 7.7. COMPLETE RELOIDS AND COMPLETION OF RELOIDS

143

Proof.

We will prove only the first as the second is similar. Let

f

=

G

Src

f

{

α

} ×

RLD

F

(

α

)

α

Src

f

=

G

Src

f

{

α

} ×

RLD

G

(

α

)

α

Src

f

for some

F, G

F

(Dst

f

)

Src

f

. We need to prove

F

=

G

. Let

β

Src

f

.

f

u

(

Src

f

{

β

} ×

RLD

>

F

(Dst

f

)

) = (proposition

461

)

G

(

Src

f

{

α

} ×

RLD

F

(

α

))

u

(

Src

f

{

β

} ×

RLD

>

F

(Dst

f

)

)

α

Src

f

=

Src

f

{

β

} ×

RLD

F

(

β

)

.

Similarly

f

u

(

Src

f

{

β

RLD

>

F

(Dst

f

)

) =

Src

f

{

β

RLD

G

(

β

). Thus

Src

f

{

β

RLD

F

(

β

) =

Src

f

{

β

} ×

RLD

G

(

β

) and so

F

(

β

) =

G

(

β

).

Definition

785

.

Completion

and

co-completion

of a reloid

f

RLD

(

A

;

B

) are

defined by the formulas:

Compl

f

= Cor

(

RLD

(

A

;

B

);

ComplRLD

(

A

;

B

))

f

; CoCompl

f

= Cor

(

RLD

(

A

;

B

);

CoComplRLD

(

A

;

B

))

f.

Theorem

786

.

Atoms of the lattice

ComplRLD

(

A

;

B

) are exactly reloidal prod-

ucts of the form

A

{

α

} ×

RLD

b

where

α

A

and

b

is an ultrafilter on

B

.

Proof.

First, it’s easy to see that

A

{

α

} ×

RLD

b

are elements of

ComplRLD

(

A

;

B

). Also

RLD

(

A

;

B

)

is an element of

ComplRLD

(

A

;

B

).

A

{

α

} ×

RLD

b

are atoms of

ComplRLD

(

A

;

B

) because they are atoms of

RLD

(

A

;

B

).

It remains to prove that if f is an atom of

ComplRLD

(

A

;

B

) then

f

=

A

{

α

RLD

b

for some

α

A

and an ultrafilter

b

on

B

.

Suppose

f

is a non-empty complete reloid. Then

A

{

α

} ×

RLD

b

v

f

for some

α

A

and an ultrafilter

b

on

B

. If

f

is an atom then

f

=

A

{

α

} ×

RLD

b

.

Obvious

787

.

ComplRLD

(

A

;

B

) is an atomistic lattice.

Proposition

788

.

Compl

f

=

F

n

f

|

Src

f

{

α

}

α

Src

f

o

for every reloid

f

.

Proof.

Let’s denote

R

the right part of the equality to be proven.

That

R

is a complete reloid follows from the equality

f

|

Src

f

{

α

}

=

Src

f

{

α

} ×

RLD

im(

f

|

Src

f

{

α

}

)

.

The only thing left to prove is that

g

v

R

for every complete reloid

g

such that

g

v

f

.

Really let

g

be a complete reloid such that

g

v

f

. Then

g

=

G

Src

f

{

α

} ×

RLD

G

(

α

)

α

Src

f

for some function

G

: Src

f

F

(Dst

f

).

We have

Src

f

{

α

} ×

RLD

G

(

α

) =

g

|

Src

f

{

α

}

v

f

|

Src

f

{

α

}

. Thus

g

v

R

.

Conjecture

789

.

Compl

f

u

Compl

g

= Compl(

f

u

g

) for every

f, g

RLD

(

A

;

B

).

Theorem

790

.

Compl

F

R

=

F

h

Compl

i

R

for every set

R

P

RLD

(

A

;

B

) for

every sets

A

,

B

.