proof of Hausdorff paradox

proof of Hausdorff paradox

We start by outlining the general ideas, followed by the strict proof.

General approach

The

unit sphere

S

2

in the

Euclidean space

ℝ

3

is
finite in the sense that it is

http://planetmath.org/node/4826

bounded

and has a finite volume:

4

3

⁢

π

.

On the other hand,

S

2

is an

infinite

point set. As everyone who has
ever visited Hilbert’s Hotel knows, it is possible to split an

infinite set

, such as the

natural numbers

, in pieces and all pieces
are, in a sense, equal to the original set, for example if

ℕ

3

:=

{

n

∈

ℕ

∣

n

⁢

is divisible by

⁢

3

}

,

then naïvely

ℕ

3

has a third of the size of

ℕ

, and
half of the size of

ℕ

∖

ℕ

3

. There exists,
however, a

bijection

from

ℕ

to

ℕ

∖

ℕ

3

, so

ℕ

3

has, again naïvely, both half and a third of
the size of

ℕ

.

To do the same thing with

S

2

, bijections won’t suffice, however: we
need

isometries

to establish

congruence

. We can

almost

reduce
this problem to the case of bijections in the following way. The group

R

of

rotations

(rotations are isometries) in

ℝ

3

acts on

S

2

(say, from the right). Given an

countably infinite

subgroup

G

of

R

which acts freely (or almost so) on

S

2

, take a set of

http://planetmath.org/node/1517

orbit
representatives

M

of the action of

G

on

S

2

(this requires the

http://planetmath.org/node/310

axiom of choice

). Then a

disjoint

decomposition

G

1

,

…

⁢

G

n

of

G

into

countable sets

acting on

M

yields a disjoint decomposition of

S

2

. Doing

similar

juggling

with

G

1

,

…

,

G

n

as we did with

ℕ

,

ℕ

3

and

ℕ

∖

ℕ

3

above should
yield the result,

if

all the

G

1

,

…

⁢

G

n

are related by fixed
isometries, for example if

G

1

=

G

2

⁢

φ

for some isometry

φ

, and similar

relations

for the other pieces. This last

restriction

will make the proof possible only in three or more

dimensions

, and indeed, Banach and Tarski later showed in

[

BT

]

that analogous

theorems

on the line or in the plane do not hold.

The proof

Let

G

be the subgroup of rotations of

ℝ

3

generated by a
half rotation

φ

and a one third rotation

ψ

around
different axes. We fix an

orthonormal basis

for the rest
of the proof, so we may identify

φ

and

ψ

with their

rotation matrices

acting from the right on the

row vectors

of

ℝ

3

, such that without restriction

ψ

=

(

λ

μ

0

-

μ

λ

0

0

0

1

)

,

φ

=

(

-

cos

⁡

ϑ

0

sin

⁡

ϑ

0

-

1

0

sin

⁡

ϑ

0

cos

⁡

ϑ

)

,

where

λ

=

cos

⁡

2

⁢

π

3

=

-

1

2

,

μ

=

sin

⁡

2

⁢

π

3

=

1

2

⁢

3

and

ϑ

is
arbitrary for now.

Since

φ

2

=

1

and

ψ

3

=

1

, there exists
for each

element

g

from

G

other than the unity,

ϕ

,

ψ

or

ψ

2

a

positive integer

n

and numbers

m

k

∈

{

1

,

2

}

,

1

≤

k

≤

n

,
such that

g

can be written in one of the following ways:

(

α

)

(

∏

k

=

1

n

φ

⁢

ψ

m

k

)

,

(

β

)

ψ

m

1

⁢

(

∏

k

=

2

n

φ

⁢

ψ

m

k

)

⁢

φ

,

(

γ

)

(

∏

k

=

1

n

φ

⁢

ψ

m

k

)

⁢

φ

,

(

δ

)

ψ

m

1

⁢

(

∏

k

=

2

n

φ

⁢

ψ

m

k

)

(

n

≥

2

).

