background image

Proof.

Y

i

n

RLD

h

CoCompl

f

i

iX

i

=

Y

i

n

RLD

Cor

h

f

i

iX

i

=

Cor

Y

i

n

RLD

h

f

i

i X

i

=

(*)

Cor

Y

i

n

RLD

h

f

i

i

Pr

i

RLD

 

Y

RLD

X

!

=

Cor

*

Y

(

A

)

f

+

Y

RLD

X

=

*

CoCompl

Y

(

A

)

f

+

Y

RLD

X

.

(*) You should verify the special case when

X

i

= 0

F

for some

i

.

Theorem 8.

Let

f

be an indexed family of funcoids.

[TODO: Reverse theorem (for non-least

funcoids).]

1.

Q

f

is directly compact if every

f

i

is directly compact.

2.

Q

f

is reversely compact if every

f

i

is reversely compact.

3.

Q

f

is compact if every

f

i

is compact.

Proof.

It is enough to prove only the first statement.

Let each

f

i

is directly compact.

Let

h

Q

f

i

a

0

. Then

h

Q

f

i

a

=

D

Q

(

A

)

f

E

a

=

Q

i

dom

f

RLD

h

f

i

i

Pr

i

RLD

a

. Thus every

h

f

i

i

Pr

i

RLD

a

0

. Consequently by compactness Cor

h

f

i

i

Pr

i

RLD

a

0

;

Q

i

dom

f

Cor

h

f

i

i

Pr

i

RLD

a

0

;

Cor

Q

i

dom

f

h

f

i

i

Pr

i

RLD

a

0

; Cor

h

Q

f

i

a

0

.

So

Q

f

is directly compact.

I will denote

the diagonal relation.

Proposition 9.

The following expressions are pairwise equal:

1.

h

f

×

f

i

;

2.

F

{h

f

×

f

i

p

|

p

atoms

}

;

3.

F

{h

f

i

x

×

RLD

h

f

i

x

|

x

F

}

;

4.

(

RLD

)

in

F

{h

f

i

x

×

FCD

h

f

i

x

|

x

F

}

;

5.

(

RLD

)

in

F

{

f

(

x

×

FCD

x

)

f

1

|

x

F

}

;

6.

F

{

((

RLD

)

in

f

)

(

x

×

RLD

x

)

((

RLD

)

in

f

)

1

|

x

F

}

.

2