# Complete Hausdorffness and Complete Regularity on Supra Topological Spaces

Authors: T. M. Al-shami
Journal: Journal of Applied Mathematics
 (2021)

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## Abstract

The supra topological topic is of great importance in preserving some topological properties under conditions weaker than topology and constructing a suitable framework to describe many real-life problems. Herein, we introduce the version of complete Hausdorffness and complete regularity on supra topological spaces and discuss their fundamental properties. We show the relationships between them with the help of examples. In general, we study them in terms of hereditary and topological properties and prove that they are closed under the finite product space. One of the issues we are interested in is showing the easiness and diversity of constructing examples that satisfy supraTi spaces compared with their counterparts on general topology.


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## Body

## 1. Introduction and Preliminaries

Topological space has been generalized in many manners. They can be classified into three main types; the first one was obtained by strengthening or weakening the conditions of a topology such as Alexendroff topology [1], supra topology [2], and generalized topology [3]. The second one was given by adding newly mathematical structures to the topology such as ordered topology [4], ideal topology [5], and bitopology [6]. The third one was established by defining a topology using one of the generalizations of crisp sets such as fuzzy topology [7] and soft topology [8]. Later on, wide generalizations were constructed by combining two or more types of the previous ones such as fuzzy bitopology [9], ideal ordered bitopology [10], and soft ordered topology [11].Mashhour et al. [2] introduced the supra topology concept by deleting only the intersection condition. In analogy with topology, they studied the concepts of interior and closure operators, continuity, and separation axioms on supra topology. One of the concrete merits of supra topology is obtaining different examples that give meaning to the concepts and properties defined on a finite set; for example, the only topology defined on a finite set which is a T1 space is the discrete topology, whereas there are several sorts of supra topologies that are a T1 space. Additionally, it is a hard work to find counterexamples which illustrate that Ti−1/2 spaces do not imply Ti in the cases of i=21/2,3,31/2,4; see for example, [12, 13]. After the emergence of supra topology, many authors have explored various topological notions in the spaces of supra topology such as compact [14, 15] and paracompact spaces [16], neighborhood system [17], separation axioms [18]), operators, and generalized open sets [19, 20]). Some published articles demonstrated that many topological findings are still true on supra topologies and illustrated that some of them are invalid like the distributive property of the closure (resp., interior) operator for the union (resp., intersection) between two sets and a compact set in a Hausdorff space is closed. In general, one can note that all topological properties and characterizations which are related to the intersection operator do not remain valid in supra topological spaces. In conclusion, we point to the existence of several studied extensions of supra topology such as supra ordered topology [21], supra fuzzy topology [22], supra soft topology [23], supra ideal topology [24], supra ordered bitopology [25], and supra soft ordered topology [26].In the field of applications, the supra topology represents a framework that is general enough to model phenomena and handle practical issues. In [27], the authors elucidated how supra topological frames induced by semiopen sets handled some digital problems. It is well known that the classes of regular sets and semiopen sets define supra topology structures. In [28], these classes were examined to fix or eliminate obstacles on the digital scope.Our goal in this article is to complete separation axioms reported in [19] by introducing the concepts of supra completely Hausdorff and supra completely regular spaces. We elaborate their master properties and demonstrate the relationships between them by some illustrative examples. Also, we investigate their behaviours with respect to subspaces, Så homeomorphism maps, and the finite product of spaces.Now, we mention some definitions and results in which we need to illustrate the obtained findings.Definition 1 (see [2]).
We callΣ as a supra topology on X provided that Σ is a subcollection Σ of 2X such that X∈Σ and the arbitrary unions of members of Σ are a member of Σ. A pair X,Σ is called a supra topological space. Every subset of X that belongs to Σ is called supra open, and a set is called supra closed if its complement belongs to Σ.To refer supra topological spaces, we use the the pairsX,Σ and Y,Ω.Definition 2 (see [19]).
For a subsetE of X,Σ, the union of all supra open sets that is contained in E is called the supra interior of E (denoted by sintE); the intersection of all supra closed sets that contains E is called the supra closure of E (denoted by sclE).Definition 3 (see [2]).
We call a mapSå continuous if the inverse image of every supra open subset is a supra open subset.Proposition 4 (see [2]).
A mapf from X,Σ to Y,Ω is an Så continuous iff sclf−1U⊆f−1sclU for every U⊆Y.Definition 5 (see [19]).
A supra relative topology on a subsetA of X,Σ is given by ΣA=A⋂G:G∈Σ. A pair A,ΣA is said to be a supra subspace of X,Σ.Lemma 6 (see[19]).
A subsetE of A,ΣA, which is a supra subspace of X,Σ, is supra closed iff there is a supra closed subset F of X,Σ such that E=A⋂F.Definition 7 (see [2, 19]).
X,Σ is said to be:
(1)
ST0 if for any a≠b∈X, there is a supra open set containing (only) one of the points a or b(2)
ST1 if for any a≠b∈X, there are two supra open sets U and V such that a∈U/V and b∈U/V. If U and V are disjoint, then, we call X,Σ supra Hausdorff (or ST2).(3)
Supra regular if for any supra closed setF and every element a∈F, there are disjoint supra open sets U and V satisfying that a∈U and F⊆V(4)
Supra normal if for any disjoint supra closed setsF and H, there are disjoint supra open sets U and V satisfying that F⊆U and H⊆VX,Σ which is ST1 and supra regular (resp., supra normal) is called ST3 (resp., ST4).Theorem 8 (see [2]).
X,Σ is ST1 iff every singleton set is supra closed.Theorem 9 (see [19]).
InX,Σ, the next statements are identical.
(1)
X,Σ is supra regular(2)
For eacha∈X and a∈U∈Σ, there is V∈Σ such that a∈V⊆sclV⊆U(3)
Every supra open setU can be written as follows: U=⋃H:H∈Σ,sclH⊆U.Proposition 10.
G×H∩U×V=G∩H×U∩V.Definition 11.
LetXi,Σi:i=1,2,⋯,n be a family of supra topological spaces. We call B=∏i=1nΣi=∏i=1nGi:Gi∈Σi a base for a supra topology Σ on X=∏i=1nXi. A pair X,Σ is called a finite product supra space.Proposition 12.
LetA and B be subsets of X,Σ and Y,Ω, respectively. Then, sclA×sclB=sclA×B.


