Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book This ﬁle is with partial proofs (with rough gibberish) about open problems I have tried to solve
but have failed. If you solve something of this please notify me by email porton@narod.ru.
Conjecture 1. (RLD)
out
(g f )w (RLD)
out
g (RLD)
out
f ( v or w) for composable funcoids f and
g.
Proof. (RLD)
out
g (RLD)
out
f =
d
G2up g
RLD
G
d
F 2up f
RLD
F v G F for every F 2 up f and G 2 up g.
Thus (RLD)
out
g (RLD)
out
f v
d
F 2up f ;G2up g
RLD
(G F ) v ?? [FIXME: Opposite inequalities!]
(RLD)
out
(g f ) =
d
RLD
up(g f ) = ?? =
d
F 2up f ;G 2up g
RLD
(G F ) w
d
G2up (RLD)
out
g;F 2up (RLD)
out
f
RLD
(G F ) =
d
G2up
d
R L D
up g;F 2up
d
R L D
up f
RLD
(G F ) =
d
RLD
up g
d
RLD
up f
Proposition 2. A product of nonempty posets is a dcpo if and only iﬀ each multiplier is a dcpo.
Proof. [TODO: More detailed proof] Suppose one multiplier is not a dcpo. Take a chain with ﬁxed
elements (thanks our posets are nonempty) from other multipliers and for this multiplier take the
values which form a chain without the join. This proves that the product is not a dcpo.
Now take that all multipliers are dcpo. Take a chain that is a set for which holds a vb ^ b v a )
a = b.
We have
??
Take an element t of the chain.
For each k we have a
i
0
=
a
i
if i = k
t
i
if i =/ k
and b
i
0
=
b
i
if i = k
t
i
if i =/ k
a
0
?? belongs to the chain?
??We have a
i
v b
i
^ b
i
v a
i
) a
i
= b
i
for all a, b in the chain. Take
Let . Take
??
It is a chain componentwise because the predicate a v b ^ b v a ) a = b holds?? componentwise.
Thus every component has a (calculated componentwise) join.
Conjecture 3. Funcoids f from A to B bijectively corresponds to the sets R of pairs (X ; Y) of
ﬁlters (on A and B correspondingly) that
1. R is nonempty.
2. R is a lower set.
3. R is a dcpo (or what is the same product of two dcpos)
4. We can add axiom: A ×
FCD
B v
F
fX ×
FCD
Y j (X ; Y) 2 R
0
g ) (A; B) 2 R
by the mutually inverse formulas:
(X ; Y) 2 R , X ×
FCD
Y v f (1)
f =
G
fX ×
FCD
Y j (X ; Y) 2 Rg: (2)
Proof. Let R be deﬁned by formula (1). That R is a nonempty lower set is obvious. Let's prove
that R is a dcpo.
Let T be a chain in R and 8(A; B) 2 T : A ×
FCD
B v f.
8(A; B) 2 T : (X / A ) B v hf iX )
taking join we have:
X / A )
F
B2Pr
1
T
B v hf iX
A ×
FCD
F
B2Pr
1
T
B v f
Repeating this, we get
F
A2Pr
0
T
A ×
FCD
F
B2Pr
1
T
B v f . Thus R is a dcpo.
It remains to prove that the formulas are mutually inverse.
Let f
0
be a funcoid, R be induced by f
0
by formula (1), f
1
be induced by R by formula (2).
We will prove that f
1
= f
0
.
Misc 1 f
1
=
F
fX ×
FCD
Y j X ×
FCD
Y v f
0
g what is equal to f
0
because the lattice of funcoids is
atomistic and every atom is a funcoidal product.
Let now R
0
be a set of pairs of ﬁlters conformung to our axioms, f be a funcoid induced by R
by formula (2), R
1
be a set of pairs of ﬁlters induced by f by formula (1). We will prove R
1
= R
0
.
(X ; Y) 2 R
1
, X ×
FCD
Y v
F
fX ×
FCD
Y j (X ; Y) 2 R
0
g ( (X ; Y) 2 R
0
X ×
FCD
Y v
F
fX ×
FCD
Y j (X ; Y) 2 R
0
g ) (X ; Y) 2 R
0
because ??. [FIXME: It seems we need
additional axioms!]
Theorem 4. If a funcoid is weakly metamonovalued, then it is monovalued.
Proof. We have (g u h) f = (g f) u (h f) for every reloids g, h. We need to prove that f is
monovalued.
Prove that exists F 2 up f such that for every g, h we have (g u h) F = (g F ) u (h F ).
it's enough to prove that (g u h) F w (g F ) u (h F )
Really, ??
thus F is monovalued.
f f
¡1
=
d
fF F
¡1
j F 2 up f g
Proposition 5. The following statements are equivalent for every endofuncoid µ and a set U:
1. U is connected regarding µ.
2. For every a; b 2 U there exists a totally ordered set P U such that min P = a, max P = b,
and for every partion fX; Y g of P into two sets X, Y such that 8x 2 X ; y 2 Y : x < y, we
have X [µ]
Y.
Proof.
(. Let A; B 2 PU are nonempty. We need to prove A [µ]
B. We can assume without loss of
generality that A \ B = ;. Because A and B are nonempty, we can take a 2 A and b 2 B. ??
). The case a =b is trivial. Assume a=/ b. Take A; B 2PU such that a 2A, b 2 B and A[B = U
and A \ B = ;. Then take orders P
A
on A with min P
A
= a and P
B
on B with max P
B
= b.
Then the poset P = P
A
+ P
B
??
1 Directed funcoids
Let [¡1; +1] be the extended real line with the complete funcoid induced by the usual topology
on this set.
Proposition 6. Every ultraﬁlter on [¡1; +1] converges to exactly one point.
Proof. It is a well known fact.
Below is wrong: http://math.stackexchange.com/q/1874451/4876
[FIXME: below is wrong] Use http://math.stackexchange.com/a/1874862/4876
[TODO:
+
!
>
or
]
[TODO: Compare without explicit formulas.]
Lemma 7. For every ultraﬁlter a and its limit point x:
0. h[¡1; +1]ia = ∆(x) if x = a
1. h[¡1; +1]ia =
+
(x) if x > a
2. h[¡1; +1]ia =
¡
(x) if x < a
Proof. 0. Obvious.
1. h[¡1; +1]ia =
d
A2up a
h[¡1; +1]i
A =
d
A2up a; Av]x;+1]
h[¡1; +1]i
A =
d
A2up a; Av]x;+1]
F
y2A
∆(y)
2 Section 1 Because every y > x, we have ∆(y) v ]x; +1] and thus h[¡1; +1]ia v ]x; +1].
It is clear that
F
y2A
∆(y) w
+
(x) and h[¡1; +1]ia v
+
(x). Thus h[¡1; +1]ia =
+
(x).
2. h[¡1; +1]ia =
d
A2up a
h[¡1; +1]i
A =
d
A2up a;Av[¡1;x[
h[¡1; +1]i
A =
d
A2up a; Av[¡1;x [
F
y2A
∆(y)
Because every y < x we have ∆(y) v [¡1; x[ and thus h[¡1; +1]ia v [¡1; x[.
