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Funcoids are Filters

by Victor Porton

Web: http://www.mathematics21.org

April 12, 2015

1 Draft status

This is a rough draft.

In this article notations are used accordingly:

http://www.mathematics21.org/binaries/rewrite-plan.pdf

Particularly hf i

∗

X =

def

fy j x 2 X ^ x f yg for a binary relation f and a set X.

The motto of this article is: “Funcoids are ﬁlters on a lattice.”

2 Rearrangement of collections of sets

Let Q is a set of sets.

Let ≡ be the relation on

S

Q deﬁned by the formula

a ≡ b , 8X 2 Q: (a 2 X , b 2 X):

[TODO: Generalize it by the formula a ≡ b , 8X 2 Q: (a 2 atoms X , b 2 atoms X):]

[TODO: Reloids RLD(A; B) between posets A and B is F(atoms

A

× atoms

B

)?]

Proposition 1. ≡ is an equivalence relation on

S

Q.

Proof.

Reﬂexivity . Obvious.

Symmetry. Obvious.

Transitivity. Let a ≡ b ^ b ≡ c. Then a 2 X , b 2 X , c 2 X for every X 2 Q. Thus a ≡ c.

Deﬁnition 2. Rearrangement R(Q) of Q is the set of equivalence classes of

S

Q for ≡.

Obvious 3.

S

R(Q) =

S

Q.

Obvious 4. ; 2/ R(Q).

Lemma 5. card R(Q) ≤ 2

card Q

.

Proof. Having an equivalence class C, we can ﬁnd the set f 2 PQ of all X 2 Q such that a 2 X

for all a 2 C. b ≡ a ,8X 2 Q: (a 2 X , b 2 X) , 8X 2 Q: (X 2 f , b 2 X). So C = fb 2

S

Q j b ≡ ag

can be restored knowing f . Consequently there are no more than card PQ = 2

card Q

classes.

Corollary 6. If Q is ﬁnite, then R(Q) is ﬁnite.

Proposition 7. If X 2 Q, Y 2 R(Q) then X \ Y =/ ; , Y ⊆ X.

Proof. Let X \ Y =/ ; and x 2 X \ Y . Then y 2 Y , x ≡ y , 8X

0

2 Q: (x 2 X

0

, y 2 X

0

) )

(x 2 X , y 2 X) , y 2 X for every y. Thus Y ⊆ X.

Y ⊆ X ) X \ Y =/ ; because Y =/ ;.

Proposition 8. If ; =/ X 2 Q then there exists Y 2 R(Q) such that Y ⊆ X ^ X \ Y =/ ;.

Proof. Let a 2 X. Then [a] = fb 2

S

Q j 8X

0

2 Q: (a 2 X

0

, b 2 X

0

)g = fb 2

S

Q j 8X

0

2 Q:

b 2 X

0

g ⊆ fb 2

S

Q j b 2 X g = X. But [a] 2 R(Q).

X \ Y =/ ; follows from Y ⊆ X by the previous proposition.

1

Proposition 9. If X 2 Q then X =

S

(R(Q) \ PX).

Proof.

S

(R(Q) \ PX) ⊆ X is obvious.

Let x 2 X. Then there is Y 2 R(Q ) such that x 2 Y . We have Y ⊆ X that is Y 2 PX by a

proposition above. So x 2 Y where Y 2 R(Q) \ PX and thus x 2

S

(R(Q) \ PX). We have

X ⊆

S

(R(Q) \ PX).

3 Finite unions of Cartesian products

Let A, B be sets.

I will denote X = A n X.

Let denote ¡(A;B) the set of all ﬁnite unions X

0

× Y

0

[::: [ X

n¡1

×Y

n¡1

of Cartesian products,

where n 2 N and X

i

2 PA, Y

i

2 PB for every i = 0; :::; n ¡ 1.

Proposition 10. The following sets are pairwise equal:

1. ¡(A; B);

2. the set of all sets of the form

S

X 2S

(X × Y

X

) where S are ﬁnite collections on A and

Y

X

2 PB for every X 2 S;

3. the set of all sets of the form

S

X 2S

(X ×Y

X

) where S are ﬁnite partitions of A and Y

X

2PB

for every X 2 S;

4. the set of all ﬁnite unions

S

(X;Y )2σ

(X × Y ) where σ is a relation between a partition of

A and a partition of B (that is dom σ is a partition of A and im σ is a partition of B).

