Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book

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Algebraic General Topology. Volume 1 addons
Victor Porton
Email address: porton@narod.ru
URL: http://www.mathematics21.org
2000 Mathematics Subject Classification. 54J05, 54A05, 54D99, 54E05, 54E15,
54E17, 54E99
Key words and phrases. algebraic general topology, quasi-uniform spaces,
generalizations of proximity spaces, generalizations of nearness spaces,
generalizations of uniform spaces
Abstract. This file contains future addons for the free e-book “Algebraic
General Topology. Volume 1”, which are yet not enough ripe to be included
into the book.
Contents
Chapter 1. About this document 5
Chapter 2. Applications of algebraic general topology 6
1. “Hybrid” objects 6
2. A way to construct directed topological spaces 6
3. Some inequalities 8
4. Continuity 9
5. A way to construct directed topological spaces 12
6. Integral curves 12
Chapter 3. More on generalized limit 16
1. Hausdorff funcoids 16
2. Restoring functions from limit 16
Chapter 4. Extending Galois connections between funcoids and reloids 18
Chapter 5. Boolean funcoids 20
1. One-element boolean lattice 20
2. Two-element boolean lattice 20
3. Finite boolean lattices 21
4. About infinite case 21
Chapter 6. Interior funcoids 23
Chapter 7. Filterization of pointfree funcoids 25
Chapter 8. Systems of sides 26
1. More on Galois connections 26
2. Definition 27
3. Concrete examples of sides 28
4. Product 30
5. Negative results 31
6. Dagger systems of sides 31
Chapter 9. Backward Funcoids 32
Chapter 10. Quasi-atoms 33
Chapter 11. Cauchy Filters on Reloids 34
1. Preface 34
2. Low spaces 34
3. Almost sub-join-semilattices 35
4. Cauchy spaces 35
5. Relationships with symmetric reloids 36
6. Lattices of low spaces 37
7. Up-complete low spaces 41
3
CONTENTS 4
8. More on Cauchy filters 42
9. Maximal Cauchy filters 43
10. Cauchy continuous functions 44
11. Cauchy-complete reloids 44
12. Totally bounded 44
13. Totally bounded funcoids 45
14. On principal low spaces 45
15. Rest 45
Chapter 12. Funcoidal groups 47
1. On “Each regular paratopological group is completely regular” article 48
Chapter 13. Micronization 53
Chapter 14. More on connectedness 54
1. For topological spaces 54
Chapter 15. Relationships are pointfree funcoids 59
Chapter 16. Manifolds and surfaces 60
1. Sides of a surface 60
2. Special points 60
Bibliography 63
CHAPTER 1
About this document
This file contains future addons for the free e-book “Algebraic General Topol-
ogy. Volume 1”, which are yet not enough ripe to be included into the book.
Theorem (including propositions, conjectures, etc.) numbers in this document
start from the last theorem number in the book plus one. Theorems references
inside this document are hyperlinked, but references to theorems in the book are
not hyperlinked (because PDF viewer Okular 0.20.2 does not support Backward
button after clicking a cross-document reference, and thus I want to avoid clicking
such links).
5
CHAPTER 2
Applications of algebraic general topology
1. “Hybrid” objects
Algebraic general topology allows to construct “hybrid” objects of “continuous”
(as topological spaces) and discrete (as graphs).
Consider for example D t T where D is a digraph and T is a topological space.
The n-th power (D tT )
n
yields an expression with 2
n
terms. So treating D tT
as one object (what becomes possible using algebraic general topology) rather than
the join of two objects may have an exponential benefit for simplicity of formulas.
2. A way to construct directed topological spaces
2.1. Some notation. I use E and ι notations from volume-2.pdf. FiXme:
Reorder document fragments to describe it before use.
I remind that f|
X
= f id
X
for binary relations, funcoids, and reloid.
f k
X
= f (E
X
)
1
.
fX = id
X
f id
1
X
.
As proved in volume-2.pdf, the following are bijections and moreover isomor-
phisms (for R being either funcoids or reloids or binary relations):
1
.
n
(f|
X
,fk
X
)
fR
o
;
2
.
n
(fX,ι
X
f)
fR
o
.
As easily follows from these isomorphisms and theorem 1293:
Proposition 2064. For funcoids, reloids, and binary relations:
1
. f C(µ, ν) f k
A
C(ι
A
µ, ν);
2
. f C
0
(µ, ν) f k
A
C
0
(ι
A
µ, ν);
3
. f C
00
(µ, ν) f k
A
C
00
(ι
A
µ, ν).
2.2. Directed line and directed intervals. Let A be a poset. We will
denote A = A {−∞, +∞} the poset with two added elements −∞ and +, such
that + is strictly greater than every element of A and −∞ is strictly less.
FiXme: Generalize from R to a wider class of posets.
Definition 2065. For an element a of a poset A
1
. J
(a) =
n
xA
xa
o
;
2
. J
>
(a) =
xA
x>a
;
3
. J
(a) =
n
xA
xa
o
;
4
. J
<
(a) =
xA
x<a
;
5
. J
6=
(a) =
n
xA
x6=a
o
.
Definition 2066. Let a be an element of a poset A.
1
. ∆(a) =
d
F
n
]x;y[
x,yA,x<ay>a
o
;
2
.
(a) =
d
F
n
[a;y[
yA,y>a
o
;
6
2. A WAY TO CONSTRUCT DIRECTED TOPOLOGICAL SPACES 7
3
.
>
(a) =
d
F
n
]a;y[
yA,x<ay>a
o
;
4
.
(a) =
d
F
n
]x;a]
xA,x<a
o
;
5
.
<
(a) =
d
F
n
]x;a[
xA,x<a
o
;
6
.
6=
(a) = ∆(a) \ {a}.
Obvious 2067.
1
.
(a) = ∆(a) u
F
@J
(a);
2
.
