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

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Victor Porton
Abstract
I deﬁne connectors and generalized connectedness which generalizes topo-
logical connectedness, path connectedness, connectedness of digraphs, proximal
connectedness, uniform connectedness, and some other kinds of connectedness.
This article also serves as a simple introduction for my future writing s where
I will consider more diﬃcult topic of ﬁlters connected regarding funcoids and
reloids.
Keywords: connected, connectedness , disconnected, disconnectedness, path
connectedness, connectivity, connected space, disconnected space
A.M.S. subject classiﬁcation: 54D05, 54A99
Contents
1 Related works 2
2 Notation 2
3 Main deﬁnition 2
4 Examples of connectedness 4
4.1 Topological connectedness . . . . . . . . . . . . . . . . . . . . . . 4
4.2 Path connectedness and similar . . . . . . . . . . . . . . . . . . . 4
4.3 Proximal connectedness . . . . . . . . . . . . . . . . . . . . . . . 6
4.4 Connectedness rega rding a digraph . . . . . . . . . . . . . . . . . 6
4.5 Weak connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.6 Uniform connectedness . . . . . . . . . . . . . . . . . . . . . . . . 9
4.6.1 Some basic properties of ﬁlters . . . . . . . . . . . . . . . 9
4.6.2 Uniform triples . . . . . . . . . . . . . . . . . . . . . . . . 10
4.6.3 Uniform connectedness . . . . . . . . . . . . . . . . . . . . 11
4.6.4 Connectors for uniform connectedness . . . . . . . . . . . 12
URL: http://www.mathematics21.org (Victor Porton)
Preprint submitted to Elsevier July 31, 2010
5 Properties 13
5.1 Extendability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.2 Criteria of connectedness . . . . . . . . . . . . . . . . . . . . . . 13
5.2.1 Connectedness of unions of sets . . . . . . . . . . . . . . . 14
5.3 Links gener ated by a connector . . . . . . . . . . . . . . . . . . . 15
5.4 Relationships of Q(U ; τ) and T (U ; τ) . . . . . . . . . . . . . . . . 19
6 Future research 20
1. Related works
In  is researched an other way to generalize connectedness. Below is
remarked how these two ways are connected.
2. Notation
I will denote hf i X = {f(x) | x X} for every function f and set X.
X [f] Y x X, y Y : x f y (X × Y ) f 6= for every binary
relation f and sets X and Y .
3. Main deﬁnition
Let U is a set.
Deﬁnition 1 I will call a connector a binary relation r P(P U × PU) for
some s et U . The connector space is the pair (U ; r). I will call U the base of
the connector space (U; r).
I will denote A (U; r) B A r B for every sets A and B.
Deﬁnition 2 Let r is a connector. I call a set A connected (regarding r) when
X, Y PA \ {∅} : (X Y = A X Y = X r Y ). (1)
I will call connectedness the set of connected sets (regarding some connector
r). I will denote
CC(U; r) = {A P U | A is connected regarding r}
the connectedness regarding the connector s pace (U; r). (“CC” is deciphered as
“connector connectedness”.)
A set is connected regarding a connector space (U ; r) i it is connected re-
garding the connector r.
Intuitively: A set is connected if for every partition of it into two components
these two components are bo und with each other (“ to be bound” mean to be
related by the relation r).
I will call the above formula ge neralized connectedness.
2
Deﬁnition 3 Normalized connector space is such a connector space (U ; r)
that
X, Y PU : (X = ∅∨Y = ¬(X r Y )) and X, Y P U : (XY 6= X r Y ).
Deﬁnition 4 Normalization of a connector space (U; r) is the connector
N(U ; r) = (U; r
) deﬁned by the formula (for every X, Y PU )
X r
Y
0 if X = Y = ,
1 if X Y 6= ,
X r Y otherwise .
Obvious 1 Normalization of a connector space is a normalized connector space.
Obvious 2 A set is connected regarding a connector space iﬀ it is connected
regarding its normalization.
Obvious 3 For a normalized connector r a set A is connected iﬀ
X, Y PA \ {∅} : (X Y = A X r Y ).
Deﬁnition 5
Restriction r|
A
of a connector r to a set A is t he connector r (P A ×
PA).
Restriction (U; r)|
A
of a connector space (U ; r) to a set A PU is the
connector space (A; r (PA × P A)).
Theorem 1 CC((U; r)|
K
) = C C(U ; r) P K for every set K PU.
Proof A CC((U ; r)|
K
) A K X, Y PA\{∅} : (X Y = AX Y =
X (r (PK × P K)) Y ) A K X , Y PA \ {∅} : (X Y =
A X Y = X r Y ) A K A CC(U; r ) A C C(U ; r) PK for
every set A.
Corollary 1 CC ((U ; r)|
K
) C C(U ; r).
I will deﬁne an order on every set of c onnectors with the same base by the
formula
(U; r
0
) (U ; r
1
) r
0
r
1
.
3
4. Examples of connectedness
4.1. Topological connectedness
Let A is a topological space. If we take
X r Y (X is not open or Y is not open)
or
X r Y (X is not closed or Y is not closed)
or
X r Y cl
XY
(X) Y 6= cl
XY
(Y ) X 6= (2)
where openness and closedness is taken on the space A restricted to the set
X Y and c l
A
means the closure on the subspa ce A, then we get the classical
deﬁnition of a set connected regarding a topology.
