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# New Topology on Symmetrized Omega Algebra

By Mesfer Alqahtani, Cenap Özel and Ibtesam Alshammari

Submitted: October 16th 2019Reviewed: May 7th 2020Published: July 13th 2020

DOI: 10.5772/intechopen.92764

## Abstract

The purpose of this paper is to define a new topology called symmetrized omega algebra topology and discuss some of its topological properties. Two different examples from an ordered infinite set of symmetrized omega topology are introduced. Furthermore, we study the relationship between symmetrized omega topology and weaker kinds of normality.

### Keywords

• tropical geometry
• idempotent semiring
• topological space
• topological properties
• omega algebra and symmetrized omega algebra

## 1. Introduction

Tropical geometry is the most recent but fast growing branch of mathematical sciences, which is analytically based on idempotent analysis and algebraically on idempotent semirings also known as tropical semirings. These are basically extended sets of real numbers R:Rand R:Rwhich are given monoidal structures by using min and max operations for addition, respectively. In order to adhere to the semiring structure, the additive operation of Ris used as the multiplication operation. By these choices, both Rand Rbecome idempotent semirings. The literature, they are also termed as min and max plus algebras, respectively. In both cases, 0of Rbecomes a multiplicative identity and and become additive identities of these semirings, respectively. Interestingly, some authors associate Rto tropical geometry, while other authors associate Rto tropical geometry (see [1, 2, 3, 4]). Omega algebra or “ωalgebra” for short, unifies the different terms and introduces an original structure, which, in fact, is an “abstract tropical algebra”. The Rand Rand their nearby structures, like minmaxand maxtimes algebras, etc., are all subsumed under omega algebra. All these are idempotent semirings, which are also called dioids. In previous studies, for the construction of all such semirings, an ordered infinite abelian group is mandatory. In ωalgebra, the definition is extended to cyclically ordered abelian groups and also to finite sets under some suitable ordering. Note that cyclically ordered abelian groups are more general than that of ordered abelian groups [5]. The aim of this paper is to define a new topology on symmetrized omega algebra, and discus some of its topological properties. Two different examples from an ordered infinite sets of symmetrized omega topology are introduced. Furthermore, we study the relationship between symmetrized omega topology and weaker kinds of normality. Our paper is organized as follows. In Section 2, we review an abstract definition for some basic facts about abstract omega algebras. In addition, we give a brief of symmetrized omega algebra and rules of calculation in omega. In Section 3, we define a new topology on symmetrized omega algebra and discuss some of its topological properties. In Section 4, we provide two different examples of symmetrized omega topology: the first and second examples are from an ordered infinite set. Finally, we study the relationship between symmetrized omega topology and weaker kinds of normality in Section 5. Throughout this paper, we do not assume T2in the definition of compactness. We also do not assume regularity in the definition of Lindelöfness.

The ideas from this paper were taken from the PhD thesis of Mr. Mesfer Hayyan Alqahtani in King Abdulaziz University.

## 2. Preliminaries

In this section, we provide an abstract definition for review some basic facts about abstract omega algebra. Furthermore, we also provide a brief of symmetrized omega algebra and rules of calculation in omega. For more details, see [6].

### 2.1 Omega algebra

Let Gebe an abelian group. Let Abe a closed subset of Gand eA.Then Aeis a submonoid of G.Assume that ωis an indeterminate (may belong to Aor G,as we will see in Examples 1 and 2). Obviously, in this case ωis no longer an indeterminate. Because the terms are generated from tropical geometry, this indeterminate can be called a tropical indeterminate.

Definition 1. [6]

We say that Aω=Aωis an omega algebra (in short ωalgebra) over the group Gin case Aωis closed under two binary operations,

,:Aω×AωAω,E1

such that a1,a2,a3A,the following axioms are satisfied:

1. a1a2=a1ora2;

ii. a1ω=a1=ωa1;
iii. ωω=ω;
iv. a1a2=a2a1A;
v. a1a2a3=a1a2a3;
vi. a1e=a1;
vii. a1ω=ωa1=ωifωea1ifω=e;
viii. ωω=ω;
ix. a1a2a3=a1a2a1a3.

