5.39. EQUIVALENT FILTERS AND REBASE OF FILTERS

131

Proof.

The backward implication is obvious. Let now

X ÷

C

=

Y ÷

C

.

Take

x

∈ X

,

y

∈ Y

. We have

X ÷

C

=

x

÷

C

= (

x

÷

B

)

÷

C

for

B

=

C

t

Base(

x

)

t

Base(

y

). Similary

Y ÷

C

= (

y

÷

B

)

÷

C

. Thus (

x

÷

B

)

÷

C

= (

y

÷

B

)

÷

C

and thus

x

÷

B

=

y

÷

B

, so

x

y

that is

X

=

Y

.

Proposition

732

.

A ÷

A

=

d

n

A

(

X

u

A

)

X

S

A

o

for every unfixed filter

A

.

Proof.

Take

a

∈ A

.

l

A

(

X

u

A

)

X

S

A

=

l

A

(

X

u

A

u

Base(

a

))

X

S

A

=

l

A

(

X

u

A

)

X

S

A ∩

P

Base(

a

)

=

l

A

(

X

u

A

)

X

S

a

P

Base(

a

)

=

l

A

(

X

u

A

)

X

a

=

a

÷

A

=

A ÷

A.

5.39.5. The diagram for unfixed filters.

Fix a set

B

.

Lemma

733

.

X 7→ X ÷

B

and

x

7→

[

x

] are mutually inverse order isomorphisms

between

unfixed filter

X

B

S

X

and

F

(

DB

).

Proof.

First,

X ÷

B

F

(

DB

) for

X

unfixed filter

X

B

S

X

and [

x

]

unfixed filter

X

B

S

X

for

x

F

(

DB

).

Suppose

X

0

unfixed filter

X

B

S

X

,

x

=

X

0

÷

B

, and

X

1

= [

x

]. We will prove

X

0

=

X

1

. Really,

x

∈ X

1

,

x

=

k

÷

B

for

k

∈ X

0

,

x

k

, thus

x

∈ X

0

. So

X

0

=

X

1

.

Suppose

x

0

F

(

DB

),

X

= [

x

0

],

x

1

=

X ÷

B

. We will prove

x

0

=

x

1

. Really,

x

1

=

x

0

÷

B

. So

x

1

=

x

0

because Base(

x

0

) = Base(

x

1

) =

B

.

So we proved that they are mutually inverse bijections. That they are order

preserving is obvious.

Lemma

734

.

S

and

X 7→ h

B

ui

X

=

X ∩

P

B

are mutually inverse order

isomorphisms between

F

(

DB

) and

n

X ∈

F

(

Z

)

B

∈X

o

.

Proof.

First,

S

x

n

X ∈

F

(

Z

)

B

∈X

o

for

x

F

(

DB

) because of theorem

728

and

h

B

ui

X ∈

F

(

DB

) obviously.

Let’s prove

h

B

ui

X

=

X ∩

P

B

. If

X

∈ h

B

ui

X

then

X

∈ X

(because

B

∈ X

)

and

X

P

B

. So

X

∈ X ∩

P

B

. If

X

∈ X ∩

P

B

then

X

=

B

X

∈ h

B

ui

X

.

Let

x

0

F

(

DB

),

X

=

S

x

, and

x

1

=

h

B

ui

X

. Then obviously

x

0

=

x

1

.

Let now

X

0

n

X ∈

F

(

Z

)

B

∈X

o

,

x

=

h

B

ui

X

0

, and

X

1

=

S

x

. Then

X

1

=

X

0

n

K

Z

K

w

B

o

=

X

0

.

So we proved that they are mutually inverse bijections. That they are order

preserving is obvious.

Theorem

735

.

The diagram at the figure

5

(with the horizontal “unnamed”

arrow

defined

as the inverse isomorphism of its opposite arrow) is a commutative

diagram (in category

Set

), every arrow in this diagram is an isomorphism. Ev-

ery cycle in this diagram is an identity (therefore “parallel” arrows are mutually

inverse). The arrows preserve order.

Proof.

It’s proved above, that all morphisms (except the “unnamed” arrow,

which is the inverse morphism by definition) depicted on the diagram are bijections

and the depicted “opposite” morphisms are mutually inverse.

That arrows preserve order is obvious.

It remains to apply lemma

193

(taking into account the proof of theorem

728

).