It is clear that
F
y2A
∆(y) w
¡
(x) and h[¡1; +1]ia v
¡
(x). Thus h[¡1; +1]ia =
¡
(x).
[TODO: More detailed proof.]
Corollary 8. For every ultraﬁlter a and its limit point x:
h[¡1; +1] u ≥ia =
+
(x) if x a
h[¡1; +1] u ≥ia = ? if x < a
Proof. Take into account h[¡1; +1] u ≥ia = h[¡1; +1]i a u h≥ia.
Lemma 9. . Then for every ultraﬁlter a and its limit point x:
0. ??
1.
D
[¡1; +1]
E
a =
+
(x) if x a
2.
D
[¡1; +1]
E
a =
¡
(x) if x < a
Proof. hRia =
d
A2up a
hRi
A =
d
A2up a
F
y2A
+
(y)
1. ??
2. hRia =
d
A2up a;Av[¡1;x[
F
y2A
+
(y) = ??
F
y2A
+
(y) w
¡
(x);
F
y2A;A v[¡1;x[
+
(y) w
¡
(x)
hRia v
¡
(x) is obvious.
Theorem 10. [¡1; +1] u
FCD
>=/ [¡1; +1].
Proof. h[¡1; +1] u ≥ia = ? if x < a but
D
[¡1; +1]
E
a =
¡
(x) if x < a.
Proposition 11. f
~
= Compl(f u >).
Proof. ??
2 Entourages of product of funcoids
Lemma 12. 8H 2 up(g f)9F 2 up f : H w g F
Proof. H w g F , H f
¡1
w g (proposition 1813). But f
¡1
w f
¡1
, thus H w g F (
H f
¡1
w g. Take G = H f
¡1
, but H f
¡1
2/ up g because g f f
¡1
w g does not hold by
http://math.stackexchange.com/a/1862976/4876
??
[FIXME: This was proof of another lemma. ]
Instead of funcoids we will assume that f and g are ﬁlters on ¡ (they are isomorphic).
g f =
d
F¡
up
¡
(g f) =
d
F¡
((up
¡
g) (up
¡
f)) =
d
F 2up
¡
f ;G 2up
¡
g
F¡
(G F ) (lemma 1274).
g f =
d
F 2up f ;G 2up g
F¡
(G F ) (follows from above or directly from theorm 781)
fG F j F 2 up f ; G 2 up gg is?? a generalized ﬁlter base on ¡.
Thus by properties of generalized ﬁlter bases (on ¡) for every H 2 up
¡
(g f) there exists
F 2 up
¡
f, G 2 up
¡
g such that G F v H.
??
Attempt to construct it: [Example: f =1, choose F
H
0
= 1 and G
H
0
= H
0
. Then
d
H
0
2up
¡
H
G
H
0
=
d
H
0
2up
¡
H
H
0
= Cor H 2 up f .]
Entourages of product of funcoids 3 Represent H as a meet of elements of ¡ (H =
d
up
¡
H). For each H
0
2up
¡
H choose F
H
0
2up
¡
f,
G
H
0
2 up
¡
g such that H
0
w G
H
0
F
H
0
. Moreover we can choose maximal F
H
0
, G
H
0
such that this
inequatily holds. Then take F =
d
H
0
2up
¡
H
F
H
0
and G =
d
H
0
2up
¡
H
G
H
0
. [FIXME: Does F 2 up f ?]
[TODO: If this does not work, then seems that there is a counter-example, because it is the
stongest.]
[TODO: We MUST take maximal rather than arbitrary F
H
0
, G
H
0
. Otherwise take f = 1, g = id
.
Then if we take F
H
0
= 1 and replace all possible G
H
0
! G
H
0
n f(a; b)g, then G = ? 2/ up g]
??
H 2 up(g f ) ) hH i
X w hg f i
X = hgihf i
X.
For every X take Y
X
2 up hf i
X such that hgiY
X
v hH i
X. If g is complete, we may assume
that Y
X
is a maximal set for which hgiY
X
v hH i
X holds.
Take F =
d
f(X × Y
X
) t (X × >) j X 2 T Src f g. But is F complete or co-complete, or if meet
is taken on Rel, do we have F 2 up f?
??
Let g be principal. Let a be an atomic ﬁlter. Take (??not always possible) Y
a
2 up hf i
a such
that hgiY
a
v hH ia
Let F =
F
a2atoms
(a × Y
a
) - also co-complete
??
Let for each b 2 atoms hf i
a deﬁne Z
b
= ??
??
Let f be complete. Replace it with principal funcoid F , such that hf i
fxg v hF i
x v Y
fxg
.
Prove g F v H
??
Split F into a join of monovalued functions. This does not work because every function produces
its own g.
??
G f v H; G Compl f = Compl(G f ) v H
??
hG f i
X = hGihf i
X = hGi
d
hhf i
i
up X v
d
hhG f i
i
up X v ?? v hg f i
X v hH iX
??
Find maximal [FIXME: there is no maximal because composition is not distributive over arbi-
trary joins.] funcoids F and G such that G F v H, then prove they are principal (or (co-)complete)
??
Replace G with G
M
mapping x 7! hGi
fxg, Then ?? consider it as an isomorphism between sets.
(Q P )
M
x = hQ P i
fxg = hQihP i
fxg = hQiP
M
x
g F v H , hgi F
M
v H
M
??
f = (FCD)(RLD)
in
f, g = (FCD)(RLD)
in
g; H 2 up(g f )
??
Use Todd Trimble's idea with ξ: H 2 up(g f ) , (AH
~
C ( A(g f)
~
C) , (AH
~
C ( 9B :
(Af
~
B ^Bg
~
C)) ,(AH
~
C ( 9B: (B w hf iA ^ C whg iB)) [FIXME: check direction of implication]
f(A; C) j C w hH iAg f(A; C) j 9B: (B w hf iA ^ C w hg iB)g.
Suppose A 2 dom f(A; C) j 9B: (B w hf iA ^ C w hgiB)g or C 2 im f(A; C) j 9B: (B w hf iA ^
C w hgiB)g.
Then 9B: (B w hf iA ^ C w hgiB). Take B
A;C
such that B
A;C
w hf iA ^ C w hg iB
A;C
Take B
A
0
=
T
C 2??
B
A;C
.
Then B
A
0
w hf iA ^ C w hgiB
A
0
. [FIXME: it does not hold, only B
A
0
w Cor hf iA]
Take co-complete funcoid hF iA = B
A
0
. It is possible?? because B
X tY
0
=
T
C 2??
B
X tY ;C
= ??
B
A;X tY
w hf i(X t Y ) ^ C w hg iB
X tY ;C
??
Instead of intersecting funcoids, consider join f =
F
X 2??
X × Y
X
or f =
F
X 2??
X × Y
X
. It is
enough to consider ultraﬁlters X .