>
(a) = ∆(a) u
F
@J
>
(a);
3
.
(a) = ∆(a) u
F
@J
(a);
4
.
<
(a) = ∆(a) u
F
@J
<
(a);
5
.
6=
(a) = ∆(a) u
F
@J
6=
(a).
Definition 2068. Given a partial order A and x A, the following defines
complete funcoids:
1
. h|A|i
{x} = ∆(x);
2
. h|A|
i
{x} =
(x);
3
. h|A|
>
i
{x} =
>
(x);
4
. h|A|
i
{x} =
(x);
5
. h|A|
<
i
{x} =
<
(x);
6
. h|A|
6=
i
{x} =
6=
(x).
Proposition 2069. The complete funcoid corresponding to the order topol-
ogy
1
is equal to |A|.
Proof. Because every open set is a finite union of open intervals, the com-
plete funcoid f corresponding to the order topology is described by the formula:
hfi
{x} =
d
F
n
]a;b[
a,bA,a<xb>x
o
= ∆(x) = h|A|i
{x}. Thus f = |A|.
Exercise 2070. Show that |A|
(in general) is not the same as “right order
topology”
2
.
Proposition 2071.
1
.
D
|A|
1
E
@X = @
n
aA
yA:(y>aX[a;y[6=)
o
;
2
.
|A|
1
>
@X = @
n
aA
yA:(y>aX]a;y[6=)
o
;
3
.
D
|A|
1
E
@X = @
n
aA
xA:(x<aX]x;a]6=)
o
;
4
.
|A|
1
<
@X = @
n
aA
xA:(x<aX]x;a[6=)
o
.
Proof. a
D
|A|
1
E
@X @{a} 6
D
|A|
1
E
@X h|A|
i
@{a} 6 @X
(a) 6 @X y A : (y > a X [a; y[6= ).
a
|A|
1
>
@X @{a} 6
|A|
1
>
@X h|A|
>
i
@{a} 6 @X
>
(a) 6
@X y A : (y > a X]a; y[6= ).
The rest follows from duality.
Remark 2072. On trivial ultrafilters these obviously agree:
1
. h|R|
i
{x} = h|R|u ≥i
{x};
2
. h|R|
>
i
{x} = h|R|u >i
{x};
3
. h|R|
i
{x} = h|R|u ≤i
{x};
4
. h|R|
<
i
{x} = h|R|u <i
{x}.
1
See Wikipedia for a definition of “Order topology”.
2
See Wikipedia
3. SOME INEQUALITIES 8
Corollary 2073.
1
. |R|
= Compl(|R|u );
2
. |R|
>
= Compl(|R|u >);
3
. |R|
= Compl(|R|u );
4
. |R|
<
= Compl(|R|u <).
Obvious 2074. FiXme: also what is the values of \ operation
1
. |R|
= |R|
>
t 1;
2
. |R|
= |R|
<
t 1.
3. Some inequalities
FiXme: Define the ultrafilter “at the left” and “at the right” of a real number.
Also define “convergent ultrafilter”.
Denote
+
=
d
xR
]x; +[ and
−∞
=
d
xR
] ; x[.
The following proposition calculates h≥ix and h>ix for all kinds of ultrafilters
on R:
Proposition 2075.
1
. h≥i{α} = [α; +[ and h>i{α} =]α; +[.
2
. h≥ix = h>ix =]α; +[ for ultrafilter x at the right of a number α.
3
. h≥ix = h>ix =
<
(α) t[α; +[=
(α)t]α; +[ for ultrafilter x at the
left of a number α.
4
. h≥ix = h>ix =
+
for ultrafilter x at positive infinity.
5
. h≥ix = h>ix = R for ultrafilter x at negative infinity.
Proof.
1
. Obvious.
2
.
h≥ix =
F
l
Xup x
h≥i(Xu]α; +[) =
F
l
Xup x
]α; +[=]α; +[;
h>ix =
F
l
Xup x
h>i(Xu]α; +[) =
F
l
Xup x
]α; +[=]α; +[.
3
.
<
(α) t [α; +[=
(α)t]α; +[ is obvious.
h>ix =
F
l
Xup x
h>iX w
F
l
Xup x
(∆
<
(α)t]α; +[) =
<
(α)t]α; +[
but h≥ix v
<
(α) t [α; +[ is obvious. It remains to take into account that
h>ix v h≥ix.
4
. h≥ix =
d
F
Xup x
h≥iX =
d
F
Xup x,inf XX
h≥i(Xu]α; +[) =
d
F
Xup x
[inf X; +[=
d
F
x>α
[x; +[=
+
; h>ix =
d
F
Xup x
h>iX =
d
F
Xup x,inf XX
h>i(Xu]α; +[) =
d
F
Xup x
] inf X; +[=
d
F
x>α
[x; +[=
+
.
5
. h≥ix w h>ix =
d
F
Xup x
h>iX but h>iX =] ; +[ for X up x
because X has arbitrarily small elements.
Lemma 2076. h|R|ix v h>ix = h≥ix for every nontrivial ultrafilter x.
Proof. h>ix = h≥ix follows from the previous proposition.
h|R|ix =
d
Xup x
h|R|iX =
d
Xup x
d
yX
∆(y).
Consider cases:
4. CONTINUITY 9
x is an ultrafilter at the right of some number α.
h|R|ix =
d
Xup x
d
yXu]α;+[
∆(y) v]α; +[= h≥ix because
d
yXu]α;+[
∆(y) v]α; +[.
x is an ultrafilter at the left of some number α.
h|R|ix v ∆(α) is obvious. But h≥ix w ∆(α).
x is an ultrafilter at positive infinity.
h|R|ix v
+
is obvious. But h≥ix =
+
.
x is an ultrafilter at negative infinity.
Because h≥ix = R.
Corollary 2077. h|R|u ≥ix = h|R|ix for every nontrivial ultrafilter x.
