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CHAPTER 14

Counter-examples about funcoids and reloids

For further examples we will use the filter defined by the formula

∆ =

l

F

(

R

)

(

;

)

R

,  >

0

.

I will denote Ω(

A

) the Fréchet filter on a set

A

.

Example

1024

.

There exist a funcoid

f

and a set

S

of funcoids such that

f

u

F

S

6

=

F

h

f

ui

S

.

Proof.

Let

f

= ∆

×

FCD

F

(

R

)

{

0

}

and

S

=

n

FCD

(

R

;

R

)

((

;+

)

×{

0

}

)

R

,>

0

o

. Then

f

u

G

S

= (∆

×

FCD

F

(

R

)

{

0

}

)

u ↑

FCD

(

R

;

R

)

((0; +

)

× {

0

}

) =

(∆

u ↑

F

(

R

)

(0; +

))

×

FCD

F

(

R

)

{

0

} 6

=

FCD

(

R

;

R

)

while

F

h

f

ui

S

=

F

{⊥

FCD

(

R

;

R

)

}

=

FCD

(

R

;

R

)

.

Example

1025

.

There exist a set

R

of funcoids and a funcoid

f

such that

f

F

R

6

=

F

h

f

◦i

R

.

Proof.

Let

f

= ∆

×

FCD

F

(

R

)

{

0

}

,

R

=

n

R

{

0

FCD

R

(

;+

)

R

,>

0

o

.

We have

F

R

=

R

{

0

FCD

R

(0; +

);

f

F

R

=

FCD

(

R

;

R

)

(

{

0

} × {

0

}

)

6

=

FCD

(

R

;

R

)

and

F

h

f

◦i

R

=

F

{⊥

FCD

(

R

;

R

)

}

=

FCD

(

R

;

R

)

.

Example

1026

.

There exist a set

R

of reloids and a reloid

f

such that

f

F

R

6

=

F

h

f

◦i

R

.

Proof.

Let

f

= ∆

×

RLD

F

(

R

)

{

0

}

,

R

=

n

R

{

0

RLD

R

(

;+

)

R

,>

0

o

.

We have

F

R

=

R

{

0

RLD

R

(0; +

);

f

F

R

=

RLD

(

R

;

R

)

(

{

0

} × {

0

}

)

6

=

RLD

(

R

;

R

)

and

F

h

f

◦i

R

=

F

{⊥

RLD

(

R

;

R

)

}

=

RLD

(

R

;

R

)

.

Example

1027

.

There exist a set

R

of funcoids and filters

X

and

Y

such that

1

.

X

[

F

R

]

Y ∧

@

f

R

:

X

[

f

]

Y

;

2

.

h

F

R

iX

A

F

n

h

f

iX

f

R

o

.

Proof.

1

Take

X

= ∆ and

Y

=

>

F

(

R

)

,

R

=

n

FCD

(

R

;

R

)

((

;+

)

×

R

)

R

,>

0

o

.

Then

F

R

=

FCD

(

R

;

R

)

((0; +

)

×

R

). So

X

[

F

R

]

Y

and

f

R

:

¬

(

X

[

f

]

Y

).

2

With the same

X

and

R

we have

h

F

R

iX

=

>

F

(

R

)

and

h

f

iX

=

F

(

R

)

for

every

f

R

, thus

F

n

h

f

iX

f

R

o

=

F

(

R

)

.

Example

1028

.

F

n

RLD

B

B∈

T

o

6

=

A ×

RLD

F

T

for some filter

A

and set of filters

T

(with a common base).

193