Making Mountains Out of Molehills
The Banach-Tarski Paradox
By
Aaron Sinz
Paul Miller
Brian Sikora
Jay Laporte
Bob Kronberger
Table of Contents
Introduction................................................................................................................................. 3
The Banach-Tarski Theorem .................................................................................................. 4
Definitions................................................................................................................................... 4
Rigid Motion........................................................................................................................... 4
Partitions of Sets ..................................................................................................................... 4
Hausdorff Paradox .................................................................................................................. 5
Piecewise Congruence ............................................................................................................ 5
Schröder-Bernstein Theorem ...................................................................................................... 7
Cardinality............................................................................................................................... 8
Schröder-Bernstein Theorem ................................................................................................ 10
Finite Case ........................................................................................................................ 11
Infinite Case ...................................................................................................................... 12
The Axiom of Choice................................................................................................................ 19
Definition .............................................................................................................................. 19
Schools of through about the Axiom of Choice.................................................................... 20
Platonism........................................................................................................................... 20
Constructivism .................................................................................................................. 21
Formalism ......................................................................................................................... 21
Should you accept the Axiom of Choice?............................................................................. 21
Against: ............................................................................................................................. 21
For:.................................................................................................................................... 21
Conclusion ................................................................................................................................ 22
References................................................................................................................................. 22
2
Introduction
The topic we chose was one none of us were familiar with, but we thought it looked like
an interesting concept. The results may seem a bit unbelievable at first, but hopefully our paper
will be able to give a better understanding of how this result proves to be true. Our report is on
The Banach-Tarski Paradox.
The Banach-Tarski Paradox has many theorems we need in proving this fact and we will
be covering a few of the main ideas and results. We will be giving a simple example to get
started and then an intuitive and precise statement of the Banach-Tarski Theorem. Then we will
be covering some of the important definitions we will need in proving the Banach-Tarski
Theorem such as rigid motions, partitions of sets, and piecewise congruence. Next, we will be
going over the Schröder-Bernstein Theorem and the Axiom of Choice, both of which are key
concepts in proving the Banach-Tarski Theorem. Finally, we will be finishing up with some of
the implications for mathematical modeling such as what is a mathematical model and what the
Banach-Tarski Paradox actually shows.
To start out, we’d like to begin with an example to give you some sort of an idea of what
the Banach-Tarski Theorem is all about. For instance, say we take an eight ball off a pool table,
cut it up into a finite number of pieces, and then reassemble it into a life-size statue of someone.
Does this sound possible?
Loosely speaking, the theorem says that if X and Y are any two objects in space that are
each small enough to be contained in some (perhaps very large) ball and each large enough to
contain some (perhaps very small) ball, then one can divide X into some finite number of pieces
and then reassemble them (using only rigid motions), which will be discussed in a moment, to
form Y.
This seems to be false if we believe the practice of confusing the “ideal” objects of
geometry with the “real” objects of the world around us. It certainly seems insane to claim that a
billiard ball can be chopped into pieces and then be put back together to form a life-size statue of
someone. We aren’t making this claim. However, in mathematics, no matter how far-fetched
this may seem, the theorem is true!
3
The Banach-Tarski Theorem
To be precise, the Banach-Tarski Theorem states:
If X and Y are bounded subsets of R^3 having nonempty interiors, then there exists a
natural number n and partitions {X j : 1 ≤ j ≤ n} and {Y j : 1 ≤ j ≤ n} of X and Y (into n
pieces each) such that X j is congruent to Y j for all j.
Definitions
To get a better understanding of the Banach-Tarski Paradox you must understand some of
the key ingredients. They are rigid motions, partitions of sets, the Hausdorff Paradox and
piecewise congruence.
Rigid Motion
A rigid motion is a mapping r of R3 → R3 having the form r ( x ) = ρ ( x ) + a for
x ∈ R 3 where ρ is a fixed rotation and a ∈ R 3 is fixed. Being a rotation ρ is a 3x3 orthogonal
matrix whose inverse is equal to its transpose and whose determinant is equal to 1. With the
determinant equal to 1 orientation is preserved.