Observe that there are several connectors which deﬁne the same c onnected-
ness (because their normalized connectors are identical).
4.2. Path connectedness and similar
Deﬁnition 6 I will cal l a ternary relation τ P(U × U × PU ) link.
I will call the pair (U ; τ) a l ink space.
I will denote a τ
A
b = τ
A
(a, b) = τ(a, b, A).
Remark 1 The expression τ (a, b, A) generaliz e s the sta tement “There exists a
path from a to b through A.” (where path may be taken in the sens e used in
topology or the sense used in graph theory ).
Deﬁnition 7 I will cal l a link space (U; τ) increasing iﬀ
A, B PU : (A B τ
A
τ
B
).
Deﬁnition 8 I will call the restriction of a link space(U ; τ) to a set A P U
the link space (A; τ (A × A × P A)).
Deﬁnition 9 I cal l a link space (U ; τ) symmetric when τ
A
is symm etric for
every A PU, transitive when τ
A
is transitive for every A PU , reﬂexive
when τ
A
is reﬂexive on A for every A PU. I will call a link space equivalence
when it is symmetric, transitive, and reﬂexive.
Deﬁnition 10 I will call a set A connected regarding a link τ when x, y
A : τ(x, y, A). I call connectedness regarding a link space (U; τ) the collection
of all connected (regarding τ) sets on U. I will denote LC(U ; τ) the connected-
ness regarding (U ; τ). (“LC” is deciphered as “link connectedness”.)
4
To get path connectedness we take (for some topology A)
τ
A
(x, y) f C([0; 1]; A|
A
) : (f (0) = x f(1) = y). (3)
Deﬁnition 11 We can deﬁne two connector spaces T (U; τ) and Q(U ; τ) with
the base U corresponding to a link space (U; τ) by the formulas:
X, Y PU : (X T (U; τ ) Y x X, y Y : τ(x, y, X Y ));
X, Y PU : (X Q(U; τ) Y x X, y Y : τ(x, y, X Y )).
Obvious 4 If τ is reﬂexive then Q(U; τ) is a normalized connector.
Obvious 5
1. (T (U; τ))|
K
= T ((U; τ )|
K
);
2. (Q(U ; τ))|
K
= Q((U; τ)|
K
).
Proposition 1 LC(U ; τ) = CC(T (U; τ)) for every reﬂexive link space (U ; τ).
Proof Let A is connected regarding T (U; τ ). Then
X, Y PA \ {∅} : (X Y = A X Y = X T (U; τ) Y )
that is
X, Y PA \ {∅} : (X Y = A X Y = x X, y Y : τ(x, y, X Y )).
Let a, b A and a 6= b. Then exist X, Y P A \ {∅} such that X Y =
A X Y = and a X, b Y . So τ(a, b, X Y ) that is τ(a, b, A). So taking
in account reﬂex ivity of τ we get that A is connected regarding τ.
Let now A is connected r e garding (U; τ). Let X, Y PA \ {∅} X Y =
A X Y = . We have τ(a, b, A) for every a X, b Y . Thus X T (U; τ) Y .
So A is connected regarding T (U; τ).
Theorem 2 For every equivalence link space (U ; τ)
LC(U; τ ) = CC(T (U; τ )) = CC(Q(U; τ)).
Proof Enough to prove LC(U; τ) = CC(Q(U ; τ)).
Let A is not connected regarding (U; τ) that is there are a, b A such that
¬(a τ
A
b). Then a K and b A \ K where K is a equivalence class regarding
τ
A
. So ¬(K Q(U ; τ) A \ K) and thus A is not connected regarding Q(U; τ).
Let A is connected regarding (U ; τ). Then for every X, Y PA\{∅} we have
x X, y Y : x τ
A
y and thus x X, y Y : x τ
A
y that is X Q (U ; τ) Y .
So A is connected regarding Q(U ; τ).
5
Remark 2 We may introduce other variants of path-connectedness replacing
topology A with a proximity or uniformity and continuity with proximal conti-
nuity or uniform continuity.
Proposition 2 The link space is an increasing equivalence for every A be it a
topology, proximity, or uniformity.
Proof Easy to prove in every of the three cases.
4.3. Proximal connectedness
The notion of proximal connectedness (also called “equiconnectedness”) is
deﬁned e.g. in , , and .
To get proximal connectedness we simply take the connector r = δ for a
proximity δ.
Remark 3 Connectedness regarding a proximity can be trivially generalized
to connectedness regarding a funcoid , but I omit this because the theory of
funcoids is not yet oﬃcially published.
Proposition 3 A set A is proximally connected iﬀ
X, Y PA \ {∅} : (X Y = A X δ Y ).
Proof Because δ is a normalized connector.
4.4. Connectedness regarding a digraph
The category of binary relations is the category whose objects are sets and
whose morphisms from a set A to a set B ar e triples (f; A; B) where f is a
binary relation a nd dom f A and im f b. Composition of morphis ms is
deﬁned in the natural way.
We will order this category by product order, that is
(f; A
0
; B
0
) (g; A
1
; B
1
) f g A
0
A
1
B
0
B
1
.
For two morphisms (f ; A
0
; B
0
) and (g; A
1
; B
1
) we have the meet of morphisms
by the formula
(f; A
0
; B
0
) (g; A
1
; B
1
) = (f g; A
0
A
1
; B
0
B
1
).