Theorem 13. 8H 2 up(g f )9F 2 up f ; G 2 up g: H w G F for every composable funcoids f and
g. [TODO: Also state it for reloids.]
4 Section 2 Proof. Let H 2 up(g f ). Then 9F 2 up f: H w g F that is H 2 up(g F ). Thus 9G 2 up g such
that H w G F .
3 Join of transitive reloids
(f t g) (f t g) w (f f ) t (g g) [need other direction]
Join of compositions of all ﬁnite sequences of f and g (in any order). It is equivalent to taking
all alternating S
(f g f ::: f) starting or ending with f or g.
“Relationships between completeness properties” in https://en.wikipedia.org/wiki/Complete-
ness_%28order_theory%29
or alternatively:
µ = (f t g) t ((f t g) (f t g)) t ((f t g) (f t g) (f t g)) t :::
µ µ w µ
“No similarly useful description of a subbase for the inﬁmum of a family of quasi-uniformities
is known.” by http://www.sciencedirect.com/science/article/pii/S0166864107000181
up
F
n2N; f
n;i
2S
(f
n;n
::: f
n;0
)
=
T
n2N;f
n;i
2S;F
n;i
2up f
n;i
(F
n;n
::: F
n;0
)
Lemma 14. µ =
F
n2N; f
n;i
2S
(f
n;n
::: f
n;0
) is a transitive reloid, for every set S of endoreloids
(on the same set).
Proof. Denote [U
n
: n 2 N] =
F
n2N
(U
n
::: U
0
).
Let U
n;X
2 up X for all n 2 N and X 2 S.
S
X 2S
U
n;X
: n 2 N
is ??
We need to prove µ µ v µ.
F
X 2S
U
n;X
: n 2 N
2 up µ.
Let U
n;X
2 X and U
n
=
S
X 2S
U
n;X
.
Then
F
X 2S
U
2n;X
: n 2 N
F
X 2S
U
2n¡1;X
: n 2 N
F
X 2S
U
n;X
: n 2 N
because??
U
2n;X
U
2n¡1;X
2
F
X 2S
U
n;X
: n 2 N
?? [TODO: No need for reindexation.]
??
U
n
is the join of all compositions of n-tuples. They form a generalized ﬁlter base. Thus it's
enough to show that every U
n
can be decomposed into smaller n-tuples. But that's obvious. (It
isn't because it is an inﬁnite join!)
Thus is above U
n
0
U
n
00
.
Thus µ is also decomposed, because every its element is minorated as shown above.
??
Take P 2 up µ. Then P 2 up A for every A 2 up(f
n;n
::: f
n;0
).
P w F
N ;N
::: F
N ;0
for some F
N ;i
2 up f
N ;i
for all N 2 N.
Thus P w
F
n2N; f
i
2S
(F
N ;N
::: F
N ;0
) where F
N ;i
2 up f
N ;i
.
Take F
N
0
= F
N ;bN /2 c
::: F
N ;0
and F
N
00
= F
N ;N
::: F
N ;bN /2 c+1
This way we exhaust all possible values?
(F
n;n
::: F
n;n
) (F
m;m
::: F
m;0
) v P . But this is obvious.
Thus easily follows P w
¡
F
n2N; f
i
2S
(F
n;n
::: F
n;0
)
¡
F
n2N; f
i
2S
(F
n;n
::: F
n;0
)
;
P 2 up(µ µ).
Alternative formula:
S
hGRi
S t Z(
S
hGRi
S) t Z(Z(
S
hGRi
S)) t :::
[TODO: Both for reloids and for Cauchy spaces ¡
i
as in the attached article in email.]
[TODO: Also for funcoids (using (FCD)).]
3.1 Exponentials in category of graphs
http://arxiv.org/pdf/math/0605275.pdf deﬁnition 2.3 deﬁnes exponential graph
Join of transitive reloids 5 Let G and H are graphs.
The exponential graph MOR(G; H) is deﬁned by the formulas:
Ob MOR(G; H) = (Ob H)
Ob G
;
(f; g) 2 GR MOR(G; H) , 8(v; w) 2 GR G: (f (v); g(w)) 2 GR H
for every f ; g 2 Ob MOR(G; H) = (Ob H)
Ob G
.
Equivalently
(f; g) 2 GR MOR(G; H) , 8(v; w) 2 GR G: g f(v; w)g f
¡1
GR H
(f; g) 2 GR MOR(G; H) , g (GR G) f
¡1
GR H
(f; g) 2 GR MOR(G; H) ,
f ×
(C)
g
GR G GR H
Evaluation ": (MOR(G; H) × G) ! H
If (f; g) 2 GR MOR(G; H) and x 2 GR G then "((f; g); x) = (fx; gx) = g f(x; x )g f
¡1
3.2 Exponentials in category Fcd
The below gives deﬁnitions for exponential object, (exponential) evaluation, and exponential trans-
pose, but no proof is given that they are really exponential object, (exponential) evaluation, and
exponential transpose. Please write porton@narod.ru if you ﬁnd a proof.
If G, H are endofuncoids, then MOR(G ; H) (exponential object) is an endofuncoid.
MOR(G; H) =
G
ft 2 atoms FCD((Ob H)
Ob G
; (Ob H)
Ob G
) j (im t) G (dom t)
¡1
v H g =
G
t 2 atoms FCD((Ob H)
Ob G
; (Ob H)
Ob G
) j
(dom t) ×
(C)
(im t)
G v H
:
Evaluation:
"
¡¡
f ×
(A)
g
×
(A)
x
= hf i(RLD)
in
x ×
FCD
hgi(RLD)
in
x:
"
¡¡
f ×
(A)
g
×
(A)
x
= g ((RLD)
in
x ×
FCD
(RLD)
in
x) f
¡1
=
f ×
(C)
g
((RLD)
in
x ×
FCD
(RLD)
in
x)
for atomic f ; g 2 FCD(Ob G; Ob H).
Evaluation ": MOR(A; B) × A ! B
"(F × x) =
G
g (x ×
FCD
x) f
¡1
j f ; g 2 atoms
FCD(Ob A;Ob B)
; f × g / F
:
"(F × x) =
F
fg
1
×
FCD
f
1
j f
0
; g
0
; f
1
; g
1
2 atoms
F
; (f
0
×
FCD
f
1
) × (g
0
×
FCD
g
1
) / F g.
Proposition 15. ": MOR(A; B) × A ! B.
Proof. We need to prove " (MOR(A; B) × A) v B ". ??
Transpose f: Z ! MOR(A; B) for a morphism f: Z × A ! B
(f )x =
G
fb ×
FCD
hf i(x × b) j b 2 atoms
FCD
g:
Proposition 16. f : Z ! MOR(A; B).
Proof. We need to prove f Z v MOR(A; B) f whenever f (Z × A) v B f . ??
Awoday 6.5 “Equational deﬁnition” gives a simple way to check cartesian closed categories.
