Generalized topology and the family of monotonic maps $\Gamma (X)$

Abstract

In this paper, interesting properties of the generalized topological spaces, generated by the monotonic maps $\sigma = (cl_{\delta }\circ int_{\delta }),$ $\alpha = (int_{\delta }\circ cl_{\delta }\circ int_{\delta }),$ $\pi = (int_{\delta }\circ cl_{\delta })$ and $\beta = (cl_{\delta }\circ int_{\delta }\circ cl_{\delta }),$ for any generalized topological space $(X,g_{\delta })$ are deduced and analyzed. Special subfamilies of the family of monotonic maps $\Gamma (X)$ are studied and interesting results regarding generalized topologies are obtained.

Introduction

In [4], $\acute{A.}$ Cs$\acute{a}$zs$\acute{a}$r introduced the generalized topological spaces. He showed that each monotonic map $\delta : P(X)\rightarrow P(X)$ $(\delta (A)\subset \delta (B),$ for each $A\subset B)$ defines a generalized topology $g_{\delta }$ on X, containing all the subsets O,  that satisfy $\delta (O)\supset O.$ The family of all monotonic maps $\delta$ is denoted by $\Gamma (X).$ Moreover, each generalized topology g on the set X defines a monotonic map $\delta _{g},$ such that $\delta _{g}(O)\supset O,$ for every $O\in g.$

To learn about the studies of the $\gamma -$generalized topological spaces $(X, g_\gamma ),$ see the references [4,5,6,7,8]. Moreover, to learn about the studies of the generalized continuity of functions on the $\gamma -$generalized topological spaces $(X, g_\gamma ),$ which is generated by the monotonic functions $\gamma \in \{int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta \},$ see the references [1,2,3, 9,10,11,12,13], where $g_\delta$ is a given generalized topology on X. In addition to that the properties of interior and closure operators are outlined in [14].

The outline of this manuscript is as follows: In the first section, some properties of the subclasses of the family of monotonic maps $\Gamma (X)$, whose elements generate the same generalized topology, are studied. Moreover, the relations between the family $\Phi \subset \Gamma (X)$ of all monotonic maps $\gamma \in \Gamma (X),$ for which there exists a function $f:X \rightarrow X$ such that $\gamma (A)=f^{-1} (A); A\in P(X),$ and the generalized topologies on X are studied.

In the second section, the study of some properties and examples on the family of all monotonic maps $\Gamma (X)$ are outlined.

In the third section, the generalized topologies generated by the monotonic maps: $\sigma = (cl_{\delta }\circ int_{\delta }),$ $\alpha = (int_{\delta }\circ cl_{\delta }\circ int_{\delta }),$ $\pi = (int_{\delta }\circ cl_{\delta }),$ $\beta = (cl_{\delta }\circ int_{\delta }\circ cl_{\delta })$ are studied.

In the fourth section, some interesting relations between the elements of the subfamily $\{int_{\delta }, cl_{\delta }, \sigma , \alpha , \pi , \beta \}\subset \Gamma (X),$ for any $\delta -$Cs$\acute{a}$zs$\acute{a}$r generalized topological space $(X,g_{\delta })$, are found.

Study of special classes of the family of all monotonic maps $\Gamma (X)$

Equivalence classes on the family of all monotonic maps $\Gamma (X)$

Definition 1

Consider the following binary operations on the family $\Gamma (X),$ where $\gamma ,\delta \in \Gamma (X)$ and A is a subset of X : 

• 1 $(\gamma \circ \delta )(A)=\gamma (\delta (A)).$

• 2 $(\gamma \cap \delta )(A)=\gamma (A) \cap \delta (A).$

• 3 $(\gamma \cup \delta )(A)=\gamma (A) \cup \delta (A).$

It is clear that for every $\gamma ,\delta \in \Gamma (X),$ the maps $\gamma \circ \delta , \gamma \cap \delta , \gamma \cup \delta$ are monotonic maps and are elements of $\Gamma (X).$

Definition 2

(Equivalent relation on the family $\Gamma (X)$) The maps $\gamma , \delta \in \Gamma (X)$ are called equivalent maps, if the family of all $\gamma -$open sets is identical with the family of all $\delta -$open sets $(g_\delta =g_\gamma ),$ and we write $\gamma \approx \delta .$

Each equivalence class of this relation is characterized by its family of open sets,which forms a generalized topology. The equivalence class, which contains the map $\delta$, will be denoted by $\Gamma _{\delta }(X)$ or simply $\Gamma _{\delta }.$

Theorem 3

The interior operator $int_{\delta }$ of the generalized topology $g_{\delta }$ is the smallest element of the class $\Gamma _{\delta }$. Moreover, for every $A\subset X:$

$$\begin{aligned} int_{\delta }(A)=\bigcap _{\gamma \in \Gamma _{\delta }}\gamma (A) \end{aligned}$$

Proof

The proof is obtained through the following steps:

• 1 Let $I_{\delta }(A)=\bigcap _{\gamma \in \Gamma _{\delta }}\gamma (A),$ for every $A\subset X.$

  It is clear that $\gamma (A)\supset I_{\delta }(A)=\bigcap _{\gamma \in \Gamma _{\delta }}\gamma (A),$ for every $\gamma \in \Gamma _{\delta }$ and $A\subset X.$ Moreover, $I_{\delta }\in \Gamma (X).$

• 2 Let A be an open set in the class $\Gamma _{\delta },$ then for every $\gamma \in \Gamma _{\delta },$ we have $A\subset \gamma (A)$ and $A \subset I_{\delta }(A).$ Therefore, every open set in $\Gamma _{\delta }$ is an open set relative to $I_{\delta }.$

  Now, if C is an open set relative to $I_{\delta },$ i.e. $C \subset I_{\delta }(C)=\bigcap _{\gamma \in \Gamma _{\delta }}\gamma (C).$ It follows that $C \subset \gamma (C);$ for each $\gamma \in \Gamma _{\delta }.$ Therefore, C is $\gamma -$open set for all $\gamma \in \Gamma _{\delta }$ and the family of open sets in $\Gamma _{\delta }$ is identical with the family of open sets of $I_{\delta }$ and so $I_{\delta }\in \Gamma _{\delta }.$ Hence $I_{\delta }$ is the smallest element in $\Gamma _{\delta }.$

• 3 It is clear that $int_{\delta }$ is a monotonic map, then $int_{\delta }\in \Gamma (X),$; moreover, the relation $O \subset int_{\delta }(O)$ is valid only for the elements of $g_{\delta }.$ Then, $int_{\delta }\in \Gamma _{\delta }.$

• 4 Since $int_{\delta }\in \Gamma _{\delta },$ then $int_{\delta }(B) \supset I_{\delta }(B);$ for every $B \subset X.$

• 5 We shall show that $int_{\delta }(B) \subset I_{\delta }(B);$ for every $B \subset X.$

  Let $B\subset X,$ then $int_{\delta }(B) \subset B.$ Since $int_{\delta } \in g_{\delta }$ and $I_{\delta }\in \Gamma _{\delta },$ then

  $int_{\delta }(B) \subset I_{\delta }(int_{\delta }(B))\subset I_{\delta }(B).$

  From 1 up to 5, it follows that $I_{\delta } = int_{\delta }.$ Therefore, $g_{\delta }$ is the smallest element of the class $\Gamma _{\delta }$.

$\square$

Definition 4

Let $\Gamma _{\delta }$ be an equivalence class for any $\delta -$generalized topology on the set X. Every $\gamma \in \Gamma _{\delta }$ defines the map

$$\begin{aligned} \theta _{\gamma }(B)=X-\gamma (X-B); B\subset X. \end{aligned}$$

Definition 5

To each equivalence class $\Gamma _{\delta },$ there exists an associated class

$$\begin{aligned} \Gamma ^{\delta }=\{\theta _{\gamma }: \gamma \in \Gamma _{\delta }\}. \end{aligned}$$

Theorem 6

Let $(X, g_{\delta })$ be a generalized topological space, then the following properties are satisfied:

• 1 $\Gamma ^{\delta }\subset \Gamma (X).$

• 2 For any $\delta -$closed subset B,  it follows that $\theta _{\gamma }(B)\subset B,$ for all $\gamma \in \Gamma _\delta .$

Proof

• 1 Let $A\subset B,$ then $X -A \supset X - B.$ Therefore, $\gamma (X-A)\supset \gamma (X-B),$ for every $\gamma \in \Gamma _{\delta },$ then

  $$\begin{aligned} \theta _{\gamma }(A)=X-\gamma (X-A)\subset X-\gamma (X-B)= \theta _{\gamma }(B). \end{aligned}$$

  Which means that $\theta _{\gamma }$ is a monotonic map and $\theta _{\gamma }\in \Gamma (X).$ Consequently, $\Gamma ^{\delta }\subset \Gamma (X).$

• 2 Let $A = X - B$ be an open set in $g_{\delta },$ then $A = X - B \subset \gamma (A)=\gamma (X- B),$ for all $\gamma \in \Gamma _\delta .$ Consequently, $B \supset X -\gamma (X - B) = \theta _{\gamma }(B).$

$\square$

Theorem 7

The closure operator $cl_{\delta }$ of the generalized topology $g_{\delta }$ is the largest element of the class $\Gamma ^{\delta }$. Moreover, for every $B\subset X:$

$$\begin{aligned} cl_{\delta }(B)=\theta _{int_{\delta }} (B) = \bigcup _{\gamma \in \Gamma _{\delta }}\theta _{\gamma }(B). \end{aligned}$$

Proof

The proof is obtained through the following two steps:

