Definition 74

We will denote

A/

(

) =

A/

((

)

A

×

A

)

for a set

A

and an

equivalence relation

on a set

B

A

. I will call

a congruence on

A

when

(

)

A

×

A

is a congruence on

A

.

Theorem 72

Let

F

be the set of filters over a boolean lattice

A

and

A ∈

F

.

Consider the function

γ

:

Z

(

D

A

)

A

/

defined by the formula (for every

p

Z

(

D

A

)

)

γp

=

X

A

|

X

F

A

=

p

.

Then:

1.

γ

is a lattice isomorphism.

2.

Q

q

:

γ

1

q

=

Q

F

A

for every

q

A

/

.

Proof

p

Z

(

D

A

) :

γp

6

=

because of the theorem 59. Thus easy to see that

γp

A

/

and that

γ

is an injection.

Let’s prove that

γ

is a lattice homomorphism:

γ

(

p

0

F

p

1

) =

X

A

|

X

F

A

=

p

0

F

p

1

;

γp

0

A

/

γp

1

=

X

0

A

|

X

0

F

A

=

p

0

A

/

X

1

A

|

X

1

F

A

=

p

1

=

X

0

F

X

1

|

X

0

, X

1

A

, X

0

F

A

=

p

0

X

1

F

A

=

p

1

X

A

|

X

F

A

=

p

0

F

p

1

=

γ

(

p

0

F

p

1

)

.

Because

γp

0

A

/

γp

1

and

γ

(

p

0

F

p

1

) are equivalence classes, thus follows

γp

0

A

/

γp

1

=

γ

(

p

0

F

p

1

).

γ

(

p

0

F

p

1

) =

X

A

|

X

F

A

=

p

0

F

p

1

;

γp

0

A

/

γp

1

=

X

0

A

|

X

0

F

A

=

p

0

A

/

X

1

A

|

X

1

F

A

=

p

1

=

X

0

F

X

1

|

X

0

, X

1

A

, X

0

F

A

=

p

0

X

1

F

A

=

p

1

X

0

F

X

1

|

X

0

, X

1

A

,

(

X

0

F

X

1

)

F

A

=

p

0

F

p

1

X

A

|

X

F

A

=

p

0

F

p

1

=

γ

(

p

0

F

p

1

)

.

Because

γp

0

A

/

γp

1

and

γ

(

p

0

F

p

1

) are equivalence classes, thus follows

γp

0

A

/

γp

1

=

γ

(

p

0

F

p

1

).

To finish the proof it’s enough to show that

Q

q

:

q

=

γ

(

Q

F

A

) for

every

q

A

/

. (From this follows that

γ

is surjective because

q

is not empty

and thus

Q

q

:

q

=

γ

(

Q

F

A

).) Really,

γ

(

Q

F

A

) =

X

A

|

X

F

A

=

Q

F

A

= [

Q

] =

q.

52