Easy to see that the right part of this formula is a morphism.
We will deﬁne A ×
C
B = (A × B; A; B).
I will deﬁne a digraph as an endomorphism of the category of binary re la-
tions. In other words, a digraph is (U; f) where U is a set and f is a binary
relation on U.
By deﬁnition a (f; A; B) b a f b (a; b) f.
By deﬁnition h(f; A; B)i X = hfi X and [(f; A; B)] = [f ].
6
Deﬁnition 12 Connectedness regarding a digraph is the connectedness for
the link (U ; τ) where U is the set of vertices of the digraph and τ(x, y, A) means
that there are a path from x to y in the subgraph restricted to A.
Obvious 6 The link space (U; τ) in the above deﬁnition is an increasing equiv-
alence.
Deﬁnition 13 S(U; f)
def
= (U; (=)|
U
f f
2
f
3
. . .) for every digraph (U; f ).
Proposition 4 There is a path from element a to element b in a set A through
a digraph µ iﬀ a S(µ (A ×
C
A)) b.
Proof
If exists a path fr om a to b, then {b}
(µ (A ×
C
A))
n
{a} where n is the
path length. Consequently {b}
S(µ (A ×
C
A))
{a}; a S(µ (A ×
C
A))
b.
If a (S(µ (A ×
C
A))) b then exists n N such that a (µ (A ×
C
A))
n
b.
By deﬁnition of composition of binary relations this means that there
exist ﬁnite sequence x
0
. . . x
n
where x
0
= a, x
n
= b for n N and
x
i
(µ (A ×
C
A)) x
i+1
for every i = 0, . . . , n 1. That is there is path
from a to b.
Lemma 1 If X Y = and ¬(X [f] Y ) then ¬(X [f
n
] Y ) for every sets X, Y ,
digraph f, and natural number n.
Proof For n = 0 it is obvious. Let’s prove by induction that it’s true for n > 1.
For n = 1 it is obvious.
Let it’s true for n = k > 0. ¬(X
f
k+1
Y ) Y
f
k+1
X =
Y
f
k
hfi X = ¬(hfi X
f
k
Y ) what is true by induction be c ause hfi X
Y = is equivalent to ¬(X [f ] Y ).
Theorem 3 The following st atemen ts are equivalent for a digraph µ and a set
A:
1. A is connected regarding the digraph µ.
2. S(µ (A ×
C
A)) A ×
C
A.
3. S(µ (A ×
C
A)) = A ×
C
A.
4. A is connected regarding the connector [µ].
5. X, Y A \ {∅} : (X Y = A X [µ] Y ).
7
Proof
(1)(2) Let for every a, b A there is a path between a and b in A through µ.
Then a (S(µ
C
(A × A))) b for every a, b A. It is possible only when
S(µ
C
(A × A)) A × A.
(3)(1) For every two vertices a and b we have a (S(µ (A ×
C
A))) b. So (by
the previous theorem) for every two vertices a and b exist path from a to
b.
(3)(4) Suppose that ¬(X
µ (A ×
C
A)
Y ) for some X, Y PU \ {∅} such
that XY = A and XY = . Then by a lemma ¬(X
(µ (A ×
C
A))
n
Y )
for every n N. Consequently ¬(X
S(µ (A ×
C
A))
Y ). So S(µ(A×
C
A)) 6= A × A.
(4)(3) If
S(µ
C
(A × A))
{v} = A for every vertex v then S(µ
C
(A ×
A)) = A×
C
A. Consider the remaining case when V
def
=
S(µ (A ×
C
A))
{v}
A for some vertex v. Let W = A\V . If card A = 1 then S(µ
C
(A×A))
(=)|
A
= A×
C
A; otherwise W 6= . Then V W = A and so V [µ] W what
is equivalent to V
µ
C
(A × A)
W that is
µ
C
(A × A)
V W 6= .
This is impos sible because
µ (A ×
C
A)
V =
µ (A ×
C
A)
S(µ (A ×
C
A))
V
S(µ (A ×
C
A))
V = V .
(2)(3) Because S(µ (A ×
C
A)) A ×
C
A.
(5)(4) Obvious.
(4)(5) Let (4) holds and let X Y = A. If X = Y = A then X [µ] Y because
A 6= . Otherwise X A or Y A. Let for example X A. Then
Y \ X 6= . So X [µ] Y \ X by (4) and consequently X [µ] Y .
Corollary 2 A set A is connected regarding a digraph µ iﬀ it is connected
regarding µ (A ×
C
A).
Theorem 4 The following statements are equivalent for each digraph µ =
(U; f ) and sets X, Y PU:
1. X T (U; τ) Y ;
2. X ×
C
Y S(µ ((X Y ) ×
C
(X Y )));
3. X ×
C
Y = S(µ ((X Y ) ×
C
(X Y ))).
8
Proof
X ×
C
Y S(µ ((X Y ) ×
C
(X Y )))
x X, y Y : x S(µ ((X Y ) ×
C
(X Y ))) y
x X, y Y : τ(x, y, X Y ) X T (U ; τ) Y.
X ×
C
Y S(µ((X Y )×
C
(X Y ))) X ×
C
Y = S(µ((X Y )×
C
(X Y )))
because S(µ ((X Y ) ×
C
(X Y ))) (X Y ) ×
C
(X Y ).