Remark 2. [6]

1. is a pairwise comparison operation such as max, min, inf, sup, up, down, lexicographic ordering, or anything else that compairs two elements of Aω.Obviously, it is associative and commutative and the tropical indeterminate ωplays the role of the identity. Hence Aωωis a commutative monoid.

2. is also associative and commutative on Aω, and eplays the role of the multiplicative identity of Aω. Hence, Aωeis also a commutative monoid.

3. The left distributive law (ix) also gives the right distributive law.

4. Every element of Aωis an idempotent under .

5. Altogether, we write both structures as: Aω=Aωωe.This is an idempotent semiring.

Remark 3. [6] A ωalgebra can similarly be defined over a commutative monoid, ring, or even a semiring. More generally, one may construct analogously such algebras on other weaker structures. In this note, we confined ourselves to only ωalgebras over abelian groups and rings.

Example 4. [6] Max-plus algebra, min-plus algebra and all such “so called” algebras are particular cases of the ωalgebra over the ring Ror its associated subrings. A simpler example is the following. In the abelian group Z+,for any integer m,we have Wm=0m2m.This is an additive submonoid of (Z,+).Let ω=,a1a2=maxa1a2and a1a2=a1+a2,a1,a2Wm.Then,

Wm=Wm0E2

is algebra over the abelian group of integers Z. Hence, we have a sequence of ωsubalgebras

WmW2m.

Example 5. [6] Cartesian products of omega algebras. In this example, we explain a construction of an omega algebra from other given omega algebras. Let Giiei:i=1nbe abelian groups and Aωiiiωiei:i=1nbe a respective family of omega algebras, where ωiare tropical indeterminate. As usual, we define the Cartesian product as

Xω=Aω1××Aωn=a1an:aiAωii=1n.E3

In order to provide a convenient technique to give an additive structure to Xω,we assume that the ntuplesa=a1an,b=b1bnXωare in lexicographic ordering. Then, to define the sum

ab=aorbE4

by using the following rules:

Ifa11b1=a1thenab=a.E5
Ifai=bifor1ikn,andak+1k+1bk+1=ak+1,then,ab=a.E6

Similarly, rules for ab=bcan be determined. Multiplication can be to define component wise. Thus,

ab=a11b1annbn.E7

The other rules of the Definition 1 can straightforwardly be verified. Hence, Xωωe, where ω=ω1ωnis the additive identity and e=e1enis the multiplicative identity of Xω,is an omega algebra over the Cartesian product of abelian groups G1××Gn..

### 2.2 The symmetrized omega algebra

Let Gebe an abelian group and Aωωean ωalgebra over the group G. Following the method used in constructing integers from the natural numbers, we consider the set of ordered pairs Pω=Aω2with component wise addition , for all ab, cdPω,

abcd=acbdE8

Because of the four possibilities ab, ad, cdor cbfor the result, the addition in (8), is ambiguous. As our goal from constructing the algebra of pairs is the construction of the symmetrized omega algebra of Aω, we are in front of two possibilities: One is to use –for n=2, and define an equivalence relation on the ωalgebra of pairs which is compatible with relevant operations, and the other is to define an equivalence relation on the set Pωthat allows the component wise addition to be defined in the quotient set.

First Construction, let be the ordering defined on Aωby the relation

abab=bE9

which gives a total order on Aωand for all aAω, we have ωa. For ab, such that ab=b, we denote by a<b. Under the ordering , rules (5) and (6) defined in Example 5, are satisfied on Pω=Aω2and so Pωis an ωalgebra under the addition defined in 1 and the component wise multiplication. Let be the relation defined on Pωas follows: for all ab, cdPω

Then is reflexive and symmetric but not transitive for Aωcontains more than 4elements. In fact, let a, b, c, dAωsuch that a<b<c<d, then we have

which give abddand ddbc, but there is no relation between aband bc. As is not an equivalence relation, we cannot use it to obtain the quotient ωalgebra Pω(like the one to obtain integers from the natural numbers).

Definition 6. [6] Let be the equivalence relation close to defined as follows: for all ab, cdPω,

abcdabcdifabandcdab=cdotherwiseE11

In addition to the class element ω¯=ωω¯; for all aAω, with aω, we have three kinds of equivalence classes:

1. aω¯=abPωb<a, called positive ωelement.

