The Islamic University of Gaza Deanery of Higher Studies Faculty of Science Department of Mathematics

Topological Properties For Digital Spaces

By Khaled G. Hegazy

Supervised By:

Dr.Hisham B.Mahdi

Submitted in Partial Fulfillment of the Requirements for M.Sc. Degree.

2014-1435

My wife,

and My Friends.

Contents

Contents i

Acknowledgements iii

Abstract iv

Introduction 1

1 Preliminaries 6

1.1 Introduction to Topological Spaces . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Alexandroff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Cubical Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Digital Spaces on Z and Z2 23

2.1 Preliminaries and Main Concepts of Digital Spaces . . . . . . . . . . . . . 24

2.2 The Digital Topology on Z . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.3 Two Digital Topologies on Z2 . . . . . . . . . . . . . . . . . . . . . . . . . 37

3 Digital Topologies on Z3 59

3.1 Five Topologies on Z3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.2 The Marcus -Wyse topology on Z3 . . . . . . . . . . . . . . . . . . . . . . 71

i

3.3 β-Space on Z3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.4 αs-Space on Z3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4 Topologies with αβ-Saddle Points 94

4.1 Product of Khalimesky Spaces on Z3 . . . . . . . . . . . . . . . . . . . . . 94

4.2 αβ-space on Z3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4.3 Alexandroff Hopf topology on Z3 . . . . . . . . . . . . . . . . . . . . . . . 103

5 Digital Label Images 114

5.1 Introduction to Label Images . . . . . . . . . . . . . . . . . . . . . . . . . 114

5.2 Covering Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Bibliography 122

ii

Acknowledgements

After Allah, I am grateful to my parents for their continuous prayers and support.

Also I am grateful to my wife for here patience, share me my complicated, and support. I

would also like to thank all my brothers, my sisters, and my friends for encouraging me to

continue. They have been there for me all of my life, and I owe them my deepest thanks.

All my thanks go to my supervisor Dr. Hisham B. Mahdi for his guidance, suggestions,

contribution, and support during this thesis. I would like to express my deep thanks to

all the faculty members in the Mathematics Department who taught me, especially Prof.

Ayman H. Esaqa., Prof. Eisa D. Elhabil and Prof. Asad Y. Asad for their special effects on

me. Finally, I would like to thank my dear university for its support and encouragement.

iii

Abstract

In this thesis, we study the digital spaces Zd, d ∈ N, and their topological properties.

We show that there are only one topology on Z, two topologies on Z2, and five topologies

on Z3 whose (topological) connected sets are graphicly connected. We prove that these

topologies are a generalized locally finite T0-Alexandroff spaces. Up to this fact, some of

the basic concepts and results that are involved in [25], [26] and [27] are studied on these

topologies. We describe the orders induced on these topologies and the relations between

them. To take advantage of our results we applied them in the digital label images.

Some of the important results are finding Marcus -Wyse function and 2,3-Alexandroff-

Hopf functions. We prove that the Alexandroff-Hopf topology on Z2 (resp. on Z3) is the

product topology of two ( resp. three) Khalimsky topologies on Z (resp. on Z3). We

define the ith-summation topology S2 (resp. S3) and find the relationship between it and

the digital topologies on Z2 (resp. on Z3). We prove that the αs-space is the dual to the

β-space.

iv

Introduction

The digital topology plays a very important role in computer vision, image processing,

medical applications and many other applications. The digital d-space Zd is the d-tuples

(x1, x2, ..., xd) of the Euclidean d-space having integer coordinates. A point with integer

coordinates is called a digital point. The digital 2- or 3-space is the most commonly

used representation space in computer graphics and computer vision, where the points

are assigned some gray-level or color values. Based on these values and the adjacency

relations, the points in the digital space are grouped into connected components, which

are used in computer graphics to display and in computer vision to recognize (parts of)

real objects. Due to the applications, the adjacency relations, which play a primary role

in the grouping process, must be consistent with human spatial intuition about nearness

of points in the digital space. Digital topology was first studied in the late of 1960s by

the computer image analysis researcher Azriel Rosenfeld (1931-2004). His publications

on this subject played a major role in establishing and developing this field. Rosenfeld

published a paper with title ”Digital Topology”[4]. He published it in 1973 for the first

time. This paper was very influential since for the first time some very difficult problems

of digital topology were treated rigorously.

The thesis depend basically on the results of Duda, Hart and Munson [34]. In [35] Eck-

hardta and Latecki search for the topologies on Zd whose connected sets are connected

in the intuitive sense. For these purpose, they put two (resp. three) conditions 1d and

1

2d (resp. 1d, 2d and 3d) for d = 1,2(resp. d ≥ 3). They showed that there are only two

topologies on Z2 and five topologies on Z3 which satisfy these conditions. Two topologies

for Z2 and Z3 are well known, one of them is the Marcus-Wyse topology and the second

is the Alexandroff-Hopf topology.

The results and the ideas in [35] will be investigated in this thesis. We get the same

results by other ways. Fortunately, any topology satisfy the conditions 1d, 2d (3d) is an

T0-Alexandroff. More precisely it is a generalized locally finite T0-Alexandroff spaces.

Alexandroff spaces were first introduced by P. Alexandroff in 1937 in [1] under the name

of Diskrete Raume (discrete space). An Alexandroff topology is a topology which has

the additional property ”arbitrary intersection of open sets is open”. Any T0-Alexandroff

space related to a corresponding poset in one to one and onto way via (Alexandroff)

specialization order such that each one can completely determine by the other. Mahdi

and Elatrash in [26] use the in proofs to illustrate the results and the concepts. The

importance of their study comes from the fact that the topological properties can be

characterized just by using the corresponding poset.

In an attempt to take advantage from our results, we study the digital label images. We

setting a new way to covering images depending on the paper [23] which published in

2014. This thesis consists of five chapters. Chapter one contains three sections about

preliminaries that will be used in the remainder of this thesis. In section one, we talk

about the basic definitions and theorems of the topology. In the second section, we give

an introduction to the T0-Alexandroff spaces and their corresponding posets. We study

the (Alexandroff) specialization pre-order, the duality, the product of Alexandroff spaces

and the Khalimsky Line. The last section focus on cubical complexes and the regular

(binary) images.

Chapter two study the topologies on Z and Z2 which satisfy the two conditions 1d and 2d

2

and their topological properties. This chapter has been divided into three sections. The

first section study the concepts of Graph-Theory. We define the direct and the indirect

neighborhoods. Also we prove that if Z2 is a digital space satisfies the two conditions

12 and 22, then Z2 is strongly irresolvable, an α-T 12

−space, and a semi-T 12

−space. In

general digital space, Zd contains an open, dense and hereditarily irresolvable subspace.

For each dense subset D of Zd, Do is dense. The induced α-space (Zd, τα) is submaximal.

Moreover we give a general characteristic definitions of the maximal, the minimal and

the saddle points with respect to any digital space. In the second section, we study the

digital space Z. We show that the only digital topology on Z which satisfies 11 and 21

is the Khalimsky topology. We prove that the Khalimsky space Z is submaximal, nodec,

T∗1 -space, T 3

4

and a semi-T2−space. In last section, we study the digital topology on Z2.

We prove that there exist two digital topology on Z2 that satisfies 12 and 22. The first

one is the Marcus- Wyse (Z2, τm), and the other is the Alexandroff-Hopf (Z2, τs). The

minimal neighborhood of the minimal point x in (Z2, τm), V (x) = N4(x) ∪ x. In the

Alexandroff-Hopf space, if x ∈ m, then V (x) = N8(x)∪x, and if x ∈ SD, then V (x) =

(M ∩ N4(x)) ∪ x. We prove that the Alexandroff-Hopf topology on Z2 is the product

topology of two Khalimsky topologies on Z. (x, y) ≤(2)

A(a, b) iff x ≤

Kha and y ≤

Khb.

Next, we define the ith-summation topology S2 and find the relationship between it and

the two topologies on Z2. Finally, we produce the two function Marcus-Wyse function

and 2-Alexandroff-Hopf function. We hope that this two functions play important rule in

the applications of the digital topology.

Chapter three is talking about the digital topologies on Z3. In the first section, we show

that there are only five topologies on Z3 that satisfy the three conditions 13, 23 and 33.

The first one is the Marcus-Wyse topology on Z3 τm where its points is either maximal or

minimal. We study this topology in section two, and prove that the minimal neighborhood

3

of any minimal point x is equal to N6(x)∪x. We prove that the Marcus-Wyse space Z3

is submaximal, nodec, T∗1 -space, T 3

4

and a semi-T2−space. Section three study the topol-

ogy contains β-saddle points with no α-saddle and call it β-topology on Z3. In this space,

the points are either maximal, minimal or β-saddle points. We split the minimal points

into two disjoint subsets m1 and m2. The first subset contains the minimal points that

covered by β-saddle points and the second subset contains the minimal points that cov-

ered by maximal points. We show that the minimal neighborhood of x whenever x in m1(

resp. m2, β-S) is equal to x∪ (N26(x) \N8(x)) (resp. x∪N6(x), x∪ (N6(x) \m1)).

In the fourth section, we study the topology with α-saddle points and no β-saddle points.

We call it αs-topology on Z3. The points in this topology are either maximal, minimal

or α-saddle points. We prove that the αs-space is the dual to the β-space. The minimal

neighborhood of the minimal point x is equal to x ∪ (N6(x) ∩M1), and the minimal

neighborhood of α-saddle point y is equal to y ∪ (N26(y) \ (m ∪ N8(y))), where M1 is

the dual of m1.

Chapter four consists of three sections. The first one is talking about the product topol-

ogy of three Khalimsky spaces. Our object from this chapter is knowing if there is one

topology of the five digital topologies on Z3 homeomorphic to the product of three dig-

ital (Khalimsky) spaces on Z. We show that the product space contains both α-saddle

and β-saddle points. So the three topologies which were studied in previous chapter are

surely not the product topology. Hence the product topology maybe one of the remain-

der topologies. In the second section, we study the fourth topology on Z3 and call it

αβ-topology. The minimal neighborhood of the saddle point x (resp. minimal point y) is

x ∪ (N6(x) \m) (resp. y ∪ (N26(y) \ (N8(y) ∪m))). We show that this topology is

not the product topology. The final section study the Alexandroff-Hopf topology, which

is the last digital topology on Z3. We show that the minimal neighborhood of x whenever

4

x ∈ α-S (resp. β-S, m) is x ∪ (N6(x) \ β-S) (resp. x ∪ (N6(x) \m) ∪ (N12(x) \ β-S)),

x ∪N26(x)). We prove that this topology is the product of three digital topology on Z.

Digital label images are studied in the final chapter. In this chapter, we use our previous

results in some applications of the digital topologies. We will focus on the Alexandroff-

Hopf topology on Z2. Specially we study the digital label images which was defined in [23].

The digital label images are the images whose domain is Zd and whose codomains are sets

on which there generally exists no meaningful order relation. But we will translate this

definition to correspond with our results. The object of this translation is to replace the

cellular-complex topology by digital space (in this chapter the Alexandroff-Hopf topology

on Z2). Importance of this issue, that this process used in the simplicity process and for

encoding the images.

Finally, we give a conclusion page contain a summary of our work and what we are

looking for to make in the future to improve this work and develop it.

5

Chapter 1

Preliminaries

1.1 Introduction to Topological Spaces

A topology on a nonempty set X is a collection τ of subset of X called the open sets

satisfying :

(1) ϕ and X belong to τ ,

(2) any union of elements of τ belongs to τ , and

(3) any finite intersection of elements of τ belongs to τ .

We say that (X, τ) is a topological space, or simply when no confusion X is a topological

space. If X is a topological space and E ⊆ X, we say that E is closed iff Ecis open.

The set that is both open and closed is called clopen set. If E ⊆ X, the closure of E

in X is the set E = cl(E)=∩K⊆ X: K is closed and E ⊆K. The interior of E in X

is the set E=Int(E)=

∪U⊆ X: U is open and U ⊆ E. The frontier of E is the set

FrX(E) = E ∩ Ec = E − E

. If X is a topological space and x ∈ X, a neighborhood

(abbreviated nhood) of x is a set U which contains an open set V containing x. Thus

6

evidently; U is a nhood of x iff x ∈ U. The collection Ux of all nhoods of x is called the

nhood system at x. A nhood base at x in X is a subcollection Bx taken from the nhood

system Ux, having the property that each U ∈ Ux contains some V ∈ Bx. That is Ux is

completely determined by Bx as follows: Ux= U ⊆ X : V ⊆ U for some V ∈ Bx. Thus

in any topological space the collection of all open nhoods of x form a nhood base at x.

If (X, τ) is a topological space, a base for τ (or a base for X) is a collection B ⊆ τ such

that τ = ∪

B∈LB : L ⊆ B. For a subset B ⊆ X, the collection τ1 = G ∩B : G ∈ τ is

a topology on B, called the relative topology for B. The set S ⊆ X is called connected, if

there is no decomposition S = S1 ∪ S2 such that S1 ∩ S2 = ∅, and S1, S2 are nonempty

and relatively open in S. It is easy to see that a space is connected if and only if the only

sets that are clopen are the empty set and the whole space X. A point x in X is a limit

point of S if every neighbourhood of x contains at least one point of S different from x

itself. A subset A of a topological space X is called dense if A = X. In the case, when

A = X every point x in X either belongs to A or is a limit point of A. If X and Y are

topological spaces, a function f from X to Y is called continuous function if f−1(V ) is

open in X whenever V is open in Y . f is called an open function if f(U) is open in Y

whenever U is open set in X. A function f from X to Y is called a homeomorphism iff

f is one to one, onto, continuous, and f−1 is also continuous. In this case, we say that

X and Y are homeomorphic. We can easy prove that a function f from X to Y is a

homeomorphism function iff f is one to one, onto, continuous and open function.

Definition 1.1.1. A subset A of a topology space (X, τ) is called :

(1) a semi-open [22] if A ⊆ A , and semi-closed set [7] if Ac is semi-open. Thus A is

semi-closed if and only if A

⊆ A.

(2) a preopen [30] if A ⊆ A

, and preclosed set [14] if Acis preopen. Thus A is preclosed

if and only if A ⊆ A.

7

(3) an α-open [32] if A ⊆ A

, and α-closed [10] if Acis α-open. Thus A is α-closed if

and only if A⊆ A.

The family of all semi-open (resp. preopen, α-open) sets is denoted by SO(X) (resp.

PO(X), τα). In [32], it was proved that τα is a topology on X. In general, SO(X) and

PO(X) need not be topologies on X.

Definitions 1.1.2. A subset A of a space (X, τ) is called :

(1) a generalized closed ( briefly, g-closed ) [21] if A ⊆ U whenever A ⊆ U and U is

open.

(2) a semi-generalized closed ( briefly, sg-closed ) [6] if sCl(A) ⊆ U whenever A ⊆ U

and U is semi-open.

(3) a nowhere dense if A

= ϕ.

Definitions 1.1.3. A topological space (X, τ) is called:

(1) a To-space if whenever x and y are distinct points in X, there is an open set which

contains one of x, y and not the other.

(2) a semi-To-space [28] if whenever x and y are distinct points inX, there is a semi-open

set contains one of x, y and not the other.

(3) an α-To-space [29] if whenever x and y are distinct points in X, there is an α-open

set contains one of x, y and not the other.

(4) a pre-To-space if whenever x and y are distinct points in X, there is an pre-open set

contains one of x, y and not the other.

(5) a T 14

-space [2] if for every finite subset F of X and every y /∈ F , there exists a set

Ay containing F and disjoint from y such that Ay is either open or closed.

8

(6) a T 13

-space [3] if for every compact subset F of X and every y /∈ F , there exists a

set Ay containing F and disjoint from y such that Ay is either open or closed.

(7) a T 12

-space [21] if every g-closed set is closed. Equivalently, if every singleton is

either open or closed [12].

(8) a T 34

-space [11] if every singleton is either regular open or closed.

(9) a T1-space if whenever x and y are distinct points in X, there are disjoint open sets

U and V such that x ∈ U and y ∈ V .

(10) a semi-T 12

space [6] if every sg-closed set is semi-closed. Equivalently, if every

singleton is either semi-open or semi-closed [8].

(11) an α-T 12

space if every singleton is either α-open or α-closed.

(12) a semi-T1-space [28] if whenever x and y are distinct points in X , there is a semi-open

set of each not containing the other. Equivalently, if every singleton is semi-closed.

(13) an α-T1-space [29] if whenever x and y are distinct points in X , there is a α-open

set of each not containing the other. Equivalently, if every singleton is α-closed.

(14) a T∗1 -space [17] if every nowhere dense subset of X is union of closed sets.

(15) a feebly T1-space [20] if every singleton is nowhere dense or clopen. Equivalently, if

every preopen singleton x is closed and hence clopen [11].

(16) a semi-TD-space [20] if every singleton is either open or nowhere dense.

(17) a resolvable space [13] if X = D ∪ Dcwhere both D and D

care dense, and an

irresolvable space if it is not resolvable. A subset A of X is resolvable(irresolvable)

if A is resolvable(irresolvable) as a subspace.

9

(18) a hereditarily irresolvable space [13] if no nonempty subset is resolvable, and strongly

irresolvable space [19] if no nonempty open set is resolvable.

(19) a submaximal space [18] if every dense subset is open, equivalent condition every

preopen subset is open.

(20) a nodec space [36] if all nowhere dense sets are closed.

Definition 1.1.4. A topological space (X, τ) is called locally finite if each element x of

X is contained in a finite open set and a finite closed set.

Definition 1.1.5. If X is a topological space, Y is a set and g: X −→ Y is an onto

mapping then the collection τg of subsets of Y defined by τg = G ⊆ Y : g−1(G) is open

in X is a topology in Y , called the quotient topology induced on Y by g.

The quotient topology induced on Y by g is the largest topology on Y making g

continuous.

Definition 1.1.6. Let X be a topological space. A decomposition D of X is a collection

of disjoint subsets of X whose union is X.

Definition 1.1.7. Let X be a topological space, and D a decomposition of X. Define a

map P of X onto D by letting P (x), for x ∈ X, be the element of D containing x, then

P is called the natural map(or decomposition map) of X onto D.

Definition 1.1.8. Let D be a decomposition of X. We define the decomposition topology

on D to be the quotient topology on D induced by the natural map P :X −→ D.

Partially Ordered Set (poset)

A partially ordered set (poset) [5] is a set P with a binary relation that satisfies the

following axioms ∀ x, y, z in P , we have that:

10

(i) x ≤ x (reflexivity).

(ii) x ≤ y and y ≤ x imply x = y (anti-symmetry).

(iii) x ≤ y and y ≤ z imply x ≤ z (transitivity).

A relation which is satisfying reflexive and transitive conditions is called a preorder. A

preorder is an order if it is in addition anti-symmetric.

A poset P is a lattice if ∀x, y ∈ P , x∨y =supx; y and x∧y = infx; y exist. An element

x ∈ P is a maximal element if whenever x ≤ z then x = z, and an element y is a minimal

element if whenever z ≤ y then y = z. The set of all maximal (resp. minimal) elements

of P is denoted by M (resp m). If A is a subset of P , then the order of P induces an

order in A in natural way, x ≤ y in A if x ≤ y in X. In this case, we define M(A) (resp.

m(A)) to be the set of all maximal (resp. minimal) elements of A under the induced

order. Two elements x, y in P are comparable if either x ≤ y or y ≤ x, otherwise they

are incomparable. A subset C ⊆ P is a chain if any two elements of C are comparable.

We say x is covered by y or y covers x and we write x y or y x if x < y and when

x ≤ z < y then z = x, where (z < y means z ≤ y and z = y ). A set O ⊆ P is a down set

if whenever x ∈ O and y ≤ x, we have y ∈ O. A set U ⊆ P is an up set if whenever x ∈ U

and x ≤ y we have y ∈ U . For x ∈ P we define the down set ↓ x := y ∈ P : y ≤ x

and the up set ↑ x := y ∈ P : y ≥ x. For a set A ⊆ P , we define the down set ↓ A

:= y ∈ P : (∃x ∈ A)y ≤ x, and the up set ↑ A := y ∈ P : (∃x ∈ A)y ≥ x. For a

point x ∈ X, we define x = ↑ x ∩M . A poset P satisfies the ascending chain conditions,

(briefly,ACC) if for any increasing sequence x1 ≤ x2 ≤ · · · xn ≤ · · · in P there exists

k ∈ N such that xk = xk+1 = · · · . A poset P satisfies the descending chain conditions,

(briefly,DCC) if for any sequence x1 ≥ x2 ≥ · · · ≥ xn ≥ · · · . there exists k ∈ N such that

xk = xk+1 = · · · . We say that P is finite chain condition (briefly, FCC), if it satisfies

both ACC and DCC. It should be noted that, in the case when P satisfies ACC (resp.

11

DCC), the set M ( resp. m) is nonempty set. There is a very useful way to depict posets

using the so called Hasse diagrams. To indicate p ≤ q, we picture p somewhere below

q and draw a line connecting p with q. To make pictures easy to draw and understand;

if p ≤ q and q ≤ r, we only draw connecting p and q, and q and r, and do not draw a

connecting line between p and r. Of course, p ≤ r by transitivity, but connecting p and r

by a line would make the diagram messy, so we avoid it. For the same reason, we do not

draw a loop connecting p with itself.

1.2 Alexandroff Spaces

In general, a topological space (X, τ) satisfies the property that a finite intersection of open

sets is open and an arbitrary intersection of open sets need not be open. An Alexandroff

space (or minimal neighborhood space) X is a space in which the arbitrary intersection

of open sets is open. Equivalently, each point of X has a minimal neighborhood base.

Examples of Alexandroff spaces are discrete spaces and finite topological spaces. In fact,

each finite topological space is locally finite, and each locally finite space is Alexandroff.

This subject was first studied in 1937 by P. Alexandroff [1] under the name of Diskrete

Raaume (discrete space). The name is not valid now, since discrete spaces are spaces

where the singletons are open.

Recall that if B is a subset of a topological space X, then the closure of B is the intersection

of all closed sets containing B, and the closure is usually denoted by B, and we write

ClX(B) for the closure of B in X. This notation allows us to specify in what space we

consider the closure, and in Alexandroff spaces ClX(B) is also a notation dual to V (B)

where VX(B) is the intersection of all open sets containing B. In general VX(B) is not an

open set, but in the Alexandroff space it is. In fact VX(B) is the smallest neighborhood

containing B. If there is no danger of ambiguity, we will just write V (B) and CL(B)

12

instead of VX(B) and CLX(B) respectively. we define V (x) = V (x) and CL(x) =

CL(x). It is easy to see that V (x) is the smallest neighborhood of x in the Alexandroff

spaces. Conversely, if every point of the space has a smallest neighborhood, then an

arbitrary intersection of open sets is open. Hence this existence could have been used as

an alternative definition of an Alexandroff space. A point x is called open point or an

isolated point if V (x) = x, that is if x is open, and is called closed point if CL(x) =

x. If x is either open or closed it is called pure, otherwise it is called mixed.

Lemma 1.2.1. Let X be an Alexandroff space and x, y ∈ X. x ∈ y if and only if

y ∈ V (x).

(Alexandroff) Specialization Pre -Prder

For each To-Alexandroff space (X, τ), there is a corresponding poset (X,≤τ ) in one to one

and onto way, where each one of them is completely determined by the other. Given a

poset (P,≤), then B = ↑ x : x ∈ X is a base for a topology on P . This topology denoted

by τ≤ is a T0-Alexandroff topology. On the other hand, if (X, τ) is Alexandroff space, we

define a pre-order ≤τ which is called (Alexandroff) specialization pre-order, by a ≤τ b if

and only if a ∈ b, or equivalently b ∈ V (a). It is proved that this pre-order is symmetric

and hence a partial order if and only if X is To. Moreover, if (X,≤) is a poset and τ(≤) its

induced To-Alexandroff topology, then the specialization order of τ(≤) is the original order

≤ itself; that is ≤τ(≤) = ≤. If (X, τ) is a To-Alexandroff space with specialization order

≤τ , then the induced topology by the specialization order is the topology τ itself; that is

τ(≤τ ) = τ . A To-Alexandroff space together with its corresponding order ≤ denoted as

(X, τ(≤)). A set U is an open set in the Alexandroff Spaces iff it is an up set (↑ U = U),

and a set W is a closed in the Alexandroff Spaces iff it is a down set (↓ W = W ).

13

Definition 1.2.2. [15] Two distinct points x and y inX are called adjacent if the subspace

x, y is connected.

Lemma 1.2.3. If X is an Alexandroff space and x, y ∈ X, then x and y are adjacent if

and only if y ∈ V (x) or x ∈ V (y) if and only if y ∈ V (x) ∪ x.

Proof. Let X be an Alexandroff space and let S = x, y. So the relative topology

τS⊆ ϕ, x, y, S. Now x and y are not adjacent iff the subspace x, y is unconnected

iff ∃ nonempty open sets in U and W in S such that U ∩W = ϕ and U ∪W = S iff ”(U

= x and W = y) or (W = x and U = y)” ( WLOG of prove assume the first) iff

x = U = S ∩U∗ and y = W = S ∩W ∗, for some U∗,W ∗ ∈ τXiff x ∈ U∗, y /∈ U∗, and

y ∈ W ∗, x /∈ W ∗ iff y /∈ V (x) and x /∈ V (y), hence x, y ∈ X, x and y are adjacent if and

only if y ∈ V (x) or x ∈ V (y). And y ∈ V (x) or x ∈ V (y) if and only if y ∈ V (x)∪CL(x)

is coming directly from x ∈ y if and only if y ∈ V (x).

So the subspace x, y is disconnected if and only if x is not in V (y) and y is not in V (x).

Separation axioms in the Alexandroff spaces

The T0 axiom, states that given two distinct points x and y, there is an open set containing

one of them but not the other. In an Alexandroff spaces, we have an equivalent formulation

that V (x) = V (y) implies x = y for every x and y. The T 12axiom states that all points

are pure. Clearly any T 12space is also T0. T1 axiom states that points are closed. In an

Alexandroff space this implies that every set is closed and hence that every set is open.

Therefore, a Alexandroff space satisfying the T1 axiom must be discrete space.

Definition 1.2.4. A T0-Alexandroff space which satisfies the ascending chain condition

(ACC) is called Artinian T0-Alexandroff spaces and a T0-Alexandroff space which satisfies

14

the descending chain condition (DCC) is called Noetherian T0-Alexandroff spaces. A T0-

Alexandroff space which satisfies both (ACC) and (DCC) is called a generalized locally

finite T0-Alexandroff spaces.

Theorem 1.2.5. Let X and Y are a T0- Alexandroff spaces, if there a bijective function f :

X −→ Y such that for all x ∈ X f(VX(x)) = VY (f(x)), then X and Y are homeomorphic.

Proof. Let f : X −→ Y be a bijective function, such that for all x ∈ X f(VX(x)) =

VY (f(x)), then f(↑Xx) = ↑

Yf(x) ∀x ∈ X. To prove that f is open and continuous

function let U is a nonempty open set in X and let f(x) ∈ f(U), then x ∈ U . Since U

is open set in X, then ↑Xx ⊆ U . Hence f(↑

Xx) = ↑

Yf(x) ⊆ f(U). Therefore f(U) is

open in Y and f is an open function. Now let U is a nonempty open set in Y and let

x ∈ f−1(U). Then f(x) ∈ U . But since U is open in Y , then f(↑

Xx) = ↑

Yf(x) ⊆ f(U).

Hence ↑Xx ⊆ U . So f is a continuous function. ThereforX and Y are homeomorphic.

Theorem 1.2.6. [26] Let (X, τ(≤)) be a T0-Alexandroff space, then all the following are

equivalent:

(1) X is submaximal.

(2) Each element of X is either maximal or minimal; i.e. each element of X is pure.

(3) X is nodec.

Theorem 1.2.7. [26] Let (X, τ(≤)) be a T0-Alexandroff space, then

(i) X contains an open, dense and hereditarily irresolvable subspace.

(ii) For A ⊆ X where Ao =ϕ, A is nowhere dense.

(iii) X is strongly irresolvable.

(iv) For each dense subset D of X, Do is dense.

15

(v) (X, τα) is submaximal.

Theorem 1.2.8. [25] Let (X, τ(≤)) be a T0-Alexandroff space, then all the following are

equivalent:

(i) X is T 12

-space.

(ii) X is T 13

-space.

(iii) X is T 14

-space.

(iv) Each element of X is pure.

(v) X is submaximal.

(vi) PO(X)=τα = τ(≤); that is, every preopen set is open.

(vii) X is T∗1 -space.

Theorem 1.2.9. [25] Let (X, τ(≤)) be a T0-Alexandroff space. Then X is T 34

if and only

if the following two conditions are satisfied

(a) X is T 12

-space.

(b) ∀x /∈M , | x |≥ 2

Theorem 1.2.10. [25] Let (X, τ(≤)) be a Artinian T0-Alexandroff space. Then

(a) X is an α-T 12

−space.

(b) X is a semi-T 12

−space.

Theorem 1.2.11. [25] Let (X, τ(≤)) be a Artinian T0-Alexandroff space, then all the

following are equivalent:

(1) X is a semi-T2−space.

16

(2) X is a semi-T1−space.

(3) ∀x /∈M , | x |≥ 2.