Partitions of Sets
A partition of a set X is a family of sets whose union is X and whose any two members
are identical of disjoint. So to say { X : 1 ≤ j ≤ n} is a partition of X into n subsets means that
j
X = X 1 ∪ X 2 ∪ ... ∪ X n and X i ∩ X j = φ if i ≠ j . In order to partition a set the set must be
countable. In other words if A is a countable set, there exists a 1 to 1 mapping into real numbers
or you could say there is a way of labeling all the points of A into a sequence. Now if we
consider the two rotations,
− cos θ
−1 2 − 3 2 0
ψ = 3 2 − 1 2 0 and φ =
0
0
0
1
sin θ
4
0
sin θ
−1
0
0 cos θ
Geometrically ψ rotates R 3 by 120 degrees about the z-axis and φ rotates R 3 by 180 degrees
about the line in the xy plane x cos 1 φ = x sin 1 φ . Now the matrix ψ 2 is the same as ψ
2
2
except
3 is replaced by a − 3 .
Hausdorff Paradox
The Hausdorff Paradox is a precursor to the Banach-Tarski Paradox. It states that their
exists
a
partition
of
the
unit
square
into
four
subsets
such
that:
(i ) P is countable, (ii )φ (S1 ) = S 2 ∪ S 3 , (iii )ψ (S1 ) = S 2 , (iv )ψ 2 (S1 ) = S 3
ψ3 =φ2 = i
(1)
If we let G be the set of all matrices that can be obtained as a product of a finite number of
factors each of which is ψ or φ then because of (1) it is clear that G is a group under matrix
multiplication. Then G can be expressed in at least one way as ρ = σ 1σ 2 ...σ n where n ≥ 1 and
each σ j is equal to θ , ψ , ψ 2 and if 1 ≤ j ≤ n then exactly one of σ j and σ j+1 is θ . We call
these expressions reduced words in the letters θ ,ψ , and ψ 2 .
For example the expression
φψ 2φφψ 2φ is not a reduced word because of the adjacent θ ’s, but is equal to φψφ . So each
member of G other than i, θ , ψ , ψ 2 can be written in one of the forms:
α = ψ P φψ P φ ...ψ P φ
1
m
2
β = φψ P φψ P ...φψ P
1
2
m
γ = φψ P φψ P ...φψ P φ
1
2
m
δ = ψ P φψ P φ ...φψ P
1
2
m
where m ≥ 1 and Pj is 1 or 2.
Piecewise Congruence
Piecewise congruence is important to the Banach-Tarski Paradox because two sets must
be piecewise congruent if you want to map the partitions of one object onto another. Two sets X
and Y are piecewise congruent, written X~Y, if {X j : 1 ≤ j ≤ n} of X into n subsets and a
5
corresponding set
{f
j
: 1 ≤ j ≤ n} of rigid motions, such that
piecewise congruent to a subset of Y we write X<~Y.
These are some simple properties of piecewise congruence:
(i )X ~ X
(ii )X ~ Y Y ~ X
(iii )X ~ Y and Y ~ Z X ~ Z
(iv )X ~ Y X <~ Y
(v )X <~ Y and Y <~ Z X <~ Z
(vi )X ⊂ Y = X <~ Y
(vii )X <~ Y and Y <~ X X ~ Y
6
{f (X ) : 1 ≤ j ≤ n}.
j
j
If X is
Schröder-Bernstein Theorem
One of the fundamental ideas used in proving the Banach-Tarski paradox is the idea of
creating one object from another object that may not be the same size. Property vii of theorem f
is one of the key properties used in the idea of creating an object from another object.
Theorem f : Given sets X and Y ∈ R 3
vii ) X <~ Y and Y < ~ X
X
X ~Y
Y
The first condition required for this property is that you can take an object X and break it
into pieces to form part of the object Y.
X1
X2
X4
X3
X4
X3
X2
X1
X
Y
The second condition required for this property is that you can take an object Y and break
it into pieces to form part of the object X.
Y1
Y4
Y3
Y2
Y2
Y3
Y1
X
Y
7
Y4
The result of this property is that you are than able to take an object X and break it into
pieces to from the entire object Y.
X1
X3
X2
X1
X4
X
X3
X2
X4
Y
Now the proof of this property follows along in similar lines as the Schröder-Bernstein
Theorem, except that the Schröder-Bernstein Theorem’s notation is simpler. So we will prove
the Schröder-Bernstein Theorem instead.