2. ωa¯=baPωb<a, called negative ωelement.

3. aa¯called balanced ωelement.

Unfortunately, the addition defined by (7) and rules (8) and (9) in Example 5 is not compatible with the equivalence relation in Pω, because for aω, ab, ωc, dcPω, such that

aωabωcdc,E12

we have

aωωcabdciffabdc=abE13
and ifabdc=dc,E14

then there is no compatibility. So the omega algebra of pairs cannot produce the symmetrized omega algebra.

Second Construction

Proposition 7. [6]

ab¯¯cd¯=acbd¯

on the quotient set Pωis well defined and satisfies the axioms i, iiand iiiof Definition 1, with the zero class element w¯=ωω¯, except this case aω¯¯ωa¯=ωa¯¯aω¯=aa¯,where aAω\ωdoes not satisfy the axiom i.

Proposition 8. [6]

1. The set Pωis closed under the binary multiplication operation ¯defined as follows: for all ab¯, cd¯Pω;

and satisfies axioms from ivto ixof Definition 1, with the unit class element e¯=eω¯.

1. In addition, we have for all a, bAω

1. aω¯¯bω¯=abω¯;

2. aω¯¯ωb¯=ωab¯;

3. aω¯¯bb¯=abab¯;

4. ωa¯¯bb¯=abab¯.

Definition 9. [6] The structure Pω¯¯ω¯e¯is called the symmetrized ωalgebra over the abelian group G×Gand we denote it by Sω.

In the coming sections just for simplicity we will only use and instead the operations ¯and ¯,respectively.

Remark 10. [6]

1. Despite the nature of the positive and the negative ωelements, they are not the inverses of each other for the additive operation ¯,

2. We have three symmetrized ωsubalgebras of Sω,

Sω+=aω¯aAω,
Sω=ωa¯aAω,
Sω0=aa¯aAω.

• The three symmetrized ωsubalgebras of Sωare connected by the zero class element ω¯.

• The positive ωelements, the negative ωelements and the balanced elements are called signed and denoted by Sω=Sω+Sω, where the zero class ωω¯corresponds to ω.

• ### 2.3 Rules of calculation in omega

Notation 11. [6] Let aAω. Then we admit the following notations:

+a=aω¯,a=ωa¯,a=aa¯.E16

By results in Proposition 7 and Proposition 8 and the above notation, it is easy to verify the rules of calculation in the following proposition:

Proposition 12. [6] For all a,bAω, we have

1. +a+b=+ab;

2. +ab=+aifb<abifb>a;aifba

3. ±ab=±aifb<abifb>a:

4. ab=ab;

5. +a+b=+ab;

6. +ab=ab;

7. ±ab=ab;

8. ab=+ab.

From the previous rules, we can notice that the sign of the result in the addition operation follows the greater element in Aω. While in the multiplication operation, the balance sign is the strong one (has priority).

## 3. Symmetrized omega topology

In this section, we define a new topology on symmetrized omega algebra and discuss some of its topological properties.

Throughout this paper, we assume that A=.

Proposition 13. Let Sω=Pω¯¯ω¯e¯be a symmetrized ωalgebra over the abelian group G×G, where Pω=Aω×Aωand A=. We define a new topology on Sωcalled a symmetrized omega topology, denoted by τωas follow:

τω=Sω{USω:Sω0Uand for any +a,aU, their multiplicative inverses exists in U, where aAω\ω}.

Proof. Condition ,Sωτωis satisfied from the definition of τω.Now let V1,V2τωbe arbitrary. If either V1or V2is equal , then V1V2=τω. Assume now, V1V2. If either V1or V2is equal Sω, then V1V2=V1orV2τω. So assume that, V1SωV2, then V1V2τω, because Sω0V1and Sω0V2, hence Sω0V1V2, also for any element +a,aV1V2, where aω, then we have +a,aV1and +a,aV2, then the multiplicative inverse of +a,amust belong to V1and V2. Hence, the multiplicative inverse of +a,abelong to V1V2, then V1V2τω. For the third condition let Sγτωfor any γΛ. If Sγ=for all γΛ, then γΛSγ=τω. So, assume that some member is nonempty, but since the empty set does not affect any union, we may assume, without loss of generality, that Sγfor all γΛ. If there exist γ1Λsuch that Sγ1=Sω, then γΛSγ=Sωτω. So, assume now that SγSωfor all γΛ. Then γΛSγτω, because Sω0Sγfor all γΛ. Hence Sω0γΛSγ. Also for any +a,aγΛSγ, where aω, there exists γ1,γ2Λsuch that +aSγ1and aSγ2. Hence the multiplicative inverse of +a,abelong to Sγ1and Sγ2respectively, then the multiplicative inverse of +a,abelong to γΛSγ. Hence γΛSγτω.