5. the set of all ﬁnite intersections

T

i=0;:::;n¡1

(X

i

× Y

i

[ X

i

× B) where n 2 N and X

i

2 PA,

Y

i

2 PB for every i = 0; :::; n ¡ 1.

Proof.

(1)⊇(2), (2)⊇(3). Obvious.

(1)⊆(2). Let Q 2 ¡(A; B). Then Q = X

0

× Y

0

[ ::: [ X

n¡1

× Y

n¡1

. Denote S = fX

0

; :::; X

n¡1

g.

We have Q =

S

X

0

2S

(X

0

×

S

fY

i

j X

i

= X

0

g) 2 (2).

(2)⊆(3). Let Q =

S

X 2S

(X × Y

X

) where S is a ﬁnite collection on A and Y

X

2 PB for every

X 2 S. Let

P =

[

X

0

2R(S)

¡

X

0

×

[

fY

X

j X 2 S ^ X

0

⊆ X g

To ﬁnish the proof let's show P = Q.

hP i

∗

fxg =

S

fY

X

j X 2 S ^ X

0

⊆ X g where x 2 X

0

.

Thus hP i

∗

fxg =

S

fY

X

j X 2 S ^ x 2 X g = hQi

∗

fxg. So P = Q.

(4)⊆(3).

S

(X;Y )2σ

(X × Y ) =

S

X 2dom σ

(X ×

S

fY 2 PB j (X; Y ) 2 σ g) 2 (3).

(3)⊆(4).

S

X 2S

(X ×Y

X

)=

S

X 2S

(X ×

S

(R(fY

X

j X 2 Sg) \PY

X

))=

S

X 2S

(X ×

S

fY

0

2

R(fY

X

j X 2 S g) j Y

0

⊆ Y

X

g) =

S

X 2S

(X ×

S

fY

0

2 R(fY

X

j X 2 S g) j (X; Y

0

) 2 σg ) =

S

(X;Y )2σ

(X × Y ) where σ is a relation between S and R(fY

X

j X 2 S g), and (X;

Y

0

) 2 σ , Y

0

⊆ Y

X

.

(5)⊆(3). Obvious.

(3)⊆(5). Let Q =

S

X 2S

(X × Y

X

) =

S

i=0;:::;n¡1

(X

i

× Y

i

) for a partition S = fX

0

; :::; X

n¡1

g

of A. Then Q =

T

i=0;:::;n¡1

(X

i

× Y

i

[ X

i

× B).

Exercise 1. Formulate the duals of these sets.

Proposition 11. ¡(A; B) is a boolean lattice, a sublattice of the lattice P(A × B).

Proof. That it's a sublattice is obvious. That it has complement, is also obvious. Distributivity

follows from distributivity of P (A × B).

2 Section 3

I will denote F¡(A; B) = f(A; B; F ) j F 2 F¡[A; B]g.

Remark 12. It should be instead be denoted as (F ◦ ¡)(A; B) but for brevity I omit ◦.

4 Before the diagram

Next we will prove the below theorem 35 (the theorem with a diagram). First we will present

parts of this theorem as several lemmas, and then then state a statement about the diagram which

concisely summarizes the lemmas (and their easy consequences).

Obvious 13. up

¡(Src f;Dst f )

f = (up f) \ ¡ for every reloid f .

Conjecture 14.

F(B)

up

A

X is not a ﬁlter for some ﬁlter X 2 F¡(A; B) for some sets A, B.

