Many concepts of generalized open sets were introduced by mathematicians during the last decades, which based on the open sets in topological space. Such as the concept of semi-open sets was introduced by Levine in 1963 (see [1]). The concepts of $\alpha$-open (in the sense $\gamma$-open) sets and regular open sets were introduced by Njastad and Császár, respectively (see [2-3]). Umehara et al. (see [4]) defined the notion of $(\gamma, \gamma')$-open sets, furthermore, they introduced some new separation axioms and investigated their relationships in 1992. In fact, the theory of generalized open sets became an important topic in general topology and obtained a number of good results (see [1-16]). In this paper, we introduce and study the concept of $(\alpha, \beta)$-$\gamma$-open sets and investigate $(\alpha, \beta)$-$\gamma$-$T_{i}$ separation axioms.

Throughout this paper, $X$ always denotes a non-empty set, $(X, \mathcal{T})$ denotes topological space. Let us denote the collection of all open sets, closed sets in $(X, \mathcal{T})$ by $\mathcal{T}$ and $\mathcal{F}$, respectively. The closure of $A$ and interior of $A$ are denoted by cl$(A)$ and int$(A)$, respectively. The terminology and symbol without definition in this paper may be found in [4-6].

Definition 1.1 [3]  Let $(X, \mathcal{T})$ be a topological space, and $P(X)$ denote the collection of the power set of $X$. Then we say a mapping $\alpha$ is an operator associated to $\mathcal{T}$ on $X$ (written in short as an operator on $\mathcal{T}$), if $\alpha:$$P(X)\rightarrow P(X)$ satisfies $V\subset\alpha(V)$ for each $V\in\mathcal{T}$. where $\alpha(V)$ denotes the value of $\alpha$ at $V$.

Definition 1.2 [3]  Let $(X, \mathcal{T})$ be a topological space and $\alpha$ be an operator on $\mathcal{T}$, A non-empty subset $A$ is said to be $\alpha$-open if for each $x\in A$, there exists an open set $U$ such that $x\in U$ and $\alpha(U)\subset A$. We suppose that the empty set is $\alpha$-open and denote the set of all $\alpha$-open sets by $\mathcal{T}_\alpha$. The complement of $\alpha$-open set is called $\alpha$-closed. We denote $cl_\alpha(A)=\{{x\in X|\alpha(U) \cap A\ne \emptyset}$ for each open set $U$ containing $x$ $\}$.

Definition 1.3 [4]  Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta$ be operators on $\mathcal{T}$, A non-empty subset is said to be $(\alpha, \beta)$-open if for each $x\in A$, there exists a $\beta$-open set $U$ such that $x\in U$ and $\alpha(U)\subset A$. We suppose that the empty set is $(\alpha, \beta)$-open and denote the set of all $(\alpha, \beta)$-open sets by $\mathcal{T}_{(\alpha, \beta)}$. The complement of $(\alpha, \beta)$-open set is called $(\alpha, \beta)$-closed.

Definition 1.4  Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, A non-empty subset is said to be $(\alpha, \beta)$-$\gamma$-open if for each $x\in A$, there exist $\gamma$-open sets $U$ and $V$ such that $x\in U$, $x\in V$ and $\alpha(U)\cup\beta(V) \subset A$. We suppose that the empty set is $(\alpha, \beta)$-$\gamma$-open and denote the set of all $(\alpha, \beta)$-$\gamma$-open sets by $\mathcal{T}_{(\alpha, \beta)-\gamma}$. The complement of $(\alpha, \beta)$-$\gamma$-open set is called $(\alpha, \beta)$-$\gamma$-closed. We denote $cl_{(\alpha, \beta)-\gamma}(A)=\{{x\in X|(\alpha(U)\cup\beta(V))\cap A\ne \emptyset}$ for any $\gamma$-open set $U$ and $V$ containing $x$ $\}$.

Furthermore, we denote the intersection of all $(\alpha, \beta)$-$\gamma$-closed sets containing $A$ by $CL_{(\alpha, \beta)-\gamma}(A)$. Namely, we defined as follows $CL_{(\alpha, \beta)-\gamma}(A)=\cap\{{F|F}$ is an $(\alpha, \beta)$-$\gamma$-closed set in $(X, \mathcal{T})$ such that $A\subset F$ $\}$. It is obvious that $cl_{(\alpha, \beta)-\gamma}(A)$$\subset$ $CL_{(\alpha, \beta)-\gamma}(A)$ for any subset $A$ of $(X, \mathcal{T})$.

From the above Definitions 1.3 and 1.4, we know that a subset $A$ is $(\alpha, \beta)$-open in $(X, \mathcal{T})$, then it is $(\alpha, \beta)$-$\gamma$-open, but the converse need not be true.

