9.6. PROPERTIES PRESERVED BY RELATIONSHIPS

212

2

. If

f

is a (co-)complete reloid then up

f

= up(

FCD

)

f

.

Proof.

By order isomorphism, it is enough to prove the first.

Because up

f

is a filter, by properties of generalized filter bases we have

F

up(

RLD

)

out

f

⇔ ∃

g

up

f

:

F

w

g

F

up

f

.

9.6. Properties preserved by relationships

Proposition

1131

.

(

FCD

)

f

is reflexive iff

f

is reflexive (for every endoreloid

f

).

Proof.

f

is reflexive

1

Rel

Ob

f

v

f

⇔ ∀

F

up

f

: 1

Rel

Ob

f

v

F

1

Rel

Ob

f

v

d

FCD

up

f

1

Rel

Ob

f

v

(

FCD

)

f

(

FCD

)

f

is reflexive.

Proposition

1132

.

(

RLD

)

out

f

is reflexive iff

f

is reflexive (for every endofun-

coid

f

).

Proof.

f

is reflexive

1

Rel

Ob

f

v

f

(corollary

925

)

⇔ ∀

F

up

f

: 1

Rel

Ob

f

v

F

1

Rel

Ob

f

v

d

RLD

up

f

1

Rel

Ob

f

v

(

RLD

)

out

f

(

RLD

)

out

f

is reflexive.

Proposition

1133

.

(

RLD

)

in

f

is reflexive iff

f

is reflexive (for every endofun-

coid

f

).

Proof.

(

RLD

)

in

f

is reflexive iff (

FCD

)(

RLD

)

in

f

if reflexive iff

f

is reflexive.

Obvious

1134

.

(

FCD

), (

RLD

)

in

, and (

RLD

)

out

preserve symmetry of the argu-

ment funcoid or reloid.

Proposition

1135

.

a

×

RLD

F

a

=

for every nontrivial ultrafilter

a

.

Proof.

a

×

RLD

F

a

= (

RLD

)

out

(

a

×

FCD

a

) =

d

RLD

up(

a

×

FCD

a

)

v

1

FCD

u

(

>

FCD

\

1

FCD

) =

FCD

.

Example

1136

.

There exist filters

A

and

B

such that (

FCD

)(

A ×

RLD

F

B

)

@

A ×

FCD

B

.

Proof.

Take

A

=

B

=

a

for a nontrivial ultrafilter

a

.

a

×

RLD

F

a

=

. Thus

(

FCD

)(

a

×

RLD

F

a

) =

@

a

×

FCD

a

.

Conjecture

1137

.

There exist filters

A

and

B

such that (

FCD

)(

A

n

B

)

@

A ×

FCD

B

.

Example

1138

.

There is such a non-symmetric reloid

f

that (

FCD

)

f

is sym-

metric.

Proof.

Take

f

= ((

RLD

)

in

(=)

|

R

)

u

(

)

R

.

f

is non-symmetric because

f

6

(

>

)

R

but

f

(

<

)

R

. (

FCD

)

f

= (=)

|

R

because (=)

|

R

v

f

v

(

RLD

)

in

(=)

|

R

.

Proposition

1139

.

If (

RLD

)

in

f

is symmetric then endofuncoid

f

is symmetric.

Proof.

Suppose (

RLD

)

in

f

is symmetric then

f

= (

FCD

)(

RLD

)

in

f

is symmet-

ric.

Conjecture

1140

.

If (

RLD

)

out

f

is symmetric then endofuncoid

f

is symmet-

ric.

Proposition

1141

.

If

f

is a transitive endoreloid, then (

FCD

)

f

is a transitive

funcoid.

Proof.

f

=

f

f

; (

FCD

)

f

= (

FCD

)(

f

f

); (

FCD

)

f

= (

FCD

)

f

(

FCD

)

f

.

Conjecture

1142

.

There exists a non-transitive endoreloid

f

such that

(

FCD

)

f

is a transitive funcoid.