Proof. h|R|u ≥ix = h|R|i u h≥ix = h|R|ix.
So h|R|u ≥i and h|R|i agree on all ultrafilters except trivial ones.
Proposition 2078. |R|
>
u >= |R|
>
u = |R|
>
.
Proof. |R|
>
v > because h|R|
>
i
x v h>i
x and |R|
>
is a complete funcoid.
Lemma 2079. h|R|
>
ix @ h|R|
ix for a nontrivial ultrafilter x.
Proof. It enough to prove h|R|
>
ix 6= h|R|
ix.
Take x be an ultrafilter with limit point 0 on im z where z is the sequence
n 7→
1
n
.
h|R|
>
ix v h|R|
>
i
im z =
l
nim z
>
1
n
v
l
nim z
1
n
;
1
n 1
1
n
im z.
Thus h|R|
>
ix im z. But h|R|
ix v h=ix 6 im z.
Corollary 2080. |R|
>
@ |R|
.
Proposition 2081. |R|
>
@ |R|
u >.
Proof. It’s enough to prove |R|
>
6= |R|
u >.
Really, h|R|
u >ix = h|R|
ix 6= h|R|
>
ix (lemma).
Proposition 2082.
1
. |R|
|R|
= |R|
;
2
. |R|
>
|R|
>
= |R|
>
;
3
. |R|
|R|
>
= |R|
>
;
4
. |R|
>
|R|
= |R|
>
.
Proof. ??
Conjecture 2083.
1
. (|R| u ) (|R| u ) = |R| u .
2
. (|R| u >) (|R| u >) = |R| u >.
4. Continuity
I will say that a property holds on a filter A iff there is A up A on which the
property holds.
FiXme: f C(A, B) f C(ι
A
|R|
, ι
B
|R|
) (f, f)
C((A, ι
A
|R|
), (B, ι
B
|R|
))
Lemma 2084. Let function f : A B where A, B PR and A is connected.
4. CONTINUITY 10
1
. f is monotone and f C(A, B) iff f C(A, B) C(ι
A
|R|
, ι
B
|R|
)
iff f C(A, B) C(ι
A
|R|
>
, ι
B
|R|
) iff f C(ι
A
|R|
, ι
B
|R|
)
C(ι
A
|R|
, ι
B
|R|
).
2
. f is strictly monotone and C(A, B) iff f C(A, B) C(ι
A
|R|
>
, ι
B
|R|
>
)
iff f C(ι
A
|R|
>
, ι
B
|R|
>
) C(ι
A
|R|
<
, ι
B
|R|
<
).
FiXme: Generalize for arbitrary posets. FiXme: Generalize for f being a funcoid.
Proof. Because f is continuous, we have hf ι
A
|R|i
{x} v hι
B
|R| fi
{x}
that is hf i
∆(x) v ∆(f (x)) for every x.
If f is monotone, we have hfi
(x) v [f(x); [. Thus
hfi
(x) v
(f(x)), that is hf ι
A
|R|
i
{x} v hι
B
|R|
fi
{x}, thus
f C(ι
A
|R|
, ι
B
|R|
).
If f is strictly monotone, we have hfi
>
(x) v]f(x); [. Thus
hfi
>
(x) v
>
(f(x)), that is hf ι
A
|R|
>
i
{x} v hι
B
|R|
>
fi
{x}, thus f
C(ι
A
|R|
>
, ι
B
|R|
>
).
Let now f C(ι
A
|R|
, ι
B
|R|
).
Take any a A and let c =
n
bB
ba,x[a;b[:f(x)f(a)
o
. It’s enough to prove that
c is the right endpoint (finite or infinite) of A.
Indeed by continuity f (a) f(c) and if c is not already the right endpoint
of A, then there is b
0
> c such that x [c; b
0
[: f(x) f(c). So we have x [a; b
0
[:
f(x) f(c) what contradicts to the above.
So f is monotone on the entire A.
f C(ι
A
|R|
, ι
B
|R|
) f C(ι
A
|R|
>
, ι
B
|R|
) is obvious. Reversely f
C(ι
A
|R|
>
, ι
B
|R|
) f ι
A
|R|
>
v ι
B
|R|
f x R : hfihι
A
|R|
>
i
{x} v
hι
B
|R|
i
hfi
{x} x R : hfi
>
(x) v
f(x) x R : hfi
>
(x) t
{f(x)} v
f(x) x R : hf i
>
(x) t {x} v
f(x) x R : hf i
(x) v
f(x) x R : hfihι
A
|R|
i
{x} v hι
B
|R|
i
hfi
{x} x R : f ι
A
|R|
v
ι
B
|R|
f f C(ι
A
|R|
, ι
B
|R|
).
Let f C(ι
A
|R|
>
, ι
B
|R|
>
). Then f C(ι
A
|R|
>
, ι
B
|R|
) and thus it is mono-
tone. We need to prove that f is strictly monotone. Suppose the contrary. Then
there is a nonempty interval [p; q] A such that f is constant on this interval. But
this is impossible because f C(ι
A
|R|
>
, ι
B
|R|
>
).
Prove that f C(ι
A
|R|
, ι
B
|R|
) C(ι
A
|R|
, ι
B
|R|
) implies f C(A, B).
Really, it implies hf i
(x) v
(fx) and hfi
(x) v
(fx) thus hfi∆(x) =
hfi(∆
(x) t {x} t
(x))
f(x) t {f(x)} t
f(x) = ∆(f(x)).
Prove that f C(ι
A
|R|
>
, ι
B
|R|
>
) C(ι
A
|R|
<
, ι
B
|R|
<
) f C(A, B). Really, it
implies hfi
<
(x) v
<
(fx) and hfi
>
(x) v
>
(fx) thus hfi∆(x) = hfi(∆
<
(x)t
{x} t
>
(x))
<
f(x) t {f(x)} t
>
f(x) = ∆(f(x)).
Theorem 2085. Let function f : A B where A, B PR.