It's enough to prove:
1. " (f × 1
A
) = f
2. " (g × 1
A
) = g
"(f × 1
A
)x = "((f )x × x) =
F
fg
1
×
FCD
f
1
j f
0
; g
0
; f
1
; g
1
2 atoms
F
; (f
0
×
FCD
f
1
) × (g
0
×
FCD
g
1
) /
(f )xg = ??
??
6 Section 3 4 Exponentials in category Rld
If now G, H are endoreloids, then MOR(G; H) is an endoreloid.
Deﬁnition 17. MOR
α
(G;H)=
F
t 2atoms
RLD((Ob H)
Ob G
;(Ob H)
Ob G
)
j (im t) G(domt)
¡1
vH
.
Proposition 18. MOR
α
(G; H) =
F
t
0
×
RLD
t
1
j t
0
; t
1
2 atoms
F((Ob H)
Ob G
)
; t
1
G t
0
¡1
v H
.
Proof. If t
0
; t
1
2 F((Ob H)
Ob G
) and t
1
G t
0
¡1
v H then there are t 2 atoms(t
0
×
RLD
t
1
). Thus
(im t) G (dom t)
¡1
v H and t 2 atoms
RLD((Ob H)
Ob G
;(Ob H )
Ob G
)
.
If t 2 atoms
RLD((Ob H)
Ob G
;(Ob H)
Ob G
)
and (im t) G (dom t)
¡1
v H then dom t = t
0
and im t = t
1
for some t
0
; t
1
2 atoms
F((Ob H)
Ob G
)
and thus t
1
G t
0
¡1
v H.
Deﬁnition 19. MOR
β
(G; H)=
F
ft 2 RLD((Ob H)
Ob G
; (Ob H)
Ob G
) j (im t) G (dom t)
¡1
vH g.
Conjecture 20. MOR
α
(G; H) = MOR
β
(G; H).
Obvious 21.
1. MOR
α
(G; H) =
F
t 2 atoms
RLD((Ob H)
Ob G
;(Ob H)
Ob G
)
j
(dom t) ×
(C)
(im t)
G v H
;
2. MOR
β
(G; H) =
F
t 2 RLD((Ob H)
Ob G
; (Ob H)
Ob G
) j
(dom t) ×
(C)
(im t)
G v H
.
Conjecture 22.
1. t 2 atoms MOR
a
(G; H) , (im t) G (dom t)
¡1
v H;
2. t 2 atoms MOR
β
(G; H) , (im t) G (dom t)
¡1
v H.
Evaluation ": (MOR(G; H) × G) ! H that is for x 2 G = RLD(Ob G; Ob G)
it is a function on objects deﬁned by the formula:
[TODO: Infer x and y from x × y.]
"((f ×
RLD
g) × x) = g (x ×
RLD
x) f
¡1
=
f ×
(C)
g
(x ×
RLD
x)
for atomic f ; g 2 F((Ob H)
Ob G
).
"(F × x) =
F
fg (x ×
RLD
x) f
¡1
j f ; g 2 F((Ob H)
Ob G
); f ×
RLD
g / F g
Let now f : Z × A ! B.
In Set f
~
(a)(b) = f (a; b); f
~
(a) = b 7! f (a; b)
(f )a =
F
fb × h(FCD)f i(a × b) j b 2 atoms
F
g [FIXME: Loss of information, see below.]
Awoday 6.5 “Equational deﬁnition” gives a simple way to check cartesian closed categories.
It's enough to prove:
1. " (f × 1
A
) = f
2. " (g × 1
A
) = g
Really, "(f × 1
A
)z = "(fz × z) = "(
F
fb × h(FCD)f i(z × b) j b 2 atoms
F
g × z) =
F
fg
0
(z ×
RLD
z) f
1
j f
0
; g
0
2 F((Ob H)
Ob G
); f
0
×
RLD
g
0
/
F
fb × h(FCD)f i(z × b) j b 2 atoms
F
gg
[FIXME: It cannot be equal to f due loss on information in (FCD).]
5 On decomposisiton of binary relations and reloids
Example 23. ρ
d
G =/
d
hρiG for some st G of reloids (with matching sources and destinations).
Proof. Take =
d
f"
R
(¡"; ") j " > 0g. Take fαg ×
RLD
p where p v is a nontrivial ultraﬁlter.
hρ
d
Gi(fαg ×
RLD
p) = (
d
G) (fαg ×
RLD
p)
h
d
hρiGi(fαg ×
RLD
p) = (because f αg ×
RLD
p is atomic)=
d
fhρgi(fαg ×
RLD
p) j g 2 Gg =
d
fg (fαg ×
RLD
p) j g 2 Gg.
On decomposisiton of binary relations and reloids 7 6 Rest
Lemma 24. Every non-empty set has a well ordering with greatest element.
Proof. Take an arbitrary well ordering of our set and move the ﬁrst element to the end of the
order.
Theorem 25. L 2 [f])[f]\
Q
i2dom A
atoms L
i
=/ ; for every pre-multifuncoid f of the form whose
elements are atomic posets.
Proof. If arity f = 0 our theorem is trivial, so let arity f =/ 0. Let v is a well-ordering of arity f
with greatest element m (it exists by the lemma).
Let Φ is a function which maps non-least elements of posets into atoms under these elements and
least elements into themselves. (Note that Φ is deﬁned on least elements only for completeness, Φ
is never taken on a least element in the proof below.) [TODO: Fix the “universal set” paradox here.]
Deﬁne a transﬁnite sequence a by transﬁnite induction with the formula
a
c
= Φhf i
c
(aj
X(c)nfcg
[Lj
(arity f )nX(c)
):
Let b
c
= aj
X(c)nfcg
[Lj
(arity f )nX(c)
. Then a
c
= Φhf i
c
b
c
.
Let us prove by transﬁnite induction
a
c
2 atoms L
c
:
a
c
= Φhf i
c
Lj
(arity f )nfcg
vhf i
c
Lj
(arity f )nfcg
. Thus a
c
v L
c
. [TODO: Is it true for pre-multifun-
coids?]
The only thing remained to prove is that hf i
c
b
c
=/ 0
that is hf i
c
(aj
X(c)nfcg
[Lj
(arity f )nX(c)
) =/ 0 that is y / hf i
c
b
c
??
L
c
/ hf i
c
b
c
, b
c
(0) / hf i
0
(b
c
j
(arity f )nf0g
[fc; L
c
g) , a
0
/ hf i
0
(aj
X(c)nf0;cg
[
Lj
(arity f )nX(c)
[fc; L
c
g) , a
0
/ hf i
0
(aj
X(c)nf0;c g
[Lj
((arity f )nX(c))[fcg
) ( a
0
/ hf i
0
(aj
X(c)nf0;cg
[
aj
((arity f )nX(c))[fcg
) , a
0
/ hf i
0
(aj
(X(c)nf0;c g) [((arity f)n X (c))[fcg
) , a
0
/ hf i
0
(aj
(arity f )nf0g
) ,
Φhf i
0
(Lj
(arity f )nf0g
) / hf i
0
(aj
(arity f )nf0g
).