## 2. Supra Completely Hausdorff Spaces

In this portion, we define a concept of supra completely Hausdorff spaces and investigate its master properties. A number of examples which clarify relationships between this concept and some separation axioms that we consider is provided.Definition 13.
X,Σ is said to be supra completely Hausdorff (or supra T21/2); if for any a≠b∈X, there are two supra open sets U and V containing a and b, respectively, such that sclU⋂sclV=∅.Proposition 14.
Every supraTi space X,Σ is supra Ti−1/2, for i=3,21/2.Proof.
We prove the proposition in case ofi=3 and the other case is obvious. Let a≠b∈X. A set b is supra closed because X,Σ is supra T1. The supra regularity of X,Σ implies that there are two disjoint supra open sets U and V containing a and b, respectively. From Theorem 9, there are supra open sets G and H such that a∈G⊆sclG⊆U and b∈H⊆sclH⊆V. The disjointness of the supra closures of G and H ends the proof.Corollary 15.
WhenX,Σ is supra regular, the next concepts are identical.
(1)
X,Σ is supra T0(2)
X,Σ is supra T1(3)
X,Σ is supra T2(4)
X,Σ is supra T21/2The converse of the above result is not true in general as the two examples clarify as follows.Example 16.
We construct a supra topologyΣ on X=ω1,ω2,ω3,ω4 as follows:
(1)Σ=∅,X,ω1,ω2,ω3,ω4,ω1,ω3,ω2,ω4,ω2,ω3,ω1,ω2,ω3,ω1,ω2,ω4,ω1ω3,ω4,ω2,ω3,ω4.
It can be checked thatX,Σ is supra T21/2. On the other hand, ω1,ω4 is a supra closed set and ω2∈ω1,ω4. X,Σ is not supra T3 because there do not exist two disjoint supra open sets such that one of them includes ω2 and the other includes ω1,ω4.Example 17.
We construct a supra topologyΣ on X=a,b,x,y,z as follows:
(2)Σ=∅,X,a,b,x,y,a,b,x,z,a,b,y,z,a,x,y,z,b,x,y,z,x,y,z,a,b,z,a,x,y,a,b,x,a,b,y,b,x,y,a,x,a,y,b,x,b,y,a,b,x,y.
Obviously,X,Σ is supra T2. In contrast, a≠b and the supra closure of any supra open set containing a and the supra closure of any supra open set containing b have a nonempty intersection, so that X,Σ is not supra T21/2.Theorem 18.
The two concepts are identical if∣X∣≤4 as follows.
(1)
X,Σ is supra T21/2(2)
X,Σ is supra T2Proof.
The implication1⟶2 is obvious.
To prove that2⟶1, let x≠y∈X. Then, there are two disjoint supra open subsets U and V of X,Σ containing x and y, respectively. It is well known that U∩sclV=∅ and sclU∩V=∅. Now, we have the following two cases:
(1)
∣U∣=1 or ∣U∣=3. Then, U is a supra clopen set, so that sclU∩sclV=∅(2)
∣U∣=2. Then, ∣V∣=2 or ∣V∣=1. If ∣V∣=2, then, U and V are supra clopen sets because they are disjoint, so that sclU∩sclV=∅. If ∣V∣=1, then, V is a supra clopen set. Thus, sclU∩V¯=∅
The above two cases end the proof thatX,Σ is supra T21/2.Theorem 19.
If for every distinct pointsa,b in X,Σ there is an Så continuous function f of X,Σ into the usual topology R such that fa≠fb; then X,Σ is supra completely Hausdorff.Proof.
