background image

Thus

g

f

establishes a bijection which proves that

A

is isomorphic to

C

.

Lemma 35.

Let card

X

=

card

Y

,

u

is an atomic f.o. on

X

and

v

is an atomic f.o. on

Y

; let

A

up

u

and

B

up

v

. Let

u

÷

A

and

v

÷

B

are directly isomorphic. Then if card

(

X

\

A

) =

card

(

Y

\

B

)

we

have

u

and

v

directly isomorphic.

Proof.

Arbitrary extend the bijection witnessing being directly isomorphic to the sets

X

\

A

and

Y

\

B

.

Theorem 36.

If card

X

=

card

Y

then being isomorphic and being directly isomorphic are the

same for atomic f.o.

u

on

X

and

v

on

Y

.

Proof.

That if two filter objects are isomorphic then they are directly isomorphic is obvious.

Let atomic f.o.

u

and

v

are isomorphic that is there is a bijection

f

:

A

B

where

A

up

u

,

B

up

v

witnessing isomorphism of

u

and

v

.

If one of the filters

u

or

v

is a trivial atomic f.o. then the other is also a trivial atomic f.o. and as

it is easy to show they are directly isomorphic. So we can assume

u

and

v

are not trivial atomic f.o.

If card

(

X

\

A

) =

card

(

Y

\

B

)

our statement follows from the last lemma.

Now assume without loss of generality card

(

X

\

A

)

<

card

(

Y

\

B

)

.

card

B

=

card

Y

because card

(

Y

\

B

)

<

card

Y

.

It is easy to show that thre exists

B

B

such that card

(

X

\

A

) =

card

(

Y

\

B

)

and

card

B

=

card

B

.

We will find a bijection

g

from

B

to

B

which witnesses direct isomorphism of

v

to

v

itself.

Then the composition

g

f

witnesses a direct isomorphism of

u

÷

A

and

v

÷

B

and by the lemma

u

and

v

are directly isomorphic.

Let

D

=

B

\

B

. We have

D

up

v

.

There exists a set

E

B

such that card

E

>

card

D

and

E

up

v

.

We have card

E

=

card

(

D

E

)

and thus there exists a bijection

h

:

E

D

E

.

Let

g

(

x

) =

x

if

x

B

\

E

;

h

(

x

)

if

x

E.

g

|

B

\

E

and

g

|

E

are bijections.

im

(

g

|

B

\

E

) =

B

\

E

; im

(

g

|

E

) =

im

h

=

D

E

;

(

D

E

)

(

B

\

E

) = (

D

(

B

\

E

))

(

E

(

B

\

E

)) =

∅ ∪ ∅

=

.

Thus

g

is a bijection from

B

to

(

B

\

E

)

(

D

E

) =

B

D

=

B

.

To finish the proof it’s enough to show that

h

g

i

v

=

v

. Indeed it follows from

B

\

E

up

v

.

Proposition 37.

1. For every

A

up

A

and

B

up

B

we have

A

>

2

B

iff

A ÷

A

>

2

B ÷

B

.

2. For every

A

up

A

and

B

up

B

we have

A

>

1

B

iff

A ÷

A

>

1

B ÷

B

.

Proof.

1.

A

>

2

B

iff there exist a bijective

Set

-morphism

f

such that

B

=

h↑

f

iA

. The equality is

obviously preserved replacing

A

with

A ÷

A

and

B

with

B ÷

B

.

2.

A

>

1

B

iff there exist a bijective

Set

-morphism

f

such that

B ⊆ h↑

f

iA

. The equality is

obviously preserved replacing

A

with

A ÷

A

and

B

with

B ÷

B

.

Rudin-Keisler order of ultrafilters is considered in such a book as [6].

Proposition 38.

For ultrafilters

>

2

is the same as Rudin-Keisler ordering.

Proof.

x

>

2

y

iff there exist sets

A

up

x

and

B

up

y

a bijective

Set

-morphism

f

:

X

Y

such

that up

(

y

÷

B

) =

{

C

P

Y

| h

f

1

i

C

up

(

x

÷

A

)

}

that is when

C

up

(

y

÷

B

)

⇔ h

f

1

i

C

up

(

x

÷

A

)

what is equivalent to

C

up

y

⇔ h

f

1

i

C

up

x

what is the definition of Rudin-Keisler ordering.

6

Section 3