Chapter 14
Counter-examples

funcoids

and

reloids

For further examples we will use the lter dened by the formula

=

l

"

F

(

R

)

(

¡

"

;

"

)

j

"

2

R

; " >

0

:

I will denote

(

A

)

the Fréchet lter on a set

A

.

Example 14.1.

There exist a funcoid

f

and a set

S

of funcoids such that

f

u

F

S

=

/

F

h

f

u i

S

.

Proof.

Let

f

FCD

"

F

(

R

)

f

0

g

and

S

=

"

FCD

(

R

;

R

)

((

"

; +

1

)

f

0

g

)

j

"

2

R

; " >

0

. Then

f

u

F

S

=

¡

FCD

"

F

(

R

)

f

0

g

u "

FCD

(

R

;

R

)

((0; +

1

)

f

0

g

) =

¡

u "

F

(

R

)

(0; +

1

)

FCD

"

F

(

R

)

f

0

g

=

/ 0

FCD

(

R

;

R

)

while

F

h

f

u i

S

=

0

FCD

(

R

;

R

)

= 0

FCD

(

R

;

R

)

.

Example 14.2.

There exist a set

R

of funcoids and a funcoid

f

such that

f

F

R

=

/

F

h

f

i

R

.

Proof.

Let

f

FCD

"

R

f

0

g

,

R

=

f"

R

f

0

FCD

"

R

(

"

; +

1

)

j

"

2

R

; " >

0

g

.

We have

F

R

=

"

R

f

0

FCD

"

R

(0; +

1

)

;

f

F

R

=

"

FCD

(

R

;

R

)

(

f

0

g  f

0

g

) =

/ 0

FCD

(

R

;

R

)

and

F

h

f

i

R

=

0

FCD

(

R

;

R

)

= 0

FCD

(

R

;

R

)

.

Example 14.3.

There exist a set

R

of reloids and a reloid

f

such that

f

F

R

=

/

F

h

f

i

R

.

Proof.

Let

f

RLD

"

R

f

0

g

,

R

=

f"

R

f

0

RLD

"

R

(

"

; +

1

)

j

"

2

R

; " >

0

g

.

We have

F

R

=

"

R

f

0

RLD

"

R

(0; +

1

)

;

f

F

R

=

"

RLD

(

R

;

R

)

(

f

0

g  f

0

g

) =

/ 0

RLD

(

R

;

R

)

and

F

h

f

i

R

=

0

RLD

(

R

;

R

)

= 0

RLD

(

R

;

R

)

.

Example 14.4.

There exist a set

R

of funcoids and lters

X

and

Y

such that

1.

X

[

F

R

]

Y ^

@

f

2

R

:

X

[

f

]

Y

;

2.

h

F

R

iX

A

F

fh

f

iX j

f

2

R

g

.

Proof.

1. Take

X

and

Y

= 1

F

(

R

)

,

R

=

"

FCD

(

R

;

R

)

((

"

; +

1

)

R

)

j

"

2

R

; " >

0

. Then

F

R

=

"

FCD

(

R

;

R

)

((0; +

1

)

R

)

. So

X

[

F

R

]

Y

and

8

f

2

R

:

:

(

X

[

f

]

Y

)

.

2. With the same

X

and

R

we have

h

F

R

iX

= 1

F

(

R

)

and

h

f

iX

= 0

F

(

R

)

for every

f

2

R

, thus

F

fh

f

iX j

f

2

R

g

= 0

F

(

R

)

.

Example 14.5.

F

fA

RLD

B j B 2

T

g

=

/

RLD

F

T

for some lter

A

and set of lters

T

(with

a common base).

Proof.

Take

R

+

=

f

x

2

R

j

x >

0

g

,

A

,

T

=

f"f

x

g j

x

2

R

+

g

where

"

=

"

R

.

F

T

=

"

R

+

;

RLD

F

T

RLD

"

R

+

.

F

fA

RLD

B j B 2

T

g

=

F

f

RLD

"f

x

g j

x

2

R

+

g

.

We'll prove that

F

f

RLD

"f

x

g j

x

2

R

+

g

=

RLD

"

R

+

.

Consider

K

=

S

ff

x

(

¡

1/

x

; 1/

x

)

j

x

2

R

+

g

.

K

2

GR

(

RLD

"f

x

g

)

and thus

K

2

GR

F

f

RLD

"f

x

g j

x

2

R

+

g

. But

K

2

/

GR

(

RLD

"

R

+

)

.

171