Example 1.1  Let $X=\{{a, b, c}\}$ and $\mathcal{T}=\{\varnothing, \{{a}\}, \{{b}\}, \{{a, b}\}, X\}$. Consider the operators $\alpha, \beta, \gamma$ defined as follow: $\alpha:$$P(X)\rightarrow P(X)$, where $\alpha={\rm id}$. that is ${\rm id}(A)=A$ for every subset $A$ of $(X, \mathcal{T})$; $\beta:$$P(X)\rightarrow P(X)$, where $\beta(A)=A$, if $b\in A$ and $\beta(A)=cl(A)$, if $b\not\in A$; $\gamma:$$P(X)\rightarrow P(X)$, where $\gamma(A)=cl(A)$, if $b\not\in A$ and $\gamma(A)=A$, if $b\in A$. We can see $\{{b, c}\}$ is $(\alpha, \beta)$-open but not $(\alpha, \beta)$-$\gamma$-open.

Furthermore, using Definitions 1.3 and 1.4, we have $A$ is an $(\alpha, \beta)$-$\gamma$-open set if and only if $A$ is $(\alpha, \gamma)$-open and $(\beta, \gamma)$-open, the proof follows.

Proof   Sufficiency. Suppose $A$ is an $(\alpha, \beta)$-$\gamma$-open set. Let $x\in A$, then there exist $\gamma$-open sets $U$ and $V$ such that $x\in U$, $x\in V$ and $\alpha(U)\cup\beta(V) \subset A$. Thus implies $\alpha(U)\subset A$ and $\beta(V)\subset A$. Hence $A$ is $(\alpha, \gamma)$-open and $(\beta, \gamma)$-open.

Necessity. Let $x\in A$, since $A$ is $(\alpha, \gamma)$-open, then there exists a $\gamma$-open set $U$ such that $x\in U$, $\alpha(U)\subset A$. Similarly, $A$ is $(\beta, \gamma)$-open, then there exists a $\gamma$-open set $V$ such that $x\in V$, $\beta(V)\subset A$, then $\alpha(U)\cup\beta(V)\subset A$. Hence $A$ is an $(\alpha, \beta)$-$\gamma$-open set.

Definition 1.5 [6]  Let $(X, \mathcal{T})$ be a topological space and $\alpha$ be an operator. $\alpha$ is said to be regular if for each $x\in X$ and any open sets $U$ and $V$ containing $x$, there exists an open set $W$ containing $x$ such that $\alpha(W)\subset\alpha(U)\cap \alpha(V)$.

Definition 1.6   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \gamma$ be operators on $\mathcal{T}$, $(\alpha, \gamma)$ is said to be $(\alpha, \gamma)$ -regular if for each $x\in X$ and any $\gamma$-open set $U$ and $V$ containing $x$, there exists a $\gamma$-open set $W$ containing $x$ such that $\alpha(W)\subset\alpha(U)\cap \alpha(V)$.

Definition 1.7   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \gamma$ be operators on $\mathcal{T}$, $(X, \mathcal{T})$ is said to be $(\alpha, \gamma)$-regular if for each $x\in X$ and every $\gamma$-open set $U$ containing $x$, there exist $\gamma$-open sets $W$ and $S$ such that $\alpha(W)\cup\alpha(S)\subset U$.

Definition 1.8   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, $(X, \mathcal{T})$ is said to be $(\alpha, \beta)$-$\gamma$-regular if for each $x\in X$ and every $\gamma$-open set $U$ containing $x$, there exist $\gamma$-open sets $W$ and $S$ such that $\alpha(W)\cup\beta(S)\subset U$.

Definition 1.9   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, $(\alpha, \beta)$-$\gamma:$$P(X)\rightarrow P(X)$ is said to be $(\alpha, \gamma)$-pre open if for each $x\in X$ and every $\gamma$-open set $U$ containing $x$, there exists an $(\alpha, \beta)$-$\gamma$-open set $W$ such that $x\in W$ and $W\subset\alpha(U)$.

Remark 1.1   By the above definition we can obtain the following statements:

Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$ and id: $P(X)\rightarrow P(X)$ be an identity operation, that is id$(A)=A$ for every subset $A$ of $(X, \mathcal{T})$.

(1) If $\alpha=\beta=\gamma=$id, we have that $A$ is an $(\alpha, \beta)$-$\gamma$-open set if and only if (written in short as iff) $A$ is an open set.

(2) If $\beta=\gamma=$id, we have that $A$ is an $(\alpha, \beta)$-$\gamma$-open set iff $A$ is a $\gamma$-open set.