• 1 Since the interior monotonic map $int_{\delta },$ where $int_{\delta }(C) =\bigcup _{C\supset A\in g_{\delta }}A$ defines the map $\theta _{int_{\delta }},$ where $\theta _{int_{\delta }} (B) = X - int_{\delta }(X - B),$ then $\theta _{int_{\delta }}\in \Gamma ^{\delta }.$ Consequently,

  $$\begin{aligned} \theta _{int_{\delta }} (B) = X - int_{\delta }(X - B)=X-\bigcap _{\gamma \in \Gamma _{\delta }}\gamma (X-B)=\bigcup _{\gamma \in \Gamma _{\delta }}(X-\gamma (X-B))=\bigcup _{\gamma \in \Gamma _{\delta }}\theta _{\gamma }(B). \end{aligned}$$

  which means that the monotonic map $\theta _{int_{\delta }}$ is the largest monotonic map in the associated class $\Gamma ^{\delta }.$

• 2 Let $B \subset X,$ then

  $$\begin{aligned} \theta _{int_{\delta }} (B) = X - int_{\delta }(X - B)=X-\bigcup _{(X-B)\supset A\in g_{\delta }}A=\bigcap _{A\in g_{\delta },B\subset X-A}(X-A)=\bigcap _{(X-D)\in g_{\delta },B\subset D}D. \end{aligned}$$

  Since

  $$\begin{aligned} cl_{\delta }(B)=\bigcap _{(X-D)\in g_{\delta },B\subset D}D, \end{aligned}$$

  then $\theta _{int_{\delta }} (B)=cl_{\delta } (B),$ for any $B\subset X,$ which implies that $\theta _{int_{\delta }} =cl_{\delta }.$

$\square$

The subfamily $\Phi \subset \Gamma (X)$ corresponding to the family of functions $X^X$

Let $X^X$ be the family of all functions $f: X \rightarrow X.$ Then, for every $f\in X^X,$ there exists $\gamma _f\in \Gamma (X),$ which is defined as follows: for each $A \in P(X)$

$$\begin{aligned} \gamma _f: P(X) \rightarrow P(X); \gamma _f (A)=f^{-1} (A). \end{aligned}$$

Definition 8

The map $\Psi : X^X \rightarrow \Gamma (X)$ is defined by $\Psi (f)=\gamma _f.$

The map $\Psi$ is an injective map: Let $\Psi (f)=\Psi (g)$ and $x\in X.$ If $f(x)=y,$ then $x\in \gamma _f (\{y\})=\gamma _g (\{y\}).$ It follows that $g(x)=y=f(x).$ But $x\in X$ is an arbitrary element, then $f=g.$

Definition 9

The subfamily $\Phi$ of the family of monotonic maps $\Gamma (X)$ is defined as:

$$\begin{aligned} \Phi =\left\{ \gamma _f\in \Gamma (X): f\in X^X \right\} =\Psi (X^X)\subset \Gamma (X). \end{aligned}$$

The subfamily $\Phi \subset \Gamma (X)$ has a close relationship to the family of continuous functions on the topological spaces on X.

Lemma 10

The monotonic map $\gamma \in \Gamma (X)$ is an element of $\Phi ,$ if and only if it satisfies the following conditions:

• (a) $\gamma (\{y_1\})\cap \gamma (\{y_2\})=\emptyset ,$ for all $y_1,y_2\in X$ and $y_1\ne y_2.$

• (b) $\bigcup _{y\in X}\gamma (\{y\})=X.$

• (c) $\gamma (\bigcup _{i\in I}A_i)=\bigcup _{i\in I}\gamma (A_i)$ and $\gamma (\bigcap _{i\in I}A_i)=\bigcap _{i\in I}\gamma (A_i),$ where I is an arbitrary indexed set.

  Then, $\gamma =f^{-1},$ where f is a function from X to itself, where $f(x)=y;$ if $x\in \gamma (\{y\}).$

Proof

Let $\gamma \in \Phi ,$ then there exists $f\in X^X$ and $\gamma =\gamma _f.$

(a):

Let $y_1,y_2\in X$ such that $y_1\ne y_2,$ then

$$\begin{aligned} \gamma (\{y_1\})\cap \gamma (\{y_2\})=\gamma _f(\{y_1\})\cap \gamma _f(\{y_2\})=f^{-1}(y_1\cap y_2)=f^{-1}(\emptyset )=\emptyset . \end{aligned}$$

(b):

$\bigcup _{y\in X}\gamma (\{y\})=\bigcup _{y\in X}\gamma _f(\{y\})=\bigcup _{y\in X}f^{-1}(\{y\})=f^{-1}(\bigcup _{y\in X}\{y\})=f^{-1}(X)=X$.

(c):

$$\begin{aligned} \gamma \left( \bigcup _{i\in I}A_i\right) =\gamma _f\left( \bigcup _{i\in I}A_i\right) =f^{-1}\left( \bigcup _{i\in I}A_i\right) =\bigcup _{i\in I}f^{-1}(A_i)=\bigcup _{i\in I}\gamma _f(A_i)=\bigcup _{i\in I}\gamma (A_i). \end{aligned}$$

Moreover,

$$\begin{aligned} \gamma \left( \bigcap _{i\in I}A_i\right) =\gamma _f\left( \bigcap _{i\in I}A_i\right) =f^{-1}\left( \bigcap _{i\in I}A_i\right) =\bigcap _{i\in I}f^{-1}(A_i)=\bigcap _{i\in I}\gamma _f(A_i)=\bigcap _{i\in I}\gamma (A_i). \end{aligned}$$

$\square$

Lemma 11

The subfamily $\Phi \subset \Gamma (X)$ is closed relative to the composition binary operation.

Proof

Let $\gamma _f, \gamma _g\in \Phi .$ Since $\gamma _f (A)=f^{-1} (A)$ and $\gamma _g (B)=g^{-1} (B),$ for all $A,B\subset X,$ then

$$\begin{aligned} (\gamma _f\circ \gamma _g)(A)=(f^{-1}\circ g^{-1})(A)=(g\circ f)^{-1} (A)=\gamma _{g\circ f} (A). \end{aligned}$$

$\square$

For each $G\subset P(X),$ define the family $\mathbf {\Omega }_G\subset P(\Phi )$ as:

$$\begin{aligned} \mathbf {\Omega }_G=\left\{ {\mathcal {H}}\subset \Phi : h(G)\subset G\quad \forall \quad h\in {\mathcal {H}}\right\} \end{aligned}$$

Definition 12

The subfamily $G\subset P(X)$ is called invariant relative to ${\mathcal {H}}_G$ if ${\mathcal {H}}_G$ is the maximal element of the family $\mathbf {\Omega }_G$ with respect to the inclusion relation.

Remark 13

The family ${\mathcal {H}}_G$ is not empty for every $G\subset P(X),$ since the identity function $id_X (A)=A,$ for all $A\in P(X)$ belongs to every ${\mathcal {H}}_G.$

For each $G\subset P(X),$ define the subfamily ${\textbf{F}}_G$ of the family $X^X$ as:

$$\begin{aligned} {\textbf{F}}_G=\left\{ f\in X^X\quad :\quad \gamma _f\in {\mathcal {H}}_G\right\} \end{aligned}$$

Theorem 14

If $G\subset P(X)$ is a generalized topology on X,  then the family ${\textbf{F}}_G$ is the family of generalized continuous functions on the topological space (X, G).

Proof

The proof is clear, since: If $f\in {\textbf{F}}_G$ and $A\in G,$ then $\gamma _f\in {\mathcal {H}}_G,$ which implies that $\gamma _f(A)=f^{-1}(A)\in G.$ Therefore, $f:(X,G)\rightarrow (X,G)$ is a generalized continuous function. $\square$

Lemma 15

All the elements $h\in \Phi$ satisfy the relations:

$$\begin{aligned} h\left( \bigcup _{i\in K} A_i\right) =\bigcup _{i\in K} h(A_i), \quad h\left( \bigcap _{i\in K} A_i\right) =\bigcap _{i\in K} h(A_i), \end{aligned}$$

for any arbitrary family $\{A_i\subset X: i\in K\}\subset P(X),$ where K is an arbitrary index set.

Proof

The proof is straightforward, since for any function $f\in X^X,$

$$\begin{aligned} f^{-1}\left( \bigcup _{i\in K} A_i\right) =\bigcup _{i\in K} f^{-1}(A_i), \quad f^{-1}\left( \bigcap _{i\in K} A_i\right) =\bigcap _{i\in K} f^{-1}(A_i). \end{aligned}$$

$\square$

Theorem 16

If G and $G_0$ are subsets of P(X) and $G\subset G_0,$ then ${\mathcal {H}}_G\subset {\mathcal {H}}_{G_0 },$ if each element of $G_0$ can be written as arbitrary unions of finite (arbitrary) intersections of elements of G.

Proof

Let $G,G_0$ be subsets of P(X),  where $G\subset G_0.$ Let $G,G_0$ be invariant relative to ${\mathcal {H}}_G$ and ${\mathcal {H}}_{G_0 }$ respectively. Then, $h(G)\subset G; h\in {\mathcal {H}}_G.$ Let $g\in G_0,$ then from the assumption, g can be written in the form: $g=\bigcup _{i\in I_0}\bigcap _{j_i\in K_i } A_{j_i};$ where $A_{j_i}\subset G,$ for all i, j. Consequently, if $h\in {\mathcal {H}}_G,$ then from Lemma (1.15), it follows that

$$\begin{aligned} h(g)= h\left( \bigcup _{i\in I_0}\bigcap _{j_i\in K_i } A_{j_i}\right) =\bigcup _{i\in I_0}\bigcap _{j_i\in K_i } h(A_{j_i})=\bigcup _{i\in I_0}\bigcap _{j_i\in K_i } A^*_{j_i}\subset G_0, \end{aligned}$$

since $A^*_{j_i}=h(A_{j_i })\in G.$ Then, $h\in {\mathcal {H}}_{G_0},$ and so ${\mathcal {H}}_G\subset {\mathcal {H}}_{G_0}.$ $\square$

Corollary 17

If G is a subset of P(X),  then ${\mathcal {H}}_G\subset {\mathcal {H}}_{\tau (G)},$ where $\tau (G)$ is the (generalized topology) topology on the set X, generated by G as a (generalized base) sub-base. Since the elements of $\tau (G)$ are obtained from the elements of G,  using (arbitrary unions) arbitrary unions and arbitrary finite intersections.