1
. f is locally monotone and f C(A, B) iff f C(A, B)C(ι
A
|R|
, ι
B
|R|
)
iff f C(A, B) C(ι
A
|R|
>
, ι
B
|R|
) iff f C(ι
A
|R|
, ι
B
|R|
)
C(ι
A
|R|
, ι
B
|R|
).
2
. f is locally strictly monotone and C(A, B) iff f C(A, B)
C(ι
A
|R|
>
, ι
B
|R|
>
) iff f C(ι
A
|R|
>
, ι
B
|R|
>
) C(ι
A
|R|
<
, ι
B
|R|
<
).
Proof. By the lemma it is (strictly) monotone on each connected component.
See also related math.SE questions:
1
. http://math.stackexchange.com/q/1473668/4876
2
. http://math.stackexchange.com/a/1872906/4876
3
. http://math.stackexchange.com/q/1875975/4876
4. CONTINUITY 11
4.1. Directed topological spaces. Directed topological spaces are defined
at
http://ncatlab.org/nlab/show/directed+topological+space
Definition 2086. A directed topological space (or d-space for short) is a pair
(X, d) of a topological space X and a set d C([0; 1], X) (called directed paths or
d-paths) of paths in X such that
1
. (constant paths) every constant map [0; 1] X is directed;
2
. (reparameterization) d is closed under composition with increasing con-
tinuous maps [0; 1] [0; 1];
3
. (concatenation) d is closed under path-concatenation: if the d-paths a,
b are consecutive in X (a(1) = b(0)), then their ordinary concatenation
a + b is also a d-path
(a + b)(t) = a(2t), if 0 t
1
2
,
(a + b)(t) = b(2t 1), if
1
2
t 1.
I propose a new way to construct a directed topological space. My way is more
geometric/topological as it does not involve dealing with particular paths.
Definition 2087. Let T be the complete endofuncoid corresponding to a topo-
logical space and ν v T be its “subfuncoid”. The d-space (dir)(T, ν) induced by
the pair (T, ν) consists of T and paths f C([0; 1], T ) C(|[0; 1]|
, ν) such that
f(0) = f(1).
Proposition 2088. It is really a d-space.
Proof. Every d-path is continuous.
Constant path are d-paths because ν is reflexive.
Every reparameterization is a d-path because they are C(|[0; 1]|
, ν) and we
can apply the theorem about composition of continuous functions.
Every concatenation is a d-path. Denote f
0
= λt [0;
1
2
] : a(2t) and f
1
= λt
[
1
2
; 1] : b(2t 1). Obviously f
0
, f
1
C([0; 1], µ) C(|[0; 1]|
, ν). Then we conclude
that a + b = f
1
t f
1
is in f
0
, f
1
C([0; 1], µ) C(|[0; 1]|
, ν) using the fact that the
operation is distributive over t.
Below we show that not every d-space is induced by a pair of an endofuncoid
and its subfuncoid. But are d-spaces not represented this way good anything except
counterexamples?
Let now we have a d-space (X, d). Define funcoid ν corresponding to the d-
space by the formula ν =
d
ad
(a |R|
a
1
).
Example 2089. The two directed topological spaces, constructed from a fixed
topological space and two different reflexive funcoids, are the same.
Proof. Consider the indiscrete topology T on R and the funcoids 1
FCD(R,R)
and 1
FCD(R,R)
t({0}×
FCD
). The only d-paths in both these settings are constant
functions.
Example 2090. A d-space is not determined by the induced funcoid.
Proof. The following a d-space induces the same funcoid as the d-space of all
paths on the plane.
Consider a plane R
2
with the usual topology. Let d-paths be paths lying inside
a polygonal chain (in the plane).
6. INTEGRAL CURVES 12
Conjecture 2091. A d-path a is determined by the funcoids (where x spans
[0; 1])
(λt R : a(x + t))|
∆(0)
.
5. A way to construct directed topological spaces
I propose a new way to construct a directed topological space. My way is more
geometric/topological as it does not involve dealing with particular paths.
Conjecture 2092. Every directed topological space can be constructed in the
below described way.
Consider topological space T and its subfuncoid F (that is F is a funcoid
which is less that T in the order of funcoids). Note that in our consideration F is
an endofuncoid (its source and destination are the same).
Then a directed path from point A to point B is defined as a continuous function
f from [0; 1] to F such that f(0) = A and f(1) = B. FiXme: Specify whether the
interval [0; 1] is treated as a proximity, pretopology, or preclosure.
Because F is less that T , we have that every directed path is a path.
Conjecture 2093. The two directed topological spaces, constructed from a
fixed topological space and two different funcoids, are different.
For a counter-example of (which of the two?) the conjecture consider funcoid
T u (Q ×
FCD
Q) where T is the usual topology on real line.We need to consider
stability of existence and uniqueness of a path under transformations of our funcoid
and under transformations of the vector field. Can this be a step to solve Navier-
Stokes existence and smoothness problems?
6. Integral curves
We will consider paths in a normed vector space V .
Definition 2094. Let D be a connected subset of R. A path is a function
D V .
Let d be a vector field in a normed vector space V .
Definition 2095. Integral curve of a vector field d is a differentiable function
f : D V such that f
0
(t) = d(f(t)) for every t D.
Definition 2096. The definition of right side integral curve is the above defi-
nition with right derivative of f instead of derivative f
0
. Left side integral curve is
defined similarly.
6.1. Path reparameterization. C
1
is a function which has continuous de-
rivative on every point of the domain.
By D
1
I will denote a C
1
function whose derivative is either nonzero at every
point or is zero everywhere.
Definition 2097. A reparameterization of a C
1
path is a bijective C
1
function
φ : D D such that φ
0
(t) > 0. A curve f
2
is called a reparametrized curve f
1
if
there is a reparameterization φ such that f
2
= f
1
φ.
