪ A2).
Finally, to show ν is left-invariant, let A ⊆ H and h ∈ G. Then
ν(hA) = µ
(⋃g∈M
(hA)g
)= µ
(⋃g∈M
h(Ag)
)= µ
(⋃g∈M
Ag
)= ν(A).
So ν is a left-invariant finitely additive measure on H of total measure 1, and thus,
H is amenable. �
Theorem 7.10. Let H be a normal subgroup of a group G. If G is amenable, then
G/H is amenable.
Proof. Let H be a normal subgroup of an amenable group G, and let µ be a left-
invariant finitely additive measure on G of total measure 1. Define ν : P (G/H) −→
[0, 1] by
ν(B) = µ(⋃B).
Observe
ν(G/H) = µ(⋃
G/H)
= µ(G) = 1.
66
To show that ν is finitely additive, let B1,B2 ⊆ G/N such that B1 ∩ B2 = ∅. Then⋃B1 ∩
⋃B2 = ∅ (as cosets form a partition of G), and
ν(B1) + ν(B2) = µ (⋃B1) + µ (
⋃B2)
= µ (⋃B1 +
⋃B2)
= µ (⋃
(B1 ∪ B2)) = ν(B1 ∪ B2).
Finally, to show ν is left-invariant, let B ⊆ G/H and h ∈ G. Let C be a selector from
B. Then, in particular, B = {gH : g ∈ C}, and
ν((hH)B) = µ(⋃
(hH)B)
= µ
(⋃g∈C
hHg
)= µ
(h⋃g∈C
Hg
)
= µ
(h⋃g∈C
gH
)= µ
(⋃g∈C
gH
)= µ
(⋃B)
= ν(B).
So ν is a left-invariant finitely additive measure on G/H of total measure 1, and thus,
G/N is amenable. �
Theorem 7.11. Let H be a normal subgroup of a group G. If H and G/H are
amenable, then G is amenable.
Proof. Let H be a normal subgroup of a group G and suppose also that H and G/H
are amenable. Then let ν1 and ν2 be left-invariant finitely additive measures of total
measure 1 on H and G/H respectively. For A ⊆ G, define fA : G −→ [0, 1] by
fA(g) = ν1(H ∩ g−1A).
Suppose g1H = g2H. Then g−12 g1H = H so g−1
2 g1 = h ∈ H. Then
fA(g2) = ν1(H ∩ g−12 A) = ν1(hH ∩ hg−1
1 A)
= ν1(h(H ∩ g−11 A)) = ν1(H ∩ g−1
1 A) = fA(g1).
Thus, fA is constant on each coset of H and we may define f ∗A : G/H −→ [0, 1] by
f ∗A(gH) = fA(g).
67
Now define µ : P (G/H) −→ [0, 1] by
µ(A) =
∫f ∗A(gH) dν2(gH).
Observe fG(g) = 1 for all g ∈ G, and hence, f ∗G(gH) = 1 for all gH ∈ G/H. Then
µ(G) =
∫f ∗G(gH) dν2(gH) =
∫1 dν2(gH) = ν2(G/H) = 1.
To show that ν is finitely additive, let A1, A2 ⊆ G such that A1 ∩ A2 = ∅. Note if
A1 ∩ A2 = ∅ then g−1A1 ∪ g−1A2 = ∅, so
fA1(g) + fA2(g) = ν1(N ∩ g−1A1) + ν1(N ∩ g−1A2)
= ν1((N ∩ g−1A1) ∪ (N ∩ g−1A2)) = ν1(N ∩ (g−1A1 ∪ g−1A2))
= ν1(N ∩ g−1(A1 ∪ A2)) = fA1∪A2(g).
Then
µ(A1) + µ(A2) =∫f ∗A1
(gH) dν2(gH) +∫f ∗A2
(gH) dν2(gH)
=∫
(f ∗A1(gH) + f ∗A2
(gH)) dν2(gH) =∫f ∗A1∪A2
(gH) dν2(gH) = µ(A1 ∪ A2).
