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14.2. RUDIN-KEISLER EQUIVALENCE AND RUDIN-KEISLER ORDER

246

Corollary

1305

.

For every two atomic filters (with possibly different bases)

A

and

B

there exists at most one bijective reloid triple from

A

to

B

.

Proof.

Suppose that

f

and

g

are two different bijective reloids from

A

to

B

. Then

g

1

f

is not the identity reloid (otherwise

g

1

f

= id

RLD

dom

f

and so

f

=

g

because

f

and

g

are isomorphisms). But

g

1

f

is a bijective reloid (as a

composition of bijective reloids) from

A

to

A

what is impossible.

14.2. Rudin-Keisler equivalence and Rudin-Keisler order

FiXme

: Define monomorphisms and epimorphisms

Theorem

1306

.

Atomic filters

a

and

b

(with possibly different bases) are iso-

morphic iff

a

b

b

a

.

Proof.

Let

a

b

b

a

. Then there are a monovalued reloids

f

and

g

such that dom

f

=

a

and im

f

=

b

and dom

g

=

b

and im

g

=

a

. Thus

g

f

and

f

g

are monovalued morphisms from

a

to

a

and from

b

to

b

. By the above

we have

g

f

= id

RLD

a

and

f

g

= id

RLD

b

so

g

=

f

1

and

f

1

f

= id

RLD

a

and

f

f

1

= id

RLD

b

. Thus

f

is an injective monovalued reloid from

a

to

b

and thus

a

and

b

are isomorphic.

The last theorem cannot be generalized from atomic filters to arbitrary filters,

as it’s shown by the following example:

Example

1307

.

A ≥

1

B ∧ B ≥

1

A

but

A

is not isomorphic to

B

for some filters

A

and

B

.

Proof.

Consider

A

=

R

[0; 1] and

B

=

d

n

R

[0;1+

[

>

0

o

. Then the function

f

=

λx

R

:

x/

2 witnesses both inequalities

A ≥

1

B

and

B ≥

1

A

. But these filters

cannot be isomorphic because only one of them is principal.

Lemma

1308

.

Let

f

0

and

f

1

be

Set

-morphisms. Let

f

(

x, y

) = (

f

0

x, f

1

y

) for a

function

f

. Then

D

FCD

(Src

f

0

×

Src

f

1

,

Dst

f

0

×

Dst

f

1

)

f

E

(

A ×

RLD

B

) =

FCD

f

0

A ×

RLD

FCD

f

1

B

.

Proof.

D

FCD

(Src

f

0

×

Src

f

1

,

Dst

f

0

×

Dst

f

1

)

f

E

(

A ×

RLD

B

) =

D

FCD

(Src

f

0

×

Src

f

1

,

Dst

f

0

×

Dst

f

1

)

f

E

l

Src

f

0

×

Src

f

1

(

A

×

B

)

A

∈ A

, B

∈ B

=

l

Dst

f

0

×

Dst

f

1

h

f

i

(

A

×

B

)

A

∈ A

, B

∈ B

=

l

Dst

f

0

×

Dst

f

1

(

h

f

0

i

A

× h

f

1

i

B

)

A

∈ A

, B

∈ B

=

l

Dst

f

0

h

f

0

i

A

× ↑

Dst

f

1

h

f

1

i

B

)

A

∈ A

, B

∈ B

= (theorem

888

)

l

Dst

f

0

h

f

0

i

A

A

∈ A

×

RLD

l

Dst

f

1

h

f

1

i

B

B

∈ B

=

FCD

f

0

A ×

RLD

FCD

f

1

B

.

Theorem

1309

.

Let

f

be a monovalued reloid. Then GR

f

is isomorphic to

the filter dom

f

.