background image

Proposition 41

S

F

X

|

X

∈ P

S

 

is closed under finite meets.

Proof

Let

R

=

S

F

X

|

X

∈ P

S

 

. Then

[

A

X

F

[

A

Y

=

[

A

((

X

Y

)

(

X

\

Y

))

F

[

A

Y

=

[

A

(

X

Y

)

F

[

A

(

X

\

Y

)

F

[

A

Y

=

[

A

(

X

Y

)

F

[

A

Y

F

[

A

(

X

\

Y

)

F

[

A

Y

=

[

A

(

X

Y

)

F

[

A

Y

F

0

=

[

A

(

X

Y

)

F

[

A

Y.

Applying the formula

S

A

X

F

S

A

Y

=

S

A

(

X

Y

)

F

S

A

Y

twice we get

[

A

X

F

[

A

Y

=

[

A

(

X

Y

)

F

[

A

(

Y

(

X

Y

))

=

[

A

(

X

Y

)

F

[

A

(

X

Y

) =

[

A

(

X

Y

)

.

But for any

A, B

R

exist

X, Y

∈ P

S

such that

A

=

S

A

X

and

B

=

S

A

Y

.

So

A

F

B

=

S

A

X

F

S

A

Y

=

S

A

(

X

Y

)

R

.

Conjecture 2

1. Every filter object on a set can be partitioned into atomic filter objects.

2. This partition is unique.

19.2. Quasidifference

Problem 1

Which of the following expressions are pairwise equal for all

a, b

F

for

each lattice

F

of filters on a set? (If some are not equal, provide counter-examples.)

1.

T

F

z

F

|

a

b

F

z

 

(quasidifference of

a

and

b

);

2.

S

F

z

F

|

z

a

z

F

b

=

 

(second quasidifference of

a

and

b

);

3.

S

F

(atoms

F

a

\

atoms

F

b

);

4.

S

F

a

F

(

U

\

B

)

|

B

up

b

 

.

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