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A Top Down Look at the Banach-Tarski Paradox Leonard M. Wapner El Camino College Infinity is where things happen that don’t. - an anonymous schoolboy Loosely Speaking ...

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a top down look at the banach tarski paradox

A Top Down Lookat the Banach-Tarski Paradox

Leonard M. Wapner

El Camino College

loosely speaking
Loosely Speaking ...

The theorem states that it is possible to partition a solid ball into finitely many pieces and reassemble them to form two balls, each of the same shape and volume as the original.

B1

B

B2

limerick
Limerick

There once were two mathematicians

Who split a lead ball with partitions

Though with only five parts

Our two masters of arts

Reassembled it into munitions!

georg cantor
Georg Cantor

I see it, but I don’t believe it!

I was forced by logic ...

henri poincare
Henri Poincare

… a disease from which

mathematics would have to

recover !

tarski obit
Tarski Obit

An irate citizen once demanded of the Illinois legislature that they outlaw the teaching of this result in Illinois schools!

From an obituary of Alfred Tarski in California Monthly, the UC Berkeley alumni magazine.

axiom of choice
Axiom of Choice

A

B

C

. . . . . . .

a, b, c, ...

rejoice
Rejoice!

Isn’t AC great? How else could we get such counterintuitive results?

great circle distance
Great-circle Distance

d

M

L

O

Find

Find d

Find 

OL & OM

definitions and notation
Definitions and Notation
  • Let S = {(x,y,z) : x2+y2+z2 = 1}. We refer to S as the unit sphere (centered at the origin).
  • Let B = {(x,y,z) : x2+y2+z2 < 1}. We refer to B as the unit ball (centered at the origin).
  • Given two sets C and D, we say C is congruent to D, and write C  D, if C can be made to coincide with D using only
definitions and notation cont
Definitions and Notation (cont)
  • … translations and rotations of C.
  • Given two sets C and D, we say C is equivalent by finite decomposition to D, and write C  D, if we can divide C into a finite number of disjoint parts, respectively translate and rotate these parts in such a way that they can be rearranged to form D.
definitions and notation cont17
Definitions and Notation (cont)
  • Given any two rotations 1 and 2 of S, define the product rotation 12 as the rotation obtained by first rotating by 2 then by 1. Similarly, for any rotation  and natural number n, define the rotation n to be the rotation equivalent to n sequential  rotations.
statement of theorem
Statement of Theorem

A closed unit ball B can be decomposed into two disjoint sets, B = B1  B2, such that B  B1 and B  B2 .

B1

B

B2

proof of theorem
Proof of Theorem
  • Form and partition a group of rotations G acting on S.
  • Use G to create two spheres, S1 and S2, from the original sphere S. This is the Hausdorff Paradox.
  • By “thickening” S1 and S2 , create B1 and B2 such that B ~ B1 and B ~ B2.
rotations
Rotations
  • Define three basic rotations , 2 and .
  • Form the group G of all products of these rotations consisting of a finite number of factors.
  • Sort these products into three subsets G1, G2 and G3.
  • Note significant properties of the subsets.
sphere
Sphere
  • Define P, the set of poles.
  • Partition S\P into three sets K1, K2 and K3
  • Show K1 K2  K3  K2  K3.
  • Use “cookie cutter” methods to create two spheres, S1 and S2, from the original sphere S.
  • Absorb the poles.
slide22
Ball
  • “Thicken” S1 and S2 to form B1 and B2.
  • Absorb the origin.
anagram
Anagram

Banach-Tarski

Banach-TarskiBanach-Tarski

equivalence by finite decomposition
Equivalence By Finite Decomposition

Square

Isosceles Triangle

C

D

C ~ D

basic rotations
Basic Rotations

 axis

(z axis)

 axis

(z = x)

: 120°

2:240

45°

: 180°

matrix definitions of and
Matrix Definitions of  and 

-1/2 -3/2 0

3/2 -1/2 0

0 0 1

0 0 1

0 -1 0

1 0 0

 =

 =

partitioning g
Partitioning G

1  G1,   G2,   G2, 2  G3

  G1   G2   G3

or

2

  G2

  G1

  G1

  G2

  G3

  G1

 begins with:

2  G3

2  G1

2  G2

properties of g 1 g 2 and g 3
Properties of G1 , G2 and G3

G1

G2

G3

, , 2, ...

1, , 2, ...

2, , 2, ...

G1 = G2

2G1 = G3

G1 = G2G3

partitioning s p
Partitioning S\P
  • We call two points in S\P equivalent if there exists a rotation in G which sends one point to the other. This is an equivalence relation and allows us to partition S\P into equivalence classes. There are uncountably many such classes.
  • Using AC, choose one point from each class and define the choice set M as the set of these chosen points.
  • Let K1 = G1M , K2 = G2M and K3 = G3M.
p the set of poles
P - The Set of Poles

Every rotation G has two poles. These are the two points of S which remain fixed under . Let P be the set of all such poles for all G . Since G is countable, the set P is also countable.

hyperbolic plane
Hyperbolic Plane

K1 K2  K3

K1 K2  K3