 A stronger conjecture:

Conjecture 14.28.

F

RLD

B

@

A

n

B

@

RLD

B

for some lters

A

,

B

. Particularly, is this

formula true for

A

=

B

u "

R

(0; +

1

)

?

The above conjecture is similar to Fermat Last Theorem as having no value by itself but being

somehow challenging to prove it (not expected to be as hard as FLT however).

Example 14.29.

A

n

B

@

RLD

B

for some lters

A

,

B

.

Proof.

It's enough to prove

A

n

B

=

/

RLD

B

.

Let

+

u "

R

(0; +

1

)

. Let

A

=

B

+

.

Let

K

= (

6

)

j

R

R

.

Obviously

K

2

/

GR

(

RLD

B

)

.

A

n

B v "

RLD

(

Base

(

A

);

Base

(

B

))

K

and thus

K

2

GR

(

A

n

B

)

because

"

FCD

(

Base

(

A

);

Base

(

B

))

K

w

+

FCD

"

Base

(

B

)

B

=

FCD

"

Base

(

B

)

B

for

B

= (0; +

1

)

.

Thus

A

n

B

=

/

RLD

B

.

Example 14.30.

F

RLD

B

@

RLD

B

for some lters

A

,

B

.

[TODO: Does it hold for some

principal lters

A

,

B

?]

Proof.

This follows from the above example.

Proposition 14.31.

(

A

n

B

)

u

(

A

o

B

) =

F

RLD

B

for every lters

A

,

B

.

Proof.

(

A

n

B

)

u

(

A

o

B

)

v

d

f"

RLD

f

j

f

2

Rel

(

Base

(

A

);

Base

(

B

))

;

"

FCD

f

w A

FCD

Bg

=

d

f"

RLD

f

j

f

2

xyGR

(

FCD

B

)

g

= (

RLD

)

out

(

FCD

B

) =

F

RLD

B

.

To nish the proof we need to show

A

n

B w A

F

RLD

B

and

A

o

B w A

F

RLD

B

. By symmetry

it's enough to show

A

n

B w A

F

RLD

B

what is proved above.

Example 14.32.

(

A

n

B

)

t

(

A

o

B

)

@

RLD

B

for some lters

A

,

B

.

Proof.

(based on [

8

]) Let

A

=

B

= (

N

)

. It's enough to prove

(

A

n

B

)

t

(

A

o

B

) =

/

RLD

B

.

Let

X

2 A

,

Y

2 B

that is

X

2

(

N

)

,

Y

2

(

N

)

.

Removing one element

x

from

X

produces a set

P

. Removing one element

y

from

Y

produces

a set

Q

. Obviously

P

2

(

N

)

,

Q

2

(

N

)

.

Obviously

(

P

N

)

[

(

N

Q

)

2

GR

((

A

n

B

)

t

(

A

o

B

))

.

(

P

N

)

[

(

N

Q

)

+

X

Y

because

(

x

;

y

)

2

X

Y

but

(

x

;

y

)

2

/ (

P

N

)

[

(

N

Q

)

.

Thus

(

P

N

)

[

(

N

Q

)

2

/

GR

(

RLD

B

)

by properties of lter bases.

Example 14.33.

(

RLD

)

out

(

FCD

)

f

=

/

f

for some convex reloid

f

.

Proof.

Let

f

=

RLD

B

where

A

and

B

are from example

14.30

.

(

FCD

)(

RLD

B

) =

FCD

B

by proposition

8.9

.

So

(

RLD

)

out

(

FCD

)(

RLD

B

) = (

RLD

)

out

(

FCD

B

) =

F

RLD

B

=

/

RLD

B

.

14.1 Second product. Oblique product

175