The following example shows that in general, if $G,G_0$ are subsets of P(X),  where $G\subset G_0.$ Then it is not necessary that ${\mathcal {H}}_G\subset {\mathcal {H}}_{G_0 }.$

Example 18

Let $X=\left\{ a,b,c\right\} , G_1=\left\{ \{a\}\right\}$ and $G_2=\{\{a\},\{b\}\}.$ Consider the function $f: X \rightarrow X,$ where $f(a)=a, f(b)=c$ and $f(c)=b.$ Then, $\gamma _f (G_1)= f^{-1} (G_1 )=\{\{a\}\}=G_1,$ which implies that $\gamma _f\in {\mathcal {H}}_{G_1}.$ But $\gamma _f (G_2)= f^{-1} (G_2 )=\{\{a\},\{c\}\}\not \subset G_2,$ which implies that $\gamma _f\not \in {\mathcal {H}}_{G_2}.$ Therefore, ${\mathcal {H}}_{G_1}\not \subset {\mathcal {H}}_{G_2 }.$

It is clear that the element $\{b\}\in G_2$ can’t be obtained from the elements of $G_1,$ using the union and intersection operations. This justifies why ${\mathcal {H}}_{G_1}$ is not contained in ${\mathcal {H}}_{G_2},$ although $G_1\subset G_2.$

The following example shows that in general, if $G\subset P(X).$ Then ${\mathcal {H}}_G\ne {\mathcal {H}}_{\tau (G)},$ where $\tau (G)$ is the generalized topology generated by G.

Example 19

Let $X=\left\{ a,b,c\right\} .$ Choose $G=\{ \{a\},\{b\}\}.$ Then, $\tau (G)=\{\emptyset , \{a\},\{b\},\{a,b\}\}.$ Consider the function $g: X \rightarrow X,$ where $g(a)=b, g(b)=b$ and $g(c)=c.$ Then, the action of $\gamma _g$ is defined as follows: $\gamma _g(G)=g^{-1}(G)=\{\emptyset ,\{a,b\}\}\not \subset G,$ then $\gamma _g\not \in {\mathcal {H}}_G,$ $\gamma _g(\tau (G))=g^{-1}(\tau (G))=\{\emptyset ,\{a,b\}\}\subset \tau (G),$ then $\gamma _g \in {\mathcal {H}}_{\tau (G)}.$ Therefore, ${\mathcal {H}}_G\ne {\mathcal {H}}_{\tau (G))}.$

Theorem 20

Let $G_1,G_2\subset P(X),$ then ${\mathcal {H}}_{G_1}\cap {\mathcal {H}}_{G_2}\subset {\mathcal {H}}_{G_1\cap G_2}.$

Proof

Let $\gamma \in {\mathcal {H}}_{G_1}\cap {\mathcal {H}}_{G_2},$ then $\gamma \in {\mathcal {H}}_{G_i}, i\in \{1,2\}.$ It follows that $\gamma (G_i)\subset G_i, i\in \{1,2\}.$ Consequently, $\gamma (G_1\cap G_2 )=\gamma (G_1)\cap \gamma (G_2)\subset G_1\cap G_2.$ It follows that $\gamma \in {\mathcal {H}}_{G_1\cap G_2 }.$ Therefore, ${\mathcal {H}}_{G_1}\cap {\mathcal {H}}_{G_2}\subset {\mathcal {H}}_{G_1\cap G_2}.$ $\square$

Example (1.17) shows that the inverse statement of Theorem (1.19) is not valid. Since $G_1\cap G_2=G_1$ and ${\mathcal {H}}_{G_1\cap G_2}={\mathcal {H}}_{G_1}\not \subset {\mathcal {H}}_{G_2}.$

Remark 21

If ${\mathcal {H}}\subset \Gamma (X)$ and $\gamma \in \Gamma (X),$ then $\gamma \circ {\mathcal {H}}$ and ${\mathcal {H}}\circ \gamma$ can be defined as:

$$\begin{aligned} \gamma \circ {\mathcal {H}}=\{\gamma \circ h: h\in {\mathcal {H}}\}, \quad {\mathcal {H}}\circ \gamma =\{ h \circ \gamma : h\in {\mathcal {H}}\}\subset \Gamma (X). \end{aligned}$$

Definition 22

Let $G\subset P(X),$ then the ordered pair $\prec G,{\mathcal {H}}_G\succ$ is called an invariant system.

Theorem 23

If $\prec G,{\mathcal {H}}_G\succ$ is an invariant system and $f\in X^X$ is a one-to-one correspondence, then $\prec f(G),\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _ f\succ$ is an invariant system and ${\mathcal {H}}_{f(G)}=\gamma _{f^{-1}} \circ {\mathcal {H}}_G\circ \gamma _ f.$

Proof

Since $\prec G,{\mathcal {H}}_G\succ$ is an invariant system, then $h(G)\subset G; h\in {\mathcal {H}}_G.$ Consequently, we have: $(\gamma _{f^{-1}} \circ h \circ \gamma _f)(f(G))=\gamma _{f^{-1}}(h(\gamma _f (f(G))))=f(h(f^{-1}(f(G))))\subset f(h(G))\subset f(G).$ Therefore, $\gamma _{f^{-1}} \circ h \circ \gamma _f\in {\mathcal {H}}_{f(G)},$ for all $h\in {\mathcal {H}}_G.$ Hence, $\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _ f\subset {\mathcal {H}}_{f(G)}.$

Now, we show that $\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _ f$ is a maximal element of the family

$$\begin{aligned} \mathbf {\Omega }_{f(G)}=\left\{ {\mathcal {H}}\subset \Phi : h(f(G))\subset f(G)\quad \forall \quad h\in {\mathcal {H}}\right\} . \end{aligned}$$

Let $\gamma _{f^{-1}}\circ {\mathcal {H}}_G\circ \gamma _ f\subset {\mathcal {H}},$ where ${\mathcal {H}}\in \mathbf {\Omega }_{f(G)},$ and let $h\in {\mathcal {H}},$ then $h(f(G))=h(\gamma _{f^{-1}}(G))\subset f(G).$ It follows that $\gamma _f(h(\gamma _{f^{-1}}(G)))\subset \gamma _f(\gamma _{f^{-1}}(G))=G.$ Hence $\gamma _f \circ h \circ \gamma _{f^{-1}}\in {\mathcal {H}}_G,$ and so

$$\begin{aligned} \gamma _{f^{-1}} \circ \gamma _f \circ h \circ \gamma _{f^{-1}} \circ \gamma _f=h\in \gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _f. \end{aligned}$$

Therefore, ${\mathcal {H}}=\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _f,$ and $\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _ f$ is a maximal element of the family $\mathbf {\Omega }_{f(G)}.$ Consequently, $\gamma _{f^{-1}} \circ {\mathcal {H}}_G \circ \gamma _f={\mathcal {H}}_{f(G)},$ which implies that $\prec f(G),\gamma _{f^{-1}}\circ {\mathcal {H}}_G\circ \gamma _ f\succ$ is an invariant system. $\square$

Corollary 24

• (1) If $f\in X^X$ is a one-to-one correspondence function, then $f^{-1}\in X^X$ is a one-to-one correspondence. Using Theorem (2.7), it follows that

  $$\begin{aligned} \prec f^{-1} (G),\gamma _f \circ {\mathcal {H}}_G \circ \gamma _{f^{-1}}\succ \end{aligned}$$

  is an invariant system.

• (2) Each one-to-one correspondence function f and invariant system $\prec G,{\mathcal {H}}_G\succ$ define a sequence of invariant systems:

  $$\begin{aligned} \left\{ \prec f^n (G),\gamma _{f^{-n}} \circ {\mathcal {H}}_G \circ \gamma _{f^n}\succ \quad : \quad n\in N\right\} . \end{aligned}$$

Remark 25

From the study of the invariant systems $\prec G,{\mathcal {H}}_G\succ ,$ it is shown that there exists a one-to-one correspondence between the family of $G-$continuous functions and the family of $f(G)-$continuous functions: $h\leftrightarrow \gamma _{f^{-1}} \circ h \circ \gamma _f; h\in {\mathcal {H}}_G,$ where $f\in X^X$ is any one-to-one correspondence function.

Study of some properties and examples on $\Gamma (X)$

Definition 26

[7]. Let (X, g) be a generalized topological space, then $\Sigma \subset g$ is called a base for g if: every $A \in g$ can be constructed as a union of some members of $\Sigma .$

Moreover, any subfamily $\Sigma \subset P(X)$ generates the unique generalized topology g on X,  where

$$\begin{aligned} g = G(\Sigma ) = \left\{ A \subset X: \exists \quad \Sigma _{\circ }\subset \Sigma ,\quad A =\bigcup _{B_{i}\in \Sigma _{\circ }}B_{i}\right\} , \end{aligned}$$

and g is the smallest generalized topology on X,  containing $\Sigma .$

Theorem 27

Let $\gamma _{1}\in \Gamma _{\gamma _{1}}$ and $\gamma _{2}\in \Gamma _{\gamma _{2}}.$ Then:

• 1 $g_{\gamma _{1}\circ \gamma _{2}}\supset g_{\gamma _{1}}\cap g_{\gamma _{2}},$ and if $\gamma _{1}, \gamma _{2}\in \Gamma _{\delta },$ then $g_{\gamma _{1}\circ \gamma _{2}}\supset g_{\delta },$ but, in some cases the equality holds.