Leta≠b∈X. By hypothesis, there exists x≠y∈Y such that fa=x and fb=y. Since R is supra completely Hausdorff, there are two disjoint supra open subsets U and V of R containing x and y, respectively, such that sclU⋂sclV=∅. Since f is an Så continuous map, f−1U and f−1V are supra open subsets of X,Σ containing a and b, respectively, such that f−1sclU⋂f−1sclV=∅. It follows from Proposition 4 that sclf−1U⊆f−1sclU and sclf−1V⊆f−1sclV, so that sclf−1U⋂sclf−1V=∅. Hence, X,Σ is supra completely Hausdorff, as required.Definition 20.
A mapf:X,Σ⟶Y,Ω is said to be Så open (resp., Så closed) if the image of any supra open (resp., supra closed) set is supra open (resp., supra closed). If a bijective map is Så open and Så continuous, then it is called an Så homeomorphism.Proposition 21.
A mapf:X,Σ⟶Y,Ω is Så closed iff sclfH⊆fsclH for every H⊆X.Proof.
Straightforward.Definition 22.
(1)
A property that passes from a supra topological space to every supra subspace is called a supra hereditary property(2)
A property that is preserved by anySå homeomorphism map is called a supra topological propertyTheorem 23.
Letf:X,Σ⟶Y,Ω be a bijective Så open map. If X,Σ is supra completely Hausdorff, then, Y,Ω is supra completely Hausdorff.Proof.
Letx≠y∈Y. Then, it follows from the bijective of f that there are a≠b∈X such that a=f−1x and b=f−1y. Since X,Σ is supra completely Hausdorff, there are two disjoint supra open subsets U and V of X,Σ containing a and b, respectively, such that sclU⋂sclV=∅. Now, fU and fV are supra open subsets of Y,Ω containing x and y, respectively, such that fsclU⋂fsclV=∅. Since f is a bijective Så open map, it is also an Så closed map, so that sclfU⊆fsclU and sclfV⊆fsclV. Thus, sclfU⋂sclfV=∅. Hence, Y,Ω is a supra completely Hausdorff, as required.Corollary 24.
The property of being a supra completely Hausdorff space is a supra topological property.Lemma 25.
Consider thatA,ΣA is a subspace of X,Σ. Then, sclUA=A⋂sclU for each U⊆A, where sclUA is the supra closure operator in A,ΣA.Proof.
sclUA=∩H:U⊆HandA/H∈ΣA=∩A∩F:U⊆A∩FandFc∈Σ=∩A∩F:U⊆FandFc∈Σ=A∩∩F:U⊆FandFc∈Σ=A∩sclUTheorem 26.
The property of being a supra completely Hausdorff space is a supra hereditary property.Proof.
Suppose thatE,ΣE is a subspace of a supra completely Hausdorff space X,Σ. Let x≠y∈E⊆X. Then, there are two disjoint supra open subsets U and V of X,Σ containing x and y, respectively, such that sclU∩sclV=∅. Now, G=E∩U and H=E∩V are two disjoint supra open subsets of E,ΣE containing x and y, respectively. From Lemma 25, we obtain sclGE=E∩sclG=E∩sclU∩E⊆E∩sclU. Similarly, sclHE⊆E∩sclV. The disjointness of sclGE and sclHE shows that E,ΣE is a supra completely Hausdorff.Theorem 27.
The finite product of supra completely Hausdorff spaces is a supra completely Hausdorff.Proof.
ConsiderX,Σ and Y,Ω as two supra topological spaces.
Suppose thatx1,y1≠x2,y2. This means that either x1≠x2 or y1≠y2. Say, x1≠x2. Therefore, there are two disjoint supra open subsets U and V of X,Σ containing x1 and x2, respectively, such that sclU⋂sclV=∅. Now, U×Y and V×Y are two disjoint supra open subsets of X×Y,Σ×Ω containing x1,y1 and x2,y2, respectively. It follows from Proposition 12 and Proposition 10 that sclU×Y⋂sclV×Y=∅. Hence, X×Y,Σ×Ω is a supra completely Hausdorff.