??
a
0
= Φhf i
0
(Lj
(arity f )nf0g
)
??
??Two ways to prove:??
L
c
/ hf i
c
b
c
, b
c
(k) / hf i
k
(b
c
j
(arity f )nfkg
[fc; L
c
g) , a
k
/ hf i
k
(aj
X(c)nfk;c g
[
Lj
(arity f )nX(c)
[fc; L
c
g) , a
k
/ hf i
k
(aj
X(c)nfk;c g
[Lj
((arity f )nX(c))[fcg
) ( a
k
/ hf i
k
(aj
X(c)nfk;c g
[
aj
((arity f )nX(c))[fcg
) , a
k
/ hf i
k
(aj
(X(c)nfk;c g)[((arity f )nX(c))[fcg
) , a
k
/ hf i
k
(aj
(arity f )nfkg
)
??
L
c
/ hf i
c
b
c
, a
k
/ hf i
k
(aj
X(c)nfc;kg
[Lj
(arity f )nX(c)
[f(c; L
c
)g) , a
k
/ hf i
k
(aj
X(c)nfc;k g
[
Lj
((arity f )nX(c))[fcg
) ( a
k
/ hf i
k
(aj
X(c)nfc;k g
[aj
((arity f )nX(c))[fcg
) , a
k
/ hf i
k
(aj
(arity f )nfkg
)
??
y / hf i
c
b
c
, a
k
/ hf i
k
(aj
X(c)nfc;k g
[Lj
(arity f )nX(c)
[f(c; y)g) ( a
k
/ hf i
k
(aj
X(k)nfk g
[
Lj
(arity f )nX(c)
[f(c; y)g)
y / hf i
c
b
c
, L
k
/ hf i
k
(aj
X(c)nfcg
[Lj
(arity f )n(X(c)[fkg)
[f(c; y)g) (
??
a
m
= Φhf i
m
(λi 2 (arity f ) n fmg: a
i
) = Φhf i
m
aj
(arity f )nfmg
; a
m
/ Φhf i
m
aj
(arity f )nfmg
;
a
m
/ hf i
m
aj
(arity f )nfmg
; a 2 [f].
Theorem 26. Let A = A
i2n
is a family of boolean lattices.
A relation δ 2 P
Q
atoms
F(A
i
)
such that for every a 2
Q
atoms
F(A
i
)
8A 2 a: δ \
Y
i2n
atoms "
A
i
A
i
=/ ; ) a 2 δ (3)
can be continued till the function f for a unique staroid f of the form λi 2 n: P(A
i
). The funcoid
f is completary.
8 Section 6 For every X 2
Q
i2n
F(A
i
)
X 2 GR f , δ \
Y
i2n
atoms X
i
=/ ;: (4)
Proof. [FIXME: ??Use of unproved conjecture ?.]
By the theorem 11 (used that it is a boolean lattice) we have X 2 GR f , GR f \
Q
i2n
atoms X
i
=/ ; and thus (4). From this also follows uniqueness.
It is left to prove that there exists a completary staroid f such that f is a continuation of δ.
Consider the relation f deﬁned by the formula X 2 f , δ \
Q
i2n
atoms "
A
i
X
i
=/ ;.
I
0
t I
1
2 f , δ \
Q
i2n
atoms "
A
i
(I
0
i t I
1
i) =/ ; , δ \
Q
i2n
(atoms "
A
i
I
0
i [ atoms "
A
i
I
1
i) =/ ;.
Thus by the lemma I
0
t I
1
2 f , 9c 2 f0; 1g
n
: δ \
Q
i2n
atoms "
A
i
I
c(i)
=/ ; , 9c 2 f0; 1g
n
:
(λi 2 n: I
c(i)
i) 2 f . Trivially if 9i 2 n: X
i
= 0 then X 2/ f . So f is a completary staroid.
Let a 2
Q
atoms
F(A
i
)
.
The reverse of (3) is obvious. So we have a 2 δ , 8A 2 a: δ \
Q
i2n
atoms "
A
i
A
i
=/ ; , 8A 2 a:
A 2 f , 8A 2 a: A 2 f , a f , a 2 f. Thus f is a continuation of δ .
hf
0
× f
1
ix =
F
fhf
0
iX ×
FCD
hf
1
iX j X 2 atoms xg
h(f
0
× f
1
)
¡1
iy = hf
0
¡1
idom y u hf
1
¡1
iim y (not only for atomic y):
h(f
0
× f
1
)
¡1
iy =
F
fh(f
0
× f
1
)
¡1
ifpg j p 2 yg =
F
fhf
0
¡1
idom fpg u hf
1
¡1
iim fpg j p 2 yg =
F
fhf
0
¡1
idom fpg j p 2 yg u
F
fhf
1
¡1
iim fpg j p 2 yg = hf
0
¡1
idom y u hf
1
¡1
iim y
It seems that these are not components of a funcoid.
Conjecture 27. Let R is a set of staroids of the form λi 2 n: F(A
i
) where every A
i
is a boolean
lattice. If x 2
Q
i2n
atoms
F(A
i
)
then x 2 GR
d
R , 8f 2 R: x 2 f .
Proof. Let denote x 2 δ , 8f 2 R: x 2 f for every x 2
Q
i2n
atoms
F(A
i
)
. For every a 2
Q
i2n
atoms
F(A
i
)
8X 2 a: δ \
Q
i2n
atoms "
A
i
X
i
=/ ; , 8X 2 a9x 2
Q
i2n
atoms "
A
i
X
i
: x 2 δ , 8X 2 a9x 2
Q
i2n
atoms "
A
i
X
i
8f 2 R: x 2 f ) 8X 2 a; f 2 R9x 2
Q
i2n
atoms "
A
i
X
i
: x 2 f ) 8X 2 a; f 2 R:
X 2 f , 8f 2 R : a f , 8f 2 R: a 2 f , a 2 δ.
So by the previous theorem δ can be contimued till p for some staroid p of the form λi 2 n:
P(f
i
).
Let's prove p =
d
R.
x 2 p,x 2δ ) x 2 f for every f 2R and x2
Q
i2n
atoms
F(A
i
)
. Thus p f . Consequently
8f 2 R: p f .
Suppose that q is a staroid of the form λi 2 n: P(A
i
) such that 8f 2 R: q f. Then for every
x 2
Q
i2n
atoms
F(A
i
)
we have x 2 q ) 8f 2 R: x 2 f , x 2 δ , x 2 p. So q p that is q p.
We have proved p =
d
R. It's remained to prove that x 2 p , 8f 2 R: x 2 f for every
x 2
Q
i2n
atoms
F(A
i
)
. Really, x 2 p , x 2 δ , 8f 2 R: x 2 f .
Example 28. There exists a multifuncoid on power sets, which is not a completary multifuncoid.
Proof. Let arity f = N and form f = ( P f0; 1g)
N
.
Characteristic function of a set D is Λ(D) = λi 2 N :
f1g if i 2 D;
f0g if i 2/ D:
GR f = fΛ(D) j D 2 g.