• 2 $g_{\gamma _{1}\cup \gamma _{2}}\supset G\{g_{\gamma _{1}}, g_{\gamma _{2}}\},$ but the equality is valid for some special cases, where $G\{g_{\gamma _{1}}, g_{\gamma _{2}}\}$ is the generalized topology, which is generated by the family $g_{\gamma _{1}}\cup g_{\gamma _{2}}.$

• 3 $g_{\gamma _{1}\cap \gamma _{2}}=g_{\gamma _{1}}\cap g_{\gamma _{2}}.$ Moreover, if $\gamma _{1}, \gamma _{2}\in \Gamma _{\delta },$ then $g_{\gamma _{1}\cap \gamma _{2}}\in \Gamma _{\delta }.$

  Then, the intersection operation forms a binary operation on $\Gamma _{\delta }.$

Proof

• 1 Let $O\in g_{\gamma _{1}}\cap g_{\gamma _{2}},$ then $O\subset \gamma _{1}(O)$ and $O\subset \gamma _{2}(O)$ Consequently, $(\gamma _{1}\circ \gamma _{2})(O)= \gamma _{1}(\gamma _{2}(O))\supset \gamma _{1}(O)\supset O,$ then $O\in g_{\gamma _{1}\circ \gamma _{2}}.$ Therefore, $g_{\gamma _{1}\circ \gamma _{2}}\supset g_{\gamma _{1}}\cap g_{\gamma _{2}}.$

  See Example (2.1), in which, $g_{\delta }\ne g_{\delta ^{2}}$ if $\delta =\gamma$ and $g_{\delta }= g_{\delta ^{2}}$ if $\delta =\gamma ^{2}$

• 2 Let $O\in g_{\gamma _{1}}\cup g_{\gamma _{2}},$ then $O\subset \gamma _{1}(O)$ or $O\subset \gamma _{2}(O).$ Consequently,

  $(\gamma _{1}\cup \gamma _{2})(O)= \gamma _{1}(O)\cup \gamma _{2}(O)\supset O,$ then $O\in g_{\gamma _{1}\cup \gamma _{2}}.$ Therefore, $g_{\gamma _{1}\cup \gamma _{2}}\supset g_{\gamma _{1}}\cup g_{\gamma _{2}}.$

  Since $G\{g_{\gamma _{1}}, g_{\gamma _{2}}\}$ is the smallest generalized topology on X,  containing $g_{\gamma _{1}\cup \gamma _{2}},$ then $g_{\gamma _{1}\cup \gamma _{2}}\supset G\{g_{\gamma _{1}}, g_{\gamma _{2}}\}.$

  See Example (2.2), for the equality case.

• 3 Let $O\in g_{\gamma _{1}}\cap g_{\gamma _{2}},$ then $O\subset \gamma _{1}(O)$ and $O\subset \gamma _{2}(O).$ Consequently,

  $O\subset \gamma _{1}(O)\cap \gamma _{2}(O)=(\gamma _{1}\cap \gamma _{2})(O),$ then $O\in g_{\gamma _{1}\cap \gamma _{2}}$ and $g_{\gamma _{1}\cap \gamma _{2}}\supset g_{\gamma _{1}}\cap g_{\gamma _{2}}.$

  Now, et $O\in g_{\gamma _{1}\cap \gamma _{2}},$ then $O\subset (\gamma _{1}\cap \gamma _{2})(O)=\gamma _{1}(O)\cap \gamma _{2}(O).$ Consequently, $O\subset \gamma _{1}(O)$ and $O\subset \gamma _{2}(O),$ then $O\in g_{\gamma _{1}}\cap g_{\gamma _{2}}$ and $g_{\gamma _{1}\cap \gamma _{2}}\subset g_{\gamma _{1}}\cap g_{\gamma _{2}}.$

  Therefore, the intersection operation is a binary operation on $\Gamma _{\delta }.$

$\square$

In the following example, a map $\gamma \in \Gamma _{\delta }$ is constructed to have the following properties:

(i):

$\gamma ^2=\gamma \circ \gamma \not \in \Gamma _{\gamma },$ but, $\gamma ^3\in \Gamma _{\gamma }.$

(ii):

$g_{\gamma ^{(2n+1)}}=g_{\gamma }=G\{O_0,O_1\cup O_2\}\in \Gamma _{\gamma }; n\in \{0,1,2,3,\ldots \}.$

(iii):

$g_{\gamma ^{2n}}=g_{\gamma ^{2}}=G\{O_0,O_1, O_2\}\supset g_{\gamma ^{(2n+1)}}=g_{\gamma }$ and $g_{\gamma ^{2}}\not \in \Gamma _{\gamma }; n\in \{1,2,3,\ldots \}.$

Example 28

Let X be a non-empty set and $O_0,O_1,O_2$ be non-empty mutually disjoint subsets of X. Define the map $\gamma : P(X)\rightarrow P(X)$ as follows:

• $\gamma (A) = O_0;$ if $A \supset O_0$ and A does not contain $O_1,O_2.$

• $\gamma (A) = O_2;$ if $A \supset O_1$ and A does not contain $O_0,O_1.$

• $\gamma (A) = O_1;$ if $A \supset O_2$ and A does not contain $O_0,O_2.$

• $\gamma (A)=\bigcup _{i\in \ell }\gamma (O_i);$ if $A\supset \bigcup _{i\in \ell }O_i,$ where $\ell \subset \{0,1,2\}.$

• $\gamma (A)=\emptyset ;$ if A does not contain $O_0$ or $O_1$ or $O_2.$

The map $\gamma$ satisfies the following:

(i):

$\gamma (O_0)=O_0.$

(ii):

$\gamma (O_1\cup O_2)=\gamma (O_1)\cup \gamma (O_2)=O_2\cup O_1.$

(ii):

$\gamma (O_0\cup O_1\cup O_2)=O_0\cup O_1\cup O_2.$

(iv):

$\gamma (A)\not \supset A;$ if $A\not \in G\{O_0,O_1\cup O_2\}.$

Therefore, the topology

$$\begin{aligned} g_{\gamma }=G\{O_0,O_1\cup O_2\}=\{\emptyset , O_0,O_1\cup O_2,O_0\cup O_1\cup O_2\}. \end{aligned}$$

The four steps (shown above) which constructs the map $\gamma$ will be denoted by the following notation:

$$\begin{aligned} \gamma : O_0 \uparrow ; O_1\rightarrow O_2\rightarrow O_1. \end{aligned}$$

The map $\gamma ^2\not \in \Gamma _{\gamma },$ since: $\gamma ^2(O_i)=O_i;$ $i\in \{0,1,2\},$ and $\gamma ^2\not \supset A,$ for any $A\subset X, A\not \in G\{O_0,O_1, O_2\}.$

Therefore,

$$\begin{aligned} g_{\gamma ^2}=G\{O_0,O_1, O_2\}=\{\emptyset , O_0,O_1,O_2,O_0\cup O_1,O_0\cup O_2,O_1\cup O_2,O_0\cup O_1\cup O_2\}. \end{aligned}$$

Therefore,

$$\begin{aligned} g_{\gamma ^2}=G\{O_0,O_1, O_2\}=\{\emptyset , O_0,O_1,O_2,O_0\cup O_1,O_0\cup O_2,O_1\cup O_2,O_0\cup O_1\cup O_2\}. \end{aligned}$$

The map $\gamma ^2$ can be constructed using the following symbols:

$$\begin{aligned} \gamma ^2:[O_0,O_1,O_2]\uparrow . \end{aligned}$$

The map $\gamma ^3 \in \Gamma _{\gamma },$ since

$\gamma ^3(O_0)=O_0,$   $\gamma ^3(O_1)=O_2,$   $\gamma ^3(O_2)=O_1,$   $\gamma ^3(O_1\cup O_2)=O_2\cup O_1,$ and $\gamma ^3(A)\not \supset A,$ for any $A\subset X$ and $A\not \in G\{O_0,O_1\cup O_2\}.$

Therefore,

$$\begin{aligned} g_{\gamma ^3}=g_{\gamma }=G\{O_0,O_1\cup O_2\}=\{\emptyset , O_0,O_1\cup O_2,O_0\cup O_1\cup O_2\}. \end{aligned}$$

Consequently,

$$\begin{aligned} g_{\gamma ^{(2n+1)}}=g_{\gamma }=G\{O_0,O_1\cup O_2\}\in \Gamma _{\gamma }; n\in \{0,1,2,3,\ldots \}. \end{aligned}$$

and

$$\begin{aligned} g_{\gamma ^{2n}}=g_{\gamma ^{2}}=G\{O_0,O_1, O_2\}\supset g_{\gamma ^{(2n+1)}}=g_{\gamma } \end{aligned}$$

and

$$\begin{aligned} g_{\gamma ^{2}}\not \in \Gamma _{\gamma }; n\in \{1,2,3,\ldots \}. \end{aligned}$$

A general construction of Example (2.3) can be illustrated in the following theorem.

