Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book Funcoids are Filters
by Victor Porton
Web: http://www.mathematics21.org
April 12, 2015
1 Draft status
This is a rough draft.
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.
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.
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 ). 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 .
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.
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. 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 .
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
 Victor Porton. Algebraic General Topology. Volume 1. 2014.
Bibliography 9 