Finally, to show ν is left-invariant, let A ⊆ G and h ∈ G. Note
fhA(g) = ν1(H ∩ g−1hA) = ν1(H ∩ (g−1g)−1A) = fA(h−1g).
Then f ∗hA(gH) = f(A((h−1g)H) = f ∗A(h−1H)(gH)), and
µ(hA) =
∫f ∗hA(gH) dν2(gH) =
∫f ∗hA((h−1H)(gH)) dν2(gH)
=
∫f ∗A(gH) dν2(gH) = µ(A).
So µ is a left-invariant finitely additive measure on G of total measure 1, and thus,
G is amenable. �
Theorem 7.12. If G is an Abelian group, then G is amenable.
68
Proof. Let G be an Abelian group and X a finite subset of G. Let [0, 1]P (G) be the
space of all functions f : P (G) −→ [0, 1] equipped with the product topology. Thus,
the basic open sets are of the form
{f ∈ [0, 1]P (G) : f(A1) ∈ U1, f(A2) ∈ U2, . . . , f(An) ∈ Un}
where A1, A2, . . . , An ∈ P (G) and U1, U2, . . . , Un are open subsets of [0, 1]. By Ty-
chonoff’s Theorem [10, p.234], the space [0, 1]P (G) is compact, and therefore has the
finite intersection property. For ε > 0 and a finite set X ⊆ G, we define
Mε,X = {f ∈ [0, 1]P (G) : f(G) = 1, for any A1, A2 ⊆ G such that A1 ∩ A2 = ∅,
f(A1) + f(A2) = f(A1 ∪ A2),
and for any A ⊆ G and g ∈ X, |f(A)− f(gA)| ≤ ε}.
Let
F = {Mε,X : ε > 0, X ⊆ G,X is finite}.
Our goal is to show that F satisfies the conditions for the finite intersection property.
To show Mε,X is non-empty, choose N ≥ 2ε. Fix an enumeration of elements of
X = {g1, g2, . . . , gn} and for any subset A ⊆ G, define
YA = {(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, 2, . . . , n}, gi11 gi22 . . . ginn ∈ A}.
Then define
µε(A) = |YA|/Nn.
First, observe YG = {(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, 2, . . . , n}
µε(G) = |YG|Nn = |{(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, 2, . . . , n}}|/Nn = Nn
Nn = 1.
69
To show that µε is finitely additive, let A1, A2 ⊆ G such that A1 ∩ A2 = ∅. Then
YA1 ∪ YA2 = YA1∪A2 , and hence
µε(A1) + µε(A2) =|YA1
|Nn +
|YA2|
Nn
=|YA1∪A2
|Nn = µε(A1 ∪ A2).
Finally, to show that µε is almost left-invariant with respect to elements of X, let
A ⊆ G and gk ∈ X. Observe now that, as G is Abelian,
YgkA ={
(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, 2, . . . , n}, gi11 gi22 . . . ginn ∈ gkA}
={
(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, 2, . . . , n}, gi11 gi22 . . . gik−1 . . . ginn ∈ A}.
Therefore,
YA4 YgkA = (YA \ YgkA) ∪ (YgkA \ YA) =
{(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, . . . , n}, gi11 gi22 . . . gik−1
k−1 gNk g
ik+1
k+1 . . . ginn ∈ A}
∪{(i1, i2, . . . , in) : ij ∈ {1, 2, . . . , N}, j ∈ {1, . . . , n}, gi11 gi22 . . . gik−1
k−1 gik+1
k+1 . . . ginn ∈ A}.
Thus,
|µε(A)− µε(gkA)| =∣∣∣∣ |YA|Nn
− |YgkA|Nn
∣∣∣∣ =
∣∣|YA| − |YgkA|∣∣Nn
≤ |YA4 YgkA|Nn
=2Nn−1
Nn=
2
N< ε.
Thus, µε ∈Mε,X , so Mε,X is not empty.
To show Mε,X is closed, let ν ∈ [0, 1]P (G)\Mε,X . Then ν does not have total measure
1, ν is not finitely additive, or ν is not almost invariant with respect to elements of
X.