Remark 15. About this conjecture see also:

• http://goo.gl/DHyuuU

• http://goo.gl/4a6wY6

Lemma 16. Let A, B be sets. The following are mutually inverse order isomorphisms between

F¡(A; B) and FCD(A; B):

1. A 7!

d

FCD

up A;

2. f 7! up

¡(A;B)

f.

Proof. Let's prove that up

¡(A;B)

f is a ﬁlter for every funcoid f. We need to prove that P \ Q 2up f

whenever

P =

\

i=0;:::;n¡1

(X

i

× Y

i

[ X

i

× B) and Q =

\

j=0;:::;m ¡1

(X

j

0

× Y

j

0

[ X

j

0

× B):

This follows from P 2 up f , 8i 2 0; :::; n ¡ 1: hf iX

i

⊆ Y

i

and likewise for Q, so having

hf i(X

i

\ X

j

0

) ⊆ Y

i

\ Y

j

0

for every i = 0; :::; n ¡ 1 and j = 0; :::; m ¡ 1. From this it follows

((X

i

\ X

j

0

) × (Y

i

\ Y

j

0

)) [ (X

i

\ X

j

0

× B) ⊇ f

and thus P \ Q 2 up f .

Let A, B be ﬁlters on ¡. Let

d

FCD

up A =

d

FCD

up B. We need to prove A = B. (The rest

follows from proof of the theorem 6.104 from my book [1]). We have: [TODO: Separate the ﬁrst

equality below from theorem 6.104 into a separate lemma.]

A =

l

FCD

fX × Y [ X × B 2 A j X 2 PA; Y 2 PBg =

l

FCD

fX × Y [ X × B j X 2 PA; Y 2 PB; 9P 2 A: P ⊆ X × Y [ X × B g =

l

FCD

fX × Y [ X × B j X 2 PA; Y 2 PB; 9P 2 A: hP i

∗

X ⊆ Y g = (*)

l

FCD

X × Y [ X × B j X 2 PA; Y 2 PB;

l

fhP i

∗

X j A 2 up Ag v Y

=

l

FCD

8

<

:

X × Y [ X × B j X 2 PA; Y 2 PB;

l

(

hP i

∗

X j A 2 up

l

RLD

up A

)

v Y

9

=

;

=

l

FCD

8

<

:

X × Y [ X × B j X 2 PA; Y 2 PB;

*

(FCD)

l

RLD

up A

+

X v Y

9

=

;

= (**)

l

FCD

8

<

:

X × Y [ X × B j X 2 PA; Y 2 PB;

*

l

FCD

up

l

RLD

up A

+

X v Y

9

=

;

=

l

FCD

8

<

:

X × Y [ X × B j X 2 PA; Y 2 PB;

*

l

FCD

up A

+

X v Y

9

=

;

:

Before the diagram 3

(*) by properties of generalized ﬁlter bases, because fhP i

∗

X j P 2 Ag is a ﬁlter base.

(**) by theorem 8.3 in [1].

Similarly

up B =

l

FCD

8

<

:

X × Y [ X × B j X 2 PA; Y 2 PB;

*

l

FCD

up B

+

X v Y

9

=

;

:

Thus A = B.

[TODO: For pointfree funcoids?]

Proposition 17. g ◦ f 2 ¡(A; C) if f 2 ¡(A; B) and g 2 ¡(B; C) for some sets A, B, C.

Proof. Because composition of Cartesian products is a Cartesian product.

Deﬁnition 18. g ◦ f =

d

F¡(A;C)

fG ◦ F j F 2 up f ; G 2 up gg for f 2 F¡(A; B) and g 2 F¡(B; C)

(for every sets A, B, C).

We deﬁne f

¡1

for f 2 F¡(A; B) similarly to f

¡1

for reloids and similarly derive the formulas:

1. (f

¡1

)

¡1

= f;

2. (g ◦ f )

¡1

= f

¡1

◦ g

¡1

.

4.1 Associativity over composition

I will denote base (A; Z) = A, core(A; Z) = Z for a ﬁltrator (A; Z). [TODO: move above in the book]

Obvious 19. P(core F) \

d

F(base F)

up

base F

f = f for f ??.