(3) If $\gamma=$id, we have that $A$ is an $(\alpha, \beta)$-$\gamma$-open set iff $A$ is an $(\alpha, \beta)$-open set.

Theorem 1.1   Let be $(X, \mathcal{T})$ a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, then the following statements hold:

(1) Any union of $\alpha$-open set is $\alpha$-open.

(2) Any union of $(\alpha, \beta)$-open set is $(\alpha, \beta)$ -open.

(3) Any union of $(\alpha, \beta)$-$\gamma$-open set is $(\alpha, \beta)$-$\gamma$-open.

Proof   Statement (1) due to [3].

(2) Let $x\in\bigcup_{i\in I}A_{i}$. Then $x\in A_{i}$ for some $i$. Since each $A_{i}$ is $(\alpha, \beta)$-open, there exists a $\beta$-open set $U_{i}$ such that $x\in U_{i}$ and $\alpha(U_{i})\subset A_{i}\subset\bigcup_{i\in I}A_{i}$. Hence $\bigcup_{i\in I}A_{i}$ is an $(\alpha, \beta)$-open set.

(3) Let $x\in\bigcup_{i\in I}A_{i}$. Then $x\in A_{i}$ for some $i$. Since each $A_{i}$ is $(\alpha, \beta)$-$\gamma$-open, there exist $\gamma$-open sets $U_{i}$ and $V_{i}$ such that $x\in U_{i}, x\in V_{i}$, $\alpha(U_{i})\cup\beta(V_{i})\subset A_{i}$, this implies that $\alpha(U_{i})\cup\beta(V_{i})\subset\bigcup_{i\in I}A_{i}$. Therefore $\bigcup_{i\in I}A_{i}$ is an $(\alpha, \beta)$-$\gamma$-open set.

Theorem 1.2   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, then the following statements hold:

(1) $(X, \mathcal{T})$ is $(\alpha, \beta)$-$\gamma$-regular iff $\mathcal{T}_{(\alpha, \beta)-\gamma}=\mathcal{T}_{\gamma}$.

(2) $(X, \mathcal{T})$ is $(\alpha, \beta)$-$\gamma$-regular iff $(X, \mathcal{T})$ is $(\alpha, \gamma)$-regular and $(\beta, \gamma)$-regular.

Proof   (1) Sufficiency. We have $\mathcal{T}_{(\alpha, \beta)-\gamma}\subset\mathcal{T}_{\gamma}$ by Definition 1.5. Now we proof that $\mathcal{T}_{(\alpha, \beta)-\gamma}\supset\mathcal{T}_{\gamma}$. Let $A\in\mathcal{T}_{\gamma}$ and $x\in A$, since $(X, \mathcal{T})$ is $(\alpha, \beta)$-$\gamma$-regular, then for $\gamma$-open set $A$ containing $x$, there exist $\gamma$-open sets $W$ and $S$ such that $\alpha(W)\cup\beta(S)\subset A$. Hence $A$ is $(\alpha, \beta)$-$\gamma$-open, that is $A\in\mathcal{T}_{(\alpha, \beta)-\gamma}$, then $\mathcal{T}_{(\alpha, \beta)-\gamma}\supset\mathcal{T}_{\gamma}$. Thus $\mathcal{T}_{(\alpha, \beta)-\gamma}=\mathcal{T}_{\gamma}$.

Necessity. Let $x\in X$ and a subset $A$ containing $x$, where $A$ is a $\gamma$-open set. Suppose $\mathcal{T}_{(\alpha, \beta)-\gamma}=\mathcal{T}_{\gamma}$, then $A$ is $(\alpha, \beta)$-$\gamma$-open, thus there exist $\gamma$-open sets $U$ and $V$ such that $x\in U, x\in V$, $\alpha(U)\cup\beta(V)\subset A$. This implies $(X, \mathcal{T})$ is $(\alpha, \beta)$-$\gamma$-regular.

(2) By using Theorem 1.2 (1) and Theorem 1.1 (3).

Proposition 1.1   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, If $A$ and $B$ are $(\alpha, \beta)$-$\gamma$-open sets, $(\alpha, \gamma)$ and $(\beta, \gamma)$ are regular operation. Then $A\cap B$ is $(\alpha, \beta)$-$\gamma$-open.