Therefore, Sωτωis topological space.

Proposition 14. If Sω=Pω¯¯ω¯e¯be a symmetrized ωalgebra over the abelian group G×G, where Pω=Aω×Aω,and A=. Then an element ahas a multiplicative inverse in Aωif and only if the elements +a,ahave a multiplicative inverses in Sω.

Proof. Let aAωbe arbitrary, which has a multiplicative inverse, denoted by a1, then

+a+a1=aω¯a1ω¯=aa1ωωaωωa1¯=aa1ω¯=aa1ω¯=eω¯=e¯,

then +a1is a multiplicative inverse of +ain Sω.Also,

aa1=ωa¯ωa1¯=ωωaa1ωa1aω¯=aa1ω¯=aa1ω¯=eω¯=e¯,

then a1is a multiplicative inverse of ain Sω.

Conversely, let +aSωbe arbitrary, which has a multiplicative inverse xy¯, where x,yAω, then we have:

+axy¯=aω¯xy¯=axωyayωx¯=axay¯=axay¯=eω¯=e¯,E17

then ax=eand ay=ω. Hence, x=a1is the multiplicative of ain Aω.

Let aSωbe arbitrary, which has a multiplicative inverse xy¯, where x,yAω, then we have:

axy¯=ωa¯xy¯=ωxayωyax¯=ayax¯=ayax¯=eω¯=e¯,E18

then ay=eand ax=ω. Hence, y=a1is the multiplicative inverse of ain Aω.

Proposition 15. For any aSω0, where ωe,then ahas no multiplicative inverse.

Proof. Suppose that, aSω0has a multiplicative inverse xy¯, where x,yAω, then

axy¯=aa¯xy¯=axayayax¯=axayayax¯=eω¯.E19

Hence, axay=eand ayax=ω, thus a contradiction.

Corollary 16. If aAω\ωhas no multiplicative inverse, then Sωis the only open set in Sωτωcontaining +aand a..

Remark 17.

1. We denote for any element aSω,by sign.aor signaa,where sign.,signa+;

2. If a=ω, then a=+a=a;

3. If a1is the multiplicative inverse of ain Aω, then +a1and a1are the multiplicative inverses of +aand a,respectively in Sω(vice versa);

4. If ahas no multiplicative inverse in Aω, then +aand ahave no multiplicative inverses in Sω(vice versa).

Proposition 18. A symmetrized omega topological space Sωτωhas a base

B=SωSω0Sω0+a+a1Sω0aa1:aAω\ωhasamultiplicative inverse.E20

Proof. For the first condition, let BBbe arbitrary. If B=Sω0or Sωthen Bτω(satisfied by the definition of τω). Assuming that, B=Sω0+a+a1orSω0aa1for any aAω\ω, which has a multiplicative inverse in Aω, then Bτω, because Sω0B,and the elements +aandain Bits multiplicative inverse +a1anda1respectively, exists in B. Thus Bτω. For the second condition, let signaaSωbe arbitrary. Let Ube any open neighborhood of signaain Sω. Then we have three cases:

Case 1: If signa=, then there exists B=Sω0B, such that aBU, because the smallest open neighborhood in Sωcontaining ais Sω0.

Case 2: If signa=+,where aω(If a=ω, then we have +ω=ω=ω, this is Case 1),

Subcase 2.1: If ahas a multiplicative inverse in Aω, then there exists B=Sω0+a+a1B, such that +aBU, because the smallest open neighborhood in Sωcontaining +ais Sω0+a+a1.

