number of ultrafilters

number of ultrafilters

Theorem

.

Let

X

be a set.
The number (

http://planetmath.org/CardinalNumber

) of

ultrafilters

(

http://planetmath.org/Ultrafilter

) on

X

is

|

X

|

if

X

is finite (

http://planetmath.org/Finite

),
and

2

2

|

X

|

if

X

is

infinite

(

http://planetmath.org/Infinite

).

Proof.

If

X

is finite then each ultrafilter on

X

is principal,
and so there are exactly

|

X

|

ultrafilters.
In the rest of the proof we will assume that

X

is infinite.

Let

F

be the set of all finite subsets of

X

,
and let

Φ

be the set of all finite subsets of

F

.

For each

A

⊆

X

define

B

A

=

{

(

f

,

ϕ

)

∈

F

×

Φ

∣

A

∩

f

∈

ϕ

}

,
and

B

A

∁

=

(

F

×

Φ

)

∖

B

A

.
For each

𝒮

⊆

𝒫

⁢

(

X

)

define

ℬ

𝒮

=

{

B

A

∣

A

∈

𝒮

}

∪

{

B

A

∁

∣

A

∉

𝒮

}

.

Let

𝒮

⊆

𝒫

⁢

(

X

)

,
and suppose

A

1

,

…

,

A

m

∈

𝒮

and

D

1

,

…

,

D

n

∈

𝒫

⁢

(

X

)

∖

𝒮

,
so that we have

B

A

1

,

…

,

B

A

m

,

B

D

1

∁

,

…

,

B

D

n

∁

∈

ℬ

𝒮

.
For

i

∈

{

1

,

…

,

m

}

and

j

∈

{

1

,

…

,

n

}

let

a

i

,

j

be such that
either

a

i

,

j

∈

A

i

∖

D

j

or

a

i

,

j

∈

D

j

∖

A

i

.
This is always possible, since

A

i

≠

D

j

.
Let

f

=

{

a

i

,

j

∣

1

≤

i

≤

m

,

1

≤

j

≤

n

}

,
and put

ϕ

=

{

A

i

∩

f

∣

1

≤

i

≤

m

}

.
Then

(

f

,

ϕ

)

∈

B

A

i

, for

i

=

1

,

…

,

m

.
If for some

j

∈

{

1

,

…

,

n

}

we have

D

j

∩

f

∈

ϕ

,
then

D

j

∩

f

=

A

i

∩

f

for some

i

∈

{

1

,

…

,

m

}

,
which is impossible,
as

a

i

,

j

is in one of these sets but not the other.
So

D

j

∩

f

∉

ϕ

,
and therefore

(

f

,

ϕ

)

∈

B

D

j

∁

.
So

(

f

,

ϕ

)

∈

B

A

1

∩

⋯

∩

B

A

m

∩

B

D

1

∁

∩

⋯

∩

B

D

n

∁

.
This shows that any finite subset of

ℬ

𝒮

has nonempty

intersection

,
and therefore

ℬ

𝒮

can be extended to an ultrafilter

𝒰

𝒮

.

Suppose

ℛ

,

𝒮

⊆

𝒫

⁢

(

X

)

are distinct.
Then, relabelling if necessary,

ℛ

∖

𝒮

is nonempty.
Let

A

∈

ℛ

∖

𝒮

.
Then

B

A

∈

𝒰

ℛ

,
but

B

A

∉

𝒰

𝒮

since

B

A

∁

∈

𝒰

𝒮

.
So

𝒰

ℛ

and

𝒰

𝒮

are distinct
for distinct

ℛ

and

𝒮

.
Therefore

{

𝒰

𝒮

∣

𝒮

⊆

𝒫

⁢

(

X

)

}

is
a set of

2

2

|

X

|

ultrafilters on

F

×

Φ

.
But

|

F

×

Φ

|

=

|

X

|

,
so

F

×

Φ

has the same

number of ultrafilters

as

X

.
So there are at least

2

2

|

X

|

ultrafilters on

X

,
and there cannot be more than

2

2

|

X

|

as each ultrafilter is an

element of

𝒫

⁢

(

𝒫

⁢

(

X

)

)

.
∎

Corollary

.

The number of topologies on an

infinite set

X

is

2

2

|

X

|

.

Proof.

Let

X

be an infinite set.
By the

theorem

, there are

2

2

|

X

|

ultrafilters on

X

.
If

𝒰

is an ultrafilter on

X

,
then

𝒰

∪

{

∅

}

is a topology on

X

.
So there are at least

2

2

|

X

|

topologies on

X

,
and there cannot be more than

2

2

|

X

|

as each topology is an element of

𝒫

⁢

(

𝒫

⁢

(

X

)

)

.
∎

Title

number of ultrafilters

Canonical name

NumberOfUltrafilters

Date of creation

2013-03-22 15:51:49

Last modified on

2013-03-22 15:51:49

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

14

Author

yark (2760)

Entry type

Theorem

Classification

msc 03E99

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