Theorem 29

Let X be a non-empty set and $O_0,O_1,O_2,\ldots ,O_n$ be non-empty mutually disjoint subsets of X. Define the map $\gamma : P(X) \rightarrow P(X)$ as follows:

• $\gamma (A) = O_0;$ if $A \supset O_0$ and A does not contain any of the subsets $O_i; 1 \le i\le n.$

• $\gamma (A) = O_{i+1};$ if $A \supset O_i$ and A does not contain any of the subsets $O_s; s\ne i, 0\le s\le n,$ and 1 $\le i\le n-1.$

• $\gamma (A) = O_1;$ if $A \supset O_n$ and A does not contain any subset of $O_s, 0\le s< n.$

• $\gamma (A)=\bigcup _{i\in \ell }\gamma (O_i);$ if $A\supset \bigcup _{i\in \ell }O_i,$ where $\ell \subset \{0,1,2,3,4,\ldots ,n\}.$

• $\gamma (A)=\emptyset ;$ if A does not contain any set $O_i, i\in \{0,1,2,3,\ldots ,n\}.$

The map $\gamma$ can be defined as:

$$\begin{aligned} \gamma : O_0 \uparrow ; O_1 \rightarrow O_2 \rightarrow O_3\rightarrow O_4\rightarrow ....\rightarrow O_n\rightarrow O_1. \end{aligned}$$

Then, for any positive integer numbers n, s, k,  it follows that:

• 1
  $$\begin{aligned} g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup \cdots \cup O_n\}. \end{aligned}$$
• 2
  $$\begin{aligned} g_{\gamma }\ne g_{\gamma ^{s}}= G\{O_0,O_1\cup O_{1+s}\cup O_{1+2s}\cup \cdots \cup O_{n-s+1},\ldots ,O_s\cup O_{2s}\cup O_{3s}\cup \cdots \cup O_n \}, \end{aligned}$$

  whenever $1 \le s \le n - 1$ and $n = ks, k > 1.$

• 3
  $$\begin{aligned} g_{\gamma }\ne g_{\gamma ^{s}}= G\{O_0,O_1,O_2,O_3,\ldots ,O_n\}, \end{aligned}$$

  whenever $s\ge n$ and $s = kn.$

• 4
  $$\begin{aligned} g_{\gamma }=g_{\gamma ^{s}}= G\{O_0,O_1\cup O_2\cup O_3\cup \cdots \cup O_n\}, \end{aligned}$$

  whenever $(s\ge n$ and $s \ne kn)$ or $(s \le n - 1$ and $n \ne ks, k > 1).$

Proof

• 1 The definition of the map $\gamma$ implies that

  $$\begin{aligned} \gamma (O_0)=O_0,\gamma (O_1\cup O_2\cup O_3\cup \cdots \cup O_n)=O_1\cup O_2\cup O_3\cup ......\cup O_n \end{aligned}$$

  and $\gamma (A)\not \supset A,$ for all $A\subset X;$

  $$\begin{aligned} A\not \in \{\emptyset , O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n,O_0\cup O_1\cup O_2\cup O_3\cup ......\cup O_n\}. \end{aligned}$$

  Therefore, the generalized topology generated by the monotonic map $\gamma$ is

  $$\begin{aligned} g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n\} \end{aligned}$$

  or

  $$\begin{aligned} g_{\gamma }=\{\emptyset , O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n,O_0\cup O_1\cup O_2\cup O_3\cup ......\cup O_n\}. \end{aligned}$$

• 2 Let s be any positive integer such that $1 \le s \le n - 1,$ then

  $$\begin{aligned} \gamma ^s(O_i)=\left\{ \begin{array}{c} O_0 \quad \quad : \quad \quad \quad i=0,\\ O_{s+i}\quad \quad \quad : \quad 1\le i\le n-s,\\ O_{i-n+s}\quad \quad \quad \quad : \quad n-s+1\le i \le n. \end{array} \right. \end{aligned}$$

  • (i) Let $n = ks, k \in \{2, 3, 4,\ldots \},$ then:

    $$\begin{aligned}{} & {} \gamma (O_0)=O_0.\\{} & {} \gamma (O_1\cup O_{1+s}\cup O_{1+2s}\cup ......\cup O_{n-s+1})=O_1\cup O_{1+s}\cup O_{1+2s}\cup ......\cup O_{n-s+1}.\\{} & {} \gamma (O_2\cup O_{2+s}\cup O_{2+2s}\cup ......\cup O_{n-s+2})=O_2\cup O_{2+s}\cup O_{2+2s}\cup ......\cup O_{n-s+2}.\\{} & {} \gamma (O_3\cup O_{3+s}\cup O_{3+2s}\cup ......\cup O_{n-s+3})=O_3\cup O_{3+s}\cup O_{3+2s}\cup ......\cup O_{n-s+3}..\\{} & {} .\\{} & {} .\\{} & {} \gamma (O_s\cup O_{2s}\cup O_{3s}\cup ......\cup O_{n})=O_s\cup O_{2s}\cup O_{3s}\cup ......\cup O_{n}. \end{aligned}$$

    Moreover, $\gamma (A)\not \supset A,$ for all $A\subset X;$

    $$\begin{aligned} A\not \in \{\emptyset , O_0,O_1\cup O_{1+s}\cup O_{1+2s}\cup ......\cup O_{n-s+1}\} \end{aligned}$$

    or

    $$\begin{aligned} A\not \in \{O_2\cup O_{2+s}\cup O_{2+2s}\cup ......\cup O_{n-s+2},O_3\cup O_{3+s}\cup O_{3+2s}\cup ......\cup O_{n-s+3}\} \end{aligned}$$

    or

    $$\begin{aligned} A\not \in \{O_4\cup O_{4+s}\cup O_{4+2s}\cup ......\cup O_{n-s+4},\ldots ,O_s\cup O_{2s}\cup O_{3s}\cup ......\cup O_{n}\}. \end{aligned}$$

    Therefore, $\gamma ^s$ can be defined as:

    $$\begin{aligned}{} & {} \gamma ^s: O_0\uparrow ; O_1\rightarrow O_{1+s}\rightarrow O_{1+2s}\rightarrow .....\rightarrow O_{n-s+1}\rightarrow O_1,\\{} & {} O_2\rightarrow O_{2+s}\rightarrow O_{2+2s}\rightarrow .....\rightarrow O_{n-s+2}\rightarrow O_2,\\{} & {} O_3\rightarrow O_{3+s}\rightarrow O_{3+2s}\rightarrow .....\rightarrow O_{n-s+3}\rightarrow O_3,\\{} & {} .\\{} & {} .\\{} & {} O_s\rightarrow O_{2s}\rightarrow O_{3s}\rightarrow .....\rightarrow O_{n}\rightarrow O_s. \end{aligned}$$

    And so, the generalized topology generated by the monotonic map $\gamma ^s$ is

    $$\begin{aligned} g_{\gamma }\ne g_{\gamma ^{s}}= G\{O_0,O_1\cup O_{1+s}\cup O_{1+2s}\cup ....\cup O_{n-s+1},....,O_s\cup O_{2s}\cup O_{3s}\cup ....\cup O_n \}. \end{aligned}$$

  • (ii) Let $n \ne ks, k \in \{2, 3, 4,\ldots \},$ then:

    $$\begin{aligned}{} & {} \gamma (O_0)=O_0.\\{} & {} \gamma (O_1\cup O_2\cup ......\cup O_{n-s}\cup O_{n-s+1}\cup O_{n-s+2}\cup ....\cup O_{n-1}\cup O_n)\\{} & {} = O_{1+s}\cup O_{2+s}\cup ......\cup O_{n}\cup O_1\cup O_2.... O_{s-1}\cup O_s \\{} & {} =O_1\cup O_2\cup ......\cup O_{n-s}\cup O_{n-s+1}\cup O_{n-s+2}\cup ....\cup O_{n-1}\cup O_n. \end{aligned}$$

    Moreover, $\gamma (A)\not \supset A,$ for all $A\subset X;$

    $$\begin{aligned} A\not \in \{\emptyset , O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n,O_0\cup O_1\cup O_2\cup O_3\cup ......\cup O_n\}. \end{aligned}$$

    Therefore, the generalized topology generated by the monotonic map $\gamma ^s$ is

    $$\begin{aligned} g_{\gamma ^s}=g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n\} \end{aligned}$$

    or

    $$\begin{aligned} g_{\gamma ^s}=\{\emptyset , O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n,O_0\cup O_1\cup O_2\cup O_3\cup ......\cup O_n\}. \end{aligned}$$

• 3 Let s be any positive integer such that $s = kn, k \in \{1,2,3,\ldots \},$ then

  $$\begin{aligned} \gamma ^s(O_i)=O_i,\quad i\in \{0,1,2,3,\ldots \}. \end{aligned}$$

  Therefore, the map $\gamma ^s$ can be defined as:

  $$\begin{aligned} \gamma ^s: [O_0,O_1,O_2,\ldots ,O_{n-1},O_n]\uparrow . \end{aligned}$$

  And so the generalized topology generated by the monotonic map $\gamma ^s$ is

  $$\begin{aligned} g_{\gamma ^s}=G\{O_0,O_1,O_2,\ldots ,O_{n-1},O_n\}. \end{aligned}$$

• 4 Let s be any positive integer such that

  $$\begin{aligned} s \ge n, \quad s = kn + r,\quad k \in \{1,2,3,\ldots \},\quad 1\le r\le n - 1, \end{aligned}$$

  then

  $$\begin{aligned} \gamma ^s(O_i)=\gamma ^{nk+r}=\gamma ^r(O_i)=O_{i+r},\quad 1\le i\le n,\quad 1\le r\le n - 1. \end{aligned}$$

  Moreover, this case is the case [2].

$\square$

Corollary 30

Using Theorem (2.4), the following results can be obtained easily:

• 1 If n is a prime number, then there exists only two generalized topologies on X,  which can be constructed as follows:

  • (i) $g_{\gamma }\ne g_{\gamma ^s}=G\{O_0,O_1,O_2,\ldots ,O_{n-1},O_n\},$ whenever s is divisible by n.

  • (ii) $g_{\gamma ^s}=g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n\},$ whenever s is not divisible by n.

• 2 If s is a prime number and $n > s,$ then there exist only two generalized topologies on X,  which can be constructed as follows:

  • (i( $g_{\gamma }\ne g_{\gamma _{s}}= G\{O_0,O_1\cup O_{1+s}\cup O_{1+2\,s}\cup ....\cup O_{n-s+1},\ldots ,O_s\cup O_{2\,s}\cup O_{3\,s}\cup ....\cup O_n \},$ whenever n is divisible by s.

