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5.39. EQUIVALENT FILTERS AND REBASE OF FILTERS

127

The following is an embedding from filters

A

on a lattice

DA

into the lattice

of filters on

Z

:

S

A

=

n

K

Z

X

∈A

:

X

v

K

o

.

Proposition

700

.

Values of this embedding are filters on the lattice

Z

.

Proof.

That

S

A

is an upper set is obvious.

Let

P, Q

S

A

. Then

P, Q

Z

and there is an

X

∈ A

such that

X

v

P

and

Y

∈ A

such that

Y

v

Q

. So

X

u

Y

∈ A

and

P

u

Q

w

X

u

Y

∈ A

, so

P

u

Q

S

A

.

5.39.1. Rebase of filters.

Definition

701

.

Rebase

for every filter

A

and every

A

Z

is

A ÷

A

=

d

n

A

(

X

u

A

)

X

∈A

o

.

Obvious

702

.

h

A

ui

S

A

is a filter on

A

.

Proposition

703

.

The rebase conforms to the formula

A ÷

A

=

h

A

ui

S

A

.

Proof.

We know that

h

A

ui

S

A

is a filter.

If

P

∈ h

A

ui

S

A

then

P

P

A

and

Y

u

A

v

P

for some

Y

∈ A

. Thus

P

w

Y

u

A

d

n

A

(

Y

u

A

)

Y

∈A

o

.

If

P

d

n

A

(

X

u

A

)

X

∈A

o

then by properties of generalized filter bases, there exists

X

∈ A

such that

P

w

X

u

A

. Also

P

P

A

. Thus

P

∈ h

A

ui

S

A

.

Proposition

704

.

X ÷

Base(

X

) =

X

.

Proof.

Because

X

u

Base(

X

) =

X

for

X

∈ X

.

Proposition

705

.

(

X ÷

A

)

÷

B

=

X ÷

B

if

B

v

A

.

Proof.

(

X ÷

A

)

÷

B

=

d

B

(

Y

u

B

)

Y

d

n

A

(

X

u

A

)

X

∈X

o

=

d

n

B

(

X

u

A

)

X

∈X

o

u ↑

B

B

=

d

n

B

(

X

u

A

u

B

)

X

∈X

o

=

d

n

B

(

X

u

B

)

X

∈X

o

=

X ÷

B

.

Proposition

706

.

If

A

∈ A

then

A ÷

A

=

A ∩

P

A

.

Proof.

A ÷

A

=

h

A

ui

S

A

=

h

A

ui

n

K

Z

X

∈A

:

X

v

K

o

=

K

Z

K

∈A∧

K

P

A

 

=

A ∩

P

A

.

Proposition

707

.

Let filters

X

and

Y

be such that Base(

X

) = Base(

Y

) =

B

.

Then

X ÷

C

=

Y ÷

C

⇔ X

=

Y

for every

Z

3

C

w

B

.

Proof.

X ÷

C

=

Y ÷

C

⇔ X ∪

n

K

P

C

K

w

B

o

=

Y ∪

n

K

P

C

K

w

B

o

⇔ X

=

Y

.

5.39.2. Equivalence of filters.

Definition

708

.

Two filters

A

and

B

(with possibly different base sets) are

equivalent (

A ∼ B

) iff there exists an

X

Z

such that

X

∈ A

and

X

∈ B

and

P

X

∩ A

=

P

X

∩ B

.

Proposition

709

.

X

and

Y

are equivalent iff (

X ∼ Y

) iff

Y

=

X ÷

Base(

Y

)

and

X

=

Y ÷

Base(

X

).

Proof.