Corollary 20.

d

F(base F)

up

base F

is an injection.

Lemma 21.

d

RLD

up

¡(A;C)

(g ◦ f ) =

¡

d

RLD

up

¡(B;C)

g

◦

¡

d

RLD

up

¡(B;C)

for every f 2F(¡(A; B)),

g 2 F(¡(B; C)) (for every sets A, B, C).

Proof. If K 2

d

RLD

up

¡(A;C)

(g ◦ f) then K ⊇ G◦ F for some F 2 f, G 2 g. But F 2up

¡(A;B)

f, thus

F 2

l

RLD

up

¡(A;B)

f

and similarly

G 2

l

RLD

up

¡(B;C)

g:

So we have

K ⊇ G ◦ F 2

l

RLD

up

¡(B;C)

g

!

◦

l

RLD

up

¡(A;B)

f

!

:

Let now

K 2

l

RLD

up

¡(B;C)

g

!

◦

l

RLD

up

¡(A;B)

f

!

:

Then there exist F 2

d

RLD

up

¡(A;B)

f and G 2

d

RLD

up

¡(B;C)

g such that K ⊇G◦ F . By properties

of generalized ﬁlter bases we can take F 2 up

¡(A;B)

f and G 2 up

¡(B;C)

g. Thus K 2 up

¡(A;C)

(g ◦ f )

and so K 2

d

RLD

up

¡(A;C)

(g ◦ f).

Lemma 22. (FCD)

d

RLD

f =

d

FCD

up f for every f 2 F¡(A; B) (where A, B are sets).

4 Section 4

Proof. (FCD)

d

RLD

f =

d

FCD

up

d

RLD

f =

d

FCD

up f.

Proposition 23. (RLD)

in

(f t g) = (RLD)

in

f t (RLD)

in

g for every funcoids f ; g 2 FCD(A; B).

[TODO: Move it above in the book.]

Proof. (RLD)

in

(f t g) =

F

RLD

a ×

RLD

b j a 2 atoms

F(A)

; b 2 atoms

F(B)

; a ×

FCD

b v f t g

=

F

RLD

a ×

RLD

b j a 2 atoms

F(A)

; b 2 atoms

F(B)

; a ×

FCD

b v f _ a ×

FCD

b v g

=

F

RLD

a ×

RLD

b j a 2

atoms

F(A)

; b 2 atoms

F(B)

; a ×

FCD

b v f

t

F

RLD

a ×

RLD

b j a 2 atoms

F(A)

; b 2 atoms

F(B)

;

a ×

FCD

b v g

= (RLD)

in

f t (RLD)

in

g.

Lemma 24. (RLD)

in

X = X for X 2 ¡(A; B).

Proof. X = X

0

× Y

0

[ ::: [ X

n

× Y

n

= (X

0

×

FCD

Y

0

) t

FCD

::: t

FCD

(X

n

×

FCD

Y

n

).

(RLD)

in

X = (RLD)

in

(X

0

×

FCD

Y

0

) t

RLD

::: t

RLD

(RLD)

in

(X

n

×

FCD

Y ) =

(X

0

×

RLD

Y

0

) t

RLD

::: t

RLD

(X

n

×

RLD

Y

n

) = X

0

× Y

0

[ ::: [ X

n

× Y

n

= X.

Lemma 25.

d

RLD

up f = (RLD)

in

d

FCD

up f for every ﬁlter f 2 F¡(A; B).

Proof. (RLD)

in

d

FCD

f =

d

RLD

h(RLD)

in

i

∗

up f = (by the previous lemma)=

d

RLD

up f.

Lemma 26.

1. f 7!

d

RLD

up f and A 7! ¡(A; B) \ up A are mutually inverse bijections between F¡(A; B)

and a subset of reloids.

2. These bijections preserve composition.

Proof. 1. That they are mutually inverse bijections is obvious.

2.

