# Some Applications of Supra Preopen Sets

Authors: M. E. El-Shafei; A. H. Zakari; T. M. Al-shami
Journal: Journal of Mathematics
 (2020)

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## Abstract

The aim of this work is to define some concepts on supra topological spaces using supra preopen sets and investigate main properties. We started this paper by correcting some results obtained in previous study and presenting further properties of supra preopen sets. Then, we introduce a concept of supra prehomeomorphism maps and discuss its main properties. After that we explore the concepts of supra limit and supra boundary points of a set with respect to supra preopen sets and examine their behaviours on the spaces that possess the difference property. Finally, we formulate the concepts of supra pre-Ti-spaces i=0,1,2,3,4 and give completely descriptions for each one of them. In general, we study their main properties in detail and show the implications of these separation axioms among themselves as well as with STi-space with the help of some interesting examples.


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## Body

## 1. Introduction and Preliminaries

A setX with a family μ of its subsets is called a supra topological space [1], denoted by X,μ, if X∈μ and the arbitrary union of members of μ is in μ. Mashhour et al. [1] generalized some topological notions such as interior and closure operators and continuity and separation axioms. Al-Shami [2] has studied the classical topological notions such as limit points of a set, compactness, and separation axioms on the supra topological spaces.Some results via topology are not still valid via supra topology such as the distribution of the closure operator between the union of two sets and the distribution of the interior operator between the intersection of two sets. Also, the property of a compact subset of aT2-space is closed and is invalid on the supra topologies. To extend a class of supra open sets, the notions of supra α-open [3], supra preopen [4], supra b-open [5], supra β-open [6], supra R-open [7], and supra semiopen sets [8] have been introduced and their main properties have been discussed. These generalizations of supra open sets were defined in a similar way of defining them on general topology. In other words, their definitions were formulated using supra interior and supra closure operators instead of interior and closure operators. These generalizations have been utilized to define new versions of compactness and connectedness, see, for example [9–14]. Mustafa and Qoqazeh [15] took advantage of supra D-sets to define separation axioms on supra topological spaces. Recently, Al-Shami and El-Shafei [16] have studied separation axioms on supra soft topological spaces.It should be noted that the supra topological frame can be more convenient to solve some practical problems and to model some phenomena as pointed out in [17]. Also, the possibility of applying semiopen sets to deal with some problems on digital topology has been demonstrated in [18].The layout of the paper is as follows. In Section2, we correct some results of [4] and investigate further properties of supra preopen sets. Also, it presents the concept of supra prehomeomorphism maps and explores main properties. The concepts of supra prelimit and supra preboundary points of a set are studied in Section 3. Section 4 introduces new types of separation axioms using supra preopen sets and elucidates the relationships between them with the help of examples. Section 5 concludes the paper with summary and further works.In what follows, we collect the relevant definitions and results from supra topology and supra preopen sets to make this paper self-contained and easy to read.Definition 1.
(see [1]). A family μ of subsets of a nonempty set X is called a supra topology provided that the following two conditions hold:(1)
X and ∅∈μ(2)
μ is closed under arbitrary union
Then, the pairX,μ is called a supra topological space. Every element of μ is called a supra open set and its complement is called a supra closed set.Remark 1.
(1)
Since∪i∈∅Gi=∅, then some authors remove the empty set ∅ from the first condition of a supra topology(2)
μ is called an associated supra topology with a topology τ if τ⊆μ(3)
Through this paper, we considerX,μ and Y,ν are associated supra topological spaces with the topological spaces X,τ and Y,θ, respectivelyDefinition 2.
(see [1]). Let A be a subset of X,μ. Then, intμA is the union of all supra open sets contained in A and clμA is the intersection of all supra closed sets containing A.
If there is no confusion, we writeintA and clA in the places of intμA and clμA, respectively.Definition 3.
A subsetA of X,μ is said to be(1)
Supraα-open [3] if A⊆intclintA(2)
Supra preopen [4] if A⊆intclA(3)
Suprab-open [5] if A⊆intclA∪clintADefinition 4.
(see [1]). For a subset A of X,μ, pintμA is the union of all supra preopen sets contained in A and pclμA is the intersection of all supra preclosed sets containing A.
If there is no confusion, we writepintA and pclA in the places of pintμA and pclμA, respectively.Definition 5.
(see [4]). A map g:X,μ⟶Y,ν is said to be(1)
Supra precontinuous ifg−1H is a supra preopen set in X for each open set H in Y(2)
Supra preopen (resp. supra preclosed) ifgH is a supra preopen (supra preclosed) set in Y for each open (resp. closed) set in XTheorem 1 (see [4]).
For a mapg:X,μ⟶Y,ν, we have the following results for every A∈X(1)
g is supra precontinuous if and only if gpclμA⊆clθgA(2)
g is supra preopen if and only if gintτA⊆pintνgclA(3)
g is supra preclosed if and only if pclνgA⊆gclτADefinition 6.
(see [1]). A map g:X,μ⟶Y,ν is said to be(1)
S⋆-continuous if g−1H is a supra open set in X for every supra open set H in Y(2)
S⋆-open (resp. S⋆-closed) if gH is a supra open (resp. supra closed) set in Y for every supra open set H in X(3)
S⋆-homeomorphism if it is bijective, S⋆-continuous, and S⋆-openTheorem 2.
For a mapg:X,μ⟶Y,ν, we have the following results for every A⊆X:(1)
g is S⋆-continuous if and only if gclμA⊆clνgA(2)
g is S⋆-open if and only if gintμA⊆intνgclA(3)
g is S⋆-closed if and only if clνgA⊆gclμADefinition 7.
(see [1]). Let A be a subset of X,μ. The family μA=A∩G:G∈μ is called a supra relative topology on A. A pair A,μA is called a supra subspace of X,μ.Definition 8.
(see [11]). β is called a basis for a supra topology X,μ if every member of μ can be expressed as a union of elements of β.Definition 9.
(see [11]). Let Xi,μi:i=1,2,…,n be the collection of supra topological spaces. Then, β=∏i=1nμi=∏i=1nGi:Gi∈μi defines a basis for a supra topology T on X=∏i=1nXi. The pair X,T is called a finite product supra spaces.Theorem 3 (see [11]).
A and B are supra preopen sets iff the product of them is supra preopen.