Schröder - Bernstein : If A ≤ B and B ≤ A , then A = B .
Cardinality
Now the symbol around the A and B is called cardinality. So before we go into the proof
of the Schröder-Bernstein Theorem, we are going to give you an idea of what cardinality of a set
is.
Cardinality : The number of elements in a set.
Now we need to explain how to determine what the relationship between the cardinality
of two sets. The relationship of the cardinality between two sets is determined by the existence
of a bijective and injective function between the two sets.
Before we go any further in explaining the relationship between the cardinality of sets we
must get an understanding of what it means to have a bijective function mapping and an injective
function mapping.
A bijective function is a function that maps one set to a second set, and the inverse of that
function will map the second set back onto the first set. A bijective mapping is both one-to-one
and onto. If one set has a bijective function that maps one set to another set, then the two sets
have the same number of elements.
8
An injective function is a function that maps one set to a second set. An injective
function is one-to-one, but is not onto. Therefore, if you apply the inverse of the injective
function to the second set, you will get more elements then the first set. So the second set has
more elements than the first set.
We will now show how the relationship between the types of mappings that exist
between two sets and the relationship of the cardinality of the two sets. From the explanations of
what bijective and injective functions are, you may have a good idea how the type of mapping
relates to the cardinality of a set. To show this relationship consider the following two cases for
a set A and a set B:
Case I ) Bijection exits between A and B.
If ∃ bijective function f | f ( A) → B, then A = B .
or
Case II ) Bijection does not exist between A and B.
a ) If ∃ injective function g | g ( A) → B, then A < B .
.b) If ∃ injective function g | g (B ) → A, then B < A .
The above cases summarize every possible relationship between the cardinality of two
sets. So if there exists a bijective function between the two sets, then the number of elements in
two sets is equal. However, if there is no bijective function, then one set has more elements than
the other set has.
The direction of the injective function determines which set has more
elements.
There are cases where you may not need to determine what type of function mapping
exists to determine the relationship between the cardinalities of two sets. For the case where the
two sets are finite, you may not need to consider what type of mapping exists between the two
functions, because the cardinality for each set is an integer. The relationship of the cardinality of
the sets is simply the relationship between the two integers. The case where one of the sets is
finite and the other set is infinite is even simpler. The infinite set is always larger than the finite
set, which is the same as comparing numbers.
However, it is important to remember that the relationship between the cardinality of two
sets is determined by these functions when both sets are infinite, because the cardinality of an
9
infinite set may be less than, greater than, or equal to the cardinality of another infinite set. You
also may not be able to tell the relationship between the cardinality of two sets simply by looking
at the elements contained in each. For example, consider the set of integers and the set of even
integers. You might think that cardinality of the set of integers is twice as large as the cardinality
of the set of even integers, after all the set of integers contain both the set of even integers and
the set of odd integers. It turns out the cardinality of these two sets is equal. The reason for this
is that there exists a bijective function from one set to the other.
f (x ) = 2 x
x
2
−1
f (2 Ζ ) → Ζ
f −1 ( x ) =
f (Ζ ) → 2 Ζ
Now that we have a feel for what the cardinality of a set is, and how we may compare the
cardinality between two sets, we may continue with the proof of the Schröder-Bernstein
Theorem.
Schröder-Bernstein Theorem
Schröder - Bernstein : If A ≤ B and B ≤ A , then A = B .
We will start our proof of the Schröder-Bernstein Theorem with examining the conditions
required for the Schröder-Bernstein Theorem.
To satisfy the conditions of the Schröder-
Bernstein Theorem, we must compare the cardinality of two sets. To accomplish this, we must
first consider all of the possible combinations that may be able to satisfy the conditions for the
theorem. The possible number of set types for A and B is two, because each may be finite or
infinite. Thus the total number of possibilities we must take into consideration is 4.
Case I:
Case II:
Case III:
Case IV:
A is finite
A is infinite
A is finite
A is infinite
B is finite
B is infinite
B is infinite
B is finite
However, we will only have to prove it for Case I and Case II. Case III and Case IV will
not need to be considered, because neither case satisfies both of the conditions required for the
Schröder-Bernstein Theorem.