Theorem 5 Q(U ; τ) and [µ] have the same normalization (for every digraph
µ = (U ; f)).
Proof Let X, Y P U, X, Y 6= , X Y = . We need to prove X Q (U ; τ)
Y X [µ] Y .
X Q(U; τ) Y X [µ] Y is obvio us.
Let X Q(U ; τ) Y . Then there exists a path in X Y from a point of X to
a point of Y . Easy to see that there ex ist consequtive points x, y of this path
such that x µ y. So X [µ] Y .
Theorem 6 Regarding every digraph (U ; µ), connectedness is the same for con-
nector spaces:
1. T (U; τ);
2. Q(U ; τ);
3. (U ; [µ]).
Proof From the theorems 2 and 3.
4.5. Weak connectedness
By deﬁnition a set A is weakly connected regarding a digraph µ iﬀ it is
connected regarding the corresponding graph (that is connected regarding the
digraph µ µ
1
). So weak connectedness is also a kind of generalized connect-
edness.
4.6. Uniform connectedness
4.6.1. Some basic properties of ﬁlters
Let F is the set of ﬁlters on some set U.
I will denote [A) the principal ﬁlter corresponding to a set A.
Note that I do not require that ﬁlters do not contain the empty set, thus [)
is well deﬁned.
Proposition 5 a
F
b = { A B | A a, B b} for every ﬁlters a and b.
9
Proof First prove that {A B | A a, B b} is a ﬁlter. Let X, Y {A B | A a, B b}.
Then X = A
1
B
1
and Y = A
2
B
2
where A
1
, A
2
a and B
1
, B
2
b. Con-
sequently X Y = (A
1
A
2
) (B
1
B
2
) where A
1
A
2
a, B
1
B
2
b;
thus X Y {A B | A a, B b }. Let X {A B | A a, B b} and
C X. We have X = A B where A a, B b. We have C = C X =
C (A B) = (C A) (C B) where C A a and C B b; thus
C {A B | A a, B b}. So {A B | A a, B b} is a ﬁlter.
We need to prove that {A B | A a, B b} is the lowest upper bound
of {a, b}. We have {A B | A a, B b} a b ecause if X a then X =
X U {A B | A a, B b}. Similarly {A B | A a, B b} b. Thus
it is an upper bound.
Let p is an uppe r bound of {a, b}. Then p a that is A a : A p and
B b : B p. Thus because p is a ﬁlter we have A a, B b : A B p
that is p {A B | A a, B b}.
Proposition 6 [A)
F
[B) = [A B) for every subsets A and B of U .
Proof We need to prove that [A B) is the least upper bound of {[A), [B)}.
That [A B) [A), [B) is obvio us.
Remained to prove that a F : (a [A), [B) a [A B)). Really,
a [A), [B) A, B a A B a a [A B).
4.6.2. Uniform triples
I will deﬁne uniform connectedness. Below I will show that my deﬁnition is
equivalent to the class ical deﬁnition of uniform connectedness.
I will call a uniform triple on a set U the triple (f ; A; B) wher e f is a ﬁlter
on P(U × U) and A, B are such sets that A × B f. Note that uniform spaces
can be considered as uniform triples with A = B. I will denote R the set of
ﬁlters on P(U × U) and U the set of uniform triples.
I will call a generalized uniform space a uniform triple with A = B.
Remark 4 In fact there ca n be deﬁned composition of uniform triples and they
thus form morphisms of certain category. But in this article I’ll not dive into
details here. See my draft article .
We will introduce order on the set of uniform triples on a set by the formula
(f; A
0
; B
0
) (g; A
1
; B
1
) f g A
0
A
1
B
0
B
1
.
Easy to see that (f; A
0
; B
0
)
U
(g; A
1
; B
1
) = (f
R
g; A
0
A
1
; B
0
B
1
).
For a morphism (f; A; B) of the ca tegory of binary relations, I will denote
[(f; A; B)) = ([f ); A; B). Easy to see that [(f; A; B)) is a uniform triple.
By abuse of notation I will denote
(f; A
0
; B
0
) (g; A
1
; B
1
) f g A
0
= A
1
B
0
= B
1
where f is a binary relation and g is a ﬁlter on P(U × U ).
10
4.6.3. Uniform connectedness
Let µ is a generalize d uniform space.
Deﬁnition 14 I will denote S
(µ) =
S
U
{[S(f)) | f µ}.
Obvious 7 S
is a monotone function.
Deﬁnition 15 A set A is (uniforml y) connected regarding µ iﬀ S
(µ
U
[A ×
C
A)) [A ×
C
A).
Proposition 7 S
([f)) = [S(f)) for every digraph f.
Proof S
([f)) =
S
U
{[S(g)) | g [f )} =
S
U
{[S(f))} = [S(f)).
Obvious 8 A set A is connected regarding a generalized uniform space µ iﬀ
S
(µ
U
[A ×
C
A)) = [A ×
C
A).
Uniform connectedness is a ge ne ralization of digraph connectedness:
Proposition 8 A set A is uniformly connected regarding [µ) iﬀ it is connected
regarding µ (for every digraph µ).
Proof S
([µ)
U
[A × A)) = S
([µ (A ×
C
A))) = [S(µ (A ×
C
A))).
Thus S
([µ)
U
[A × A)) = [A ×
C
A) S(µ (A ×
C
A)) = A ×
C
A.
