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CHAPTER 15

Counter-examples about funcoids and reloids

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

∆ =

F

(

R

)

l

]

;

[

R

,  >

0

.

I will denote Ω(

A

) the Fréchet filter on a set

A

.

Example

1333

.

There exist a funcoid

f

and a set

S

of funcoids such that

f

u

d

S

6

=

d

h

f

ui

S

.

Proof.

Let

f

= ∆

×

FCD

F

(

R

)

{

0

}

and

S

=

n

FCD

(

R

,

R

)

(]

;+

[

×{

0

}

)

R

,>

0

o

. Then

f

u

l

S

= (∆

×

FCD

F

(

R

)

{

0

}

)

u ↑

FCD

(

R

,

R

)

(]0; +

[

×{

0

}

) =

(∆

u ↑

F

(

R

)

]0; +

[)

×

FCD

F

(

R

)

{

0

} 6

=

FCD

(

R

,

R

)

while

d

h

f

ui

S

=

d

{⊥

FCD

(

R

,

R

)

}

=

FCD

(

R

,

R

)

.

Example

1334

.

There exist a set

R

of funcoids and a funcoid

f

such that

f

d

R

6

=

d

h

f

◦i

R

.

Proof.

Let

f

= ∆

×

FCD

F

(

R

)

{

0

}

,

R

=

n

R

{

0

FCD

R

]

;+

[

R

,>

0

o

.

We have

d

R

=

R

{

0

FCD

R

]0; +

[;

f

d

R

=

FCD

(

R

,

R

)

(

{

0

} × {

0

}

)

6

=

FCD

(

R

,

R

)

and

d

h

f

◦i

R

=

d

{⊥

FCD

(

R

,

R

)

}

=

FCD

(

R

,

R

)

.

Example

1335

.

There exist a set

R

of reloids and a reloid

f

such that

f

d

R

6

=

d

h

f

◦i

R

.

Proof.

Let

f

= ∆

×

RLD

F

(

R

)

{

0

}

,

R

=

n

R

{

0

RLD

R

]

;+

[

R

,>

0

o

.

We have

d

R

=

R

{

0

RLD

R

]0; +

[;

f

d

R

=

RLD

(

R

,

R

)

(

{

0

} × {

0

}

)

6

=

RLD

(

R

,

R

)

and

d

h

f

◦i

R

=

d

{⊥

RLD

(

R

,

R

)

}

=

RLD

(

R

,

R

)

.

Example

1336

.

There exist a set

R

of funcoids and filters

X

and

Y

such that

1

.

X

[

d

R

]

Y ∧

@

f

R

:

X

[

f

]

Y

;

2

.

h

d

R

iX

A

d

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

d

R

=

FCD

(

R

,

R

)

(]0; +

[

×

R

). So

X

[

d

R

]

Y

and

f

R

:

¬

(

X

[

f

]

Y

).

2

With the same

X

and

R

we have

h

d

R

iX

=

>

F

(

R

)

and

h

f

iX

=

F

(

R

)

for

every

f

R

, thus

d

n

h

f

iX

f

R

o

=

F

(

R

)

.

Example

1337

.

d

B∈

T

(

A ×

RLD

B

)

6

=

A ×

RLD

d

T

for some filter

A

and set of

filters

T

(with a common base).

252