Duality

If (X, τ) is an Alexandroff space with collection F of closed sets, then F is itself a topology

on X called the dual topology of τ on X and usually denoted by τ d. Moreover, if (X, τ)

is an Alexandroff space, then (X, τ d) is also an Alexandroff space. If (X, τ(≤)) is a T0-

Alexandroff space, then it’s Alexandroff dual is also T0-space. Suppose that ≤dis the

induced specialization order of the dual Alexandroff space τ d, then x ≤dy iff y ≤ x. This

implies that the nhood V (x) of x in (X, τ(≤)) is the same of the set of closure set of x

in (X, τ d(≤d)), and vice versa. That is V (x) = cl

d(x) and V

d(x) = cl(x). and we get the

following corollary :

Lemma 1.2.12. Let (X, τ(≤)) be a T0-Alexandroff spaces, then the dual space (X, τ d(≤d))

is a T0-Alexandroff spaces such that ∀x ∈ X, ↑ x (= the up of x in (X,≤)) equals ↓dx

(= the down of x in (X,≤d)).

Proof. (⇒) for all x, y ∈ X; y ∈↑ x if and only if x ∈↓ y if and only if ↑dx ⊆↑

dy if and only

if y ≤dx if and only if y ∈↓

dx, hence ↑ x= ↓

dx for all x ∈ X

(⇐) First let we prove that if ↑ x = ↓dx, then ↓ x = ↑

dx. y ∈↓ x if and only if x ∈↑ y

= ↓dy if and only if y ∈↑

dx, so ↓ x = ↑

dx whenever ↑ x = ↓

dx. Now assume that U is

open set in (X, τ(≤)), then ↑ x ⊆ U ∀x ∈ U , but ↓dx = ↑ x ⊆ U ∀x ∈ U . Hence U is a

closed set in (X, τ(≤d)), and if U is closed set in (X, τ(≤)), then ↓ x ⊆ U ∀x ∈ U . But

↓ x = ↑dx ⊆ U ∀x ∈ U . Therefore U is open set in (X, τ(≤

d)) space. Hence (X, τ(≤))

and (X, τ(≤d)) are dual spaces.

17

The Product of Alexandroff Spaces

Theorem 1.2.13. [16] Let X and Y be Alexandroff spaces, and for x ∈ X and y ∈ Y let

VX(x) and V

Y(y) be minimal open neighborhoods of x and y respectively. Then X × Y is

an also an Alexandroff space with minimal open neighborhood VP(x, y) = V

X(x) × V

Y(y).

Let (Pi,≤i), i = 1, 2, · · · , n be posets, then many types of partial orders can be formed

on the set of cartesian product∏n

iPi = P1 × P2 · · · × Pn. The most famous order is the

Coordinatewise order ≤c. For two elements a = (a1 , a2 , · · · , an) and b = (b1 , b2 , · · · , bn) in∏n

iPi, we have that a ≤c b⇐⇒ a

i≤

ibi∀ i = 1, 2, · · · , n. If X and Y are two topological

spaces, then the product spaces X × Y is T0 if and only if both X and Y are T0 . Hence

by previous theorem if X and Y are two T0-Alexandroff spaces, then the product spaces

X×Y is T0-Alexandroff space, so it has a specialization order ≤p induced on the cartesian

product X × Y . The following theorem gives us the characteristic of this order.

Theorem 1.2.14. [27] Let (X, τ1(≤1)) and (Y, τ2(≤2)) be two T0-Alexandroff spaces with

corresponding posets (X,≤1), (Y,≤2) respectively. Then the specialization order ≤p of

the T0-Alexandroff space X × Y coincides with the Coordinatewise order induced by the

product of the corresponding posets.

The Khalimsky Line

Recall that a space is connected if the only sets which are clopen are the empty set and the

whole space. If X and Y are topological spaces and f : X → Y is a continuous mapping,

then the image f (X) is connected if X is connected, since if B is clopen in f(X), then

f−1(B) is clopen in X, and hence is either X or the empty set, and then B is either f(X)

or the empty set; which means that f(X) is connected.

Proposition 1.2.15. [31] Let f:X → Y be a surjective mapping from a connected topo-

18

logical space onto a set Y. Suppose Y is equipped with a topology such that f is continuous,

then Y is connected.

We will primarily use the above proposition to define connected topologies on Z. Let

X = R and Y = Z, and let f be a surjective mapping from R to Z. If we equip Z with the

strongest topology such that f is continuous, then Z is connected. Indeed there are many

surjective mappings R → Z. It is natural to think of Z as an approximation of the real

line, and so to consider mappings expressing this idea. If f is an increasing surjection,

then f−1(n) is an interval for every integer n.

If m is an odd integer, let Pm = (m−1,m+1), and if m is an even integer, let Pm = m.

The family Pmm∈Z forms a partition of the Euclidean line R and thus we may consider

the quotient space. Moreover if we identify each Pm with the integer it contains (i.e

f−1(m) = Pm). The inverse image of an even number n is the closed (with respect to the

Euclidean topology), so that n is closed. Whereas the inverse image of an odd number

m is open, so that m is open. This topology was introduced by Efim Khalimsky and

is called the Khalimsky topology. Z with this topology is called the Khalimsky line. It is

immediate that the Khalimsky line is connected since R is connected, the Khalimsky line

is a connected space. The Khalimsky topology is an Alexandroff space [31]. In terms of

smallest neighborhoods, we have VZ(m) = m if m is odd and VZ(n) = n− 1, n, n+ 1

if n is even.

1.3 Cubical Complexes

Let F1

0= a : a ∈ Z and F1

1= a, a+ 1 : a ∈ Z. Let f ⊆ Zd If f is the Cartesian

product of m elements of F1

1and d−m elements of F1

0, we say that f is a face or an m-face

(of Zd), m is the dimension of f , and we write dim(f) = m (c.f. [24]). The (d-D) space of

cubical complexes Fdis the set composed of all faces of Zd

. We denote by Fd

k(0 ≤ k ≤ d)

19

the set composed of all k-faces of Zd. Clearly that Fd

k⊆ Fd

. The couple (Fd,⊆) is a poset.

Thus there is a corresponding To-Alexandroff space (Fd, τ⊆). Indeed this topology is the

cellular-complex topology for the digital plane introduced by Alexandroff and Hopf [33].

Let F ⊆ Fdbe a set of faces, and let f ∈ F be a face. Then the face f is a facet of F if

f is a maximal in F . Actually, if x = (x1 , x2 , · · · , xd) ∈ Zd

, the set x =∏d

i=1x

i, x

i+1 is

a facet of Fdand x is called the leader of x and we write L(x) = x. Some faces of Z2 are

depicted in Figure 1.1

Figure 1.1: Two represntations of a set of faces F = f, g, h

Example 1.3.1. Let A = (0, 0), (0, 1), (1, 0), (1, 1) ⊆ Z2, and let the set of all faces of A

is denoted by F2

A. Then F2

A= (0, 0), (0, 1), (1, 0), (1, 1), (0, 0), (0, 1), (0, 0), (1, 0),

(0, 1), (1, 1), (1, 0), (1, 1), (0, 0), (0, 1), (1, 0), (1, 1). Let f1 = (0, 0), f2 = (0, 1),

f3 = (1, 0), f4 = (1, 1), h1 = (0, 0), (1, 0), h2 = (0, 0), (0, 1), h3 = (0, 1), (1, 1),

h4 = (1, 0), (1, 1), and g = (0, 0), (0, 1), (1, 0), (1, 1). Hence dim(g) = 2, and for

any i ∈ 1, 2, 3, 4, dim(hi) = 1, dim(f

i) = 0. For instance g = 0, 1 × 0, 1, h2 =

0 × 0, 1 and f4 = 1 × 1. Moreover, the face g is a facet of F2

A(and also of F2

).

Also we can consider the relative topology (F2

A, τ⊆) by the following Hasse diagram (see

Figure 1.2 )

Let f, h, g ∈ F2, such that dim(g) = 2, dim(h) = 1 and dim(f) = 0. It is clearly

that g is a maximal face while f is a minimal face. On F2the minimal neighborhood

of the 2-face g contains only the face g itself, the minimal neighborhood for the 1-face

h is contains the face h itself and two 2-faces, and the minimal neighborhood for the

20

Figure 1.2: Up of faces on cellular-complex topology F2

A

0-face f contains f itself and four 2-faces together with four 1-faces. Moreover, on F3

the minimal neighborhood for the 3-face is containing the face itself. Consider Figure

1.3. Also if f, h, g ∈ F3such that dim(g) = 2, dim(h) = 1 and dim(f) = 0, then the

minimal neighborhood for 2-face g is containing three faces; the face g and two 3-faces.

The minimal neighborhood for 1-face h is containing h together with four 2-faces and

four 3-faces, and the minimal neighborhood for 0-face f is containing f together with six

1-faces, twelve 2-faces and eight 3-faces.

Figure 1.3: Part of cellular-complex space F2

Definition 1.3.2. [24] When two faces g, h ∈ Fd covers a face f ∈ Fd (with respect to

cellular-complex topology) and ↑ g∩ ↑ h = ϕ, we say that their are opposite with respect

21

to the face f . We denoted opp(f) the set of all g, h for g opposite to h with respect to

f .

Definition 1.3.3. [24] For a sequence ϵ of d elements in −1, 1, A function µ(f) :

Fd \ Fdd−→ 0, 1 is an ε-regular (binary) image if foll all f ∈ Fd

mwith m < d we have,

µ(f) =

∧

a,b∈opp(x)µ(a) ∨ µ(b), if ε(m+ 1) = 1;∨

a,b∈opp(x)µ(a) ∧ µ(b), if ε(m+ 1) = -1.

22

Chapter 2

Digital Spaces on Z and Z2

Eckhardt and Latecki in [35] put two conditions called them 1dand 2

d, such that any

topology on Zd satisfies these conditions will be topologically connected whenever it is

graphicly connected. They said that if d ≥ 3, then there exist infinitely topologies on Zd

satisfy the conditions 1dand 2

d. So they put another condition 3

d. These three conditions

satisfy all the definitions of adjacency relations that are most commonly used in the digital

spaces. Up to these conditions, they proved that there are only two topologies on Z2 and

five topologies on Z3. Using these same conditions with different technique, we get the

same result. Moreover, we will prove that any topology on Zd satisfy these conditions

is a g-locally finite T0-Alexandroff space. We study the topological properties for these

topologies and find the minimal neighborhood for any point in the digital space Zd, for d

= 1,2,3. In this chapter, we will focus on the topologies on Z and Z2 which satisfies the

conditions 1dand 2

d.

23

2.1 Preliminaries and Main Concepts of Digital Spaces

The digital d-space Zd is the d-tuples (x1, x2, · · · , xd) of the Euclidean d-space having

integer coordinates. A point with integer coordinates is called a digital point. The digital

2- or 3-space is the most commonly used representation space in computer graphics and

computer vision, where the points are assigned some gray-level or color values. Based on

these values and the adjacency relations, the points in the digital space are grouped into

connected components, which are used in computer graphics to display and in computer

vision to recognize (parts of) real objects. Due to the applications, the adjacency relations,

which play a primary role in the grouping process, must be consistent with human spatial

intuition about nearness of points in the digital space. The following two conditions reflect

this intuition with respect to nearness of two points in the digital plane having the same

color:

(1d) If a set in Zd is 2d-connected (defined below), then it is (topologically) connected.

(2d) If a set in Zd is not (3d−1)−connected (defined below), then it is not (topologically)

connected.

These conditions are satisfied by all definitions of adjacency relations that are most com-

monly used in computer graphics and computer vision. In this thesis, we will search about

all topologies in Z, Z2 and Z3 which satisfying the two conditions 1dand 2

dand study

their topological properties. We search about topologies that are topologically connected

whenever they are graphicly connected.

Graph-Theoretic Concepts

Definition 2.1.1. Let a = (a1 , a2 , · · · , ad), b = (b1 , b2 , · · · , bd) ∈ Zd. The length of the i’th

coordinate between a and b is ri(a, b) = |a

i− b

i|, and the total coordinates length between

24

a and b is R(a, b) =∑d

i=1 |ai − bi|.

Definition 2.1.2. For a given point x ∈ Zd, we define:

(1) the (3d − 1)-neighborhood N3d−1(x) of x to be the set of all points y ∈ Zd such that

x = y and ri(x, y) ≤ 1 for every i ∈ 1, 2, · · · , d. The elements in N3d−1(x) are called

(3d− 1)−neighbors(or indirect neighbors) of x. If y in (3d− 1)-neighborhood of x, then x

is also (3d − 1)-neighborhood of y.

(2) the 2d-neighborhood N2d(x) of x to be the set of all points with integer coordinates y ∈

Zd such that x = y and R(x, y) = 1. The elements in N2d(x) are called 2d−neighbors(or

direct neighbors) of x. If y in 2d-neighborhood of x, then x is also 2d-neighborhood of y.

(3) If two different points x, y ∈ Zd are n-neighbors for n = 2d or n = (3d− 1), then x and

y are also said to be n-adjacent.

Figure 2.1: 4- and 8-neighborhoods of a given point in Z2

The 2d- or (3d − 1)-connected sets are defined based on the graph structure induced by

2d- or (3d − 1)-adjacency relations.

Definition 2.1.3.

(1) An n-path from x to y is a sequence of points x = x1, x2, · · · , xk = y such that for

1 < i ≤ k, xi is n-adjacent to xi−1 for 1 < i ≤ k. The length of the path equals k-1.

25

(2) A digital set X ⊆ Zd is n-connected for n=2d, (3d − 1) if for every pair of points

x, y ∈ X, there is n-path contained in X from x to y.

Eckhardt and Latecki in [35] proved the Proposition 2.1.7, but the prove has a problem

in the special case d = 1. We will solve this problem in our proof, but before this consider

the following lemmas.

Lemma 2.1.4. For any point x ∈ Zd, let A = Zd \ N3d−1(x) ∪ x. Then A is

2d-connected for all d ≥ 2.

Proof. Let a = (a1 , a2 , · · · , ad), b = (b1 , b2 , · · · , bd) ∈ A, where x = (x1 , x2 , · · · , xd

), and let

ci= mina

i, b

i, c′

i= maxa

i, b

i, t

i= r

i(a, b) where i = 1, 2, · · · , d, and let t = R(a, b). So

we have two cases:

Case(1): xi/∈ [c

i, c

′

i] for all i ∈ 1, 2, · · · , d. If a1 ≤ b1(res. b1 ≤ a1) define y

j=

(a1 + j, a2 , · · · , ad), j = 0, 1, 2, · · · , t1 (res. y

j= (a1 − j, a2 , · · · , ad

), j = 0, 1, 2, · · · , t1).

Hence yt1= (b1 , a2 , · · · , ad

). And if a2 ≤ b2 (res. b2 ≤ a2) define ys = (b1 , a2 +

s, · · · , ad), s = 0, 1, 2, · · · , t2 (res. ys = (b1 , a2 − s, · · · , ad

), s = 0, 1, 2, · · · , t2). So yt1+t2

= (b1 , b2 , a3 , · · · , ad),· · · , yt = (b1 , b2 , · · · , bd) = b. It is clear that r

i(y

k1, x) ≥ 2 for some

i ∈ 1, 2, · · · , d and for all k1 = 0, 1, 2, · · · , t. Then there exist 2d-path connected in A

from a to b.

Case(2): There exists xi∈ [c

i, c

′

i] for some i ∈ 1, 2, .., d. Let q′

1= min

i=1,2,··· ,dri(a, x),

q′

2= min

i=1,2,··· ,dri(b, x), q1 = q′

1+ 2, and q2 = q

′

2+ 2. Assume that q

′

1= rα(a, x) and q

′

2

= rβ(b, x) for some α, β ∈ 1, 2, · · · , d. Define a = y0 , yr = (a1 , a2 , · · · , aα + r, · · · , a

d),

b = y′

0, y

′

s= (b1 , b2 , · · · , bβ + s, · · · , b

d) where r = 1, 2, · · · , q1 and s = 1, 2, · · · , q2 . Since

d ≥ 2 there exist i ∈ 1, 2, · · · , d \ α such that ri(yr , x) ≥ 2, and so yr ∈ A for all r

= 1, 2, · · · , q1 . But since rα(yq1, x) ≥ 2 and r

β(yq2

, x) ≥ 2, then any digital point has the

number aα + q1 or greater than it in the α’th coordinate is belong to A, and any digital

point has the number bβ+ q2 or greater than it in the β’th coordinate is belong to A.

26

Now let k1 = maxi=1,2,··· ,dri(a, x), k2 = max

i=1,2,··· ,dri(b, x), and k = maxk1+2, k2+2.

Define the sequence yq1+l1

= (a1 + l1 , a2 , · · · , ad) such as yq1+1 is a direct neighborhood

of yq1, where l1 = 1, 2, · · · , k, and define the sequence y

q1+k+l2= (a1 + k, a2 + l2 , · · · , ad

),

where l2 = 1, 2, · · · , k, · · · · · · , yq+l

d= (a1 + k, a2 + k, · · · , a

d+ l

d), where l

d= 1, 2, · · · , k

and q = q1 + k.(d−1). And define the sequence y′

q2+p1= (b1 +p1 , b2 , · · · , bd) such as y

′

q2+1

is a direct neighborhood of y′

q2, where p1 = 1, 2, · · · , k, and define the sequence y

′

q2+k+p2=

(b1 +k, b2 +p2 , · · · , bd), where p2 = 1, 2, · · · , k, · · · · · · , y′

q′+p

d

= (b1 +k, b2 +k, · · · , ad+p

d),

where pd= 1, 2, · · · , k, and q′ = q2 + k.(d−1). Let v = q1+k.d and u = q2+k.d, it is clear

that the two sequences y0 , y1 , · · · , yv−1 , yv and y′

0, y

′

1, · · · , y′

u−1, y

′

umake independently an

2d-path connected in A from a to yv and from b to y′

urespectively.

Now let yv = (a′

1, a

′

2, · · · , a′

d), y

′

u= (b

′

1, b

′

2, · · · , b′

d), e

i= mina′

i, b

′

i, and e′

i= maxa′

i, b

′

i,

then it is clear that xi/∈ [e

i, e

′

i] for all i ∈ 1, 2, · · · , d, Therefore by Case(1) above there

exists a 2d-path connected in A from a to b. Hence A is 2d-connected.

Lemma 2.1.5. For any point x ∈ Zd, let A = Zd \ N3d−1(x) ∪ x. Then A ∪ x is

not (3d − 1)-connected.

Proof. Let B =A ∪ x, and let y be an arbitrary point in A. Suppose that there exist

a sequence y = x0 , x1 ,· · · ,xk,x

k+1= x in B where x

iis (3d − 1) adjacent to x

i−1for

0 < i ≤ k+1, then xk= x and x

k∈ N

3d−1(x)∩A which is a contradiction. Hence A∪x

is not (3d − 1)- connected.

Lemma 2.1.6. For any topology satisfying conditions 11and 2

1, every point x in Z has

an open neighborhood that is contained in N2(x) ∪ x.

Proof. Assume that there exists a point x in Z such that none of whose open neighbor-

hoods is contained in N2(x) ∪ x. Let B = N2(x) ∪ x = x − 1, x, x + 1, and let

A1 , A2 ⊆ Z such that A1 = y ∈ Z : y ≥ x + 2 and A2 = y ∈ Z : y ≤ x − 2. The

27

set A1 is a topologically connected to prove this let a, b be any two distinct points in A1.

WLOG let a < b and b − a = t. Let a = x0 , and xi= a + i, i = 1, 2, · · · , t. Hence

xt = b. So there exist an 2-path connected in A1 from a to b. By condition 11A1 is a

topologically connected set. Similarly we can prove that A2 is a topologically connected

set. But A1 ∪ x is not 2-connected set since any 2-path connected from x to any point

in A1 will contain the point x+ 1 which is not in A1 ∪ x. By condition 12, A1 ∪ x is

not a topologically connected set. Similarly A2 ∪x is not a topologically connected set.

Now since x has no open neighborhoods contained in B, so x is not open. Assume to

contrary that x has two open neighborhoods U1 , U2 such as U1 ∩ A1 = ϕ, U1 ∩ A2 = ϕ,

U2 ∩A2 = ϕ, and U2 ∩A1 = ϕ. Then U3 = U1 ∩U2 is an open neighborhood of x contained

in B which is a contradiction. So we have three cases:

Case(1): For any U ∈ Ux, U ∩A1 = ϕ and U ∩A2 = ϕ. Hence A1 ∪x is a topologically

connected set which is a contradiction.

Case(2): For any U ∈ Ux, U ∩A2 = ϕ and U ∩A1 = ϕ. Hence A2 ∪x is a topologically

connected set which is a contradiction.

Case(3): For any U ∈ Ux, U ∩ A1 = ϕ, and U ∩ A2 = ϕ. Hence A1 ∪ x and A2 ∪ x

are topologically connected sets which is a contradiction.

Hence for any topology satisfying conditions 11and 2

1, every point x in Z has an open

neighborhood that is contained in N2(x) ∪ x.

Proposition 2.1.7. [35] For any topology of Zd satisfying conditions 1dand 2

d, every

point x in Zd has an open neighborhood that is contained in N3d−1(x) ∪ x.

Proof. If d = 1, then by Lemma 2.1.6 every point x in Z has an open neighborhood

that is contained in N2(x) ∪ x. Now if d ≥ 2. Assume that there exists a point x

in Zd such that none of whose open neighborhoods is contained in N3d−1(x) ∪ x. Let

A = Zd \ (N3d−1(x) ∪ x). By Lemma 2.1.4, A is 2d−connected set. By condition

28

1d, it is a topologically connected set. Since x has no open neighborhood contained in

N3d−1(x)∪ x, x is not open. By Lemma 2.1.5, A∪ x is not (3d− 1)-connected. By

condition 2d, the set A ∪ x is not topologically connected set. Since there is no open

neighborhood of x contained in N3d−1(x)∪ x, every open set containing x intersects A,

this implies that A ∪ x is a topologically connected set, which is a contradiction.

Lemma 2.1.8. Any topology on Zd satisfying the conditions 1dand 2

d, is an Alexandroff

topology.

Proof. Proposition 2.1.7 implies that for each x ∈ Zd there exist an open (topological)

neighborhood of x which is contained in N3d−1(x)∪x. Since N3d−1(x)∪x is a finite set,

there is a smallest neighborhood of x which is the intersection of all open sets containing

x in N3d−1(x) ∪ x. So any topology satisfying conditions 1dand 2

dis an Alexandroff

topology.

Notation 2.1.9. Henceforth, we will consider the digital topology(topologies) to mean the

topology(topologies) on Zd which satisfying the conditions 1dand 2

dwhere d = 1,2,3,· · · .

Definition 2.1.10. For any pair of points x, y ∈ Zd, and with respect to any digital

topology on Zd we define x→ y if and only if every open set containing x also contains y.

Recall we define the (Alexandroff) specialization order ”≤” of an Alexandroff topology

by x ≤ y if x ∈ y.

Observation 2.1.11. If x, y ∈ Zd, and if τ is any digital topology on Zd, then x→ y if and

only if y ∈ V (x) if and only if x ∈ y. Hence the relation ”→” is the same relation ”≤”.

That is ”→” = ”≤” on Zd.

Let y ← x mean x → y, and x 9 y denote the negation of x → y. Observe that

x 9 y holds if and only if there exists an open set containing x that does not contain y.

It is easy to check that the relation ”→” is reflexive and transitive.

29

Theorem 2.1.12. Let Zd be a digital space. Then:

(1) if two points x, y ∈ Zd are 2d-neighbors then either x→ y or y → x.

(2) if two distinct points x, y ∈ Zd are not (3d − 1)-neighbors then x9 y and y 9 x.

Proof. (1) Assume that the two points x, y ∈ Zd are 2d-neighbors. By the condition 1d,

the set x, y is topologically connected. Then x and y are adjacent. By Lemma 1.2.3,

x ∈ V (y) or y ∈ V (x). So either x→ y or y → x.

(2) Suppose that x, y ∈ Zd, are not (3d − 1)-neighbors. Then by the condition 2d, the

set x, y is not topologically connected. By Lemma 1.2.3, x /∈ V (y) and y /∈ V (x). So

x9 y and y 9 x.

Definition 2.1.13. The points x= (x1 , x2 , · · · , xd), y = (y1 , y2 , · · · , yd

),· · · , z = (z1 , z2 , · · · , zd)

in Zd are said to be in the same level if there exist i ∈ 1, 2, .., d such that xj= y

j= · · ·

= zjfor all j = i.

Theorem 2.1.14. Let Zd be a digital space. Suppose that x,y and z are three distinct

points in Zd in the same level such that y ∈ N2d(x) and z ∈ N2d(y). Then x → y if and

only if z → y. Analogously, y → x if and only if y → z (Thus, we have either x→ y ← z

or x← y → z).

Proof. Suppose that x = (x1 , x2 , · · · , xd), y = (y1 , y2 , · · · , yd

),· · · ,z = (z1 , z2 , · · · , zd), then

∃i ∈ 1, 2, · · · , d such as xj= y

j= z

jfor all j = i. Since y ∈ N2d(x) and z ∈ N2d(y),

then ri(x, y) = 1 and r

i(y, z) = 1. Therefore r

i(x, z) = 0 or r

i(x, z) = 2. Since x = z, we

have ri(x, z) = 2. Hence x /∈ N

3d−1(z). So, x9 z and z 9 x.

Now, by Theorem 2.1.12(1), since y ∈ N2d(x), either x → y or y → x. Suppose that

x → y. Suppose to contrary that y → z, then by the transitivity of ”→”, we get x → z

which is a contradiction. And since z ∈ N2d(y), either y → z or z → y. Suppose that

30

z → y. Suppose to contrary that y → x, then by the transitivity of ”→”, we get z → x

which is a contradiction. Therefore x → y if and only if z → y. Similarly y → x if and

only if y → z.

Lemma 2.1.15. If Zd is a digital space, then the relation ”→” is antisymmetric relation,

and hence a partial order on Zd.

Proof. Suppose that there exist two distinct points x=(x1, x2, · · · , xd) and y=(y1, y2, · · · , yd)

in Zd such that x → y and y → x. By Theorem 2.1.12 part (2), x ∈ N(3d−1)

(y) and so

there exist i ∈ 1, 2, · · · , d such that ri(x, y) = 1. Define z

j= x

jfor all j = i, ri(x, z)

= 1 and zj= y

j(In this case, z

i= x

i+ 1 if y

i= x

i− 1 and z

i= x

i− 1 if y

i= x

i+ 1)

clearly, z = y, z ∈ N(2d)

(x) and ri(z, y) = 2. Hence z /∈ N(3d−1)

(y). Therefore z 9 y and

y 9 z. Since z ∈ N(2d)

(x), then by Theorem 2.1.12 part (1), x → z or z → x. If x → z,

then by the transitivity of ”→”, y → z which is contradiction. And if z → x, then by the

transitivity of ”→”, z → y which is a contradiction.

Theorem 2.1.16. Any digital topology on Zd is a generalized locally finite T0-Alexandroff

space.

Proof. We prove above that any digital topology on Zd is an Alexandroff space. Now

suppose to contrary that it is not a T0-space. So there exist two distinct points x , y

such that V (x) = V (y), so x ∈ V (x) = V (y) and y ∈ V (y) = V (x). This implies

that x → y and y → x which contradicts Lemma 2.1.15. So any digital topology on Zd

is a T0-Alexandroff space. Finally we want to prove that Zd with the binary relation

”→” satisfies both (ACC) and (DCC). Assume that Zd dose not satisfy (ACC). This

implies that there exists a sequence x1 → x2 → · · · xn → · · · with no k ∈ N such that

31

xk = xk+1 = · · ·. Without loss of generality, suppose that xi= x

i+1∀ i. Now by Theorem

2.1.12, x1 → x2 implies x2 ∈ N(3d−1)(x1). By the transitivity of ”→” x1 → x

i∀ i ≥ 2, so

xi∈ N

(3d−1)(x1) ∀ i ≥ 2 which is a contradiction, since N

(3d−1)(x1) is finite set. Similarly

we can prove that Zd satisfies (DCC).

Theorem 2.1.17. Let Zd be a digital space, then

(1) Zd contains an open, dense and hereditarily irresolvable subspace.

(2) Zd is strongly irresolvable.

(3) For each dense subset D of Zd, Do is dense.

(4) (Zd, τα) is submaximal.

(5) For A ⊆ Zd where Ao =ϕ, A is nowhere dense.

(6) Zd is an α-T 12

−space.

(7) Zd is a semi-T 12

−space.

Proof. Follows from, Theorem 1.2.7, Theorem 1.2.10 and Theorem 2.1.16.

That the set which is an open, dense and hereditarily irresolvable subspace in Zd is

the set of all maximal points M . A point x ∈ Zd is a maximal point if whenever x → y

then x = y. A point x is minimal if whenever z → x then x = z. Otherwise it is called

saddle point, we denoted the set of all saddle points by SD.

Lemma 2.1.18. If Zd is a digital space and x ∈ Zd, then :

(1) x is a maximal point if and only if y → x ∀y ∈ N2d(x).

(2) x is a minimal point if and only if x→ y ∀y ∈ N2d(x).