If ν(G) < 1, then
{f ∈ [0, 1]P (G) : f(G) ∈
[0, 1− 1− ν(G)
2
)}is an open set disjoint from Mε,X and ν belongs to it.
70
If ν is not finitely additive, then there exist A1, A2 ⊆ G such that A1 ∩A2 = ∅ and
ν(A1) + ν(A2) 6= ν(A1 ∪ A2). Then let
ε0 =∣∣[ν(A1) + ν(A2)]− ν(A1 ∪ A2)
∣∣.Then {
f ∈ [0, 1]P (G) : ν(A1) ∈(ν(A1)− ε0
4, ν(A1) + ε0
4
),
ν(A2) ∈(ν(A2)− ε0
4, ν(A2) + ε0
4
),
ν(A1 ∪ A2) ∈(ν(A1 ∪ A2)− ε0
4, ν(A1 ∪ A2) + ε0
4
)}is an open set disjoint from Mε,X and ν belongs to it.
If ν is not almost invariant with respect to elements of X, then there exist A ⊆ G
and gk ∈ X such that |ν(A)− ν(gkA)| > ε. Let ε0 = |ν(A)− ν(gkA)| − ε > 0. Then
{f ∈ [0, 1]P (G) : ν(A) ∈
(ν(A)− ε
4, ν(A) + ε
4
), ν(gkA) ∈
(ν(gkA)− ε
4, ν(gkA) + ε
4
)}is an open set disjoint from Mε,X and ν belongs to it.
In any case, ν is contained in an open set disjoint from Mε,X , so Mε,X is closed.
Finally, to show that any finite intersection is non-empty, let
Mε1,X1 ,Mε2,X2 , . . . ,Mεn,Xn ∈ F ,
where εi > 0 and Xi ∈ X for all i ∈ {1, 2, . . . , n}. Then for ε = min{ε1, ε2, . . . , εn}
and X = X1 ∪X2 ∪ . . . ∪Xn,
∅ 6= Mε,X ⊆Mε1,X1 ∩Mε2,X2 ∩ . . . ∩Mεn,Xn .
Hence we have shown that the family F of all sets Mε,X satisfies the conditions for
the finite intersection property, so
⋂F 6= ∅.
71
Since every element of⋂F is a left-invariant finitely additive measure µ on G of total
measure 1, the group G is amenable. �
During the preceding proof, it is noteworthy that we only had to consider finitely
generated subgroups of G. The following theorem states that explicitly.
Theorem 7.13. If G =⋃α∈I Gα where Gα is an amenable subgroup of G and for any
α, β ∈ I, there exists γ ∈ I such that Gα ≤ Gγ and Gβ ≤ Gγ, then G is amenable.
Proof. Let G =⋃α∈I Gα where Gα is an amenable subgroup of G and for any α, β ∈ I,
there exists γ ∈ I such that Gα ≤ Gγ and Gβ ≤ Gγ. As each Gα is amenable, let µα
be a left-invariant finitely additive measure on Gα of total measure 1. Let [0, 1]P (G)
be the space of all functions f : P (G) −→ [0, 1] equipped with the product topology
so that the basic open sets are of the form
{f ∈ [0, 1]P (G) : f(A1) ∈ U1, f(A2) ∈ U2, . . . , f(An) ∈ Un}
where A1, A2, . . . , An ∈ P (G) and U1, U2, . . . , Un are open subsets of [0, 1]. By Ty-
chonoff’s Theorem, [0, 1]P (G) is compact, and therefore has the finite intersection
property. For α ∈ I, let
Mα = {f ∈ [0, 1]P (G) : f(G) = 1, for any A1, A2 ⊆ G such that A1 ∩ A2 = ∅,
f(A1) + f(A2) = f(A1 ∪ A2),
and for any g ∈ Gα and A ⊆ G, f(gA) = f(A)}.
We must now show that the family FM of all sets Mα satisfies the conditions
for the finite intersection property. Let A ⊆ G. Then define fα(A) = µα(A ∩ Gα).