Obvious 9 A set A is connected regarding a generalized uniform space µ iﬀ
X S
(µ
U
[A × A)) : X A ×
C
A.
Obvious 10 A set A is connected regarding a generalized uniform space µ iﬀ
it is connected regarding µ
U
[A × A).
Proposition 9 A set A is connected regarding a generalized uniform s pace µ
iﬀ A is connected regarding every digraph f µ.
Proof
Let a set A is connected regarding µ and f µ. Then [f) µ; consequently
[f)
U
[A×
C
A) µ
U
[A×
C
A) and so S
([f)
U
[A×
C
A)) S
(µ
U
[A×
C
A)) [A×
C
A). Thus S
([f (A×
C
A))) [A×
C
A); [S(f (A×
C
A)))
[A ×
C
A); S(f (A ×
C
A)) A ×
C
A that is A is c onnected regarding f .
S
(µ
U
[A×
C
A)) =
S
U
[S(f)) | f µ
U
[A ×
C
A)
=
S
U
[S(g h)) | g µ, h [A ×
C
A)
S
U
[S(g (A ×
C
A))) | g µ
=
S
U
[A ×
C
A) | g µ
= [A ×
C
A).
11
4.6.4. Connectors for u niform connectedness
Let’s ﬁnd a connector which generates the same connectedness as the de-
scribed above uniform connectedness.
Proposition 10 x U : [{x} ×
C
{x}) S
(µ) for every generalized uniform
space µ = (U; f).
Proof S
(µ) =
S
U
{[S(f)) | f µ}. But {x} ×
C
{x} S(f); thus [{x} ×
C
{x}) [S(f )) and consequently
S
U
{[S(f)) | f µ} [{x} ×
C
{x}).
Lemma 2 [
S
S) F X S : [X) F for every collection S of sets and
every ﬁlt er F.
Proof
Obvious.
Let X S : [X) F that is X S, Y F : X Y . Then Y F :
S
S Y that is [
S
S) F .
From the above lemma follows that
[A ×
C
A) S
(µ
U
[A ×
C
A))
x A : [{x} ×
C
(A \ {x} )) S
(µ
U
[A ×
C
A)) [{x} ×
C
{x}) S
(µ
U
[A ×
C
A)).
Because x A : [{x} ×
C
{x}) S
(µ
U
[A ×
C
A)), we have
[A×
C
A) S
(µ
U
[A×
C
A)) x A : [{x}×
C
(A\{x})) S
(µ
U
[A×A))
Consequently
[A ×
C
A) S
(µ
U
[A ×
C
A))
X, Y PA : (X Y = X Y = A [X ×
C
Y ) S
(µ
U
[A ×
C
A)).
So, our sought-for c onnector is deﬁned (for example) by the formula
X r Y [X ×
C
Y ) S
(µ
U
[(X Y ) ×
C
(X Y ))).
A is connected regarding µ iﬀ f µ, X, Y PU : (X Y = A X [f ] Y )
X, Y PU : (X Y = A f µ : X [f] Y ). Thus
X r Y f µ : X [f ] Y f µ : X × Y f 6= (4)
is also a connector which induces uniform connectedness.
If µ is a uniformity, X r Y X δ Y where δ is the proximity induced
by µ. T hus my deﬁnition of uniform connectedness is equivalent to traditional
deﬁnition of uniform connectedness. (See theorem 1 in .)
12
5. Properties
5.1. Extendability
Deﬁnition 16 I will call a connector space (U ; r) up-directed when
X
0
, Y
0
, X
1
, Y
1
PU : (X
0
r Y
0
X
1
X
0
Y
1
Y
0
X
1
r Y
1
).
Deﬁnition 17 I will call a connector space (U ; r) extendable when
X
0
, Y
0
, X
1
, Y
1
PU : (X
1
Y
1
= X
0
r Y
0
X
1
X
0
Y
1
Y
0
X
1
r Y
1
).
Obvious 11 Every up-directed connector space is extendable.
Example 1 The fo llowing connector space s a re up-directed (and thus extend-
able):
1. the c onnector space deﬁned by the for mula (2);
2. (U ; [f]) for every digraph (U; f );
3. Q(U; τ) for an increasing link space (U; τ );
4. the c onnector space deﬁned by the for mula (4);
5. A proximity space (U; δ).
Proposition 11 A connector space is extendable iﬀ its normalization is up-
directed.
Proof
Let X N(r) Y and X
X, Y
Y . We have X
6= , Y
6= . If X
Y
6=
then X
N(r) Y
. Otherwise by extendabilty X
r Y
and consequently
X
N(r) Y
. Thus N(r) is up-directed.
Let X
1
Y
1
= X
0
r Y
0
X
1
X
0
Y
1
Y
0
. Then X
0
N(r) Y
0
and
consequently X
1
N(r) Y
1
. So X
1
r Y
1
.
5.2. Criteria of connectedness
Obvious 12 Empty set is connected regarding every connector.
Obvious 13 Every singleton is connected regarding every connector.
13
5.2.1. Connectedness of unions of sets
Lemma 3 If X Y = A B and X, Y 6= and X Y = then either
{X, Y } = {A, B} or A intersects both X and Y or B intersects both X and Y
(for every sets A, B, X, Y ).