It is well known that this defines an equivalence relation of functions.
Proposition 2098. Reparameterization of D
1
function is D
1
.
Proof. If the function has zero derivative, it is obvious.
Let f
1
has everywhere nonzero derivative. Then f
0
2
(t) = f
0
1
(φ(t))φ
0
(t) what is
trivially nonzero.
6. INTEGRAL CURVES 13
Definition 2099. Vectors p and q have the same direction (p q) iff there
exists a strictly positive real c such that p = cq.
Obvious 2100. Being same direction is an equivalence relation.
Obvious 2101. The only vector with the same direction as the zero vector is
zero vector.
Theorem 2102. A D
1
function y is some reparameterization of a D
1
integral
curve x of a continuous vector field d iff y
0
(t) d(y(t)) that is vectors y
0
(t) and
d(y(t)) have the same direction (for every t).
Proof. If y is a reparameterization of x, then y(t) = x(φ(t)). Thus y
0
(t) =
x
0
(φ(t))φ
0
(t) = d(x(φ(t)))φ
0
(t) = d(y(t))φ
0
(t). So y
0
(t) d(y(t)) because φ
0
(t) > 0.
Let now x
0
(t) d(x(t)) that is that is there is a strictly positive function c(t)
such that x
0
(t) = c(t)d(x(t)).
If x
0
(t) is zero everywhere, then d(x(t)) = 0 and thus x
0
(t) = d(x(t)) that is x
is an Integral curve. Note that y is a reparameterization of itself.
We can assume that x
0
(t) 6= 0 everywhere. Then F (x(t)) 6= 0. We have that
c(t) =
||x
0
(t)||
||d(x(t))||
is a continuous function. FiXme: Does it work for non-normed
spaces?
Let y(u(t)) = x(t), where
u(t) =
Z
t
0
c(s)ds,
which is defined and finite because c is continuous and monotone (thus having
inverse defined on its image) because c is positive.
Then
y
0
(u(t))u
0
(t) = x
0
(t)
= c(t)d(x(t)), so
y
0
(u(t))c(t) = c(t)d(y(u(t)))
y
0
(u(t)) = d(y(u(t)))
and letting s = u(t) we have y
0
(s) = d(y(s)) for a reparameterization y of x.
6.2. Vector space with additional coordinate. Consider the normed vec-
tor space with additional coordinate t.
Our vector field d(t) induces vector field
ˆ
d(t, v) = (1, d(v)) in this space. Also
ˆ
f(t) = (t, f (t)).
Proposition 2103. Let f be a D
1
function. f is an integral curve of d iff
ˆ
f is
a reparametrized integral curve of
ˆ
d.
Proof. First note that
ˆ
f always has a nonzero derivative.
ˆ
f
0
(t)
ˆ
d(
ˆ
f(t))
(1, f
0
(t)) (1, d(f(t))) f
0
(t) = d(f(t)).
Thus we have reduced (for D
1
paths) being an integral curve to being a
reparametrized integral curve. We will also describe being a reparametrized in-
tegral curve topologically (through funcoids).
6.3. Topological description of C
1
curves. Explicitly construct this fun-
coid as follows:
R(d, φ) =
n
vV
b
vd<φ,v6=0
o
for d 6= 0 and R(0, φ) = {0}, where
b
ab is the angle
between the vectors a and b, for a direction d and an angle φ.
6. INTEGRAL CURVES 14
Definition 2104. W (d) =
d
RLD
n
R(d,φ)
φR,φ>0
o
u
d
RLD
r>0
B
r
(0). FiXme: This is
defined for infinite dimensional case. FiXme: Consider also FCD instead of RLD.
Proposition 2105. For finite dimensional case R
n
we have W (d) =
d
RLD
n
R(d,φ)
φR,φ>0
o
u
(RLD)n
where
(RLD)n
= ×
RLD
· · · ×
RLD
| {z }
n times
.
Proof. ??
Finally our funcoids are the complete funcoids Q
+
and Q
described by the
formulas
hQ
+
i
@{p} = hp+iW (d(p)) and hQ
i
@{p} = hp+iW (d(p)).
Here is taken from the “counter-examples” section.
In other words,
Q
+
=
l
pR
(@{p} ×
FCD
hp+iW (d(p))); Q
=
l
pR
(@{p} ×
FCD
hp+iW (d(p))).
That is hQ
+
i
@{p} and hQ
i
@{p} are something like infinitely small spherical
sectors (with infinitely small aperture and infinitely small radius).
FiXme: Describe the co-complete funcoids reverse to these complete funcoids.
Theorem 2106. A D
1
curve f is an reparametrized integral curve for a direc-
tion field d iff f C(ι
D
|R|
>
, Q
+
) C(ι
D
|R|
<
, Q
).
Proof. Equivalently transform f C(ι
D
|R|, Q
+
); f ι
D
|R| v Q
+
f;
hf ι
D
|R|i
@{t} v hQ
+
fi
@{t}; hfi
>
(t) u D v hQ
+
i
f(t); hf i
>
(t) v
hQ
+
i
f(t); hf i
>
(t) v f (t) + W(D(f(t))); hf i
>
(t) f(t) v W (D(f(t)));
r > 0, φ > 0δ > 0 : hf i
(]t; t + δ[) f (t) R(d(f (t)), φ) B
r
(f(t));
r > 0, φ > 0δ > 00 < γ < δ : hf i
(]t; t + γ[) f(t) R(d(f(t)), φ) B
r
(f(t));
r > 0, φ > 0δ > 00 < γ < δ :
hfi
(]t; t + γ[) f(t)
γ
R(d(f(t)), φ)B
r/δ
(f(t));
r > 0, φ > 0δ > 0 :
+
f(t) R(d(f(t)), φ) B
r/δ
(f(t));
r > 0, φ > 0 :
+
f(t) R(d(f(t)), φ);
+
f(t) d(f(t))
where
+
is the right derivative.