Clearly fα(G) = µα(Gα) = 1. To show fα is finitely additive, let A1, A2 ⊆ G such that
A1∩A2 = ∅. Then (A1∪A2)∩Gα = (A1∩Gα)∪(A2∩Gα) and (A1∩Gα)∩(A2∩Gα) = ∅.
So
fα(A1 ∪ A2) = µα((A1 ∪ A2) ∩Gα = µα(A1 ∩Gα) + µα(A2 ∩Gα = fα(A1) + fα(A2).
72
To show fα is left-variant with respect to G, for g ∈ Gα,
fα(gA) = µα(gA ∩Gα) = µα(gA ∩ gGα)
= µα(g(A ∩Gα)) = µα(A ∩Gα) = fα(A).
Thus, fα ∈Mα, so Mα is non-empty.
To show Mα is closed, let ν ∈ [0, 1]P (G) \Mα. Then either ν(G) < 1, ν(A ∪ B) 6=
ν(A) + ν(B) for some disjoint A,B ⊆ G, or ν(gA) 6= ν(A). If ν(G) < 1, then
{f ∈ [0, 1]P (G) : ν(G) ∈
[0, ν(G) +
1− ν(G)
2
]}is an open set disjoint from Mα and ν belongs to it. If ν is not finitely additive, then
there exist A1, A2 ⊆ G such that A1 ∩A2 = ∅ and ν(A1) + ν(A2) 6= ν(A1 ∪A2). Then
let
ε0 =∣∣[ν(A1) + ν(A2)]− ν(A1 ∪ A2)
∣∣.Then {
f ∈ [0, 1]P (G) : ν(A1) ∈(ν(A1)− ε0
4, ν(A1) + ε0
4
),
ν(A2) ∈(ν(A2)− ε0
4, ν(A2) + ε0
4
),
ν(A1 ∪ A2) ∈(ν(A1 ∪ A2)− ε0
4, ν(A1 ∪ A2) + ε0
4
)}is an open set disjoint from Mα and ν belongs to it. If for some g ∈ G and A ⊆ G,
ν(gA) 6= ν(A), then let ε = |ν(gA)− ν(A)|. Then
{f ∈ [0, 1]P (G) : ν(gA) ∈
(ν(gA)− ε
4, ν(gA) + ε
4
), ν(A) ∈
(ν(A)− ε
4, ν(A) + ε
4
)}is an open set disjoint from Mα and ν belongs to it. Thus Mα is closed.
Finally, for any α1, α2, . . . , αn ∈ I there exists a γ ∈ I such that Gα1 ≤ Gγ,
Gα2 ≤ Gγ, . . ., Gαn ≤ Gγ, so
∅ 6= Mγ ⊆Mα1 ∩Mα2 ∩ . . . ∩Mαn ,
73
so ⋂k∈{1,2,...,n}
Mαk6= ∅.
Thus, the family FM satisfies the conditions for the finite intersection property, so
⋂FM 6= ∅.
Thus there exists a µ : G −→ [0, 1] that is left-invariant, finitely additive, and has
total measure 1, so G is amenable. �
74
8. Examples of Amenable Groups
In this final section, we will show that the group of isometries in R and R2 are
amenable, and by extending measures in R and R2, we will show that no bounded
subset of in R and R2 with non-empty interior is paradoxical.
8.1. Isometries in R and R2.
Definition 8.1. An isometry f : Rn −→ Rn is a distance-preserving function.
Therefore, an isometry f : Rn −→ Rn is of the form
f
x1
x2
...
xn
=
a1,1 a1,2 · · · a1,n
a2,1 a2,2 · · · a2,n
......
. . ....
an,1 an,2 · · · an,n
x1
x2
...
xn
+
b1
b2
...
bn
,
where the determinant of a1,1 a1,2 · · · a1,n
a2,1 a2,2 · · · a2,n
......
. . ....
an,1 an,2 · · · an,n
is -1 or 1 [6, p.312].
Let Gn be the group of isometries in Rn. Also, let Tn be the group of translations
of Gn, i.e. those isometries of the form
t
x1
x2
...
xn
=
1 0 · · · 0
0 1 · · · 0
......
. . ....
0 0 · · · 1
x1
x2
...
xn
+
b1
b2
...
bn
.
