Transcript Powerpoint template for scientific posters (Swarthmore

An Overview of The Banach-Tarski Paradox
Dan Sullivan ‘07
Swarthmore College, Department of Mathematics & Statistics
Abstract
Isometries
The Banach-Tarski “Paradox”
Continuation of the Proof
In 1924, Stefan Banach and Albert Tarski published the paper
Sur la décomposition des ensembles de points en aprties
respectivement congruentes in Fundamenta Mathematicae
in which they detailed the famous paradox on which this
poster is based. Using real analysis, modern algebra and,
perhaps most importantly, the Axiom of Choice formulated
by Ernst Zermelo in the development of the ZermeloFrankel set theory axioms, Banach and Tarski proved a
theorem that has come to be one of the most interesting and
controversial results in modern mathematics. In what would
become known as the “Banach-Tarski Paradox,” the two
showed that one could decompose the unit ball into 5
disjoint sets and, using only rotations and translations,
create two unit balls from the resulting pieces. This poster
outlines the argument and the background information
necessary to proceed with the proof.
An isometry is a distance-preserving isomorphism between
metric spaces. In order to create two unit balls from our
original, we require two types of isometries: translation and
rotation.
Statement of the Theorem:
Part 2:
The unit ball B = {(x, y, z) | x2 + y2 + z2 = 1} can be partitioned
into 2 sets B1 and B2 such that B ~ B1 and B ~ B2, where
“~” means “is equidecomposable to.”
Translation:
Some Initial Remarks:
Translation maps move every point of a set a constant direction
in a specified distance. They can be characterized by a
single vector that provides both a direction and a magnitude
by which to move each point of the original set.
Note that the statement of the Banach-Tarski Paradox does not
in any way guarantee the ability to partition a physical ball
in such a way as to lead to the creation of two equivalent
balls in space. Rather, the sets we use to partition B possess
no measure and thus the doubling described by the BanachTarski Paradox can only be realized in purely settheoretical terms.
Since the beauty of the Banach-Tarski paradox lies in the
simplicity of the ideas motivating it, a sketch of the proof is
provided so as to convey the logic and flow of the argument
without allowing the reader to become bogged down in
some of the details.
In order to aid in the processing of the proof of the BanachTarski Paradox, the proof has been partitioned into 3
sections, each with it’s own individual goal. The sections
can be described as follows:
1. Partitioning the Group of Rotations acting on B
2. Partitioning the Unit Sphere into 2 Copies of Itself
3. Partitioning the Unit Ball into 2 Copier of Itself
They proceed as follows.
Note that each rotation in G has 2 poles, i.e. it leaves 2 points
fixed on the Unit Sphere, S. Let P denote the set of all
poles determined by rotations in G. Since G is countably
infinite, it follows naturally that P is as well.
Let S – P denote the set of all points on the sphere not in P.
Now, partition S – P into the disjoint orbits of points as
determined by the rotations in G. Using the AC, we form
the choice set C by selecting one element from each orbit
of S – P. It is important to note the following concerning C:
C is uncountably infinite, C ∩ P = Ø, no point in C can be
rotated to another point in C and C can generate S – P
using the rotations of G.
Now, we partition S – P into the sets K1 = G1C, K2 = G2C and
K3 = G3C. Note also that by the relations given at the end
up part 1, we have that K1 ≈ K2 ≈ K3 ≈ K2 U K3. This
allows us to split each Ki into 2 subsets congruent to K2 and
K3. Using these 6 subsets, we
form two equidecomposable
copies of S – P by splitting
separating the subsets into 2
collections. Finally, we attach
the set P to one of the
Fig. 6. Plugging the holes on the
collections and, by rotating surface of the sphere.
circles on the surface of the sphere as explained in
Example #2, plug the holes in our other copy so as to form
2 equidecomposable copies of the unit sphere, S.
Part 1:
Part 3:
Fig. 3. Examples of translation and rotation in the plane.
Rotation:
Fig. 1. A visual representation of the Banach-Tarski paradox.
Rotation maps move every point of a set in a circular motion. In
2 dimensions, points are moved about a center point of
rotation, while in 3 dimensions points are rotated around an
axis. Unlike translations, points in the set can be fixed by
this type of isometry if they fall on the center or axis of
rotation.
The Axiom of Choice (AC)
Formal Definition:
For any collection C of non-empty sets, we can choose a
member from each set in that collection. That is, there exists
a choice function f defined on C such that for each set S in
C, f(S) is in S.
