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14. COUNTER-EXAMPLES ABOUT FUNCOIDS AND RELOIDS

195

Proof.

Note that

D

id

FCD

Ω(

U

)

E

X

=

X u

Ω(

U

) for every filter

X

on

U

.

Let

f

= id

FCD

(

U

)

,

g

=

FCD

(

U

;

U

)

((

U

×

U

)

\

id

U

).

Let

x

be a non-trivial ultrafilter on

U

. If

X

x

then card

X

2 (In fact,

X

is infinite but we don’t need this.) and consequently

h

g

i

X

=

>

F

(

U

)

. Thus

h

g

i

x

=

>

F

(

U

)

. Consequently

h

f

u

g

i

x

=

h

f

i

x

u h

g

i

x

=

x

u >

F

(

U

)

=

x.

Also

D

id

FCD

Ω(

U

)

E

x

=

x

u

Ω(

U

) =

x

.

Let now

x

be a trivial ultrafilter. Then

h

f

i

x

=

x

and

h

g

i

x

=

>

F

(

U

)

\

x

. So

h

f

u

g

i

x

=

h

f

i

x

u h

g

i

x

=

x

u

(

>

F

(

U

)

\

x

) =

F

(

U

)

.

Also

D

id

FCD

Ω(

U

)

E

x

=

x

u

Ω(

U

) =

F

(

U

)

.

So

h

f

u

g

i

x

=

D

id

FCD

Ω(

U

)

E

x

for every ultrafilter

x

on

U

. Thus

f

u

g

= id

FCD

Ω(

U

)

.

Example

1036

.

There exist binary relations

f

and

g

such that

FCD

(

A

;

B

)

f

u ↑

FCD

(

A

;

B

)

g

6

=

FCD

(

A

;

B

)

(

f

g

) for some sets

A

,

B

such that

f, g

A

×

B

.

Proof.

From the proposition above.

Example

1037

.

There exists a principal funcoid which is not a complemented

element of the lattice of funcoids.

Proof.

I will prove that quasi-complement of the funcoid id

FCD

(

N

)

is not its

complement. We have:

(id

FCD

(

N

)

)

=

G

c

FCD

(

N

;

N

)

c

id

FCD

(

N

)

w

G

(

N

{

α

FCD

N

{

β

}

α, β

N

,

N

{

α

FCD

N

{

β

id

FCD

(

N

)

)

=

G

N

{

α

FCD

N

{

β

}

α, β

N

, α

6

=

β

=

FCD

(

N

;

N

)

G

{

α

} × {

β

}

α, β

N

, α

6

=

β

=

FCD

(

N

;

N

)

(

N

×

N

\

id

N

)

(used corollary

674

). But by proved above (id

FCD

(

N

)

)

u

id

FCD

(

N

)

6

=

F

(

N

)

.

FiXme

:

Stronger conjecture: (

FCD

f

)

= ¯

f

for every binary relation

f

.

Example

1038

.

There exists a funcoid

h

such that GR

h

is not a filter.

Proof.

Consider the funcoid

h

= id

FCD

Ω(

N

)

. We have (from the proof of proposi-

tion

1035

that

f

GR

h

and

g

GR

h

, but

f

g /

GR

h

.

Example

1039

.

There exists a funcoid

h

6

=

FCD

(

A

;

B

)

such that (

RLD

)

out

h

=

RLD

(

A

;

B

)

.

Proof.

Consider

h

= id

FCD

Ω(

N

)

. By proved above

h

=

f

u

g

where

f

=

id

FCD

(

N

)

=

FCD

(

N

;

N

)

id

N

,

g

=

FCD

(

N

;

N

)

(

N

×

N

\

id

N

).

We have id

N

,

N

×

N

\

id

N

GR

h

.

So

(

RLD

)

out

h

=

l

D

RLD

(

N

;

N

)

E

GR

h

v↑

RLD

(

N

;

N

)

(id

N

(

N

×

N

\

id

N

)) =

RLD

(

N

;

N

)

;