Case III will not satisfy the second condition, because the
cardinality of an infinite set is always larger than the cardinality of a finite set. Similarly, Case
IV will not satisfy the first condition.
10
Finite Case
Now we will prove the Schröder-Bernstein Theorem for the case where A and B are both
finite. Proving it for this case I is relatively easy. Since the cardinality of set A and the
cardinality of set B are each integers representing the number of elements in each set.
If A ≤ B and B ≤ A , then A ? B .
Let r , s ∈ Ζ | r = A , s = B .
Substituting r and s into the Schröder - Bernstein Theorem.
If r ≤ s and s ≤ r , then r = s.
Substituting A and B back into the Schröder - Bernstein Theorem.
If A ≤ B and B ≤ A , then A = B .
We have now proven the Schröder-Bernstein Theorem for the case where A and B are
both finite.
11
Infinite Case
Finally we must prove it for the case where A and B are both infinite. This proves to be
more interesting, because we are not able to directly compare the number of elements of A and
the number of elements of B. To show this we will walk through the statement of the SchröderBernstein Theorem step by step.
A ≤ B
What this condition is stating is that there exists an injective function from set A into set
B. Showing graphically, there is an injective function f mapping the entire set of A into some
subset of B.
f
f(A)
A
B
12
B ≤ A
What this condition is stating is that there exists an injective function from set B into set
A. Showing graphically, there is an injection function g mapping the entire set of B into some
subset of A.
g(B)
g
A
B
A = B
What this result is stating is that there exists a bijective function between A and B.
Showing graphically, we need to show there is a bijective function h mapping the entire set of A
into the entire set of B.
h(B)
h
h(A)
A
B
13
From the injective functions f and g, we should be able to show there exist a bijective
function h. Combining all f and g would give us.
f
g
g(B)
A
f(A)
B
There are more than enough mappings between the two sets to get a bijective function
from A to B. To help us construct the bijective function h, we will start by using the injective
function f, which maps every element of A into B. Since f is an injective function f will not map
to every element in B.
f(A)
f
A
B
14
Now the only elements in B that are not mapped from A to B are simply B \ f(A).
f
B \ f(A)
A
B
To get these mappings, we already know that each element in B is mapped to an element
in A by the injective function g. So if we were to take the inverse of g for all of the elements in
B, which were not already mapped from A to B by the injective function f, we would then have a
mapping from A to every element in B.
A \ g(B)
g(B)
g(B \ f(A))
f
f(A)
g
B \ f(A)
g ¹
A
B
However, the elements in A which are also members of g (B \ f(A)) are mapped to two
elements in B.
15
To rid of this catastrophe, we introduce a subset of A, E.
Where we define E = A \ g(B \ f(E)).
A \ (E E )
E
g
E
A
f
B \ f(E)
g
f(E)
B
We now want to consider the subset E | E
E.
g
g
E
E
B \ f(E)
f(E)
f
A
B
16
We now will let
={E|E
E }, and we will also let D equal the union of every E in
and D equal the union of every E in .
g
D
D
f(D)
f
A
From the above statements D
g
B
D . Now there also will exist a D
therefore D = D .
17
D ) . So
We now can define a piecewise definition for the bijective function h.
h( x ) =
A\D
f ( x)
g −1 ( x )
g ¹
x∈D
x∈ A\ D
B \ f(D)
D=D
f(D)
f
A
B
We have know shown there exists a bijective function from A to B, so we have proved
the Schröder-Bernstein Theorem for the case where A and B are both infinite and this concludes
our proof of the Schröder-Bernstein Theorem.
18
The Axiom of Choice
The Axiom of choice is one of the most discussed axioms of mathematics. But virtually
all mathematicians believe that the Axiom of Choice is a correct principle that should be
accepted as an axiom. Yet it is still generally viewed as special and less obvious that the other
axioms.
More precisely, the Axiom of Choice is typically used to prove the existence of a set that
cannot be explicitly defined in set-builder notation. That is the Axiom of Choice allows you to
define sets that you cannot describe or see directly. This can cause a sort of vagueness which
some mathematicians find troublesome.
Definition
The definition of the Axiom of Choice is as follows:
For every collection A of nonempty sets, there is a function f such that, for every B in
A, f (B ) ∈ B . Such a function is called a choice function for A.