¡

d

RLD

up g

◦

¡

d

RLD

up f

=

d

RLD

G ◦ F j F 2

d

RLD

f ; G 2

d

RLD

g

=

d

RLD

fG ◦

F j F 2 f ; G 2 g g =

d

RLD

d

F¡(Src f ;Dst g)

fG ◦ F j F 2 f ; G 2 gg =

d

RLD

(g ◦ f). So

d

RLD

preserves composition. That A 7! ¡(A; B) \ up A preserves composition follows from properties of

bijections.

Lemma 27. Let A, B, C be sets.

1.

¡

d

FCD

up g

◦

¡

d

FCD

up f

=

d

FCD

up(g ◦ f) for every f 2 F¡(A; B), g 2 F¡(B; C);

2.

¡

up

¡(B;C)

g

◦

¡

up

¡(A;B)

f

= up

¡(A;B)

(g ◦ f) for every funcoids f 2 FCD(A; B) and

g 2 FCD(B: C).

Proof. It's enough to prove only the ﬁrst formula, because of the bijection from thereom 16.

Really:

d

FCD

up(g ◦ f ) =

d

FCD

up

d

RLD

up(g ◦ f ) =

d

FCD

up

¡

d

RLD

up g ◦

d

RLD

up f

=

(FCD)

¡

d

RLD

up g ◦

d

RLD

up f

=

¡

(FCD)

d

RLD

up g

◦

¡

(FCD)

d

RLD

up f

=

¡

d

FCD

up

d

RLD

up g

◦

¡

d

FCD

up

d

RLD

up f

=

¡

d

FCD

up g

◦

¡

d

FCD

up f

.

Corollary 28. (h ◦ g) ◦ f = h ◦ (g ◦ f ) for every f 2 F(¡(A; B)), g 2 F¡(B; C), h 2 F¡(C; D) for

every sets A, B, C, D.

Lemma 29. ¡(A; B) \ GR f is a ﬁlter on the lattice ¡(A; B) for every reloid f 2 RLD(A; B).

Proof. That it is an upper set, is obvious. If A; B 2 ¡(A; B) \ GR f then A; B 2 ¡(A; B) and A;

B 2 GR f. Thus A \ B 2 ¡(A ; B) \ GR f .

Proposition 30. If Y 2 up hf iX for a funcoid f then there exists A 2upX such that Y 2 up hf iA.

Proof. Y 2 up

d

F

fhf iA j A 2 up ag.

Before the diagram 5

So by properties of generalized ﬁlter bases, there exists A 2 up a such that Y 2 up hf iA.

Lemma 31. (FCD)f =

d

FCD

(¡(A; B) \ GR f) for every reloid f 2 RLD(A; B).

Proof. Let a be an ultraﬁlter. We need to prove

h(FCD)f ia =

*

l

FCD

(¡(A; B) \ GR f)

+

a

that is

*

l

FCD

up f

+

a =

*

l

FCD

(¡(A; B) \ GR f)

+

a

that is

l

F 2up f

F

hF ia =

l

F 2¡(A;B)\up f

F

hF ia:

For this it's enough to prove that Y 2 up hF ia for some F 2 up f implies Y 2 up hF

0

ia for some

F

0

2 ¡( A; B) \ GR f .

Let Y 2 up hF ia. Then (proposition above) there exists A 2 up a such that Y 2 up hF iA.

Y 2 up hA ×

FCD

Y t A ×

FCD

1ia; hA ×

FCD

Y t A ×

FCD

1iX = Y 2 up hF iX if 0 =/ X v A and

hA ×

FCD

Y t A ×

FCD

1iX = 1 2 up hF iX if X v A.

Thus A ×

FCD

Y t A ×

FCD

1 w F . So A ×

FCD

Y t A ×

FCD

1 is the sought for F

0

.

4.2 Relationships between (FCD) and (RLD)

¡

Deﬁnition 32. (RLD)

¡

f =

d

RLD

up

¡(Src f ;Dst f )

f for every funcoid f. I call (RLD)

¡

as ¡-reloid

or Gamma-reloid.