## 2. Some Corrections and New Results of Supra Preopen Sets

We begin this section with Proposition1 below, originally proposed as Proposition 2.1 in [4].Proposition 1.
(1)
The intersection of supra open and supra preopen is supra preopen(2)
The intersection of supraα-open and supra preopen is supra preopenWe give the below example to demonstrate that the above proposition need not be true in general.Example 1.
Letμ=∅,X,1,4,1,2,3,2,3,4 be a supra topology on X=1,2,3,4. Since intcl2,4=X, then 2,4 is a supra preopen set and since intclint1,3,4=X, then 1,3,4 is a supra α-open set. Now, 1,4∩2,4=4 and 1,3,4∩2,4=4. Since intcl4=∅ and intclint4=∅, then 4 is neither a supra preopen set nor a supra α-open set. This emphasizes that the above proposition is erroneous.The above proposition is true in the case ofX,μ is a topological space because the following two properties are satisfied on topological spaces, but they do not valid in supra topological spaces:(1)
intA∩B=intA∩intB for every subsets A,B of a topological space X,τ(2)
IfA is an open set, then clA∩B⊆clA∩B for every subset B of a topological space X,τProposition 2.
LetA be a supra b-open subset of X,μ such that intA=∅. Then, A and Ac are supra preopen.Proof.
SinceA is a supra b-open set, then A⊆intclA∪clintA. Since intA=∅, then clintA=∅. Therefore, A⊆intclA. Thus, A is a supra preopen set. Also, intAc=X. This implies that clAc=X. Therefore, Ac is supra preopen set.In general, there does not exist a relationship between supra preopen sets inX,μ and its subspaces as the next two example show.Example 2.
Letμ=∅,X,1,2,3,1,2,4 be a supra topology on X=1,2,3,4. If Z=1,3, then μZ=∅,Z is a relative supra topology Z. In X,μ, a set B=3 is not supra preopen. On the contrary, it can be seen that B is a supra preopen set in Z,μZ.Theorem 4.
The property of being a supra preopen set is preserved underS⋆-continuous and S⋆-open map.Proof.
Let a mapg:X,μ⟶Y,ν be S⋆-continuous and S⋆-open and let A be a supra preopen subset of X,μ. Obviously, gA⊆gintμclμA. Since g is S⋆-open, then it follows from Theorem 2 that gA⊆gintμclμA⊆intνgclμA. Since g is S⋆-continuous, then it follows from Theorem 2 that gA⊆intνgclμA⊆intνclνgA. Thus, gA⊆intνclνgA. Hence, gA is a supra preopen set.Definition 10.
For a nonempty subsetA of X,μ, the family μA=A∩G:G is a supra preopen subset of X,μ is called a relative pretopology on A. A pair A,μA is called a presubspace of X,μ.One can easily prove that a presubspaceA,μA of X,μ is a supra topological space.Proposition 3.
LetY,μY be a presubspace of X,μ. A subset H of Y is supra preclosed in Y,μY iff there exists a supra preclosed subset F of X,μ such that H=Y∩F.Proof.
Necessity: let H be a supra preclosed subset of Y,μY. Then, there exists a supra preopen subset W of Y,μY such that H=Y\W. Now, there exists a supra preopen subset V of X,μ such that W=Y∩V. Therefore, H=Y\Y∩V=Y∩Vc. By taking F=Vc, the proof of the necessary part is complete.Sufficiency: let H=Y∩F such that F is a supra preclosed subset of X,μ. Then, Y\H=Y\Y∩F=Y∩X\Y∩F=Y∩X\F. Since X\F is a supra preopen subset of X,μ, then Y\H is a supra preopen subset of Y,μY. Thus, H is a supra preclosed subset of Y,μY.In the rest part of this section, we present a concept of supra prehomeomorphism and supra pre⋆-homeomorphism maps and discuss some basic properties (⋆denotes another type of homeomorphism maps).Definition 11.
A bijective mapg:X,μ⟶Y,ν is said to be a supra prehomeomorphism if it is supra precontinuous and supra preopen.Since every supra open set is supra preopen, then every supra homeomorphism map is a supra prehomeomorphism. However, the converse is not always true as it is illustrated in the following example.Example 3.
Assume thatτ=∅,G⊆ℝ:1∈G and θ=∅,1,ℝ are two topologies on the real numbers set ℝ. Let μ=∅,G⊆ℝ:1∈ℝ or 2∈ℝ and ν=∅,1,2,3,2,4,ℝ be two associated supra topologies with τ and θ, respectively. Then, the identity map g:ℝ,μ⟶ℝ,ν is a supra prehomeomorphism, but it is not a supra homeomorphism because the image of a supra open set 1,3 is on a supra open set.Theorem 5.
The equivalence of the following properties hold if a mapg:X,μ⟶Y,ν is bijective and supra precontinuous:(1)
g is a supra prehomeomorphism(2)
g−1 is supra precontinuous(3)
g is supra preclosedProof.
Straightforward.Theorem 6.
A bijective mapg:X,μ⟶Y,ν is a supra prehomeomorphism if and only if pclνgA⊆gclτA and gpclμA⊆clθgA for every A⊆X.Proof.
⟹′′ If g is a supra prehomeomorphism map. Then, g is supra precontinuous and supra preclosed. It follows from Theorem 1 that pclνgA⊆gclτA and gpclμA⊆clθgA.
⟹′′ If g is a bijective map such that pclνgA⊆gclτA and gpclμA⊆clθgA, then g is supra precontinuous and supra preclosed. It follows from Theorem 1 that g is a supra prehomeomorphism map.Definition 12.
A mapg:X,μ⟶Y,ν is said to be(1)
Supra pre⋆-continuous if g−1H is a supra preopen set in X for every supra preopen set in Y(2)
Supra pre⋆-open (resp. supra pre⋆-closed) if gH is a supra preopen (resp. supra preclosed) set in Y for every supra preopen (resp. supra preclosed) set in X(3)
Supra pre⋆-homeomorphism if it is bijective, supra pre⋆-continuous, and supra pre⋆-open


