proof of Banach-Tarski paradox

proof of Banach-Tarski paradox

We deal with some technicalities first, mainly concerning the
properties of equi-decomposability. We can then prove the

paradox

in a
clear and unencumbered line of

argument

: we show that, given two

unit
balls

U

and

U

′

with arbitrary origin,

U

and

U

∪

U

′

are

equi-decomposable

, regardless whether

U

and

U

′

are

disjoint

or
not. The original proof can be found in

[

BT

]

.

Technicalities

Theorem 1

.

Equi-decomposability gives rise to an

equivalence relation

on the
subsets of

Euclidean space

.

Proof.

The only non-obvious part is

transitivity

. So let

A

,

B

,

C

be
sets such that

A

and

B

as well as

B

and

C

are
equi-decomposable. Then there exist disjoint decompositions

A

1

,

…

⁢

A

n

of

A

and

B

1

,

…

⁢

B

n

of

B

such that

A

k

and

B

k

are

congruent

for

1

≤

k

≤

n

. Furthermore, there exist
disjoint decompositions

B

1

′

,

…

,

B

m

′

of

B

and

C

1

,

…

⁢

C

m

of

C

such that

B

l

′

and

C

l

are congruent for

1

≤

l

≤

m

. Define

B

k

,

l

:=

B

k

∩

B

l

′

⁢

for

⁢

1

≤

k

≤

n

,

1

≤

l

≤

m

.

Now if

A

k

and

B

k

are congruent via some

isometry

θ

:

A

k

→

B

k

, we obtain a disjoint decomposition of

A

k

by setting

A

k

,

l

:=

θ

-

1

⁢

(

B

k

,

l

)

. Likewise, if

B

l

′

and

C

l

are
congruent via some isometry

θ

′

:

B

l

′

→

C

l

, we obtain a
disjoint decomposition of

C

l

by setting

C

k

,

l

:=

θ

′

⁢

(

B

k

,

l

)

. Clearly, the

A

k

,

l

and

C

k

,

l

define a disjoint decomposition of

A

and

C

, respectively, into

n

⁢

m

parts. By transitivity of

congruence

,

A

k

,

l

and

C

k

,

l

are
congruent for

1

≤

k

≤

n

and

1

≤

l

≤

m

. Therefore,

A

and

C

are equi-decomposable.
∎

Theorem 2

.

Given

disjoint sets

A

1

,

,

…

A

n

,

B

1

,

…

⁢

B

n

such that

A

k

and

B

k

are equi-decomposable for

1

≤

k

≤

n

, their
unions

A

=

⋃

k

=

1

n

A

k

and

B

=

⋃

k

=

1

n

B

k

are
equi-decomposable as well.

Proof.

By definition, there exists for every

k

,

1

≤

k

≤

n

an

integer

l

k

such that there are disjoint decompositions

A

k

=

⋃

i

=

1

l

k

A

k

,

i

,

B

k

=

⋃

i

=

1

l

k

B

k

,

i

such that

A

k

,

i

and

B

k

,

i

are congruent for

1

≤

i

≤

l

k

. Rewriting

A

and

B

in the form

A

=

⋃

k

=

1

n

⋃

i

=

1

l

k

A

k

,

i

,

B

=

⋃

k

=

1

n

⋃

i

=

1

l

k

B

k

,

i

gives the result.
∎

Theorem 3

.

Let

A

,

B

,

C

be sets such that

A

and

B

are equi-decomposable
and

C

⊊

A

, then there exists

D

⊊

B

such that

C

and

D

are equi-decomposable.

Proof.

Let

A

1

,

…

,

A

n

and

B

1

,

…

⁢

B

n

be disjoint decompositions of

A

and

B

, respectively, such that

A

k

and

B

k

are congruent via
an isometry

θ

k

:

A

k

→

B

k

for all

1

≤

k

≤

n

. Let

θ

:

A

→

B

a map such that

θ

⁢

(

x

)

=

θ

k

⁢

(

x

)

if

x

∈

A

k

. Since the

A

k

are disjoint,

θ

is well-defined
everywhere. Furthermore,

θ

is obviously

bijective

. Now set

C

k

:=

C

∩

A

k

and define

D

k

:=

θ

k

⁢

(

C

k

)

, so that

C

k

and

D

k

are congruent for

1

≤

k

≤

n

, so the

disjoint union

D

:=

D

1

∪

⋯

∪

D

n

and

C

are equi-decomposable. By
construction,

θ

⁢

(

C

)

=

D

. Since

C

is a

proper subset

of

A

and

θ

is bijective,

D

is a proper subset of

B

.
∎

Theorem 4

.