Subcase 2.2: If ahas no multiplicative inverse in Aω, then there exists B=Sω, such that +aBU, because the smallest open neighborhood in Sωcontaining +ais Sω.

Case 3: If signa=,where aω.

Subcase 3.1: If ahas a multiplicative inverse in Aω, then there exists B=Sω0aa1B, such that aBU, because the smallest open neighborhood in Sωcontaining ais Sω0aa1.

Subcase 3.2: If ahas no multiplicative inverse in Aω, then there exists B=Sω, such that aBU, because the smallest open neighborhood in Sωcontaining ais Sω.

Therefore, Bis a base for the symmetrized omega topological space Sωτω.

Corollary 19. If Aω\ωbe a group, then the symmetrized omega topological space Sωτωhas a base,

B=Sω0Sω0+a+a1Sω0aa1:aAω\ω.E21

Corollary 20. Let UAω, then Uτωif and only if for each signaaU, there exists basic open set BB, such that signaaBU.

Proposition 21. If Aωhas a finite number of elements, which have a multiplicative inverses, then the symmetrized omega topological space Sωτωis second countable.

Proof. Suppose that a1,a2,,am, where mZ+are the finite number of elements in Aω, which have a multiplicative inverses. Then.

B=SωSω0Sω0+a1+a11Sω0a1a11Sω0+am+am1Sω0amam1is a countable base for Sωτω.

Proposition 22. The symmetrized omega topological space Sωτωis first countable.

Proof. Let signaaSωbe arbitrary. Then we have three cases:

Case 1: If signa=, then Ba=Sω0is a countable local base at a.

Case 2: If signa=+,where aω(If a=ω, then +ω=ω=ω, this is Case 1),

Subcase 2.1: If ahas a multiplicative inverse in Aω, then B+a=Sω0+a+a1is a countable local base at +a.

Subcase 2.2: If ahas no multiplicative inverse in Aω, then B+a=Sωis a countable local base at +a.

Case 3: If signa=,where aω,

Subcase 3.1: If ahas a multiplicative inverse in Aω, then Ba=Sω0aa1is a countable local base at a.

Subcase 3.2: If ahas no multiplicative inverse in Aω, then Ba=Sωis a countable local base at a. Hence, for any signaaSω, there exists a countable local base at signaa.

Therefore, Sωτωis first countable.

Proposition 23. The symmetrized omega topological space Sωτωis separable.

Proof. There exists ω=ωω¯Sω, such that for any Uτω, we have Uω, because any open set in Sωτωmust be containing Sω0,and ωSω0. Then ωis countable dense subset of Sω.Therefore, Sωτωis separable.

Let us recall this definition.

Definition 24. A topological space Xis said to be hyperconnected space if every non-empty open set of Xis dense in Xor there exists no disjoint non-empty open sets in X.

Proposition 25. The symmetrized omega topological space Sωτωis hyperconnected.

Proof. If Sωis singleton, then it is hyperconnected. Suppose that Sω, which has more than one element. Since any nonempty open set in Sωis containing Sω0, then Sωhas no disjoint nonempty open sets. Hence, Sωτωis hyperconnected.

Since any hyperconnected space is connected and locally connected, then we conclude the following corollaries.

Corollary 26. The symmetrized omega topological space Sωτωis connected.

Corollary 27. The symmetrized omega topological space Sωτωis locally connected.

Proposition 28. Let Aω\ωbe a group, has more than one element. Then the symmetrized omega topological space Sωτωis not T0.

Proof. If Aω=ω=e, then Sω=ωis singleton, we are done (because some of omega algebra, has ω=e). Suppose that Aωhas more than one element. Let aω, then there exist aωin Sω.Let Ube any open set in Sω, containing either aor ω, by the definition of τωwe have Sω0U,but ω,aSω0.Then there is no open set containing only ωor a.Hence, Sωτωis not T0.

Proposition 29. If Aω\ωbe a group, has more than one element, then the symmetrized omega topological space Sωτωis not regular.

Proof. There exists K=Sω\Sω0is a closed subset of Sωand there exists aω, such that aK. We cannot separate a,and Kby any open sets (because any open sets in Sωis containing Sω0, where aSω0). Therefore, Sωτωis not regular.