## 3. Limit and Boundary Points of a Set with Respect to Supra Preopen Sets

This section defines the concepts of supra prelimit and supra preboundary points of a set and studies the interrelations between them. It provides some examples to show the obtained results and examines some properties of the supra prederived set on the spaces that possess the difference property.Definition 13.
A subsetA of X,μ is said to be a supra preneighbourhood of x∈X provided that there is a supra preopen set F containing x such that x∈F⊆A.Definition 14.
A pointx∈X is said to be a supra prelimit point of a subset A of X,μ provided that every supra preneighborhood of x contains at least one point of A other than x itself.All supra prelimit points ofA is said to be a supra prederived set of A and is denoted by Ap′.Proposition 4.
IfA⊆B, then Ap′⊆Bp′ for every subsets A and B of X,μ.Proof.
Straightforward.Corollary 1.
We have the following results for any two subsetsA and B of X,μ:(1)
Ap′∪Bp′⊆A∪Bp′(2)
A∩Bp′⊆Ap′∩Bp′The following example illustrates that the converse of the above proposition and corollary fails.Example 4.
Letμ=∅,X,1,2,1,3,1,2,3,1,2,4,1,3,4,2,3,4 be a supra topology on X=1,2,3,4. Then, ∅,X,1,2,1,3,1,4,2,3,1,2,3, 1,2,4,1,3,4,2,3,4 is the collection of all supra preopen subsets of X,μ. If A=1,2,3 and B=1,4, then Ap′=2,3,4 and Bp′=4. Obviously, Xp′=X. Now, we have the following cases:(i)
Bp′⊆Ap′, but B⊆A(ii)
Ap′∪Bp′=2,3,4 and A∪Bp′=Xp′=X(iii)
Ap′∩Bp′=4 and A∩Bp′=∅Proposition 5.
LetA be a subset of X,μ and x∈X. Then, x∈Ap′ if and only if x∈A∖xp′.Proof.
Necessity: let x∈Ap′. Then, for every supra preopen set G containing x, we have G\x∩A≠∅. Therefore, G\x∩A\x≠∅. Thus, x∈A\xp′.Sufficiency: it follows from Proposition 4.Theorem 7.
LetA be a subset of X,μ. Then, the following results hold.(i)
A is a supra preclosed set iff Ap′⊆A(ii)
A∪Ap′ is a supra preclosed set(iii)
pclA=A∪Ap′Proof.
(i)
Suppose thatA is a supra preclosed set and x∉A. Then, Ac is a supra preopen set containing x. In this case, Ac∩A=∅ leads to x∉Ap′. Therefore, Ap′⊆A. Conversely, let x∈Ac and let Ap′⊆A. Then, x∉Ap′. Therefore, there is a supra preopen set Gx such that Gx\x∩A=∅. Since x∈Ac, then Gx∩A=∅. Now, Gx⊆Ac. Therefore, Ac=∪Gx:x∈Ac. Thus, A is supra preclosed.(ii)
Letx∉A∪Ap′. Then, x∉A and x∉Ap′. Therefore, there is a supra preopen set G such that(1)G∩A=∅.Now, for eachx∈G, we have x∉Ap′. This means that(2)G∩Ap′=∅.From (1) and (2), we obtain G∩A∪Ap′=∅. This implies that x∉A∪Ap′p′. Hence, A∪Ap′p′⊆A∪Ap′. By i, A∪Ap′ is a supra preclosed set, as required.(iii)
SinceA⊆pclA and Ap′⊆pclAp′⊆pclA, then A∪Ap′⊆pclA. Since A∪Ap′ is a supra preclosed set containing A and pclA is the smallest supra preclosed set containing A, then pclA⊆A∪Ap′. Therefore, pclA=A∪Ap′.Corollary 2.
IfA is a supra preclosed subset of X,μ, then Ap′, Ap′p′, Ap′p′p′, …, are supra preclosed sets.Theorem 8.
Ifg:X,μ⟶Y,ν is a supra pre⋆-homeomorphism map, then gAp′=gAp′ for each A⊆X.Proof.
Leta∉gAp′. Then, there is a supra preopen set H containing a such that H\a∩gA=∅. So, g−1H\a∩gA=g−1∅. This implies that g−1H\g−1a∩A=∅. Thus, g−1a∉Ap′. Since g is bijective, then a∉gAp′. Therefore, gAp′⊆gAp′. By reversing the preceding steps, we find that gAp′⊆gAp′. Hence, the proof is complete.Definition 15.
For a nonempty setX, a subcollection Λ of 2X is said to have the difference property provided that G∈Λ implies that G\x∈Λ.The following two examples illustrate the existence and uniqueness of the difference property.Example 5.
Letμ=∅,G⊆ℕ:Gis infinite be a supra topology on the set of natural numbers ℕ. It is clear that the infinity of G implies the infinity of G\x. That is, G∈μ implies G\x∈μ. Then, ℕ,μ has the difference property. Also, it can be seen that the collection of supra preopen subsets of ℕ,μ coincides with the collection of supra open sets. Hence, ℕ,μ has the difference property for the collection of supra preopen sets.Example 6.