Proof   Suppose $A$ and $B$ are $(\alpha, \beta)$-$\gamma$-open sets. Let $x\in A\cap B$. By Definition 1.4, there exist $\gamma$-open sets $H$ and $K$ such that $x\in H, x\in K$, $\alpha(H)\cup\beta(K)\subset A$. In the same way, there exist $\gamma$-open sets $W$ and $S$ such that $x\in W, x\in S$, $\alpha(W)\cup\beta(S)\subset B$. Furthermore, $(\alpha(H)\cup\beta(K))\cap(\alpha(W)\cup\beta(S))\subset A\cap B$, that is$(\alpha(H)\cap(\alpha(W))\cup(\beta(K)\cap\beta(S))\subset A\cap B$. Since $(\alpha, \gamma)$ and $(\beta, \gamma)$ be regular operation. There exist $\gamma$-open sets $U$ and $V$ such that $\alpha(U)\subset\alpha(H)\cap(\alpha(W)$ and $\beta(V)\subset\beta(K)\cap\beta(S)$, respectively. Hence $\alpha(U)\cup\beta(V)\subset A\cap B$, this implies $A\cap B$ is $(\alpha, \beta)$-$\gamma$-open.

Remark 1.2   If $(\alpha, \gamma)$ and $(\beta, \gamma)$ are regular operation, it is obvious that $\mathcal{T}_{(\alpha, \beta)-\gamma}$ can form a topology on $X$ by the above Theorem 1.1 (3) and Proposition 1.1.

Theorem 1.3   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$. Then the following statements hold:

(1) $x\in cl_{(\alpha, \beta)-\gamma}(A)$ iff $V\cap A\ne\emptyset$ for every $(\alpha, \beta)$-$\gamma$-open $V$ containing $x$.

(2) $A\subset cl_{(\alpha, \beta)-\gamma}(A)$ for every subset $A$ of $(X, \mathcal{T})$.

(3) $cl_{(\alpha, \beta)-\gamma}(A)\subset cl_{(\alpha, \beta)-\gamma}(B)$ if $A\subset B$.

(4) $A$ is $(\alpha, \beta)-\gamma$-closed iff $cl_{(\alpha, \beta)-\gamma}(A)=A$.

(5)  $cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))=cl_{(\alpha, \beta)-\gamma}(A)$, that is $cl_{(\alpha, \beta)-\gamma}(A)$ is $(\alpha, \beta)$-$\gamma$-closed.

(6) $cl_{(\alpha, \beta)-\gamma}(A)\cap G\subset cl_{(\alpha, \beta)-\gamma}(A\cap G)$ if $(\alpha, \gamma)$ and $(\beta, \gamma)$ are regular operation for every subset $A$ of $(X, \mathcal{T})$ and every $(\alpha, \beta)$-$\gamma$-open set $G$.

(7) $cl_{(\alpha, \beta)-\gamma}(A\cup B)=cl_{(\alpha, \beta)-\gamma}(A) \cup cl_{(\alpha, \beta)-\gamma}(B)$ if $(\alpha, \gamma)$ and $(\beta, \gamma)$ are regular operation for subsets $A, B\subset X$.

Proof   Proof of (1)-(3) is obvious from Definition 1.4.

(4) Sufficiency. First we claim that $cl_{(\alpha, \beta)-\gamma}(A)\subset A$. Let $x\not\in A$, by assumption $A$ is $(\alpha, \beta)$-$\gamma$-closed, then $x\in X-A$ and $X-A$ is $(\alpha, \beta)$-$\gamma$-open. Hence there exist $\gamma$-open sets $U$ and $V$ such that $x\in U, x\in V$, $\alpha(U)\cup\beta(V)\subset X-A$. This implies $(\alpha(U)\cup\beta(V))\cap A =\emptyset$. Hence, by definition 1.4 we have that $x\not\in cl_{(\alpha, \beta)-\gamma}(A)$. Thus $cl_{(\alpha, \beta)-\gamma}(A)\subset A$. From Theorem 1.3(2), therefore $cl_{(\alpha, \beta)-\gamma}(A)=A$.

Necessity. Let $x\in X-A$, that is $x\not\in A$. By the assumption that $cl_{(\alpha, \beta)-\gamma}(A)=A$, then $x\not\in cl_{(\alpha, \beta)-\gamma}(A)$. Hence there exist $\gamma$-open sets $U$ and $V$ containing $x$ such that $(\alpha(U)\cup\beta(V))\cap A =\emptyset$, i.e., $\alpha(U)\cup\beta(V)\subset X-A$. Namely $X-A$ is $(\alpha, \beta)$-$\gamma$-open. Therefore $A$ is $(\alpha, \beta)$-$\gamma$-closed.

(5) It is enough to proof $cl_{(\alpha, \beta)-\gamma}(A)\subset cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))$ by Theorem 1.3 (2) and Theorem 1.3 (3). Then we proof that $cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))\subset cl_{(\alpha, \beta)-\gamma}(A)$. Suppose $x\in cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))$, then for every $(\alpha, \beta)$-$\gamma$-open set $V$ containing $x$ such that $V\cap cl_{(\alpha, \beta)-\gamma}(A)\ne\emptyset$. This implies $V\cap A\ne\emptyset$. Hence $x\in cl_{(\alpha, \beta)-\gamma}(A)$, this shows that $cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))\subset cl_{(\alpha, \beta)-\gamma}(A)$. Hence $cl_{(\alpha, \beta)-\gamma}(cl_{(\alpha, \beta)-\gamma}(A))=cl_{(\alpha, \beta)-\gamma}(A)$.