Proposition 30. If Aω\ωbe a group, has more than one element, then the symmetrized omega topological space Sωτωis not normal.

Proof. If Aω=ω, then Sω=ωis singleton, we are done (because some of omega algebra, we have ω=e). Suppose that Aωhas more than one element. Let aAω\ω. Then we have two cases:

Case 1: If a=e, then we have K=+e,and H=eare two disjoint closed subsets of Sω, such that we cannot separate them by any open sets (because any nonempty open sets in Sωis containing Sω0).

Case 2: If ae, then we have K=+a+a1,and H=aa1are two disjoint closed subsets of Sω, such that we cannot separate them by any open sets (because any nonempty open sets in Sωis containing Sω0). Therefore, Sωτωis not normal.

Proposition 31. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis normal.

Proof. Suppose that, Vbe any non-empty closed subset of Sω.Then +aV.Suppose not, +aV,then +aSω\V.By the definition of τω,Sω\Vis not open, thus a contradiction. Hence, +abelong to any non-empty closed subsets of Sω.Let Kand Hbe any two disjoint closed subsets of Sω. Then Kor His equal . If K=, then there exists U=and V=Sωare two disjoint open sets in Sωcontaining Kand H,respectively. If H=, then there exists U=and V=Sωare two disjoint open sets in Sωcontaining Hand K,respectively. Therefore, Sωτωis normal.

Proposition 32. If Aω\ωbe a group and Ais uncountable infinite set, then the symmetrized omega topological space Sωτωis not compact (Lindelöf).

Proof. There exists Sω0Sω0+a+a1Sω0aa1:aAω\ω,which is an open cover of Sω, and has no finite (countable) subcover of Sω.

Proposition 33. Let aAω\ωhas no multiplicative inverse. Then the symmetrized omega topological space Sωτωis compact.

Proof. Let Cα:αΛbe any open cover of Sω. Since +aSω,then for some βΛ, there exists Cβcontaining +a. But Cβ=Sω, because Sωis the only open set containing +a. Hence, Cβis a finite subcover of Cα:αΛ,which cover Sω. Therefore Sωτωis a compact space.

Since any compact space is Lindelo¨f and countably compact, then we conclude the following corollaries.

Corollary 34. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis Lindelo¨f.

Corollary 35. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis countably compact.

Remark 36. Since every nonempty open sets in Sωτωcontains Sω0. Then the closure of any nonempty open sets is equal Sω

## 4. Some of the fundamental properties for different examples on symmetrized omega topology

In this section, we give two different examples of symmetrized omega topologies. The examples are from an ordered infinite set.

Example 37. By Example 4, we set W=0,1,2,3.Then Sτ,which is topological space, where S=P¯¯¯0¯be a symmetrized algebra over the abelian group Z×Zand P=W×W. Let aW\0be arbitrary. Then +a1and a1are not exists in S, where +a1and a1are the multiplicative inverses of +aand ain Srespectively (because ain Whas no multiplicative inverse). If a=0,then +01=+0and 01=0(because the multiplicative inverse of 0in Wis 0, that is 01=0). Hence,

τ=SS0S0+0S00S0+00.E22

A direct check shows that Sτis a topological space.

Proposition 38. The symmetrized omega topological space Sτis Second countable.

Proof. There exists only one element 0W, which has a multiplicative inverse, then by Proposition 21, Sτis second countable.

Since any second countable space is first countable and separable, then we conclude the following corollaries.

Corollary 39. The symmetrized omega topological space Sτis first countable.

Corollary 40. The symmetrized omega topological space Sτis separable.

Proposition 41. The symmetrized omega topological space Sτis not T0.

Proof. There exists +2+3in S. Let Ube any open set, which either containing +2or +3.However, there exists only one open set U=Scontaining +2,+3. Hence, Sτis not T0.

Proposition 42. The symmetrized omega topological space Sτis not regular.

Proof. There exists a closed set K=S\S0+0and +0K, such that +0and Kcannot separate by any two disjoint open sets. Hence, Sτis not regular.

Proposition 43. The symmetrized omega topological space Sτis normal.

Proof. There exists an element 2W\, which has no multiplicative inverse, then by Proposition 31, Sτis a normal space.

Proposition 44. The symmetrized omega topological space Sτis hyperconnected.