To have

complete

control over the

structure

of

G

, we fix

ϑ

in such a way, that

g

can be written

uniquely

in one of the
ways (

α

)–(

δ

), with fixed

n

and

m

k

. In other words: we
fix

ϑ

such that the unity

1

cannot be written in
one of the ways (

α

)–(

δ

).

To see how to do that, let us see to where the vector

(

0

,

0

,

1

)

is
transported by a

transformation

of type (

α

). It is easily
verified that

φ

⁢

ψ

=

(

-

λ

⁢

cos

⁡

ϑ

-

μ

⁢

cos

⁡

ϑ

sin

⁡

ϑ

μ

-

λ

0

λ

⁢

sin

⁡

ϑ

μ

⁢

sin

⁡

ϑ

cos

⁡

ϑ

)

⁢

and

⁢

φ

⁢

ψ

2

=

(

λ

⁢

cos

⁡

ϑ

-

μ

⁢

cos

⁡

ϑ

sin

⁡

ϑ

μ

λ

0

-

λ

⁢

sin

⁡

ϑ

μ

⁢

sin

⁡

ϑ

cos

⁡

ϑ

)

,

so that

(

0

,

0

,

1

)

⁢

φ

⁢

ψ

=

(

λ

⁢

sin

⁡

ϑ

,

μ

⁢

sin

⁡

ϑ

,

cos

⁡

ϑ

)

and

(

0

,

0

,

1

)

⁢

φ

⁢

ψ

2

=

(

-

λ

⁢

sin

⁡

ϑ

,

μ

⁢

sin

⁡

ϑ

,

cos

⁡

ϑ

)

.
More generally, let

n

be a positive integer,

p

1

,

p

2

polynomials

of degree

n

-

1

and

p

3

a polynomial of degree

n

, then

(

p

1

⁢

(

cos

⁡

ϑ

)

⁢

sin

⁡

ϑ

,

p

2

⁢

(

cos

⁡

ϑ

)

⁢

sin

⁡

ϑ

,

p

3

⁢

(

cos

⁡

ϑ

)

)

⁢

φ

⁢

ψ

=

(

(

λ

⁢

p

3

⁢

(

cos

⁡

ϑ

)

+

μ

⁢

p

2

⁢

(

cos

⁡

ϑ

)

-

λ

⁢

cos

⁡

ϑ

⁢

p

1

⁢

(

cos

⁡

ϑ

)

)

⁢

sin

⁡

ϑ

,

(

μ

⁢

p

3

⁢

(

cos

⁡

ϑ

)

-

λ

⁢

p

2

⁢

(

cos

⁡

ϑ

)

-

μ

⁢

cos

⁡

ϑ

⁢

p

1

⁢

(

cos

⁡

ϑ

)

)

⁢

sin

⁡

ϑ

,

(

1

-

cos

2

⁡

ϑ

)

⁢

p

1

⁢

(

cos

⁡

ϑ

)

+

p

3

⁢

(

cos

⁡

ϑ

)

)

,

(

p

1

⁢

(

cos

⁡

ϑ

)

⁢

sin

⁡

ϑ

,

p

2

⁢

(

cos

⁡

ϑ

)

⁢

sin

⁡

ϑ

,

p

3

⁢

(

cos

⁡

ϑ

)

)

⁢

φ

⁢

ψ

2

=

(

(

-

λ

⁢

p

3

⁢

(

cos

⁡

ϑ

)

+

μ

⁢

p

2

⁢

(

cos

⁡

ϑ

)

+

λ

⁢

cos

⁡

ϑ

⁢

p

1

⁢

(

cos

⁡

ϑ

)

)

⁢

sin

⁡

ϑ

,

(

μ

⁢

p

3

⁢

(

cos

⁡

ϑ

)

+

λ

⁢

p

2

⁢

(

cos

⁡

ϑ

)

-

μ

⁢

cos

⁡

ϑ

⁢

p

1

⁢

(

cos

⁡

ϑ

)

)

⁢

sin

⁡

ϑ

,

(

1

-

cos

2

⁡

ϑ

)

⁢

p

1

⁢

(

cos

⁡

ϑ

)

+

p

3

⁢

(

cos

⁡

ϑ

)

)

.