Let

A

,

B

and

C

be sets such that

A

and

C

are
equi-decomposable and

A

⊆

B

⊆

C

. Then

B

and

C

are
equi-decomposable.

Proof.

Let

A

1

,

…

,

A

n

and

C

1

,

…

⁢

C

n

disjoint decompositions of

A

and

C

, respectively, such that

A

k

and

C

k

are congruent via
an isometry

θ

k

for

1

≤

k

≤

n

. Like in the proof of

theorem

3

, let

θ

:

A

→

C

be the well-defined,
bijective map such that

θ

⁢

(

x

)

=

θ

k

⁢

(

x

)

if

x

∈

A

k

. Now,
for every

b

∈

B

, let

𝒞

⁢

(

b

)

be the

intersection

of all sets

X

⊆

B

satisfying

•

b

∈

X

,

•

for all

x

∈

X

, the

preimage

θ

-

1

⁢

(

x

)

lies in

X

,

•

for all

x

∈

X

∩

A

, the image

θ

⁢

(

x

)

lies in

X

.

Let

b

1

,

b

2

∈

B

such that

𝒞

⁢

(

b

1

)

and

𝒞

⁢

(

b

2

)

are not
disjoint. Then there is a

b

∈

𝒞

⁢

(

b

1

)

∩

𝒞

⁢

(

b

2

)

such that

b

=

θ

r

⁢

(

b

1

)

=

θ

s

⁢

(

b

2

)

for suitable integers

s

and

r

. Given

b

′

∈

𝒞

⁢

(

b

1

)

, we have

b

′

=

θ

t

⁢

(

b

1

)

=

θ

t

+

s

-

r

⁢

(

b

2

)

for a suitable integer

t

,
that is

b

′

∈

𝒞

⁢

(

b

2

)

, so that

𝒞

⁢

(

b

1

)

⊆

𝒞

⁢

(

b

2

)

. The reverse

inclusion

follows
likewise, and we see that for arbitrary

b

1

,

b

2

∈

B

either

𝒞

⁢

(

b

1

)

=

𝒞

⁢

(

b

2

)

or

𝒞

⁢

(

b

1

)

and

𝒞

⁢

(

b

2

)

are
disjoint. Now set

D

:=

{

b

∈

B

∣

𝒞

⁢

(

b

)

⊆

A

}

,

then obviously

D

⊆

A

. If for

b

∈

B

,

𝒞

⁢

(

b

)

consists
of the

sequence

of

elements

…

,

θ

-

1

⁢

(

b

)

,

b

,

θ

⁢

(

b

)

,

…

which is

infinite

in both
directions, then

b

∈

D

. If the sequence is infinite in only one
direction, but the final element lies in

A

, then

b

∈

D

as well.
Let

E

:=

θ

⁢

(

D

)

and

F

:=

B

∖

D

, then clearly

E

∪

F

⊆

C

.

Now let

c

∈

C

. If

c

∉

E

, then

θ

-

1

⁢

(

c

)

∉

D

, so

𝒞

⁢

(

θ

-

1

⁢

(

c

)

)

consists of a sequence

…

,

θ

-

2

⁢

(

c

)

,

θ

-

1

⁢

(

c

)

,

…

which is infinite in
only one direction and the final element does not lie in

A

. Now

θ

-

1

⁢

(

c

)

∈

𝒞

⁢

(

θ

-

1

⁢

(

c

)

)

, but since

θ

-

1

⁢

(

c

)

does

lie in

A

, it is not the final element. Therefore the
subsequent element

c

lies in

𝒞

⁢

(

θ

-

1

⁢

(

c

)

)

, in particular

c

∈

B

and

𝒞

⁢

(

c

)

=

𝒞

⁢

(

θ

-

1

⁢

(

c

)

)

⊈

A

, so

c

∈

B

∖

D

=

F

. It follows that

C

=

E

∪

F

, and furthermore

E

and

F

are disjoint.