Proof. Since any nonempty open set in Sis containing S0, then Shas no disjoint nonempty open sets. Hence, Sτis hyperconnected.

Since any hyperconnected space is connected and locally connected, then we conclude the following corollaries.

Corollary 45. The symmetrized omega topological space Sτis connected.

Corollary 46. The symmetrized omega topological space Sτis locally connected.

Proposition 47. The symmetrized omega topological space Sτis compact.

Proof. There exists an element 2W\, which has no multiplicative inverse. Hence by Proposition 33, Sτis compact.

Since any compact space is Lindelo¨f and countably compact, then we conclude the following corollaries.

Corollary 48. The symmetrized omega topological space Sτis countably compact.

Corollary 49. The symmetrized omega topological space Sτis Lindelöf.

Example 50. In the ring R+, we have R+is an additive submonoid of an abelian group R+.Let ω=,ab=maxaband ab=a+b,a,bR. Then R=R0is algebra over the ring R+. We have S=P¯¯¯0¯be a symmetrized algebra over the abelian group R×Rand P=R×R. Then, using the same proof as that Proposition 13. Therefore, Sτis a topological space.

Remark 51. The symmetrized omega topological space Sτis first countable, separable, hyperconnected, connected and locally connected and does not satisfy any of these T0, regular, normal, Lindelo¨f and compact.

Example 52. In the ring R+, we have R+is an additive submonoid of an abelian group R+.Let ω=+,ab=minaband ab=a+b,a,bR. Then, R+=R++0is +algebra over the ring R+. We have S+=P+¯¯+¯0¯be a symmetrized +algebra over the abelian group R×Rand P+=R+×R+. Then, using the same proof as that Proposition 13. Therefore, S+τ+is a topological space.

Proposition 53. The symmetrized omega topological spaces Sτand S+τ+are homeomorphic.

Proof. There exists a map h:SτS+τ+is defined by:

hsignaa=signaaifaRsign+ifsignaa=sign;E23

Let signaa,signbbSbe arbitrary. Let hsignaa=hsignbb, then signaa=signbb. Hence, his an injective. Let signaaS+is arbitrary, then there exists a signaaS, such that hsignaa=signaa. Hence, his surjective.

Let Bτ+be any basic open set. By Proposition 18, we have B=S0S0+a+a1S0aa1:aRand B=S+0S+0+a+a1S+0aa1:aRare a base for Rand R+,respectively.

To prove that his continuous, we have three cases:

Case 1: If B=S+0, then h1B=h1S+0=S0τ.

Case 2: If B=S+0+a+a1, then h1B=h1S+0+a+a1=S0+a+a1τ.

Case 3: If B=S+0aa1, then h1B=h1S+0aa1=S0aa1τ. Hence, his continuous.

To prove that h1is continuous, we have three cases: (since his one to one and onto, then h11B=hB).

Case 1: If B=S0, then h11B=hB=hS0=S+0τ+.

Case 2: If B=S0+a+a1, then h11B=hB=hS0+a+a1=S+0+a+a1τ+.

Case 3: If B=S0aa1, then h11B=hB=hS0aa1=S+0aa1τ+. Hence h1is continuous (which means his open).

Therefore, his homeomorphism, then Sτand S+τ+are homeomorphic.

## 5. Symmetrized omega topology and other properties

Recall that a subset Aof a space Xis said to be regularly-open or an open domain if it is the interior of its own closure (see [7]). A set Ais said to be a regularly-closed or a closed domain if its complement is an open domain. A subset Aof a space Xis called a π-closed if it is a finite intersection of closed domain sets (see [8]). A subset Ais called a π-open if its complement is a π-closed. If Tand Tare two topologies on a set Xsuch that T'T, then Tis called the coarser topology than T, and Tis called the finer. A space Xis π-normal [9] if any pair of disjoint closed subsets Aand Bof X, one of which is π-closed, can be separated by two disjoint open subsets. A space Xis almost-normal [9] if any pair of disjoint closed subsets Aand Bof X, one of which is a closed domain, can be separated by two disjoint open subsets. A space Xis mildly normal [10] if any pair of disjoint closed domain subsets Aand Bof Xcan be separated by two disjoint open subsets. A space XTis epi-mildly normal [11] if there exists a coarser topology Ton Xsuch that XTis T2and mildly normal space. A space XTis epi-almost normal [12] if there exists a coarser topology Ton Xsuch that XTis T2and almost normal space.

