# More onDα-Closed Sets in Topological Spaces

Authors: Xiao-Yan Gao; Ahmed Mostafa Khalil
Journal: Journal of Mathematics
 (2021)

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## Abstract

The aim of this paper is to present and study topological properties ofDα-derived, Dα-border, Dα-frontier, and Dα-exterior of a set based on the concept of Dα-open sets. Then, we introduce new separation axioms (i.e., Dα−R0 and Dα−R1) by using the notions of Dα-open set and Dα-closure. The space of Dα−R0 (resp., Dα−R1) is strictly between the spaces of α−R0 (resp., α−R1) and g−R0 (resp., g−R1). Further, we present the notions of Dα-kernel and Dα-convergent to a point and discuss the characterizations of interesting properties between Dα-closure and Dα-kernel. Finally, several properties of weakly Dα−R0 space are investigated.


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## Body

## 1. Introduction and Preliminaries

Many researchers (see [1–9]) were interested in general topology-like family (e.g., the family of all α-open sets) and also the notion of generalized closed (briefly, g-closed) subset of a topological space [10–14]. In 1982, Dunham [14] used the generalized closed sets to define a novel closure operator and consequently a novel topology τ∗, on the space, and discussed several of the properties of this novel topology. Sayed and Khalil [15] introduced and studied a novel type of sets called Dα-open sets in topological spaces and studied the notions of Dα-continuous, Dα-open, and Dα-closed functions between topological spaces. Further, they investigated several properties of Dα-closed and strongly Dα-closed graphs. In fact, research on spaces analogous to topological spaces and generalized closed sets among topological spaces may have certain driving effect on research on theory of rough set, soft set, spatial reasoning, implicational spaces and knowledge spaces, and logic (see [16–18]). For this reason, we will define the notions of Dα-derived, Dα-border, Dα-frontier, and Dα-exterior of a set based on the notion of Dα-open sets. We will also discuss new separation axioms (Dα−R0 and Dα−R1) by using the notions of Dα-open set and Dα-closure operator.The rest of this article is arranged as follows. In this section, we briefly recall several notions:α-open set, an α-closed set, generalized open set, generalized closed set, α−R0 space, g−R0 space, α−R1 space, g−R0 space, α-derived, α-border, α-frontier, and α-exterior of a set, which are used in the sequel. In Section 2, we define the notions of Dα-derived, Dα-border, Dα-frontier, and Dα-exterior of a set based on Dα-open sets. In Section 3, we present the notions Dα−R0, Dα−R1, Dα-kernel, and Dα-convergent to a point and introduce the characterizations of interesting properties between Dα-closure and Dα-kernel. In Section 4, we define the weakly Dα−R0 space and investigate some properties of weakly Dα−R0 space.Throughout the present paper, two subsetsA of a space X,τ, CA and ℐA, denote the closure and the interior of A, respectively. Since we require the following known definitions, notations, and some properties, we recall in this section.Definition 1.
LetX,τ be a topological space and A⊆X. Then,(1)
A is α-open [1] if A⊆ℐCℐA and α-closed [1] if CℐCA⊆A(2)
A is generalized closed (briefly, g-closed) [10] if CA⊆U whenever A⊆U and U is open in X(3)
A is generalized open (briefly, g-open) [10] if X\A is g-closed(4)
A is Dα-open [15] if A⊆ℐ∗Cℐ∗A and Dα-closed [15] if C∗ℐC∗A⊆A
Theα-closure of a subset A of X [2] is the intersection of all α-closed sets containing A and is denoted by CαA. The α-interior of a subset A of X [2] is the union of all α-open sets contained in A and is denoted by ℐαA. The intersection of all g-closed sets containing A [14] is called the g-closure of A and is denoted by C∗A and the g-interior of A [19] is the union of all g-open sets contained in A and is denoted by ℐ∗A. The intersection of all Dα-closed sets containing A [15] is called the Dα-closure of A and is denoted by CαDA and the Dα-interior of A [15] is the union of all Dα-open sets contained in A and is denoted by ℐαDA.
We need the following notations:(i
)αOX (resp., αCX) denotes the family of all α-open sets (resp., α-closed sets) in X,τ(ii)
GOX (resp., GCX) denotes the family of all generalized open sets (resp., generalized closed sets) in X,τ(iii)
DαOX (resp., DαCX) denotes the family of all Dα-open sets (resp., Dα-closed sets) in X,τ(iv)
αOX,x=U|x∈U∈αOX,τ, OX,x=U|x∈U∈τ, and αCX,x=U|x∈U∈αCX,τ(v)
DαOX,x=U|x∈U∈DαOX,τ and DαCX,x=U|x∈U∈DαCX,τDefinition 2.