What does this mean?
Given a set of any number of non-empty sets, we can create a
choice set that possesses one element from each individual
set within the larger set. This choice set has a non-trivial
intersection with each element of the original set and
contains exactly one element from each of the sets that
compose the collection.
Why do we care?
Without an obvious ordering of the elements of each set within
the collection, we can not easily define the function that
chooses an element from each member of the collection.
However, it is intuitively obvious that we can create the
choice set as defined above. Thus, the Axiom of Choice tells
us in a non-constructive manner that what we inherently
believe is in fact true.
Fig. 2. This picture illustrates a
practical example of the AC. While an
infinite collection of pairs of shoes
possesses an inherent ordering to
form a choice set (i.e. one can always
choose the left shoe), a collection of
socks requires the AC to form such a
set as there is no obvious means of
selecting a sock from each pair.
Equidecomposability
Formal Definition:
Consider two sets, A and B. A and B are equidecomposable
if they can be represented as finite unions of the same
number of disjoint subsets, i.e.
n
n
B  i1 Bi
A  i1 Ai
and
where Ai ∩ Aj = Bi ∩ Bj = Ø for all i ≠ j, such that, for any i,
the subset Ai is congruent to Bi.
Example #1:
Let A = {1, 2, 3, 4, 5, …} and B = {1, 2}U{5, 6, 7, …}. We can
partition A and B as follows:
A1 = {1, 2}, A2 = {3, 4, 5, …}, B1 = {1, 2}, B2 = {5, 6, 7, …}
A1 = B1 by definition and by shifting each element of B2
down 2 we get that A2 is congruent to B2. Thus A and B are
equidecomposable.
Example #2:
We can show a circle is equidecomposable with a
circle missing a point on the circumference. We
partition the circle into 2 sets, specifically
Fig. 4.
those points of integer distance away from a Formation of
fixed point (i.e. the missing point), A1,
the subset, A1.
and all other points, A2. We do the same for the circle
missing a point, only this time we obviously do not include
the point denoted 0 in the set B1. A2 = B2 and we show A1 is
congruent to B2 by simply shifting the elements of A1 up 1
to get that a circle is equidecomposable to a circle missing a
point on the circumference.
A Sketch of the Proof:
We begin by defining two rotations on the unit ball, σ and τ,
such that σ is a rotation of 2π/3 radians about the z axis and τ
is a rotation of π radians about the line z = x. Note here that
τ2 = σ3 = I, the identity rotation, and that the only rotation
beginning with τ and followed by any number of σ’s that is
equivalent to τ is στ (be advised rotation compositions are
being written from right to left).
Here it is important to note that the Uniqueness Theorem states
that in the group G of all rotations formed by solely by σ and
τ, every rotation in G has a unique, reduced form
representation. As a result, the list of all rotations in G
created using τ and σ gives us a countably infinite list of
rotations of the unit ball.
At this stage in the proof, it is necessary that we take the group
of rotations G and partition it into 3 disjoint subsets, dubbed
G1, G2 and G3. Begin
by placing I in G1,
τ and σ in G2 and
τ2 in G3. Then, determine
the placement of all other
rotations based on the
chart to the right.
Finally, before moving on
to the next part of the proof,
note that we have defined G1,
Fig. 5. Guideline for the placement of
G2 and G3 such that the
following relations are true: each rotation of G into G1, G2 or G3.
τG1 = G2, τ2G1 = G3 and σG1 = G2 U G3.
We have partitioned the group of rotations, G, into 3 disjoint
sets and we now proceed to the next stage of the proof,
We now “thicken” the surface by extending our sets inward
toward the center of the ball. All of the relations we have
previously defined continue to hold, so it remains only to
tackle the issue of the center of the ball.
We attach the center of the original ball to one of our 2
copies, and we select a sphere contained within the other
copy that includes in its surface the center of the larger
ball. We plug the center by using the surface shifting
technique seen in Fig. 6, and consequently we have finally
created 2 equidecomposable copies of the unit ball.
References
Wapner, Leonard M. The Pea & The Sun: A Mathematical
Paradox. Wellesley, MA: A K Peters, Ltd., 2005.
Wagon, Stan. The Banach-Tarski Paradox. New York: Press
Syndicate of the University of Cambridge, 1985.
Acknowledgements
I want to thank Walter Stromquist and the entire Swarthmore
mathematics department for their help in my mathematical
education and my friends on Wharton EF 1st for providing
constant sources of encouragement and inspiration.