(6) Let $x\in cl_{(\alpha, \beta)-\gamma}(A)\cap G$, then $x\in cl_{(\alpha, \beta)-\gamma}(A)$ and $x\in G$. Let $V$ be an $(\alpha, \beta)$-$\gamma$-open set containing $x$. Then by Proposition 1.1, $V\cap G$ is also an $(\alpha, \beta)$-$\gamma$-open set containing $x$. Since $x\in cl_{(\alpha, \beta)-\gamma}(A)$, then$(V\cap G)\cap A\ne\emptyset$. This implies that $V\cap( G\cap A)\ne\emptyset$ for every $(\alpha, \beta)$-$\gamma$-open set $V$ containing $x$. By Theorem 1.3 (1), we have $x\in cl_{(\alpha, \beta)-\gamma}(A\cap G)$. Hence $cl_{(\alpha, \beta)-\gamma}(A)\cap G\subset cl_{(\alpha, \beta)-\gamma}(A\cap G)$

(7) It is obvious that $cl_{(\alpha, \beta)-\gamma}(A) \cup cl_{(\alpha, \beta)-\gamma} B)\subset cl_{(\alpha, \beta)-\gamma}(A\cup(B)$ by Theorem 1.3 (3). Let $x\not\in cl_{(\alpha, \beta)-\gamma}(A) \cup cl_{(\alpha, \beta)-\gamma}(B)$, this implies that $x\not\in cl_{(\alpha, \beta)-\gamma}(A) $ and $x\not\in cl_{(\alpha, \beta)-\gamma}(B)$. Then there exist $\gamma$-open sets $U$ and $V$ containing $x$ such that $(\alpha(U)\cup\beta(V))\cap A =\emptyset$ and there exist $\gamma$-open sets $W$ and $S$ containing $x$ such that $(\alpha(W)\cup\beta(S))\cap B =\emptyset$. Thus, we have $((\alpha(U)\cap\alpha(W))\cup(\beta(V)\cap\beta(S)))\cap (A\cup B) =\emptyset$. Since $(\alpha, \gamma)$ and $(\beta, \gamma)$ are regular operation, then there exists a $\gamma$-open set $E$ containing $x$ such that $ \alpha(E)\subset \alpha(U)\cap\alpha(W)$ and there exists $\gamma$-open set $F$ containing $x$ such that $\beta(F)\subset \beta(V)\cap\beta(S)$, respectively. Hence $((\alpha(E)\cup\beta(F))\cap (A\cup B) =\varnothing$. That is $x\not\in cl_{(\alpha, \beta)-\gamma}(A\cup B)$, then $ cl_{(\alpha, \beta)-\gamma}(A\cup(B)\subset cl_{(\alpha, \beta)-\gamma}(A) \cup cl_{(\alpha, \beta)-\gamma} B)$. Hence $cl_{(\alpha, \beta)-\gamma}(A) \cup cl_{(\alpha, \beta)-\gamma}(B)=cl_{(\alpha, \beta)-\gamma}(A\cup B)$.

Remark 1.3   From Definition 1.4, we have the following statements:

(1) $A\subset CL_{(\alpha, \beta)-\gamma}(A)$ for every subset $A$ of $(X, \mathcal{T})$.

(2) $CL_{(\alpha, \beta)-\gamma}(A)\subset CL_{(\alpha, \beta)-\gamma}(B)$ if $A\subset B$.

(3) $A$ is $(\alpha, \beta)$-$\gamma$-closed iff $CL_{(\alpha, \beta)-\gamma}(A)=A$.

(4)  $CL_{(\alpha, \beta)-\gamma}(CL_{(\alpha, \beta)-\gamma}(A))=CL_{(\alpha, \beta)-\gamma}(A)$, that is $CL_{(\alpha, \beta)-\gamma}(A)$ is $(\alpha, \beta)$-$\gamma$-closed.

Definition 2.1   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$.