It now follows similarly as in the preceding proofs that

D

and

E

=

θ

⁢

(

D

)

are equi-decomposable. By theorem

2

,

B

=

D

∪

F

and

C

=

E

∪

F

are equi-decomposable.
∎

The proof

We may assume that the unit ball

U

is centered at the origin, that
is

U

=

𝔹

3

, while the other unit ball

U

′

has an arbitrary
origin.
Let

S

be the surface of

U

, that is, the

unit sphere

. By the

Hausdorff paradox

, there exists a disjoint decomposition

S

=

B

′

∪

C

′

∪

D

′

∪

E

′

such that

B

′

,

C

′

,

D

′

and

C

′

∪

D

′

are congruent, and

E

′

is

countable

. For

r

>

0

and

A

⊆

ℝ

3

, let

r

⁢

A

be the set of
all vectors of

A

multiplied by

r

. Set

B

:=

⋃

0

<

r

<

1

r

⁢

B

′

,

C

:=

⋃

0

<

r

<

1

r

⁢

C

′

,

D

:=

⋃

0

<

r

<

1

r

⁢

D

′

,

E

:=

⋃

0

<

r

<

1

r

⁢

E

′

.

These sets give a disjoint decomposition of the unit ball with the
origin deleted, and obviously

B

,

C

,

D

and

C

∪

D

are congruent (but

E

is of course

not

countable). Set

A

1

:=

B

∪

E

∪

{

0

}

where

0

is the origin.

B

and

C

∪

D

are trivially
equi-decomposable. Since

C

and

B

as well as

D

and

C

are
congruent,

C

∪

D

and

B

∪

C

are equi-decomposable. Finally,

B

and

C

∪

D

as well as

C

and

B

are congruent, so

B

∪

C

and

B

∪

C

∪

D

are equi-decomposable. In total,

B

and

B

∪

C

∪

D

are equi-decomposable by theorem

1

, so

A

1

and

U

are
equi-decomposable by theorem

2

. Similarly, we conclude that

C

,

D

and

B

∪

C

∪

D

are equi-decomposable.

Since

E

′

is only countable but there are uncountably many

rotations

of

S

, there exists a rotation

θ

such that

θ

⁢

(

E

′

)

⊊

B

′

∪

C

′

∪

D

′

, so

F

:=

θ

⁢

(

E

)

is a proper
subset of

B

∪

C

∪

D

. Since

C

and

B

∪

C

∪

D

are
equi-decomposable, there exists by theorem

3

a proper subset

G

⊊

C

such that

G

and

F

(and thus

E

) are
equi-decomposable. Finally, let

p

∈

C

∖

G

an arbitrary point
and set

A

2

:=

D

∪

G

∪

{

p

}

,

a disjoint union by construction. Since

D

and

B

∪

C

∪

D

,

G

and

E

as well as

{

0

}

and

{

p

}

are equi-decomposable,

A

2

and

U

are equi-decomposable by theorem

2

.

A

1

and

A

2

are disjoint (but

A

1

∪

A

2

≠

U

!).

Now

U

and

U

′

are congruent, so

U

′

and

A

2

are
equi-decomposable by theorem

1

. If

U

and

U

′

are
disjoint, set

H

:=

A

2

. Otherwise we may use theorem

3

and
choose

H

⊊

A

2

such that

H

and

U

′

∖

U

are
equi-decomposable. By theorem

2

,

A

1

∪

H

and

U

∪

(

U

′

∖

U

)

=

U

∪

U

′

are equi-decomposable. Now we have

A

1

∪

H

⊆

U

⊆

U

∪

U

′

,

so by theorem

4

,

U

and

U

∪

U

′

are equi-decomposable.

References

BT

St. Banach, A. Tarski

, Sur la dé

composition

des ensembles de points en parties respectivement congruentes,

Fund. math.

6, 244–277, (1924).

Title

proof of Banach-Tarski paradox

Canonical name

ProofOfBanachTarskiParadox

Date of creation

2013-03-22 15:19:03

Last modified on

2013-03-22 15:19:03

Owner

GrafZahl (9234)

Last modified by

GrafZahl (9234)

Numerical id

7

Author

GrafZahl (9234)

Entry type

Proof

Classification

msc 51M25

Classification

msc 03E25

Related topic

HausdorffParadox

Generated on Fri Feb 9 22:09:14 2018 by

LaTeXML