Proof Let {X, Y } 6= {A, B}. Suppos e that A inters ects both X and Y do es
not hold (for example suppose that A X = 0) and prove B intersects both
X and Y .
We have X B and thus B X 6= 0. If also B Y = 0 then B X. So
X = B and thus either Y = A what contradicts to our supposition or A Y in
which case A intersects both X and Y .
Theorem 7 If sets A, B PU are connected regarding an exten dable connec-
tor space (U; r) and A r B then A B is also connected regarding (U; r).
Proof We need to prove that
X, Y P(A B) \ {∅} : (X Y = A B X Y = X r Y ).
Let X, Y P(A B) \ {∅} and X Y = A B X Y = . Then by the
lemma either {X, Y } = {A, B} and thus X r Y A r B so having X r Y , or A
inters ects both X and Y or B intersects b oth X and Y . Consider for example
then case X A 6= and Y A 6= .
In this case we have (X A) (Y A) = (X Y ) A = (A B) A = A
and (X A) (Y A) X Y = . Thus X A r Y A and consequently
X r Y (taken in account extendability).
Corollary 3 If sets A, B PU are connected regarding an extendable connec-
tor space (U; r) and A B 6= then A B is also connected regarding (U ; r ).
Proof Replace r with its normalization N(r). This preserves the same con-
nectedness. A B 6= A N (r) B. Thus we can apply the theorem.
There holds also inﬁnite version of the previous corollary:
Theorem 8 If S PP U is a collection of connected (regarding an extendable
connector space (U; r)) sets and
T
S 6= then
S
S is connected (regarding this
connector space).
Proof Let {X, Y } is a partition of
S
S. Then exist a point p
T
S such
that p X or p Y . Without lost of generality we may assume p X. Since
Y 6= , we have q Y for some q
S
S that is q A for some A S. So
A X, A Y 6= and thus {A X, A Y } is a partition of A. Since A is
connected, we have A X r A Y and thus (taken in account extendability)
X r Y . So
S
S is connected.
14
Corollary 4 Connectedness generated by an extendable connector space is a
c-structure in the sense of .
Remark 5 Connectedness generated by an extendable connector space is not
necessarily a connective structure in the sense of . A counter-example is
proximal connectedness on the set R \ {0}. (Take A = (−∞; 0), B = (0; +)
to violate the axiom (iii) in the main deﬁnition of .)
5.3. Links generated by a connector
Deﬁnition 18 a ρ(E) b K E : a, b K for every collection E of sets.
Deﬁnition 19 L(E)
A
= ρ(PA E) for every collection E of sets and a set A.
Let (U; r) is a connector space.
Deﬁnition 20 ζ
(U;r)
() is the link space deﬁned by the formula ζ
(U;r)
()
A
=
(U;
(U;r)|
A
).
Deﬁnition 21 Let (
(U;r)
) = ρ (CC(U ; r )).
Proposition 12 ζ
(U;r)
()
K
= (
(U;r)|
K
) = L(CC(U ; r))
K
= ρ(CC((U; r)|
K
))
for every connector space (U; r) and set K PU.
Proof (
(U;r)|
K
) = ρ (CC((U ; r)|
K
)) = ρ(CC(U; r ) P K) = L(CC(U; r))
K
.
ζ
(U;r)
()
K
= (
(U;r)|
K
) by deﬁnition.
Obvious 14 ζ
(U;r)
() is an increasing link space.
Obvious 15 (
(U;r)
) is s ymm etric for every connector space (U ; r).
Proposition 13 (
(U;r)
) is reﬂexive on U for every connector space (U; r).
Proof Follows from the fact that singletons are connected.
Theorem 9 (
(U;r)
) is an equivalence relation on U for every ext endable con-
nector space (U ; r ).
Proof We need to prove only transitivity. Let a
(U;r)
b and b
(U;r)
c. Then
exist X, Y CC(U; r) such tha t a, b X and b, c Y . Because X Y 6= we
have X Y CC(U ; r). So a
(U;r)
c.
Deﬁnition 22 A connected component (regarding a connectedness space (U; r))
is a n on-empty maximal connected s et.
Proposition 14 A set A PU is connected regarding a connector space (U; r)
iﬀ there are exactly one connected component of the connector space (U; r)|
A
.
15
Proof If A is connected regarding (U ; r) then A is connected regarding (U; r)|
A
and thus is a connected component regarding (U; r)|
A
.
If A is a connected component regarding (U; r)|
A
then A is c onnected re-
garding (U; r)|
A
and thus is connected regarding (U; r).
Theorem 10 Equivalence classes regarding
(U;r)
are exactly connected com-
ponents for every extendable connector space (U; r).
Proof Let K is a connected component. Then K is connected and thus
a
(U;r)
b for every a , b K. If a 6≡
(U;r)
b then there are no connected set X
such that a, b X and thus a / K b / K. Thus K is an e quivalence class of
(U;r)
.
Let now K is an equivalence class of
(U;r)
. Let choose arbitrary k K. For
every x K exists a connec ted set X
x
such that k, x X. Having a commo n
point k the union A of all X
x
is a connected set. It’s imp ossible A K because
otherwise y
(U;r)
k for some y 6∈ K. So A = K is the maximal connected set.
Corollary 5 For every extendable connector space (U ; r) its connectedness is
equal to connectedness regarding the link ζ
(U;r)
().