Furthermore, we will also be concerned with those isometries that preserve orien-
tation. We will denote the group of operations that preserve orientation by SGn, and
75
we note
SGn = {f ∈ Gn : f(x) = Ax+ t, det(A) = 1}.
Example 8.2. The group of isometries G1 is amenable.
Proof. First we note
G1 = {f : f(x) = ax+ b, |a| = 1, b ∈ R}.
Also,
T1 = {f : f(x) = x+ b}.
We will argue that T1 is a normal subgroup of G1 by showing that fT1 = T1f for any
f ∈ G1. First, let f ′ ∈ G1 so f ′(x) = a′x+ b′, and let f1 ∈ T1 so f1(x) = x+ b. Then
f ′(f1(x)) = a′(x+ b) + b′
= (a′x+ b′) + a′b
= f2(f ′(x)),
where f2(x) = x+ a′b, so f2 ∈ T1. Thus f ′T1 ⊆ T1f′.
Similarly, let f ′ ∈ G1 so f ′(x) = a′x+ b′ and let f1 ∈ T1 so f1(x) = x+ b. Then
f1(f ′(x)) = (a′x+ b′) + b
= a′(x+ a′b) + b′
= f ′(f2(x)),
where f2(x) = x+ a′b, so f2 ∈ T1. Thus T1f′ ⊆ f ′T1, and hence f ′T1 = T1f
′, so T1 is
a normal subgroup of G1.
Observe now that T1∼= R by the isomorphism φ : R −→ T1 defined by φ(n) = x+n.
As R is an Abelian group, T1 is also Abelian, so by Theorem 7.12, T1 is amenable.
Next observe G1/T1 = {x,−x}. Therefore G1/T1∼= Z2, so G1/T1 is also Abelian,
and therefore, G1/T1 is amenable.
Finally, we may conclude G1 is amenable by Theorem 7.11 as T1 and G1/T1 are
amenable. �
76
Example 8.3. The group of isometries G2 is amenable.
Proof. To show G2 is amenable, we will argue that SG2 and G2/SG2 are amenable.
First, observe
G2 =
cos θ − sin θ
sin θ cos θ
xy
+
ab
: 0 ≤ θ < 2π, a, b ∈ R
∪
cos θ sin θ
sin θ − cos θ
xy
+
ab
: 0 ≤ θ < 2π, a, b ∈ R
[6, p.314].
Also note
SG2 =
cos θ − sin θ
sin θ cos θ
xy
+
ab
: 0 ≤ θ < 2π, a, b ∈ R
.
To show SG2 is amenable, we will argue that T2 and SG2/T2 are amenable. First
we show that T2 is a normal subgroup of SG2 by showing that fT2 = T2f for any
f ∈ SG2. First, let f ′ ∈ SG2 and f1 ∈ T2. Then
f ′
xy
=
cos θ − sin θ
sin θ cos θ
xy
+
a′b′
,
and
f1
xy
=
xy
+
ab
.
Then
f ′ ◦ f1
xy
=
cos θ − sin θ
sin θ cos θ
xy
+
ab
+
a′b′
=
cos θ − sin θ
sin θ cos θ
xy
+
cos θ − sin θ
sin θ cos θ
ab
+
a′b′
=
cos θ − sin θ
sin θ cos θ
xy
+
a′b′
+
cos θ sin θ
− sin θ cos θ
ab
= f2 ◦ f ′xy
,
77
where
f2
xy
=
xy
+
cos θ sin θ
− sin θ cos θ
ab
.
Thus f ′T2 ⊆ T2f′.
Similarly, let f ′ ∈ SG2 and f1 ∈ T2, so
f ′
xy
=
cos θ − sin θ
sin θ cos θ
xy
+
a′b′
and
f1
xy
=
xy
+
ab
.
