background image

1.

a

is prime.

2. For every

A

A

exactly one of

A, A

 

is in

up

a

.

3.

a

is an atom of

F

.

Proof

(1)

(2)

Let

a

be prime. Then

A

A

A

= 1

up

a

. Therefore

A

up

a

A

up

a

. But since

A

A

A

= 0

6∈

up

a

it is impossible

A

up

a

A

up

a

.

(2)

(3)

Obviously

a

6

= 0. Let f.o.

b

a

. So up

b

up

a

. Let

X

up

b

\

up

a

.

Then

X

6∈

up

a

and thus

X

up

a

and consequently

X

up

b

. So

0 =

X

A

X

up

b

and thus

b

= 0. So

a

is atomic.

(3)

(1)

By the previous proposition (taking into account the corollary 10 and

the theorem 23).

11. Some criteria

Theorem 53

For a semifiltered, star-separable, down-aligned filtrator

(

A

;

Z

)

with finitely meet closed and separable core where

Z

is a complete boolean lattice

and both

Z

and

A

are atomistic lattices the following conditions are equivalent

for any

F ∈

A

:

1.

F ∈

Z

;

2.

S

∈ P

A

: (

F ∩

A

S

A

S

6

= 0

⇒ ∃K ∈

S

:

F ∩

A

K 6

= 0)

;

3.

S

∈ P

Z

: (

F ∩

A

S

A

S

6

= 0

⇒ ∃

K

S

:

F ∩

A

K

6

= 0)

.

Proof

Our filtrator is with join-closed core.

(1)

(2)

Let

F ∈

Z

. Then (taking in account the proposition 27)

F ∩

A

[

A

S

6

= 0

⇔ F

+

[

A

S

⇒ ∃K ∈

S

:

F

+

K ⇔ ∃K ∈

S

:

F ∩

A

K 6

= 0

.

(2)

(3)

Obvious.

(3)

(1)

Let the formula (3) be true. Then for

L

Z

and

S

= atoms

Z

L

it takes

the form

F ∩

A

S

A

atoms

Z

L

6

= 0

⇒ ∃

K

S

:

F ∩

A

K

6

= 0 that is

F ∩

A

L

6

=

0

⇒ ∃

K

S

:

F ∩

A

K

6

= 0 because

S

A

atoms

Z

L

=

S

Z

atoms

Z

L

=

L

.

That is

F ∩

A

L

6

= 0

⇒ F ∩

A

K

L

6

= 0 where

K

L

S

. Thus

K

L

is an atom

of both

A

and

Z

(see the theorem 50), so having

F ∩

A

L

6

= 0

⇒ F ⊇

K

L

.

Let

F

=

[

Z

K

L

|

L

Z

,

F ∩

A

L

6

= 0

 

.

39