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14.1. ORDERING OF FILTERS

240

Lemma

1279

.

Let card

X

= card

Y

,

u

be an ultrafilter on

X

and

v

be an

ultrafilter on

Y

; let

A

u

and

B

v

. Let

u

÷

A

and

v

÷

B

be 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

X

\

B

.

Theorem

1280

.

If card

X

= card

Y

then being isomorphic and being directly

isomorphic are the same for ultrafilters

u

on

X

and

v

on

Y

.

Proof.

That if two filters are isomorphic then they are directly isomorphic is

obvious.

Let ultrafilters

u

and

v

be isomorphic that is there is a bijection

f

:

A

B

where

A

u

,

B

v

witnessing isomorphism of

u

and

v

.

If one of the filters

u

or

v

is a trivial ultrafilter then the other is also a trivial

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

u

and

v

are not trivial ultrafilters.

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 otherwise card(

X

\

A

) = card(

Y

\

B

).

It is easy to show that there exists

B

0

B

such that card(

X

\

A

) = card(

Y

\

B

0

)

and card

B

0

= card

B

.

We will find a bijection

g

from

B

to

B

0

which witnesses direct isomorphism of

v

to

v

itself. Then the composition

g

f

witnesses a direct isomorphism of

u

÷

A

and

v

÷

B

0

and by the lemma

u

and

v

are directly isomorphic.

Let

D

=

B

0

\

B

. We have

D /

v

.

There exists a set

E

B

such that card

E

card

D

and

E /

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

0

.

To finish the proof it’s enough to show that

h

g

i

v

=

v

. Indeed it follows from

B

\

E

v

.

Proposition

1281

.

1

. For every

A

∈ A

and

B

∈ B

we have

A ≥

2

B

iff

A ÷

A

2

B ÷

B

.

2

. For every

A

∈ A

and

B

∈ 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

=

FCD

f

A

. 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 ⊆

FCD

f

A

. The equality is obviously preserved replacing

A

with

A ÷

A

and

B

with

B ÷

B

.