Then
f1 ◦ f ′xy
=
cos θ − sin θ
sin θ cos θ
xy
+
a′b′
+
ab
=
cos θ − sin θ
sin θ cos θ
xy
+
cos θ − sin θ
sin θ cos θ
cos θ sin θ
− sin θ cos θ
ab
+
a′b′
=
cos θ − sin θ
sin θ cos θ
xy
+
cos θ sin θ
− sin θ cos θ
ab
+
a′b′
= f ′ ◦ f2
xy
,
where
f2
xy
=
xy
+
cos θ sin θ
− sin θ cos θ
ab
.
Thus T2f′ ⊆ f ′T2, and hence f ′T2 = T2f
′, so T2 is a normal subgroup of SG2.
78
We want to argue now that T2 is amenable. If f1, f2 ∈ T2, then
f1
xy
=
xy
+
a1
b1
f2
xy
=
xy
+
a2
b2
.
Then
f1 ◦ f2
xy
=
xy
+
a2
b2
+
a1
b1
=
xy
+
a1
b1
+
a2
b2
= f2 ◦ f1
xy
.
Thus, T2 is Abelian, and by Theorem 7.12, T2 is amenable.
Next, we want to argue that SG2/T2 is amenable. Observe now that SG2/T2∼= SO2
where
SO2 =
cos θ − sin θ
sin θ cos θ
xy
: 0 ≤ θ < 2π
.
by the isomorphism φ : SG2/T2 −→ SO2 defined by
φ
cos θ − sin θ
sin θ cos θ
xy
+
ab
T2
=
cos θ − sin θ
sin θ cos θ
xy
.
Observe now that SO2 is Abelian ascos θ1 − sin θ1
sin θ1 cos θ1
cos θ2 − sin θ2
sin θ2 cos θ2
=
cos θ1 cos θ2 − sin θ1 sin θ2 − cos θ1 sin θ2 − sin θ1 cos θ2
sin θ1 cos θ2 + cos θ1 sin θ2 − sin θ1 sin θ2 + cos θ1 cos θ2
=
cos θ2 − sin θ2
sin θ2 cos θ2
cos θ1 − sin θ1
sin θ1 cos θ1
.
Thus, SG2/T2 is Abelian as well, and by Theorem 7.12, SG2/T2 is amenable.
79
Then, by Theorem 7.11, SG2 is amenable, as T2 and SG2/T2 are amenable.
Now, we will argue that SG2 is a normal subgroup of G2 by showing that fSG2 =
SG2f for any f ∈ G2. So let f ∈ G2. Then either
f ∈ SG2 =
cos θ − sin θ
sin θ cos θ
xy
+
ab
: 0 ≤ θ < 2π, a, b ∈ R
or
f ∈
cos θ sin θ
sin θ − cos θ
xy
+
ab
: 0 ≤ θ < 2π, a, b ∈ R
As SG2 is a group, if f ∈ SG2, then fSG2 = SG2f . So suppose f /∈ SG2 and let
f ′ ∈ SG2. Then
f
xy
=
cos θ1 sin θ1
sin θ1 − cos θ1
xy
+
ab
f ′
xy
=
cos θ2 − sin θ2
sin θ2 cos θ2
xy
+
a′b′
.
Then
f ◦ f ′xy
=
cos θ1 sin θ1
sin θ1 − cos θ1
cos θ2 − sin θ2
sin θ2 cos θ2
xy
+
a′b′
+
ab
=
cos θ1 sin θ1
sin θ1 − cos θ1
cos θ2 − sin θ2
sin θ2 cos θ2
xy
+
a′′b′′
=
cos θ1 cos θ2 + sin θ1 sin θ2 sin θ1 cos θ2 − cos θ1 sin θ2
sin θ1 cos θ2 − cos θ1 sin θ2 −(cos θ1 cos θ2 + sin θ1 sin θ2)
xy
+
a′′b′′
= A
xy
+
a′′b′′
.
80
Observe
det(A) = −(cos2 θ1 cos2 θ2 + 2 cos θ1 cos θ2 sin θ1 sin θ2 + sin2 θ1 sin2 θ2)
−(sin2 θ1 cos2 θ2 − 2 cos θ1 cos θ2 sin θ1 sin θ2 + cos2 θ1 sin2 θ2)
= −(cos2 θ1 cos2 θ2 + cos2 θ1 sin2 θ2) + sin2 θ1 cos2 θ2 + sin2 θ1 sin2 θ2)
= −[cos2 θ1(cos2 θ2 + sin2 θ2) + sin2 θ1(cos2 θ2 + sin2 θ2)]
= −(cos2 θ1 + sin2 θ1)
= −1.