Lemma 33. (FCD)(RLD)

¡

f = f for every funcoid f.

Proof. For every ﬁlter X 2 F(Src f) we have h (FCD)(RLD)

¡

f iX =

d

F 2up (RLD)

¡

f

F

hF iX =

d

F 2up

¡(Src f ;Dst f )

f

F

hF iX .

Obviously

d

F 2up

¡(Src f;Dst f)

f

F

hF iX w hf iX . So (FCD)(RLD)

¡

f w f .

Let Y 2 up hf iX . Then (propositiona above) there exists A 2 up X such that Y 2 up hf iA.

Thus A × Y [ A × 1 2 up f. So h(FCD)(RLD)

¡

f iX =

d

F 2up

¡(Src f;Dst f)

f

F

hF iX v hA ×

Y [ A × 1iX = Y . So Y 2 up h(FCD)(RLD)

¡

f iX that is hf iX w h(FCD)(RLD)

¡

f iX that is

f w (FCD)(RLD)

¡

f.

Proposition 34. (RLD)

¡

is neither upper nor lower adjoint of (FCD) (in general).

Proof. It is not upper adjoint because (RLD)

in

is the upper adjoint of (FCD) and (RLD)

in

=/ (RLD)

¡

.

If (RLD)

¡

is the lower adjoint of (FCD), then f w (RLD)

¡

(FCD) f and thus f w (RLD)

in

(FCD) f.

But f v (RLD)

in

(FCD) f, thus having (RLD)

in

(FCD) f = f what is not an identity (take f = (=)j

A

for an inﬁnite set A).

5 The diagram

Theorem 35 . The following is a commutative diagram (in category Set), every arrow in this

diagram is an isomorphism. Every cycle in this diagram is an identity (therefore “parallel” arrows

are mutually inverse). The arrows preserve order, composition, and reversal (f 7! f

¡1

).

6 Section 5

funcoidal reloids

(RLD)

in

(FCD)

funcoids

d

RLD

ﬁlters on ¡

f 7! f \ ¡

up

¡

d

FCD

Proof. First we need to show that

d

RLD

f is a funcoidal reloid. But it follows from lemma 25.

Next, we need to show that all morphisms depicted on the diagram are bijections and the

depicted “opposite” morphisms are mutually inverse.

That (FCD) and (RLD)

in

are mutually inverse was proved above in the book.

That

d

RLD

and f 7! f \ ¡ are mutually inverse was proved above.

That

d

FCD

and up

¡

are mutually inverse was proved above.

It remains to prove that three-element cycles are identities. But this follows from lemma 25.

That the morphisms preserve order and composition was proved above. That they preserve

reversal is obvious.

6 Some additional properties

Proposition 36. For every funcoid f 2 FCD(A; B) (for sets A, B):

1. dom f =

d

F(A)

hdomi

∗

up

¡(A;B)

f;

2. im f =

d

F(A)

himi

∗

up

¡(A;B)

f.

Proof. Take fX × Y j X 2 PA; Y 2 PA; X × Y ⊇ f g ⊆ up

¡(A;B)

f. I leave the rest reasoning as

an exercise.

Proposition 37. (RLD)

¡

f w (RLD)

in

f w (RLD)

out

f for every funcoid f.

Proof. We already know that (RLD)

in

f w (RLD)

out

f (see above in the book).

The formula (RLD)

¡

f w (RLD)

in

f follows from 8G 2 up

¡(Src f;Dst f )

f: G w f.

Example 38. (RLD)

¡

f A (RLD)

in

f A (RLD)

out

f for some funcoid f.

Proof. Take f = (=)j

R

. We already know that (RLD)

in

f A (RLD)

out

f (see above in the book).

It remains to prove (RLD)

¡

f =/ (RLD)

in

f.

Take F =

S

i2Z

([i; i + 1[ × [i; i + 1[).

Then F 2 f = up (RLD)

in

f (because hF ia w hf ia for both principal ultraﬁlter a = fig and every

other ultraﬁlter a).