32

Proof. (1) (⇒) Assume that x is a maximal point, and let y ∈ N2d(x). Then by Theorem

2.1.12, x → y or y → x. If x → y, then x = y. So y /∈ N2d(x) which is a contradiction.

Therefore y → x.

(⇐) Suppose that y → x ∀y ∈ N2d(x), and suppose to contrary that x is not a maximal

point. So ∃ w ∈ N3d−1

(x) \ N2d(x) such that x → w. Suppose that x = (x1 , x2 , · · · , xd

)

and w = (w1 , w2 , · · · ., wd). Since w ∈ N

3d−1(x) and w /∈ N

2d(x), ∃ i, j ∈ 1, 2, · · · , d,

i = j and ri(x,w) = r

j(x,w) = 1. Take z = (z1 , z2 , · · · , zd) ∈ Zd such that r

k(x, z) =

0 ∀ k = j, rj(x, z) = 1, and r

j(z, w) = 0 (In this case z

j= x

j∓ 1 in the case w

j=

xj± 1). Hence r

j(z, w) = 2, then z /∈ N

3d−1(w). Thus z 9 w and w 9 z. Moreover,

z ∈ N2d(x), so z → x. Since x→ w, we get by transitivity of ”→ ” that z → w which is

a contadiction.

(2)(⇒) Assume that x is a minimal point, and let y ∈ N2d(x), then by Theorem 2.1.12

x→ y or y → x. If y → x, then x = y. So y /∈ N2d(x) which is a contradiction. So y → x.

(⇐) Suppose that x → y ∀y ∈ N2d(x), and assume to contrary that x is not a minimal

point, this implies that ∃ w ∈ N3d−1

(x) \ N2d(x) such that w → x. Suppose that x

= (x1 , x2 , · · · , xd) and w = (w1 , w2 , · · · , wd

). Since w ∈ N3d−1

(x) and w /∈ N2d(x), ∃

i, j ∈ 1, 2, · · · , d, i = j and ri(x,w) = r

j(x,w) = 1. Take z = (z1 , z2 , · · · , zd) ∈ Zd such

that rk(x, z) = 0 ∀ k = j, r

j(x, z) = 1, and r

j(z, w) = 0 (In this case z

j= x

j∓ 1 in the

case wj= x

j± 1). Hence r

j(z, w) = 2, then z /∈ N

3d−1(w). Thus z 9 w and w 9 z.

Moreover, z ∈ N2d(x), so x → z. Since w → x, we get by transitivity of ” → ” that

w → z which is a contadiction.

Corollary 2.1.19. If Zd is a digital space and x ∈ Zd, then x is a saddle point if and

only if ”x→ y and z → x for some distinct points y, z ∈ N2d(x)”.

Proof. (⇒) Let x be a saddle point. Suppose that y → x ∀y ∈ N2d(x), then by Lemma

33

2.1.18, x is a maximal point which is a contradiction. And if x→ y ∀y ∈ N2d(x), then by

Lemma 2.1.18, x is a minimal point which is a contradiction. Hence x → y and z → x

for some distinct points y, z ∈ N2d(x).

(⇐) Suppose that x → y and z → x for some distinct points y, z ∈ N2d(x). Assume to

contrary that x is not a saddle point. Then x is a maximal or minimal point which is

contract with Lemma 2.1.18. So x is a saddle point.

2.2 The Digital Topology on Z

In this section, we prove that there exists only one topology on Z (up to the homeomor-

phic) satisfying the two conditions 11and 2

1which is the Khalimsky topology. We discuss

some of its topological proprieties. For d = 1 the two conditions 1dand 2

dtranslate to :

(11) If a set in Z is 2-connected, then it is (topologically) connected.

(21) If a set in Z is not 2-connected, then it is not (topologically) connected.

Definition 2.2.1. Define the subsets O and E of Z as the following :

O = x ∈ Z : x is odd number, and E = x ∈ Z : x is even number.

Theorem 2.2.2. The only topology on Z (up to the homeomorphic) satisfying the two

conditions 11and 2

1is the Khalimsky topology.

Proof. Suppose that τ is a digital topology on Z, then τ is a generalized locally finite

T0-Alexandroff space. Hence M = ϕ. Pick x ∈ M . Since d = 1, N3d−1

(x) = N2(x) =

N2d(x) = x−1, x+1, and for y /∈ N2(x), x9 y and y 9 x. Since x is maximal we have

by Lemma 2.1.18, x− 1→ x← x+ 1. Consider the point x− 1 ∈ Z, we have N2(x− 1)

= x − 2, x. So x − 2 ∈ N2(x − 1) but x − 1 ∈ N2(x) and x − 1 → x. By Theorem

2.1.14, we must have x − 2 ← x − 1 → x. Hence by Lemma 2.1.18, x − 1 is a minimal

34

point and V (x − 1) = ↑ (x − 1) = x − 2, x − 1, x. Similarly, x + 1 is a minimal point

and V (x+ 1) = x, x+ 1, x+ 2. Finally, for x− 2 we have that x− 3 ∈ N2(x− 2), but

x− 2 ∈ N2(x− 1) and x− 2← x− 1. By Theorem 2.1.14, x− 1← x− 2→ x− 3. Thus

x− 2 is a maximal point and V (x− 2) = x− 2. Similarly we have x+ 2 is a maximal

point and V (x+2) = x+2. Therefore if x is maximal in (Z, τ), then the set of maximal

elements M = x + 2k : k ∈ Z and the set of minimal elements m = (x + 1) + 2n :

n ∈ Z. Moreover V (y) = y − 1, y, y + 1 ∀ y ∈ m. Hence τ is a Khalimsky topology

on Z if x is odd number, and τ is the dual of the Khalimsky topology on Z if x is even

number.

Using Hasse diagram we can consider the corresponding poset of the Khalimsky topology

by the following diagram

Figure 2.2: Khalimsky Topology Via Poset

The set of maximal elementsM is the set of odd numbers and the set of minimal elements

m is the set of even numbers. Since the Khalimsky topology is a T0-Alexandroff space,

then there exists an induced order, denote it by ≤Kh

and denote the up of the point x

with respect to this topology by VKh(x). The up of the point x in the Khalimsky dual

topology is VdK(x) = ↑

dKx. In this case x ≤

khy if and only if x→ y.

Lemma 2.2.3. The Khalimsky topology is homeomorphic to the Khalimsky dual topology.

Proof. Let f : Z −→ Z such that f(x) = x+1. It is clear that f is one to one and onto. Let

35

the topology on the domain the Khalimsky topology, and on the rang the Khalimsky dual

topology.

If x is odd, then f(↑Kh

x) = fx = x + 1 = ↑dK

(x + 1) = ↑dK

f(x). If x is

even, then f(↑Kh

x) = fx − 1, x, x + 1 = x, x + 1, x + 2 = ↑dK

(x + 1) = ↑dK

f(x).

Then by Theorem 1.2.5, the Khalimsky topology is homeomorphic to the Khalimsky dual

topology.

The Khalimsky space is the set Z together with the Khalimsky topology. It is also

called the digital space Z.

Theorem 2.2.4. Let Z be the Khalimsky Space. Then

(1) Z is submaximal.

(2) Z is nodec.

(3) PO(Z)= τα = τ(≤); that is, every preopen set is open.

(4) Z is T∗1 -space.

(5) Z is T 34

-space.

(6) Z is a semi-T2−space.

Proof. Since each element of Z is pure, then by Theorem 2.1.16 and Theorem 1.2.8, we

get the parts (1),(2),(3) and (4).

Since Z is submaximal, then by Theorem 1.2.8 it is T 12

-space. And since for any even

number x, VKh(x) = ↑

Khx = x − 1, x, x + 1, then | x | = 2 ∀x /∈ M . By Theorem

1.2.9, Z is T 34

-space. Part(6) is coming directly from Theorem 1.2.11, Theorem 2.1.16,

and | x | = 2 ∀x /∈ M .

36

2.3 Two Digital Topologies on Z2

In this section, we prove that there exist two topologies (up to the homeomorphic) on

Z2 that are satisfying the two conditions 12and 2

2. We discuss some of its topological

properties. For d = 2 the two conditions 12and 2

2translate to :

(12) If a set in Z2 is 4-connected, then it is (topologically) connected.

(22) If a set in Z2 is not 8-connected, then it is not (topologically) connected.

As a direct consequence of Theorem 2.1.18 , we obtain that there exist only three types

of neighborhoods N4(x) of any point x ∈ Z2 which we called them the N4 neighborhood

of a maximal, minimal, and saddle points (the last type is up to rotations by 90):

Figure 2.3: The N4 neighborhood of a maximal, minimal, and saddle points

Lemma 2.3.1. Let x ∈ Z2. If x is maximal (resp. minimal, saddle) point, then no point

in N4(x) is maximal (resp. minimal, saddle) point.

Proof. Let y ∈ N4(x). Suppose that x and y are maximal points. Then x→ y and y → x.

Hence x = y ∈ N4(x) which is a contradiction. Similarly if both x and y are minimal

points. Finally if x and y are saddle points, then either x → y or y → x. Suppose that

x → y. Choose a, b ∈ N4(x) in the same level different from y, and z, w ∈ N4(y) in the

same level different from x as shown in Figure 2.4 :

37

Figure 2.4: Points in different levels

Since x and y are saddle points and x → y, we must have a → x ← b and z ← y → w.

Then a → x → y → w. By transitivity of ”→”, a → w. Hence by Theorem 2.1.12 a ∈

N8(w) which is a contradiction. Similarly if y → x.

By Theorem 2.1.16, Z2 is a generalized locally finite T0-Alexandroff space. Hence Z2

contains maximal and minimal points. In fact, we have two cases : Z2 contains saddle

points or Z2 dose not contain any of saddle points.

In case(1), if there exist saddle points, then by one single saddle point x and using

Lemma 2.3.1, we can configure all ”→”-relations in the digital plane, and we get exactly

one possible relation. In fact, there is four homeomorphic topologies on Z2 contains saddle

points depend on choosing x, and up to homeomorphic, they are looking as one topology

on Z2 denote it by τs (see the Figure 2.5)

Simply and using the transitvity of ” → ”, we give Figure 2.5 without cross relation

” ” as follows ( See Figure 2.6) :

In case(2) if there is no saddle point, then by choosing one minimal (maximal) point

x and by Lemma 2.3.1, we can configure the ”→”-relation in the whole digital plane to

get exactly one possible relation. In fact there exist two homeomorphic topologies on Z2

with no saddle points depend on choosing x as maximal or as minimal. They are looking

38

Figure 2.5: The topology on Z2 with saddle points

Figure 2.6: The topology on Z2 with saddle points without cross relation ” ”

as one topology (up to homeomorphic) denote it by τm. See Figure 2.7

Observation 2.3.2. In the digital space Z2 with τm. If x ∈ Z2 is a maximal (resp. minimal),

then all points in N8(x) \N4(x) are maximal (resp. minimal) points.

Lemma 2.3.3. In the digital space Z2 with τm the smallest open set containing a minimal

point x is V (x) = N4(x) ∪ x.

Proof. Let x be a minimal point, and let y ∈ N4(x). Since x is a minimal point, then

by Lemma 2.1.18, x → y. So y ∈ V (x). Since x ∈ V (x), then x ∪ N4(x) ⊆ V (x).

Conversely if y ∈ V (x) and y is not in x∪N4(x), then x→ y, and so y ∈ N8(x)\N4(x).

By Observation 2.3.2, y is a minimal point and this is a contradiction (since x → y and

x = y). Thus, V (x) ⊆ x ∪N4(x). Hence V (x) = x ∪N4(x).

39

Figure 2.7: The topology on Z2 with no saddle points

Note that if x is a maximal point, then V (x) = x is true in any generalized locally finite

T0-Alexandroff space.

Proposition 2.3.4. Let τs be the topology on Z2 with saddle points and x ∈ Z2. Then :

(1) if x is a maximal or minimal point, ∀ y ∈ N4(x), y is a saddle point.

(2) if x is a saddle point, then ∀ y ∈ N4(x), either y is maximal or minimal point.

(3) if x is a maximal(resp. minimal) point, then ∀ y ∈ N8(x) \ N4(x), y is a mini-

mal(resp. maximal) point.

(4) if x is a saddle point, ∀ y ∈ N8(x) \N4(x), y is a saddle point.

Proof. See Figure 2.5.

Proposition 2.3.5. In the digital space (Z2, τs) with saddle points, the smallest open set

containing a minimal point x is V (x) = N8(x) ∪ x. The smallest open set of a saddle

point x is V (x) = (M ∩N4(x)) ∪x.

Proof. Suppose that x is a minimal point. Let y ∈ V (x), then x → y. If x = y, then by

Theorem 2.1.12 part(2), y ∈ N8(x), hence V (x) ⊆ N8(x) ∪ x.

40

Conversely, if y ∈ N8(x) we have two cases either y ∈ N4(x) or y ∈ N8(x) \N4(x). If y ∈

N4(x), then x→ y since x is a minimal point, and hence y ∈ V (x). If y ∈ N8(x) \N4(x),

then by Proposition 2.3.4, y is a maximal point. If x = (x1 , x2), then y = (x1 + i, x2 + j)

where i, j ∈ 1,−1. Let z = (x1 + i, x2), so z ∈ N4(x) and z ∈ N4(y). By Lemma 2.1.18,

x→ z → y. By the transitivity of ”→”, we have x→ y. Then y ∈ V (x). Since x ∈ V (x),

N8(x) ∪ x ⊆ V (x). Thus, V (x) = N8(x) ∪ x.

For a saddle point x if y /∈ N8(x) ∪ x, then y 9 x and x 9 y. If y ∈ N8(x) \ N4(x),

then by Proposition 2.3.4 part(4), y is a saddle point. This implies that x and y are

incomparable. Hence y /∈ V (x). If y ∈ N4(x), then by Proposition 2.3.4 part(2), either

y is minimal or y is maximal. If y is minimal then y → x and y /∈ V (x). If y is maximal

then x→ y and y ∈ V (x). Thus, V (x) = (M ∩N4(x)) ∪x.

Definition 2.3.6. An elementary square of Z2 is a set A(i,j)

= i, i + 1 × j, j + 1 =

(i, j), (i+1, j), (i+1, j+1), (i, j+1) where i, j ∈ Z. In the digital plane, any elementary

square form a square whose vertices are the elements in the elementary square.

Observation 2.3.7. [35] A topology in Z2 is completely determined by the relation →

on a single elementary square. There are exactly two possibilities (up to rotations and

translations):

(a) Antiparallel case: In this case, the arrows of parallel sides point into opposite direc-

tions. This is holding on (Z2, τm). (See Figure 2.7).

(b) Parallel case: In this case, the arrows of parallel sides of the elementary square have

the same direction. This is holding on (Z2, τs). (See Figure 2.6).

For instead see Figure 2.8. We note that in the parallel case a diagonal arrow is implied

by transitivity. This arrow was not shown in the Figure 2.8. It should be noted that if

we know the direction of three sides in the elementary square, then we can determine the

41

direction of the fourth side. To see this, note that if we determine the direction of three

side in an elementary square, then the direction of two parallel sides are determined and

so by Observation 2.3.7, we can determine the fourth side, using the three relations.

Figure 2.8: Antiparallel, parallel sides

Observation 2.3.8. Let us numeric the vertices of an elementary square 1,2,3 and 4. If

1→ 2 and 2→ 3 as been shown in Figure2.9, then 4→ 3 and 1→ 4.

Figure 2.9: An elementary square

Proof. Assume that 3→ 4 then by the transitive 1→ 4 which is contradiction since this

elementary square becomes neither antiparallel nor parallel elementary square. Hence

4→ 3. By Observation 2.3.7, 1→ 4.

42

Marcus-Wyse Topology τm on Z2

The Marcus Wyse topology is a special connected T 12

-Alexandroff space on Z2. This

topology was described by Marcus and Wyse in [9]. They defined this topology by its

minimal neighborhood as the following : for any point x = (a, b) ∈ Z2, V (x) = x∪N4(x)

if x+y is even and V (x) = x otherwise. Hence τm is the Marcus Wyse topology on Z2.

This topology is a T0-Alexandroff space, and its specialization order is denoted by (≤2

m).

So in this topology x→ y iff x ≤2

my.

Definition 2.3.9. Define two subsets EVd, ODd of Zd as follows

(i) EVd = (x1, x2, · · · , xd) ∈ Zd :∑d

i=1 xi is even number .

(ii) ODd = (x1, x2, · · · , xd) ∈ Zd :∑d

i=1 xi is odd number .

Lemma 2.3.10. (a) If u ∈ EV2, then N4(u) ⊆ OD2 and N8(u) \N4(u) ⊆ EV2.

(b) If u ∈ OD2, then N4(u) ⊆ EV2 and N8(u) \N4(u) ⊆ OD2.

Proof. (a) If u = (a, b) ∈ EV2, then (a± 1, b± 1) ∈ EV2 and (a± 1, b), (a, b± 1) ∈ OD2.

Thus, N4(u) ⊆ OD2 and N8(u) \N4(u) ⊆ EV2. (b) is similar to (a).

Proposition 2.3.11. If u, v ∈ EV2 (resp. OD2), then there exists a finite sequence u =

u0,u1,u2,· · · ,un = v in EV2 (resp. OD2) such that ui−1∈ N8(ui

) \N4(ui) ∀ i = 1,2,· · · ,n.

Proof. Suppose that u, v ∈ EV2 and ui= (ai, bi). Since u0 = u, un = v ∈ EV2, both

”a0, b0” and ”an, bn” are even or both odd. Suppose that a0, b0 are both even, then we

have two cases for v = un = (an, bn).

Case(1) an, bn are both even. Suppose that a0 ≤ an and b0 ≥ bn . Then an − a0 = 2k, and

b0 − bn = 2m for some k,m ∈ N ∪ 0. Take u1 = (a0 + 1, b0 − 1), u2 = (a0 + 2, b0), u3

= (a0 + 3, b0 − 1), u4 = (a0 + 4, b0),· · · , u2k= (a0 + 2k, b0) = (an , b0). And take u

2k+1=

(an +1, b0 − 1), u2k+2

= (an , b0 − 2), u2k+3

= (an +1, b0 − 3),· · · , u2k+2m

= (an , b0 − 2m) =

43

(an , bn) = v. clearly ui∈ EV2 and ui−1

∈ N8(ui)\N4(ui

) ∀i = 1, 2, · · · , 2k+2m. Similarly

if (a0 ≤ an and b0 ≤ bn), (a0 ≥ an and b0 ≤ bn) or (a0 ≥ an and b0 ≥ bn).

Case(2) an, bn are both odd. In this case take un−1 = (an−1, bn−1). Since an−1, bn−1 are

both even and un−1 ∈ N8(un)\N4(un), then by case(1) above there exists a finite sequence

u = u0 , u1 ,· · · ,un−1 , un = v in EV2 such that ui−1∈ N8(ui

)\N4(ui) i = 1,2,· · · ,n. Similarly

if both a0, b0 are both odd. Similar proof if u, v ∈ OD2.

Theorem 2.3.12. Let (Z2, τm) be the Marcus-Wyse space. If there exists a minimal (resp.

a maximal) point u such that u ∈ EV2, then EV2 = m and OD2 = M (resp. M = EV2

and m = OD2).

Proof. Suppose that there exists a point u ∈ EV2. Let v ∈ EV2 be arbitrary. By

Proposition 2.3.11, there exists a finite sequence u = u0, u1, · · · , un = v in EV2 such that

ui−1∈ N8(ui

) \ N4(ui), i ∈ 1, 2, · · · , n. By Observation 2.3.2, u is minimal (resp.

maximal) point. This implies that v is minimal (resp. maximal) point. If w ∈ OD2 where

w = (a, b), then (a + 1, b) ∈ EV2 and so (a + 1, b) is minimal (resp. maximal). Since

w ∈ N4(a+ 1, b), by Lemma 2.3.1, w is maximal (resp. minimal). Hence m = EV2(res.

M = EV2). Similarly we can prove that M = OD2 (res. m = OD2) whenever u ∈ EV2

and u is a minimal (resp. maximal) point.

So in the Marcus-Wyse on Z2, we have two possible cases: ”m = EV2 and M = OD2”

or ”M = OD2 and m = EV2”. It is clear that the topology in the first is homeomorphic

to that in the second case. By convention we will take the topology in first case to be the

Marcus -Wyse topology on Z2 and the top in the second case will be its dual. The up of

a point x ∈ Z2 in Marcus -Wyse topology denoted by ↑m x, and in Marcus-Wyse dual

topology by ↑dmx.

Let Z2 be the Marcus -Wyse space, then we can consider a part of Z2 as in Fig. 2.10:

Also we can consider this part by using Hasse diagram. (see Fig. 2.11):

44

Figure 2.10: Part of Marcus-Wyse Topology on Z2

Figure 2.11: a part of Marcus-Wyse topology on Z2

Since the Marcus -Wyse topology on Z2 is a T0-Alexandroff space, then it has a special

order, denote it by (≤2

m). Using Lemma 2.3.3, if (x, y) ∈ EV2 = m, then ↑m (x, y) =

V (x, y) = N4(x, y) ∪ (x, y). And if (x, y) ∈ OD2 = M , then ↑m (x, y) = V (x, y) =

(x, y).

Theorem 2.3.13. Let (Z2, τm) be the Marcus-Wyse topology on Z2, a ∈ Z be a fixed

point. If A = (a, y) : y ∈ Z, and B = (y, a) : y ∈ Z, then the two subspaces A and

B are homeomorphic to the Khalimsky topology.

Proof. For a point a, we have two cases:

Case(1) a ∈ Z is a fixed even number. Let f : A −→ Z be such that f(a, x) = x.

45

It is clear that f is a bijective function. If x is odd number, then f(↑m (a, x) ∩ A) =

f(a, x) = x = ↑Kh

x = ↑Kh

f(a, x). If x is even number, then f(↑m (a, x) ∩ A) =

f(a, x− 1), (a, x), (a, x+ 1) = x− 1, x, x+ 1 = ↑Khx = ↑

Khf(a, x).

Case(2) a ∈ Z is a fixed odd number. Let f : A −→ Z be such that f(a, x) = x+ 1. If x

is odd number, then f(↑m (a, x)∩A) = f(a, x− 1), (a, x), (a, x+ 1) = x, x+ 1, x+ 2

= ↑Kh

(x + 1) = ↑Kh

f(a, x). If x is even number, then f(↑m (a, x) ∩ A) = f(a, x)

= x + 1 = ↑Kh

(x + 1) = ↑Kh

f(a, x). By Theorem 1.2.5, A is homeomorphic to the

Khalimsky topology. Similarly we can prove that B is homeomorphic to the Khalimsky

topology.

Definition 2.3.14. Let Zd be a topological space. We define the ith-summation set Sdi

on Zd as the following: Sdi = (x1, x2, · · · , xd) ∈ Zd :

∑d

j=1xj = i. The summation

family set Sd on Zd is Sd = Sdi : i ∈ Z.

Definition 2.3.15. For any pair of elements S2i , S

2j ∈ S2, we define S2

i ∼ S2j if and only

if whenever x ∈ S2i , there exists y ∈ S2

j such that x ≤2

my.

Lemma 2.3.16. The summation family set S2 with the binary relation ” ∼ ” is a partially

ordered set (poset).

Proof. For S2i , S

2j , S

2r ∈ S2.

(i) S2i ∼ S2

i , since x ≤2

mx, ∀x ∈ S2

i . (” ∼ ” is reflexive).

(ii) Let S2i ∼ S2

j and S2j ∼ S2

i . If x = (x1, x2) ∈ S2i , then ∃ y = (y1, y2) ∈ S2

j such that,

x ≤2

my. So we have two cases :

Case(1) If x ∈ OD2 (in the case when i is odd number), then x is a maximal point, and

hence x = y. Therefore S2i = S2

j .

Case(2): If x ∈ EV2 (in the case when i is even number), then x is a minimal point,

and hence ↑m x = (x1, x2), (x1 − 1, x2), (x1 + 1, x2), (x1, x2 − 1), (x1, x2 + 1). Then

46

j ∈ i− 1, i, i + 1. Suppose that i = j, then j ∈ i− 1, i + 1. Hence j is odd number.

By Case(1) and since S2j ∼ S2

i , we have S2i = S2

j , which implies that i = j and this is

impossible since i is even and j is odd. Thus i = j and S2i = S2

j (” ∼ ” is anti-symmetry).

(iii) Let S2i ∼ S2

j , and S2j ∼ S2

r . If x ∈ S2i , then there exists y ∈ S2

j such that x ≤2

my.

Since S2j ∼ S2

r and y ∈ S2j , then there exists z ∈ S2

r such that y ≤2

mz. By transitivity of

≤2

m”, we have x ≤2

mz. Hence S2

i ∼ S2r (” ∼ ”is transitivity ).

For the poset S2, let τΩbe the T0-Alexandroff space on S2 induced by the order ∼.

We call it the summation topology on S2. For S2i , VΩ

(S2i ) = ↑Ω S2

i = S2j ∈ S2 : S2

i ∼ S2j .

when i is odd number, VΩ(S2

i ) = ↑Ω S2i = S2

i and when i is even number, and S2i ∼ S2

j ,

then i ∈ i− 1, i, i+ 1, thus VΩ(S2

i ) = ↑Ω S2i = S2

i−1, S2i , S

2i+1.

Theorem 2.3.17. The summation topology τΩon S2 is homeomorphic to the Khalimsky

topology.

Proof. Let f : S2 −→ Z be a function defined by f(S2i ) = i. It is clear that f is a

bijective function. If i is odd number, then f(↑ΩS2i ) = fS2

i = i = ↑Khi = ↑

Khf(S2

i ).

If i is even number, then f(↑ΩS2i ) = fS2

i−1, S2i , S

2i+1 = i − 1, i, i + 1 = ↑

Khi =

↑Kh

f(S2i ). Then by Theorem 1.2.5 the summation topology τ

Ωon S2 is homeomorphic

to the Khalimsky topology.

Recall that F2is the set of all faces of Z2

, F2

kis the set of all k-faces of Z2

, and if f =

a, a+ 1 × b, b+ 1, then the leader of f , L(f) = (a, b). Let F2

0,2= F2 \ F2

1.

Definition 2.3.18. The Marcus-Wyse function κ : F2

0,2−→ Z2

is a bijection function

define as :

47

κ(f) =

(a+b,a-b), if dim(f) = 0 and f = (a, b);

(a+b+1,a-b), if dim(f) = 2 and L(f) = (a, b).

Using the Marcus-Wyse function and Theorem 1.2.5, we have the following theorem.

Theorem 2.3.19. Let (Z2, τm) be the Marcus-Wyse topology on Z2, then (Z2, τm) is

homeomorphic to (F2

0,2, τ⊆).

Theorem 2.3.20. Let (Z2, τm) be the Marcus-Wyse topology on Z2, then

(1) Z2 is submaximal.

(2) Z2 is nodec.

(3) PO(Z2)=τα = τ(≤); that is, every preopen set is open.

(4) Z2 is T∗1 -space.

(5) Z2 is T 34

.

(6) Z2 is a semi-T2−space.

Proof. Mimic to the proof of Theorem 2.2.4.

Alexandroff-Hopf topology on Z2

Eckhardt in [35] said that the digital topology on Z2 which has saddle points is home-

omorphic to the cellular-complex topology (Alexandroff-Hopf topology). We will latter

define a very useful function and using it to prove this fact.

Definitions 2.3.21. Define the subsets of Z2 EE, OO, EO and OE as following:

(1) EE = E × E = (x, y) ∈ Z2 : x, y are even numbers .

48

(2) OO = O ×O = (x, y) ∈ Z2 : x, y are odd numbers .

(3) EO = E ×O = (x, y) ∈ Z2 : x is even number and y is odd number .

(4) OE = O × E = (x, y) ∈ Z2 : x is odd number and y is even number .

Lemma 2.3.22. Let (Z2, τs) be the Alexandroff-Hopf space (= the topology on Z2 with

saddle points), and let (x, y) ∈ Z2.

(1) If (x, y) is a maximal (resp. minimal) point, then (x+2, y), (x, y+2), (x−2, y) and

(x, y − 2) are maximal (resp. minimal) points.

(2) If (x, y) is a saddle point, then (x+1, y+1), (x−1, y−1), (x+1, y−1), (x−1, y+1),

(x+ 2, y), (x, y + 2), (x− 2, y) and (x, y − 2) are saddle points.

Proof. (1) Let Z2 be the Alexandroff-Hopf topology, and let (x, y) ∈ Z2 is a maximal (resp.

minimal) point. Since (x+1, y+1), (x−1, y−1) ∈ N8(x, y)\N4(x, y), then by Proposition

2.3.4 part(3), (x+ 1, y + 1) and (x− 1, y − 1) are minimal (resp. maximal) points. But

(x+2, y), (x, y+2) ∈ N8(x+1, y+1)\N4(x+1, y+1), and (x−2, y), (x, y−2) ∈ N8(x−

1, y−1)\N4(x−1, y−1). Then by Proposition 2.3.4 part(3), (x+2, y), (x, y+2), (x−2, y)

and (x, y − 2) are maximal (resp. minimal) points.