Obviously f is an anchored relation.
Let k 2 N .
(val f)
k
L = fX 2 (form f)
k
j L [ f(k; X)g 2 GR f g = fX 2 (P f0g)
N
j 9D 2 Ω: L [ f(k;
X)g = Λ(D)g = fX 2 (P f0; 1g)
N
j 9D 2 Ω: (L = Λ(D)j
N nfkg
^X = Λ(D)(k))g.
X 2 (val f)
k
L, 9D 2 Ω: (L = Λ(D)j
N nfkg
^X
k
= Λ(D)(k)) , ?? , 9D 2
N nfkg
; P 2 ff0g; f1gg:
(L = Λ(D) ^ X
k
= P ) , 9D 2
N nfkg
: L = Λ(D) ^ 9P 2 ff0g; f0; 1gg: X
k
= P , 9D 2
N nfkg
:
L = Λ(D) ^ X 2 ff0g; f1gg (Note it does not depend on X.)
Let X ; Y 2(P f0; 1g)
N
. Then X t Y 2 (val f )
k
L, 9D 2
N nfkg
: L = Λ(D) ^ X tY 2f f0g; f1gg
Rest 9 If X 2 ff0gg, Y 2 ff1gg then X t Y 2/ ff0g; f0; 1gg
??
That X
i
= ; ) X 2/ (val f )
k
L is obvious. So f is a pre-multifuncoid.
??
Conjecture 29. If a is a completary multifuncoid and Dst f
i
is a starrish poset for every i 2 n
then StarComp(a; f) is a completary multifuncoid.
Proof. Let 8K 2
Q
form f: (K w L
0
^ K w L
1
) K 2 StarComp(a; f)) that is 8K 2
Q
form f:
¡
K w L
0
^ K w L
1
) 9y 2
Q
i2n
atoms A
i
: (8i 2 n: y
i
[f
i
] K
i
^ y 2 a)
that is
8K 2
Q
form f9y 2
Q
i2n
atoms A
i
: (K w L
0
^ K w L
1
) (8i 2 n: y
i
[f
i
] K
i
^ y 2 a)) that is
8K 2
Q
form f9y 2
Q
i2n
atoms A
i
: ((K w L
0
^ K w L
1
) 8i 2 n: y
i
[f
i
] K
i
) ^ y 2 a) that is??
8K 2
Q
form f9y 2
Q
i2n
atoms A
i
: (9c 2 f0; 1g
n
8i 2 n: y
i
[f
i
] L
c(i)
i ^ y 2 a) that is ??
Conjecture 30.
Q
(D)
F is a pre-multifuncoid if every F
i
is a pre-multifuncoid.
Proof. Let X ; Y 2
form
Q
(D)
F
(i;j)
.
8Z 2
form
Q
(D)
F
(i;j)
:
Z X ^ Z Y ) Z 2
val
Q
(D)
F
(i;j)
L
, 8Z 2
form
Q
(D)
F
(i;j)
: (Z X ^ Z Y ) 9K 2 (form F
i
)j
(arity F
i
)nfj g
: Z 2 (val F
j
)K) , 8Z 2
form
Q
(D)
F
(i;j)
: (9K 2 (form F
i
)j
(arity F
i
)nfj g
: (Z X ^ Z Y ) Z 2 (val F
j
)K)) ,
?? , 8Z 2
form
Q
(D)
F
(i;j)
: (9K 2 (form F
i
)j
(arity F
i
)nfj g
: (X 2 (val F
j
)K _ Y 2 (val F
j
)K)) ,
8Z 2
form
Q
(D)
F
(i;j)
: (9K 2 (form F
i
)j
(arity F
i
)nfj g
: X 2 (val F
j
)K _ 9 K 2 (form F
i
)j
(arity F
i
)nfj g
:
Y 2 (val F
j
)K) ,
Let f is a funcoid.
Then there exists a reloid g such that ??
=============
(RLD)
in
f =
S
a ×
RLD
b j a 2 atoms 1
F(Src f )
; b 2 atoms 1
F(Dst f )
; a ×
FCD
b f
(RLD)
in
(g f) =
S
a ×
RLD
b j a 2 atoms 1
F(Src f )
; b 2 atoms 1
F(Dst f )
; a ×
FCD
b g f
=
S
a ×
RLD
b j a 2 atoms 1
F(Src f )
; b 2 atoms 1
F(Dst f )
; a ×
RLD
b (RLD)
in
(g f )
(RLD)
in
(FCD)((RLD)
in
g (RLD)
in
f) = (RLD)
in
((FCD)(RLD)
in
g (FCD)(RLD)
in
f) = (RLD)
in
(g
f)
Lemma 31. 8Y 2 up hf i
X9F 2 up f: hF iX Y for every funcoid f.
Proof. ??
(RLD)
in
(g f ) =
Theorem 32. g f =
T
"
FCD(Src f ;Dst g)
(G F ) j F 2 up f ; G 2 up g
Proof. It's enough?? to prove that 8H 2 up(g f)9F 2 up f ; G 2 up g: H G F .
X [g f]
Y , hf i
X / hg
¡1
i
Y , 8X
0
2 up hf i
X ; Y
0
2 up hg
¡1
i
Y : X
0
/ Y
0
, 9F 2 up f ;
G 2 up g: hF iX / hGiY , 9F 2 up f ; G 2 up g: X [G F ] Y (used the lemma).
Let H 2 up(g f). Then X [H]
Y ) X [g f ]
Y ) 9F 2 up f ; G 2 up g: X [G F ] Y for every
X, Y . Thus ??(it does not work because F amd G depend on X and Y ).
Lemma 33. f r =
T
ff "
FCD
R j R 2 up rg
Proof. Obviously f r
T
ff "
FCD
R j R 2 up rg.
hf ri
X ??
T
fhf ihRiX j R 2 up rg =
T
fhf ih"
FCD
Ri
X j R 2 up rg =
T
fhf
"
FCD
Ri
X j R 2 up rg h
T
ff "
FCD
R j R 2 up rgi
X
10 Section 6 W = fhRiX j R 2 up rg is?? a generalized ﬁlter base?? (No, it isn't, up r isn't a ﬁlter.)
Let P
0
; P
1
2 W . Then P
0
= hR
0
iX, P
1
= hR
1
iX
Let F 2 up(f r). Then ?? F 2 up
T
ff "
FCD
R j R 2 up rg.
So f r
T
ff "
FCD
R j R 2 up rg
X [f "
FCD
R] Y , ["
FCD
R]X / hf
¡1
iY
??
Let W = fh"
FCD
Ria j R 2 up rg. W is a generalized ﬁlter base??. Really If X ; Y 2 W . Then
X = h"
FCD
X ia and Y = h"
FCD
Y ia for some X 2 up r, Y 2 up r.
\
ff "
FCD
R j R 2 up rg
a =
\
fhf "
FCD
Ria j R 2 up rg = (because fh"
FCD
Ria j R 2 up rg is a g.f.b.??)
hf i
\
fh"
FCD
Ria j R 2 up rg =
hf ihria
Counter-example attempt: Let r = id
FCD
. ??