It remains to prove F 2/ up (RLD)

¡

f.

Some additional properties 7

Suppose F 2 up (RLD)

¡

= up

d

RLD

up

¡(Src f ;Dst f )

f. Then by properties of generalized ﬁlter

bases, there is F

0

2 up

¡(Src f;Dst f )

f such that F ⊇ F

0

. Because F

0

⊆

S

i2Z

([i; i + 1[ × [i; i + 1[) and

F

0

⊇ (=)j

R

, there is a point q 2 [i; i + 1[ × [i; i + 1[ for each i 2 Z ; thus, F

0

2/ ¡(Src f ; Dst f).

Thus F 2/ up (RLD)

¡

f.

Theorem 39. For every reloid f and X 2 F(Src f), Y 2 F(Dst f):

1. X [(FCD) f ] Y , 8F 2 up

¡(Src f;Dst f )

f: X [F ] Y;

2. h(FCD)f iX =

d

F 2up

¡(Src f ;Dst f )

f

F

hF iX .

Proof. 1. 8F 2 up

¡(Src f;Dst f )

f: X [F ] Y , (by properties of generalized ﬁlter bases, taking into

account that funcoids are isomorphic to ﬁlters),X

d

FCD

up

¡(Src f;Dst f )

f

Y , X [(FCD)f] Y.

2.

d

F 2up

¡(Src f ;Dst f )

f

F

hF ia =

d

FCD

up

¡(Src f;Dst f )

f

a = h(FCD)f ia for every ultraﬁlter a.

It remains to prove that the function

' = λ X 2 F(Src f ):

l

F 2up

¡(Src f ;Dst f )

f

F

hF iX

is a component of a funcoid (from what follows that ' = h(FCD)f i). To prove this, it's enough to

show that it preserves ﬁnite joins and ﬁltered meets. [TODO: Deﬁnition of ﬁltered meets.]

'0 = 0 is obvious. '(I t J ) =

d

F 2up

¡(Src f ;Dst f )

f

F

(hF iI t hF iJ ) =

d

F 2up

¡(Src f ;Dst f )

f

F

hF iI t

d

F 2up

¡(Src f ;Dst f )

f

F

hF iJ = ' I t ' J . If S is a generalized ﬁlter base of Src f , then '

d

F

S =

d

F 2up

¡(Src f ;Dst f )

f

F

hF i

d

F

S =

d

F 2up

¡(Src f ;Dst f )

f

F

d

F

hhF ii

∗

S =

d

F 2up

¡(Src f ;Dst f )

f

F

d

X 2S

F

hF iX =

d

X 2S

F

d

F 2up

¡(Src f ;Dst f )

f

F

hF iX =

d

X 2S

F

' X =

d

F

h'i

∗

S.

So ' is a component of a funcoid.

Deﬁnition 40. f =

d

RLD

up

¡(Src f;Dst f )

f for reloid f.

Conjecture 41. For every reloid f :

1. f = (RLD)

in

(FCD) f ;

2. f = (RLD)

¡

(FCD) f.

Obvious 42. f w f for every reloid f .

Example 43. (RLD)

¡

f =/ (RLD)

out

f for some funcoid f .

Proof. Take f = id

Ω(N )

FCD

. Then, as it was shown above, (RLD)

out

f = 0 and thus (RLD)

out

f = 0.

But (RLD)

¡

f w (RLD)

in

f =/ 0. So (RLD)

¡

f =/ (RLD)

out

f.

Conjecture 44. (RLD)

¡

f = (RLD)

in

f for every funcoid f.

Proposition 45. [TODO: Move it above in the book.] f v A ×

FCD

B , dom f v A ^ im f v B for

every funcoid f and ﬁlters A 2 F(Src f), B 2 F(Dst f).

Proof. f v A ×

FCD

B ) dom f v A because dom(A ×

FCD

B) v A.