(1) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space if for every pair of distinct points $x, y\in X$, there exist two $\gamma$ -open sets $P_1$ and $P_2$ such that $x\in P_1$, $y\not\in \alpha (P_1)$ or $y\in P_2$, $x\not\in\beta(P_2)$;

(2) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T_{1/2}$ space if every singleton set $\{{x}\}$ is $(\alpha, \beta)$-$\gamma$-open or $(\alpha, \beta)$-$\gamma$-closed where $x\in X$;

(3) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space if for every pair of distinct points $x, y\in X$, there exist two $\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $x\not\in\alpha (P_2)$ and $y\not\in\beta(P_1)$;

(4) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T_{2}$ space if for every pair of distinct points $x, y\in X$, there exist two distinct $\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $\alpha (P_1)\cap\beta(P_2)=\emptyset$;

(5) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T_{5/2}$ space if for every pair of distinct points $x, y\in X$, there exist two distinct $\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $cl_{(\alpha, \beta)-\gamma}\alpha(P_1) \cap cl_{(\alpha, \beta)-\gamma}\beta(P_2)=\emptyset$.

Remark 2.1   From the above definition the following statements hold:

(1) If a space $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space, then it is an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space;

(2) If a space $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{5/2}$ space, then it is an $(\alpha, \beta)$-$\gamma$-$T_{2}$ space.

Theorem 2.1   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$. Suppose that $(\alpha, \beta)$-$\gamma: $$P(X)\rightarrow P(X)$ is $(\alpha, \gamma)$-preopen and $(\beta, \gamma)$-preopen. Then $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space iff for every pair of distinct points $x, y$, we have $cl_{(\alpha, \beta)-\gamma}(\{{x}\}) \ne cl_{(\alpha, \beta)-\gamma}(\{{y}\})$.

Proof   Sufficiency. Suppose that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space, then for every pair of distinct points $x, y$, there exist two $\gamma$ -open sets $P_1$ and $P_2$ such that $x\in P_1$, $y\not\in \alpha (P_1)$ or $y\in P_2$, $x\not\in\beta(P_2)$ by Definition 2.1. If $x\in P_1$, $y\not\in \alpha (P_1)$, it follows from Definition 1.9 that there exists an $(\alpha, \beta)$-$\gamma$-open set $W$ such that $x\in W$ and $W\subset\alpha (P_1)$. Hence $y\in X-\alpha (P_1)\subset X-W$. Because $X-W$ is an $(\alpha, \beta)$-$\gamma$-closed set, we have that $cl_{(\alpha, \beta)-\gamma}(\{{y}\})\subset cl_{(\alpha, \beta)-\gamma}(X-W)$ and so $cl_{(\alpha, \beta)-\gamma}(\{{x}\}) \ne cl_{(\alpha, \beta)-\gamma}(\{{y}\})$. Similarly, If $y\in P_2$, $x\not\in\beta(P_2)$, it follows from Definition 1.9 that there exists an $(\alpha, \beta)$-$\gamma$-open set $S$ such that $y\in S$ and $S\subset\beta(P_2)$. Hence $x\in X-\beta(P_2)\subset X-S$. Because $X-S$ is an $(\alpha, \beta)$-$\gamma$-closed set, we have that $cl_{(\alpha, \beta)-\gamma}(\{{x}\})\subset cl_{(\alpha, \beta)-\gamma}(X-S)$ and so $cl_{(\alpha, \beta)-\gamma}(\{{y}\})\ne cl_{(\alpha, \beta)-\gamma}(\{{x}\})$.

Necessity. Suppose that $x, y\in X$ and $x\ne y$, we have that $cl_{(\alpha, \beta)-\gamma}(\{{x}\})\ne cl_{(\alpha, \beta)-\gamma}(\{{y}\})$. Then there exists a point $z\in cl_{(\alpha, \beta)-\gamma}(\{{x}\})$ but $z\not\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$ or $z\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$ but $z\not\in cl_{(\alpha, \beta)-\gamma}(\{{x}\})$. If $x\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$, then $cl_{(\alpha, \beta)-\gamma}(\{{x}\})\subset cl_{(\alpha, \beta)-\gamma}(\{{y}\})$ by Theorem 1.3 (5). This implies that $z\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$. This contradiction shows that $x\not\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$, i.e., there exist $\gamma$-open sets $U, V$ such that $x\in U, V$ and $(\alpha (U)\cup\beta(V))\cap\{{y}\}=\emptyset$ by Definition 1.4. Hence $x\in U, V$ and $y\not\in\alpha (U), \beta(V)$. This implies that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space.

Theorem 2.2   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$. Then $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space iff every singleton set $\{{x}\}$ is $(\alpha, \beta)$-$\gamma$-closed.