Letμ=∅,ℕ\1,3,ℕ\1,4,ℕ\2,3,ℕ\2,4∪G⊆ℕ:G such that 1,2⊆G or 3,4⊆G be a supra topology on the set of natural numbers ℕ. Then, 1,2∈μ, but 1,2\2=1∉μ. Therefore, X,μ does not have the difference property for the collection of supra open sets. Hence, it does not have the difference property for the collection of supra preopen sets.Theorem 9.
IfX,μ has the difference property for the collection of supra preopen sets, then the following properties hold for A⊆X:(1)
Ap′p′⊆Ap′(2)
pclAp′=Ap′=pclAp′(3)
Ap′=∅ if A is finiteProof.
(1)
Letx∉Ap′. Then, there is a supra preopen set G containing x such that G\x∩A=∅. Since X,μ has the difference property for the collection of supra preopen sets, then G\x is a supra preopen set. Therefore, G\x∩Ap′=∅. Since x∉Ap′, then G∩Ap′p′=∅. Thus, x∉Ap′p′. Hence, Ap′p′⊆Ap′.(2)
SinceAp′p′⊆Ap′, then it follows from Theorem 7 that Ap′ is a supra preclosed set. Therefore,(3)pclAp′=Ap′.Also,Ap′⊆pclAp′ because A⊆pclA. On the contrary, let x∉Ap′. Then, it follows from 1 above that Gx∩A=∅ and G\x∩Ap′=∅. This means that G\x∩pclA=∅. Therefore, x∉pclAp′. Thus, pclAp′⊆Ap′. Hence,(4)pclAp′=Ap′.From (3) and (4), the desired result is proved.(3)
LetA be a finite subset of X. Suppose that there exists an element x∈X such that x∈Ap′. Then, for every supra preopen set G containing x, we have G\x∩A≠∅. Therefore, for every y∈A such that y≠x, we have G\x,y is a supra preopen set. Thus, G\A∪x is a supra preopen set such that G\A∪x∩A=∅. This implies that x∉Ap′. However, this is a contradiction. Hence, it must be that Ap′=∅.We explain that the three properties mentioned in the above theorem need not be true ifX,μ does not have the difference property for the collection of supra preopen sets. Let A=2,4,5 be a subset of supra topological space given in Example 6. Note that the collection of supra open sets coincides with the collection of supra preopen sets. By calculating, we find that Ap′=ℕ\2,4, Ap′p′=ℕ\1,3, and clAp′=ℕ. This leads to the following three properties:(1)
Ap′p′⊆Ap′(2)
pclAp′≠Ap′(3)
Ap′≠∅ in spite of A is finiteDefinition 16.
LetA be a subset of X,μ. The supra preboundary of A (denoted by pbA) is the set of all elements which belongs to pintA∪pintAcc.Lemma 1.
LetA be a subset of X,μ. Then,(1)
pintAc=pclAc(2)
pclAc=pintAcProof.
We prove (1) and (2) is proved in a similar way.
SincepintA⊆A, then Ac⊆pintAc. Therefore, pclAc⊆pclpintAc=pintAc. Conversely, Ac⊆pclAc. Then, pclAcc⊆A. Therefore, pintpclAcc=pclAcc⊆pintA. Thus, pintAc⊆pclAc. Hence, the desired result is proved.Proposition 6.
LetA be a subset of X,μ. Then, pbA=pclA∩pclAc.Proof.
pbA=x∈X:x∉pintA and x∉pintAc=x∈X:x∉pclAcc and x∉pclAc=x∈X:x∈pclAc and x∈pclA=pclA∩pclAc.Corollary 3.
LetA be a subset of X,μ. Then,(1)
pbA=pbAc(2)
pbA is a supra preclosed set(3)
pbA=pclA\pintA(4)
pclA=pintA∪pbA(5)
pclA=A∪pbA(6)
pintA=A\pbA(7)
pbpclA⊆pbAProof.
(1)
pbA=pclA∩pclAc=pclAcc∩pclAc=pbAc(2)
It follows from noting thatpbA is the intersection of two supra preclosed sets which is supra preclosed(3)
pbA=pclA∩pclAc=pclA⋅∩pintAc=pclA\pintA(4)
pintA∪pbA=pintA∪pclA\pintA=pclA(5)
A∪pbA=A∪pclA\pintA=pclA(6)
A\pbA=A∩pclA\pintAc=A∩pclAc∪pintA=pintA(7)
pbpclA=pclpclA∩pclpclAc=pclA∩pclpclAc; Since pclAc⊆Ac, then the result holdsProposition 7.
LetA be a subset of X,μ. Then,(1)
A is a supra preopen set iff pbA∩A=∅(2)
A is a supra preclosed set iff pbA⊆AProof.
(1)
Necessity: since A is supra preopen, then pbA∩A=pbA∩pintA=∅.Sufficiency: let x∈A. Then, x∈pintA or x∈pbA. Since pbA∩A=∅, then x∈pintA. Therefore, A⊆pintA. Thus, A is supra preopen.(2)
Necessity: since A is supra preclosed, then pbA=pclA\pintA=A\pintA. Therefore, pbA⊆A.Sufficiency: suppose that pbA⊆A and let x∈Ac. Then, x∉pbA=pclA\pintA. Since x∉pintA, then x∉pclA so that for each supra preopen set G containing x, we obtain G∩A=∅. This implies that x∈G⊆Ac. Thus, Ac is a union of supra preopen sets. Hence, Ac is supra preopen.Corollary 4.
We have the following properties for a subsetA of X,μ:(1)
A is both supra preopen and supra preclosed iff pbA=∅(2)
pbpintA=∅(3)
pintpbA=∅