  • (ii) $g_{\gamma ^s}=g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n\},$ whenever n is not divisible by s.

• 3 If s, n are prime numbers, and $n \ne s,$ then there exists only one generalized topology on X,  which can be constructed as follows:

  $$\begin{aligned} g_{\gamma ^s}=g_{\gamma }=G\{O_0,O_1\cup O_2\cup O_3\cup ......\cup O_n\}. \end{aligned}$$

The following example shows that in general $g_{\gamma _1\cup \gamma _2}\ne G\{\gamma _1,\gamma _2\},$ for some $\gamma _1\in \Gamma _{\gamma _1}$ and $\gamma _2\in \Gamma _{\gamma _2}.$ Moreover, the equality will be valid for some special cases.

Example 31

Let X be non-empty set and $O_1,O_2,O_3,O_4$ be non-empty mutually disjoint subsets of X. Suppose that $O_1,O_2$ are disjoint, $O_2 = O_3 \cup O_4$ and $O_3 \cap O_4 = \emptyset .$

Define the three monotonic maps $\gamma _1,\gamma _2,\gamma _3: P(X) \rightarrow P(X)$ as follows:

• $\gamma _1(A) = O_1,$ for all $A \supset O_1$ and $A\cap O_2=\emptyset .$

• $\gamma _1(A) = O_3,$ for all $A \supset O_2$ and $A\cap O_1=\emptyset .$

• $\gamma _1(A)=\bigcup _{O_i\subset A, i\in \ell }\gamma _1(O_i),$ where $\ell \subset L=\{1,2\}.$

• $\gamma _1(A)=\emptyset ,$ for all $A\subset X$ and A does not contain any set $O_j, j\in \{1,2\}.$

• $\gamma _2(A) = O_1,$ for all $A \supset O_1$ and A does not contain $O_2.$

• $\gamma _2(A) = O_4,$ for all $A \supset O_2$ and A does not contain $O_1.$

• $\gamma _2(A)=\bigcup _{O_i\subset A, i\in \ell }\gamma _2(O_i),$ where $\ell \subset L=\{1,2\}.$

• $\gamma _2(A)=\emptyset ,$ for all $A\subset X$ and A does not contain any set $O_j, j\in \{1,2\}.$

• $\gamma _3(A) = O_1,$ for all $A \supset O_1.$

• $\gamma _3(A)=\emptyset ,$ for all $A\subset X$ and A does not contain $O_1.$

Therefore,

$\gamma _1(O_1) =\gamma _2(O_1)=\gamma _3(O_1)=O_1,$ and $\gamma _i(A)\not \supset A,$ for all $A \subset X, A\not \in \{\emptyset , O_1\}, i \in \{1,2,3\}.$

which implies that $g_{\gamma _1}=g_{\gamma _2}=g_{\gamma _3}=\{\emptyset , O_1\}.$

Hence, $g_{\gamma _1},g_{\gamma _2}$ and $g_{\gamma _3}$ are different monotonic maps, defining the same generalized topology. At the same time, we have the following:

• $(\gamma _1\cup \gamma _2)(O_1)=\gamma _1(O_1)\cup \gamma _2(O_1)=O_1.$

• $(\gamma _1\cup \gamma _2)(O_2)=\gamma _1(O_2)\cup \gamma _2(O_2)=O_3\cup O_4=O_2.$

• $(\gamma _1\cup \gamma _3)(O_1)=\gamma _1(O_1)\cup \gamma _3(O_1)=O_1.$

• $(\gamma _1\cup \gamma _3)(A)\not \supset A,$ for all $A\subset X$ and $A\not \in \{\emptyset ,O_1\}.$

Therefore,

$$\begin{aligned} g_{\gamma _1\cup \gamma _2}=G\{O_1,O_2\}=\{\emptyset ,O_1,O_2,O_1\cup O_2\},\quad g_{\gamma _1\cup \gamma _3}=\{\emptyset ,O_1\}. \end{aligned}$$

Then,

$$\begin{aligned} G\{g_{\gamma _1},g_{\gamma _2}\}=g_{\gamma _1}\cup g_{\gamma _2}=g_{\gamma _1\cup \gamma _3}=\{\emptyset ,O_1\}\subset g_{\gamma _1\cup \gamma _2}. \end{aligned}$$

And

$$\begin{aligned} g_{\gamma _1}\cup g_{\gamma _2}\ne g_{\gamma _1\cup \gamma _2},\quad g_{\gamma _1}\cup g_{\gamma _3}=g_{\gamma _1\cup \gamma _3}. \end{aligned}$$

Obtainment of generalized topologies generated by special monotonic maps

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta ,$ whose generalized interior and generalized closure are denoted by $int_{\delta }$ and $cl_{\delta }$ respectively. It is known that the monotonic maps $int_{\delta }\circ int_{\delta } = int_{\delta }$ and $cl_{\delta }\circ cl_{\delta } = cl_{\delta }.$ But the composition of the monotonic functions $int_{\delta }\circ cl_{\delta }$ and $cl_{\delta }\circ int_{\delta }$ have different behaviors.

$\sigma -$generalized topological space $(X, g_{\sigma })$

Theorem 32

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta .$ The non-empty elements of the generalized topology $g_{\sigma }$ which is defined by the map $\sigma = cl_{\delta }\circ int_{\delta }$ consists of all subsets $A \subset X,$ having non-empty $int_{\delta }(A)$ and (each $\delta -$open subset $O\in g_{\delta }$ intersects $int_{\delta }(A)$) or (each $\delta -$open subset $O\in g_{\delta },$ which does not intersect $int_{\delta }(A),$ does not intersect A also).

Proof

Let $A \subset X.$ Then:

• 1 If $int_{\delta }(A)=\emptyset ,$ then $\sigma (A) = \emptyset .$

• 2 If $int_{\delta }(A)\ne \emptyset ,$ then

  $$\begin{aligned} \sigma (A)= & {} cl_{\delta }(int_{\delta }(A)=\bigcap _{O\in g_{\delta }}\left\{ X-O \quad : \quad int_{\delta }(A)\subset X-O\right\} \\= & {} \bigcap _{O\in g_{\delta }}\left\{ X-O \quad : \quad int_{\delta }(A)\cap O=\emptyset \right\} \\= & {} X-\bigcup _{O\in g_{\delta }}\left\{ O \quad : \quad int_{\delta }(A)\cap O=\emptyset \right\} \\= & {} \left\{ \begin{array}{c} X\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad :\quad \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset \\ \\ \bigcap _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }(X-O)\quad :\quad \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O\ne \emptyset \\ \end{array} \right. \end{aligned}$$

  Therefore, the set $A \subset X$ is $\sigma -$open subset ( the class of all $g-$semi-open sets) if $\sigma (A) \supset A.$ This statement is valid in the following cases:

  • (a) For every A,  for which $int_{\delta }(A)\ne \emptyset$ and $\bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset ,$ which means that each $\delta -$open subset O intersects $int_{\delta }(A).$

  • (b) For every A,  for which $int_{\delta }(A)\ne \emptyset$ and $\bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O\ne \emptyset ,$ and

    $$\begin{aligned} A\subset \bigcap _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }(X-O)=X-\bigcap _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O. \end{aligned}$$

    Therefore,

    $$\begin{aligned} A\cap \bigcap _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset . \end{aligned}$$

    This means that each $\delta -$open subset O,  which does not intersect $int_{\delta }(A),$ does not intersect also with A.

$\square$

Remark 33

The set X is $\sigma -$open, if X is $\delta -$open or there exists a subset $A\subset X$ with non-empty $\delta -$interior, and intersects each $\delta -$open subset $O\in g_{\delta }.$

Notation 34

Throughout the rest of our study, the three special generalized topological spaces $(X, g_{i\delta }), i\in \{1,2,3\},$ will be defined as follows:

• 1 Let $(X, g_{1\delta })$ be a $1\delta -$generalized topological space on the non-empty set X,  where $g_{1\delta }$ is generated by the non-empty subsets $O_1,O_2,O_3,$satisfying the following conditions:

  $O_1 \cap O_2 = \{x_1\}, O_2 \cap O_3 = \{x_2\}$ and $O_3 \cap O_1 = \{x_3\}.$ Moreover, $x_1\not \in O_3,$ $x_2\not \in O_3$ and $x_3\not \in O_2.$

• 2 Let $(X, g_{2\delta })$ be a $2\delta -$generalized topological space on the non-empty set X,  where $g_{2\delta }$ is generated by the disjoint non-empty subsets $\{O_1,O_2\}.$

• 3 Let $(X, g_{3\delta })$ be a $3\delta -$generalized topological space on non-empty set X,  where $g_{3\delta }$ is generated by the non-empty subsets $O_1,O_2,O_3,O_4,$ satisfying the below:

  $O_1 \cap O_2 = \{x_1\},$ $O_2 \cap O_3 = \{x_2\},O_3 \cap O_4 = \{x_3\}$ and $O_4 \cap O_1 = \{x_4\}.$ Moreover, $O_1\cap O_3=\emptyset$ and $O_2\cap O_4=\emptyset .$

$\sigma -$generalized topological spaces which are defined by special generalized topological spaces $(X, g_{i\delta }), i\in \{1,2,3\}$

By using Theorem 32, the $\sigma -$generalized topological spaces can be constructed as follows:

• 1 The $\sigma -$generalized topology $g_{\sigma }$ on $g_{1\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\sigma -$open subset, then $int_{1\delta }(A)\ne \emptyset ,$ and it contains at least one $1\delta -$open subset $O_{i_0 }\in g_{1\delta }.$ Consequently, $int_{1\delta }(A)$ intersects all the elements of $g_{1\delta }.$ Therefore, the family of $\sigma -$open subset of $g_{1\delta }$ consists of each subset of X,  containing at least one of the non-empty elements of $g_{1\delta }.$