(2) Let (x, y) is a saddle point. Since (x+1, y+1), (x−1, y−1), (x+1, y−1), (x−1, y+1) ∈

N8(x, y)\N4(x, y), then by Proposition 2.3.4 part(4), (x+1, y+1), (x−1, y−1), (x+1, y−1)

and (x − 1, y + 1) are saddle points. and since (x + 2, y), (x, y + 2) ∈ N8(x + 1, y + 1) \

N4(x + 1, y + 1), and (x − 2, y), (x, y − 2) ∈ N8(x − 1, y − 1) \ N4(x − 1, y − 1), then

by Proposition 2.3.4 part(4), (x + 2, y), (x, y + 2), (x − 2, y) and (x, y − 2) are saddle

points.

Theorem 2.3.23. Let (Z2, τs) be the Alexandroff-Hopf topology, and let (x, y) ∈ EE:

(i) If (x, y) is a minimal point, then m = EE, M = OO, and SD = EO∪OE = OD2.

49

(ii) If (x, y) is a maximal point, then m = OO, M = EE, and SD = EO∪OE = OD2.

(iii) If (x, y) is a saddle point, then SD = EE ∪OO = EV2, and either ”M = OE and

m = EO” or ”M = EO and m = OE”.

Proof. Suppose that (x, y) ∈ EE.

(i) Let (x, y) be a minimal point, and assume that (a, b) ∈ EE be arbitrary. Suppose

that x ≤ a and y ≥ b. So a = x+ 2n and b = y − 2r for some n, r ∈ N ∪ 0. Using the

induction and Lemma 2.3.22, we have that if (x, y) is a minimal point then (x+ 2n, y) =

(a, y) is a minimal point. By the same way if (a, y) is a minimal point, then (a, y − 2r)

= (a, b) is a minimal point. Similarly if (x ≤ a and y ≤ b), (x ≥ a and y ≥ b) or

(x ≥ a and y ≤ b). Hence EE ⊆ m. If (a, b) ∈ OO, then (a + 1, b + 1) ∈ EE ⊆ m

and (a + 1, b + 1) ∈ N8(a, b) \ N4(a, b). By Proposition 2.3.4 part(3), (a, b) ∈ M .

Therefore OO ⊆ M . If (a, b) ∈ OD2, then (a+ 1, b) ∈ EV2 ⊆ m ∪M . Using Proposition

2.3.4 part(1), (a, b) ∈ SD. So OD2 ⊆ SD. Thusm = EE,M = OO, and SD = EO∪OE

= OD2.

(ii) Similar to (i).

(iii) Let (x, y) ∈ SD, and assume that (a, b) ∈ EE be arbitrary. By Lemma 2.3.22 part(2)

and by similar way to part(i) above, EE ⊆ SD. Assume that (a, b) ∈ OO be arbitrary

and x ≤ a and y ≥ b. Hence a = x+ 2n+ 1 and b = y − 2r − 1 for some n, r ∈ N ∪ 0.

Using the induction and Lemma 2.3.22 part(2), we have that if (x, y) ∈ SD, then

(x + 2n, y) ∈ SD. By the same way (x + 2n, y − 2r) ∈ SD. By Lemma 2.3.22 part(2),

(a, b) = (x+ 2n+ 1, y − 2r − 1) ∈ SD. Therefore OO ⊆ SD, moreover EV2 ⊆ SD. Now

let (a, b) ∈ OE be arbitrary. Then (a+ 1, b) ∈ EV2 ⊆ SD. By Proposition 2.3.4 part(2),

either (a, b) ∈M or (a, b) ∈ m. Hence either OE ⊆M orOE ⊆ m. Assume that OE ⊆M

(resp. OE ⊆ m). Let (a, b) ∈ EO be arbitrary, then (a+ 1, b) ∈ SD. By Theorem 2.3.1,

(a, b) /∈ SD. Suppose to contrary that (a, b) ∈ M (resp. (a, b) ∈ m). So by Lemma

50

2.3.22 part(1), (a+ 2, b) ∈M (resp. (a+ 2, b) ∈ m). Moreover, (a+ 1, b+ 1) ∈ OE and

(a+1, b+1) ∈M (resp. (a+1, b+1) ∈ m). By Lemma 2.3.22, (a+1, b− 1) ∈M (resp.

(a+1, b− 1) ∈ m). So N4(a+1, b) ⊂M (resp. N4(a+1, b) ⊂ m) which is a contradiction

since (a + 1, b) ∈ SD. Therefore if OE ⊆ M (resp. OE ⊆ m), then EO ⊆ m (resp.

EO ⊆M). Thus SD = EE ∪OO = EV2, and either ”M = OE and m = EO” or ”M =

EO and m = OE”.

Corollary 2.3.24. Let (Z2, τs) be the Alexandroff-Hopf topology.

(a) If m = EE, then M = OO and SD = EO ∪OE.

(b) If m = EO, then M = OE and SD = EE ∪OO.

(c) If m = OE, then M = EO and SD = EE ∪OO.

(d) If m = OO, then M = EE and SD = EO ∪OE.

Proof. (a) Directly from Theorem 2.3.23 part(i).

(b) Let (x, y) ∈ EO, then (x, y + 1) ∈ N4(x, y). Hence (x, y + 1) ∈ SD. Since

(x, y + 1) ∈ EE, then by Theorem 2.3.23 part(iii), M = OE, SD = EE ∪OO.

(c) Let (x, y) ∈ OE, then (x+1, y) ∈ N4(x, y). So (x+1, y) ∈ SD. Since (x+1, y) ∈ EE,

then by Theorem 2.3.23 part(iii), M = EO, and SD = EE ∪OO.

(d) Let (x, y) ∈ OO, then (x + 1, y + 1) ∈ N8(x, y) \ N4(x, y). So by Proposition

2.3.4 part(3), (x + 1, y + 1) ∈ M . Since (x + 1, y + 1) ∈ EE, then by Theorem

2.3.23 part(ii), M = EE and SD = EO ∪OE.

It is clearly that if Z2 has the topology τs and (x, y) ∈ Z2, such that (x, y) is a maximal

or a minimal point, then we can determined the type (maximal, minimal, saddle) of any

point in Z2. Moreover, if (x, y) is a saddle point in Z2, then we have at most two cases for

any point in Z2. In fact there exist four topologies in Z2. All of them can be considered

as a τs topology, and all of them are homeomorphic to each others.

51

Remark 2.3.25. (1) By Corollary 2.3.24, we consider the Alexandroff-Hopf topology with

m = EE, M = OO and SD = EO ∪OE.

(2) The other three types of homeomorphic topologies on Z2 are also called Alexandroff-

Hopf topology on Z2. But without mention, the Alexandroff-Hopf topology on Z2 is the

topology with minimal points EE, and when we take any one of the three types as an

Alexandroff-Hopf topology on Z2, we must say so.

(3) If Z2 is the Alexandroff-Hopf topology and (a, b) ∈ Z2, then we have the following:

(a) If (a, b) ∈ OO, then ↑ (a, b) = (a, b) = a × b.

(b) If (a, b) ∈ EE, then ↑ (a, b) = N8(a, b)∪(a, b) = a−1, a, a+1×b−1, b, b+1.

(c) If (a, b) ∈ EO, then ↑ (a, b) = (a− 1, b), (a, b), (a+ 1, b) = a− 1, a, a+ 1 × b.

(d) If (a, b) ∈ OE, then ↑ (a, b) = (a, b− 1), (a, b), (a, b+ 1) = a × b− 1, b, b+ 1.

Definition 2.3.26. Let (Z, τkh) be the Khalimsky space. The product Khalimsky space

on Z2 denoted by τp is the topology on Z2 whose minimal neighborhood of each elements

(a, b) ∈ Z2 is ↑kha × ↑

khb.

Theorem 2.3.27. Let (Z, τkh) be the Khalimsky space and (Z2, τs) be the Alexandroff-Hopf

topology on Z2. Then Z2 = Z × Z, i.e τs = τp.

Proof. Firstly the two topologies on Z2; τs and τp are T0-Alexandroff on Z2. For all

(a, b) ∈ Z2, ↑p (a, b) = ↑kh a × ↑kh b, if a is even, ↑kha = a− 1, a, a+ 1 and if a is odd,

↑kha = a. Now if (a, b) ∈ OO, then ↑p (a, b) = ↑

kha × ↑

khb = (a, b) = ↑s (a, b). If

(a, b) ∈ EE, then ↑p (a, b) = ↑kh a × ↑kh b = a− 1, a, a+1×b− 1, b, b+1 = ↑s (a, b).

If (a, b) ∈ EO, then ↑p (a, b) = ↑kha × ↑

khb = a − 1, a, a + 1 × b = ↑s (a, b). If

(a, b) ∈ OE, then ↑p (a, b) = ↑kh a × ↑kh b = a × b− 1, b, b+ 1 = ↑s (a, b).

52

Remark 2.3.28. The other three homeomorphic types of the Alexandroff-Hopf topologies

are the product of (Z, τkh)× (Z, τ d

kh), (Z, τ d

kh)× (Z, τ

kh) and (Z, τ d

kh)× (Z, τ d

kh).

Fore example, in the Hopf topology (Z, τ d

kh)× (Z, τ

kh), m = OE.

Let Z2 be the Alexandroff-Hopf space, then we can consider a part of Z2 as the

following (see Fig. 2.12):

Figure 2.12: Part of Alexandroff-Hopf topology on Z2

Also we can consider this part using Hasse diagram ( see Fig. 2.13):

Figure 2.13: Part of Alexandroff-Hopf topology on Z2

53

Description of Specialization Order of Alexandroff-Hopf Topol-

ogy on Z2

Suppose that ≤s is the induced order on Z2 by the Alexandroff-Hopf topology Previously,

we denoted this order by ” → ” symbol. That is, ”(a, b) ≤s (c, d)” ≡ ”(a, b) → (c, d)”.

According to this notation, we can use Theorem 2.3.27 and Theorem 1.2.14, to get that

(a, b) ≤s (c, d) iff a ≤khc and b ≤

khd. For this end and for a point (x, y) ∈ OE, consider

N8(x, y) = (x− 1, y− 1), (x− 1, y), (x− 1, y+1), (x, y− 1), (x, y+1), (x+1, y− 1), (x+

1, y), (x+1, y+1). Since x is odd and y is even, then x is a maximal and y is a minimal

in Zkh. Thus we have that x ≤

khx, y ≤

khy − 1, y ≤

khy and y ≤

khy + 1. Therefore

(a, y) ≤s (x, y) where a ∈ x− 1, x, x+1, (x, y) ≤s (x, b) where b ∈ y− 1, y, y+1 and

(x, y) is incomparable with (r, s) for all r ∈ x− 1, x + 1 and s ∈ y − 1, y + 1. which

is clearly that the same result we got above. In general, and for (x, y) ∈ Z2, we have four

cases:

Case(1) : (x, y) ∈ OO, (x, y) ≤s (a, b) iff (x, y) = (a, b) iff x ≤Kh

a and y ≤Kh

b iff x = a

and y = b.

Case(2) : (x, y) ∈ EE, (x, y) ≤s (a, b) iff x ≤Kh

a and y ≤Kh

b iff a ∈ x − 1, x, x + 1

and b ∈ y − 1, y, y + 1.

Case(3) : (x, y) ∈ EO, (x, y) ≤s (a, b) iff x ≤Kh

a and y ≤Kh

b iff a ∈ x − 1, x, x + 1

and b = y.

Case(4) : (x, y) ∈ OE, (x, y) ≤s (a, b) iff x ≤Kh

a and y ≤Kh

b iff a = x and b ∈

y − 1, y, y + 1.

Remark 2.3.29. For the other types of Alexandroff-Hopf topology on Z2, we may use

the order of the Khalimsky dual topology ≤d

khto description them. For example in the

topology that has EO as minimal set (a, b) ≤s (c, d) iff a ≤Khc and b ≤d

Khd.

Theorem 2.3.30. If B = Z2 \ OO, then the relative topology on B with respect to

54

Alexandroff-Hopf topology is equal to the relative topology on B with respect to Marcus-

Wyse topology. That is, (B,≤s |B) = (B,≤m |B).

Proof. Suppose that U is a nonempty open set in (B, τm). Let x = (x1, x2) ∈ U . If

x ∈ EE, then (↑m x) ∩ B = N4(x) = (↑s x) ∩ B ⊆ U (since U is open in (B, τm)). If

x ∈ EO ∪ OE, then (↑m x) ∩ B = x = (↑s x) ∩ B ⊆ U (since U is open in (B, τm)).

Thus, (↑s x) ∩ B ⊆ U for all x ∈ U . So U is open in (B, τs) and τm|B ⊆ τs |B . Similarly

we prove that τs |B ⊆ τm |B . Hence τm |B = τs |B .

Corollary 2.3.31. If C = Z2 \ EE, then the relative topology on C with respect to

the Alexandroff-Hopf topology is equal to the relative topology on C with respect to the

Marcus-Wyse dual topology.

Proof. It suffices to note that the relative topology on C with respect to the dual Alexandroff-

Hopf topology is equal to dual relative topology on C with respect to Alexandroff-Hopf

topology. Then by Theorem 2.3.30, τm|C = τs |C .

Theorem 2.3.32. Let Z2 be the Alexandroff-Hopf space, and let a ∈ Z be a fixe integer.

If B1, B2 ⊆ Z2 such that B1 = (x, a) : x ∈ Z and B2 = (a, x) : x ∈ Z, then B1, B2

are homeomorphic to the Khalimsky topology.

Proof. Let f :B1 −→ Z, such that f(x, a) = x. it is clearly that f is a bijective function.

If x is odd number, then ↑s (x, a)∩B1=(x, a). Since ↑KHx=x, then f(↑s (x, a)∩B)

= ↑Khf(x, a). If x is even number, then ↑s (x, a)∩B1=(x, a), (x−1, a), (x+1, a). Since

↑KH

x=x, x+1, x− 1, then f(↑s (x, a)∩B1) = ↑Khf(x, a). Hence B1 is homeomorphic

to the Khalimsky topology. Similarly for B2.

Recall that ∀i ∈ Z, S2i = (a, b) : a+ b = i and S2 = S2

i : i ∈ Z.

55

Definition 2.3.33. We define the binary relation ”.” on the summation family set S2

by: S2i . S2

j if and only if whenever y ∈ S2j there exist x, z ∈ S2

i such that x ≤s y ≤s z.

For example S24 . S2

5 . To see this, suppose that y = (y1 , y2) ∈ S25 . Then y1 + y2 = 5 and

so y ∈ EO ∪OE. WLOG, suppose that y ∈ EO. Take x1 = y1 , x2 = y2 − 1, z1 = y1 − 1

and z2 = y2 . Then x = (x1 , x2) ∈ EE, z = (z1 , z2) ∈ OO and x ≤s y ≤s z.

Lemma 2.3.34. The summation family set S2 with the binary relation ”.” is a partially

ordered set (poset).

Proof. Let S2i , S

2j , S

2r ∈ S2 for i, j, r ∈ Z. If x ∈ S2

i , then x ≤s x ≤s x, and hence S2i . S2

i .

So ”.” is reflexive. Suppose that S2i . S2

j and S2j . S2

i . Let y ∈ S2j . Then there exist

x, z ∈ S2i such that x ≤s y ≤s z. If y ∈ OO, then y is a maximal point and hence y =

z. Then S2i = S2

j ( since S2i ∩ S2

j = ϕ ∀i = j). If y ∈ EE, then y is a minimal point.

Hence x = y. Therefore S2i = S2

j . Now if y ∈ EO ∪OE = OD2 , assume to contrary that

S2i = S2

j . Then x ∈ EE and z ∈ OO. Thus x ∈ m and y ∈M . Since x ∈ S2i and S2

j . S2i ,

then ∃ a, b ∈ S2j , such that a ≤s x ≤s b, which is a contradiction since x is a minimal point

and a = x. Hence S2i = S2

j and ”.” is anti-symmetry. Finally suppose that S2i . S2

j

and S2j . S2

r . Let y ∈ S2r , then ∃ x, z ∈ S2

j such that x ≤s y ≤s z. Since x, z ∈ S2j and

S2i . S2

j , then ∃ a, b ∈ S2i such that a ≤s x ≤s y ≤s z ≤s b. Hence a ≤s y ≤s b. Thus

S2i . S2

r and ”.” is transitive.

There exists a T0-Alexandroff topology on S2 induced by the order .. Let we denote

it by τ≈ , and denote the up of any element S2i in S2 by ↑≈ S2

i . Then we have the following

theorem:

Theorem 2.3.35. (S2, τ≈) is homeomorphic to the Khalimsky topology.

Proof. Let f : S2 −→ Z such that f(S2i ) = i. From the fact that S2

i ∩ S2j = ϕ ∀ i = j, we

get that f is one to one and onto. Consider the following two cases:

56

Case(1): If i is odd, then S2i ⊆ EO∪OE. Assume that S2

i . S2j , and suppose to contrary

that S2i = S2

j . If j is odd number, then S2j ⊆ EO∪OE. Let y ∈ S2

j , then ∃ x, z ∈ S2i such

that x ≤s y ≤s z, which is contradiction since x, y are two distinct saddle points. If j is

even number, then S2j ⊆ OO ∪ EE. Let y ∈ S2

j , then ∃ x, z ∈ S2i such that x ≤s y ≤s z.

If y ∈ OO then y is a maximal point and y = z ∈ OO which is contradiction. If y ∈ EE,

then y is a minimal point, and so x = y ∈ EE. Again this is a contradiction. Hence S2i

= S2j and S2

i is maximal. So if i is an odd number, then ↑≈ S2i = S2

i . Hence f(↑≈ S2i )

= f(S2i ) = i = ↑

Khi.

Case(2): If i is even, then S2i ⊆ OO ∪ EE. Suppose that S2

i . S2j and i = j. Let y =

(y1 , y2) ∈ S2j . So that y1 + y2 = j. Now there exist x = (x1 , x2), z = (z1 , z2) ∈ S2

i ⊆

OO∪EE such that x ≤s y ≤s z. If x, z are both in EE or both in OO, then both x, z are

minimal or both are maximal and hence x = y ∈ S2j which is a contradiction. If x ∈ OO

and z ∈ EE, then x is a maximal and z is a minimal which is also a contradiction since

x ≤s z. So we must have x ∈ EE, z ∈ OO and y is a saddle point. Since (x1 , x2) =

x ≤s y = (y1 , y2), then the possible values of y are (x1 − 1, x2), (x1 + 1, x2), (x1 , x2 − 1)

and (x1 , x2 + 1). Thus, j = y1 + y2 = x1 + x2 ± 1 = i ± 1. Therefore S2i . S2

j for

j ∈ i−1, i, i+1. So if i is an even number, then ↑≈ S2i = S2

i−1, S2i , S

2i+1, hence f(↑≈ S2

i )

= f(S2i−1, S

2i , S

2i+1) = i, i − 1, i + 1 = ↑

Khi. Therefore (S2, τ≈) is homeomorphic to

the Khalimsky topology.

Definition 2.3.36. The 2-Alexandroff-Hopf function ψ2 : F2 −→ Z2 is a bijection func-

tion define as:

ψ2(f) =

(2a,2b), if dim(f) = 0 and f = (a, b);

(2a+1,2b), if dim(f) = 1 and f = a, a+ 1 × b;

(2a,2b+1), if dim(f) = 1 and f = a × b, b+ 1;

(2a+1,2b+1), if dim(f) = 2 and f = a, a+ 1 × b, b+ 1.

Using the 2-Alexandroff-Hopf function and Theorem 1.2.5, we have the following theorem.

57

Theorem 2.3.37. The Alexandroff-Hopf topology (Z2, τs) is homeomorphic to the cellular-

complex topology (F2, τ⊆).

Theorem 2.3.38. Let Z2 be the Alexandroff-Hopf Space, then:

(1) Z2 is a semi-T2-space.

(2) Z2 is not submaximal space and hence it is not a T 14-space.

Proof. (1) Let x = (a, b) ∈ Z2 \ OO. If x ∈ EE. Then x = (a − 1, b − 1), (a + 1, b −

1), (a− 1, b+ 1), (a+ 1, b+ 1). If x ∈ EO, then x = (a− 1, b), (a+ 1, b). And if

x ∈ OE, then x = (a, b− 1), (a, b+1). Hence |x| ≥ 2 ∀x /∈M . Then by Theorem

1.2.11, Z2 is a semi-T2-space.

(2) Since (2, 3) ∈ Z2 and (2, 3) is neither maximal nor minimal point, then by Theorem

1.2.8, Z2 is not submaximal space and hence it is not a T 14-space.

58

Chapter 3

Digital Topologies on Z3

In this chapter, we will study the digital topologies on Z3. In [35] it was proved that

there are five digital topologies satisfy the three conditions 13, 2

3and 3

3. Also in [35],

it was claimed that there exist infinitely digital topology on Z3 which satisfy the two

conditions 13, 2

3. Hence they added the third condition to get the useful topologies which

used in the applications of the digital topology. We will use the same conditions to have

the same result using another way. After that we will study each of these five topology

separately and discuss their topological properties. Moreover we will find the minimal

neighborhood for each point in Z3 with respect to each topology of these five topologies.

One of the five topology will be the Marcus -Wyse topology and one of them will be the

cellular-complex(Allexandroff-Hopf) topology.

3.1 Five Topologies on Z3

Simply in Z3, we denote that EEE to be the set E×E×E of all triple points (x, y, z) ∈ Z3

such that x, y and z are even numbers. In similar way, we have the sets EEO, EOE,

OEE, EOO, OEO, OOE and OOO. For example, the set OEO = (x, y, z) ∈ Z3

59

: y is even number and x, z is odd numbers . Define the collection A to be A =

EEE,EEO,EOE,OEE, EOO,OEO,OOE,OOO. Note that Z3 =∪

A∈AA.

Definition 3.1.1. An elementary cube in Z3 is a set consists of a point (i, j, k) ∈ Z3

together with all of its neighbors of the form (r, s, t) ∈ Z3 such that r ∈ i, i + 1,

s ∈ j, j +1, t ∈ k, k+1. The point (i, j, k) is called the generated point for the cube.

From this definition note that any elementary cube contains 8 -points and has ex-

actly one generated point. Moreover any point (x, y, z) ∈ Z3 is a generated point for

exactly one cube. Therefore we can denote any cube by its generated point in the sym-

bol CU(x, y, z). For example, the cube CU(2, 3, 1) is the set of 8 -points CU(2, 3, 1) =

(2, 3, 1), (2, 3, 2), (2, 4, 1), (2, 4, 2), (3, 3, 1), (3, 3, 2), (3, 4, 1), (3, 4, 2), where its generated

point is (2, 3, 1). The elements of any elementary cube are called the vertices of the

cube. Clearly that if xi= (a

i

1, a

i

2, a

i

3), i = 1,2,· · · ,8 are the vertices of the elementary

cube CU(xj), then

∑3

r=1a

j

r<

∑3

r=1a

i

rfor all i = j. Note that in Z3 the 2d-adjacent and

3d − 1-adjacent are translate to 6-adjacent and 26-adjacent respectively. Observe that

the elementary squares in Z2 (cubes in Z3) are the largest subsets with the property that

each two points in them are 8-adjacent (26-adjacent). By this condition, it is possible to

define n-dimensional cells for n = 2,3,· · · .

Definition 3.1.2. [35] An n-dimensional cells in Zd is the set whose vertices consist of a

point (i1 , i2 , · · · , id) ∈ Zd and all its neighbors of the form (r1 , r2 , · · · , rd) ∈ Zd such that

rj∈ i

j, i

j+ 1, j = 1, 2, · · · , d.

It should be noted that if d = 2 (resp. d = 3), then an elementary square (resp. elementary

cube) is a 2-dimensional (resp. 3-dimensional) cell.

Remark 3.1.3. If V is an elementary cube, then V ∩ A = ϕ for each A ∈ A.

To see this, suppose that the generated point of V is (i, j, k). Without loss of generality,

suppose that (i, j, k) ∈ OEO. Then (i + 1, j + 1, k) ∈ V ∩ EOO and so V ∩ EOO = ϕ.

60

Similarly V ∩A = ϕ for each A ∈ A. In fact, any element cube in Z3 intersects A (A ∈ A)

in exactly one point.

Recall that a point x ∈ Z3 is a maximal (resp. minimal) with respected to a digital space

satisfies 13and 2

3if y → x (resp. x → y) ∀y ∈ N6(x). x is a saddle point if x is neither

maximal nor minimal.

Definition 3.1.4. A saddle point x ∈ Z3 is called an α−saddle point (resp. a β−saddle

point) if there exist four direct neighbors x1, x2, x3, x4 ∈ N6(x) such that xi → x (resp.

x→ xi) for i = 1,2,3,4.

We denote the set of all α-saddle points by α-S and the set of all β-saddle points by β-S.

Using Lemma 2.1.18, it is not difficult to prove that the set of all saddle points is the

union of the two sets α-S and β-S.

It is clearly that any vertex of an elementary cube has exactly only three direct neighbors

vertex.

Lemma 3.1.5. Let x be a point in Z3. Suppose that x1, x2, x3 are the three direct neighbors

vertex of x in an elementary cube in Z3. Then with respect to any digital space on Z3, we

have the following:

(1) x is a maximal point on Z3 if and only if xi → x ∀ i =1,2,3.

(2) x is a minimal point on Z3 if and only if x→ xi ∀ i =1,2,3.

(3) x is an α-saddle point on Z3 if and only if x→ x1 and xi → x, i =2,3.

(4) x is a β-saddle point on Z3 if and only if x1 → x and x→ xi, i =2,3.

Proof. (1) (⇒) Let x be a vertex in an elementary cube, and assume that x1, x2, x3 be

the three direct neighbors of x. By Lemma 2.1.18, xi → x ∀ i =1,2,3.

(⇐) There are three points x′1, x

′2, x

′3 in N6(x) such that ∀ i =1,2,3 x

′i, x, xi are on the

61

same grid line and x′i = xi. By Theorem 2.1.12, x

′i → x ∀ i =1,2,3. Hence by Lemma

2.1.18, x is a maximal point on Z3. Similarly for parts (2),(3) and (4).

When we investigate the first and the second-dimensional digital spaces we search

about the topologies that satisfy the two conditions 1dand 2

d. But we claimed in Chapter

two that there exist infinitely many topologies on Z3 that fulfilling the two conditions 13

and 23. To prove this, consider the two following examples (Note that Example 3.1.6 is

similar to an example in [35] which was without prove):

Example 3.1.6. Define a topology τ on Z3 to be the smallest neighborhood base as the

following: the smallest neighborhood for each x ∈ EV3 is given by V (x) = x, and the

smallest neighborhood for each y ∈ OD3 is given by V (y) = y ∪ (N26(y) ∩ EV3). Then

(Z3, τ) satisfies the two conditions 13and 2

3.

Proof. To prove that Z3 satisfies 13, suppose to contrary that (Z3, τ) dose not satisfy

the condition 13. Then there exists a nonempty set U ⊆ Z3 which is 6-connected but

not (topologically) connected. Therefore there exists a proper subset A from U which is

clopen with respect to the relative topology τU. Since A is a proper subset from U , then

∃ x, y such that x ∈ U \ A and y ∈ A. Since U is 6-connected set, there is a sequence of

points x = x1, x2, · · · , xk, xk+1 = y in U , such that xi is 6-adjacent to xi−1 for each i =

2,3,4,· · · ,k+1.

Case(1): If y ∈ EV3, then xk ∈ OD3 and y ∈ ↑U (xk). So xk ∈ ↓U (y). Then xk ∈ A (since

A is closed with respect to the relative topology τU). Since xk−1 ∈ N6(xk), xk−1 ∈ EV3

and xk−1 ∈ ↑U xk. Hence xk−1 ∈ A (since A is open with respect to the relative topology

τU). Continuous this process k times to get that x ∈ A which is contradiction.

Case(2): If y ∈ OD3, then xk ∈ EV3 and xk ∈ ↑U y. So xk ∈ A. Similarly to Case(1), we

can prove that x ∈ A which is a contradiction. So (Z3, τ) satisfies the condition 13.

62

Now we are going to prove that (Z3, τ) is satisfy the condition 23. Let U ⊆ Z3 is a

nonempty set which it is not 26-connected, then there exist x, y ∈ U such that in U there

exist no 26-path from x to y. Let A = u ∈ U : there exist a 26-path from u to x and

let B =∪

a∈A(N26(a) ∩ U). Hence y /∈ B and B is a proper subset from U . Let b ∈ B.

If b ∈ EV3, then ↑U b = b ⊆ B. If b ∈ OD3, then ↑U b = b ∪ (N26(b) ∩ EV3 ∩ U)

⊆ b ∪ (N26(b) ∩ U) ⊆ B. Therefore ↑Ub ⊆ B for all b ∈ B. So B is open set with

respect to the relative topology τU. Let r ∈ U \B. If r ∈ EV3, then ↑U r = r ⊆ U \B.

If r ∈ OD3, then ↑U r = r∪(N26(r)∩EV3∩U) ⊆ r∪(N26(r)∩U) ⊆ U \B. Therefore

B is closed set with respect to the relative topology τU. Then U is not connected.

Example 3.1.7. For any x ∈ Z3 define A(x) = N26(x) ∪ (0, 0, 0). Let we define a

topology τ on Z3 by the smallest neighborhood base as the following: the smallest neigh-

borhood for each x ∈ EV3 is given by V (x) = ↑ x = x, and the smallest neighborhood

for each y ∈ OD3 is given by V (y) = ↑ y = y ∪ (A(y) ∩ EV3), then (Z3, τ) satisfy the

two conditions 13and 2

3.

Proof. Is similar to the proof in the previous example.

We can generate in this way an infinite number of different topologies by redefine A(x)

to be the union of N26(x) and some elements from EV3.

Now we are going to add a conditions 33to the two conditions 1

3, 2

3for the digital space

Z3. The three conditions 13, 2

3, 3

3translate into:

(13) If a set in Z3 is 6-connected, then it is (topologically) connected.

(23) If a set in Z3 is not (26)−connected, then it is not (topologically) connected.

(33) For each x ∈ Z3 the smallest open set containing x is 6-connected.

Observation 3.1.8. [35] Given a topology in Z3 fulfilling conditions 13and 2

3. Then the

relative topology induced in each principal plane is a topology of Z2 fulfilling conditions

63

12and 2

2. This because the intersection of this plane and N6(x) in Z3 is N4(x) in the

plane, and the intersection of this plane and N26(x) in Z3 is N8(x) in the plane for each

x.

In any elementary cube and for any two direct neighbor vertices a and b, we denote (a, b)

to be side join the two vertices a and b in the geometric representation.

Figure 3.1: An Elementary cube

Remark 3.1.9. In any elementary cube, the direction of arrows on the sides of one face

and on the fourth sides which are geometrically perpendicular to this face determine the

direction of arrows on all sides of this elementary cube.

To see this, consider an elementary cube as shown in the Figure 3.1 with generated

point vertex 1. Suppose we have the direction of arrows on the sides of the face whose

vertices are 1,4,5 and 8 and on the fourth sides (1, 2),(5, 6),(8, 7) and (4, 3). Then by

Observation 2.3.7, we can determine the direction of arrows on this elementary cube.

Theorem 3.1.10. [35] A digital topology on Z3 can be determined by the topology on an

elementary cube.

Proof. Let we determine the direction of arrows of an elementary cube, numeric its vertices

by 1,2,··,8 and called it the a-cube, and determine the cube which share the face whose

64

Figure 3.2: Two Adjacent Elementary Cube

vertices are 2,3,6 and 7 as shown in a Figure 3.2 and called it the b-cube. It is clearly that

a-cube is CU(1) and the b-cube is CU(2). By Theorem 2.1.14, if the direction of arrows

on the sides of a-cube are diteremined we can determine the direction on (6, 11), (7, 12),

(2, 9) and (3, 10). Hence by Observation 3.1.9, the direction of arrows on all sides of the

b-cube are determined. This way implies that a topology on single cube determines the

topology on a line of cubes in which each cube shares a side with exactly two others which

we will call it the a-line cube. Now consider the cube which share the fourth points 5,6,7

and 8 the c-cube. Clearly, the c-cube is CU(5). As the previous way we can determine

the topology on the c-cube, and hence the topology on a line of cubes in which contain

the a-cube and the c-cube. By doing the last step to each cube in the a-line cube we get

the whole space Z3.

From this theorem, any topology on any elementary cube that satisfies 13, 2

3and 3

3

determines a topology of Z3 that satisfies 13, 2

3and 3

3.

Lemma 3.1.11. Each elementary cube in Z3 has at least one maximal point.

Proof. Numeric an elementary cube vertices 1, 2, · · · , 8 as in Figure 3.1. Since the points

1 and 2 are direct neighbors then either 1 → 2 or 2 → 1. Without forget the generality

65

of proof suppose that 1→ 2, so we have four cases :

Case(1): If 2 → 3 and 6 → 2, then by Observation 2.3.8, 7 → 3 and 4 → 3. Thus the

vertex 3 is a maximal point.

Case(2): If 2→ 3 and 2→ 6, then by Observation 2.3.8, 5→ 6. If 7→ 6, then the vertex

6 is maximal point. If 6 → 7, then by Observation 2.3.7, 3 → 7, and by Observation

2.3.8, 8→ 7 so the vertex 7 is a maximal point.

Case(3): If 3→ 2 and 2→ 6, then by the same way in Case(1) we have the vertex 6 is a

maximal point.

Case(4): If 3 → 2 and 6 → 2, then the vertex 2 is a maximal point. So any elementary

cub in Z3 has at least one maximal point.

Theorem 3.1.12. [35] There are five topologies on the elementary cube that are satisfying

the conditions 13, 23 and 33.

Proof. By previous lemma there exists a maximal point in any elementary cube. Up to

rotations and translations let this maximal point be the generated point which is a vertex

1 in Figure 3.1.

Case(1): If the vertex 2 is a minimal point, then by Observation 2.3.7, 4→ 3 and 5→ 6

(see Figure 3.3 (i)). In this case either 7 → 3 or 3 → 7. Then we have more than one

possible case. So consider the following subcases:

Subcase(i): If 7→ 3 and 7→ 8 (see Figure 3.3(ii)), then by Observation 2.3.7, 7→ 6,

5 → 8 and 4 → 8 (see Figure 3.3(iii)). This determine a topology on the cube that has

four maximal points which are 1, 3, 6, 8 and four minimal points which are 2, 4, 5, 7.

Subcase(ii): If 7→ 3 and 8→ 7 (see Figure 3.4(i)), then by Observation 2.3.7, 7→ 6,

8 → 4 and 8 → 5 (see Figure 3.4(ii),(iii)). This determine a topology on the cube that

has three maximal points 1, 3, 6, three β-saddle points 4, 5, 7 and two minimal points

2, 8.

66

Figure 3.3: Elementary cube with four maximal points

Figure 3.4: Elementary cube which has three maximal points

Subcase(iii): If 3 → 7 (see Figure 3.5(i)), then by Observation 2.3.8, 8 → 7, and hence

by Observation 2.3.7, 6→ 7, 4→ 8 and 5→ 8 (see Figure 3.5(ii)). This also determine a

topology on the cube that has two maximal points 1, 7, three α-saddle points 3, 6, 8

and three minimal points 2, 4, 5.

Case(2): If the vertex 2 is an α-saddle point, then 3→ 2 and 6→ 2 (see Figure 3.6(i)).

Therefore by Observation 2.3.7, 3 → 4 and 6 → 5 (see Figure 3.6(ii)). In this case

consider the following subcases:

Subcase(i): If 3 → 7 and 8 → 7 (see Figure 3.6 (iii)). This implies that 6 → 7, 8 → 5

67

Figure 3.5: Elementary cube with two maximal points

and 8 → 4 (see Figure 3.6 (iv)). This determine a topology on the cube that has two

maximal points 1, 7, three α-saddle points 2, 4, 5 and three minimal points 3, 6, 8.

This topology is the same topology in Subcase(iii) of Case 1 (up to homeomorphic).

Subcase(ii): If 3 → 7 and 7 → 8 (see Figure 3.7 (i)), then 6 → 7, 5 → 8 and 4 → 8 (see

Figure 3.7 (ii)). This determine a new topology on the cube that has two maximal points

1, 8, two α-saddle points 2, 7, two β-saddle points 4, 5, and two minimal points

3, 6.

Subcase(iii): If 7 → 3 (see Figure 3.7 (iii)). By Observation 2.3.7, 7 → 6. Hence by

Observation 2.3.8, 8→ 5. Therefore by Observation 2.3.7, 8→ 4 and 7→ 8 (see Figure

3.7 (iv)). This determine a new topology on the cube that has one maximal point 1,

three α-saddle points 2, 4, 5, three β-saddle points 3, 6, 8, and one minimal point 7.

Case(3): If the vertex 2 is a β-saddle point, then we have the following subcases :

Subcase(i): If 2 → 6 and 3 → 2 and 7 → 8 (see Figure 3.8(i)). Then by Observation

2.3.8, 7 → 6. Hence by Observation 2.3.7, 3 → 7, 3 → 4, 4 → 8, 5 → 6 and 5 → 8

(see Figure 3.8 (ii)). This topology is same topology in subcase (ii) of case 1 (up to

homeomorphic) since it has three maximal points 1, 6, 8, three β-saddle points 2, 4, 7

and two minimal points 3, 5.

68

Figure 3.6: Elementary cube with three α-saddle and no β-saddle points

Subcase(ii): If 2 → 6 and 3 → 2 and 8 → 7 (see Figure 3.8 (iii)), then by Observation

2.3.8, 7 → 6. So by Observation 2.3.7, 3 → 7, 5 → 6, 8 → 5, 8 → 4 and 3 → 4

(see Figure 3.8(iv)). In this topology there exist two maximal points 1, 6, two α-saddle

points 4, 7, two β-saddle points 2, 5, and two minimal points 3, 8. So this topology

is the same topology in subcase (ii) of case 2.

Subcase(iii): If 6 → 2 and 2 → 3 and 7 → 8 (see Figure 3.9(i)), then 7 → 3, 6 → 7,

4 → 3, 4 → 8, 5 → 8 and 6 → 5 (see Figure 3.9(ii)). This topology has three maximal

points 1, 3, 8, three β-saddle points 2, 5, 7 and two minimal points 4, 6. Thus this

topology is the same topology in subcase (ii) of case 1.

Subcase(iv): If 6 → 2 and 2 → 3 and 8 → 7 (see Figure 3.9(iii)), then 7 → 3, 6 → 7,

69

Figure 3.7: Elementary cube with α and β-saddle points

4 → 3, 8 → 4, 8 → 5 and 6 → 5 (see Figure 3.9(iv)). This topology has two maximal

points 1, 3, two α-saddle points 5, 7, two β-saddle points 2, 4, and two minimal

points 6, 8. Therefore this topology is the same topology in subcase(ii) of case 2.

So up to homeomorphic there are only five different topology on the elementary cube

which satisfy the conditions 13, 23, 33. Using Theorem 3.1.10, we get the following theo-

rem.

Theorem 3.1.13. There are only five topology (up to the homeomorphic) on Z3 satisfy

the conditions 13, 23, 33.

We are going to study each one of them in the following sections.

70

Figure 3.8: Redendance cases

3.2 The Marcus -Wyse topology on Z3

As in chapter two, the space where its points are either maximal or minimal is called

the Marcus -Wyse space. The topology on Z3 induced by the subcase(i) of case 1 is

the Marcus -Wyse topology since any point of Z3 is either maximal or minimal. In this

section, we will study this topology on Z3 and denote it by τm .

Definition 3.2.1. Given a point x = (x1, x2, x3) ∈ Z3, the 18-neighbors of x (denoted by

N18(x)) are all points y = (y1, y2, y3) ∈ N26(x) such that∑3

i=1 |xi − yi| = j, j ∈ 1, 2,

the 12-neighbors of x (denoted by N12(x)) are all points z = (z1, z2, z3) ∈ N26(x) such

that∑3

i=1 |xi − zi| = 2, the 8-neighbors of x (denoted by N8(x)) are all points u =

(u1, u2, u3) ∈ N26(x) such that∑3

i=1 |xi − ui| = 3. In Figure 3.10, the black squares

denote N6(x), the white squares denote N12(x), the white and the black squares denote

to N18(x), the gray squares denote to N8(x), where the gray circle denote the point x.

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Figure 3.9: Redendance cases

Lemma 3.2.2. Let (Z3, τm) be the Marcus -Wyse topology space, where m and M are

respectively the sets of minimal points and maximal points in Z3. For x ∈ Z3, we have

the following:

(1) if x is a maximal point, then N6(x) ⊆ m and N12(x) ⊆M .

(2) if x is a minimal point, then N6(x) ⊆M and N12(x) ⊆ m.

(3) if (r, s, t) ∈ M (resp. (r, s, t) ∈ m), then (r + i, s, t), (r, s + i, t), (r, s, t + i) ⊆ M

(resp. (r + i, s, t), (r, s+ i, t), (r, s, t+ i) ⊆ m) for i ∈ −2, 2.

Proof. For part(1), let x be a maximal point. For y ∈ N6(x) if y is a maximal point then

x → y, which is a contradiction since x is a maximal point. Hence y is a minimal point

(any point in this topology is either maximal or minimal point). Therefore N6(x) ⊆ m.

Now let z ∈ N12(x), then there exists y ∈ N6(x) such that y ∈ N6(z). Since x is a maximal

point, then by above y is a minimal point. But z ∈ N6(y), then y → z. So z is not a

72

Figure 3.10: The Different Neighbors of x

minimal point. Therefore z is a maximal point. Thus N12(x) ⊆ M . Similarly we prove

part(2).

(3) Let (r, s, t) ∈ M . Then (r + 1, s, t) ∈ N6(r, s, t) and (r + 1, s, t) ∈ N6(r + 2, s, t).

Hence by parts (1) and (2) above if (r, s, t) ∈M (res.(r, s, t) ∈ m), then (r + 1, s, t) ∈ m

(res.(r+1, s, t) ∈M), and similarly (r+2, s, t) ∈M (res.(r+2, s, t) ∈ m). We can prove

the other points in similar way.

It is easy to show that EV3 = EEE ∪EOO ∪OEO ∪OOE and OD3 = OOO ∪EEO ∪

EOE ∪OEE. In the following theorem, we prove that one of the two sets EV3 and OD3

in equal to M and the other is equal to m.

Theorem 3.2.3. Let (Z3, τm) be the Marcus -Wyse topology space and let x ∈ Z3. Suppose

that x is a maximal point (resp. a minimal point).

(1) If x ∈ EV3, then M = EV3 and m = OD3 (resp. m = EV3 and M = OD3).

(2) If x ∈ OD3, then M = OD3 and m = EV3 (resp. m = OD3 and M = EV3).

Proof. (1) let x = (q1 , q2 , q3) be a maximal point in EV3. If e = (e1 , e2 , e3) ∈ N12(x),

then∑3

i=1ei=

∑3

i=1qi± 2. So N12(x) ⊆ EV3 and (x ∪ N12(x)) ∩ A = ϕ ∀ A ∈

EEE,EOO,OEO,OOE. Let y ∈ EV3, then y ∈ A0 for someA0 ∈ EEE,EOO,OEO,OOE.

73

Then there exists a0 = (x1, x2, x3) ∈ x∪N12(x) and a0 ∈ A0. Let r1(a0, y) = 2n, r2(a0, y)

= 2m and r3(a0, y) = 2r. Hence we can found a sequence of points a0 = (x1, x2, x3),

a1 = (x1 + 2i, x2, x3), a2 = (x1 + 4i, x2, x3),· · · , an = (x1 + 2ni, x2, x3), an+1 = (x1 +

2ni, x2 + 2j, x3), an+2 = (x1 + 2ni, x2 + 4j, x3),· · · ,an+m = (x1 + 2ni, x2 + 2mj, x3),

an+m+1 = (x1+2ni, x2+2mj, x3+2r), an+m+2 = (x1+2ni, x2+2mj, x3+4r),· · · ,an+m+k

=

(x1+2ni, x2+2mj, x3+2tr) = y such that i, j, r ∈ 1,−1. Then a0, a1, · · · , as ∈ A0; where

s= n+m+k. By the induction and by Lemma 3.2.2, if x is a maximal point (resp. minimal

point), then a0 is a maximal point(res. minimal point). Hence a1, a2, · · · , as = y ⊆M ,

(res. a1, a2, · · · , as = y ⊆ m). Therefore EV3 ⊆ M (resp. EV3 ⊆ m). Now we want to

prove that M ⊆ EV3 (resp. m ⊆ EV3). To prove this assume that z = (a, b, c) /∈ EV3,

so z = (a, b, c) ∈ OD3, but (a + 1, b, c) ∈ N6(z) and (a + 1, b, c) ∈ EV3 ⊆ M (resp.

(a+1, b, c) ∈ m). Hence z ∈ m (resp. z ∈M). So M ⊆ EV3 (resp. m ⊆ EV3). Therefore

M = EV3 (resp. m = EV3). Since in this topology any point is either maximal or mini-

mal, then m = OD3 (resp. M = OD3).

(2) Similarly to part (1).

There exist two topologies on Z3, each of them can be produced by the Subcase (i) of

case(1); one of them has m = EV3 and M = OD3, and the other one has m = OD3 and

M = EV3. In fact, each of them is the dual topology of the other and any one of them

can by the Marcus -Wyse dual topology on Z3. But by convention the first will be the

Marcus -Wyse topology on Z3 and the other will be the Marcus -Wyse dual topology on

Z3. The up of a point x in the Marcus -Wyse topology on Z3 is denoted by ↑m x with

respect to the Marcus -Wyse dual topology on Z3 is denoted by ↑dmx.

Theorem 3.2.4. The Marcus -Wyse topology on Z3 and the Marcus -Wyse dual topology

on Z3 are homeomorphic.

Proof. Let f :Z3 −→ Z3, such that f(x, y, z) = (x+1, y, z). It is clear that f is a bijective

74

function. Let the topology on the domain be the Marcus -Wyse topology, and on the rang

the Marcus -Wyse dual topology. If (x, y, z) ∈ OD3, then f(↑m (x, y, z)) = f((x, y, z))

= (x+1, y, z) = ↑dm(x+1, y, z) = ↑d

mf(x, y, z). If (x, y, z) ∈ EV3, then f(↑m (x, y, z)) =

f(N6(x, y, z)∪(x, y, z)) = N6(x+1, y, z)∪(x+1, y, z) = ↑dm(x+1, y, z) = ↑d

mf(x, y, z).

By Theorem 1.2.5, the Marcus -Wyse topology on Z3 is homeomorphic to the Marcus

-Wyse dual topology on Z3.

Theorem 3.2.5. In the Marcus -Wyse topology on Z3, the smallest open set containing

a minimal point x ∈ Z3 is equal to N6(x) ∪ x and the smallest open set containing a

maximal point y ∈ Z3 is equal to y.

Proof. Suppose that x ∈ Z3 is a minimal point and z ∈ N6(x). Then x → z. Hence

N6(x) ⊆ V (x). Since x ∈ V (x), N6(x)∪x ⊆ V (x). Let a ∈ V (x), then x→ a. Suppose

to contrary that a /∈ N6(x). Then there exists b ∈ N6(x) such that x → b → a. This

implies that b is a saddle point which is a contradiction (since each point in τm is either a

maximal or a minimal point). Hence V (x) ⊆ N6(x)∪x. Therefore V (x) = N6(x)∪x.

In any Alexandroff space if y is a maximal point, then V (y) = y.

In Figure 3.11, we give a part of the Hasse diagam of the Marcus -Wyse topology on

Z3 via poset.

Figure 3.11: Part of The Marcus -Wyse Topology on Z3

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Notation 3.2.6. In Figure 3.11, some of the up for some minimal points and some of the

down for some maximal points did not appear.

Since τm is a T0-Alexandroff space, it has a special order, denote it by ≤3

m. Using Theorem

3.2.5, if x ∈ EV3 = m, then ↑m x = V (x) = N6(x) ∪ x. And if x ∈ OD3 = M , then

↑m x = V (x) = x.

Theorem 3.2.7. Let (Z3,≤3

m) be the Marcus -Wyse topology on Z3, and let a, b be fixed

integer numbers.

(1) If A1=(a, x, y): x, y ∈ Z, A2=(x, a, y): x, y ∈ Z and A3=(x, y, a): x, y ∈ Z,

then for each i ∈ 1, 2, 3, Ai is homeomorphic to the Marcus-Wyse topology on Z2.

(2) If B1 = (a, b, x): x ∈ Z, B2 = (a, x, b): x ∈ Z and B3 = (a, b, x): x ∈ Z, then

for each i ∈ 1, 2, 3, Bi is homeomorphic to the Khalimsky topology.

Proof. (1) If a ∈ Z is a fixed even number, then take f : A1 −→ Z2 be such that

f(a, x, y) = (x, y). It is clear that f is a bijective function. If x + y is odd number,

then f(↑m (a, x, y) ∩ A1) = f((a, x, y)) = (x, y) = ↑m (x, y) = ↑m f(a, x, y). If

x + y is even number, then f(↑m (a, x, y) ∩ A1) = f((a, x, y), (a, x + 1, y), (a, x − 1, y),

(a, x, y + 1), (a, x, y − 1)) = N4(x, y) ∪ (x, y) = ↑m (x, y) = ↑m f(a, x, y).

If a ∈ Z is a fixed odd number, then take g : A1 −→ Z2 be such that g(a, x, y) = (x+1, y).

It is clear that g is a bijective function. If x+y is odd number, then g(↑m (a, x, y)∩A1) =

g((a, x, y), (a, x+1, y), (a, x−1, y), (a, x, y+1), (a, x, y−1)) = N4(x+1, y)∪(x+1, y)

= ↑m (x + 1, y) = ↑m g(a, x, y). If x + y is even number, then g(↑m (a, x, y) ∩ A1) =

g((a, x, y)) = (x + 1, y) = ↑m (x + 1, y) = ↑m g(a, x, y). By Theorem 1.2.5, A1 is

homeomorphic to the Marcus-Wyse topology on Z2. The prove of A2, A3 is similar to A1.

(2) If a+ b is even number, then let f : B1 −→ Z be such that f(a, b, x) = x and if a+ b

is odd number, then let g : B1 −→ Z be such that g(a, b, x) = x+ 1. The complement of

proof is similar to the proof of part (1) above.

76

Recall that the ith-summation set S3i on Z3 is defined by S3

i = (x1, x2, x3) ∈ Z3 :∑3

j=1xj = i and the summation family set S3 on Z3 is S3 = S3

i : i ∈ Z.

Definition 3.2.8. For any pair of elements S3i , S

3j ∈ S3, we say that S3

i - S3j if whenever

x ∈ S3i , there exists y ∈ S3

j such that x ≤3

my.

Lemma 3.2.9. The summation family set S3 with the binary relation ” - ” is a partially

ordered set (poset).

Proof. For S3i , S

3j , S

3r ∈ S3 for some i, j, r ∈ Z. Then:

(i) S3i - S2

i , since x ≤3

mx, ∀x ∈ S3

i . (” - ” is reflexive).

(ii) Suppose that S3i - S3

j and S3j - S3

i . If x = (x1, x2, x3) ∈ S3i , then ∃ y = (y1, y2, y3) ∈

S3j such that, x ≤3

my. If x ∈ OD3 (in the case when i is odd number), then x is a

maximal point, and hence x = y. Therefore S3i = S3

j . If x ∈ EV3 (in the case when i

is even number), then x is a minimal point, and hence y ∈↑m x = (x1, x2, x3), (x1 − 1,

x2, x3), (x1+1, x2, x3), (x1, x2− 1, x3), (x1, x2+1, x3), (x1, x2, x3− 1)(x1, x2, x3+1). This

implies that j ∈ i, i+1, i− 1. If j ∈ i+1, i− 1, then j is odd number. By above and

since S3j - S3

i , we have S3i = S3

j , which implies that i = j and this is impossible since i

is even and j is odd. Thus i = j and S3i = S3

j (” - ” is anti-symmetry).

(iii) Suppose that S3i - S3

j , and S3j - S3

r . If x ∈ S3i , then there exists y ∈ S3

j such that

x ≤3

my. Since S3

j - S3r and y ∈ S3

j , then there exists z ∈ S3r such that y ≤3

mz, then by

transitivity of ≤3

m”, we have x ≤3

mz. Hence S2

i - S2r (” - ”is transitivity ).

For the poset S3, let τξbe the corresponding T0-Alexandroff space on S3 induced by

the order -. We call it the summation topology on S3. For S3i , Vξ

(S3i ) = ↑ξ S3

i = S3j ∈ S3

: S3i - S3

j . When i is odd number, Vξ(S3

i ) = ↑ξ S3i = S3

i and when i is even number,

and S3i - S2

j , then i ∈ i − 1, i, i + 1. Thus Vξ(S3

i ) = ↑ξ S3i = S3

i−1, S3i , S

3i+1, and we

get the following theorem:

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Theorem 3.2.10. The summation topology τξon S3 is homeomorphic to the Khalimsky

topology.

Proof. Let f : S3 −→ Z be a function defined by f(S3i ) = i. It is clear that f is a bijective

function. If i is odd number, then f(↑ξS3i ) = fS3

i = i = ↑Khi = ↑

Khf(S3

i ). If i is even

number, then f(↑ξS3i ) = fS3

i−1, S3i , S

3i+1 = i− 1, i, i+ 1 = ↑

Khi = ↑

Khf(S3

i ). Then

by Theorem 1.2.5, the summation topology space S3 is homeomorphic to the Khalimsky

topology on Z.

Theorem 3.2.11. Let (Z3, τm) be the Marcus-Wyse topology on Z3, then

(1) Z3 is submaximal.

(2) Z3 is nodec.

(3) PO(Z3) = τα = τ(≤); that is, every preopen set is open.

(4) Z3 is T∗1 -space.

(5) Z3 is T 34

.

(6) Z3 is a semi-T2−space.

Proof. Similar to the proof of Theorem 2.2.4.

3.3 β-Space on Z3

In this section, we are going to study the topology induced on an elementary cube by the

subcase(ii) of case(1) and its related topology on Z3. We will call it by β-topology on Z3

and denoted it by τβ. This topology depends on the choice of the generated point, and

there are several types of homeomorphic β-Space on Z3, as we will see latter in Theorem

78

3.3.8. Recall that we say that x is covered by y or y covers x and we write x y or

y x if x < y and when x ≤ z < y then z = x. If we consider the β-topology τβ

on an elementary cube, we will find in each elementary cube two minimal points 2, 8,

three β-saddle points 4, 5, 7 and three maximal points 1, 3, 6. But one of the minimal

points 2 is covered by maximal points and the other is covered by β-saddle points 8.

The down of the maximal points is two β-saddle points and a minimal point (see the

Figure 3.12).

Figure 3.12: β-space on Elementary Cube

Definition 3.3.1. Let Z3 be a β-space. We define two subsetsm1 andm2 by the following:

m1 = x ∈ m : If x y, then y ∈ β-S, m2 = x ∈ m : If x y, then y ∈M .

It is clearly that m = m1 ∪m2 and m1 ∩m2 = ϕ.

Remark 3.3.2. Let Z3 be a β-space. If x ∈ m1, x y and z y, then z ∈ m1.

Proof. Let x ∈ m1 and x y. Then y ∈ β-S. Now if z y, then z ∈ m1.

Theorem 3.3.3. Let Z3 be a β-space and let x = (a, b, c) ∈ Z3. Then the following

statements are equivalents.

(1) x ∈ m1.

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(2) N6(x) ⊆ β-S.

(3) N12(x) ⊆M .

(4) N8(x) ⊆ m2.

(5) (a+ i, b+ j, c+ k) ∈ m2 for i, j, k ∈ −1, 1.

Proof. (1⇒ 2) Suppose that x ∈ m1. Let y ∈ N6(x), hence either y x or x y. But

x ∈ m, then y x. Therefore y ∈ β-S (since x ∈ m1). Then N6(x) ⊆ β-S.

(2 ⇒ 3) Suppose that N6(x) ⊆ β-S. Thus x ∈ m1 ∪M . Now let y ∈ N12(x). Then ∃

z ∈ Z3 such that z ∈ N6(y) and z ∈ N6(x) ⊆ β-S. By Theorem 2.1.12, either y z or

z y. Thereby y ∈ m1∪M . If y ∈ m1, then x ∈ β-S which is a contradiction. Therefore

y ∈M and N12(x) ⊆M .

(3⇒ 4) Suppose that N12(x) ⊆M and x = (a, b, c). Let y ∈ N8(x). Thus y = (a+ i, b+

j, c+ k) where i, j, k ∈ −1, 1. Define z1 = (a+ i, b+ j, c), z2 = (a+ i, b, c+ k) and z3 =

(a, b+ j, c+ k). Hence z1, z2, z3 ∈ N12(x) ⊆M and z1, z2, z3 ∈ N6(y). By Theorem 2.1.12,

y zr for all r = 1,2,3. Since there exists an elementary cube contains z1, z2, z3 and y,

y ∈ m2. Then N8(x) ⊆ m2.

(4⇒ 5) It is coming directly.

(5⇒ 1) Suppose that (a+ i, b+ j, c+ k) ∈ m2 for i, j, k ∈ −1, 1 and x = (a, b, c). Let y

= (a+i, b+j, c+k) where i, j, k ∈ −1, 1. Define z1 = (a+i, b+j, c), z2 = (a+i, b, c+k),

z3 = (a, b + j, c + k) and w = (a + i, b, c). Thus, z1, z2, z3 ∈ N6(y). By Theorem 2.1.12,

z1, z2, z3 ∈ M . w ∈ N6(z1) and w ∈ m2 ∪ β-S. w ∈ β-S since there exists an elementary

cube contains w and y and since in the β-space any elementary cube has only one point

contains in m2. But w ∈ N6(x), thus x ∈ m1 ∪M . Since there exists an elementary cube

contains z1, z2, z3 and x, then x ∈ m1 (since in the β-space any elementary cube has only

three maximal points).

80

Theorem 3.3.4. Let Z3 be a β-space, B ∈ m1,m2, β-S,M, and let x = (a, b, c) ∈ Z3.

If (a, b, c) ∈ B, then (a+ 2i, b+ 2j, c+ 2k) ∈ B, for all i, j, k ∈ Z.

Proof. If B =m1. Let (a, b, c) ∈ m1 and r ∈ −1, 1. Then (a+r, b+r, c+r) ∈ N8(x). By

Theorem 3.3.3, (a+r, b+r, c+r) ∈ m2 and (a+2r, b, c) = (a+r+r, b+r−r, c+r−r) ∈ m1.