## 4. Separation Axioms with Respect to Supra Preopen Sets

In this section, we use a class of supra preopen sets to introduce new types of separation axioms, namely,SpTi-spaces i=0,1,2,3,4. We establish some characterizations of each one of these spaces and study them in terms of hereditary and topological properties and product space. Illustrative examples are provided to show the obtained results.Definition 17.
A supra topological spaceX,μ is said to be(1)
SpT0 if for every a≠b∈X, there exists a supra preopen set containing only one of them.(2)
SpT1 if for every a≠b∈X, there exist two supra preopen sets one of them contains a but not b and the other contains b but not a(3)
Supra pre-Hausdorff (orSpT2) if for every a≠b∈X, there exist two disjoint supra preopen sets U and V containing a and b, respectively(4)
Supra preregular if for every supra preclosed setF and each a∉F, there exist disjoint supra preopen sets U and V containing F and a, respectively(5)
Supra prenormal if for every disjoint supra preclosed setsF and H, there exist disjoint supra preopen sets U and V containing F and H, respectively(6)
SpT3 (resp. SpT4) if it is both supra preregular (resp. supra prenormal) and SpT1Theorem 10.
The following three statements are equivalent:(1)
X,μ is an SpT0-space(2)
pcla≠pclb for each a≠b∈X(3)
For eacha∈X, we have ap′ is a union of supra preclosed setsProof.
1⟶2: for each a≠b∈X, there exists a supra preopen set G containing a but not b or containing b but not a. Say a∈G and b∉G. Then, a∉pclb because G is a supra preopen set containing a such that G∩b=∅. Since a∈pcla, then pcla≠pclb.
2⟶3: let b∈ap′. Then, b≠a and b∈a∪ap′=pcla. Therefore, pclb⊆pcla. Thus, b∈pclb⊆ap′. Hence, ap′=∪pclb: for each b∈ap′.
3⟶1: let a≠b. Then, we have two cases:(i)
Eitherb∈ap′. Then, there is a supra preclosed set F such that b∈F⊆ap′. Since a∉ap′, then a∉F. Therefore, Fc is a supra preopen set containing a such that b∉Fc.(ii)
Orb∉ap′. Then, there is a supra preopen set G containing b such that a∉G.
In the both cases above, we infer thatX,μ is an SpT0-space.Corollary 5.
AnSpT0-space X,μ contains at most a supra predense singleton set a is a supra predense set if pcla=X.Proof.
LetX,μ be an SpT0-space. Suppose that there are two distinct singleton set a and b such that pcla=pclb=X. Then, X,μ is not an SpT0-space, a contradiction. Hence, X,μ contains at most a supra predense singleton set.Theorem 11.
The following there statements are equivalent:(1)
X,μ is an SpT1-space(2)
Every singleton subset ofX,μ is supra preclosed(3)
The intersection of all supra preopen sets containing a setA is exactly A(4)
ap′=∅ for each a∈XProof.
1⟶2: consider X,μ is an SpT1-space and let a⊆X. For all b∈X such that a≠b, there exists a supra preopen set G containing b such that G∩a=∅. Then, b∉pcla. Therefore, pcla=a. Thus, a is a supra preclosed set.
2⟶3: let A be a subset of X,μ. Then, for each a∈Ac, we have ac is a supra preopen set containing A. Now, A⊆G:G is a supra preopen set containing A⊆ac:a∈Ac⊆A. Thus, A=G:G is a supra preopen set containing A, as required.
3⟶4: suppose that there exists a∈X such that ap′≠∅. Then, there exists b≠a such that b∈ap′. Therefore, G\b∩a≠∅ for every supra preopen set G containing b. This implies that any supra preopen set containing b contains a as well. Thus, the intersection of all supra preopen sets containing b is not equal b. However, this contradicts 3. Hence, it must be ap′=∅.
4⟶1: let a≠b. Since ap′=∅ and bp′=∅, then a and b are supra preclosed sets. Therefore, ac and bc are supra preopen sets containing b and a, respectively. Thus, X,μ is an SpT1-space.Proposition 8.
EveryX,μ satisfying the difference property for the collection of supra preopen sets is an SpT1-space.Proof.
Leta≠b∈X. Since X is a supra preopen set and X,μ satisfies the difference property for the collection of supra preopen sets, then X\a and X\b are supra preopen sets containing b and a, respectively, such that a∉X\a and b∉X\b. Hence, X,μ is an SpT1-space.We show by the following example that the converse of the above proposition is not always true.Example 7.
Letμ=∅,X,1,2,1,3 be a supra topology on X=1,2,3. Then, the collection of all supra preopen subsets of X,μ is ∅,X,1,1,2,1,3,2.3. Therefore, X,μ is an SpT1-space. On the contrary, X,μ does not satisfy the difference property for the collection of supra preopen sets because 2,3 is a supra preopen set, but 2,3\3 is not supra preopen.We need the following definition to obtain the equivalence betweenSpT0 and SpT1.Definition 18.