A topological spaceX,τ is said to be(1)
α−R0 space [20] (resp., g−R0 space [21]) if every α-open (resp., g-open) set contains the α-closure (resp., g-closure) of each of its singletons(2)
α−R1 space [20] (resp., g−R1 space [21]) if, for x,y in X with Cαx≠Cαy (resp., C∗x≠C∗y), there exist disjoint α-open (resp., g-open) sets U and V such that Cαx (resp., C∗x) is a subset of U and Cαy (resp., C∗y) is a subset of VDefinition 3 (see [22]).
A pointx∈X is said to be α-limit point of A in topological space X,τ if, for each α-open set U containing x,U∩A∖x≠ϕ. The set of all α-limit points of A is called an α-derived set of A.Definition 4 (see [22]).
LetA be a subset of a space X:(1)
Anα-border of A is defined by bαA=A∖ℐαA(2)
Anα-frontier of A is defined by FαA=CαA∖ℐαA(3)
Anα-exterior of A is defined by ExtαA=ℐαX∖A


## 2. ADα-Derived, Dα-Border, Dα-Frontier, and Dα-Exterior of a Set

Definition 5.
LetA be a subset of a space X. A point x∈X is said to be Dα-limit point of A if it satisfies the following assertion:(1)∀U∈DαOXx∈U⇒U∩A∖x≠ϕ
The set of allDα-limit points of A is called a Dα-derived set of A and is denoted by dαDA.
Note that, for a subsetA of X, a point x∈X is not a Dα-limit point of A if and only if there exists a Dα-open set U in X such that(2)x∈U,U∩A∖x=ϕ,or, equivalently,(3)x∈U,U∩A=ϕ,U∩A=x,or equivalently,(4)x∈U,U∩A⊆x.Theorem 1.
LetA and B be subsets of a topological space X. Then, the following results hold:(1)
dαDA⊆dαA, where dαA is the α-derived set ([22], Definition 2.1) of A(2)
IfA⊆B, then dαDA⊆dαDB(3)
dαDA∪dαDB⊆dαDA∪B and dαDA∩B⊆dαDA∩dαDB(4)
dαDdαDA∖A⊆dαDA(5)
dαDA∪dαDA⊆A∪dαDAProof.
(1)
It follows from ([15], Theorem 3.6 (i)).(2)
Letx∈dαDA and U∈DαOX with x∈U. Then U∩A∖x≠ϕ. Since A⊆B, it follows that U∩B∖x≠ϕ. Therefore x∈dαDB.(3)
It follows from (2) above.(4)
Letx∈dαDdαDA∖A and U∈DαOX with x∈U. Then U∩dαDA∖x≠ϕ. Let y∈U∩dαDA∖x. Then y∈U and y∈dαDA, and so U∈A∖y≠ϕ. If we take z∈U∩A∖y, then z≠x for z∈A and x∉A. Hence, U∈A∖y≠ϕ. Therefore x∈dαDA.(5)
Letx∈dαDA∪dαDA. If x∈A, the result is obvious. Suppose that x∉A. Then U∩A∪dαDA∖x≠ϕ for all U∈DαOX with x∈U. Hence, U∩A∖x≠ϕ or U∩dαDA∖x≠ϕ. The first case implies that x∈dαDA. If U∩dαDA∖x≠ϕ, then x∈dαDdαDA. Since x∉A, it follows similarly from (4) that x∈dαDdαDA∖A⊆dαDA. Therefore, dαDA∪dαDA⊆A∪dαDA holds.Theorem 2.
LetA be a subset of a topological space X. Then CαDA=A∪dαDA.Proof.
Letx∈CαDA. If x∈A, then the proof is complete. If x∉A and U∈DαOX with x∈U, then U∩A∖x≠ϕ, and so x∈dαDA. Hence, CαDA⊆A∪dαDA. The converse follows from ([15], Theorem 2.14 (i)) and dαDA⊆CαDA. Thus, A∪dαDA⊆CαDA. Therefore CαDA=A∪dαDA.Corollary 1.
A subsetA is a Dα-closed set if and only if it contains the set of the Dα-limit pointsTheorem 3.
LetA and B be subsets of X. If A is Dα-closed, then CαDA∩B⊆A∩CαDB.Proof.
It follows from Theorems 2.13 and 2.14 (vi) in [15].Lemma 1.
LetA be a subset of a topological space X. If A is Dα-closed set, then dαDA⊆A.Proof.