By the induction (a + 2i, b, c) ∈ m1 for any i ∈ Z. By the same way, we can prove that

(a+ 2i, b+ 2j, c) ∈ m1, and after that (a+ 2i, b+ 2j, c+ 2k) ∈ m1, for all i, j, k ∈ Z.

If B = m2. Let (a, b, c) ∈ m1 and r ∈ −1, 1. Then (a + r, b + r, c + r) ∈ N8(x).

By Theorem 3.3.3, (a + r, b + r, c + r) ∈ m1 and (a + 2r, b, c) ∈ m2. By the induction

(a+2i, b, c) ∈ m2 for any i ∈ Z. By the same way, we can prove that (a+2i, b+2j, c) ∈ m2,

and after that (a+ 2i, b+ 2j, c+ 2k) ∈ m2, for all i, j, k ∈ Z.

If B = β-S. Let (a, b, c) ∈ β-S and r ∈ −1, 1. Then either (x + r, y, z) ∈ m1 or

(x + r, y, z) ∈ M . If (x + r, y, z) ∈ m1, then by Theorem 3.3.3, (x + 2r, y, z) ∈ β-S. If

(x+ r, y, z) ∈M , then either (x+ 2r, y, z) ∈ β-S or (x+ 2r, y, z) ∈ m1. By the above, if

(x + 2r, y, z) ∈ m1, then (x, y, z) ∈ m1 which is a contradiction. So (x + 2r, y, z) ∈ β-S.

By the induction (a + 2i, b, c) ∈ β-S for any i ∈ Z. By the same way, we can prove that

(a+ 2i, b+ 2j, c+ 2k) ∈ β-S, for all i, j, k ∈ Z.

If B = M . Let (a, b, c) ∈ M and r ∈ −1, 1. Then either (x + r, y, z) ∈ m2 or

(x+ r, y, z) ∈ β-S. If (x+ r, y, z) ∈ m2, then by Theorem 3.3.3, (x+2r, y+ r, z+ r) ∈ m1.

Hence (x+ 2r, y, z) = (x+ 2r, y + r− r, z + r− r) ∈M . If (x+ r, y, z) ∈ β-S, then either

(x + 2r, y, z) ∈ m1 or (x + 2r, y, z) ∈ M . If (x + 2r, y, z) ∈ m1, then (x, y, z) ∈ m1 which

is a contradiction. Hence (x+2r, y, z) ∈M . Therefore (a+2i, b+2j, c+2k) ∈M , for all

i, j, k ∈ Z.

Theorem 3.3.5. Let Z3 be a β-space, and A ∈ A. Then A ⊆ B, for some B ∈ m1, m2,

β-S, M.

Proof. Pick (x, y, z) ∈ A and fix it. Then there is B in m1, m2, β-S, M such that

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(x, y, z) ∈ B. By the definition of the set A, we can write it in the form (x + 2i, y +

2j, z + 2k) : i, j, k ∈ Z. So by Theorem 3.3.4, A ⊆ B.

Corollary 3.3.6. Let Z3 be a β-space, and let A ∈ A, B ∈ m1, m2, β-S,M. If A ∩ B

= ϕ, then A ⊆ B.

Proof. Suppose that A ∩ B = ϕ. By Theorem 3.3.5, A ⊆ C for some C ∈ m1,m2, β-

S,M. Hence B ∩ C = ϕ. Thus B = C since for any two different elements in m1, m2,

β-S, M are disjoint.

Corollary 3.3.7. Let Z3 be a β-space, then m1 ∈ A.

Proof. Let x = (a, b, c) ∈ Z3. Then CU(x) ∩ m1 = ϕ. So m1 = ϕ. Let y ∈ m1 ∩ Z3.

Thus y ∈ A for some A ∈ A. By Corollary 3.3.6, A ⊆ m1. So we need to prove m1 ⊆ A.

For this end, suppose to contrary that there exists b ∈ m1 such that b /∈ A. Consider the

elementary cube CU(b) and let U be the set of all vertices for CU(b). By Remak 3.1.3,

A ∩ U = ϕ. Let c ∈ A ∩ U , then c = b (since b /∈ A). Since A ⊆ m1, then c ∈ m1. Hence

b /∈ m1 (since in this topology there exists only one point which belong to m1 in each

elementary cube) which is a contradiction. Then m1 ⊆ A and A = m1.

Theorem 3.3.8. Let Z3 be a β-space. Then we have the following:

(1) m1 = EEE iff m2 = OOO iff β-S = OEE∪EOE∪EEO iffM = EOO∪OEO∪OOE.

(2) m1 = EEO iff m2 = OOE iff β-S = OEO∪EOO∪EEE iffM = EOE∪OEE∪OOO.

(3) m1 = EOE iff m2 = OEO iff β-S = OOE∪EEE∪EOO iffM = EEO∪OOO∪OEE.

(4) m1 = OEE iff m2 = EOO iff β-S = EEE∪OOE∪OEO iffM = OOO∪EEO∪EOE.

(5) m1 = EOO iff m2 = OEE iff β-S = OOO∪EEO∪EOE iffM = EEE∪OOE∪OEO.

(6) m1 = OEO iff m2 = EOE iff β-S = EEO∪OOO∪OEE iffM = OOE∪EEE∪EOO.

(7) m1 = OOE iff m2 = EEO iff β-S = EOE∪OEE∪OOO iffM = OEO∪EOO∪EEE.

(8) m1 = OOO iff m2 = EEE iff β-S = EOO∪OEO∪OOE iffM = OEE∪EOE∪EEO.

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Proof. (1) Suppose that m1 = EEE. Let (a, b, c) ∈ EEE = m1, then (a+1, b+1, c+1) ∈

OOO. By Theorem 3.3.3, (a+1, b+1, c+1) ∈ m2. Hence by Corollary 3.3.6, OOO ⊆ m2.

Let x /∈ OOO and let U be the set of all vertices of CU(x). By Remak 3.1.3, U ∩OOO =

ϕ. Let y ∈ U ∩OOO. Then x = y and y ∈ m2 (since OOO ⊆ m2). So x /∈ m2 (since any

elementary cube in this topology has only one point belong to m2). Hence m2 ⊆ OOO.

Therefore m2 = OOO.

Suppose that m2 = OOO. Let (a, b, c) ∈ OOO = m2. If y = (a + 1, b + 1, c + 1), then

y ∈ N8(x)∩ ∈ EEE. By Theorem 3.3.3 and Corollary 3.3.7, y ∈ m1, m1 = EEE, and

N6(y) ⊆ β-S. Since N6(y) ∩ A = ϕ for all A ∈ OEE,EOE,EEO, then by Corollary

3.3.6, OEE ∪EOE ∪EEO ⊆ β-S. Now let z /∈ OEE ∪EOE ∪EEO. Hence there exists

u ∈ EEE = m1, such that z ∈ u∪ (N26(u) \N6(u)). By Theorem 3.3.3, z /∈ β-S. Thus

β-S⊆ OEE ∪ EOE ∪ EEO and β-S = OEE ∪ EOE ∪ EEO.

Suppose that β-S = OEE ∪EOE ∪EEO. Let x ∈ EEE. Then N6(x) ⊆ OEE ∪EOE ∪

EEO = β-S. By Theorem 3.3.3 and Corollary 3.3.7, x ∈ m1,m1 = EEE, andN12(x) ⊆M .

But N12(x) ∩ A = ϕ for all A ∈ EOO,OEO,OOE. Thus EOO ∪ OEO ∪ OOE ⊆ M .

Now let y /∈ EOO ∪ OEO ∪ OOE. Then there exists z ∈ EEE = m, such that y /∈

N12(z). By Theorem 3.3.3, y /∈ M . Therefore M ⊆ EOO ∪ OEO ∪ OOE and M =

EOO ∪OEO ∪OOE.

Suppose that M = EOO ∪OEO ∪OOE. Let x ∈ EEE. Then N12(x) ⊆ EOO ∪OEO ∪

OOE = M . By Theorem 3.3.3, EEE ∩m1 = ϕ. By Corollary 3.3.7, m = EEE. The

others have similar proof.

As a direct result of the above theorem, there are eight homeomorphic topologies on Z3, all

of them satisfy the Subcase(ii) of case(1). If (Z3, τβ1) and (Z3, τ

β2) are two β-spaces with

corresponding m(1)

1 , m(2)

1 respectively. By Corollary 3.3.7, m(1)

1 , m(2)

1 ∈ A. For example if

m(1)

1 = EOO and m(2)

1 = OOE, then a homeomorphism function f :(Z3, τβ1) −→ (Z3, τ

β2)

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can be defined by f(a, b, c) = (a+ 1, b, c+ 1) for (a, b, c) ∈ Z3.

Now we are going to determine the order of this topology. We denote this order by ≤β

and denote the up of x ∈ Z3 with respect to this order by Vβ(x) = ↑

βx.

Theorem 3.3.9. Let (Z3, τβ) be a βspace and x, y ∈ Z3.

(1) If x ∈M , then Vβ(x) = x.

(2)If x ∈ β-S, then Vβ(x) = x ∪ (N6(x) \m1).

(3) If x ∈ m2, then Vβ(x) = x ∪N6(x).

(4) If x ∈ m1, then Vβ(x) = x ∪ (N26(x) \N8(x)).

Proof. (1) In any Alexandroff space if x is a maximal point, then Vβ(x) = x.

(2) Suppose that x ∈ β-S. Then there exist two minimal points x1, x2 and four maximal

points x3, x4, x5, x6 in N6(x). Hence xi → x and x → xj where i ∈ 1, 2 and j ∈

3, 4, 5, 6. Thereby N6(x) \m1 ⊆ Vβ(x). Since x ∈ V

β(x), x ∪ N6(x) \m1 ⊆ V

β(x).

Assume that y /∈ N6(x) ∪ x and x → y. Hence there exists x0 ∈ N6(x) such that

x → x0 → y which is a contradiction since x0 will be a maximal point. Then Vβ(x) ⊆

N6(x)∪x. Assume that w ∈ Vβ(x)∩m1. Then x→ w which is a contradiction since w

is a minimal point. Therefore Vβ(x) ⊆ x∪N6(x)\m1. Then Vβ

(x) = x∪ (N6(x)\m1)

whenever x ∈ β-S.

(3) Suppose that x ∈ m2 ⊆ m. Whenever y ∈ N6(x), then x → y. Since x ∈ Vβ(x),

x∪N6(x) ⊆ Vβ(x). Assume that z /∈ x∪N6(x) and x→ z. So there exists y ∈ N6(x)

such that x → y → z which is a contradiction since y will be a maximal point. Thereby

Vβ(x) = x ∪N6(x) whenever x ∈ m2.

(4) Suppose that x = (a, b, c) ∈ m1. By Theorem 2.1.12 part(2), Vβ(x) ⊆ x ∪N26(x).

By Theorem 3.3.3, (a + i, b + j, c + k) ∈ m2, where i, j, k ∈ 1,−1. Hence y /∈ Vβ(x)

for all y ∈ N8(x). Therefore Vβ(x) ⊆ x ∪N26(x) \N8(x). Conversely, x ∈ Vβ

(x). Since

x = (a, b, c) ∈ m1 ⊆ m, then x∪N6(x) ⊆ Vβ(x). By Theorem 2.1.12 part(3) and part(4),

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N6(x) ⊆ β-S and (a+i, b+j, c), (a+i, b, c+j), (a, b+i, c+j) ⊆M , where i, j ∈ 1,−1.

Then (a, b, c) → (a + i, b, c) → (a + i, b + j, c) and (a + i, b + j, c) ∈ Vβ(x). Similarly we

prove (a + i, b, c + j), (a, b + i, c + j) ∈ Vβ(x). Hence x ∪ (N26(x) \ N8(x)) ⊆ V

β(x).

Therefor Vβ(x) = x ∪ (N26(x) \N8(x)).

Example 3.3.10. In the β-space where m1 = OOO, we can determine the up of the two

minimal points (2, 2, 2) and (1, 3, 3) by the two following Hasse figures(Figure 3.13 and

Figure 3.14).

Figure 3.13: Up of Point (2,2,2)

Figure 3.14: Up of Point (1,3,3)

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Theorem 3.3.11. Let (Z3, τβ) be a β-space. For arbitrary point x ∈ m1 (resp. x ∈ m2),

if we determine the type of x (as minimal, maximal or β-saddle), then we can determine

the type of any point in Z3.

Proof. Let x ∈ Z3 be a fixed point, and assume that x ∈ m1(resp. x ∈ m2). Since x ∈ Z3,

then x ∈ A for some A ∈ A, then by Corollary 3.3.7, A = m1 (resp. A = m2). Hence

by Theorem 3.3.8 (resp. by Theorem 3.3.3 and Theorem 3.3.8), we determine the type of

each point on Z3.

Example 3.3.12. In a β-space if (3, 4, 5) ∈ m2 the types of (10, 8, 13),(−10, 3, 0), and

(2, 8, 6) as following:

Since (3, 4, 5) ∈ m2, then by Theorem 3.3.3, (4, 5, 6) ∈ m1. But (4, 5, 6) ∈ EOE. Hence by

Corollary 3.3.7, m1 = EOE. By Theorem 3.3.8, m2 = OEO, β-S = OOE∪EEE∪EOO

and M = EEO∪OOO∪OEE. So (10, 8, 13) ∈M , (−10, 3, 0) ∈ m1, and (2, 8, 6) ∈ β-S.

Corollary 3.3.13. Let (Z3, τβ) be a β-space with corresponding poset (Z3,≤

β), then Z3

is a semi-T2-space.

Proof. Since ∀ x /∈M , |x| ≥ 4, then by Theorem 1.2.11, Z3 is a semi-T2-space.

Theorem 3.3.14. Let (Z3, τβ) be the β-space with m1 = EEE, and let a, b be a fixed

integer numbers.

(1) Assume that A1=(a, x, y): x, y ∈ Z, A2=(x, a, y): x, y ∈ Z and A3=(x, y, a):

x, y ∈ Z. If a is even ( resp. odd), then for each i=1,2,3, Ai is homeomorphic to the

Alexandroff Hopf ( resp. Marcus-Wyse) space on Z2.

(2) If B1 = (a, b, x): x ∈ Z, B2 = (a, x, b): x ∈ Z and B3 = (a, b, x): x ∈ Z, then

for each i=1,2,3, Bi is homeomorphic to the Khalimsky space.

Proof. (1) Assume that a ∈ Z be a fixed even number. Simply, let f : A1 −→ Z2 be such

that f(a, x, y) = (x, y). It is clearly that f is a bijective function. If (x, y) ∈ EE, then

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(a, x, y) ∈ EEE = m1 . By Theorem 3.3.9, f(↑m1(a, x, y) ∩ A1) = f(a, x, y), (a, x −

1, y), (a, x+ 1, y), (a, x, y − 1), (a, x, y + 1), (a, x− 1, y − 1), (a, x− 1, y + 1), (a, x+ 1, y −

1), (a, x+ 1, y + 1) = (x, y) ∪N8(x, y) = ↑s (x, y) = ↑s f(a, x, y). If (x, y) ∈ EO, then

(a, x, y) ∈ EEO ⊆ β-S. Hence by Theorem 3.3.9, f(↑m1(a, x, y)∩A1) = f(a, x, y), (a, x−

1, y), (a, x + 1, y) = x − 1, x, x + 1 × y = ↑s (x, y) = ↑s f(a, x, y). If (x, y) ∈ OE,

then (a, x, y) ∈ EOE ⊆ β-S. Therefore f(↑m1(a, x, y) ∩ A1) = f(a, x, y), (a, x, y −

1), (a, x, y + 1) = x × y − 1, y, y + 1 = ↑s (x, y) = ↑s f(a, x, y). If (x, y) ∈ OO, then

(a, x, y) ∈ EOO ⊆ M and f(↑m1(a, x, y) ∩ A1) = f(a, x, y) = ↑s (x, y) = ↑s f(a, x, y).

Then by Theorem 1.2.5, A1 is homeomorphic to the Alexandroff Hopf space on Z2.

For the other case, suppose that a ∈ Z be a fixed odd number. If (x, y) ∈ EE ⊆ EV2 , then

(a, x, y) ∈ OEE ⊆ β-S. So by Theorem 3.3.9, f(↑m1(a, x, y) ∩ A1) = f(a, x, y), (a, x −

1, y), (a, x+1, y), (a, x, y−1), (a, x, y+1) = (x, y)∪N4(x, y) = ↑m (x, y) = ↑m f(a, x, y).

If (x, y) ∈ EO ⊆ OD2 , then (a, x, y) ∈ OEO ⊆ M . Hence by Theorem 3.3.9, f(↑m1

(a, x, y)∩A1) = f(a, x, y) = (x, y) = ↑m (x, y) = ↑m f(a, x, y). If (x, y) ∈ OE ⊆ OD2 ,

then (a, x, y) ∈ OOE ⊆ M . Therefore f(↑m1(a, x, y) ∩ A1) = f(a, x, y) = (x, y) =

↑m (x, y) = ↑m f(a, x, y). If (x, y) ∈ OO ⊆ EV2 , then (a, x, y) ∈ OOO ⊆ m2 . Then

f(↑m1(a, x, y) ∩ A1) = f(a, x, y), (a, x − 1, y), (a, x + 1, y), (a, x, y − 1), (a, x, y + 1) =

(x, y) ∪ N4(x, y) = ↑m (x, y) = ↑m f(a, x, y). Therefore by Theorem 1.2.5, A1 is

homeomorphic to the Marcus-Wyse space on Z2. The proofs of A2 and A3 are similar.

(2) Direct result using Part (1) above, Theorem 2.3.32 and Theorem 2.3.13.

3.4 αs-Space on Z3

In this section, we are going to study the topology induced on an elementary cube by the

subcase(iii) of case(1) and its related topology on Z3. This topology contains α-saddle

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points with no β-saddle points, so we call it by αs-topology on Z3 and denoted by ταs.

There are more than one topology generated by the subcase(iii) of case(1), all of them

are homeomorphic αs-space on Z3, as we will see latter in Theorem 3.4.6. If we consider

the αs-topology ταson an elementary cube, we find three minimal points 2, 4, 5, three

α-saddle points 3, 6, 8, and two maximal points 1, 7. But one of the maximal points

1 covers minimal points and the other 7 covers α-saddle points (see the Figure 3.15).

Figure 3.15: αs-Space on Elementary Cube

Definition 3.4.1. Let (Z3, ταs) be an αs-space. Define the subsets M1 and M2 by the

following:

M1 = x ∈M : If y x, then y ∈ α-S, M2 = x ∈M : If y x, then y ∈ m .

It is clearly that M = M1 ∪M2 and M1 ∩M2 = ϕ.

We can see that the αs-space on the elementary cube in Figure 3.15 using the rotation by

90 counterclockwise. By rotation and fix the name of points, we will have the following

figure (see Figure 3.16).

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Figure 3.16: α-Space on Elementary Cube With rotation

Now if we compare between Figure 3.12 and Figure 3.16 we will find that the up (resp.

the down) of every point in the first cube is equal the down (resp. the up) in the second

cube (see Figure 3.17).

Figure 3.17: The β-space and αs-space on Some Cube

Theorem 3.4.2. The αs-topology is the dual to the topology β-topology on Z3.

Proof. It is coming directly by the previous comparison in Figure 3.17 (up to rotations

by 90), Corollary 1.2.12 and Theorem 3.1.10.

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Notation 3.4.3. It is clearly that the set M of maximal points (resp. the sets m1, m2,

β-S) with respect to τβis exactly the same of the set m of minimal points (resp. the sets

M1, M2, α-S) with respect to ταs.

Corollary 3.4.4. Let Z3 be an αs-space. Then M1 ∈ A.

Proof. By Notation 3.4.3 and Corollary 3.3.7, the set M1 in ταsis the same of m1 in τ

β

and m1 ∈ A.

Remark 3.4.5. As in Corollary 3.4.4 above, and using Notation 3.4.3, together with cor-

responding theorems in section 3.3, we get the following theorem.

Theorem 3.4.6. Let Z3 be an αs-space.

(1)M1 = EEE iffM2 = OOO iff α-S = OEE∪EOE∪EEO iff m = EOO∪OEO∪OOE.

(2)M1 = EEO iffM2 = OOE iff α-S = OEO∪EOO∪EEE iff m = EOE∪OEE∪OOO.

(3)M1 = EOE iffM2 = OEO iff α-S = OOE∪EEE∪EOO iff m = EEO∪OOO∪OEE.

(4)M1 = OEE iffM2 = EOO iff α-S = EEE∪OOE∪OEO iff m = OOO∪EEO∪EOE.

(5)M1 = EOO iffM2 = OEE iff α-S = OOO∪EEO∪EOE iff m = EEE∪OOE∪OEO.

(6)M1 = OEO iffM2 = EOE iff α-S = EEO∪OOO∪OEE iff m = OOE∪EEE∪EOO.

(7)M1 = OOE iffM2 = EEO iff α-S = EOE∪OEE∪OOO iff m = OEO∪EOO∪EEE.

(8)M1 = OOO iffM2 = EEE iff α-S = EOO∪OEO∪OOE iff m = OEE∪EOE∪EEO.

As a direct result of the above theorem, there are eight topologies on Z3, all of them

satisfy the Subcase(iii) of case(1). If (Z3, ταs1) and (Z3, ταs2

) are two αs-spaces, then their

dual spaces are homeomorphic with respect to the β-space (as we shown in the previous

section). Thus (Z3, ταs1) and (Z3, ταs2

) are homeomorphic with respect to the αs-space.

Therefore these eight topologies are homeomorphic.

Theorem 3.4.7. Let (Z3, ταs) be an αs-space and x, y ∈ Z3. Then x ≤αs

y iff y ≤βx.

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Proof. Directly from Corollary 1.2.12 and Theorem 3.4.2.

Corollary 3.4.8. Let Z3 be an αs-space and x ∈ Z3.

(1) If x ∈M , then Vαs(x) = x.

(2) If x ∈ α-S, then Vαs(x) = x∪ (N6(x) ∩M1).

(3) If x ∈ m, then Vαs(x) = x∪ (N26(x) \ (m ∪N8(x))).

Proof. (1) This is true in any T0-Alexandroff space.

(2) Suppose that x ∈ α-S. Let y ∈ N6(x) ∩M1. Then x ≤αsy and y ∈ Vαs

(x). Since

x ∈ Vαs(x), then x∪ (N6(x) ∩M1) ⊆ Vαs

(x). Let y = x and y /∈ (N6(x) ∩M1), then

y /∈ N6(x) or y /∈ m1 with respect to β-space. If y /∈ m1, then y ∈ m2 or y ∈ β-S or y ∈M .

Hence y βx. So by Theorem 3.4.7, x αs

y. Then y /∈ Vαs(x). If y /∈ N6(x). Assume

that y ≤βx. Hence y ∈ m1. By Theorem 3.3.9, x ∈ V

β(y) = y ∪ (N26(y) \ N8(y)).

But since x /∈ N6(y) ∪ y, then x ∈ N12(y). By Theorem 3.3.3, x ∈ M which is

a contradiction. So if y /∈ N6(x), then y βx. Thus, Vαs

(x) ⊆ x∪ (N6(x) ∩ M1).

Therefore Vαs(x) = x∪ (N6(x) ∩M1) whenever x ∈ α-S.

(3) Suppose that x ∈ m. Let y ∈ N26(x) \ (m ∪N8(x)). Then y ∈ N26(x) \ (M ∪N8(x))

with respect to β-space. Hence y /∈ M and y ∈ N26(x) \ N8(x). So y ∈ β-S or y ∈ m2

or y ∈ m1. If y ∈ β-S assume that x ∈ N12(y). Then there exists a point z ∈ Z3 such

that z ∈ N6(y) and z ∈ N6(x). Since y ∈ β-S and x ∈ M with respect to β-space and

by Theorem 2.1.12, we have either ”z ≤βy and z ≤

βx” or y ≤

βz ≤

βx. The first

case is a contradiction since if it is hold then z ∈ m1 ∩ m2 = ϕ. If y ≤βz ≤

βx, then

z ∈ α-S which is also a contradiction since no α saddle point in the β-space. Therefore

x ∈ N6(y) and x ∈ N6(y) \m1. Then by Theorem 3.3.9, x ≤βy. So y ≤αs

x. If y ∈ m2.

Assume that x ∈ N12(y), then there exists a point z ∈ Z3 such that z ∈ N6(y) and

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z ∈ N6(x). So by Theorem 2.1.12, y ≤βz ≤

βx which is a contradiction since y ∈ m2.

Hence x ∈ N6(y). By Theorem 3.3.9, x ≤βy. So y ≤αs

x. If y ∈ m1, then by Theorem

3.3.9, x ≤βy. So y ≤αs

x. If y ∈ N26(x) \ (m ∪ N8), then y ∈ Vαs(x). Therefore x∪

(N26(x) \ (m ∪N8(x))) ⊆ Vαs(x) since x ∈ Vαs

(x).

Now let x = y and y /∈ N26(x) \ (m ∪ N8(x)). So y /∈ N26(x) or y ∈ m or y ∈ N8(x). If

y /∈ N26(x), then by Theorem 2.1.12, y /∈ Vαs(x). If y ∈ m assume that x ≤αs

y. This

is a contradiction since y ∈ m. Hence y /∈ Vαs(x). If y ∈ N8(x) assume that x ≤αs

y, so

y ≤βx. Since x ∈M in the β-space, y ∈ m1 or y ∈ m2 or y ∈ β-S which is a contradiction

with Theorem 3.3.9, since x ∈ N8(y). Hence if x = y and y /∈ N26(x) \ (m∪N8(x)), then

y /∈ Vαs(x). This implies that Vαs

(x) ⊆ x∪ (N26(x) \ (m ∪N8(x))). Therefor Vαs(x) =

x∪ (N26(x) \ (m ∪N8(x))) whenever x ∈ m.

Example 3.4.9. Let M1 = EOO, then we can consider the up of the point (2,2,2) with

respect to ≤αsby the following Hasse diagram (see Figure 3.18 )

Figure 3.18: Up of The point (2,2,2) in (Z3,≤αs)

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Theorem 3.4.10. Let (Z3, ταs) be an αs-space. For arbitrary point x ∈ M1 (resp.

x ∈ M2), if we determine the type of x (as minimal, maximal, α-saddle), then we can

determine the types of all points in Z3.

Proof. The proof is coming directly from Notation 3.4.3 and Theorem 3.3.11.

Example 3.4.11. In an αs-space, if (2, 3, 5) ∈ M2, then we can determine the types of

(1,2,13),(-10,3,0), and (2,7,3) as follows:

Since (2, 3, 5) ∈M2, then (2, 3, 5) ∈ m2 in the β-space. By Theorem 3.3.3, (3, 4, 6) ∈ m1.

Hence (3, 4, 6) ∈ M1. But (3, 4, 6) ∈ OEE, so by Corollary 3.4.4, M1 = OEE. And by

Theorem 3.4.6, M2 = EOO, α-S = EEE ∪EEO∪OEO and m = OOO∪EEO∪EOE.

Therefore (1, 2, 13) ∈ α-S, (−10, 3, 0) ∈ m and (2, 7, 3) ∈M2.

Theorem 3.4.12. Let (Z3, ταs) be an αs-space with M1 = EEE, and let a, b be a fixed

integer numbers.

(1) Assume that A1=(a, x, y): x, y ∈ Z, A2=(x, a, y): x, y ∈ Z and A3=(x, y, a):

x, y ∈ Z. If a is even ( resp. odd), then for each i=1,2,3 Ai is homeomorphic to the

Alexandroff Hopf ( resp. Marcus-Wyse) topology on Z2.

(2) If B1 = (a, b, x): x ∈ Z, B2 = (a, x, b): x ∈ Z and B3 = (a, b, x): x ∈ Z, then

for each i=1,2,3 Bi is homeomorphic to the Khalimsky topology.

Proof. Recall that on Z2, the dual of an Alexandroff Hopf topology is again an Alexandroff

Hopf topology and the dual of a Marcus-Wyse topology is also a Marcus-Wyse topology.

Moreover, from Theorem 3.4.2, αs-space is the dual of β-space, so for any i ∈ 1, 2, 3

the relative topologies on Aiand B

iwith respect to αs-space are the dual topologies with

respect to β-space. Using this fact together with Theorem 3.3.14, we get the result.

Corollary 3.4.13. Let (Z3, ταs) be an αs-space. Then Z3 is a semi-T2-space.

Proof. Since ∀ x /∈M |x| ≥ 2. So by Theorem 1.2.11, Z3 is a semi-T2-space.

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Chapter 4

Topologies with αβ-Saddle Points

We considered in previous section three digital spaces the Marcus -Wyse topology on Z3,

β-space on Z3, and αs-space on Z3. All of them do not have together α-saddle and β-

saddle points. The remainder digital topologies on Z3 are different since their points have

both α-saddle and β-saddle points. We will prove that the points of the product topology

of three Khalimesky spaces has both α-saddle and β-saddle points. So the question arises

: is the product topology one of the five digital spaces?. If yes, who is that?.