X,μ is called a supra presymmetric space if a∈pclb implies that b∈pcla for a≠b∈X.Theorem 12.
LetX,μ be a supra presymmetric space. Then, it is SpT1 iff it is SpT0.Proof.
The necessary condition is obvious.
To prove the sufficient condition, leta≠b. Then, there exists a supra preopen set G containing only one of them. Say, a∈G and b∉G. Therefore, a∉pclb. By the supra presymmetry of X,μ, we have b∉pcla. Thus, pclac is a supra preopen set containing b. Hence, X,μ is SpT1.Theorem 13.
The following three statements are equivalent:(1)
X,μ is an SpT2-space(2)
a=∩Fi:Fi is a supra preclosed neighborhood of a for each a∈X(3)
The diagonalΔ=a,a:a∈X is supra preclosed in the product supra space X×XProof.
1⟶2: consider X,μ is an SpT2-space. Then, for a≠b, there exist two disjoint supra preopen sets Gi and Hi such that a∈Gi and b∈Hi. Obviously, Gi⊆Hic. Therefore, a∈pclGi⊆Hic=Fi. Thus, Fi is a supra preclosed neighborhood of a such that b∉Fi. Hence, a=∩Fi:Fi is a supra preclosed neighborhood of a.
2⟶1: to prove that X,μ is an SpT2-space, let a≠b. Since a=∩Fi:Fi is a supra preclosed neighborhood of a, then there exists a supra preclosed neighborhood Fi0 of a such that b∉Fi0. Therefore, there exists a supra preopen set G containing a such that a∈pclG⊆Fi0. It is clear that pclGc is a supra preopen set containing b and G∩pclGc=∅. Hence, X,μ is an SpT2-space.
1⟶3: suppose that X,μ is SpT2 and let a,b∈X×X−Δ. Then, a≠b. Therefore, there exist two disjoint supra preopen sets G and H containing a and b, respectively. Thus, a,b∈G×H⊆X×X−Δ, proving that X×X−Δ is a supra preneighbourhood of any of its points. Hence, Δ is supra preclosed.
3⟶1: suppose that Δ is a supra preclosed subset of X×X and let a≠b∈X. Then, X×X−Δ is a supra preopen set containing a,b. Therefore, there exist two supra preopen subsets G and H of X,μ such that a,b∈G×H⊆X×X−Δ. This implies that G and H are two disjoint supra preopen sets containing a and b, respectively. Hence, X,μ is SpT2.Theorem 14.
The following three statements are equivalent:(1)
X,μ is supra preregular(2)
For each supra preopen subsetU of X,μ containing a, there exists a supra preopen subset V of X,μ such that a∈V⊆pclV⊆U(3)
Every supra preopen subsetU of X,μ can be represented as follows: U=∪H:H is a supra preopen subset of X,μ and pclH⊆UProof.
1⟶2: let X,μ be a supra preregular space and U be a supra preopen set such that a∈U. Then, there exist disjoint supra preopen sets V and W containing a and Uc, respectively. Therefore, a∈V⊆Wc⊆U. Thus, a∈V⊆pclV⊆U.
2⟶3: suppose that U is a supra preopen set. By hypothesis, for each a∈U, there exists a supra preopen set H such that a∈H⊆pclH⊆U. Then, U=∪H:H is supra preopen and pclH⊆U.
3⟶1: let F be a supra preclosed set such that a∉F. Then, Fc=∪H:H is supra preopen and pclH⊆Fc. Since a∈Fc, then there exists a supra preopen set Ha containing a such that pclHa⊆Fc. Take V=pclHac. Then, V is a supra preopen set containing F and V∩Ha=∅. This completes the proof.Theorem 15.
ConsiderX,μ is a supra preregular space. Then, the following concepts are equivalent:(1)
X,μ is an SpT2-space(2)
X,μ is an SpT1-space(3)
X,μ is an SpT0-spaceProof.
The implications1⟶2⟶3 are obvious.
3⟶1: let a,b∈X such that a≠b. Since X,μ is an SpT0-space, then from Theorem 10, we obtain pcla≠pclb. Therefore, a∉pclb or b∉pcla. Say, a∉pclb. Since X,μ is supra preregular, then there exist disjoint supra preopen sets G and H containing a and pclb, respectively. Thus, X,μ is an SpT2-space.Theorem 16.
The following statements are equivalent:(1)
X,μ is supra prenormal(2)
For each supra preclosed setF and each supra preopen set U containing F, there exists a supra preopen set V such that F⊆V⊆pclV⊆U(3)
For every supra preopen setsU and V such that U∪V=X, there are two supra preclosed sets F and H contained in U and V, respectively, such that F∪H=XProof.
1⟶2: consider X,μ is supra prenormal and F is a supra preclosed subset of a supra preopen set U. Then, Uc and F are disjoint supra preclosed sets. Therefore, there exist two disjoint supra preopen sets W and V containing Uc and F, respectively. Thus, F⊆V⊆Wc=pclWc⊆U. Hence, F⊆V⊆pclV⊆U.
2⟶3: consider U and V are supra preopen sets such that U∪V=X. Then, Uc is a supra preclosed sets such that Uc⊆V. By 2, there is a supra preopen set G such that Uc⊆G⊆pclG⊆V. Thus, Gc⊆U and pclG⊆V are supra preclosed sets such that Gc∪pclG=X.
3⟶1: consider F and H are disjoint supra preclosed sets. Since Fc and Hc are supra open sets such that Fc∪Hc=X, then there are two supra preclosed sets M and N such that M⊆Fc, N⊆Hc, and M∪N=X. Thus, Mc and Nc are two disjoint supra preopen sets containing F and H, respectively. Hence, X,μ is supra prenormal.Theorem 17.