Proof A CC(U; r) A C C ((U ; r)|
A
) what is equivalent to A being a
connected component re garding (U; r)|
A
what is equivalent to A being a n equiv-
alence class regarding
(U;r)|
A
that is regarding ζ
(U;r)
()
A
that is equivalent to
A be ing connected regarding ζ
(U;r)
().
Corollary 6 The set U is partitioned into connected components for every ex-
tendable connector space (U; r).
Corollary 7 If a set is connected then it is a subset of a connected component
(for extendable connector spaces).
Theorem 11 For every extendable connector space exists a link space with the
same connectedness.
Proof Let (U; r) is an extendable connector space. Let A P U. Then
A is co nnec ted regarding (U; r) iﬀ there are one connected component of the
connector s pace (U; r)|
A
. Thus A is connected regarding (U; r) iﬀ A is connected
regarding τ where τ
A
is the equivalence relation deﬁned by the partition of the
set A into connected components by the c onnector space (U ; r)|
A
. (Taken in
account that connected components of an extendable connector space are a
partition.)
Theorem 12 Let (U ; τ) is an increasing equivalence link space. Then L(LC(U; τ)) =
τ.
16
Proof K is connected regarding (U; τ) iﬀ every two points of K are linked by
τ
K
.
a L(LC(U ; τ))
A
b K PA : (a, b K K LC(U; τ)) K PA :
(a, b K x, y K : x τ
K
y).
a L(LC(U ; τ))
A
b K PA : (a, b K a τ
K
b) K PA : a τ
K
b a τ
A
b.
Reversely, if a τ
A
b then a and b are in the same connected component K
and thus a L(LC(U; τ))
A
b.
Deﬁnition 23 For a connectedness space (U; r):
a
(U;r)
b X, Y PU : (a X b Y X Y = U X Y = X r Y ).
Obvious 16 a ζ
(U;r)
()
K
b a
(U;r)|
K
b X, Y PU : (a X b
Y X Y = K X Y = X r Y ) for every K PU .
Deﬁnition 24 is deﬁned by the formula
a
(U;r)
b a
(U;r)
b b
(U;r)
a.
Obvious 17 a
(U;r)
b X, Y PU : (a X b Y X Y = U X Y =
X r Y Y r X).
Obvious 18 a ζ
(U;r)
()
K
b a
(U;r)|
K
b X, Y P U : (a X b
Y X Y = K X Y = X r Y Y r X) for every K PU .
Remark 6 bears less information about the connector than . For example
for the connector T (U; τ) of a graph consisting of two connected components
|
T (U ;τ )
is just the diago nal relation.
Proposition 15 x
(U;r)
x and x
(U;r)
x for every x U.
Proof x
(U;r)
x follows fr om that a X b Y X Y = U X Y =
is always false if a = b. x
(U;r)
x follows fr om x
(U;r)
x.
Proposition 16
(U;r)
is transitive.
Proof Let a
(U;r)
b and b
(U;r)
c. Le t a X, c Z, X Z = U , X Z = .
We need to prove X r Z.
Obviously b X b Z. We can assume b X.
Then X r Z because b
(U;r)
c.
Theorem 13
(U;r)
is an equivalence relation.
Proof
17
Reﬂexivity Follows from reﬂexivity of
(U;r)
.
Symmetry Obvious.
Transitivity Let a
(U;r)
b and b
(U;r)
c. Then a
(U;r)
b and b
(U;r)
c.
So by transitivity of
(U;r)
we have a
(U;r)
c. Similarly c
(U;r)
a. So
a
(U;r)
c.
Theorem 14 The following statements are equivalent for every connector space
(U; r) and set K PU :
1. The set K is connected regarding (U; r).
2. x, y K : x
(U;r)|
K
y.
3. x, y K : x
(U;r)|
K
y.
4. x, y K : x
(U;r)|
K
y.
Proof
(1)(3) Let K is connected. Then we have X r Y and Y r X for every
X, Y P K \ {∅} such that X Y = K X Y = and consequently
a
(U;r)|
K
b for every a, b K.
(3)(2) Obvious.
(3)(1) Let x, y K : x
(U;r)|
K
y. Then if X, Y PK \ {∅} X Y =
K X Y = , we have some x X and y Y thus X r Y because
x
(U;r)
y. So K is connected.
(4)(1) If x, y K : x
(U;r)|
K
y then K is a subset of a connected com-
ponent regar ding (U ; r)|
K
. This component cannot be greater than K, so
K is connected r egarding (U; r)|
K
and consequently connected regarding
(U; r).
(1)(4) If K is connected reg arding (U; r) then K is connected regarding
(U; r)|
K
and thus K is a connected component regarding (U; r)|
K
so hav-
ing x, y K : x
(U;r)|
K
y.
Theorem 15 ζ
Q(U;τ )
() = ζ
T (U ;τ )
() = ζ
Q(U;τ )
() = ζ
Q(U;τ )
() = τ for
every equivalence link space (U ; τ).
18
Proof a ζ
Q(U;τ )
()
K
b iﬀ a and b are in the same connected component
regarding Q((U ; τ)|
K
).
Let’s prove that a ζ
Q(U;τ )
()
K
b = a ζ
Q(U;τ )
()
K
b.