## 3. Supra Completely Regular Spaces

In this portion, we first investigate some properties of supra regular and supra normal spaces. Then, we define a concept of supra completely regular spaces. Compared with completely regular on topological spaces, the continuous functions are replaced bySå continuous functions; however, the codomain remains 0,1 which is a subspace of the usual topology. We discuss some rudiments of a supra completely regular space with the help of examples and characterize it in many ways.Theorem 28.
If∣X∣≤4, then, every supra regular space X,Σ is supra normal.Proof.
LetF and H be two disjoint nonempty supra closed subsets of a supra regular space X,Σ. Now, we have the following four cases:
(1)
∣F∣=4. Then, H is the empty set. Hence, the proof is trivial(2)
∣F∣=3. Since F and H are disjoint, it must be that H is a singleton, so that F and H are disjoint supra open sets as well(3)
∣F∣=2. Then, ∣H∣=1 or ∣H∣=2. If ∣H∣=1, say H=x. Then, x∈F. It follows by the regularity of X,Σ that there are two disjoint supra open subsets U and V of X,Σ containing F and H, respectively. If ∣H∣=2, then, F and H are disjoint supra open sets as well(4)
∣F∣=1, say F=x. Then, x∈H. It follows by the regularity of X,Σ that there are two disjoint supra open subsets U and V of X,Σ containing F and H, respectively(5)
∣F∣=0. Then, ∅ and X are two disjoint supra open subsets of X,Σ containing F and H, respectively
The above five cases end the proof thatX,Σ is supra normal.Corollary 29.
The next two concepts are identical if∣X∣≤4.
(1)
X,Σ is supra T4(2)
X,Σ is supra T3By the next examples, we illustrate that the converse of Theorem28 fails and elucidate that the concepts of supra normal and supra regular spaces are independent of each other when ∣X∣>4.Example 30.
Consider thatΣ=∅,X,a,a,b,b,c is a supra topology on X=a,b,c. Obviously, X,Σ is supra normal. In contrast, b∈c and c is a supra closed set. X,Σ is not supra regular because there do not exist two disjoint supra open sets separate b and c.Example 31.
Consider thatΣ=PX\a,x,a,b,x,y is a supra topology on X=a,b,x,y,z. Now, a,b and x,y are supra closed subsets of X,Σ. One can check that X,Σ is supra regular. In contrast, X,Σ is not supra normal because there do not exist two disjoint supra open sets separate a,b and x,y.The following result is the version of Urysohn’s lemma on supra topology and has a similar proof to Urysohn’s lemma on general topology, so that the proof is omitted.Theorem 32.
X,Σ is supra normal iff for each disjoint supra-closed sets A,B there exists an Så continuous map g:X⟶0,1 such that ga=0 for each a∈A and gb=1 for each b∈B.Take into consideration that a supra topology which is defined on0,1 is a subspace of the usual topological space R,Σ.Definition 33.
X,Σ is said to be
(1)
Supra completely regular if for eachx∈X and supra closed set F such that if x∈F, there exists an Så continuous map g:X⟶0,1 such that gx=0 and gF=1(2)
SupraT31/2 if it is supra completely regular and supra T1Proposition 34.
The following three statements are identical:(1)
X,Σ is supra completely regular(2)
For eachx∈X and supra closed set F such that x∈F, there exists an Så continuous map g:X⟶0,1 such that gx=1 and gF=0(3)
For eachx∈X and supra open set U containing x, there exists an Så continuous map g:X⟶0,1 such that gx=1 and gUc=0Proof.