4.1 Product of Khalimesky Spaces on Z3

Recall that a saddle point x ∈ Z3 is called an α−saddle point (resp. a β−saddle point) if

there exist four direct neighbors x1, x2, x3, x4 ∈ N6(x) such that xi → x (resp. x → xi),

for i = 1,2,3,4. For the Khalimesky space (Z,≤Kh), the specialization order ≤p of the

product T0-Alexandroff space Z × Z × Z satisfies that (a1, a2, a3), (b1, b2, b3) in (Z3,≤p),

then (a1, a2, a3) ≤p (b1, b2, b3) iff ai ≤Khbi for all i ∈ 1, 2, 3. For this end, we can get

that, for example (2, 2, 2) ≤p (1, 2, 2) ≤p (1, 1, 2). So (1, 2, 2) is a saddle point. Moreover,

the points (1, 1, 2), (1, 3, 2), (1, 2, 1), (1, 2, 3) are direct neighbors and cover (1, 2, 2). So

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(1, 2, 2) is β-saddle point. Similarly, the point (1, 1, 2) is an α-saddle point. This proves

that the product space (Z3, τp) contains both β-saddle points and α-saddle points. In

this section, we are going to determine if the product space on Z3 is one of the digital

spaces. Surely, no one of the three spaces studied in chapter 3 is the product space. We

will search if it is one of the remainder two digital spaces or not.

Recall that if a is even number, then ↑Kh

a = a − 1, a, a + 1 and if a is odd number,

then ↑Kha = a. Hence we have the following lemma.

Proposition 4.1.1. Let (Z3,≤p) be the product topology of three Khalimesky Spaces and

let x = (a, b, c) ∈ Z3.

(1) If x ∈ OOO, then Vp(x) = x.

(2) If x ∈ EOO, then Vp(x) = a− 1, a, a+ 1 × b × c.

(3) If x ∈ OEO, then Vp(x) = a × b− 1, b, b+ 1 × c.

(4) If x ∈ OOE, then Vp(x) = a × b × c− 1, c, c+ 1.

(5) If x ∈ EEO, then Vp(x) = a− 1, a, a+ 1 × b− 1, b, b+ 1 × c.

(6) If x ∈ EOE, then Vp(x) = a− 1, a, a+ 1 × b × c− 1, c, c+ 1.

(7) If x ∈ OEE, then Vp(x) = a × b− 1, b, b+ 1 × c− 1, c, c+ 1.

(8) If x ∈ EEE, then Vp(x) = a− 1, a, a+ 1 × b− 1, b, b+ 1 × c− 1, c, c+ 1.

4.2 αβ-space on Z3

In this section, we are going to study the topology induced on an elementary cube by

the subcase(ii) of case(2) and its related topology on Z3. This topology contains both

α-saddle points and β-saddle points, so we call it αβ-topology on Z3 and denoted by ταβ.

Actually there are twelve homeomorphic topologies on Z3, all of them determined by the

subcase(ii) of case(2) as we will see latter. If we consider αβ-topology on the elementary

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cube of Figure 4.1, we will find two minimal points 3, 6, two β-saddle points 4, 5,

two α-saddle points 2, 7 and two maximal points 1, 8. In each elementary cube, the

minimal points is covered by one β-saddle point and two α-saddle points. Each β-saddle

point is covered by two maximal points and covers a minimal point. Each α-saddle point

is covered by a maximal point and covers two minimal points. And Each maximal point

covers two β-saddle points and an α-saddle point. Moreover if x and y are the two minimal

( resp. α-saddle, β-saddle, maximal ) points, then y ∈ N12(x). The elementary square

witch has the two minimal (resp. maximal) points as vertices has also the two α-saddle

(resp. β-saddle) points as vertices and vice versa. ( See Figure 4.1).

Figure 4.1: αβ-space on an Elementary Cube

Theorem 4.2.1. Let Z3 be an αβ-space, and x, y ∈ Z3.

(1) If x, y ∈ m and x ∈ N12(y), then N6(x) ∩N6(y) ⊆ α-S.

(2) x ∈ m if and only if N6(x) \ α-S ⊆ β-S.

(3) If x ∈ m, then N12(x) \m ⊆M .

(4) If N12(x) \m ⊆M , then either x ∈ m or x ∈M .

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(5) If x ∈ α-S, then N12(x) \ α-S ⊆ β-S.

(6) If x, y ∈ β-S and x ∈ N12(y), then N6(x) ∩N6(y) ⊆M .

(7) If x ∈M , then N12(x) \M ⊆ m.

Proof. (1) Suppose that x, y be minimal points and x ∈ N12(y). Then N6(x) ∩ N6(y) =

ϕ. Let z ∈ N6(x) ∩ N6(y), then x → z and y → z. Hence either z ∈ M or z ∈ α-S.

But in this topology, the maximal points dose not cover the minimal points. So z ∈ α-S.

Therefore N6(x) ∩N6(y) ⊆ α-S.

(2) (⇒) Suppose that x ∈ m. Let y ∈ N6(x) \α-S, then x→ y. Hence y ∈ β-S or y ∈M .

So y ∈ β-S Since in this topology the minimal points dose not covered by the maximal

points. Therefore N6(x) \ α-S ⊆ β-S.

(⇐) Let y ∈ N6(x) \ α-S, then x y and y ∈ β-S. Hence x ∈ m.

(3) Suppose that x ∈ m. Let y ∈ N12(x) \m and let z ∈ N6(x) such that z ∈ N6(y), then

by Theorem 2.1.12, x z and either y z or z y. Hence z ∈ α-S or z ∈ β-S. If

z ∈ α-S, then y ∈ m or y ∈M . But y /∈ m, hence y ∈M . Similarly if z ∈ β-S. Therefore

N12(x) \m ⊆M .

(4) Let y ∈ N12(x) \ m and let z ∈ N6(x) such that z ∈ N6(y). Then z y. Hence

z ∈ α-S or z ∈ β-S. Since z ∈ N6(x), then by Theorem 2.1.12, either x ∈ m or x ∈M .

(5) Let y ∈ N12(x) \ α-S. Hence there exist z ∈ Z3 such that z ∈ N6(x) and z ∈ N6(y).

Since z ∈ N6(x), then either z ∈ m or z ∈ M . Therefore either y ∈ α-S or y ∈ β-S since

z ∈ N6(y). But y /∈ α-S. Hence y ∈ β-S and N12(x) ⊆ α-S ⊆ β-S.

(6) Suppose that x, y ∈ β-S and x ∈ N12(y). Then N6(x) ∩N6(y) = ϕ. Let z ∈ N6(x) ∩

N6(y), then either z ∈ m or z ∈M . Assume to contrary that z ∈ m. Hence there exist an

elementary square contains two β-saddle and a minimal point which is a contradiction.

Therefore z ∈M and N6(x) ∩N6(y) ⊆M .

(7) Suppose that x ∈ M . Let y ∈ N12(x) \M and let z ∈ N6(x) such that z ∈ N6(y).

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Hence z ∈ α-S or z ∈ β-S. Since z ∈ N6(y), then either y ∈ m or y ∈ M . But y /∈ M ,

then y ∈ m and N12(x) \M ⊆ m.

Theorem 4.2.2. Let Z3 be an αβ-space, and B ∈ m,β-S,α-S,M. If x = (a, b, c) ∈ B,

then (a+ 2r, b+ 2s, c+ 2t) ∈ B for all r, s, t ∈ Z.

Proof. Suppose that B = m, i ∈ −1, 1 and x = (a, b, c) ∈ m. Since x ∈ m and

(a + i, b, c) ∈ N6(x), then by Theorem 2.1.12, x (a + i, b, c). So (a + i, b, c) ∈ α-S

or (a + i, b, c) ∈ β-S. By Theorem 2.1.14, x (a + i, b, c) (a + 2i, b, c). Hence

(a+2i, b, c) ∈ m. By the induction (a+2r, b, c) ∈ m for any r ∈ Z. By the same way, we

can prove that (a+ 2r, b+ 2s, c) ∈ m and after that (a+ 2r, b+ 2s, c+ 2t) ∈ m.

Suppose that B = β-S, i ∈ −1, 1 and x = (a, b, c) ∈ β-S. Since (a + i, b, c) ∈ N6(x),

then either (a + i, b, c) x or x (a + i, b, c). If (a + i, b, c) x, then by Theorem

2.1.14, x (a+i, b, c) (a+2i, b, c). Hence (a+i, b, c) ∈ m and either (a+2i, b, c) ∈ β-S

or (a+2i, b, c) ∈ α-S. Suppose to contrary that (a+2i, b, c) ∈ α-S. If i = 1 ( resp. i = −1),

then in the elementary cube CU(a+1, b, c) ( resp. CU(a− 2, b, c) ) either (a+1, b+1, c)

or (a+1, b, c+1) ( resp. (a− 2, b+1, c) or (a− 2, b, c+1) ) in β-S. Hence the elementary

cube CU(x) ( resp. CU(a − 1, b, c) ) has an elementary square which has two β-saddle

points and a minimal point which is a contradiction.

Now if x (a + i, b, c), then by Theorem 2.1.14, x (a + i, b, c) (a + 2i, b, c). So

(a+i, b, c) ∈M and either (a+2i, b, c) ∈ β-S or (a+2i, b, c) ∈ α-S. Suppose to contrary that

(a+2i, b, c) ∈ α-S. If i = 1 ( resp. i = −1), then in the elementary cube CU((a+1, b, c)) (

resp. CU((a−2, b, c)) ) (a+1, b+1, c), (a+1, b, c+1) ( resp. (a−2, b+1, c), (a−2, b, c+1)

) in β-S. Therefore the elementary cube CU(x) ( resp. CU(a − 1, b, c) ) has three β-

saddle points which is a contradiction. Therefore (a + 2i, b, c) ∈ β-S. By the induction

(a+2r, b, c) ∈ β-S for any r ∈ Z. By the same way, we can prove that (a+2r, b+2s, c) ∈ β-

S and after that (a+ 2r, b+ 2s, c+ 2t) ∈ β-S for all r, s, t ∈ Z.

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Suppose that B = α-S, i ∈ −1, 1 and x = (a, b, c) ∈ α-S. Since (a + i, b, c) ∈ N6(x),

then either (a+i, b, c) x or x (a+i, b, c). If (a+i, b, c) x, then x (a+i, b, c)

(a + 2i, b, c). Hence (a + i, b, c) ∈ m and either (a + 2i, b, c) ∈ β-S or (a + 2i, b, c) ∈ α-

S. Suppose to contrary that (a + 2i, b, c) ∈ β-S. If i = 1 (resp. i = −1), then in the

elementary cube CU(a+ 1, b, c) ( resp. CU(a− 2, b, c) ) (a+ 1, b+ 1, c), (a+ 1, b, c+ 1) (

resp. (a−2, b+1, c), (a−2, b, c+1) ) in α-S. Therefore the elementary cube CU(x) ( resp.

CU(a−1, b, c) ) has three α-saddle points which is a contradiction. Now if x (a+i, b, c),

then x (a+ i, b, c) (a+2i, b, c). So (a+ i, b, c) ∈M and either (a+2i, b, c) ∈ β-S or

(a+ 2i, b, c) ∈ α-S. Suppose to contrary that (a+ 2i, b, c) ∈ β-S. If i = 1 ( resp. i = −1),

then in the elementary cube CU(a+1, b, c) ( resp. CU(a− 2, b, c) ) either (a+1, b+1, c)

or (a + 1, b, c + 1) ( resp. (a − 2, b + 1, c) or (a − 2, b, c + 1) ) in α-S. So the elementary

cube CU(x) ( resp. CU(a − 1, b, c) ) has an elementary square which has two α-saddle

points and a maximal point, which is a contradiction. Therefore (a + 2i, b, c) ∈ α-S. By

the same way and by the induction, (a+ 2r, b+ 2s, c+ 2t) ∈ β-S for all r, s, t ∈ Z.

Suppose that B = M , i ∈ −1, 1 and x = (a, b, c) ∈M . Since x ∈M , then by Theorem

2.1.14, x (a+i, b, c) (a+2i, b, c). Hence either (a+i, b, c) ∈ α-S or (a+i, b, c) ∈ β-S.

So (a+ 2i, b, c) ∈M . Therefore (a+ 2r, b+ 2s, c+ 2t) ∈M for all r, s, t ∈ Z.

Corollary 4.2.3. Let Z3 be an αβ-space. Suppose that A ∈ A and B ∈ m,β-S,α-S,M.

then A ⊆ B.

Proof. Similar to the proof of Theorem 3.3.5.

Corollary 4.2.4. Let Z3 be an αβ-space. Suppose that A ∈ A and B ∈ m,β-S,α-S,M.

If A ∩B = ϕ, then A ⊆ B.

Proof. Assume that (a, b, c) ∈ B ∩ A. If (u, v, w) ∈ A, then there exist n,m, r ∈ Z such

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that u = 2n+ a, v = 2m+ b and w = 2r + c. By Theorem 4.2.2, (a+ 2n, b+ 2m, c+ 2r)

= (u, v, w) ∈ B. Hence A ⊆ B.

Corollary 4.2.5. Let Z3 be an αβ-space. For each V ∈ m,β-S,α-S,M, there exists

sets A,B ∈ A such A = B and V = A ∪B.

Proof. Suppose that V = m. For any x ∈ Z3, CU(x) has two minimal points u, v. By

Remak 3.1.3, there exist two sets A,B ∈ A such that A = B, u ∈ A and v ∈ B. By

Corollary 4.2.4, A ∪ B ⊆ m. So if we fixed A and B, we need to prove that m ⊆ A ∪ B.

Suppose that r /∈ A∪B. Hence r ∈ C for some C ∈ A, such that A = C = B. Thus there

exists w ∈ CU(x), such that u = w = v. Then w /∈ m (since in this topology there exist

only two minimal points in any elementary cube). By Theorem 4.2.2 r /∈ m. Therefore

m ⊆ A ∪B and m = A ∪B. Similarly if V = β-S or V = α-S or V = M .

Notation 4.2.6. Since in each elementary cube if x and y are the two minimal points

(resp. α-saddle, β-saddle, maximal), then y ∈ N12(x). By Corollary 4.2.5, the possible

cases of the set of all minimal (resp. α-saddle, β-saddle, maximal) points are EEE∪EOO,

EEE ∪OEO, EEE ∪OOE, EOO ∪OEO, EOO ∪OOE, OEO ∪OOE, OOO ∪OEE,

OOO ∪ EOE, OOO ∪ EEO, OEE ∪ EOE, OEE ∪ EEO and EOE ∪ EEO.

Theorem 4.2.7. Let Z3 be the αβ-space. Then

(1) m = EEE∪EOO iff α-S = EOE∪EEO iff β-S = OOO∪OEE iffM = OEO∪OOE.

(2) m = EEE∪OEO iff α-S = OEE∪EEO iff β-S = OOO∪EOE iffM = EOO∪OOE.

(3) m = EEE∪OOE iff α-S = OEE∪EOE iff β-S = OOO∪EEO iffM = EOO∪OEO.

(4) m = EOO∪OEO iff α-S = EEO∪OOO iff β-S = OEE∪EOE iffM = OOE∪EEE.

(5) m = EOO∪OOE iff α-S = EEO∪OOO iff β-S = OEE∪EOE iffM = OOE∪EEE.

(6) m = OEO∪OOE iff α-S = OEE∪OOO iff β-S = EOE∪EEO iffM = EOO∪EEE.

(7) m = OOO∪OEE iff α-S = OEO∪OOE iff β-S = EEE∪EOO iffM = EOE∪EEO.

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(8) m = OOO∪EOE iff α-S = EOO∪OOE iff β-S = EEE∪OEO iffM = OEE∪EEO.

(9) m = OOO∪EEO iff α-S = EOO∪OEO iff β-S = EEE∪OOE iffM = OEE∪EOE.

(10)m = OEE∪EOE iff α-S = EEE∪OOE iff β-S = EOO∪OEO iffM = OOO∪EEO.

(11)m = OEE∪EEO iff α-S = OEO∪EEE iff β-S = EOO∪OOE iffM = EOE∪OOO.

(12)m = EOE∪EEO iff α-S = EEE∪EOO iff β-S = OEO∪OOE iffM = OOO∪OEE.

Proof. (1) Suppose that m = EEE ∪ EOO. Let (a, b, c) ∈ EEE, then (a, b+ 1, c+ 1) ∈

EOO. By Theorem 4.2.1 part(1), (a, b+1, c), (a, b, c+1)=N6(a, b, c)∩N6(a, b+1, c+1) ⊆

α-S. Since (a, b+1, c), (a, b, c+1) ⊆ EOE∪EEO, then by Corollary 4.2.5, EOE∪EEO

= α-S.

Suppose that α-S = EOE ∪ EEO and i, j ∈ −1, 1. Let x = (a, b, d) ∈ EOE, then

by Theorem 4.2.1 part(5), N12(x) \ α-S = (a + i, b + j, c), (a + i, b, c + j) ⊆ β-S. Since

(a + i, b + j, c), (a + i, b, c + j) ⊆ OEE ∪ OOO, hence by Corollary 4.2.5, β-S =

OOO ∪OEE.

Suppose that β-S = OOO∪OEE. Let x = (a, b, d) ∈ OOO. Hence (a, b+1, c+1) ∈ OEE

and (a, b+1, c+1) ∈ N12(a, b, d). By Theorem 4.2.1 part(6), (a, b+1, c), (a, b, c+1) =

N6(a, b, c) ∩N6(a, b + 1, c + 1) ⊆ M . But (a, b + 1, c), (a, b, c + 1) ⊆ OEO ∪ OOE. By

Corollary 4.2.5, M = OEO ∪OOE.

Finally suppose that M = OEO ∪ OOE and i, j ∈ −1, 1. Let x = (a, b, d) ∈ OEO,

then by Theorem 4.2.1 part(7), N12(x) \M = (a+ i, b+ j, c), (a+ i, b, c+ j) ⊆ m. Since

(a+ i, b+ j, c), (a+ i, b, c+ j) ⊆ EOO ∪ EEE, then m = EEE ∪ EOO. Therefore m

= EEE ∪ EOO iff α-S = EOE ∪ EEO iff β-S = OOO ∪OEE iff M = OEO ∪OOE.

The other parts proved in similar way.

As a direct result for Theorem 4.2.7, there are twelve topologies on Z3 of αβ-space. It

is clearly that all of them are homeomorphic. Now we are going to determine the order

of this topology. Let we denote this order by ≤αβ

and denote the up of x ∈ Z3 under this

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topology by Vαβ(x) = ↑

αβx.

Theorem 4.2.8. Let (Z3, ταβ) be the αβ-space and x ∈ Z3.

(1) If x ∈M , then Vαβ(x) = x.

(2) If x ∈ β-S ∪ α-S, then Vαβ(x) = x ∪ (N6(x) \m).

(3) If x ∈ m, then Vαβ(x) = x ∪ (N26(x) \ (N8(x) ∪m)).

Proof. (1) True in any T0-Allexandroff space.

(2) Suppose that x ∈ β-S ∪ α-S. Let y ∈ N6(x) \m. By Theorem 2.1.14, either x ≤αβy

or y ≤αβ

x. If y ≤αβ

x, then y ∈ m which is a contradiction. Hence x ≤αβ

y and

N6(x) \ m ⊆ Vαβ(x). Thus ((N6(x) \ m) ∪ x) ⊆ V

αβ(x). For the reverse inclusion

Vαβ(x) ⊆ N6(x) \ m ∪ x. Let z = x and z /∈ N6(x) \ m. If z ∈ m, then z /∈ V

αβ(x).

suppose to contrary that x ≤αβ

z and z /∈ N6(x). Then z ∈ M and z ∈ N12(x) or

z ∈ N8(x). If z ∈ N12(x), then there exists u ∈ Z3 such that u ∈ N6(x) and u ∈ N6(z).

Hence u ≤αβz and u ∈ β-S ∪ α-S. But since u ∈ N6(x), then u, x are comparable which is

a contradiction. If z ∈ N8(x), then there exist u, v ∈ Z3 such that u ∈ N6(x), u ∈ N6(v),

v ∈ N6(z) and x ≤αβu ≤

αβv ≤

αβz. Hence u ∈M which is a contradiction (since u ≤

αβv

and u = v). So Vαβ(x) ⊆ (N6(x) \m) ∪ x and V

αβ(x) = x ∪ (N6(x) \m).

(3) Suppose that x ∈ m. Let y ∈ N26(x)\(N8(x)∪m), then y ∈ N6(x)\m or y ∈ N12(x)\m.

If y ∈ N6(x) \ m, then x ≤αβ

y and y ∈ Vαβ(x). If y ∈ N12(x) \ m, then there exist

z ∈ Z3 such that z ∈ N6(x) and z ∈ N6(y). Hence x ≤αβz and z ∈ β-S ∪ α-S. Since

z ∈ N6(y) and y /∈ m, then z ≤αβ

y. By transitivity x ≤αβ

y. Therefore y ∈ Vαβ(x)

and N26(x) \ (N8(x) ∪ m) ⊆ Vαβ(x). So x ∪ (N26(x) \ (N8(x) ∪ m)) ⊆ V

αβ(x) (since

x ≤αβ

x). Now let z /∈ N26(x) \ (N8(x) ∪ m). If z /∈ N26(x) or z ∈ m, then x and z

are incomparable. Assume that z ∈ N8(x) and x ≤αβz. So z ∈ β-S ∪ α-S. Therefore

there exists u ∈ Z3 such that u ∈ N6(x), u ∈ N12(z) and x ≤αβu ≤

αβz. Then u ∈ β-S

∪ α-S. Hence z, u ∈ β-S ∪ α-S and are comparable which is a contradiction. Therefore

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Vαβ(x) ⊆ x ∪ (N26(x) \ (N8(x) ∪m)) and V

αβ(x) = x ∪ (N26(x) \ (N8(x) ∪m)).

By the Proposition 4.1.1, we easy prove the following corollary.

Corollary 4.2.9. The αβ-space on Z3 is not the product topology on Z3.

Notation 4.2.10. Let U ∈ m,β-S,α-S,M and A,B ∈ A. By two points x, y ∈ U we can

consider the type of any point in Z3 provided that x ∈ A, y ∈ B and A = B. Consider

the following example.

Example 4.2.11. If (3, 9,−11), (7, 2, 4) ∈ α-S in an αβ-space. We can determine the

type of (4, 1, 8), (2, 8, 98) and (1, 7, 4) as the follows:

Since (3, 9,−11) ∈ OOO and (7, 2, 4) ∈ OEE. So by Corollary 4.2.5, α-S = OOO∪OEE.

By Theorem 4.2.7 part(6), m = OEO∪OOE, β-S = EOE∪EEO andM = EOO∪EEE.

Hence (4, 1, 8) ∈ β-S, (2, 8, 98) ∈M and (1, 7, 4) ∈ m.

Example 4.2.12. Suppose (Z3, ταβ) be the αβ-space with m = EEE ∪ EOO. We can

find the up of the point (2, 2, 2) in the following Figure. (see Figure 4.2).

Corollary 4.2.13. Let (Z3, ταβ) be an αβ-space. Then Z3 is a semi-T2-space.

Proof. Since ∀ x /∈M |x| ≥ 2, then by Theorem 1.2.11, Z3 is a semi-T2-space.

4.3 Alexandroff Hopf topology on Z3

In this section, we are going to study the topology induced on an elementary cube by the

subcase(iii) of case(2) and its related topology on Z3. We will prove latter this topology is

the Alexandroff Hopf topology on Z3, denoted by τs . This topology depends on the choice

of the generated point, and there are several types of homeomorphic τs on Z3. Indeed, on

Z3 there exist eight homeomorphic topologies induced by subcase(iii) of case(2), as we will

103

Figure 4.2: Up of (2, 2, 2) on the αβ-space with m = EEE ∪ EOO

see latter. If we consider this topology we will find in the elementary cube of Figure4.3

only one minimal point 7, three α-saddle points 2, 4, 5, three β-saddle points 3, 6, 8,

and one maximal point 1. On this topology the minimal point is covered by three β-

saddle points, each β-saddle point is covered by two α-saddle points and covers a minimal

point, and each α-saddle point is covered by a maximal point and covers two α-saddle

points (See Figure 4.3).

Theorem 4.3.1. Let Z3 be an Alexandroff Hopf topology, and x = (a, b, c) ∈ Z3. Then

the following statements are equivalent:

(1) x ∈ m.

(2) N6(x) ⊆ β-S.

(3) N12(x) ⊆ α-S.

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Figure 4.3: The Alexandroff -Hopf Topology on an Elementary Cube

(4) N8(x) ⊆M .

(5) (a+ i, b+ j, c+ k) ∈M for i, j, k ∈ −1, 1.

Proof. (1 ⇒ 2) Suppose that x ∈ m. Let y ∈ N6(x). Then by Theorem 2.1.12, either

x → y or y → x. But since x ∈ m, then x y. Hence y ∈ β-S since the minimal point

in this topology covered by β-saddle points. Therefore N6(x) ⊆ β-S.

(2 ⇒ 3) Suppose that N6(x) ⊆ β-S and x = (a, b, c). Let y ∈ N12(x), then there exist

z ∈ Z3 such that y ∈ N6(z) and z ∈ N6(x) ⊆ β-S. Hence y ∈ α-S∪m. Assume to

contrary that y ∈ m. Since there exists an elementary cube contains both x and y and

any elementary cube has only one minimal point, so x ∈ α-S. Then (a + 1, b, c), (a, b +

1, c), (a, b, c+1) ⊆ (CU(x)∩N6(x)) ⊆ β-S. By Theorem 2.1.12, x covers three β-saddle

points in CU(x) which is a contradiction. Therefore y ∈ α-S and N12(x) ⊆ α-S.

(3 ⇒ 4) Suppose that N12(x) ⊆ α-S and x = (a, b, c). Let y ∈ N8(x). Thus y =

(a+ i, b+j, c+k) for some i, j, k ∈ −1, 1. Define z1 = (a+ i, b+j, c), z2 = (a+ i, b, c+k)

and z3 = (a, b + j, c + k). Hence z1 , z2 , z3 ⊆ N12(x) ⊆ α-S and y ∈ β-S ∪M . Suppose

to contrary that y ∈ β-S. Since z1 , z2 , z3 ⊆ α-S ∩N6(y), this implies that in some

elementary cube a β-saddle point y is covered by three α-saddle points z1 , z2 , z3 , which

105

is a contradiction. Thereby y ∈M and N8(x) ⊆M .

(4⇒ 5) This is coming directly.

(5⇒ 1) Suppose that (a+ i, b+ j, c+k) ∈M for i, j, k ∈ −1, 1 and x = (a, b, c). Define

u1 = (a + i, b + j, c), u2 = (a, b + j, c + k), z1 = (a + i, b, c), z2 = (a, b + j, c) and z3 =

(a, b, c + k). Then u1 , u2 ⊆ α-S. Since z1 ∈ N6(u1), then z1 ∈ β-S ∪M . Since on this

topology any elementary cube has only one maximal point, then z1 /∈ M . So z1 ∈ β-S.

Similarly z2 , z3 ∈ β-S. But z1 , z2 , z3 ⊆ N6(x), hence x ∈ m ∪ α-S. If x ∈ α-S, then

x is covering three β-saddle points in some elementary cube which is a contradiction.

Therefore x ∈ m.

Theorem 4.3.2. Let Z3 be an Alexandroff Hopf topology, and B ∈ m,M, β-S,α-S. If

(a, b, c) ∈ B, then (a+ 2i, b+ 2j, c+ 2k) ∈ B ∀i, j, k ∈ Z.

Proof. If B =m. Let (a, b, c) ∈ m and r ∈ −1, 1. Then (a+r, b+r, c+r) ∈ N8((a, b, c)).

By Theorem 4.3.1, (a+r, b+r, c+r) ∈M and (a+2r, b, c) = (a+r+r, b+r−r, c+r−r) ∈ m.

By the induction, (a + 2i, b, c) ∈ m for any i ∈ Z. By the same way, we can prove that

(a+ 2i, b+ 2j, c) ∈ m, and then (a+ 2i, b+ 2j, c+ 2k) ∈ m ∀i, j, k ∈ Z.

If B =M . Let (a, b, c) ∈ m and r ∈ −1, 1. Then by Theorem 4.3.1, (a+r, b+r, c+r) ∈

m and (a+2r, b, c) = (a+r+r, b+r−r, c+r−r) ∈M . By the induction, (a+2i, b, c) ∈M

for any i ∈ Z. Similarly (a+ 2i, b+ 2j, c), (a+ 2i, b+ 2j, c+ 2k) ∈M ∀i, j, k ∈ Z.

If B = β-S. Let (a, b, c) ∈ β-S and r ∈ −1, 1. Then (a + r, b, c) ∈ m ∪ α-S. If (a +

r, b, c) ∈ m, then by Theorem 4.3.1, (a + 2r, b, c) ∈ β-S. And if (a + r, b, c) ∈ α-S, then

(a+2r, b, c) ∈ β-S ∪M . If (a+2r, b, c) ∈M , then by Theorem 4.3.1, (a+r, b+r, c+r) ∈ m

and (a, b, c) ∈ M which is a contradiction. Hence (a + 2r, b, c) ∈ β-S. By the induction

(a+ 2i, b, c) ∈ β-S for any i ∈ Z. Similarly (a+ 2i, b+ 2j, c), (a+ 2i, b+ 2j, c+ 2k) ∈ β-S

∀i, j, k ∈ Z.