If all members ofμ are supra clopen, then X,μ is an SpT4-space.Proof.
Leta∈X such that cla=A⊆X. By hypothesis, A is a supra clopen set. Then, intcla=intA=A. Therefore, a is a supra preopen set. The arbitrary selection of x implies that every singleton subset of X is supra preopen. Therefore, every subset of X is supra preopen. Hence, the desired result is proved.To explain that the converse of the above theorem is not always true, we provide the following example.Example 8.
Letμ=∅,X,3,1,2,2,3 be a supra topology on X=1,2,3. Then, the collection of all supra preopen subsets of X,μ is ∅,X,2,3,1,2,1,3,2,3. It easily checks that X,μ is an SpT4-space. On the contrary, a set 2,3∈μ is not a supra clopen set.Now, we show the implications of these separation axioms among themselves as well as withSTi-space.It should be noted that the concepts ofSTi-space which were defined by replacing “supra preopen” by “supra open” in Definition 4.1, see [1, 2].Theorem 18.
EverySpTi-space is SpTi−1 for i=1,2,3,4.Converse of this theorem is not necessary true as seen from the following examples.Example 9.
Letμ=∅,X,4,1,3,4,2,3,4,1,2,4 be a supra topology on X=1,2,3,4. Then, the collection of all supra preopen subsets of X,μ is ∅,X,4,1,4,2,4,3,4,1,3,4,2,3,4,1,2,4. Therefore, X,μ is not an SpT1-space because 1≠4 and every supra preopen set containing 1 contains 4 as well. On the contrary, it can be checked that X,μ is SpT0.Example 10.
Assume thatX,μ is the same as in Example 7. Then, X,μ is not an SpT2-space because 2≠3 and there do not exist disjoint supra preopen sets such that one of them contains 2 and the other contains 3. On the contrary, it can be checked that X,μ is SpT1.Example 11.
Letμ=∅,X,1,2,3,4,1,3,2,4,2,3,1,2,3,1,2,4,1,3,4,2,3,4 be a supra topology on X=1,2,3,4. In X,μ, every set is supra open iff it is supra preopen. Now, 1,4 is a supra preclosed set and 2∉1,4. Since there do not exist two disjoint supra preopen sets such that one of them contains 2 and the other contains 1,4, then X,μ is not supra SpT3. On the contrary, it can be checked that X,μ is SpT2.Example 12.
Letμ=∅,X,2,4,2,4,1,3,1,4,1,5,2,3,2,5,3,5, 4,5,1,2,3,1,2,4,1,2,5,1,3,4,1,3,5,1,4,5,2,3,4,2,3,5,2,4,5, 3,4,5,1,2,3,4,1,2,3,5,1,2,4,5,1,3,4,5,2,3,4,5 be a supra topology on X=1,2,3,4,5. In X,μ, every set is supra open iff it is supra preopen. Now, 1,2 and 3,4 are disjoint supra preclosed subsets of X,μ. Since there do not exist two disjoint supra preopen sets such that one of them contains 1,2 and the other contains 3,4, then X,μ is not supra prenormal. Therefore, it is not SpT4. On the contrary, one can check that X,μ is SpT3.Theorem 19.
EverySTi-space X,μ is SpTi for i=0,1,2.Proof.
It follows from the fact that every supra open set is supra preopen.Converse of this theorem is not necessary true as is seen in the following examples.Example 13.
Letμ=∅,X,1,2,3 be a supra topology on X=1,2,3. Then, X,μ is not an ST0-space. On the contrary, it follows from Theorem 17 that X,μ is SpT4.We complete this section by discussing these separation axioms in terms of hereditary and topological properties and finite product space.Definition 19.
A property is said to be a relative prehereditary property if the property passes from a supra topological space to every relative presubspace.Theorem 20.
A property of being anSpTi-space is a relative prehereditary for i=0,1,2,3.Proof.
We prove the theorem in the case ofi=3 and the other cases follow similar lines.
Suppose thatA,μA is a relative presubspace of an SpT3-space X,μ. We first show that A,μA is an SpT1-space. Let x≠y∈A⊆X. Then, there are two supra preopen subsets U and V of X,μ containing x and y, respectively, such that x∉V and y∉U. Now, G=A∩U and H=A∩V are two supra preopen subsets of A,μA containing x and y, respectively, such that x∉H and y∉G. Thus, A,μA is SpT1. Second, we show that A,μA is supra preregular. Let H be a supra preclosed subset of A,μA and a∈A such that a∉H. It follows from Proposition 3 that there is a supra preclosed subset F of X,μ such that H=F∩A. Since a∉F, then there exist disjoint supra preopen subsets U and V of X,μ containing F and a, respectively. Now, M=U∩A and N=V∩A are disjoint supra preopen subsets of A,μA containing H and a, respectively. Thus, A,μA is supra preregular. Hence, the proof is complete.Proposition 9.
Letg:X,μ⟶Y,θ be an injective supra precontinuous map. If Y,θ is Ti, then X,μ is SpTi for i=0,1,2.Proof.