We need to prove that a ζ
Q(U;τ )
()
K
b iﬀ a and b are in the same connected
component regarding Q((U ; τ)|
K
). (Then also a ζ
Q(U;τ )
()
K
b iﬀ a and b are
in the same connected component regar ding Q((U; τ)|
K
).) If a and b are in the
same connected component then x
(Q(U;τ ))|
K
y that is a ζ
Q(U;τ )
()
K
b. Let
now a ζ
Q(U;τ )
()
K
b. Suppose a X and b Y where X a nd Y are distinct
connected components regarding Q((U ; τ)|
K
). Then b U \X, X (U \X) = U
and X (U \ X) = . Thus X Q((U; τ)|
K
) (U \ X) that is for some x X and
y U \ X we have x τ
K
y what is impossible because x and y lie in diﬀerent
connected components.
ζ
Q(U;τ )
()
K
= ( |
(U;r)|
K
);
a |
Q(U;τ )|
K
b a L(CC(Q(U ; τ)))
K
b a L(CC(T (U; τ)))
K
b a
|
T (U ;τ )|
K
b (used the theorem 2).
L(CC(Q(U; τ)))
K
= ρ(PKCC(Q(U; τ))) = ρ(CC((Q(U; τ))|
K
)) = ρ(CC(Q((U; τ)|
K
))) =
ρ(LC((U; τ)|
K
)). So if a |
Q(U;τ )|
K
b then a and b lie in the same connected
component regarding (U; τ)|
K
. Thus a τ
K
b.
Let now a τ
K
b. Suppose that a and b lie in diﬀerent connected components
regarding (U; τ)|
K
. Then by equivalence every po ints of these components ar e
linked and thus they are one connected component. By contradiction a and b
lie in the same connected component regarding (U; τ)|
K
.
So we proved a |
Q(U;τ )|
K
b a τ
K
b.
5.4. Relationships of Q(U; τ) and T (U; τ)
Let ﬁnd a formula which allows to ﬁnd T (U; τ) knowing Q(U; τ) (to the
extent of equal normalization).
Let (U; r) is a connector space.
Deﬁnition 25 I will deﬁne the connector space β(U; r) = (U; r
) by the formula
(for every A, B P U)
A r
B A B CC(r).
Lemma 4 Let X, Y , A, B are sets. If X, Y, A, B 6= and X Y = A B then
X A 6= Y B 6= or X B 6= Y A 6= .
Proof If a X then a A or a B. Let for example a A. Thus X A 6= .
If Y B = then B X and Y A, so having X B 6= Y A 6= .
Theorem 16 N (T (U ; τ)) = N(β(Q(U; τ))) for every increasing equivalence
19
Proof Let A, B 6= and A B = . We need to prove that A T (U ; τ) B
A β(Q(U; τ)) B.
Let A β(Q(U; τ)) B. Then A B CC(Q(U; τ)) that is by the theorem 2
we have A B LC(U; τ). So x, y A B : x τ
AB
y that is A T (U; τ) B.
Let now A T (U; τ) B. Then a A, b B : τ (a, b, A B).
Let X Y = A B and X Y = and X, Y 6= .
By the lemma there e xist a X, b Y such that a A, b B (or a X,
b Y such that a B, b A what is analog ous). So τ(a, b, A B) and
consequently X Q (U ; τ) Y .
Thus A B CC(Q(U; τ)) that is A β(Q(U; τ)) B.
Proposition 17 N (β(U; r)) N(U; r) for every connector space (U ; r).
Proof Let A N(β(U ; r)) B for some A, B 6= , A B = , then A B
CC(U; r). Then A r B and thus A N (U; r) B.
Theorem 17 CC(β(U; r)) CC(U ; r) for every connector space (U; r).
Proof From the previous proposition.
Proposition 18 N (β(β(U; r))) = N (β(U ; r)) for every connector space (U ; r).
Proof If A N (β(β(U; r))) B then either A B 6= and thus A N(β(U ; r)) B
or A B = and A, B 6= and A β(β(U; r)) B. Then A B CC(β(U ; r))
and thus A β(U ; r ) B with consequence A N(β(U; r)) B.
Let now A N(β(U ; r)) B. Then either AB 6= and thus A N(β(β(U; r))) B
or A B = and A, B 6= and A β(U; r) B. So A B CC(U; r).
X, Y P(AB)\{∅} : (X Y = AB X Y = X Y CC(U; r));
X, Y P(A B) \ {∅} : (X Y = A B X Y = X β(U; r) Y ).
So AB CC(β(U ; r)) that is A β(β(U; r)) B and thus A N(β(β(U; r))) B.
Remark 7 CC(β(U; r)) = CC(U; r) if (U; r) = T (U; τ) or (U; r) = Q(U; τ) for
every equivalence link space (U ; τ).
Question 1 β(β(U ; r)) = β(U; r)?
Question 2 Under which conditions CC(β(U; r)) = CC(U; r) in general?
6. Future research
How connectedness is related with continuity?
Research the lattice of connector s and the lattice of links.
To deﬁne product o f two connectors is not trivial if possible at all.
We also may attempt to deﬁne quotient spa c es for connectors.
In my further research I am go ing to study generalized connectedness of
ﬁlters.
20
References
 Proximity spa c e. In Michiel Hazewinkel, editor, Encyclopaedia of Mathe-
matics. Spinger, 2002.
 M. Hazewinkel, editor. Soviet Encyclopaedia of Mathematics, volume 7. 