  It is clear that X is $\sigma -$open set, but X is $1\delta -$open only if $X = O_1\cup O_2\cup O_3.$

• 2 The $\sigma -$generalized topology $g_{\sigma }$ on $g_{2\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\sigma -$open subset, then $int_{2\delta }(A)\ne \emptyset ,$ and it contains one element of $\{O_1,O_2,O_1\cup O_2\}.$ Therefore, the family of $\sigma -$open subset of $g_{2\delta }$ consists of the following subfamilies:

  • (i) Any subset of X,  containing $O_1 \cup O_2.$

  • (ii) Any subset of X,  containing $O_1$ and does not intersect $O_2.$

  • (iii) Any subset of X,  containing $O_2$ and does not intersect $O_1.$

  It is clear that X is $\sigma -$open set, but X is $2\delta -$open only, if $X = O_1\cup O_2.$

• 3 The $\sigma -$generalized topology $g_{\sigma }$ on $g_{3\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\sigma -$open subset, then $int_{3\delta }(A)\ne \emptyset ,$ and it contains at least one element of $\{O_1,O_2,O_3,O_4\}.$ Therefore, the family of $\sigma -$open subset of $g_{3\delta }$ consists of the following subfamilies:

  • (i) A is $\sigma -$open subset if $int_{3\delta }(A)$ contains at least two elements of $\{O_1,O_2,O_3,O_4\},$ since $int_{3\delta }(A)$ intersects all the elements of $g_{3\delta }.$

  • (ii) If $int_{3\delta }(A)$ contains $O_1$ only (or $O_3$ only), then A is $\sigma -$open subset if it contains $O_1$ and does not intersect $O_3$ (or if it contains $O_3$ and does not intersect $O_1).$

  • (iii) If $int_{3\delta }(A)$ contains $O_2$ only (or $O_4$ only), then A is $\sigma -$open subset if it contains $O_2$ and does not intersect $O_4$ (or if it contains $O_4$ and does not intersect $O_2).$

  It is clear that X is $\sigma -$open set, but X is $3\delta -$open only, if $X = O_1\cup O_2\cup O_3\cup O_4.$

$\alpha -$generalized topological space $(X, g_{\alpha })$

Theorem 35

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta .$ The non-empty elements of the generalized topology $g_{\alpha }$ on X,  which is defined by the map $\alpha = int_{\delta }\circ cl_{\delta }\circ int_{\delta }$ consists of all subsets $A \subset X,$ satisfying the following conditions:

• 1 If $X\not \in g_{\delta }$ and $O\cap int_{\delta }(A)\ne \emptyset ,$ for all $O\in g_{\delta }.$ Then, $A\in g_\alpha$ if:

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta }}O\supset A\supset int_{\delta }(A). \end{aligned}$$

• 2 If $X \in g_{\delta }$ and $O\cap int_{\delta }(A)\ne \emptyset ,$ for all $O\in g_{\delta }.$ Then, $A\in g_\alpha$ if:

  $$\begin{aligned} \alpha (A)=X\supset A\supset int_{\delta }(A). \end{aligned}$$

  This means that $X\in g_\delta \Rightarrow X\in g_\sigma \cap g_\alpha .$

• 3 If $int_{\delta }(A)\ne \emptyset ,$ and

  $$\begin{aligned} \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,\exists U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset ,U\cap O\ne \emptyset }O=\emptyset . \end{aligned}$$

  Then, $A\in g_\alpha$ if:

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset }O\supset A\supset int_{\delta }(A). \end{aligned}$$

• 4 If $int_{\delta }(A)\ne \emptyset ,$ $\bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset }O\supset A,$

  $$\begin{aligned} \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,\exists U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset ,U\cap O\ne \emptyset }O\ne \emptyset \end{aligned}$$

  and

  $$\begin{aligned} A\cap \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,\exists U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset ,U\cap O\ne \emptyset }O=\emptyset . \end{aligned}$$

  Then, $A\in g_\alpha$ if:

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,U\cap O=\emptyset \forall U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset }O\supset A\supset int_{\delta }(A). \end{aligned}$$

Proof

(a):

Let $A \subset X$ such that $int_{\delta }(A)= \emptyset ,$ then $\alpha (A)=\emptyset .$ Hence, $\emptyset \in g_\alpha .$

(b):

Let $A \subset X$ such that $int_{\delta }(A)\ne \emptyset ,$ then:

$$\begin{aligned} \alpha (A)= & {} int_{\delta }(\sigma (A))=int_{\delta }\left\{ \begin{array}{c} X\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad :\quad \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset \\ \\ \bigcap _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }(X-O)\quad :\quad \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O\ne \emptyset \\ \end{array} \right. \\= & {} \left\{ \begin{array}{c} \bigcup _{O\in g_{\delta },int_{\delta }\cap O\ne \emptyset }O\quad \quad \quad \quad \quad \quad \quad : \quad X\not \in g_\delta , \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset \\ \\ X\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad :\quad X\in g_\delta ,\bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset \\ \\ \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,U\cap O=\emptyset \forall U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset }O\quad :\quad \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O\ne \emptyset \\ \end{array} \right. \end{aligned}$$

The subset A is $\alpha -$open subset, if $\alpha (A) \supset A.$ This statement is valid in the following cases:

• 1 For every $A\subset X,$ for which $int_{\delta }(A)\ne \emptyset ,$ $X\not \in g_\delta , \bigcup _{O\in g_{\delta },int_{\delta }\cap O=\emptyset }O=\emptyset$ (i.e. each $\delta -$open subset O intersects $int_{\delta }(A)),$ and

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta }}O\supset A\supset int_{\delta }(A). \end{aligned}$$

  (Therefore, X is not $\alpha -$open set if it is not $\delta -$open set).

• 2 Let $X\in g_\delta ,$ then A is $\alpha -$open subset for all $A \subset X,$ since $\alpha (A) = X \supset A \supset int_{\delta }(A).$ Therefore, X is $\alpha -$open set if it is $\delta -$open set.

• 3 For every $A\subset X,$ for which $int_{\delta }(A)\ne \emptyset ,$ and

  $$\begin{aligned} \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,\exists U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset ,U\cap O\ne \emptyset }O=\emptyset . \end{aligned}$$

  (i.e. each $\delta -$open subset O,  intersecting $int_{\delta }(A)$ does not intersect any $\delta -$open subset U,  for which $U \cap int_{\delta }(A)=\emptyset ).$

  Then, $A\in g_\alpha$ if:

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset }O\supset A\supset int_{\delta }(A). \end{aligned}$$

• 4 For every $A\subset X,$ for which $int_{\delta }(A)\ne \emptyset ,$ and

  $$\begin{aligned} \bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,\exists U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset ,U\cap O\ne \emptyset }O\ne \emptyset . \end{aligned}$$

  (i.e. each $\delta -$open subset O,  intersecting $int_{\delta }(A),$ intersects some $\delta -$open subset U,  for which $U \cap int_{\delta }(A)=\emptyset ).$

  Then, $A\in g_\alpha$ if $int_{\delta }(A)\ne \emptyset$ and

  $$\begin{aligned} \alpha (A)=\bigcup _{O\in g_{\delta },O\cap int_{\delta }(A)\ne \emptyset ,U\cap O=\emptyset \forall U\in g_\delta ,U\cap int_{\delta }(A)= \emptyset }O\supset A\supset int_{\delta }(A). \end{aligned}$$

$\square$

Remark 36

The map $\alpha = int_{\delta }\circ cl_{\delta }\circ int_{\delta }$ is called controlled by the generalized topology $g_\delta .$ It can be denoted by $\alpha _\delta .$

$\alpha -$generalized topological spaces which are defined by special generalized topological spaces $(X, g_{i\delta }), i\in \{1,2,3\}$

By using Theorem 35, the $\alpha -$generalized topological spaces can be constructed as follows:

• 1 The $\alpha -$generalized topology $g_{\alpha }$ on $g_{1\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\alpha -$open subset, then $int_{1\delta }(A)\ne \emptyset ,$ then it contains at least one $1\delta -$open subset $O_{i_0 }\in g_{1\delta }.$ Consequently, $int_{1\delta }(A)$ intersects all the elements of $g_{1\delta }.$ Therefore, the family of $\alpha -$open subset of $g_{1\delta }$ consists of:

  • (i) Each subset A of X,  containing at least one of the non-empty elements of $g_{1\delta }$ (if X is $1\delta -$open set).

  • (ii) Each subset A,  containing at least one $1\delta -$open subset and $A \subset O_1\cup O_2\cup O_3$ (if X is not $1\delta -$open set).

• 2 The $\alpha -$generalized topology $g_{\alpha }$ on $g_{2\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\sigma -$open subset, then $int_{2\delta }(A)\ne \emptyset ,$ then it contains one element of $\{O_1,O_2,O_1\cup O_2\}.$ Therefore, the family of $\alpha -$open subsets of $g_{2\delta }$ consists of the following subfamilies:

  • (i) $A = O_1 \cup O_2,$ if X is not $1\delta -$open set.

  • (ii) Any subset A of X,  containing $O_1\cup O_2,$ if X is $2\delta -$open set.

  • (iii) The subset A of X,  if $A=O_1$ or $A=O_2.$

  Therefore, $g_\alpha = g_{2\delta },$ if X is not $2\delta -$open set. This result is true for every $g_\delta$ generated by a family of disjoint subsets, when X is not $\delta -$open set.