Suppose thatA is a Dα-closed set. Let x∉A; that is, x∈X∖A. Since it is a Dα-open, x is not a Dα-limit point of A, that is, x∉dαDA, because X∖A∩A∖x=ϕ. Hence, dαDA⊆A.Theorem 4.
LetA be a subset of a topological space X. If F is a Dα-closed set of A, then dαDA⊆F.Proof.
By Theorem1 (2) and Lemma 1, A⊆F implies that dαDA⊆dαDF⊆F.Theorem 5.
LetA be a subset of a topological space X. If a point x∈X is a Dα-limit point of A, then x is also a Dα-limit point of A∖x.Proof.
The proof is obvious.Definition 6.
LetA be a subset of a topological space X. The Dα-border of A, denoted by bαDA, is defined as bαDA=A∖ℐαDA.Theorem 6.
LetA be a subset of a topological space X. Then, the following results hold:(1)
bαDA⊆bαA, where bαA is the α-border ([22], Definition 2.8) of A(2)
A=ℐαDA∪bαDA(3)
ℐαDA∩bαDA=ϕ(4)
A is a Dα-open set if and only if bαDA=ϕ(5)
bαDℐαDA=ϕ(6)
bαDA=A∩CαDX∖A(7)
bαDA=A∩dαDX∖AProof.
(1)
SinceℐαA⊆ℐαDA ([1], Theorem 3.15 (i)), we have(5)bαDA=A∖ℐαDA⊆A∖ℐαA=bαA,(2)
and (3) are obvious.(4)
Itfollows from Theorems 3.14 and 3.15 (i) in [15].(5)
SinceℐαDA is a Dα-open, it follows from (4) that bαDℐαDA=ϕ.(6)
Using ([15], Lemma 3.13 (i)), we have(6)bαDA=A∖ℐαDA=A∖X∖CαDX∖A=A∩CαDX∖A.(7)
Applying (6) and Theorem3, we have(7)bαDA=A∩CαDX∖A=A∩X∖A∪dαDX∖A=A∩dαDX∖A.The converse of (1) of Theorem 6 is not true in general as shown in the following example.Example 1.
LetX,τ be a topological space, where X=a,b,c and τ=ϕ,a,X. Then, FX=ϕ,b,c,X,αOX=ϕ,a,a,b,a,c,X,αCX=ϕ,b,c,b,c,X,DαOX=DαCX=PX. Let A=c. Then bαA=c⊈bαDA=ϕ.Definition 7.
LetA be a subset of a topological space X. The Dα-frontier of A, denoted by FrαDA, is defined as FrαDA=CαDA∖ℐαDA.Lemma 2.
LetA be a subset of X. If A is a Dα-closed subset of X, then bαDA=FrαDA.Proof.
It follows from ([15], Theorem 2.13).Theorem 7.
LetA be a subset of a topological space X. Then the following results hold:(1)
FrαDA⊆FrαA, where FrαA is the α-frontier ([22], Definition 2.11) of A.(2)
CαDA=ℐαDA∪FrαDA.(3)
ℐαDA∩FrαDA=ϕ.(4)
bαDA⊆FrαDA.(5)
FrαDA=bαDA∪dαDA∖ℐαDA.(6)
IfA is a Dα-open set, then FrαDA=bαDX∖A.(7)
FrαDA=CαDA∩CαDX∖A.(8)
FrαDA=FrαDX∖A.(9)
FrαDA is a Dα-closed set.(10)
FrαDFrαDA⊆FrαDA.(11)
FrαDℐαDA⊆FrαDA.(12)
FrαDCαDA⊆FrαDA.(13)
ℐαDA=A∖FrαDA.(14)
FrαDA=CαDA∖ℐαDA.(15)
CαDA⊆A∪FrαDA.(16)
X∖FrαDA=ℐαDA∪ℐαDX∖A.Proof.
(1)
SinceCαDA⊆CαA ([15], Theorem 2.14 (i)) and ℐαA⊆ℐαDA ([15], Theorem 3.15 (i)), we have FrαDA=CαDA∖ℐαDA⊆CαA∖ℐαA=FrαA.(2)
It is obvious.(3)
ℐαDA∩FrαDA=CαDA∖ℐαDA=ϕ.(4)
SinceA⊆CαDA ([15], Theorem 2.14 (i)), we have bαDA=A∖ℐαDA⊆CαDA∖ℐαDA=FrαA.(5)
Using Theorem2, we have(8)FrαDA=CαDA∖ℐαDA=A∪dαDA∩X∖ℐαDA=A∖ℐαDA∪dαDA∖ℐαDA=bαDA∪dαDA∖ℐαDA.(6)
It follows from (5) above, Theorem6 (4), (7), and ([15], Theorem 3.14).(7)
It follows from ([15], Lemma 3.13 (ii)).(8)
It follows from (7) above.(9)
CαDFrαDA=CαDCαDA∩CαDX∖A⊆CαDCαDA∩CαDCαDX∖A=CαDA∩CαDX∖A=FrαDAObviously,FrαDA⊆CαDFrαDA ([15], Theorem 2.14 (i)), and so FrαDA=CαDFrαDA. Hence, FrαDA is a Dα-closed set.(10)
It follows from (9) above and Lemma2.(11)
It follows from Definition7 and ([15], Theorem 3.15 (vi)).(12)
It follows from Definition7 and ([15], Theorem 2.14 (vi)).(13)
A∖FrαD=A∖CαDA∖ℐαDA=A∩X∖CαDA∪ℐαDA=ϕ∪A∪ℐαDA=ℐαDA(14)
It follows from (7) above and ([15], Lemma 3.13 (ii)).(15)
A∪FrαDA=A∪CαDA∩CαDX∖A=A∪CαDA∩A∪CαDX∖A=CαDA∩A∪CαDX∖A⊇CαDA∩A∪X∖A=CαDA∩X=CαDA(16)
ℐαDA∪ℐαDX∖A=X∖X∖ℐαDA∪X∖X∖ℐαDX∖A=X∖X∖ℐαDA∩ℐαDX∖A=X∖CαDX∖A∩CαDA=X∖FrαDAThe converse of (1) and (4) of Theorem7 is not true as shown in the following examples.Example 2.
Consider the topological spaceX,τ which is given in Example 1. Let A=c. Then FrαA=c⊈FrαDA=ϕ.Example 3.
LetX,τ be a topological space, where X=a,b,c and τ=ϕ,a,b,X. Then FX=ϕ,c,X,DαOX=ϕ,a,b,a,b,a,c,b,c,X,DαCX=ϕ,a,b,c,a,c,b,c,X. Let A=a,b. Then FrαDA=c⊈brαDA=ϕ.Theorem 8.
LetA be a subset of a topological space X. Then FrαDA=ϕ if and only if A is a Dα-closed set and a Dα-open set.Proof.
Suppose thatFrαDA=ϕ. First, we prove that A is a Dα-closed set. We have FrαDA=ϕ or CαDA∩CαDX∖A=ϕ. Hence, CαDA⊆X∖CαDX∖A=ℐαDA. Therefore, CαDA⊆A and so A is a Dα-closed set. Now, we prove that A is a Dα-open set. Indeed, we have FrαDA=ϕ or CαDA∩CαDX∖A=ϕ. Hence, A∩X∖ℐαDA=ϕ and so A⊆ℐαDA. Therefore, A is a Dα-open set. Conversely, suppose that A is a Dα-closed set and a Dα-open set. Then FrαDA=CαDA∩CαDX∖A=CαDA∩X∖ℐαDA=A∩X∖A=ϕ.Theorem 9.
LetA be a subset of a topological space X. Then,(1)
A is a Dα-open set if and only if A∩FrαDA=ϕ;(2)
A is a Dα-closed set if and only if FrαDA⊆A.Proof.
(1)
LetA be a Dα-open set. Then ℐαDA=A implies that A∩FrαDA=ℐαDA∩FrαDA=ϕ (by Theorem 7 (3)). Conversely, suppose that A∩FrαDA=ϕ. Then A∩CαDA∩CαDX∖A=ϕ or A∩CαDA∩CαDX∖A=ϕ⇒A∩CαDX∖A=ϕ, which implies that A⊆X∖CαDX∖A=ℐαDA. Moreover, ℐαDA⊆A. Therefore, ℐαDA=A and thus A is a Dα-open set.(2)
LetA be a Dα-closed set. Then CαDA=A. Now, FrαDA=CαDA∩CαDX∖A⊆CαDA=A. That is, FrαDA⊆A. Conversely, suppose that FrαDA⊆A. Then FrαDA∩X∖A=ϕ. Since FrαDA=FrαDX∖A (by Theorem 7 (8)), we have FrαDX∖A∩X∖A=ϕ. By (1), X∖A is a Dα-open set. Hence, A is a Dα-closed set.Lemma 3.
LetA be a subset of a topological space X. If A is a Dα-closed set, then A∖ℐαDA=FrαDA.Proof.
It follows from ([15], Theorem 2.13) and Theorem 7 (14).Theorem 10.
LetA and B be subsets of X. Then, the following results hold:(1)
FrαDA∪B⊆FrαDA∪FrαDB.(2)
FrαDA∩B⊆FrαDA∩CαDB∪FrαDB∩CαDA.(3)
FrαDFrαDFrαDA=FrαDFrαDA.Proof.
(9)1FrαDA∪B=CαDA∪B∩CαDX∖A∪B=CαDA∪CαDB∩CαDX∖A∩X∖B⊆CαDA∪CαDB∩CαDX∖A∩CαDX∖B=CαDA∩CαDX∖A∩CαDX∖B∪CαDB∩CαDX∖A∩CαDX∖B=FrαDA∩CαDX∖B∪FrαDB∩CαDX∖A⊆FrαDA∪FrαDB2FrαDA∩B=CαDA∩B∩CαDX∖A∩B⊆CαDA∩CαDB∩CαDX∖A∪X∖B=CαDA∩CαDB∩CαDX∖A∪CαDX∖B=CαDA∩CαDB∩CαDX∖A∪CαDA∩CαDB∩CαDX∖B=FrαDA∩CαDB∪CαDA∩FrαDB3FrαDFrαDFrαDA=CαDFrαDFrαDA∩CαDX∖FrαDFrαDA=FrαDFrαDA∩CαDX∖FrαDFrαDAi.
Now consider(10)X∖FrαDFrαDA=X∖CαDFrαDA∩CαDX∖FrαDA=X∖FrαDA∩CαDX∖FrαDA=X∖FrαDA∪X∖CαDX∖FrαDA.
Therefore,(11)CαDX∖FrαDFrαDA=CαDCαDX∖FrαDA∪X∖CαDX∖FrαDA=CαDCαDX∖FrαDA∪CαDX∖CαDX∖FrαDA=B∪X∖CαDX∖FrαDB=Xii,where B=CαDCαDX∖FrαDA. From (i) and (ii), we have(12)FrαDFrαDFrαDA=FrαDFrαDA∩X=FrαDFrαDA.Definition 8.
LetA be a subset of a topological space X. The Dα-exterior of A, denoted by ExtαDA, is defined as ExtαDA=ℐαDX∖A.Theorem 11.
LetA and B be subsets of X. Then the following results hold:(1)
ExtαA⊆ExtαDA, where ExtαA is the α-exterior ([22], Definition 2.16) of A.(2)
ExtαDA=X∖CαDA.(3)
ExtαDExtαDA=ℐαDCαDA.(4)
IfA⊆B, then ExtαDB⊆ExtαDA.(5)
ExtαDA∪B⊆ExtαDA∩ExtαDB.(6)
ExtαDA∩B⊇ExtαDA∪ExtαDB.(7)
ExtαDX=ϕ and ExtαDϕ=X.(8)
ExtαDA=ExtαDX∖ExtαDA.(9)
X=ℐαDA∪ExtαDA∪FrαDA.Proof.
(1)
It follows from ([15], Theorem 3.15 (i)).(2)
It follows from ([15], Lemma 3.13 (i)).(3)
It follows from ([15], Lemma 3.13 (ii)).(4)
It follows from ([15], Theorem 3.15 (iii)).(5)
It follows from ([15], Theorem 3.15 (vi)).(6)
It follows from ([15], Theorem 3.15 (v)).(7)
It is obvious.(8)
It follows from ([15], Theorem 3.15 (iv)).(9)
It is obvious.The opposite of (1) and (4) of Theorem11 is not true as shown in the following examples.Example 4.
Consider the topological spaceX,τ which is given in Example 1. Let A=a. Then ExtαDA=b,c⊈ExtαA=ϕ.Example 5.
Consider the topological spaceX,τ which is given in Example 3. Let A=a and B=a,b. Then ExtαDA=b,c⊈ExtαDB=ϕ.Remark 1.
The equality in statements (5) of Theorem11 need not be true as seen from Example 3. Let A=a,B=b, and A∪B=a,b. Then ExtαDA∪B=ϕ≠c=ExtαDA∩ExtαDB. Furthermore, the equality in statement (6) of the above theorem need not be true as seen from Example 3. Let A=a,b,B=c, and A∩B=ϕ. Then ExtαDA∩B=X≠a,b=ExtαDA∪ExtαDB.


## 3.Dα−R0 and Dα−R1 Spaces

Definition 9.
LetA be a subset of a topological space X. The Dα-kernel of A, denoted by KerαDA, is defined as KerαDA=∩U∈DαOX|A⊂U.Definition 10.
Letx be a point of a topological space X. The Dα-kernel of x, denoted by KerαDx, is defined as KerαDx=∩U∈DαOX|x∈U.Lemma 4.
LetX,τ be a topological space and x∈X.Then,(1)
y∈KerαDx if and only if x∈CαDy;(2)
KerαDA=∩x∈X|CαDx∩A≠ϕ.Proof.
(1)
Suppose thaty∉KerαDx. Then there exists a Dα-open set V containing x such that y∉V. Therefore, we have x∉CαDy. The proof of the opposite case can be done similarly.(2)
Letx∈KerαDA and CαDx∩A=ϕ. Hence, x∉X−CαDx which is a Dα-open set containing A. This is impossible, since x∈KerαDA. Consequently, CαDx∩A≠ϕ. Let CαDx∩A≠ϕ and x∉KerαDA. Then, there exists a Dα-open set W containing A and x∉W. Let y∉CαDx∩A. Hence, W is a Dα-neighborhood of y where x∉W. By this contradiction, x∈KerαDA and the proof is completed.Lemma 5.
The following statements are equivalent for any pointsx and y in a topological space X,τ:(1)
KerαDx≠KerαDy.(2)
CαDx≠CαDy.Proof.
(i)
1⇒2 Suppose that KerαDx≠KerαDy. Then there exists a point z in X such that z∈KerαDx and z∉KerαDy. It follows from z∈KerαDx that x∩CαDz≠ϕ. This implies that x∈CαDz. By z∉KerαDy, we have y∩CαDz=ϕ. Since x∈CαDz,CαDx⊂CαDz and y∩CαDx=ϕ, CαDx≠CαDy. Now, KerαDx≠KerαDy implies that CαDx≠CαDy.(ii)
2⇒1 Suppose that CαDx≠CαDy. Then there exists a point z in X such that z∈CαDx and z∉CαDy. Then, there exists a Dα-open set containing z and therefore x but not y, that is, y∉KerαDx. Hence, KerαDx≠KerαDy.Definition 11.
A topological spaceX,τ is said to be a Dα−R0 space if every Dα-open set contains the Dα-closure of each of its singletons.Theorem 12.
LetX,τ be a topological space. Then,(1)
everyα−R0 space is Dα−R0(2)
everyg−R0 space is Dα−R0Proof.
It is obvious from ([15], Theorem 3.6).From the above discussions, we have the following diagram in which the opposites of implications need not be true.(13)α−R0⟶Dα−R0⟵g−R0.Theorem 13.
A topological spaceX,τ is a Dα−R0 space if and only if, for any x and y in X,CαDx≠CαDy implies that CαDx∩CαDy=ϕ.Proof.
Necessity. Suppose thatX,τ is a Dα−R0 and x,y∈X such that CαDx≠CαDy. Then, there exists z∈CαDx such that z∉CαDy (or z∈CαDy such that z∉CαDx). There exists U∈DαOX such that y∉U and z∈U; hence, x∈U. Therefore, we have x∉CαDy. Thus, x∈X−CαDy∈DαOX, which implies that CαDx⊂X−CαDy and CαDx∩CαDy=ϕ. The proof for otherwise is similar.(i)
Sufficiency. LetU∈DαOX and x∈U. We will show that CαDx⊂U. Let y∉U; that is, y∈X−U. Then x≠y and x∉CαDy. This shows that CαDx≠CαDy. By assumption, CαDx∩CαDy=ϕ. Hence, y∉CαDx and therefore CαDx⊂U.Theorem 14.
A topological spaceX,τ is a Dα−R0 space if and only if, for any x and y in X,KerαDx≠KerαDy implies that KerαDx∩KerαDx=ϕ.Proof.
Suppose thatX,τ is a Dα−R0 space. Then, by Lemma 5, for any points x and y in X if KerαDx≠KerαDy, then CαDx≠CαDy. Now, we prove that KerαDx∩KerαDy=ϕ. Assume that z∈KerαDx∩KerαDy. By z∈KerαDx and Lemma 4 (1), it follows that x∈CαDz. Since x∈CαDx, by Theorem 13CαDx=CαDz. Similarly, we have CαDy=CαDz=CαDx. This is a contradiction. Therefore, we have KerαDx∩KerαDy=ϕ. Conversely, let X,τ be a topological space such that, for any points x and y in X,KerαDx≠KerαDy implies that KerαDx∩KerαDy=ϕ. If CαDx≠CαDy, then, by Lemma 5, KerαDx≠KerαDx. Hence, KerαDx∩KerαDy=ϕ, which implies that CαDx∩CαDy=ϕ. Because z∈CαDx implies that x∈KerαDz, KerαDx∩KerαDz≠ϕ. By hypothesis, we have KerαDx=KerαDz. Then z∈CαDx∩CαDx implies that KerαDx=KerαDz=KerαDy. This is a contradiction. Hence, CαDx∩CαDy=ϕ. By Theorem 13, we have that X,τ is a Dα−R0 space.Theorem 15.
For a topological spaceX,τ, the following properties are equivalent:(1)
X,τ is a Dα−R0 space.(2)
For anyA≠ϕ and G∈DαOX such that A∩G≠ϕ, there exists F∈DαCX such that A∩F≠ϕ and F⊂G.(3)
AnyA∈DαOX,G=∪F∈DαCX|F⊂G.(4)
AnyF∈DαOX,F=∩G∈DαOX|F⊂G.(5)
For anyx∈X,CαDx⊂KerαDx.Proof.
1⇒2 Let A be a nonempty set of X and G∈DαOX such that A∩G≠ϕ. There exists x∈A∩G. Since x∈G∈DαOX,CαDx⊂G. Set F=CαDx; then F∈DαCX,F⊂G, and A∩F≠ϕ.2⇒3 Let G∈DαOX. Then G⊃∪F∈DαCX|F⊂G. Let x be any point of G. There exists F∈DαCX such that x∈F and F⊂G. Therefore, we have x∈F⊂∪F∈DαCX|F⊂G and hence G=∪F∈DαCX|F⊂G.3⇒4 This is obvious.4⇒5 Let x be any point of X and y∉KerαDx. There exists U∈DαOX such that x∈U and y∉U. Hence, CαDy∩U=ϕ. By (4) ∩G∈DαOX|CαDy⊂G∩U=ϕ. There exists G∈DαOX such that x∉G and CαDy⊂G. Therefore, CαDx∩G=ϕ and y∉CαDx. Consequently, we obtain CαDx⊂KerαDx.5⇒1 Let G∈DαOX and x∈G. Suppose that y∈KerαDx. Then x∈CαDy and y∈G. This implies that CαDx⊂KerαDx⊂G. Therefore, X,τ is a Dα−R0 space.Corollary 2.
For a topological spaceX,τ, the following properties are equivalent:(1)
X,τ is a Dα−R0 space.(2)
CαDx=KerαDx for all x∈X.Proof.
1⇒2 Suppose that X,τ is a Dα−R0 space. By Theorem 15, CαDx⊂KerαDx for each x∈X. Let y∈KerαDx. Then x∈CαDx and so CαDx=CαDy. Therefore, y∈CαDx and hence KerαDx⊂CαDx. This shows that CαDx=KerαDx.2⇒1 This is obvious by Theorem 15.Theorem 16.
For a topological spaceX,τ, the following properties are equivalent:(1)
X,τ is a Dα−R0 space.(2)
x∈CαDy if and only if y∈CαDx, for any points x and y in X.Proof.
1⇒2 Assume that X,τ is a Dα−R0 space. Let x∈CαDy and let W be any Dα-open set such that y∈W. Now, by hypothesis, x∈W. Therefore, every Dα-open set containing y contains x. Hence, y∈CαDx.2⇒1 Let U be a Dα-open set and x∈U. If y∉U, then x∉CαDy and hence y∉CαDx. This implies that CαDx⊂U. Hence, X,τ is a Dα−R0 space.Theorem 17.
For a topological spaceX,τ, the following properties are equivalent:(1)
X,τ is a Dα−R0 space.(2)
IfF is Dα-closed, then F=KerαDF.(3)
IfF is Dα-closed and x∈F, then KerαDx⊂F.(4)
Ifx∈X, then KerαDx⊂CαDx.Proof.
1⇒2 Let F be Dα-closed and x∉F. Thus, X−F is Dα-open and contains x. Since X,τ is Dα−R0,CαDx⊂X−F. Thus CαDx∩F=ϕ and by Lemma 4 (2) x∉KerαDF. Therefore, KerαDF=F.2⇒3 In general, A⊂B implies that KerαDA⊂KerαDB. Therefore, it follows from (2) that KerαDx⊂KerαDF=F.3⇒4 Since x∈CαDx and CαDx is Dα-closed, by (3), KerαDx⊂CαDx.4⇒1 We show the implication by using Theorem 16. Let x∈CαDx. Then, by Lemma 4 (1), y∈KerαDx. Since x∈CαDx and CαDx is Dα-closed, by (4), we obtain y∈KerαDx⊂CαDx. Therefore, x∈CαDy implies that y∈CαDx. The opposite is obvious and X,τ is a Dα−R0 space.Definition 12.
A filter baseF is called Dα-convergent to a point x in X, if, for any Dα-open set U of X containing x, there exists B in F such that B is a subset of U.Lemma 6.
LetX,τ be a topological space and x and y are any two points in X such that every net in XDα-converging to yDα-converges to x. Then x∈CαDy.Proof.
Suppose thatxn=y for each n∈N. Then xnn∈N is a net in CαDy. Since xnn∈NDα-converges to y, xnn∈NDα-converges to x and this implies that x∈CαDy.Theorem 18.
For a topological spaceX,τ, the following statements are equivalent:(1)
X,τ is a Dα−R0 space.(2)
Ifx,y∈X, then y∈CαDx if and only if every net in XDα-converging to yDα-converges to x.Proof.
1⇒2 Let x,y∈X such that y∈CαDx. Suppose that xαα∈N is a net in X such that xαα∈NDα-converges to y. Since y∈CαDx and by Theorem 15, we have CαDx=CαDy. Therefore, x∈CαDy. This means that xαα∈ΛDα-converges to x. Conversely, let x,y∈X such that every net in XDα-converging to yDα-converges to x. Then x∈CαDy by Lemma 4 (2). By Theorem 15, we have CαDx=CαDy. Therefore, y∈CαDx.2⇒1 Assume that x and y are any two points of X such that CαDx∩CαDy≠ϕ. Let z∈CαDx∩CαDy. So there exists a net xαα∈Λ in CαDx such that xαα∈ΛDα-converges to z. Since z∈CαDy, we have xαα∈ΛDα-converges to y. It follows that y∈CαDx. Similarly, we obtain x∈CαDy. Therefore, CαDx=CαDy and, by Theorem 15, X,τ is a Dα−R0 space.Definition 13.
A topological spaceX,τ is said to be Dα−R1 space if, for x,y in X with CαDx≠CαDy, there exist disjoint Dα-open sets U and V such that CαDx is a subset of U and CαDy is a subset of V.Theorem 19.
LetX,τ be a topological space. Then,(1)
everyα−R1 space is Dα−R1(2)
everyg−R1 space is Dα−R1Proof.
It is obvious.From the above discussions, we have the following diagram in which the opposite of implications need not be true.(14)α−R1⟶Dα−R1⟵g−R1.Theorem 20.
IfX,τ is a Dα−R1 space, then X,τ is a Dα−R0 space.Proof.
LetU be Dα-open such that x∈U. If y∉U, then since x∉CαDy,CαDx≠CαDy. Hence, there exists Dα-open Vy such that CαDy⊂Vy and x∉Vy, which implies that y∉CαDx. Thus, CαDx⊂U. Therefore, X,τ is a Dα−R0 space.Theorem 21.
A topological spaceX,τ is said to be a Dα−R1 space if and only if x,y∈X,KerαDx≠KerαDy, and there exist disjoint Dα-open sets U and V such that CαDx⊂U and CαDy⊂V.Proof.
It follows from Lemma4 (1).


## 4. WeaklyDα−R0 Space

Definition 14.
A topological spaceX,τ is said to be weakly Dα−R0 space if ∩x∈XCαDx=ϕ.Theorem 22.
A topological spaceX,τ is weakly Dα−R0 space if and only if KerαDx≠X for every x∈X.Proof.
Assume that the spaceX,τ is weakly Dα−R0 space. Suppose that there is a point y in X such that KerαDy=X. Then y∉O, where O is some proper Dα-open subset of X. This implies that y∈∩x∈XCαDx. But this is a contradiction. Now suppose that KerαDx≠X for every x∈X. If there exists a point y∈X such that y∈∩x∈XCαDx, then every Dα-open set containing y must contain every point of X. This implies that the space X is the unique Dα-open set containing y. Thus, KerαDx=X, which is a contradiction. Hence, X,τ is a weakly Dα−R0 space.Theorem 23.
A topological spaceX,τ is a weakly Dα−R0 space if and only if KerαDx≠X for every x∈X.Definition 15.
A functionf:X⟶Y is said to be always Dα-closed if the image of every Dα-closed subset of X is Dα-closed in Y.Theorem 24.
Iff:X⟶Y is an always injective Dα-closed function and X is a weakly Dα−R0 space, then Y is a weakly Dα−R0 space.Proof.
The proof is clear.Theorem 25.
If the topological spaceX is weakly Dα−R0 and Y is any topological space, then the product X×Y is weakly Dα−R0.Proof.
If we show that∩x,y∈X×YCαDx,y=ϕ, then we are done. Observe that ∩x,y∈X×YCαDx,y⊂∩x,y∈X×YCαDx×CαDy=∩x∈XCαDx×∩y∈YCαDy⊂ϕ×Y=ϕ and hence the proof is completed.



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