In the same way we derive that C(|R|
<
, Q
)
f(t) d(f(t)).
Thus f
0
(t) d(f (t)) iff f C(|R|
>
, Q
+
) C(|R|
<
, Q
).
6.4. C
n
curves. We consider the differential equation f
0
(t) = d(f(t)).
We can consider this equation in any topological vector space V (https://en.
wikipedia.org/wiki/Frechet_derivative), see also https://math.stackexchange.com/
q/2977274/4876. Note that I am not an expert in topological vector spaces and
thus my naive generalizations may be wrong in details.
n-th derivative f
(n)
(t) = d
n
(f(t)); f
(n+1)
(t) = d
0
n
(f(t)) f
0
(t) = d
0
n
(f(t))
d(f(t)). So d
n+1
(y) = d
0
n
(y) d(y).
Given a point y V define
R
n
(y) =
v V
\
vd
0
(y) <
d
1
1!
(y)|v| +
d
2
(y)
2!
|v|
2
+ · · · +
d
n1
(y)
(n1)!
|v|
n1
+ O(|v|
n
), v 6= 0
for d
0
(y) 6= 0 and R
n
= {0} if d
0
(y) = 0.
6. INTEGRAL CURVES 15
Definition 2107. R
(y) = R
0
(y) u R
1
(y) u R
2
(y) u . . . .
FiXme: It does not work: https://math.stackexchange.com/a/2978532/4876.
Definition 2108. W
n
(y) = R
n
(y) u
d
RLD
r>0
B
r
(0); W
(y) = R
(y) u
d
RLD
r>0
B
r
(0).
Finally our funcoids are the complete funcoids Q
n
+
and Q
n
described by the
formulas
Q
n
+
@{p} = hp+iW
n
(p) and
Q
n
@{p} = hp+iW
n
(p)
where W
is W for the reverse vector field d(y).
FiXme: Related questions: http://math.stackexchange.com/q/1884856/4876
http://math.stackexchange.com/q/107460/4876 http://mathoverflow.net/q/88501
Lemma 2109. Let for every x in the domain of the path for small t > 0 we
have f(x + t) R
n
(F (f(x))) and f(x t) R
n
(F (f(x))). Then f is C
n
smooth.
Proof. FiXme: Not yet proved!
See also http://math.stackexchange.com/q/1884930/4876.
Conjecture 2110. A path f is conforming to the above differentiable equa-
tion and C
n
(where n is natural or infinite) smooth iff f C(ι
D
|R|
>
, Q
n
+
)
C(ι
D
|R|
<
, Q
n
).
Proof. Reverse implication follows from the lemma.
Let now a path f is C
n
. Then
f(x + t) =
n
X
i=0
f
(i)
(x)
t
i
i!
+ o(t
i
) = f(x) + f
0
(x)t +
n
X
i=2
f
(i)
(x)
t
i
i!
+ o(t
i
)
6.5. Plural funcoids. Take I
+
and Q
+
as described above in forward direc-
tion and I
and Q
in backward direction. Then
f C(I
+
, Q
+
) f C(I
, Q
) f × f C(I
+
×
(A)
I
, Q
+
×
(A)
Q
)?
To describe the above we can introduce new term plural funcoids. This is
simply a map from an index set to funcoids. Composition is defined component-
wise. Order is defined as product order. Well, do we need this? Isn’t it the same
as infimum product of funcoids?
6.6. Multiple allowed directions per point.
hQi
@{p} =
l
dd(p)
hp+iW (d).
It seems (check!) that solutions not only of differential equations but also of
difference equations can be expressed as paths in funcoids.
CHAPTER 3
More on generalized limit
Definition 2111. I will call a permutation group fixed point free when every
element of it except of identity has no fixed points.
Definition 2112. A funcoid f is Kolmogorov when hfi
{x} 6= hf i
{y} for
every distinct points x, y dom f.
1. Hausdorff funcoids
Definition 2113. Limit lim F = x of a filter F regarding funcoid f is such a
point that hfi
{x} w F.
Definition 2114. Hausdorff funcoid is such a funcoid that every proper filter
on its image has at most one limit.
Proposition 2115. The following are pairwise equivalent for every funcoid f:
1
. f is Hausdorff.
2
. x 6= y hfi
{x} hfi
{y}.
Proof.
1
2
. If 2
does not hold, then there exist distinct points x and y such that
hfi
{x} 6 hfi
{y}. So x and y are both limit points of hf i
{x}u hfi
{y},
and thus f is not Hausdorff.
2
1
. Suppose F is proper.
hfi
{x} w F hfi
{y} w F hfi
{x} 6 hfi
{y} x = y.
Corollary 2116. Every entirely defined Hausdorff funcoid is Kolmogorov.
Remark 2117. It is enough to be “almost entirely defined” (having nonempty
value everywhere except of one point).
Obvious 2118. For a complete funcoid induced by a topological space this
coincides with the traditional definition of a Hausdorff topological space.
2. Restoring functions from limit
Consider alternative definition of generalized limit:
xlim f = λr G : ν f r.
Or:
xlim
a
f =
(
(
r
1
a, ν f r)
r G
)
(note this requires explicit filter in the definition of generalized limit).
Operations on the set of generalized limits can be defined (twice) pointwise.
FiXme: First define operations on funcoids.
Proposition 2119. The above defined xlim
hµi
{x}
f is a monovalued function
if µ is Kolmogorov and G is fixed point free.
16
2. RESTORING FUNCTIONS FROM LIMIT 17
Proof. We need to prove
r
1
hµi
{x} 6=
s
1
hµi
{x} for r, s G, r 6= s.
Really, by definition of generalized limit, they commute, so our formula is equivalent
to hµi
r
1
{x} 6= hµi
s
1
{x}; hµi
r
1
s
s
1
{x} 6= hµi
s
1
{x}.
But r
1
s 6= e, so because it is fixed point free,
r
1
s
s
1
{x} 6=
s
1
{x}
and thus by kolmogorovness, we have the thesis.
Lemma 2120. Let µ and ν be Hausdorff funcoids. If function f is defined at
point x, then
fx = lim
(xlim
hµi
{x}
f)hµi
{x}
{x}
Remark 2121. The right part is correctly defined because xlim
a
f is monoval-
ued.
Proof. lim
(xlim
hµi
{x}
f)hµi
{x}
{x} = limhν fi
{x} = limhνi
fx = fx.
Corollary 2122. Let µ and ν be Hausdorff funcoids. Then function f can
be restored from values of xlim
hµi
{x}
f.
CHAPTER 4
Extending Galois connections between funcoids
and reloids
Definition 2123.
1
. Φ
f = λb B :
d
n
xA
fxvb
o
;
2
. Φ
f = λb A :
d
n
xB
fxwb
o
.
Proposition 2124.
1
. If f has upper adjoint then Φ
f is the upper adjoint of f.
2
. If f has lower adjoint then Φ
f is the lower adjoint of f .
Proof. By theorem 131.
Lemma 2125. Φ
(RLD)
out
= (FCD).
Proof.
(RLD)
out
)f =
d
n
gFCD
(RLD)
out
gwf
o
=
d
FCD
n
gRel
(RLD)
out
gwf
o
=
d
FCD
n
gRel
gwf
o
= (FCD)f.
Lemma 2126. Φ
(RLD)
out
6= (FCD).
Proof.
(RLD)
out
)f =
d
n
gFCD
(RLD)
out
gvf
o
(RLD)
out
) 6= .
Lemma 2127. Φ
(FCD) = (RLD)
out
.
Proof.
(FCD))f =
d
n
gRLD
(FCD)gwf
o
=
d
RLD
n
gRel
(FCD)gwf
o
=
d
RLD
n
gRel
gwf
o
=
(RLD)
out
f.
Lemma 2128. Φ
(RLD)
in
= (FCD).
Proof.
(RLD)
in
)f =
d
n
gFCD
(RLD)
in
gvf
o
=
d
n
gFCD
gv(FCD)f
o
= (FCD)f.
Theorem 2129. The picture at figure 1 describes values of functions Φ
and Φ
.
All nodes of this diagram are distinct.
Proof. Follows from the above lemmas.
Figure 1
(FCD) (RLD)
in
(RLD)
out
other
Φ
Φ
Φ
Φ
18
4. EXTENDING GALOIS CONNECTIONS BETWEEN FUNCOIDS AND RELOIDS 19
Question 2130. What is at the node “other”?
Trying to answer this question:
Lemma 2131.
(RLD)
out
) =
FCD
.
Proof. We have (RLD)
out
FCD
= . x 6v
FCD
(RLD)
out
x w Cor x A .
Thus max
n
xFCD
(RLD)
out
x=
o
=
FCD
.
So
(RLD)
out
) =
FCD
.
Conjecture 2132.
(RLD)
out
)f =
FCD
t (FCD)f.
The above conjecture looks not natural, but I do not see a better alternative
formula.
Question 2133. What happens if we keep applying Φ
and Φ
to the node
“other”? Will we this way get a finite or infinite set?
CHAPTER 5
Boolean funcoids
1. One-element boolean lattice
Let A be a boolean lattice and B = P 0. It’s sole element is .
f pFCD(A; B) X A : (hfiX 6 hf
1
i⊥ 6 X) X A : (0
f
1
6 X) X A :
f
1
X X A :
f
1
=
A
f
1
=
A
f
1
= {(;
A
)}.
Thus card pFCD(A; P0) = 1.
2. Two-element boolean lattice
Consider the two-element boolean lattice B = P 1.
Let f be a pointfree protofuncoid from A to B (that is (A; B; α; β) where
α B
A
, β A
B
).
f pFCD(A; B) X A, Y B : (hfiX 6 Y
f
1
Y 6 X) X
A, Y B : ((0 hfiX 0 Y ) (1 hfiX 1 Y ) hf
1
iY 6 X).
T =
n
XA
0∈hfiX
o
is an ideal. Really: That it’s an upper set is obvious. Let
P Q
n
XA
0∈hfiX
o
. Then 0 hfi(P Q) = hfiP hfiQ; 0 hfiP 0 hfiQ.
Similarly S =
n
XA
1∈hfiX
o
is an ideal.
Let now T, S PA be ideals. Can we restore hfi? Yes, because we know
0 hf iX and 1 hfiX for every X A.
So it is equivalent to X A, Y B : ((X T 0 Y ) (X S 1 Y )
hf
1
iY 6 X).
f pFCD(A; B) is equivalent to conjunction of all rows of this table:
Y equality
f
1
=
{0} X T
f
1
{0} 6 X
{1} X S
f
1
{1} 6 X
{0,1} X T X S
f
1
{0, 1} 6 X
Simplified:
Y equality
f
1
=
{0} T =
f
1
{0}
{1} S =
f
1
{1}
{0,1} T S =
f
1
{0, 1}
From the last table it follows that T and S are principal ideals.
So we can take arbitrary either
f
1
{0}, hf
1
i{1} or principal ideals T and
S.
In other words, we take
f
1
{0}, hf
1
i{1} arbitrary and independently. So
we have pFCD(A; B) equivalent to product of two instances of A. So it a boolean
lattice. FiXme: I messed product with disjoint union below.)
20
4. ABOUT INFINITE CASE 21
3. Finite boolean lattices
We can assume B = PB for a set B, card B = n. Then
f pFCD(A; B) X A, Y B : (hfiX 6 Y
f
1
Y 6 X) X
A, Y B : (i Y : i hfiX
f
1
Y 6 X).
Having values of
f
1
{i} we can restore all hf
1
iY . [need this paragraph?]
Let T
i
=
n
XA
i∈hfiX
o
.
Let now T
i
PA be ideals. Can we restore hfi? Yes, because we know
i hfiX for every X A.
So, it is equivalent to:
X A, Y B : (i Y : X T
i
f
1
Y 6 X). (1)
Lemma 2134. The formula (1) is equivalent to:
X A, i B : (X T
i
f
1
{i} 6 X). (2)
Proof. (1)(2). Just take Y = {i}.
(2)(1). Let (2) holds. Let also X A, Y B. Then hf
1
iY 6 X
S
iY
hf
1
i{i} 6 X i Y : hf
1
i{i} 6 X i Y : X T
i
.
Further transforming: i B : T
i
= hf
1
i{i}.
So
f
1
{i} are arbitary elements of B and T
i
are corresponding arbitrary
principal ideals.
In other words, pFCD(A; B)
=
AΠ . . . ΠA (card B times). Thus pFCD(A; B) is
a boolean lattice.
4. About infinite case
Let A be a complete boolean lattice, B be an atomistic boolean lattice.
f pFCD(A; B) X A, Y B : (hfiX 6 Y
f
1
Y 6 X) X
A, Y B : (i atoms Y : i atomshf iX
f
1
Y 6 X).
Let T
i
=
n
XA
iatomshfiX
o
.
T
i
is an ideal: Really: That it’s an upper set is obvious. Let P Q
n
XA
iatomshfiX
o
. Then i atomshfi(P Q) = atomshfiP atomshfiQ; i hfiP i
hfiQ.
Let now T
i
PA be ideals. Can we restore hfi? Yes, because we know
i atomshfiX for every X A and B is atomistic.
So, it is equivalent to:
X A, Y B : (i atoms Y : X T
i
f
1
Y 6 X). (3)
Lemma 2135. The formula (3) is equivalent to:
X A, i atoms
B
: (X T
i
hf
1
ii 6 X). (4)
Proof. (3)(4). Let (3) holds. Take Y = i. Then atoms Y = {i} and thus
X T
i
i atoms Y : X T
i
f
1
Y 6 X
f
1
i 6 X.
(4)(3). Let (2) holds. Let also X A, Y B. Then
f
1
Y 6 X
hf
1
i
d
atoms Y 6 X
d
iatoms Y
f
1
i 6 X i atoms Y :
f
1
i 6 X i atoms Y : X T
i
.
Further equivalently transforming: i atoms
B
: T
i
=
f
1
i.
So
f
1
i are arbitary elements of B and T
i
are corresponding arbitrary prin-
cipal ideals.
4. ABOUT INFINITE CASE 22
In other words, pFCD(A; B)
=
Q
icard atoms
B
A. Thus pFCD(A; B) is a boolean
lattice.
So finally we have a very weird theorem, which is a partial solution for the
above open problem (The weirdness is in its partiality and asymmetry):
Theorem 2136. If A is a complete boolean lattice and B is an atomistic
boolean lattice (or vice versa), then pFCD(A; B) is a boolean lattice.
[4] proves “THEOREM 4.6. Let A, B be bounded posets. A B is a com-
pletely distributive complete Boolean lattice iff A and B are completely distributive
Boolean lattices. (where A B is equivalent to the set of Galois connections be-
tween A and B) and other interesting results.
CHAPTER 6
Interior funcoids
Having a funcoid f let define interior funcoid f
.
Definition 2137. Let f FCD(A, B) = pFCD(T A, T B) be a co-complete
funcoid. Then f
pFCD(dual T A, dual T B) is defined by the formula hf
i
X =
hfiX.
Proposition 2138. Pointfree funcoid f
exists and is unique.
Proof. X 7→ hfiX is a component of pointfree funcoid dual T A dual T B
iff hfi is a component of the corresponding pointfree funcoid T A T B that is
essentially component of the corresponding funcoid FCD(A, B) what holds for a
unique funcoid.
It can be also defined for arbitrary funcoids by the formula f
= (CoCompl f)
.
Obvious 2139. f
is co-complete.
Theorem 2140. The following values are pairwise equal for a co-complete
funcoid f and X T Src f:
1
. hf
i
X;
2
.
n
yDst f
hf
1
i
{y}vX
o
3
.
d
n
Y T Dst f
hf
1
i
Y vX
o
4
.
d
n
YF Dst f
hf
1
iYvX
o
Proof.
1
=2
.
n
yDst f
hf
1
i
{y}vX
o
=
n
xDst f
hf
1
i
{x}X
o
=
n
xDst f
{x}hfiX
o
= hfiX = hf
i
X.
2
=3
. If
f
1
Y v X then (by completeness of f
1
) Y =
n
yY
hf
1
i
{y}vX
o
and
thus
l
Y T Dst f
hf
1
i
Y v X
v
y Dst f
hf
1
i
{y} v X
.
The reverse inequality is obvious.
3
=4
. It’s enough to prove that if
f
1
Y v X for Y F Dst f then exists
Y up Y such that hf
1
i
Y v X. Really let
f
1
Y v X. Then
d
hhf
1
i
i
up Y v X and thus exists Y up Y such that hf
1
i
Y v X
by properties of generalized filter bases.
This coincides with the customary definition of interior in topological spaces.
Proposition 2141. f
◦◦
= f for every funcoid f.
Proof. hf
◦◦
i
X = ¬¬hfi¬¬X = hfiX.
Proposition 2142. Let g FCD(A, B), f FCD(B, C), h FCD(A, C) for
some sets A. B, C.
g v f
h f
1
g v h, provided f and h are co-complete.
23
6. INTERIOR FUNCOIDS 24
Proof. g v f
h X A : hgi
X v hf
hi
X X A :
hgi
X v hf
i
hhi
X X A : hgi
X v ¬hfi
¬hhi
X X A : hgi
X
hfi
¬hhi
X X A :
f
1
hgi
X ¬hhi
X X A :
f
1
hgi
X v
hhi
X X A :