Theorem 50.

If card

X

=

card

Y

then isomorphism and direct isomorphism 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.

[TODO: Carefully choose

a

is isomorphic to

b

vs.

b

is isomorphic to

a

.]

Let atomic f.o.

u

and

v

are isomorphic that is there are 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

.

Let

B

B

such that card

(

X

\

A

) =

card

(

Y

\

B

)

and card

B

=

card

B

.

[TODO: Detailed proof

that

B

exists.]

We will ﬁnd a bijection

g

from

B

and

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

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 51.

[TODO: Can be generalized for

A

S

up

A

and likewise for

B

?]

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 ÷

A

>

2

B ÷

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 ÷

A

>

1

B ÷

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 52.

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

|

f

1

C

up

(

x

÷

A

)

that is when

C

up

(

y

÷

B

)

f

1

C

up

(

x

÷

A

)

what is equivalent to

C

up

y

f

1

C

up

x

what is the deﬁnition of Rudin-Keisler ordering.

Remark 53.

The relation of being isomorphic for ultraﬁlters is traditionally called

Rudin-Keisler

equivalence

.

Obvious 54.

(

>

1

)

(

>

2

)

.

Definition 55.

Let

Q

and

R

are binary relations on the set of ﬁlter objects. I will denote

MonRld

Q,R

the directed multigraph with objects being ﬁlter objects and morphisms such mono-

valued trans-reloids

f

that

(

dom

f

)

Q

A

and

(

im

f

)

R

B

.

8

Section 3