Thus, f ◦ f ′ ∈ G2 \ SG2, and as SO2 is Abelian, f ′ ◦ f ∈ G2 \ SG2. In any case,
fSG2 = SG2f , so SG2 is a normal subgroup of G2.
Finally, observe
G2/SG2 =
1 0
0 1
,
1 0
0 −1
∼= Z2.
Thus, G2/SG2 is Abelian as Z2 is Abelian, and therefore, by Theorem 7.12, G2/SG2
is amenable.
Thus, by Theorem 7.11, G2 is amenable, as SG2 and G2/SG2 are amenable. �
8.2. No Paradoxes in R or R2. Having shown that G1 and G2 are amenable, our
goal now is to find left-invariant measures on P (R) and P (R2) to show no bounded
subset with non-empty interior of R1 or R2 can be paradoxical with respect to G1
and G2 respectively.
Let R1 ⊆ P (R) be the ring formed by ∅ and the union of any finite number of
disjoint intervals (open, closed, or half open and closed).
Proposition 8.4. R1 is a ring.
Proof. Immediate, as the union of two intervals is an interval or a union of intervals,
and the difference of two intervals is either an interval, the union of intervals, or
empty. �
81
Definition 8.5. Let µ1 : R1 −→ [0,∞] be defined by
µ1(I1 ∪ I2 ∪ . . . ∪ In) =n∑i=1
(bi − ai),
where Ii has endpoints ai ≤ bi and Ii ∩ Ij = ∅ for i 6= j.
Observe
• for any X, Y ∈ R1 such that X ∩ Y = ∅,
µ(X) + µ(Y ) = µ(X ∪ Y )
• for any g ∈ G1 and X ∈ R1,
µ(gX) = µ(X)
Corollary 8.6. No bounded subset X ⊆ R with non-empty interior is paradoxical.
Proof. By Theorem 7.7, there exists a left-invariant finitely additive measure µ :
P (R) −→ [0,∞] that extends µ1 defined in 8.5. Let X ⊆ R be a bounded set with
non-empty interior. Then for some a, b, c, d ∈ R,
[a, b] ⊆ X ⊆ [c, d].
Thus,
0 < µ1([a, b]) ≤ µ1(X) ≤ µ1([c, d]) <∞,
so µ1(X) ∈ (0,∞). Hence, by Theorem 7.5, X is not paradoxical. �
Corollary 8.7. No bounded subset X ⊆ R2 with non-empty interior is paradoxical.
Proof. Let c be the Jordan measure, and let M(c) ⊆ P (R2) be the class of all Jordan
measurable sets. Then M(c) forms a sub-ring of P (X). Also, c is left-invariant with
respect to G2 (see [7]). Thus, by Theorem 7.7, there exists a left-invariant finitely
additive measure c : P (R2) −→ [0,∞] that extends c. So let X ⊆ R2 be a bounded
82
set with non-empty interior. Then
[a1, b1]× [a2, b2] ⊆ X ⊂ [c1, d1]× [c2, d2].
Thus,
0 < c([a1, b1]× [a2, b2]) ≤ c(X) ≤ µ1([c1, d1]× [c2, d2]) <∞,
so c(X) ∈ (0,∞). Hence, by Theorem 7.5, X is not paradoxical. �
83
9. Conclusion
In this thesis, we have given a detailed decomposition of the sphere into two iden-
tical spheres. This decomposition of a sphere runs contrary to every intuitive notion
we have concerning volume, and therefore is aptly named the Banach-Tarski paradox.
We are able to reconcile this seemingly false result by acknowledging the existence of
non-measurable sets. In 3-dimensional space, sets can be formed, as we have shown,
that are not measurable. The second half of this thesis showed that a similar paradox
is not possible on the number line or the plane because every set with non-empty
interior can be measured.
84
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