If B = α-S. Let (a, b, c) ∈ α-S and r ∈ −1, 1. Then (a + r, b, c) ∈ M ∪ β-S. If

106

(a + r, b, c) ∈ M , then by Theorem 4.3.1, (a + 2r, y + r, z + r) ∈ m and (a + 2r, b, c) ∈

N12((a + 2r, b + r, c + r)) ⊆ α-S. Therefore (a + 2r, b, c) ∈ α-S. If (a + r, b, c) ∈ β-S,

then (a + 2r, b, c) ∈ α-S ∪m. Suppose to contrary (a + 2r, b, c) ∈ m. By Theorem 4.3.1,

(a, b, c) ∈ m which is a contradiction. Hence (a + 2r, b, c) ∈ α-S. By the induction,

(a+ 2i, b, c) ∈ α-S for any i ∈ Z. Similarly (a+ 2i, b+ 2j, c), (a+ 2i, b+ 2j, c+ 2k) ∈ α-S

∀i, j, k ∈ Z.

Corollary 4.3.3. Let Z3 be an Alexandroff Hopf topology, and A ∈ A. Then A ⊆ B for

some B ∈ m,M, β-S,α-S.

Proof. Pick (x, y, z) ∈ A and fix it. Then there is B in m1,m2, β-S,M such that

(x, y, z) ∈ B. By the definition of the set A, we can write it in the form (x + 2i, y +

2j, z + 2k) : i, j, k ∈ Z. So by Theorem 4.3.2, A ⊆ B.

Corollary 4.3.4. Assume that Z3 an Alexandroff Hopf topology, and A ∈ A. If A∩m = ϕ,

then A = m.

Proof. If A ∩ m = ϕ, by Corollary 4.3.3, A ⊆ m. So we need to prove that m ⊆ A.

Suppose that b /∈ A. By Remak 3.1.3, A ∩ CU(b) = ϕ. If c ∈ A ∩ CU(b), then c = b.

Since A ⊆ m, then c ∈ m. Since there exists only one minimal point in CU(b) , b /∈ m.

Therefore m ⊆ A and A = m.

Theorem 4.3.5. Let Z3 be an Alexandroff Hopf space.

(1) m = EEE iffM = OOO iff β-S = OEE∪EOE∪EEO iff α-S = EOO∪OEO∪OOE.

(2) m = EEO iffM = OOE iff β-S = OEO∪EOO∪EEE iff α-S = EOE∪OEE∪OOO.

(3) m = EOE iffM = OEO iff β-S = OOE∪EEE∪EOO iff α-S = EEO∪OOO∪OEE.

(4) m = OEE iffM = EOO iff β-S = EEE∪OOE∪OEO iff α-S = OOO∪EEO∪EOE.

107

(5) m = EOO iffM = OEE iff β-S = OOO∪EEO∪EOE iffα-S = EEE∪OOE∪OEO.

(6) m = OEO iffM = EOE iff β-S = EEO∪OOO∪OEE iff α-S = OOE∪EEE∪EOO.

(7) m = OOE iffM = EEO iff β-S = EOE∪OEE∪OOO iff α-S = OEO∪EOO∪EEE.

(8) m = OOO iffM = EEE iff β-S = EOO∪OEO∪OOE iff α-S = OEE∪EOE∪EEO.

Proof. (1) Suppose that m = EEE. Let (a, b, c) ∈ EEE = m, then (a + 1, b + 1, c +

1) ∈ OOO. By Theorem 4.3.1, (a + 1, b + 1, c + 1) ∈ M . Hence by Corollary 4.3.3,

OOO ⊆M . Let (u, v, w) /∈ OOO, then (u+1, v+1, w+1) /∈ EEE = m. So by Theorem

4.3.1, (u, v, w) /∈M . Therefore M ⊆ OOO and M = OOO.

Suppose that M = OOO. Let (a, b, c) ∈ OOO = M . If y = (a + 1, b + 1, c + 1), then

y ∈ N8(x)∩ ∈ EEE. By Theorem 4.3.1, y ∈ m and N6(y) ⊆ β-S. Since N6(y) ∩ A = ϕ

for all A ∈ OEE,EOE,EEO, then by Corollary 4.3.3, OEE ∪ EOE ∪ EEO ⊆ β-

S. Now let z /∈ OEE ∪ EOE ∪ EEO. Hence there exists u ∈ EEE = m, such that

z ∈ u∪ (N26(u)\N6(u)). By Theorem 4.3.1, z /∈ β-S. Thus β-S⊆ OEE ∪EOE ∪EEO

and β-S = OEE ∪ EOE ∪ EEO.

Suppose that β-S = OEE ∪EOE ∪EEO. Let x ∈ EEE. Then N6(x) ⊆ OEE ∪EOE ∪

EEO = β-S. By Theorem 4.3.1, x ∈ m and N12(x) ⊆ α-S. But N12(x)∩A = ϕ for all A ∈

EOO,OEO,OOE. Thus EOO∪OEO∪OOE ⊆ α-S. Now let y /∈ EOO∪OEO∪OOE.

Then there exists z ∈ EEE = m, such that y /∈ N12(z). By Theorem 4.3.1, y /∈ α-S.

Thereby α-S⊆ EOO ∪OEO ∪OOE and α-S = EOO ∪OEO ∪OOE.

Suppose that α-S = EOO∪OEO∪OOE. Let x ∈ EEE. Then N12(x) ⊆ EOO∪OEO∪

OOE = α-S. By Theorem 4.3.1, EEE ∩m = ϕ. By Corollary 4.3.4, m = EEE.

As a direct result of the above theorem, there are eight topologies on Z3, all of them sat-

isfy the Subcase(iii) of case(2). To show that all of them are homeomorphic, let (Z3, τs1 ),

(Z3, τs2 ) are two Alexandroff Hopf spaces with corresponding of all minimal sets m1, m2

108

respectively. m1,m2 ∈ A. For example if m1 = EOE m2 = EEO, then a homeomor-

phism function f : (Z3, τs1 ) −→ (Z3, τs2 ) can be defined by f(a, b, c) = (a, b+1, c+1) for

(a, b, c) ∈ Z3.

Recall that we denoted the order of the Khalimsky topology by ”≤Kh” and the Khalimsky

dual topology by ”≤d

KH”. We are going to determine the order of the Alexandroff Hopf

topology and the minimal nhood of each point x ∈ Z3 with respect to it. Let we denote

them by ≤(3)

sand V

(3)

s(x) respectively.

Theorem 4.3.6. Let Z3 be an Alexandroff Hopf space and let (a, b, c), (x, y, z) ∈ Z3.

(1) If x ∈M , then V(3)

s(x) = x.

(2) If x ∈ α-S, then V (3)

s(x) = x ∪ (N6(x) \ β-S) = x ∪ (N6(x) ∩M).

(3) If x ∈ β-S, then V (3)

s(x) = x ∪ (N6(x) \m) ∪ (N12(x) \ β-S) = x ∪ (N6(x) ∩ α-

S)∪(N12(x) ∩M).

(4) If x ∈ m,then V(3)

s(x) = x ∪N26(x).

Proof. (1) True in any T0-Alexandroff space.

(2) Suppose that x ∈ α-S. Let y ∈ V (3)

s(x) and y = x. Then x ≤(3)

sy and y ∈M . Suppose

to contrary that y /∈ N6(x). By Theorem 2.1.12, y ∈ N12(x) or y ∈ N8(x). Since y ∈M ,

if y ∈ N8(x), then by Theorem 4.3.1, x ∈ m which is a contradiction. Moreover, if

y ∈ N12(x), then there exists z ∈ N6(y)∩N6(x). Hence z ∈ α-S and x ∈M ∪β-S which is

also a contradiction. So y ∈ N6(x). But y ∈M , then y ∈ N6(x)\β-S and y ∈ N6(x)∩M .

Therefore V(3)

s(x) ⊆ x ∪ (N6(x) \ β-S) and V

(3)

s(x) ⊆ x ∪ (N6(x) ∩ M). Now let

y ∈ N6(x) \ β-S, then y ∈ N6(x) and y /∈ β-S. Hence either y ≤(3)

sx or x ≤(3)

sy. If the

first one holds, then y ∈ β-S which is a contradiction. Then x ≤(3)

sy and y ∈ V (3)

s(x). So

x∪ (N6(x)\β-S) ⊆ V(3)

s(x) since x ∈ V (3)

s(x). And if z ∈ x∪ (N6(x)∩M), then either

109

z = x or z ∈ N6(x)∩M . Hence x ≤3

sz. Thus z ∈ V (3)

s(x). So x∪(N6(x)∩M) ⊆ V

(3)

s(x).

Therefore V(3)

s(x) = x ∪ (N6(x) \ β-S) = x ∪ (N6(x) ∩M).

(3) Suppose that x ∈ β-S.

Claim : N6(x)∩M = ϕ, N12(x)∩α-S = ϕ, and (N6(x)\m)∪ (N12(x)\β-S) = (N6(x)∩α-

S)∪(N12(x) ∩M). To prove this suppose that y ∈ N6(x) ∩M . So x ≤(3)

sy and x ∈ α-S

which is a contradiction. Then N6(x)∩M = ϕ. Suppose to contrary that y ∈ N12(x)∩α-S.

So there exists z ∈ N6(x) ∩N6(y). Since z ∈ N6(x) and x ∈ β-S, then z ∈ α-S ∪ m. But

z ∈ N6(y) and y ∈ α-S, hence z ∈ β-S∪M which is a contradiction. Therefore N12(x)∩α-S

= ϕ. Since x ∈ β-S,N6(x)∩β-S = ϕ. Let y ∈ N12(x)∩m. By Theorem 4.3.1, x ∈ α-S which

is a contradiction. Hence (N6(x)\m)∪ (N12(x)\β-S) = (N6(x)\ (β-S∪m)) ∪(N12(x)\ (β-

S∪m)) = (N6(x) ∪ N12(x)) \ (β-S∪m) = (N6(x) ∪ N12(x)) ∩ (α-S ∪M) = (N6(x) ∩ α-

S)∪(N12(x) ∩M). Now let y ∈ V (3)

s(x) and x = y, then x ≤(3)

sy and y ∈ M ∪ α-S. Since

x ≤(3)

sy, then y ∈ N26(x)∩ (α-S ∪M). Assume that y ∈ N8(x). Then there exists distinct

points z, w ∈ Z3 such that x ≤(3)

sz ≤(3)

sw ≤(3)

sy. This implies that x ∈ m which is a

contradiction. Thus y ∈ (N6(x)∪N12(x))∩(α-S ∪M) and V(3)

s(x) ⊆ x∪x∪(N6(x)∩α-

S)∪(N12(x) ∩M). Now let y ∈ (N6(x) \ m) ∪ (N12(x) \ β-S). Then y ∈ N6(x) \ m or

y ∈ N12(x) \ β-S. Since x ∈ β-S and y /∈ m, if y ∈ N6(x) \m, then y ∈ α-S and x ≤(3)

sy.

If y ∈ N12(x) \ β-S, then there exists z ∈ Z3 such that z ∈ N6(x) ∩ N6(y). Therefore

z ∈ m ∪ α-S. If z ∈ m ∩N6(y), then by Theorem 4.3.1, y ∈ β-S which is a contradiction.

Hence z ∈ α-S and y ∈ M . Thus x ≤(3)

sz ≤(3)

sy and y ∈ V (3)

s(x). Therefore V

(3)

s(x) =

x ∪ (N6(x) \m) ∪ (N12(x) \ β-S) = x ∪ (N6(x) ∩ α-S)∪(N12(x) ∩M).

(4) Suppose that x ∈ m. Let y ∈ V3

s(x) and y = x. Then x ≤(3)

sy. By Theorem

2.1.12, y ∈ N26(x). Thus V(3)

s(x) ⊆ x ∪ N26(x). Since x ∈ m, if y ∈ N6(x), then

x ≤3

sy. Hence y ∈ V

3

s(x). If y ∈ N12(x), then by Theorem 4.3.1, y ∈ α-S. But there

exist z ∈ Z3 such that z ∈ N6(x) ∩ N6(y). Hence z ∈ β-S. Then x ≤(3)

sz ≤(3)

sy and

110

y ∈ V(3)

P(x). If y ∈ N8(x), then by Theorem 4.3.1, y ∈ M . Then there exist z ∈ Z3

such that z ∈ N12(x) ∩ N6(y). As above x ≤(3)

sz. Since z ∈ N6(y) and y ∈ M , then

z ≤(3)

sy. Hence x ≤(3)

sy and y ∈ V (3)

s(x). Therefore x∪N26(x) ⊆ V

(3)

s(x). Then V

(3)

s(x)

= x ∪N26(x).

Recall that x ≤Kh

y iff ” x = y whenever x is odd number, and y ∈ N2(x) ∪ x

=x, x− 1, x+ 1 whenever x is even number”. Moreover, x ≤d

Khy iff ” x = y whenever

x is even number, and y ∈ N2(x) ∪ x =x, x− 1, x+ 1 whenever x is odd number”.

Theorem 4.3.7. Suppose that Z3 is an Alexandroff Hopf space with set of minimal point

m, and (a, b, c), (x, y, z) ∈ Z3.

(1) If m = EEE, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤

Khx, b ≤

Khy, and c ≤

Khz.

(2) If m = EEO, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤

Khx, b ≤

Khy, and c ≤d

Khz.

(3) If m = EOE, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤

Khx, b ≤d

Khy, and c ≤

Khz.

(4) If m = OEE, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤d

Khx, b ≤

Khy, and c ≤

Khz.

(5) If m = EOO, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤

Khx, b ≤d

Khy, and c ≤d

Khz.

(6) If m = OEO, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤d

Khx, b ≤

Khy, and c ≤d

Khz.

(7) If m = OOE, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤d

Khx, b ≤d

Khy, and c ≤

Khz.

(8) If m = OOO, then (a, b, c) ≤3

s(x, y, z) if and only if a ≤d

Khx, b ≤d

Khy, and c ≤d

Khz.

Proof. (1)Suppose that m = EEE.

Case(i) : If x = (a, b, c) ∈ OOO, then by Theorem 4.3.5, (a, b, c) ∈M . By Theorem 4.3.6,

V(3)

P(x) = x. Now (a, b, c) ≤(3)

s(x, y, z) iff (a, b, c) = (x, y, z) iff ”a = x, b = y, and c =

z” iff a ≤KH

x, b ≤KH

y, and c ≤KH

z.

111

Case(ii): If x = (a, b, c) ∈ EOO, then by Theorem 4.3.5, x ∈ α-S. By Theorem 4.3.6

V(3)

s(x) = x ∪ N6(x) ∩ M = (a, b, c), (a − 1, b, c), (a + 1, b, c). Now (a, b, c) ≤(3)

s

(x, y, z) iff (x, y, z) ∈ (a, b, c), (a+ 1, b, c), (a− 1, b, c) = a− 1, a, a+ 1 × b × c iff

x ∈ a − 1, a, a + 1 and b = y and c = z iff a ≤KH

x, b ≤KH

y, and c ≤KH

z (since a is

an even number, b and c are odd numbers).

The prove of the remaining cases (whenever (a, b, c) ∈ OEO, OOE, EEO, EOE, OEE,

EEE) is similar prove of the two cases (i) and (ii). The others have similar proof.

Corollary 4.3.8. The Alexandroff Hopf space on Z3 is the product space.

Proof. It is coming directly from Theorem 4.3.7.

Example 4.3.9. Let Z3 be the Alexandroff Hopf space with m = EOO. Then we can

determine the up of the point (2,3,3) under the τs topology by the following Hasse diagram:

(see Figure 4.4 )

Definition 4.3.10. The 3-Alexandroff-Hopf function ψ3 : F3 −→ Z3 is a bijection func-

tion define as:

ψ3(f) =

(2a,2b,2c), if dim(f) = 0 and f = (a, b, c);

(2a+1,2b,2c), if dim(f) = 1 and f = a, a+ 1 × b × c;

(2a,2b+1,2c), if dim(f) = 1 and f = a × b, b+ 1 × c;

(2a,2b,2c+1), if dim(f) = 1 and f = a × b × c, c+ 1;

(2a+1,2b+1,c), if dim(f) = 2 and f = a, a+ 1 × b, b+ 1 × c;

(2a+1,2b,2c+1), if dim(f) = 2 and f = a, a+ 1 × b × c, c+ 1;

(2a,2b+1,2c+1), if dim(f) = 2 and f = a × b, b+ 1 × c, c+ 1;

(2a+1,2b+1,2c+1), if dim(f) = 3 and f = a, a+ 1 × b, b+ 1 × c, c+ 1.

112

Figure 4.4: Up of the point (2,3,3) on an Alexandroff-Hopf space

Using the 3-Alexandroff-Hopf function and Theorem 1.2.5, we can prove that The Alexandroff-

Hopf topology with m = EEE is homeomorphic to the cellular-complex topology, and

hence we have the following theorem.

Theorem 4.3.11. The Alexandroff-Hopf topology (Z3, τs) is homeomorphic to the cellular-

complex topology (F3, τ⊆).

Theorem 4.3.12. Let Z3 be an Alexandroff-Hopf space, then:

(1) Z3 is a semi-T2-space.

(2) Z3 is not submaximal space and hence it is not a T 14-space.

113

Chapter 5

Digital Label Images

In this chapter, we use our previous results in some applications of the digital topologies.

We will focus on the Alexandroff-Hopf topology on Z2. Specially we study the digital

label images which was defined in [23]. The digital label images are the images whose

domain is Zd and whose codomains are sets on which there generally exists no meaningful

order relation. But we will translate this definition to correspond with our results. The

object of this translation is to replace the cellular-complex topology by digital space (in

this chapter the Alexandroff-Hopf topology on Z2). Importance of this issue, that this

process used in the simplicity process and for encoding the images.

5.1 Introduction to Label Images

Recall that a poset P is a lattice if ∀x, y ∈ P , x ∨ y = supx, y and x ∧ y = infx, y

exist. Let (P,≤) be a poset. Whenever there exists t ∈ P such that q ≤ t for each q ∈ P ,

we call t the top element of P and denote it by ⊤. Similarly, whenever there exists b ∈ P

such that b ≤ q for each q ∈ P , we call b the bottom element of P and denote it by ⊥. The

atom is an element of a poset that covers ⊥. A lattice is atomistic if any element, but the

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minimum is a supremum of atoms. For n ∈ 2d, 3d− 1, if x, y ∈ Zd, an n-path from x to

y (denoted by n-p(x, y) ) is a sequence of points x = x1, x2, · · · , xk = y from Zd such that

for 1 < i ≤ k, xi is n-adjacent to xi−1 and the length of this path (dented by L(n-p(x, y)))

is equal to k − 1. The length between two points x, y ∈ Zd (denoted by L(x, y)) is equal

to minL(2d-p(x, y)). For any x in the digital topology Zd, the dimension of x dim(x)

= minL(x, y) : y ∈ m. For instance, if a, b, c in (Z2, τs) such that a ∈ m, b ∈ SD

and c ∈ M , then dim(a) = 0, dim(b) = 1 and dim(c) = 2. When two points x, z ∈ Zd

cover a point y ∈ Zd and ↑x ∩ ↑z = ϕ, we say that x and z are opposite with respect

to y. We denote opp(y) the set of all x, z for x is opposite to z with respect to y. If

y ∈ M , then opp(y) = ϕ. For example, let Z3 be the Alexandroff Hopf space with m

= EOO. Then opp(2, 3, 3) = (1, 3, 3), (3, 3, 3), (2, 2, 3), (2, 4, 3), (2, 3, 2), (2, 3, 4),

opp(1, 3, 3) = (1, 2, 3), (1, 4, 3), (1, 3, 2), (1, 3, 4), opp(1, 2, 3) = (1, 2, 2), (1, 2, 4),

and opp(1, 2, 2) = ϕ (see Figure 4.4). Let [[1, n]] = [1, n]∩Z; i.e [[1, n]] = 1, 2, · · · , n. We

define the binary digital image η to be a function from the subset A (A ⊆ Zd) to 0, 1.

The foreground (resp. the background) associated to the image η : A −→ 0, 1 (with

A ⊆ Zd) is the set η−1(1)(resp. η−1

(0)). Often the foreground, background denoted

by X1, X0 respectively. Recall that, the set U ⊆ Zd is n-connected if for all x, y ∈ A

there exists n-path in A from x to y such that n ∈ 2d, 3d − 1. If V ⊆ U , we say V

is n-component of U if V is n-connected and there exists no n-path between a ∈ U \ V

and any point b ∈ V . Often used two different adjacency relations on the components of

the foreground and background. For instance let Z2 be the domain of the binary digital

image η. If the components of the foreground is 4-adjacency (resp. 8-adjacency), then

the components of the background is 8-adjacency(resp. 4-adjacency) and we say η has

(4, 8)-adjacency pair (resp. (8, 4)-adjacency pair).

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In order to model all the topological relations that can be found in a digital label

image ω (defined on Zd), we propose the following steps (c.f.[23] for more instance):

(1) We divide the image to the facets (elementary cells). For instance if the image is 2-

dimensional (resp. 3-dimensional) image we divide it to elementary squares(elementary

cube).

(2) Transpose the elementary cells to the maximal points (with respect to some digital

topology) by a bijection function θ from Fd

dto M . For example if the image is

2-dimensional and we choose the Allexandoff-Hopf topology on Z2 to be M respect

to it, then θ : F2

2−→M such that θ(f) = (2a+ 1, 2b+ 1) if L(f) = (a, b).

(3) Define a binary image λ : M −→ 0, 1 such that λ(x) = 1 if θ−1(x) has a collar

and λ(x) = 0 otherwise. This binary image is called the preserver image.

(4) We split the foreground (X1) of the binary image λ in a collection of binary images

such that each binary image represents a region of interest (a region of collar), this

region will called a initial label. The label is a union of some initial labels. Hence

the unions are labeled an atomistic finite lattice structure on labels (≤=⊆) whose

atoms are the initial labels. This lattice noted by T . For any t ∈ T we write λt

for the binary image associated to the label t in the collection built from the digital

label image λ, such that λt : M −→ 0, 1.

(5) In each binary digital image λt created in the previous step, we introduce ↓ a for

every a ∈M . Note that ↓M = Zd.

(6) Let ε : [[1, n]] −→ −1, 1 be a function called connectivity function. We define the

ε-regular (binary) image µt(x) : Zd −→ 0, 1, such that µt(x) = λt(x) if x ∈ M

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and, otherwise, we define:

µt(x) =

∧

a,b∈opp(x)µt(a) ∨ µt(b), if ε(r + 1) = 1;

∨a,b∈opp(x)

µt(a) ∧ µt(b), if ε(r + 1) = -1.

where r = dim(x). For instance when we extend M to have Z2 (finding ↓ M), we

set ε = (−1,−1) (resp. ε = (1, 1)) if the image has to be interpreted with the (4, 8)

(resp.(8, 4))-adjacency pair. In this step, we find separately µt(x) for all t ∈ T (note

that x ∈ Zd : µt(x) = 1 is called the sheet of t).

(7) After doing the previous six steps, a collection of ε-regular binary images (sheets)

(µt)t∈Tis given. To see this collection of binary images as a unique image ω (defined

on Zd), let we setting ω(x) = t ∈ T : µt(x) = 1. ω(x) is called the fiber of x.

Note that if µt(x) = 0 for all t ∈ T , then ω(x) = ϕ.

Remark 5.1.1. From the definition of the ε-regular image, when ε = 1 and y ∈ Zd, y ∈ X1

if and only if ∀x, z ∈ opp(y), x, y ∩ X1 = ϕ. Moreover when ε = -1, y ∈ X1 if and

only if ∃x, z ∈ opp(y) such that x, y ⊆ X1.

5.2 Covering Images

Proposition 5.2.1. Let Zd be a digital space, µt be ε-regular image and x ∈ Zd.

(1) If ε = 1 and x ∈ X1, then ↓ x ⊆ X1. So the sheets is closed sets.

(2) If ε = -1 and x ∈ X1, then ↑ x ⊆ X1. So the sheets is open sets.

Proof. The proof is coming directly from Remark 5.1.1.

Recall that if P is a poset and x ∈ P , then x = ↑ x ∩M . So from Proposition 5.2.1, we

can setting the following corollary.

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Corollary 5.2.2. Let Zd be a digital space, µt be the ε-regular image and x ∈ Zd.

(1) When ε = 1, ”x ∈ X1 if and only if x ∩X1 = ϕ”.

(2) When ε = -1, ”x ∈ X1 if and only if x ⊆ X1”.

Now for the lattice (T,⊆) of labels, there is a corresponding To-Alexandroff space (T, τT),

we will call this topology the fiber topology. Clearly that for any x ∈ Z2 and t ∈ T , if

t ∈ ω(x), then ↑ t ⊆ ω(x). Hence for all x ∈ Z2, ω(x) ⊆ τT=

∪t∈A ↑ t : A ⊆ T; so the

fibers are open set with respect to (T, τT) space.

To be clear all the previous steps consider the following example :

Example 5.2.3. Given the image setting in Figure 5.1(i). In order to covering this image

by the Allexandoff-Hopf topology on Z2, let we follow the previous steps:

(1) Splitting this image to the elementary squares as we see in Figure 5.1(ii).

(2) Transpose the elementary squares to the maximal points. as seen in Figure 5.1(iii).

(3) The foreground of the preserver image is the following points X1 = (3, 3), (5, 3), (7, 3),

(7, 5), (3, 5), (5, 7), (7, 7), (5, 11), (7, 11), (11, 7), (13, 7), (15, 7), (13, 11), (11, 9), (11, 11), (13, 9),

(15, 9), (15, 11).

(4) We have the following initial labels y = (3, 3), (5, 3), (7, 3), (7, 5), s = (5, 11), (7, 11),

r = (3, 5), (5, 7), (7, 7), b = (11, 7), (13, 7), (15, 7), (13, 11), c = (11, 9), (11, 11), (13, 9),

(15, 9), (15, 11). Also we have the following labels y, r, s, b, c, u, v, w such that u = c∪b,

v = g ∪ r ∪ y and w = u ∪ v ∪ s. This labels together with the empty set form a lattice T

( with order ” ≤ ” = ” ⊆ ”). Consider this lattice via the Hasse diagram in Figure 5.2.

Note that, for all t ∈ T , µt(x) = 1 whenever x ∈ t, and µt(x) = 0 otherwise.

(5) Find ↓ a for all a ∈M = OO. For example ↓ (3, 3) = N8(3, 3) ∪ (3, 3).

(6) In this step, we find separately µt(x) for all t ∈ T . For illustration, we will find the

sheet µr(x) whenever ε = 1 and the same sheet whenever ε = -1. Consider the label r

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in Figure 5.3(i), and the sheet µr(x) when ε = 1 in Figure 5.3 (ii), also the sheet µr(x)

when ε = -1 in Figure 5.3 (iii) (using Proposition 5.2.1 and Corollary 5.2.2). Observe

that the sheet when ε = 1 (resp. ε = -1) is closed (resp. open) set with respect to the

Allexandoff-Hopf topology on Z2.

(7) Here we want to find the fibers ω(x) for all x ∈ Z2. For example if ε = 1 (resp. ε =

-1), then ω(4, 7) = r, g, v, w (resp. ω(4, 7) = v, w). And if ε = 1 (resp. ε = -1), then

ω(4, 6) = r, g, v, w (resp. ω(4, 6) = ϕ). Moreover ω(5, 9) = ϕ anywhere.

Eventually, we can define a covering image as the following (this definition is similar

to a definition in [23]).

Definition 5.2.4. Let T be an atomistic finite lattice. A covering image ω is a function

from Zd to τT(fiber topology). For any t ∈ T , the sheet µt is the binary image defined on

Zd by µt(x) = 1 if t ∈ ω(x) and µt(x) = 0 otherwise. For any x ∈ Zd, the set ω(x) is the

fiber over x. A covering image is ε-regular if for any t ∈ T the sheet µt is ε-regular.

Proposition 5.2.5. [23] Let T be an atomistic lattice and ϵ a sequence of n elements

in −1, 1. A covering image ω : Zd −→ τTis ϵ-regular iff, for all x ∈ Zd such that

dim(x) < d:

ω(x) =

∩

a,b∈opp(x)ω(a) ∪ ω(b), if ε(m+ 1) = 1;

∪a,b∈opp(x)

ω(a) ∩ ω(b), if ε(m+ 1) = -1.

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Figure 5.1: Covering Image

120

Figure 5.2: A lattice of Labels

Figure 5.3: The Sheet µr(x)

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Conclusion

From our research, we can say that there exists only one digital topology on Z, two

digital topologies on Z2 and five digital topologies on Z3 which is (topologically) con-

nected whenever it is a graphically connected. All of these topologies are generalized

locally finite T0-Alexandroff spaces. Also The Alexandroff-Hopf topology on Z2 and The

Alexandroff-Hopf topology on Z3 are the product topology of the Khalimsky topology.

The Alexandroff-Hopf topology is homeomorphic to the cellular-complex topology. This

thesis will open new way for how we can use its results in the segmentation of digital

image, repairing digital images, digital label image process, and coding the digital image

process. We hope that this thesis help the researches to deal with the fourth dimension

space (or greater) more easily.

122

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