We only prove the proposition in the case ofi=2 and the other cases can be made similarly.
Leta≠b∈X. Then, it follows from the injectivity of g, that there are x≠y∈Y such that x=fa and y=fb. Since Y,θ is T2, then there are two disjoint open subsets U and V of Y,θ containing a and b, respectively. Now, g−1U and g−1V are disjoint supra preopen subsets of X,μ containing a and b, respectively. Hence, X,μ is SpT2, as required.
In a similar way, one can prove the following results.Proposition 10.
Letg:X,τ⟶Y,ν be a bijective supra preopen map. If X,τ is Ti, then Y,ν is SpTi for i=0,1,2.Proposition 11.
Letg:X,τ⟶Y,ν be an injective supra pre⋆-continuous map. If X,τ is SpTi, then Y,ν is SpTi for i=0,1,2.Proposition 12.
Letg:X,τ⟶Y,ν be a bijective supra pre⋆-open map. If X,τ is SpTi, then Y,ν is SpTi for i=0,1,2.Proposition 13.
Letg:X,τ⟶Y,ν be a supra pre⋆-homeomorphism map. Then, X,τ is SpTi iff Y,ν is SpTi for i=0,1,2,3,4.Theorem 21.
The finite product ofSpTi-spaces is SpTi for i=0,1,2.Proof.
We prove the theorem for two supra topological spacesX,μ and Y,ν in the case of i=2. One can prove the other cases similarly.
LetX×Y,T be the product supra space of X,μ and Y,ν. Suppose that x1,y1≠x2,y2. Then, either x1≠x2 or y1≠y2. Without loss of generality, suppose that x1≠x2. Therefore, there exist two disjoint supra preopen subsets U and V of X,μ containing x1 and x2, respectively. It follows from Theorem 3 that U×Y and V×Y are two supra preopen subsets of X×Y,T containing x1,y1 and x2,y2, respectively, such that U×Y∩˜V×Y=∅. Hence, X×Y,T is SpT2.Definition 20.
LetX,μ and Y,ν be two supra topological spaces and X×Y,T be their product supra space such that C1 and C2 are the collections of all supra preopen subsets of X,μ and Y,ν, respectively. Then, β=G×H:G∈C1 and H∈C2 defines a basis for a supra topology C on X×Y. We called X×Y,C a prefinite product supra space.Lemma 2.
LetX,μ and Y,ν be two supra topological spaces and X×Y,C be their preproduct supra space. If E is a supra closed subset of X×Y,C, then E=∩i∈IFi×Y∪X×Hi, where Fi and Hi are supra preclosed subsets of X,μ and Y,ν, respectively.Theorem 22.
The prefinite product ofSpTi-spaces is STi for i=0,1,2,3.Proof.
We prove the theorem for two supra topological spacesX,μ and Y,ν in the case of i=3. One can prove the other cases similarly.
LetX×Y,C be the preproduct supra space of X,μ and Y,ν. We first prove that X×Y,C is ST1. Suppose that x1,y1≠x2,y2. Then, either x1≠x2 or y1≠y2. Without loss of generality, suppose that x1≠x2. Therefore, there exist two supra preopen subsets U and V of X,μ containing x1 and x2, respectively. According to Definition 20, U×Y and V×Y are two supra open subsets of X×Y,C containing x1,y1 and x2,y2 such that x1,y1∉V×Y and x2,y2∉U×Y. Hence, X×Y,C is ST1. Second, we prove that X×Y,C is supra regular. Suppose that x,y∈X×Y and E is a supra closed subset of X×Y,C such that x,y∉E=∩i∈IFi×Y∪X×Hi, where Fi and Hi are supra preclosed subsets of X,μ and Y,ν, respectively. Then, there exists j∈I such that x,y∉Fj×Y∪X×Hj. This means that x∉Fj and y∉Hj. Since X,μ and Y,ν are supra pre regular, then there exist disjoint supra preopen subsets U and V of X,μ containing x and Fj, respectively, and there exist disjoint supra preopen subsets M and N of Y,ν containing y and Hj, respectively. Therefore, U×M and V×Y∪X×N are two supra open subsets of X×Y,C containing x,y and Fj×Y∪X×Hj, respectively. Obviously, E⊆Fj×Y∪X×Hj and U×M∩V×Y∩X×N=∅. Thus, X×Y,T is supra regular. Hence, the proof is complete.


## 5. Conclusion

We began this work by correcting some results of [4]. Then, we have presented the concept of supra pre homeomorphism and investigated main properties. Also, we have introduced and studied the concepts of supra limits and supra boundary points with respect to preopen sets. Finally, we have defined the concepts of supra preregular, supra prenormal, and SpTi-spaces i=0,1,2,3,4 and discussed their basic properties. From the concrete thoughts given in this work, more investigations can be carried out on the theoretical parts of these generalized ideas which are valuable by studying the following themes:(1)
Define weak types of supra preregular and supra prenormal spaces(2)
StudySpTi-spaces for 1/2, 2 1/2, 3 1/2, 5(3)
Explore the concepts introduced herein using the classes of supraα-open sets, supra semiopen sets, supra b-open sets, and supra β-open sets(4)
Investigate of the possibility of applying these concepts on information system, especially, separation axioms



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