1⟶2. Since X,Σ is supra completely regular, then for each x∈X and supra closed set F such that x∈F, there exists an Så continuous map g:X⟶0,1 such that gx=0 and gF=1. Therefore, a map hx=1−gx of X,Σ into 0,1 is Så continuous. Hence, hx=1 and hF=0, as required.
2⟶3, obviously.
3⟶1. Let F be a supra closed set and x∈X such that x∈F. Then, Fc is a supra open set. By hypothesis, there exists an Så continuous map g:X⟶0,1 such that gx=1 and gFcc=gF=0. Therefore, a map hx=1−gx of X,Σ into 0,1 is Så continuous. Since hx=0 and hF=1, then, X,Σ is supra completely regular.Theorem 35.
For every distinct pointsa,b in a supra T31/2 space X,Σ, there is an Så continuous function f:X,Σ⟶R such that fa≠fb.Proof.
Leta≠b∈X. Then, b is a supra closed set because X,Σ is supra T1. Since a∈b, it follows from the supra complete regularity of X,Σ that there is an Så continuous map g:X,Σ⟶0,1 such that ga=0 and gb=1. Hence, ga≠gb.Definition 36.
For anSå continuous map g of X into R, we define
(1)
g+ of X into R by g+x=supgx,0. It is obvious that g+ is Så continuous(2)
The zero setZg by Zg=x∈X:gx=0Theorem 37.
X,Σ is supra completely regular iff for every x∈X and supra closed set F such that x∈F, there are Så continuous maps f and g of X,Σ into R such that x∈sintZf, F⊆sintZg, and Zf⋂Zg=∅.Proof.
(1)
Necessity. Suppose thatF is a supra closed set and x∈X such that x∈F. Then, there is an Så continuous map h:X,Σ⟶R such that hx=0 and hF=1. Let fx=h−2/5+ and gx=3/5−h+. Obviously, the two sets Zf and Zg are disjoint. Now, x∈h−1−∞,2/5⊆Zf and F⊆h−13/5,∞⊆Zg. By the Så continuity of h, it follows that h−1−∞,2/5 and h−13/5,∞ are supra open subsets of X,Σ. Hence, the necessary part is proved(2)
Sufficiency. LetF be a supra closed set and x∈X such that x∈F. Take two maps f and g as described in the theorem. Then, we have f2x+g2x≠0 for each x∈X. Therefore, hx=f2xf2x+g2x−1 is an Så continuous map such that hx=0 and hF=1. Hence, X,Σ is supra completely regularTheorem 38.
Every supra completely regular spaceX,Σ is supra regular.Proof.
LetF be a supra closed subset of X,Σ and let x∈F. Since X,Σ is supra completely regular, there is an Så continuous map g:X⟶0,1 such that gx=0 and gF=1. Since 0,1 is a subspace of the usual topological space R,Σ, it is supra T2. So that there are two disjoint supra open sets U and V containing 0 and 1, respectively. It follows from the Så continuity of g that g−1U and g−1V are two disjoint supra open sets containing x and F, respectively. Hence, X,Σ is supra regular.Corollary 39.
Every supraT31/2 space X,Σ is supra T3.Theorem 40.
A supra normal spaceX,Σ is supra completely regular iff it is supra regular.Proof.
(1)
Necessity. It follows from the above theorem(2)
Sufficiency. LetF be a supra closed subset of X,Σ and let x∈F. By hypothesis, there are two disjoint supra open sets U and V such that x∈U and F⊆V. Now, F and Vc are two disjoint supra closed sets. By using Theorem 32, we can find an Så continuous map g:X⟶0,1 such that ga=0 for each a∈Vc and gb=1 for each b∈F. This means that gx=0 and gb=1 for each b∈F. Hence, X,Σ is supra completely regular.Corollary 41.
Every supraT4 space X,Σ is supra T31/2.Lemma 42.
The restricted map of anSå continuous map is Så continuous.Proof.