Corollary 34. g f =
T
"
FCD(Src f ;Dst g)
(G F ) j F 2 up f ; G 2 up g
.
Proof. x
"
FCD(Src f ;Dst g)
(G F )
z , 9y 2 atoms Dst f: (x ["F ] y ^ y ["G] z)
x
T
"
FCD(Src f ;Dst g)
(G F ) j F 2 up f ; G 2 up g
z , 8F 2 up f ; G 2 up g:
x
"
FCD(Src f ;Dst g)
(G F )
z , 8F 2 up f ; G 2 up g9y 2 atoms Dst f :
¡
x
"
FCD(Src f ;Dst g)
F
y ^ y
"
FCD(Src f ;Dst g)
G
z
Conjecture 35. Let f be a set, F be the set of f.o. on f, P be the set of principal f.o. on f,
let n be an index set. Consider the ﬁltrator ( F
n
; P
n
). Then if f is a multifuncoid of the form P
n
,
then E
f is a multifuncoid of the form F
n
.
Proof. (val E
f)
i
L = ?? = fX 2 A
i
j 8K 2 up L: K [ f(i; X)g 2 f g = fX 2 A
i
j 8K 2 up L:
X 2 (val f )
i
K g = fX 2 A
i
j 8K 2 up L: K [ f(i; X)g 2 f g = ?? = fX 2 A
i
j L [ f(i; X)g 2 E
f g
X 2 (val E
f)
i
L , ?? , L [ f(i; X)g 2 f
A [ B 2 (val E
f)
i
L = ?? = L [ f(i; A [ B)g 2 f , (futher trivial)
??
(val E
f)
i
L = fX 2 A
i
j L [ f(i; X)g 2 E
f g = fX 2 A
i
j up(L [ f(i; X)g) f g = ?? =
fX 2 A
i
j up L × X f g = fX 2 A
i
j 8K 2 up L; x 2 up X: K [ f(i; x)g 2 f g , fX 2 A
i
j 8K 2 up L;
x 2 up X: x 2 (val f)
i
K g = fX 2 A
i
j 8K 2 up L: up X (val f )
i
K g. [TODO: The same formula as
below only with other variable names.] [TODO: Correct order of coords.]
up(L [ f(i; X)g) f , 8K 2 up(L [ f(i; X)g): K 2 f , 8P 2 up L; X
0
2 up X:
P [ f(i; X
0
)g 2 f , 8P 2 up L; X
0
2 up X: X
0
2 (val f)
i
P , 8P 2 up L: up X (val f)
i
P
Thus (val E
f)
i
L = fX 2 A
i
j 8P 2 up L: up X (val f)
i
P g =
X 2 A
i
j up X
T
P 2up L
(val f)
i
P
. [TODO: First try to prove for the binary case.]
A [ B 2 (val E
f)
i
L ,
X 2 A
i
j up (A [ B)
T
P 2up L
(val f )
i
P
,
X 2 A
i
j up A \ up B
T
P 2up L
(val f)
i
P
(the lemma does not work, try to use a ﬁlter base)
A [ B 2 (val E
f)
i
L , 8P 2 up L: up(A [ B) (val f)
i
P , 8P 2 up L: (up A (val f )
i
P _
up B (val f )
i
P ) ( 8P 2 up L: (up A (val f)
i
P _ 8P 2 up L: up B (val f)
i
P ) (used the lemma).
[TODO: Reverse implication.]
8P 2 up L: (up A (val f )
i
P _ up B (val f )
i
P ) ) 8P 2 up L: (up A \ up B (val f)
i
P ) ,
8P 2 up L: up(A [ B) (val f )
i
P .
Thus?? A [ B 2 (val E
f)
i
L , A 2 (val E
f)
i
L _ B 2 (val E
f)
i
L.
Consider f \
RLD
S
S =
S
hf \
RLD
iS.
1. If f is not required to be complete this formula fails even for set-valued S.
2. Let f is complete. Then it fails for speciﬁcally choosen S.
Rest 11 Theorem 36. If f is a complete reloids and S is a set of complete reloids. Then [TODO: The
same for funcoids?]
f \
RLD
[
S =
[
hf \
RLD
iS:
Theorem 37. Composition with a (co?)complete reloid is an adjoint:
Proof. F is a lower adjoint. Let ξ is its upper adjoint. Then
F x y , x ξ(y)
x ξ(F x) , F ξ(y)
ξ(b) = max fx 2 RLD j F x bg (a proof circle follows this)
x F
¡1
b ) F x F b
F
S
RLD
R =
T
RLD
fF K j K 2 up
S
RLD
Rg
S
RLD
hF iR =
S
RLD
f
T
RLD
fF G j G 2 up g g j g 2 Rg
Conjecture 38. Compl f \
RLD
Compl g = Compl(f \
RLD
g) for every reloids f and g.
Proof. Compl(f \
RLD
g) =
S
RLD
(f \
RLD
g)j
fαg
RLD
j α 2 f
Compl f \
RLD
Compl g =
S
RLD
f j
fαg
RLD
j α 2 f
\
RLD
S
RLD
gj
fαg
RLD
j α 2 f
(Compl(f \
RLD
g))j
fβ g
RLD
=(f \
RLD
g)j
fβg
RLD
(Compl f \
RLD
Compl g)j
fβg
RLD
=(f \
RLD
g)j
fβg
RLD
.
So enough to prove that Compl f \
RLD
Compl g is complete.
Let A = atoms
ComplRLD
f and B = atoms
ComplRLD
g. Then ??
Obviously Compl f \
RLD
Compl g Compl f \
ComplRLD
Compl g
Suppose it exists a 2 atoms
RLD
(Compl f \
RLD
Compl g) such that a 2/
toms
RLD
(Compl f \
ComplRLD
Compl g). Then ??
Conjecture 39. If f and g are reloids, then
g f =
[
RLD
fG F j F 2 atoms
RLD
f ; G 2 atoms
RLD
gg:
Proof. g f = ?? = g
S
RLD
atoms
RLD
f = ?? =
S
RLD
hg iatoms
RLD
f
S
RLD
fG F j F 2 atoms
RLD
f ; G 2 atoms
RLD
ggs
S
RLD
fg F j F 2 atoms
RLD
f g g f
??
Theorem 40. dom (RLD)
in
f = dom f and im (RLD)
in
f = im f for every funcoid f.
Proof. Let an atomic f.o. a dom f . Then exists atomic f.o. b im f such that a ×
FCD
b f .
Consequently
a ×
RLD
b (RLD)
in
f ) 8K 2 up (RLD)
in
f: a ×
RLD
b K ) 8K 2 up (RLD)
in
f: a dom K ,
a
T
F
hdomiup (RLD)
in
f , a dom (RLD)
in
f.
Let now an atomic f.o. a dom (RLD)
in
f. Then 8K 2 up (RLD)
in
f: a dom K
What is equivalent to
8K 2
\
fup(a ×
RLD
b) j a; b 2 atoms
F
f; a ×
FCD
b f g: a dom K
Let K 2 up f . Then K a ×
FCD
b for every a; b 2 atoms
F
f where a ×
FCD
b f that is exist ??
K 2 up(a ×
RLD
b) for ??
??
from what follows?? [FIXME: b is depended on K] that exist b im f such that 8K 2
up (RLD)
in
f: a ×
RLD
b K that is 8K 2 up (RLD)
in
f: K 2up(a ×
RLD
b) and thus (RLD)
in
f a ×
RLD
b
and consequently dom (RLD)
in
f dom(a ×
RLD
b) = a.
Thus a dom f , a dom (RLD)
in
f for each atomic f.o. a from what follows dom (RLD)
in
f =
dom f.
12 Section 6 im (RLD)
in
f = im f is similar.
Theorem 41. A reloid f is monovalued iﬀ 8g 2 RLD: (g f ) g = f j
dom g
RLD
).
Proof.
). Let f is monovalued. Then exists F 2 up f such that f = F j
dom f
RLD
. Let g 2 RLD
and g f . Then exists G 2 up g such that g = Gj
dom g
RLD
. We have g = g \
RLD
f =
Gj
dom g
RLD
\
RLD
F j
dom f
RLD
=Gj
dom g
RLD
\
RLD
F j
dom g
RLD
=(G \
RLD
F )j
dom g
RLD
f j
dom g
RLD
. But obviously g
f j
dom g
RLD
. So g = f j
dom g
RLD
.
Conjecture 42. Compl f = f n
FCD
(Ω ×
FCD
f) for every funcoid f.
This conjecture may be proved by considerations similar to these in the section “Fréchet ﬁlter”
in [?].
Example 43. (RLD)
in
is not a lower adjoint (in general).
Proof. Enough to prove one of the following:
Enough to prove non-existence of max ff 2 FCD j (RLD)
in
f gg for some reloid g.
??Enough to prove (RLD)
in
S
FCD
S =/
S
FCD
h(RLD)
in
iS for some set S of funcoids.
??
Theorem 44. A ﬁlter A is connected regarding a reloid f iﬀ it is connected regarding the funcoid
(FCD)f.
Proof. A is connected regarding f iﬀ A is connected regarding every element of F 2 up f
(considered as reloids) that is iﬀ S
(F \
RLD
(A ×
RLD
A)) = A ×
RLD
A.
??
Theorem 45. (RLD)
out
(A ×
FCD
B) =/ A ×
RLD
B in general.
Proof. (RLD)
out
(A ×
FCD
B) =
T
RLD
up(A ×
FCD
B)
If the equality holds, 8F 2 up(A ×
FCD
B)9A 2 up A; B 2 up B: A × B F .
Let L, B are f.o. and the cardinality of the set H = fY 2 F j B Y Lg is inﬁnite but not
greater than cardinality of a set A. (We can always assume existence of such set A, extending the
base set f above if necessary.)
For every f.o. Y 2 H choose a set Y
Y
2 up B such that Y
Y
2/ up Y
Let z is a surjection from A to H.
Consider the binary relation F =
S
ffαg × Y
zα
j α 2 Ag.
We have F A ×
FCD
B that is F 2 up(A ×
FCD
B).
But if B 2 up B then ?? A × B * F because Fα = Y
zα
2/ up zα that is 8Y B: Y
zα
2/ up Y
[Not needed] Consider funcoid f =
S
FCD
ffαg ×
FCD
zα j α 2 Ag.
??
Theorem 46.
T
Z(DA)
fX \
F
A j X 2 T g = A \
F
T
T.
Proof. That A \
F
T
T is a lower bound of fX \
F
A j X 2 T g is obvious.
We need to prove that it is the greatest lower bound, that is that for every lower bound
B 2 Z( D A) of fX \
F
A j X 2 T g we have A \
F
T
T B.
Let B = B \
F
A is a lower bound of fX \
F
A j X 2 T g that is 8X 2 T : B \
F
A X \
F
A. Left
to prove that A \
F
T
T B \
F
A.
(X [ B) \
F
A = (X \
F
A) [
F
(B \
F
A) = X \
F
A
B [
F
(A \
F
T
T ) = (B [
F
A) \
F
(B [
T
T )
Rest 13 T
Z(DA)
fX \
F
A j X 2 T g =
T
Z(DA)
f(X [ B) \
F
A j X 2 T g B \
F
A.
Suppose that A \
F
T
T + B \
F
A. Then exists K 2 up(A \
F
T
T ) such that K 2/ up(B \
F
A).
That is 9L 2 up A: K = L \
T
T and @L 2 up A: K = L \ B.
For every X 2 T we have up(B \
F
A) up(X \
F
A) that is 8Y 2 up( X \
F
A): Y 2 up(B \
F
A);
8P 2 up A: P \ X 2 up(B \
F
A); 8P 2 up A9Q 2 up A: P \ X = B \ Q.
Thus 9Q 2 up A: L \ X = B \ Q
(B \
F
X) \
F
A = (B \
F
A) \
F
X = B \
F
A.
8X 2 T : (B \
F
X) \
F
A X \
F
A
??
(B \
F
A) \
F
X = (B \
F
A) \
F
(X \
F
A) = B \
F
A
??
Theorem 47.
S
Z(DA)
fX \
F
A j X 2 T g = A \
F
S
T.
Proof. ??
Lemma 48. If staroid 0 =/ f v a
Strd
n
for an ultraﬁlter a and an index set n, then n × fag 2 GR f.
[TODO: Can be generalized for arbitrary staroidal products?]
Proof. If K
i
w a
i
for some i 2 n then K 2/ GR a
Strd
n
and thus K 2/ GR f .
Suppose that for every K such that K
i
w a
i
for every i 2 n we have K 2/ GR f . Then GR f = ;
what is impossible.
Thus there exists K such that K
i
w a
i
for every i 2 n and K 2 GR f.
By the previous lemma, λi 2 n: K
i
u a 2 GR f that is n × fag 2 GR f .
Theorem 49. id
a[n]
Strd
is an atomic ?? if a is an atomic ﬁlter.
Proof. Suppose 0 =/ f v id
a[n]
Strd
. Then f v a
Strd
n
and thus by the lemma n × fag 2 GR f.
We need to prove that f w id
a[n]
Strd
that is L 2 GR f ( L 2 id
a[n]
Strd
that is L 2 GR f (
d
i2n
Z
L
i
2 @ a.
Really,
d
i2n
Z
L
i
2 @ a ) 8i 2 n: L
i
2 @ a ) 8i 2 n: L
i
w a ) L 2 GR f .
7 Misc
Lemma 50. x ×
FCD
y F ^ x ×
FCD
y G ) x ×
FCD
y F \ G for any atomic ﬁlters x and y and
binary relations F and G.
Proof. x ×
FCD
y F , hF ix y , 8X 2 up x: hF iX y , 8X 2 up <