The axioms of set theory provide a foundation for modern Mathematics, but there are
questions that surround the Axiom of Choice.
1.
Can it be derived from other axioms?
2.
Is it consistent with the other axioms?
3.
Should we accept it as an axiom?
First I would like to go over the first six axioms of set theory. There are as follows:
Axiom 1.
Two sets are equal if they contain the same members.
Axiom 2.
For any two different objects alb, there exists the set {a, b} which contains
just a and b .
Axiom 3.
For a set s and a “definite” predicate P, there exists the set Sp, which
contains jus those x ∈ s , which satisfy P.
Axiom 4.
For any set s, there exists the union of the members of s-that is, the set
containing just the members of the members of s.
Axiom 5.
For any set s , there exists the power set of s – that is, the set whose
members are just all the subsets of s .
19
Axiom 6.
There exists a set Z with the properties (a) v ∈ Z and (b) if x ∈ Z , the
{x}∈ Z .
For many sets, including any finite set, the first six axioms of set theory are enough to
guarantee the existence of a choice function but there do exist sets for which the Axiom of
Choice is required to show the existence of a choice function. Paul Cohen proved the existence
of such sets in 1963. This means that the Axiom of Choice cannot be derived from the other six
axioms in other words the Axiom of choice is independent of the first six axioms. This answers
question one posted above.
Goedel answered the question of whether the Axiom of Choice is consistent with the
other axioms in1938. Goedel showed that if the other axioms are consistent then The Axiom of
Choice is consistent with them. This answers question 2 from above.
To try and explain the third question of whether we should accept the Axiom of Choice
takes use to the realm of philosophy. Today there are three major schools of through concerning
the use of the Axiom of Choice. They are as follows:
1.
2.
3.
Accept it as an axiom and use it without hesitation.
Accept is as an axiom but use it only when you cannot find a proof without it.
The Axiom of Choice is unacceptable.
Most mathematicians today belong to school A, but there are a growing number of people
moving to school C, especially computer scientist who work on automated reasoning using
constructive type theories.
Schools of through about the Axiom of Choice
Underlying the schools of through about the use of the Axiom of Choice are views about
the truth and nature of mathematical objects.
The three major views are Platonism,
Constructivism, and Formalism. The following will give a general description of each one.
Platonism
A Platonist believes that mathematical objects exist independent of the human mind, and
a mathematical statement, such as the Axiom of Choice is objectively either true or false. A
20
Platonist accepts the Axiom of Choice only if it is objectively true, and probably falls into
schools A or C depending on her belief.
Constructivism
A constructivist believes that the only acceptable mathematical objects are ones that can
be constructed by the human mind, and the only acceptable proofs are constructive proofs. Since
the Axiom of Choice gives no method for constructing a choice set constructivist belong to
school C.
Formalism
A formalist believes that mathematics is strictly symbol manipulation and any consistent
theory is reasonable to study.
Should you accept the Axiom of Choice?
The question still remains “Should you accept the Axiom of Choice?” Here are some
arguments for and against it.
Against:
•
•
•
It’s not as simple, Aesthetically pleasing, and intuitive as the other axioms.
With it you can derive non-intuitive results, such as the existence of a discontinuous
additive function, the existence of a non-measurable set of reals, and the BanachTarski Paradox.
It is non constructive it conjures up a set without providing any sort of procedure for
its construction.
For:
•
•
•
Every vector space has a basis.
The union of countable many countable sets is countable.
Every infinite set has a denumerable subset.
21
Conclusion
When we think of a math modeling we usually think of it as a using some mathematical
principle to explain a real world phenomena. Something we can illustrate with visual objects.
The Banach-Tarski Paradox is a well-known mathematical principle but because of the
Axiom of choice, outlined above and the lack of a specific set of values it is just that a paradox.
So although it is possible in the mathematical world, the Banach-Tarski Paradox cannot be
modeled in the real world.
References
•
•
•
•
Dr. Steve Deckelman
“The Banach-Tarski Paradox”
o By Karl Stromberg
“The Axiom of Choice”
o By Alex Lopez-Ortiz
“ Proof, Logic and Cojecture: The Mathematicians”
o By Robert S. Wolf
22