Letg:X,Σ⟶Y,Ω be an Så continuous map and let A⊆X. To prove that g∣A:A,ΣA⟶Y,Ω is Så continuous, let G be a supra open subset of Y. Then, g−1G is a supra open subset of X,Σ. Now, A∩g−1G is a supra open subset of A,ΣA. Thus, g∣A is Så continuous.Theorem 43.
The property of being a supra completely regular space is a supra hereditary property.Proof.
Consider thatA,ΣA is a supra subspace of a supra completely regular space X,Σ. Let E be a supra closed subset of A,ΣA and x∈A such that x∈E. Then, it follows from Lemma 6 that there is a supra closed subset F of X,Σ such that E=A⋂F. Obviously, x∈F. By the supra complete regularity of X,Σ, there is an Så continuous map g:X⟶0,1 such that gx=0 and gF=1. Thus, g∣A:A,ΣA⟶0,1 is an Så continuous map such that g∣Ax=0 and g∣AE=g∣AA⋂F=1. This ends the proof that a supra completely regular space is a supra hereditary property.Corollary 44.
A supraT31/2 space is a supra hereditary property.Theorem 45.
A supra completely regular space is a supra topological property.Proof.
Considerg:X,Σ⟶Y,Ω as an Så homeomorphism and let X,Σ be a supra completely regular space. Suppose that F is a supra closed subset of Y,Ω and y∈Y such that y∈F. Then, g−1F is a supra closed subset of X,Σ and g−1y∈g−1F. Since X,Σ is supra completely regular, there is an Så continuous map h:X⟶0,1 such that hg−1y=0 and hg−1F=1. Since g is an Så homeomorphism, then, g−1 is Så continuous. Therefore, p=h∘g−1:Y,Ω⟶0,1 is Så continuous such that py=h∘g−1y=hg−1y=0 and pF=h∘g−1F=hg−1F=1. Hence, Y,Ω is a supra completely regular space.Corollary 46.
A supraT31/2 space is a supra topological property.Theorem 47.
The finite product of supra completely regular spaces is supra completely regular.Proof.
LetX×Y,Σ×Ω be the product spaces of two supra completely regular spaces X,Σ and Y,Ω. Suppose that x,y∈X×Y and G=⋃i∈IUi×Vi be a supra open subset of X×Y such that x,y∈G. Now, there exists α∈I such that x,y∈Uα×Vα. Then, for each x∈X and supra open set Uα containing x, there exists an Så continuous map g1:X⟶0,1 such that g1x=1 and g1Uαc=0, and for each y∈Y and supra open set Vα containing y, there exists an Så continuous map g2:X⟶0,1 such that g2y=1 and g2Vαc=0. Now, we define a map f:X×Y⟶0,1 by fu,v=g1u×g2v. It can be checked that f is Så continuous, we have fx,y=g1x×g2y=1×1=1 and fUα×Vαc=fUαc×Y⋃Y×Vαc=fUαc×Y⋃fY×Vαc=g1Uαc×Y⋃g2Y×Vαc=g1Uαc×g1Y⋃g2Y×g2Vαc=0×g1Y⋃g2Y×0=0. Since ⋃i∈IUi×Vic⊆Uα×Vαc, then, f⋃i∈IUi×Vic=0. Hence, X×Y,Σ×Ω is supra completely regular.Corollary 48.
The property of being a supraT31/2 space is preserved under a finite product.


## 4. Conclusion

In this paper, we have highlighted two concepts on supra topological spaces, namely, supra completely Hausdorff and supra completely regular spaces. We have investigated some of their characterizations and provided some examples to show the relationships between them. Also, we have proved that they are hereditary and topological properties as well as they are closed under the finite product of spaces.In future works, we plan to study an application of supra topology on the information system and apply the concepts given herein and those in [19] to categorize the approximation spaces. Also, we will study functionally separation axioms [29] in the frame of supra topologies and elaborate their relationships with the concepts presented herein.



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