By

induction

, it follows that

(

0

,

0

,

1

)

is transported by a
transformation of type (

α

) to a vector whose third

component

is
a nonconstant polynomial

p

in

cos

⁡

ϑ

. If we restrict

ϑ

such that

0

≤

ϑ

≤

π

, there are only finitely many values for

ϑ

such that

p

⁢

(

cos

⁡

ϑ

)

=

1

. Given all possible

combinations

of

n

and

m

k

,

1

≤

k

≤

n

, there are in total
only countably many problematic values for

ϑ

, so we can
easily fix hereby an unproblematic one. Now that the case (

α

)
has been dealt with, so have been the others automatically. For assume
one could write a

1

of type (

γ

), one could convert it to type
(

δ

) by

1

=

φ

⁢

1

⁢

φ

, and from type

δ

to type
(

α

) or type (

β

) by

1

=

ψ

3

-

m

1

⁢

1

⁢

ψ

m

1

, and
lastly from type (

β

) to type (

α

) by

1

=

φ

⁢

1

⁢

φ

,
completing the

contradiction

.

Now that we know the structure of

G

, how does it act on

S

2

?
Certainly not freely, since every rotation other than the unity has
its axis as

fixed point

set, which in case of the sphere makes two
fixed points per rotation. The

product

of two rotations is again a
rotation. Furthermore

G

is

finitely generated

and so is

countable

. So there are only countably many points of

S

2

where the
action of

G

fails to be free. Denote the set of these points by

D

,
so that

G

acts freely on

E

:=

S

2

∖

D

. The action creates a

partition

on

E

into orbits. The allows us to choose
a set

M

, such that

M

meets any orbit in precisely one element. We
have then the

disjoint union

E

=

⋃

g

∈

G

M

⁢

g

,

where

M

⁢

g

is the image of

M

under the action of the group element

g

. We now define three sets

A

,

B

,

C

to be the smallest sets satisfying:

•

M

⁢

1

=

M

⊆

A

;

•

if

M

⁢

g

is a subset of

A

,

B

or

C

, then

M

⁢

g

⁢

φ

is a
subset of

B

,

A

or

A

, respectively;

•

if

M

⁢

g

is a subset of

A

,

B

or

C

, then

M

⁢

g

⁢

ψ

is a
subset of

B

,

C

or

A

, respectively;

•

if

M

⁢

g

is a subset of

A

,

B

or

C

, then

M

⁢

g

⁢

ψ

2

is a
subset of

C

,

A

or

B

, respectively.

The sets

A

,

B

and

C

are well-defined because we ensured the
uniqueness of the

representations

(

α

)–(

δ

) above.

The sets

A

,

B

,

C

and

B

∪

C

are all

congruent

by virtue of

A

⁢

ψ

=

B

,

A

⁢

ψ

2

=

C

and

A

⁢

φ

=

B

∪

C

.

Since

S

2

=

A

∪

B

∪

C

∪

D

, a disjoint union, and

D

is
countable, the theorem is proven.

References

BT

St. Banach, A. Tarski

, Sur la dé

composition

des ensembles de points en parties respectivement congruentes,

Fund. math.

6, 244–277, (1924).

H

F. Hausdorff

, Bemerkung über den Inhalt von
Punktmengen,

Math. Ann.

75, 428–433, (1915).

Title

proof of Hausdorff paradox

Canonical name

ProofOfHausdorffParadox

Date of creation

2013-03-22 15:16:18

Last modified on

2013-03-22 15:16:18

Owner

GrafZahl (9234)

Last modified by

GrafZahl (9234)

Numerical id

6

Author

GrafZahl (9234)

Entry type

Proof

Classification

msc 03E25

Classification

msc 51M04

Classification

msc 20F05

Related topic

ChoiceFunction

Related topic

BanachTarskiParadox

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