Let now dom f v A ^ im f v B. Then hf iX =/ 0 ) X / A that is f v A ×

FCD

1. Similarly

f v 1 ×

FCD

B. Thus f v A ×

FCD

B.

Theorem 46. dom (RLD)

in

f = dom f and im (RLD)

in

f = im f for every funcoid f . [TODO: Move

it above in the book, remove the conjecture which this statement proves.]

Proof. We have for every ﬁlter X 2 F(Src f):

X w dom (RLD)

in

f , X ×

RLD

1 w ( RLD)

in

f , 8a 2 F(Src f ); b 2 F(Dst f): (a ×

FCD

b v f )

a ×

RLD

b v X ×

RLD

1) , 8a 2 F(Src f); b 2 F(Dst f ): (a ×

FCD

b v f ) a v X );

8 Section 6

X w dom f , X ×

RLD

1 w f , X ×

FCD

1 w f , 8a 2 F(Src f) ; b 2 F(Dst f ): (a ×

FCD

b v f )

a ×

FCD

b v X ×

FCD

1) , 8a 2 F(Src f ); b 2 F(Dst f ): (a ×

FCD

b v f ) a v X ).

Thus dom (RLD)

in

f = dom f. The rest follows from symmetry.

Proposition 47. dom (RLD)

¡

f = dom f and im ( RLD)

¡

f = im f for every funcoid f .

Proof. dom (RLD )

¡

f w dom f and im (RLD)

¡

f w im f because (RLD)

¡

f w (RLD)

in

and

dom (RLD)

in

f = dom f and im (RLD)

in

f = im f .

It remains to prove (as the rest follows from symmetry) that dom (RLD)

¡

f v dom f.

Really, dom (RLD)

¡

f v

d

F

fX 2 up dom f j X × 1 2 up f g =

d

F

fX 2 up dom f j X 2

up dom f g =

d

F

up dom f = dom f.

Conjecture 48. For every funcoid g we have Cor (RLD)

¡

g = (RLD)

¡

Cor g.

7 More on properties of funcoids

Proposition 49. ¡(A; B) is the center of lattice FCD(A; B).

Proof. See theorem 4.139 in [1].

Proposition 50. up

¡(A;B)

(A ×

FCD

B) is deﬁned by the ﬁlter base fA × B j A 2 up A; B 2 up Bg

on the lattice ¡(A; B).

Proof. It follows from the fact that A ×

FCD

B =

d

FCD

fA × B j A 2 up A; B 2 up Bg.

Proposition 51. up

¡(A;B)

(A ×

FCD

B) = F(¡(A; B)) \ (A ×

RLD

B).

Proof. It follows from the fact that A ×

FCD

B =

d

FCD

fA × B j A 2 up A; B 2 up Bg.

Proposition 52. For every f 2 F(¡(A; B)):

1. f ◦ f is deﬁned by the ﬁlter base fF ◦ F j F 2 up f g (if A = B);

2. f

¡1

◦ f is deﬁned by the ﬁlter base fF

¡1

◦ F j F 2 up f g;

3. f ◦ f

¡1

is deﬁned by the ﬁlter base fF ◦ F

¡1

j F 2 up f g.

Proof. I will prove only (1) and (2) because (3) is analogous to (2).

1. It's enough to show that 8F ; G 2 up f9H 2 up f: H ◦ H vG ◦ F . To prove it take H = F u G.

2. It's enough to show that 8F ; G 2 up f9H 2 up f : H

¡1

◦ H v G

¡1

◦ F . To prove it take

H = F u G. Then H

¡1

◦ H = (F u G)

¡1

◦ (F u G) v G

¡1

◦ F .

Theorem 53. For every sets A, B, C if g; h 2 F¡(A; B) then

1. f ◦ (g t h) = f ◦ g t f ◦ h;

2. (g t h) ◦ f = g ◦ f t h ◦ f.

Proof. It follows from the order isomorphism above, which preserves composition.

Bibliography

[1] Victor Porton. Algebraic General Topology. Volume 1. 2014.

Bibliography 9