Theorem 54. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis π-normal.

Proof. Since the only π-closed sets are the ground set Sωand the empty set, then Sωτωis a π-normal.

It is clear from the definitions that

normalπnormalalmost normalmildly normal.E24

By (24) and Theorem 54, we conclude the following Corollaries.

Corollary 55. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis almost normal.

Corollary 56. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis mildly normal.

If Aω\ωbe a group has more than one element, then Sωτωis not T0(see Proposition 28), we have the following Propositions:

Proposition 57. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis not Epi-mildly Normal.

Proof. Suppose that, Sωτωis Epi-mildly Normal. Then there exists a coarser topology Ton Sωsuch that SωTis T2and mildly normal space. Hence Sωτωis T2, thus a contradiction. Then Sωτωis not Epi-mildly Normal.

Proposition 58. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis not Epi-almost Normal.

Proof. Using the same proof of Proposition 57.

Definition 59. Let Xbe a space. Then:

1. A space Xis called a C-normal if there exist a normal space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each compact subspace AX, [13].

2. A space Xis called a CC-normal if there exists a normal space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each countably compact subspace AX. [14].

3. A space Xis called an L-normal if there exist a normal space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each lindelöf subspace AX, [15].

4. A space Xis called an S- normal if there exist a normal space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each separable subspace AX, [16].

5. A space Xis called a C-paracompact if there exist a paracompact space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each compact subspace AX, [17].

6. A space Xis called a C2-paracompact if there exist a Hausdorff paracompact space Yand a bijective function f:XYsuch that the restriction function fA:AfAis a homeomorphism for each compact subspace AX, [17].

Proposition 60. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis C-normal.

Proof. By Proposition 31, Sωis a normal space. Then there exist Y=Sωis a normal space and the identity function id:SωSωis bijective. Let Cbe any compact subset of Sωτω.Then the restriction function idC:CfCis a homeomorphism. Therefore, Sωτωis a Cnormal.

Since any normal space is CC-normal, L-normal and S-normal, just by taking X=Yand fto be the identity function. Hence, we conclude the following Propositions.

Proposition 61. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis CC-normal.

Proof. Using the same proof of Proposition 60.

Proposition 62. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis L-normal.

Proof. Using the same proof of Proposition 60.

Proposition 63. If aAω\ωhas no multiplicative inverse, then the symmetrized omega topological space Sωτωis S-normal.

Proof. Using the same proof of Proposition 60.

Example 64. By Example 37, Sτis C-normal, CC-normal, L-normal and S-normal.

Theorem 65. If Aω\ωbe a group has more than one element, then the symmetrized omega topological space Sωτωis not S-normal.

Proof. From the proposition any separable S-normal must be normal (see [16]) and since Sωτωis separable and not normal (see Propositions 30, 23, respectively), then Sωτωis not S-normal.

Example 66. By Example 50, Sτis not a S-normal.

Theorem 67. The symmetrized omega topological space Sωτωis not C2-paracompact.

Proof. Since any C2-paracompact Frechet space is Hausdorff (see [17]) and Sωτωis First countable and not a Hausdorff space, Sωτωcannot be C2-paracompact.

Theorem 68. Let aAω\ωhas no multiplicative inverse. Then the symmetrized omega topological space Sωτωis not C-paracompact.

Proof. Assume that Sωτωis C-paracompact. Let Ybe a paracompact space and f:SωYbe bijective such that the restriction fC:CfCis a homeomorphism for all compact subspace Cof Sωτω. Hence, SωY, since Sωis compact (see Proposition 33). However, Sωis paracompact, thus a contradiction. Because any paracompact space is Hausdorff space and Sωis not a Hausdorff space. Therefore, Sωτωis not a C-paracompact.

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Mesfer Alqahtani, Cenap Özel and Ibtesam Alshammari (July 13th 2020). New Topology on Symmetrized Omega Algebra [Online First], IntechOpen, DOI: 10.5772/intechopen.92764. Available from: