Some applications of Dα-closed sets in topological spaces

Abstract

In this paper, a new kind of sets called Dα-open sets are introduced and studied in a topological spaces. The class of all Dα-open sets is strictly between the class of all α-open sets and g-open sets. Also, as applications we introduce and study Dα-continuous, Dα-open, and Dα-closed functions between topological spaces. Finally, some properties of Dα-closed and strongly Dα-closed graphs are investigated.

2010 Mathematics subject classification:

• 54A05
• 54C08
• 54C10
• 54D10

Keywords:

• α-closed
• Dα-closed
• Dα-continuous
• Dα-closed graph
• Strongly Dα-closed graph

1 Introduction and preliminaries

Generalized open sets play a very important role in General Topology, and they are now the research topics of many topologies worldwide. Indeed a significant theme in General Topology and Real Analysis is the study of variously modified forms of continuity, separation axioms, etc. by utilizing generalized open sets. One of the most well-known notions and also inspiration source are the notion of α-open [1] sets introduced by Njåstad in 1965 and generalized closed (g-closed) subset of a topological space [2] introduced by Levine in 1970. Since then, many mathematicians turned their attention to the generalization of various concepts in General Topology by considering α-open sets [3–^{[4][5][6][7][8][9]}10] and generalized closed sets [11–^[12]13]. In 1982 Dunham [14] used the generalized closed sets to define a new closure operator, and thus a new topology τ*, on the space, and examined some of the properties of this new topology. Throughout the present paper $(X,\tau ),(Y,\sigma )$ and (Z, ν) denote topological spaces (briefly X, Y and Z) and no separation axioms are assumed on the spaces unless explicitly stated. For a subset A of a space (X, τ), Cl(A) and Int(A) 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 1.1

Let (X, τ) be a topological space and A ⊆ X. Then

(i)	A is α-open [1] if $A\subseteq Int(Cl(Int(A))$ and α-closed [1] if $Cl(Int(Cl(A))\subseteq A$.

(ii)	A is generalized closed (briefly g-closed) [2] if Cl(A) ⊆ U whenever A ⊆ U and U is open in X.

(iii)

A is generalized open(briefly g-open) [2] if X\A is g-closed.

The α-closure of a subset A of X [3] is the intersection of all α-closed sets containing A and is denoted by Cl_α(A). The α-interior of a subset A of X [3] is the union of all α-open sets contained in A and is denoted by Int_α(A). The intersection of all g-closed sets containing A [14] is called the g-closure of A and denoted by Cl*(A), and the g-interior of A [15] is the union of all g-open sets contained in A and is denoted by Int*(A).

We need the following notations:

•	αO(X) (resp. αC(X)) denotes the family of all α-open sets (resp. α-closed sets) in (X, τ).

•	GO(X) (resp. GC(X)) denotes the family of all generalized open sets (resp. generalized closed sets) in (X, τ).

•	$\alpha O(X,x)=\left\{U\right|x\in U\in \alpha O(X,\tau )\}$, $O(X,x)=\left\{U\right|x\in U\in \tau \}$ and $\alpha C(X,x)=\left\{U\right|x\in U\in \alpha C(X,\tau )\}$.

Definition 1.2

A function f : X → Y is said to be:

(i)	α-continuous [16] (resp. g-continuous [17]) if the inverse image of each open set in Y is α-open (resp. g-open) in X.

(ii)	α-open [16] (resp. α-closed [16]) if the image of each open (resp. closed ) set in X is α-open (resp. α-closed) in Y.

(iii)

g-open [18] (resp. g-closed [18]) if the image of each open (resp. closed) set in X is g-open (resp. g-closed) in Y.

Definition 1.3

Let f : X → Y be a function:

(i)	The subset $\{(x,\mathrm{f}(x))|x\in X\}$ of the product space X × Y is called the graph of f [19] and is usually denoted by G(f).

(ii)	a closed graph [19] if its graph G(f) is closed sets in the product space X × Y.

(iii)

a strongly closed graph [20] if for each point $(x,y)\notin G(f)$, there exist open sets U ⊂ X and V ⊂ Y containing x and y, respectively, such that $(U\times Cl(V))\cap G(f)=\varphi$.

(iv)	a strongly α-closed graph [21] if for each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in \alpha O(X,x)$ and V ∈ O(Y, y) such that $(U\times Cl(V))\cap G(f)=\varphi$.

Definition 1.4

A topological space (X, τ) is said to be:

(i)	α-T1 [9] (resp. g-T1 [22]) if for any distinct pair of points x and y in X, there exist α-open (resp. g-open) set U in X containing x but not y and an α-open (resp. g-open) set V in X containing y but not x.

(ii)	α-T2 [8] (resp. g-T2 [22]) if for any distinct pair of points x and y in X, there exist α-open (resp. g-open) sets U and V in X containing x and y, respectively, such that U ∩ V = ϕ.

Lemma 1.5

Let A ⊆ X, then

(i)	$X\backslash Cl*(A)=Int*(X\backslash A)$.

(ii)	$X\backslash Int*(A)=Cl*(X\backslash A)$.

Lemma 1.6

A function $f:(X,\tau )\to (Y,\sigma )$ has a closed graph [19] if for each $(x,y)\in (X\times Y)\backslash G(f)$, there exist U ∈ O(X, x) and V ∈ O(Y, y) such that f(U) ∩ V = ϕ.

Lemma 1.7

The graph G(f) is strongly closed [23] if and only if for each point $(x,y)\notin G(f)$, there exist open sets U ⊂ X and V ⊂ Y containing x and y, respectively, such that $f(U)\cap Cl(V)=\varphi$.

2 Dα-closed sets

In this section we introduce Dα-closed sets and investigate some of their basic properties.

Definition 2.1

A subset A of a space X is called Dα-closed if $Cl*(Int(Cl*(A)))\subseteq A$.

The collection of all Dα-closed sets in X is denoted by DαC(X).

Lemma 2.2

If there exists an g-closed set F such that $Cl*(Int(F))\subseteq A\subseteq F$, then A is Dα-closed.

Proof

Since F is g-closed, Cl*(F) = F. Therefore, $Cl*(Int(Cl*(A)))\subseteq Cl*(Int(Cl*(F)))=Cl*(Int(F))\subseteq A$. Hence A is Dα-closed.

Remark 2.3

The converse of above lemma is not true as shown in the following example.

Example 2.4

Let (X, τ) be a topological space, where X = {a, b, c} and $\tau =\{\varphi ,\{a\},\{a,b\},X\}$. Then $F_X=\{\varphi ,\{c\},\{b,c\},X\}$, $GC(X)=\{\varphi ,\{c\},\{a,c\},\{b,c\},X\}$, $GO(X)=\{\varphi ,\{a\},\{b\},\{a,b\},X\}$, $D\alpha C(X)=\{\varphi ,\{b\},\{c\},\{a,c\},\{b,c\},X\}$, $D\alpha O(X)=\{\varphi ,\{a\},\{b\},\{a,b\},\{a,c\},X\}$. Therefore $\left\{c\right\}\in D\alpha C(X)$ and $\{a,c\}\in GC(X)$ but $Cl*(Int\{a,c\})=\{a,c\}\not\subset \left\{c\right\}\subset \{a,c\}$.

Theorem 2.5

Let (X, τ) be a topological space. Then

(i)	Every α-closed subset of (X, τ) is Dα-closed.

(ii)	Every g-closed subset of (X, τ) is Dα-closed.

Proof

(i) Since closed set is g-closed, $Cl*(A)\subseteq Cl(A)$ [14]. Now, suppose A is α-closed in X, then $Cl(Int(Cl(A)))\subseteq A$. Therefore, $Cl*(Int(Cl*(A)))\subseteq Cl(Int(Cl(A)))\subseteq A$. Hence A is Dα-closed in X.

(ii) Suppose A is g-closed. Then Cl*(A) = A [14]. Therefore, $Int(Cl*(A))\subseteq Cl*(A)$. Then $Cl*(Int(Cl*(A)))\subseteq Cl*(Cl*(A))\subseteq Cl*(A)=A$ [14]. Hence A is Dα-closed.

Remark 2.6

The converse of above theorem is not true as shown in the following example.

(i)	Dα-closed set need not be α-closed. (see Example 2.7 below)

(ii)	Dα-closed set need not be g-closed. (see Example 2.7 below)

Example 2.7

Let (X, τ) be a topological space, where $X=\{a,b,c\}$ and $\tau =\{\varphi ,\{a,b\},X\}$. Then $F_X=\alpha C(X)=\{\varphi ,\{c\},X\}$, $\alpha O(X)=\{\varphi ,\{a,b\},X\}$, $GC(X)=\{\varphi ,\{c\},\{a,c\},\{b,c\},X\}$, $GO(X)=\{\varphi ,\{a\},\{b\},\{a,b\},X\}$, $D\alpha C(X)=\{\varphi ,\{a\},\{b\},\{c\},\{a,c\},\{b,c\},X\}$, $D\alpha O(X)=\{\varphi ,\{a\},\{b\},\{a,b\},\{a,c\},\{b,c\},X\}$. Therefore $\left\{a\right\}\in D\alpha C(X)$, but $\left\{a\right\}\notin \alpha C(X)$ and $\left\{a\right\}\notin GC(X)$.

From the above discussions we have the following diagram in which the converses of implications need not be true.$\alpha \text{-closed set}\to D\alpha \text{-closed set}\leftarrow g\text{-closed set}$

Theorem 2.8

Arbitrary intersection of Dα-closed sets is Dα-closed.

Proof

Let {F_i : i ∈ Λ} be a collection of Dα-closed sets in X. Then $Cl*(Int(Cl*(F_i)))\subseteq F_i$ for each i. Since ∩ F_i ⊆ F_i for each i, $Cl*(\cap F_i)\subseteq Cl*(F_i)$ for each i. Hence $Cl*(\cap F_i)\subseteq \cap Cl*(F_i),i\in \Lambda$. Therefore $Cl*(Int(Cl*(\cap F_i)))\subseteq Cl*(Int(\cap Cl*(F_i)))\subseteq Cl*(\cap Int(Cl*(F_i)))\subseteq \cap Cl*(Int(Cl*(F_i)))\subseteq \cap F_i$. Hence ∩F_i is Dα-closed.

Remark 2.9

The union of two Dα-closed sets need not to be Dα-closed as shown in Example 2.7, where both {a} and {b} are Dα-closed sets but $\left\{a\right\}\cup \left\{b\right\}=\{a,b\}$ is not Dα-closed.

Corollary 2.10

If a subset A is Dα-closed and B is α-closed, then A ∩ B is Dα-closed.

Proof

Follows from Theorem 2.5 (i) and Theorem 2.8.

Corollary 2.11

If a subset A is Dα-closed and F is g-closed, then A ∩ F is Dα-closed.

Proof

Follows from Theorem 2.5 (ii) and Theorem 2.8.

Definition 2.12

Let A be a subset of a space X. The Dα-closure of A, denoted by $C{l}_{\alpha }^D(A)$, is the intersection of all Dα-closed sets in X containing A. That is $C{l}_{\alpha }^D(A)=\cap \{F:A\subseteq F\text{and}F\in D\alpha C(X)\}$.

Theorem 2.13

Let A be a subset of X. Then A is Dα-closed set in X if and only if $C{l}_{\alpha }^D(A)=A$.

Proof

Suppose A is Dα-closed set in X. By Definition 2.12, $C{l}_{\alpha }^D(A)=A$. Conversely, suppose $C{l}_{\alpha }^D(A)=A$. By Theorem 2.8 A is Dα-closed.

Theorem 2.14

Let A and B be subsets of X. Then the following results hold.

(i)	$A\subseteq C{l}_{\alpha }^D(A)\subseteq C{l}_{\alpha }(A),C{l}_{\alpha }^D(A)\subseteq Cl*(A)$.

(ii)	$C{l}_{\alpha }^D(\varphi )=\varphi$ and $C{l}_{\alpha }^D(X)=X$.

(iii)

If A ⊆ B, Then $C{l}_{\alpha }^D(A)\subseteq C{l}_{\alpha }^D(B)$.

(iv)	$C{l}_{\alpha }^D(C{l}_{\alpha }^D(A))=C{l}_{\alpha }^D(A)$.

(v)	$C{l}_{\alpha }^D(A)\cup C{l}_{\alpha }^D(B)\subseteq C{l}_{\alpha }^D(A\cup B)$.

(vi)	$C{l}_{\alpha }^D(A\cap B)\subseteq C{l}_{\alpha }^D(A)\cap C{l}_{\alpha }^D(B)$.

Proof

(i) Follows From Theorem 2.5 (i) and (ii), respectively.

(ii) and (iii) are obvious.

(iv) If A ⊆ F, F ∈ DαC(X), then from (iii) and Theorem 2.13, $C{l}_{\alpha }^D(A)\subseteq C{l}_{\alpha }^D(F)=F$. Again $C{l}_{\alpha }^D(C{l}_{\alpha }^D(A))\subseteq C{l}_{\alpha }^D(F)=F$. Therefore $C{l}_{\alpha }^D(C{l}_{\alpha }^D(A))\subseteq \cap \{F:A\subseteq F,F\in D\alpha C(X)\}=C{l}_{\alpha }^D(A)$.

(v) and (vi) follows from (iii).

Remark 2.15

The equality in the statements (v) of the above theorem need not be true as seen from Example 2.7, where A = {a}, B = {b}, and $A\cup B=\{a,b\}$. Then one can have that, $C{l}_{\alpha }^D(A)=\left\{a\right\}$; $C{l}_{\alpha }^D(B)=\left\{b\right\}$; $C{l}_{\alpha }^D(A\cup B)=X$; $C{l}_{\alpha }^D(A)\cup C{l}_{\alpha }^D(B)=\{a,b\}$. Further more the equality in the statements (iv) of the above theorem need not be true as shown in the following example.

Example 2.16

Let (X, τ) be a topological space, where X = {a, b, c} and $\tau =\{\varphi ,\{b\},\{c\},\{b,c\},X\}$. Then $F_X=GC(X)=D\alpha C(X)=\{\varphi ,\{a\},\{a,b\},\{a,c\},X\}$, $GO(X)=\{\varphi ,\{b\},\{c\},\{b,c\},X\}$. Let A = {a}, B = {b}, and A ∩ B = ϕ. Then one can have that, $C{l}_{\alpha }^D(A)=\left\{a\right\}$; $C{l}_{\alpha }^D(B)=\{a,b\}$; $C{l}_{\alpha }^D(A\cap B)=\varphi$; $C{l}_{\alpha }^D(A)\cap C{l}_{\alpha }^D(B)=\left\{a\right\}$.

3 Dα-open sets

In this section we introduce Dα-open sets and investigate some of their basic properties.

Definition 3.1

A subset A of a space X is called an Dα-open if X \ A is Dα-closed. Let DαO(X) denote the collection of all an Dα-open sets in X.

Lemma 3.2

Let A ⊆ X, then

(i)	$X\backslash Cl*(X\backslash A)=Int*(A)$.

(ii)	$X\backslash Int*(X\backslash A)=Cl*(A)$.

Proof

Obvious.

Theorem 3.3

A subset A of a space X is Dα-open if and only if $A\subseteq Int*(Cl(Int*(A)))$.

Proof

Let A be Dα-open set. Then X \ A is Dα-closed and $Cl*(Int(Cl*(X\backslash A)))\subseteq X\backslash A$. By Lemma 3.2 $A\subseteq Int*(Cl(Int*(A)))$. Conversely, suppose $A\subseteq Int*(Cl(Int*(A)))$. Then $X\backslash Int*(Cl(Int*(A)))\subseteq X\backslash A$. Hence $(Int*(Cl(Int*(X\backslash A))))\subseteq X\backslash A$. This shows that X \ A is Dα-closed. Thus A is Dα-open.

Lemma 3.4

If there exists g-open set V such that $V\subseteq A\subseteq Int*(Cl(V))$, then A is Dα-open.

Proof

Since V is g-open, X \ V is g-closed and $X\backslash Int*(Cl(V))\subseteq X\backslash A\subseteq X\backslash V$. Therefore From Lemma 3.2 $Cl*(Int(X\backslash V))\subseteq X\backslash A\subseteq X\backslash V$. From Lemma 2.2 we have X \ A is Dα-closed. Hence A is Dα-open.

Remark 3.5

The converse of Lemma 3.4 need not to be true as seen from Example 2.4, where $\{a,b\}\in D\alpha O(X)$ and $\left\{b\right\}\in GO(X)$ but $\left\{b\right\}\subset \{a,b\}\not\subset \left\{b\right\}$.

Theorem 3.6

Let (X, τ) be a topological space. Then

(i)	Every α-open subset of (X, τ) is Dα-open.

(ii)	Every g-open subset of (X, τ) is Dα-open.

Proof

From Theorem 2.5, the proof is obvious.

Remark 3.7

The converse of the above theorem is not true as seen from Example 2.7, where $\{b,c\}\in D\alpha O(X)$ but $\{b,c\}\notin \alpha O(X)$ and $\{b,c\}\notin GO(X)$.

From the above discussions we have the following diagram in which the converses of implications need not be true.$\alpha \text{-open set}\to D\alpha \text{-open set}\leftarrow g\text{-open set}$

Theorem 3.8

Arbitrary union of Dα-open set is Dα-open.

Proof

Follows from Theorem 2.8.

Remark 3.9

The intersection of two Dα-open sets need not be Dα-open as seen from Example 2.7, where both {b, c} and {a, c} are Dα-open sets but $\{b,c\}\cap \{a,c\}=\left\{c\right\}$ is not Dα-open.

Corollary 3.10

If a subset A is Dα-open and B is α-open, then A ∪ B is Dα-open.

Proof

Follows from Theorem 3.6 (i) and Theorem 3.8.

Corollary 3.11

If a subset A is Dα-open and U is g-open, then A ∪ U is Dα-open.

Proof

Follows from Theorem 3.6 (ii) and Theorem 3.8.

Definition 3.12

Let A be a subset of a space X. The Dα-interior of A is denoted by $In{t}_{\alpha }^D(A)$, is the union of all an Dα-open sets in X contained in A. That is $In{t}_{\alpha }^D(A)=\cup \{U:U\subseteq A,U\in D\alpha O(X)\}$.

Lemma 3.13

If A is a subset of X, then

(i)	$X\backslash C{l}_{\alpha }^D(A)=In{t}_{\alpha }^D(X\backslash A)$.

(ii)	$X\backslash In{t}_{\alpha }^D(A)=C{l}_{\alpha }^D(X\backslash A)$.

Proof

Obvious.

Theorem 3.14

Let A be a subset of X. Then A is Dα-open if and only if $In{t}_{\alpha }^D(A)=A$.

Proof

Follows from Theorem 2.13 and Lemma 3.13.

Theorem 3.15

Let A and B be subsets of X. Then the following results hold.

(i)	$In{t}_{\alpha }(A)\subseteq In{t}_{\alpha }^D(A)\subseteq A$,$Int*(A)\subseteq In{t}_{\alpha }^D(A)$.

(ii)	$In{t}_{\alpha }^D(\varphi )=\varphi$ and $In{t}_{\alpha }^D(X)=X$.

(iii)

If A ⊆ B, then $In{t}_{\alpha }^D(A)\subseteq In{t}_{\alpha }^D(B)$.

(iv)	$In{t}_{\alpha }^D(In{t}_{\alpha }^D(A))=In{t}_{\alpha }^D(A)$.

(v)	$In{t}_{\alpha }^D(A)\cup In{t}_{\alpha }^D(B)\subseteq In{t}_{\alpha }^D(A\cup B)$.

(vi)	$In{t}_{\alpha }^D(A\cap B)\subseteq In{t}_{\alpha }^D(A)\cap In{t}_{\alpha }^D(B)$.

Proof

Obvious.

Remark 3.16

The equality in the statements (v) of Theorem 3.15 need not be true as seen from Example 2.7, where A = {b, c}, B = {a, c}, and A ∪ B = X. Then one can have that, $In{t}_{\alpha }^D(A)=\{b,c\}$; $In{t}_{\alpha }^D(B)=\left\{c\right\}$;$In{t}_{\alpha }^D(A)\cup In{t}_{\alpha }^D(B)=\{b,c\}$; $In{t}_{\alpha }^D(A\cup B)=X$. Furthermore the equality in the statements (iv) of the above theorem need not be true as seen from Example 2.7, where A = {b, c}, B = {a, c}, and A ∩ B = {c}. Then one can have that, $In{t}_{\alpha }^D(A)=\{b,c\}$; $In{t}_{\alpha }^D(B)=\{a,c\}$; $In{t}_{\alpha }^D(A\cap B)=\varphi$; $In{t}_{\alpha }^D(A)\cap In{t}_{\alpha }^D(B)=\left\{c\right\}$.

Theorem 3.17

Let x ∈ X. Then $x\in C{l}_{\alpha }^D(A)$ if and only if U ∩ A ≠ ϕ for every Dα-open set U containing x.

Proof

Let $x\in C{l}_{\alpha }^D(A)$ and there exists Dα-open set U containing x such that U ∩ A = ϕ. Then $A\subseteq X\backslash U$ and X\U is Dα-closed. Therefore $C{l}_{\alpha }^D(A)\subseteq C{l}_{\alpha }^D(X\backslash U)=X\backslash U$. This implies $x\notin C{l}_{\alpha }^D(A)$, which is a contradiction. Conversely, assume that U ∩ A ≠ ϕ for every Dα-open set U containing x and $x\notin C{l}_{\alpha }^D(A)$. Then there exists Dα-closed subset F containing A such that $x\notin F$. Hence x ∈ X \ F and X \ F is Dα-open. Therefore A ⊆ F, $(X\backslash F)\cap A=\varphi$ This is a contradiction to our assumption.

Lemma 3.18

Let A be any subset of (X, τ). Then

(i)	$A\cap Int*(Cl(Int*(A)))$ is Dα-open;

(ii)	$A\cup Cl*(Int(Cl*(A)))$ is Dα-closed.

Proof

(i)	$Int*(Cl(Int*(A\cap Int*(Cl(Int*(A))))))=Int*(Cl(Int*(A)\cap Int*(Cl(Int*(A)))))=Int*(Cl(Int*(A)))$. This implies that $A\cap Int*(Cl(Int*(A)))=A\cap Int*(Cl(Int*(A\cap Int*(Cl(Int*(A))))))\subseteq Int*(Cl(Int*(A\cap Int*(Cl(Int*(A))))))$. Therefore $A\cap Int*(Cl(Int*(A)))$ is Dα-open.

(ii)	From (i) we have $X\backslash (A\cup Cl*(Int(Cl*(A)))=(X\backslash A)\cap Int*(Cl(Int*(X\backslash A)))$ is Dα-open that further implies $A\cup Cl*(Int(Cl*(A)))$ is Dα-closed.

Theorem 3.19

If A is a subset of a topological space X,

(i)	$In{t}_{\alpha }^D(A)=A\cap Int*(Cl(Int*(A)))$.

(ii)	$C{l}_{\alpha }^D(A)=A\cup Cl*(Int(Cl*(A)))$.

Proof

(i)

Let $B=In{t}_{\alpha }^D(A)$. Clearly B is Dα-open and B ⊆ A. Since B is Dα-open, $B\subseteq Int*(Cl(Int*(B)))\subseteq Int*(Cl(Int*(A)))$. This proves that $B\subseteq A\cap Int*(Cl(Int*(A)))$. By Lemma 3.18, $A\cap Int*(Cl(Int*(A)))$ is Dα-open. By the definition of $In{t}_{\alpha }^D(A),A\cap Int*(Cl(Int*(A)))\subseteq B$. Then it follows that $B=A\cap Int*(Cl(Int*(A)))$. Therefore $In{t}_{\alpha }^D(A)=A\cap Int*(Cl(Int*(A)))$.

(ii)	By Lemma 3.13 we have $C{l}_{\alpha }^D(A)=X\backslash In{t}_{\alpha }^D(X\backslash A),=X\backslash ((X\backslash A)\cap Int*(Cl(Int*(X\backslash A))))$, using (i)

• $=X\backslash (X\backslash A)\cup (X\backslash Int*(Cl(Int*(X\backslash A)))=A\cup Cl*(Int(Cl*(A)))$.

4 Dα-continuous functions

In this section we introduce Dα-continuous functions and investigate some of their basic properties.

Definition 4.1

A function f : X → Y is called Dα-continuous if the inverse image of each open set in Y is Dα-open in X.

Theorem 4.2

(i)	Every α-continuous function is Dα-continuous.

(ii)	Every g-continuous function is Dα-continuous.

Proof

It is obvious from Theorem 3.6.

Remark 4.3

(i)	Dα-continuous function need not be α-continuous.

• (see Example 4.4 (i) below)

(ii)	Dα-continuous function need not be g-continuous.

• (see Example 4.4 (ii) below)

Example 4.4

(i) Let X = {a, b, c} associated with the topology $\tau =\{\varphi ,\{a\},X\}$ and Y = {x, y, z} associated with the topology $\sigma =\{\varphi ,\{x,y\},\{z\},Y\}$. Let f : X → Y be a function defined by f(a) = f(b) = x, f(c) = z. One can have that $F_X=\{\varphi ,\{b,c\},X\}$, $GC(X)=\{\varphi ,\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},X\}$, $GO(X)=\{\varphi ,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},X\}$, $\alpha O(X)=\{\varphi ,\{a\},\{a,b\},\{a,c\},X\}$, DαC(X) = DαO(X) =P(X). Since {z} is open in Y, $f^{-1}(\{z\})=\left\{c\right\}\in D\alpha O(X)$, but $\left\{c\right\}\notin \alpha O(X)$. Therefore f is Dα-continuous but not α-continuous.

(ii) Let (X,τ) and (Y,σ) be the topological spaces in (i) and f : X → Y be a function defined by f(a) = x, f(b) = f(c) = z. Since {z} is open in Y, $f^{-1}(\{z\})=\{b,c\}\in D\alpha O(X)$, but $\{b,c\}\notin GO(X)$. Therefore f is Dα-continuous but not g-continuous.

From the above discussions we have the following diagram in which the converses of implications need not be true.$\alpha \text{-continuity}\to D\alpha \text{-continuity}\leftarrow g\text{-continuity}$

Theorem 4.5

Let f : X → Y be a function. Then the following are equivalent:

(i)	f is Dα-continuous.

(ii)	For each x ∈ X and each open set V ⊂ Y containing f(x), there exists Dα-open set W ⊂ X containing x such that f(W) ⊂ V.

(iii)

The inverse image of each closed set in Y is Dα-closed in X.

(iv)	$f(C{l}_{\alpha }^D(A))\subseteq Cl(f(A))$ for every subset A of X.

(v)	$C{l}_{\alpha }^D(f^{-1}(B))\subseteq f^{-1}(Cl(B))$ for every subset B of Y.

(vi)	$f^{-1}(Int(B))\subseteq In{t}_{\alpha }^D(f^{-1}(B))$ for every subset B of Y.

Proof

(i)⇒(ii) Since V ⊂ Y containing f(x) is open, then $f^{-1}(V)\in D\alpha O(X)$. Set $W=f^{-1}(V)$ which contains x, therefore f(W) ⊂ V.

(ii)⇒(i) Let V ⊂ Y be open, and let $x\in f^{-1}(V)$, then f(x) ∈ V and thus there exists $W_x\in D\alpha O(X)$ such that x ∈ W_x and f(W_x) ⊂ V. Then $x\in W_x\subset f^{-1}(V)$, and so $f^{-1}(V)={\displaystyle \underset{x\in f^{-1}(V)}{\cup }{W}_x}$ but ${\displaystyle \underset{x\in f^{-1}(V)}{\cup }{W}_x}\in D\alpha O(X)$ by Theorem 3.8. Hence $f^{-1}(V)\in D\alpha O(X)$, and therefore f is Dα-continuous.

(i)⇒(iii) Let F ⊂ Y be closed. Then Y\F is open and $f^{-1}(Y\backslash F)\in D\alpha O(X)$, i.e. $X-f^{-1}(F)\in D\alpha O(X)$. Then $f^{-1}(F)$ is Dα-closed of X.

(iii)⇒(iv) Let $A\subseteq X$ and F be a closed set in Y containing f(A). Then by (iii), $f^{-1}(F)$ is Dα-closed set containing A. It follows that $C{l}_{\alpha }^D(A)\subseteq C{l}_{\alpha }^D(f^{-1}(F))=f^{-1}(F)$ and hence $f(C{l}_{\alpha }^D(A))\subseteq F$. Therefore $f(C{l}_{\alpha }^D(A))\subseteq Cl(f(A))$.

(iv)⇒(v) Let $B\subseteq Y$ and $A=f^{-1}(B)$. Then by assumption, $f(C{l}_{\alpha }^D(A))\subseteq Cl(f(A))\subseteq Cl(B)$. This implies that $C{l}_{\alpha }^D(A)\subseteq f^{-1}(Cl(B))$. Hence $C{l}_{\alpha }^D(f^{-1}(B))\subseteq f^{-1}(Cl(B))$.

(v)⇒(vi) Let $B\subseteq Y$. By assumption, $C{l}_{\alpha }^D(f^{-1}(Y\backslash B))\subseteq f^{-1}(Cl(Y\backslash B))$. This implies that, $C{l}_{\alpha }^D(X\backslash f^{-1}(B))\subseteq f^{-1}(Y\backslash Int(B))$ and hence $X\backslash In{t}_{\alpha }^D(f^{-1}(B))\subseteq X\backslash f^{-1}(Int(B))$. By taking complement on both sides we get $f^{-1}(Int(B))\subseteq In{t}_{\alpha }^D(f^{-1}(B))$.

(vi)⇒(i) Let U be any open set in Y. Then Int(U) = U. By assumption, $f^{-1}(Int(U))\subseteq In{t}_{\alpha }^D(f^{-1}(U))$ and hence $f^{-1}(U)\subseteq In{t}_{\alpha }^D(f^{-1}(U))$. Then $f^{-1}(U)=In{t}_{\alpha }^D(f^{-1}(U))$. Therefore by Theorem 3.14, $f^{-1}(U)$ is Dα-open in X. Thus f is Dα-continuous.

Theorem 4.6

Let f : X → Y be Dα-continuous and let g : Y → Z be continuous. Then gof : X → Z is Dα-continuous.

Proof

Obvious.

Remark 4.7

Composition of two Dα-continuous functions need not be Dα-continuous as seen from the following example.

Example 4.8

Let $X=\{a,b,c\}$ associated with the topology $\tau =\{\varphi ,\{b\},\{a,b\},X\}$, $Y=\{x,y,z\}$ associated with the topology $\sigma =\{\varphi ,\{x\},Y\}$ and $Z=\{p,q,r\}$ associated with the topology $\nu =\{\varphi ,\{r\},Z\}$ and $f:(X,\tau )\to (Y,\sigma )$ by f(a) = y, f(b) = x, f(c) = z. Define $g:(Y,\sigma )\to (Z,\nu )$ by $g(x)=g(y)=p$, g(z) = r. One can have that $F_X=\{\varphi ,\{c\},\{a,c\},X\}$, $GC(X)=\{\varphi ,\{c\},\{a,c\},\{b,c\},X\}$, $GO(X)=\{\varphi ,\{a\},\{b\},\{a,b\},X\}$, $D\alpha C(X)=\{\varphi ,\{a\},\{c\},\{a,c\},\{b,c\},X\},D\alpha O(X)=\{\varphi ,\{a\},\{b\},\{a,b\},\{b,c\},X\}$, and $F_Y=\{\varphi ,\{y,z\},Y\},GC(Y)=\{\varphi ,\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},Y\}$, $GO(Y)=\{\varphi ,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},Y\}$, $D\alpha C(Y)=D\alpha O(Y)=P(X)$. Clearly, f and g are Dα-continuous. {r} is open in Z. But ${(gof)}^{-1}(\{r\})=f^{-1}(g^{-1}(\left\{r\right\}))=f^{-1}(\{z\})=\left\{c\right\}$, which is not Dα-open in X. Therefore gof is not Dα-continuous.

5 Dα-open functions and Dα-closed functions

In this section we introduce Dα-open functions and Dα-closed functions and investigate some of their basic properties.

Definition 5.1

A function f : X → Y is said to be Dα-open (resp. Dα-closed) if the image of each open (resp. closed) set in X is Dα-open (resp. Dα-closed) in Y.

Theorem 5.2

(i)	Every α-open function is Dα-open.

(ii)	Every g-open function is Dα-open.

Proof

It is obvious from Theorem 3.6.

Remark 5.3

(i)	Dα-open function need not be α-open. (see Example 5.4 below)

(ii)	Dα-open function set need not be g-open. (see Example 5.5 below)

Example 5.4

(i) Let X = {x,y,z} associated with the topology $\tau =\{\varphi ,\{x\},X\}$ and Y = {a,b,c} associated with the topology $\sigma =\{\varphi ,\{a,b\},\{c\},Y\}$. Let $f:(X,\tau )\to (Y,\sigma )$ be a function defined by f(x) = a, f(y) = b and f(z) = c. One can have that $F_Y=\alpha O(Y)=\{\varphi ,\{a,b\},\{c\},Y\}$, GC(Y) = GO(Y) = DαC(Y) = DαO(Y) = P(X). Since {x} is open in X, $f(\{x\})=\left\{a\right\}\in D\alpha O(Y)$, but $\left\{a\right\}\notin \alpha O(Y)$. Therefore f is Dα-open function but not α-open.

Example 5.5

(ii) Let X = {x,y,z} associated with the topology $\tau =\{\varphi ,\{y\},\{x,y\},X\}$ and Y = {a, b, c} associated with the topology $\sigma =\{\varphi ,\{a\},Y\}$. Let $f:(X,\tau )\to (Y,\sigma )$ be a function defined by f(x) = b, f(y) = c and f(z) = a. One can have that $F_Y=\{\varphi ,\{b,c\},Y\}$, $GC(Y)=\{\varphi ,\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},Y\}$, $GO(Y)=\{\varphi ,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},Y\}$, $D\alpha C(Y)=D\alpha O(Y)=P(X)$. Since {x,y} is open in X, $f(\{x,y\})=\{b,c\}\in D\alpha O(Y)$, but $\{b,c\}\notin GO(Y)$. Therefore f is Dα-open function but not g-open.

From the above discussions we have the following diagram in which the converses of implications need not be true.$\alpha \text{-open function}\to D\alpha \text{-open function}\leftarrow g\text{-open function}$

Theorem 5.6

Let f : X → Y be a function. The following statements are equivalent.

(i)	f is Dα-open.

(ii)	For each x ∈ X and each neighborhood U of x, there exists Dα-open set W ⊆ Y containing f(x) such that W ⊆ f(U).

Proof

(i)⇒(ii) Let x ∈ X and U is a neighborhood of x, then there exists an open set V ⊆ X such that x ∈ V ⊆ U. Set W = f(V). Since f is Dα-open, $f(V)=W\in D\alpha O(Y)$ and so $f(x)\in W\subseteq f(U)$.

(ii)⇒(i) Obvious.

Theorem 5.7

Let f : X → Y be Dα-open (resp. Dα-closed) function and W ⊆ Y. If A ⊆ X is a closed (resp. open) set containing $f^{-1}(W)$, then there exists Dα-closed (resp. Dα-open) set H ⊆ Y containing W such that $f^{-1}(H)\subseteq A$.

Proof

Let $H=Y\backslash f(X\backslash A)$. Since $f^{-1}(W)\subseteq A$, we have $f(X\backslash A)\subseteq Y\backslash W$. Since f is Dα-open (resp. Dα-closed), then H is Dα-closed (resp. Dα-open) set and $f^{-1}(H)=X\backslash f^{-1}(f(X\backslash A))\subset X\backslash (X\backslash A)=A$.

Corollary 5.8

If f : X → Y is Dα-open, then $f^{-1}(C{l}_{\alpha }^D(B))\subseteq Cl(f^{-1}(B))$ for each set B ⊂ Y.

Proof

Since $Cl(f^{-1}(B))$ is closed in X containing $f^{-1}(B)$ for a set B ⊆ Y. By Theorem 5.7, there exists Dα-closed set H ⊆ Y, B ⊆ H such that $f^{-1}(H)\subseteq Cl(f^{-1}(B))$. Thus, $f^{-1}(C{l}_{\alpha }^D(B))\subseteq f^{-1}(C{l}_{\alpha }^D(H))\subseteq f^{-1}(H)\subseteq Cl(f^{-1}(B))$.

Theorem 5.9

A function f : X → Y is Dα-open if and only if $f(Int(A))\subseteq In{t}_{\alpha }^D(f(A))$ for every subset A of X.

Proof

Suppose f : X → Y is Dα-open function and A ⊆ X. Then Int(A) is open set in X and $f(Int(A))$ is Dα-open set contained in f(A). Therefore $f(Int(A))\subseteq In{t}_{\alpha }^D(f(A))$. Conversely, let be $f(Int(A))\subseteq In{t}_{\alpha }^D(f(A))$ for every subset A of X and U is open set in X. Then Int(U) = U, $f(U)\subseteq In{t}_{\alpha }^D(f(U))$. Hence $f(U)=In{t}_{\alpha }^D(f(U))$. By Theorem 3.14 f(U) is Dα-open.

Theorem 5.10

For any bijective function $f:(X,\tau )\to (Y,\sigma )$ the following statements are equivalent.

(i)	$f^{-1}$ is Dα-continuous function.

(ii)	f is Dα-open function.

(iii)

f is Dα-closed function.

Proof

(i)⇒(ii) Let U be an open set in X. Then X \ U is closed in X. Since $f^{-1}$ is Dα-continuous, ${(f^{-1})}^{-1}(X\backslash U)$ is Dα-closed in Y. That is $f(X\backslash U)=Y\backslash f(U)$ is Dα-closed in Y. This implies f(U) is Dα-open in Y. Hence f is Dα-open function.

(ii)⇒ (iii) Let F be a closed set in X. Then X \ F is open in X. Since f is Dα-open, f(X \ F) is Dα-open in Y. That is $f(X\backslash F)=Y\backslash f(F)$ is Dα-open in Y. This implies f(F) is Dα-closed in Y. Hence f is Dα-closed function.

(iii)⇒ (i) Let F be closed set in X. Since f is Dα-closed function, f(F) is Dα-closed in Y. That is ${(f^{-1})}^{-1}(F)$ is Dα-closed in Y. Hence $f^{-1}$ is Dα-continuous function.

Remark 5.11

Composition of two Dα-open functions need not be Dα-open as seen from the following example.

Example 5.12

Let X = {x, y, z} associated with the topology $\tau =\{\varphi ,\{x,y\},\{z\},X\}$, Y = {p, q, r} associated with the topology $\sigma =\{\varphi ,\{p\},Y\}$ and Z = {a,b,c} associated with the topology, $\nu =\{\varphi ,\{b\},\{a,b\},Z\}$. Define $f:(X,\tau )\to (Y,\sigma )$ by f(x) = p, f(y) = q, f(z) = r and $g:(Y,\sigma )\to (Z,\nu )$ by g(p) = b, g(q) = a, g(r) = c. One can have that; $F_Y=\{\varphi ,\{q,r\},Y\}$, $GC(Y)=\{\varphi ,\{q\},\{r\},\{p,q\},\{p,r\},\{q,r\},Y\}$, $GO(Y)=\{\varphi ,\{p\},\{q\},\{r\},\{p,q\},\{p,r\},Y\}$, $D\alpha C(Y)=D\alpha O(Y)=P(X)$ and $F_Z=\{\varphi ,\{c\},\{a,c\},Z\}$, $GC(Z)=\{\varphi ,\{c\},\{a,c\},\{b,c\},Z\}$, $GO(Z)=\{\varphi ,\{a\},\{b\},\{a,b\},Z\}$, $D\alpha C(Z)=\{\varphi ,\{a\},\{c\},\{a,c\},\{b,c\},Z\}$, $D\alpha O(Z)=\{\varphi ,\{a\},\{b\},\{a,b\},\{b,c\},Z\}$. Clearly, f and g are Dα-open function. {z} is open in X. But $gof(\{z\})=g(f(\left\{z\right\}))=g(\{r\})=\left\{c\right\}$ which is not Dα-open in Z. Therefore gof is not Dα-open function.

6 Dα-closed graph and strongly Dα-closed

In this section we introduce Dα-closed graph and strongly Dα-closed and investigate some of their basic properties.

Definition 6.1

A function f : X → Y has Dα-closed graph if for each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $(U\times Cl*(V))\cap G(f)=\varphi$.

Remark 6.2

Evidently every closed graph is Dα-closed. That the converse is not true is seen from the following example.

Example 6.3

Let X = {a, b, c} associated with the topology $\tau =\{\varphi ,\{a,b\},X\}$ and Y = {x, y, z} associated with the topology $\sigma =\{\varphi ,\{x\},\{x,y\},Y\}$. Let $f:(X,\tau )\to (Y,\sigma )$ be a function defined by f(a) = f(c) = x, f(b) = y. One can have that $F_X=\{\varphi ,\{c\},X\}$, $GC(X)=\{\varphi ,\{c\},\{a,c\},\{b,c\},X\}$, $GO(X)=\{\varphi ,\{a\},\{b\},\{a,b\},X\}$, $D\alpha O(X)=\{\varphi ,\{a\},\{b\},\{a,b\},\{a,c\},\{b,c\},X\}$ and $F_Y=\{\varphi ,\{z\},\{y,z\},Y\}$, $GC(Y)=\{\varphi ,\{z\},\{x,z\},\{y,z\},Y\}$, $GO(Y)=\{\varphi ,\{x\},\{y\},\{x,y\},Y\}$. Since $\{a,c\}\in D\alpha O(X,c)$ and $\left\{y\right\}\in GO(Y,y)$ but $\{a,c\}\notin O(X)$ and $\left\{y\right\}\notin O(Y)$. Therefore G(f) is Dα-closed but not closed.

Theorem 6.4

Let $f:(X,\tau )\to (Y,\sigma )$ be a function and

(i)	f is Dα-closed graph;

(ii)	For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $f(U)\cap Cl*(V)=\varphi$.

(iii)

For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in D\alpha O(Y,y)$ such that $(U\times C{l}_{\alpha }^D(V))\cap G(f)=\varphi$.

(iv)	For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in D\alpha O(Y,y)$ such that $f(U)\cap C{l}_{\alpha }^D(V)=\varphi$. Then (1) (i)⇔ (ii) (2) (i)⇒ (iii) (3) (ii)⇒ (iv) (4) (i)⇒ (iv)

Proof

(i)⇒(ii) Suppose f is Dα-closed graph. Then for each $(x,y)\in (X\times Y)\backslash G(f)$, there exists $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $(U\times Cl*(V))\cap G(f)=\varphi$. This implies that for each $f(x)\in f(U)$ and y ∈ Cl*(V). Since y ≠ f(x), $f(U)\cap Cl*(V)=\varphi$.

(ii)⇒(i) Let $(x,y)\in (X\times Y)\backslash G(f)$. Then there exists $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $f(U)\cap Cl=*(V)=\varphi$. This implies that f(x) ≠ y for each x ∈ U and y ∈ Cl*(V). Therefore $(U\times Cl*(V))\cap G(f)=\varphi$.

(i)⇒(iii) Suppose f is Dα-closed graph. Then for each $(x,y)\in (X\times Y)\backslash G(f)$, there exists $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $(U\times Cl*(V))\cap G(f)=\varphi$. Since g-open set is Dα-open, $C{l}_{\alpha }^D(V)\subseteq Cl*(V)$. Therefore $(U\times C{l}_{\alpha }^D(V))\cap G(f)=\varphi$.

(ii)⇒(iv) Let $(x,y)\in (X\times Y)\backslash G(f)$. Then there exists $U\in D\alpha O(X,x)$ and $V\in GO(Y,y)$ such that $f(U)\cap Cl*(V)=\varphi$. Since $C{l}_{\alpha }^D(V)\subseteq Cl*(V)$, $f(U)\cap C{l}_{\alpha }^D(V)\subseteq f(U)\cap Cl*(V)=\varphi$.

(i)⇒(iv) From (ii).

Definition 6.5

A topological space (X,τ) is said to be Dα-T₁ if for any distinct pair of points x and y in X, there exist Dα-open U in X containing x but not y and an Dα-open V in X containing y but not x.

Theorem 6.6

(i)	Every α-T1 space is Dα-T1.

(ii)	Every g-T1 space is Dα-T1.

Proof

It is obvious from Theorem 3.6.

Remark 6.7

The converse of the above theorem is not true as seen from Example 2.7.

Theorem 6.8

Let f : X → Y be any surjection with G(f) Dα-closed. Then Y is g-T₁.

Proof

Let $y_1\text{},y_2(y_1\ne y_2)\in Y$. The subjectivity of f gives the existence of an element x_o ∈ X such that f(x_o) = y₂. Now $(x_o\text{},y_1)\in (X\times Y)\backslash G(f)$. The Dα-closeness of G(f) provides $U_1\in D\alpha O(X,x_o)$, $V_1\in GO(Y,y_1)$ such that $f(U_1)\cap Cl*(V_1)=\varphi$. Now $x_o\in U_1\Rightarrow f(x_o)=y_2\in f(U_1)$. This and the fact that $f(U_1)\cap Cl*(V_1)=\varphi$ guarantee that $y_2\notin V_1$. Again from the subjectivity of f gives a x₁ ∈ X such that f(x₁) = y₁. Now $(x_1\text{},y_2)\in (X\times Y)\backslash G(f)$ and the Dα-closedness of G(f) provides $U_2\in D\alpha O(X,x_1)$, $V_2\in GO(Y,y_2)$ such that $f(U_2)\cap Cl*(V_2)=\varphi$. Now $x_1\in U_2\Rightarrow f(x_1)=y_1\in f(U_2)$ so that $y_1\notin V_2$. Thus we obtain sets $V_1\text{},V_2\in GO(Y)$ such that y₁ ∈ V₁ but $y_2\notin V_1$ while y₂ ∈ V₂ but $y_1\notin V_2$. Hence Y is g-T₁.

Corollary 6.9

Let f : X → Y be any surjection with G(f) Dα-closed. Then Y is Dα-T₁.

Proof

Follows From Theorems 6.6 (i) and 6.8.

Theorem 6.10

Let f : X → Y be any injective with G(f) Dα-closed. Then X is Dα-T₁.

Proof

Let $x_1\text{},x_2(x_1\ne x_2)\in X$. The injectivity of f implies $f(x_1)\ne f(x_2)$ whence one obtains that $(x_1\text{},f(x_2))\in (X\times Y)\backslash G(f)$. The Dα-closedness of G(f) provides $U_1\in D\alpha O(X,x_1)$, $V_1\in GO(Y,f(x_2))$ such that $f(U_1)\cap Cl*(V_1)=\varphi$. Therefore $f(x_2)\notin f(U_1)$ and a fortiori $x_2\notin U_1$. Again $(x_2\text{},f(x_1))\in (X\times Y)\backslash G(f)$ and Dα-closedness of G(f) as before gives $U_2\in D\alpha O(X,x_2)$, $V_2\in GO(Y,f(x_1))$ with $f(U_2)\cap Cl*(V_2)=\varphi$, which guarantees that $f(x_1)\notin f(U_2)$ and so $x_1\notin U_2$. Therefore, we obtain sets U₁ and $U_2\in D\alpha O(X)$ such that x₁ ∈ U₁ but $x_2\notin U_1$ while x₂ ∈ U₂ but $x_1\notin U_2$. Hence X is Dα-T₁.

Corollary 6.11

Let f : X → Y be any bijection with G(f) Dα-closed. Then both X and Y are Dα-T₁.

Proof

It readily follows from Corollary 6.9 and Theorem 6.10.

Definition 6.12

A topological space (X, τ) is said to be Dα-T₂ if for any distinct pair of points x and y in X, there exist Dα-open sets U and V in X containing x and y, respectively, such that U ∩ V = ϕ.

Theorem 6.13

(i)	Every α-T2 space is Dα-T2.

(ii)	Every g-T2 space is Dα-T2.

Proof

Obvious.

Remark 6.14

The converse of the above theorem is not true as seen from Example 2.7.

Theorem 6.15

Let f : X → Y be any surjection with G(f) Dα-closed. Then Y is g-T₂.

Proof

Let $y_1\text{},y_2(y_1\ne y_2)\in Y$. The subjectivity of f gives a x₁ ∈ X such that f(x₁) = y₁. Now $(x_1\text{},y_2)\in (X\times Y)\backslash G(f)$. The Dα-closedness of G(f) provides $U\in D\alpha O(X,x_1)$, $V\in GO(Y,y_2)$ such that $f(U)\cap Cl*(V)=\varphi$. Now $x_1\in U\Rightarrow f(x_1)=y_1\in f(U)$. This and the fact that $f(U)\cap Cl*(V)=\varphi$ guarantee that $y_1\notin Cl*(V)$. This mean that there exists $W\in GO(Y,y_1)$ such that W ∩ V = ϕ. Hence Y is g-T₂.

Corollary 6.16

Let f : X → Y be any surjection with G(f) Dα-closed. Then Y is Dα-T₂.

Proof

Follows from Theorems 6.13 (ii) and 6.15.

Definition 6.17

A function f : X → Y has a strongly Dα-closed graph if for each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and V ∈ O(Y, y) such that $(U\times Cl(V))\cap G(f)=\varphi$.

Corollary 6.18

A strongly Dα-closed graph is Dα-closed. That the converse is not true is seen from Example 6.3, where $\left\{y\right\}\in GO(Y,y)$ but $\left\{y\right\}\notin O(Y)$. Therefore G(f) is Dα-closed but not strongly Dα-closed.

Remark 6.19

Evidently every strongly α-closed graph (resp. strongly closed graph) is strongly Dα-closed graph. That the converse is not true is seen from the following example.

Example 6.20

Let X = {a, b, c} associated with the topology $\tau =\{\varphi ,\{a,b\},X\}$ and Y = {x, y, z} associated with the topology $\sigma =\{\varphi ,\{x,y\},\{z\},Y\}$. Let $f:(X,\tau )\to (Y,\sigma )$ be a function defined by f(a) = f(c) = x, f(b) = y. One can have that $F_X=\{\varphi ,\{c\},X\}$, $GC\left(X\right)=\{\varphi ,\{c\},\{a,c\},\{b,c\},X\}$, $GO\left(X\right)=\{\varphi ,\{a\},\{b\},\{a,b\},X\}$, $\alpha O\left(X\right)=\{\varphi ,\{a,b\},X\}$, $D\alpha O\left(X\right)=\{\varphi ,\{a\},\{b\},\{a,b\},\{a,c\},\{b,c\},X\}$. Since $\{a,c\}\in D\alpha O(X,c)$ and $\left\{z\right\}\in O(Y,z)$ but $\{a,c\}\notin \alpha O(X)$ (resp. $\{a,c\}\notin O(X)$). Therefore G(f) is strongly Dα-closed but not strongly α-closed (resp. strongly closed).

Theorem 6.21

For a function $f:(X,\tau )\to (Y,\sigma )$, the following properties are equivalent:

(i)	f has strongly Dα-closed graph.

(ii)	For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and V ∈ O(Y, y) such that $f(U)\cap Cl(V)=\varphi$.

(iii)

For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in \alpha O(Y,y)$ such that $(U\times C{l}_{\alpha }(V))\cap G(f)=\varphi$.

(iv)	For each $(x,y)\in (X\times Y)\backslash G(f)$, there exist $U\in D\alpha O(X,x)$ and $V\in \alpha O(Y,y)$ such that $f(U)\cap C{l}_{\alpha }(V)=\varphi$.

Proof

Similar to the proof of Theorem 6.4.

Theorem 6.22

If f : X → Y is a function with a strongly Dα-closed graph, then for each x ∈ X, $f(x)=\cap \{C{l}_{\alpha }(f(U)):U\in D\alpha O(X,x)\}$.

Proof

Suppose the theorem is false. Then there exists a y ≠ f(x) such that $y\in \cap \{C{l}_{\alpha }(f(U)):U\in D\alpha O(X,x)\}$. This implies that $y\in C{l}_{\alpha }(f(U))$ for every $U\in D\alpha O(X,x)$. So V ∩ f(U) ≠ ϕ for every $V\in \alpha O(Y,y)$. This, in its turn, indicates that $C{l}_{\alpha }(V)\cap f(U)\supset V\supset f(U)\ne \varphi$, which contradicts the hypothesis that f is a function with Dα-closed graph. Hence the theorem holds.

Theorem 6.23

If f : X → Y is Dα-continuous function and Y is T₂. Then G(f) is strongly Dα-closed.

Proof

Let $(x,y)\in (X\times Y)\backslash G(f)$. Since Y is T₂, there exists a set V ∈ O(Y, y) such that $f(x)\notin Cl(V)$. But Cl(V) is closed. Now $Y\backslash Cl(V)\in O(Y,f(x))$. By Theorem 4.5 there exists $U\in D\alpha O(X,x)$ such that $f(U)\subseteq Y\backslash Cl(V)$. Consequently, $f(U)\cap Cl(V)=\varphi$ and therefore G(f) is strongly Dα-closed.

Theorem 6.24

Let f : X → Y be any surjection with G(f) strongly Dα-closed. Then Y is T₁ and α-T₁.

Proof

Similar to the proof of Theorem 6.8 and T₁-ness always guarantees α-T₁-ness. Hence Y is α-T₁.

Corollary 6.25

Let f : X → Y be any surjection with G(f) strongly Dα-closed. Then Y is Dα-T₁.

Proof

Follows From Theorems 6.6 (i) and 6.24.

Theorem 6.26

Let f : X → Y be any injective with G(f) strongly Dα-closed. Then X is Dα-T₁.

Proof

Similar to the proof of Theorem 6.10.

Corollary 6.27

Let f : X → Y be any bijection with G(f) strongly Dα-closed. Then both X and Y are Dα-T₁.

Proof

It readily follows from Corollary 6.25 and Theorem 6.26.

Theorem 6.28

Let f : X → Y be any surjection with G(f) strongly Dα-closed. Then Y is T₂ and α-T₂.

Proof

Similar to the proof Theorem 6.15 and T₂-ness always guarantees α-T₂-ness. Hence Y is α-T₂.

Corollary 6.29

Let f : X → Y be any surjection with G(f) strongly Dα-closed. Then Y is Dα-T₂.

Proof

Follows From Theorems 6.13 (i) and 6.28.