8.3. RELOIDAL PRODUCT OF FILTERS

187

Theorem

1014

.

A ×

RLD

B

=

d

n

a

×

RLD

b

a

atoms

A

,b

atoms

B

o

for every filters

A

and

B

.

Proof.

Obviously

A ×

RLD

B w

d

n

a

×

RLD

b

a

atoms

A

,b

atoms

B

o

.

Reversely, let

K

up

d

n

a

×

RLD

b

a

atoms

A

,b

atoms

B

o

. Then

K

up(

a

×

RLD

b

) for every

a

atoms

A

,

b

atoms

B

.

K

w

X

a

×

Y

b

for some

X

a

up

a

,

Y

b

up

b

;

K

w

l

X

a

×

Y

b

a

atoms

A

, b

atoms

B

=

l

X

a

a

atoms

A

×

l

Y

b

b

atoms

B

w

A

×

B

where

A

up

A

,

B

up

B

;

K

up(

A ×

RLD

B

).

Theorem

1015

.

If

A

0

,

A

1

F

(

A

),

B

0

,

B

1

F

(

B

) for some sets

A

,

B

then

(

A

0

×

RLD

B

0

)

u

(

A

1

×

RLD

B

1

) = (

A

0

u A

1

)

×

RLD

(

B

0

u B

1

)

.

Proof.

(

A

0

×

RLD

B

0

)

u

(

A

1

×

RLD

B

1

) =

RLD

l

P

u

Q

P

up(

A

0

×

RLD

B

0

)

, Q

up(

A

1

×

RLD

B

1

)

=

RLD

l

(

A

0

×

B

0

)

u

(

A

1

×

B

1

)

A

0

up

A

0

, B

0

up

B

0

, A

1

up

A

1

, B

1

up

B

1

=

RLD

l

(

A

0

u

A

1

)

×

(

B

0

u

B

1

)

A

0

up

A

0

, B

0

up

B

0

, A

1

up

A

1

, B

1

up

B

1

=

RLD

l

K

×

L

K

up(

A

0

u A

1

)

, L

up(

B

0

u B

1

)

=

(

A

0

u A

1

)

×

RLD

(

B

0

u B

1

)

.

Theorem

1016

.

If

S

P

(

F

(

A

)

×

F

(

B

)) for some sets

A

,

B

then

l

A ×

RLD

B

(

A

,

B

)

S

=

l

dom

S

×

RLD

l

im

S.

Proof.

Let

P

=

d

dom

S

,

Q

=

d

im

S

;

l

=

d

n

RLD

B

(

A

,

B

)

S

o

.

P ×

RLD

Q v

l

is obvious.

Let

F

up(

P ×

RLD

Q

). Then there exist

P

up

P

and

Q

up

Q

such that

F

w

P

×

Q

.

P

=

P

1

u · · · u

P

n

where

P

i

dom

S

and

Q

=

Q

1

u · · · u

Q

m

where

Q

j

im

S

.

P

×

Q

=

d

i,j

(

P

i

×

Q

j

).

P

i

×

Q

j

up(

A ×

RLD

B

) for some (

A

,

B

)

S

.

P

×

Q

=

d

i,j

(

P

i

×

Q

j

)

up

l

.

So

F

up

l

.

Corollary

1017

.

d

RLD

T

=

A ×

RLD

d

T

if

A

is a filter and

T

is a set

of filters with common base.

Proof.

Take

S

=

{A} ×

T

where

T

is a set of filters.

Then

d

n

RLD

B

B∈

T

o

=

A ×

RLD

d

T

that is

d

RLD

T

=

A ×

RLD

d

T

.

Definition

1018

.

I will call a reloid

convex

iff it is a join of direct products.