Proof   Sufficiency. Suppose that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space, then for every pair of distinct points $x, y\in X$, there exist two $\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $x\not\in\alpha (P_2)$ and $y\not\in\beta(P_1)$ by Definition 2.1. Similarly then for every pair of distinct points $y, x$, there exist two $\gamma$-open sets $P_3$ and $P_4$ containing $x, y$ respectively such that $y\not\in \alpha (P_3)$ and $x\not\in\beta(P_4)$. Therefore we have $y\in P \alpha (P_2)\cup\beta(P_4)\subset X-\{{x}\}$. This implies $X-\{{x}\}$ is an $(\alpha, \beta)$-$\gamma$-open set, i.e., $\{{x}\}$ is $(\alpha, \beta)$-$\gamma$-closed.

Necessity. Suppose that $x, y\in X$ and $x\ne y$. By the assumption $X-\{{x}\}$ is an $(\alpha, \beta)$-$\gamma$-open set containing $y$ and $X-\{{y}\}$ is an $(\alpha, \beta)$-$\gamma$-open set containing $x$. Then there exist two-open sets $P_1$ and $P_2$ containing $y$ such that $y\in \alpha (P_1)\cup\beta(P_2)\subset X-\{{x}\}$ by Definition 1.4. Similarly there exist two -open sets $P_3$ and $P_4$ containing $x$ such that $x\in \alpha (P_3)\cup\beta(P_4)\subset X-\{{y}\}$. This implies $x\not\in \alpha (P_1)$, $x\not\in\beta(P_2)$ and $y\not\in \alpha (P_3)$, $y\not\in\beta(P_4)$. Therefore there exist two $\gamma$-open sets $P_1$ and $P_4$ or $P_2$ and $P_3$ containing $x, y$ respectively such that $x\not\in \alpha (P_1)$ and $y\not\in\beta(P_4)$ or $x\not\in\beta(P_2)$ and $y\not\in \alpha (P_3)$. Hence $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space.

Theorem 2.3   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, the following relations hold: $(\alpha, \beta)$-$\gamma$-$T_{5/2}\Rightarrow(\alpha, \beta)$-$\gamma$-$T_{2}\Rightarrow(\alpha, \beta)$-$\gamma$-$T_{1}\Rightarrow(\alpha, \beta)$-$\gamma$-$T_{1/2}\Rightarrow(\alpha, \beta)$-$\gamma$-$T_{0}, $ where $ A\Rightarrow B$ represents $A$ implies $B$.

Proof   It is clear by above Theorems 2.1, 2.2 and Remark 2.1.

Definition 2.2   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$.

(1) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T^{*}_{0}$ space if for every pair of distinct points $x, y\in X$, there exists an $(\alpha, \beta)$-$\gamma$-open set $P$ such that $x\in P$, $y\not\in P$ or $y\in P$, $x\not\in P$;

(2) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T^{*}_{1}$ space if for every pair of distinct points $x, y\in X$, there exist two $(\alpha, \beta)$-$\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $x\in P_1, y\not\in P_1$ or $y\in P_2, x\not\in P_2$;

(3) $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T^{*}_{2}$ space if for every pair of distinct points $x, y\in X$, there exist two $(\alpha, \beta)$-$\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $P_1\cap P_2=\emptyset$;

(4)  $(X, \mathcal{T})$ is said to be an $(\alpha, \beta)$-$\gamma$-$T^{*}_{5/2}$ space if for every pair of distinct points $x, y\in X$, there exist two $(\alpha, \beta)$-$\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $CL_{(\alpha, \beta)-\gamma}(P_1)\cap CL_{(\alpha, \beta)-\gamma}(P_2)=\emptyset$.

Theorem 2.4   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$, Suppose that $(\alpha, \beta)-\gamma:$ $ P(X)\rightarrow P(X)$ is $(\alpha, \gamma)$-preopen and $(\beta, \gamma)$-preopen. Then $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{0}$ space iff for every pair of distinct points $x, y$, we have $CL_{(\alpha, \beta)-\gamma}(\{{x}\})\ne CL_{(\alpha, \beta)-\gamma}(\{{y}\})$.

Proof   Sufficiency. Suppose that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{0}$ space, then for every pair of distinct points $x, y$, there exists an $(\alpha, \beta)$-$\gamma$-open set $P$ such that $x\in P$, $y\not\in P$ or $y\in P$, $x\not\in P$ by Definition 2.2. If $x\in P$, $y\not\in P$, then $y\in X-P$, that is $\{{y}\}\subset X-P$. We have that $X-P$ is $(\alpha, \beta)$-$\gamma$-closed. Thus $CL_{(\alpha, \beta)-\gamma}(\{{y}\})\subset X-P$. This implies $CL_{(\alpha, \beta)-\gamma}(\{{x}\}) \ne CL_{(\alpha, \beta)-\gamma}(\{{y}\})$. Similarly, if $y\in P$, $x\not\in P$, it follows that $\{{x}\}\subset X-P$. We have that $X-P$ is $(\alpha, \beta)$-$\gamma$-closed. Thus $CL_{(\alpha, \beta)-\gamma}(\{{x}\})\subset X-P$. This implies $CL_{(\alpha, \beta)-\gamma}(\{{x}\})\ne CL_{(\alpha, \beta)-\gamma}(\{{y}\})$.

Necessity. Suppose that $x, y\in X$ and $x\ne y$, we have that

$ CL_{(\alpha, \beta)-\gamma}(\{{x}\})\ne CL_{(\alpha, \beta)-\gamma}(\{{y}\}). $

Then there exists a point $z\in CL_{(\alpha, \beta)-\gamma}(\{{x}\})$ but $z\not\in CL_{(\alpha, \beta)-\gamma}(\{{y}\})$ or $z\in CL_{(\alpha, \beta)-\gamma}(\{{y}\})$ but $z\not\in CL_{(\alpha, \beta)-\gamma}(\{{x}\})$. If $x\in CL_{(\alpha, \beta)-\gamma}(\{{y}\})$, then $CL_{(\alpha, \beta)-\gamma}(\{{x}\})\subset CL_{(\alpha, \beta)-\gamma}(\{{y}\})$ by Remark 1.3 (2), (3). This implies that $z\in cl_{(\alpha, \beta)-\gamma}(\{{y}\})$. This contradiction shows that $x\not\in CL_{(\alpha, \beta)-\gamma}(\{{y}\})$, then $X-CL_{(\alpha, \beta)-\gamma}(\{{y}\})$ is an $(\alpha, \beta)$-$\gamma$-open set containing $x$ but not $y$. This implies that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{0}$ space.

Theorem 2.5   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$. Then $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{1}$ space iff every singleton set $\{{x}\}$ is $(\alpha, \beta)$-$\gamma$-closed.

Proof   Sufficiency. Suppose that $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{1}$ space, then for every pair of distinct points $x, y\in X$, there exist two $(\alpha, \beta)$-$\gamma$-open sets $P_1$ and $P_2$ containing $x, y$, respectively, such that $x\in P_1$ and $y\not\in P_1$ or $y\in P_2$ and $x\not\in P_2$. Hence $y\in P_2\subset X-\{{x}\}$. This implies $X-\{{x}\}$ is an $(\alpha, \beta)$-$\gamma$-open set, i.e., $\{{x}\}$ is $(\alpha, \beta)$-$\gamma$-closed.

Necessity. Suppose that $x, y\in X$ and $x\ne y$. By the assumption $X-\{{x}\}$ is an $(\alpha, \beta)$-$\gamma$-open set containing $y$ and $X-\{{y}\}$ is an $(\alpha, \beta)$-$\gamma$-open set containing $x$. Then $X-\{{x}\}$ and $X-\{{y}\}$ are the satisfied $(\alpha, \beta)$-$\gamma$-open sets such that $y\in X-\{{x}\}$, $x\not\in X-\{{x}\}$ and $x\in X-\{{y}\}$, $y\not\in X-\{{y}\}$. Hence $(X, \mathcal{T})$ is an $(\alpha, \beta)$-$\gamma$-$T^{*}_{1}$ space.

Remark 2.2   From Theorem 2.2 and Theorem 2.5 we obtain that $(\alpha, \beta)$-$\gamma$-$T_{1}$ space is an $(\alpha, \beta)$-$\gamma$-$^{*}T_{1}$ space, the converse is true.

Theorem 2.6   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$. The following relations hold: $(\alpha, \beta)$-$\gamma$-$T^{*}_{5/2}\Rightarrow(\alpha, \beta)$-$\gamma$-$T^{*}_{2}\Rightarrow(\alpha, \beta)$-$\gamma$-$T^{*}_{1}\Rightarrow(\alpha, \beta)$-$\gamma$-$T^{*}_{0}. $

Proof   It is clear by above Theorem 2.4, 2.5 and Definition 2.2.

Theorem 2.7   Let $(X, \mathcal{T})$ be a topological space and $\alpha, \beta, \gamma$ be operators on $\mathcal{T}$.The following statements hold:

(1) Every $(\alpha, \beta)$-$\gamma$-$T^{*}_{0}$ space is an $(\alpha, \beta)$-$\gamma$-$T_{0}$ space;

(2) Every $(\alpha, \beta)$-$\gamma$-$T^{*}_{1}$ space is an $(\alpha, \beta)$-$\gamma$-$T_{1}$ space;

(3) Every $(\alpha, \beta)$-$\gamma$-$T^{*}_{2}$ space is an $(\alpha, \beta)$-$\gamma$-$T_{2}$ space.

Proof   (1), (3) proofs are obvious by Definition 2.2. The proof for (2) is obtained by Remark 2.2.