• 3 The $\alpha -$generalized topology $g_\alpha$ on $g_{3\delta }$ is defined as follows:

  Let $A \ne \emptyset$ be $\alpha -$open subset, then $int_{3\delta }(A)\ne \emptyset ,$ then it contains at least one element of $\{O_1,O_2,O_3,O_4\}.$ Therefore, the family of $\alpha -$open subset of $g_{3\delta }$ consists of the following subfamilies:

  • (i) The subset A of X,  if $int_{3\delta }(A)$ contains at least two elements of $\{O_1,O_2,O_3,O_4\},$ since $int_{3\delta }(A)$ intersects all the elements of $g_{3\delta }$ if: X is $3\delta -$open set, or $A\subset O_1 \cup O_2 \cup O_3 \cup O_4$ and X is not $3\delta -$open set.

  • (ii) The subset A of X,  if $A=O_1$ or $A=O_2$ or $A=O_3$ or $A=O_4.$

$\pi -$generalized topological space $(X, g_\pi )$

Notation 37

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta .$ Each subset $A \subset X$ divides the elements of the generalized topology $g_\delta$ into two classes:

$$\begin{aligned} \triangle _A=\{O\in g_\delta : A\cap O\ne \emptyset \},\quad \nabla _A=\{U\in g_\delta : A\cap U= \emptyset \}. \end{aligned}$$

For each $O\in \triangle _A,$ we define

$$\begin{aligned} U_O=\bigcup _{U\in \nabla _A,U\cap O\ne \emptyset }U. \end{aligned}$$

And

$$\begin{aligned} \varepsilon _A=\left\{ x\in A: x\not \in \bigcup _{O\in g_\delta }O\right\} . \end{aligned}$$

It is clear that the family $\{\triangle _A,\nabla _A\},$ for all $A\subset X$ forms a partition for the $\delta -$generalized topology $g_\delta$ on X. Moreover, $(\triangle _A=\emptyset \Rightarrow \nabla _A=g_\delta )$ and $(\nabla _A=\emptyset \Rightarrow \triangle _A=g_\delta ).$

Theorem 38

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta .$ The non-empty elements of the generalized topology $g_{\pi }$ on X ( the family of all $\pi -$preopen sets), which is defined by the monotonic map $\pi = int_\delta \circ cl_\delta$ consists of all non-empty subsets $A \subset X,$ satisfying the following conditions:

• (i) $\nabla _A=\emptyset$ and X is $\delta -$open set.

• (ii) If $\nabla _A =\emptyset ,$ and X is not $\delta -$open set and $A\subset \bigcup _{O\in g_\delta }O.$

• (iii) If $\nabla _A \ne \emptyset ,$ $\triangle _A \ne \emptyset ,$ and $A\subset \bigcup _{O\in \triangle _A,O\cap U=\emptyset ; U\in \nabla _A}O.$

Proof

Consider the action of the map $\pi$ on the subset A of X : 

Let $A =\emptyset ,$ then $\pi (A)=\emptyset .$ Hence, $\emptyset \in g_\pi .$

Let $A \ne \emptyset ,$ then we get

$$\begin{aligned} \pi (A)= & {} int_{\delta }(cl_\delta (A))=int_{\delta }\left\{ \begin{array}{c} \bigcap _{U\in g_\delta }(X-U)\quad :\quad \quad \quad \triangle _A=\emptyset .\\ \\ \bigcap _{U\in \nabla _A}(X-U)\quad \quad :\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset .\\ \\ X\quad \quad \quad \quad \quad \quad :\quad \quad \quad \nabla _A=\emptyset ,. \end{array} \right. \\= & {} \left\{ \begin{array}{c} \emptyset \quad \quad \quad :\quad \triangle _A=\emptyset \quad or\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,U_O\ne \emptyset ;O\in \triangle _A.\\ \\ \bigcup _{U_O=\emptyset ,O\in g_\delta }O \quad \quad \quad \quad \quad \quad \quad \quad \quad :\quad \quad \quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,\exists O\in \triangle _A,U_O= \emptyset .\\ \\ X\quad \quad \quad \quad \quad :\quad \quad \quad \nabla _A=\emptyset ,X\in g_\delta .\\ \\ \bigcup _{O\in g_\delta }O\quad \quad :\quad \quad \quad \nabla _A=\emptyset , X\not \in g_\delta . \end{array} \right. \end{aligned}$$

The nonempty subset A is $\pi -$open subset, if $\pi (A)\supset (A).$ Therefore, the subset A is $\pi -$open subset in the following cases:

(i):

If $\nabla _A=\emptyset$ and $X\in g_\delta .$ Then, A is $\pi -$open subset, since $\pi (A)=X\supset A.$

(ii):

If $\nabla _A =\emptyset ,$ and X is not $\delta -$open set. Then, A is $\pi -$open subset if $A\subset \bigcup _{O\in g_\delta }O.$

(iii):

If $\triangle _A\ne \emptyset$ and $\nabla _A\ne \emptyset ,$ then the nonempty subset A is $\pi -$open subset, if:

$$\begin{aligned} \pi (A)=\bigcup _{U_O=\emptyset ,O\in g_\delta }O=\bigcup _{O\in \triangle _A,O\cap U=\emptyset ; U\in \nabla _A}O\supset A. \end{aligned}$$

$\square$

$\pi -$generalized topological spaces which are defined by special generalized topological spaces $(X, g_{i\delta }), i\in \{1,2,3\}$

By using Theorem 38, the $\pi -$generalized topological spaces can be constructed as follows:

• 1 The $\pi -$generalized topology $g_\pi$ on $g_{1\delta }$ is defined as follows:

  A is $\pi -$open subset:

  • (i) If A intersects each of the subsets $\{O_1,O_2,O_3\}$ and X is $1\delta -$open set.

  • (ii) If A intersects each of the subsets $\{O_1,O_2,O_3\}$ and X is not $1\delta -$open set, then $A\subset O_1\cup O_2\cup O_3.$

• 2 The $\pi -$generalized topology $g_\pi$ on $g_{2\delta }$ is defined as follows:

  A is $\pi -$open subset in the following cases:

  • (i) $A \cap O_1 \ne \emptyset , A \cap O_2 =\emptyset$ and $\pi (A) = O_1 \supset A.$

  • (ii) $A \cap O_2 \ne \emptyset , A \cap O_1 =\emptyset$ and $\pi (A) = O_2 \supset A.$

  • (iii) $A \cap O_1 \ne \emptyset , A \cap O_2 \ne \emptyset$ and $\pi (A) = O_1\cup O_2 \supset A.$

• 3 The $\pi -$generalized topology $g_\pi$ on $g_{3\delta }$ is defined as follows:

  A is $\pi -$open subset in the following cases:

  • (i) If it is included in $O_1$ and intersects $O_2,O_4.$

  • (ii) If it is included in $O_2$ and intersects $O_1,O_3.$

  • (iii) If it is included in $O_3$ and intersects $O_2,O_4.$

  • (iv) If it is included in $O_4$ and intersects $O_3,O_1.$

  • (v) If it is included in $O_1\cup O_2\cup O_3\cup O_4$ and intersects all the elements $\{O_1,O_2,O_3,O_4\}.$

Remark 39

Consider the following case: If $\nabla _A\ne \emptyset$ and $U_O \ne \emptyset ,$ for some $O \in \triangle _{1A}\subset \triangle _A.$ It follows that A is not $\pi -$open subset. Since if $U_O \ne \emptyset ,$ for some $O \in \triangle _{1A}\subset \triangle _A,$ then the points of A in O are not included in $\pi (A)=\bigcup _{U_O=\emptyset ,O\in \triangle _A}O.$ Then, A is not included in $\pi (A)$ and is not $\pi -$open subset.

$\beta -$generalized topological space $(X, g_\beta )$

Theorem 40

Let $g_{\delta }$ be a given generalized topology on X,  which is generated by the monotonic map $\delta .$ The non-empty elements of the generalized topology $g_{\beta }$ on X,  which is defined by the monotonic map $\beta = cl_\delta \circ int_\delta \circ cl_\delta$ consists of all non-empty subsets $A \subset X,$ satisfying the following conditions:

• 1 If for some $O_0 \in g_\delta ,$ A intersects $O_0$ and A intersects each $O \in g_\delta ,$ intersecting $O_0.$ Moreover, $A \subset \bigcup _{U\cap A\ne \emptyset ,U\in g_\delta }(X-U).$

• 2 If A intersects every $O \in g_\delta .$

Proof

Consider the action of the map $\beta$ on the subset A of X : 

Let $A =\emptyset ,$ then $\beta (A)=\emptyset .$ Hence, $\emptyset \in g_\beta .$

Let $A \ne \emptyset ,$ then we get:

$$\begin{aligned} \beta (A)= & {} cl_{\delta }(\pi (A))=cl_{\delta }\left\{ \begin{array}{c} \emptyset \quad :\quad \triangle _A=\emptyset \quad or\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,U_O\ne \emptyset ;O\in \triangle _A.\\ \\ \bigcap _{U\in \nabla _A}(X-O) \quad :\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,\exists O\in \triangle _A,U_O= \emptyset .\\ \\ X\quad \quad \quad \quad \quad :\quad \nabla _A=\emptyset ,X\in g_\delta .\\ \\ \bigcup _{O\in g_\delta }O\quad \quad :\quad \nabla _A=\emptyset , X\not \in g_\delta . \end{array} \right. \\= & {} \left\{ \begin{array}{c} \emptyset \quad :\quad \triangle _A=\emptyset \quad or\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,U_O\ne \emptyset ;O\in \triangle _A.\\ \\ \bigcup _{U_O=\emptyset ,O\in g_\delta }O \quad :\quad \triangle _A\ne \emptyset , \nabla _A\ne \emptyset ,\exists O\in \triangle _A,U_O= \emptyset .\\ \\ X\quad :\quad \nabla _A=\emptyset . \end{array} \right. \end{aligned}$$

The nonempty subset A is $\beta -$open subset, if $\beta (A)\supset (A).$ Therefore, the subset A is $\beta -$open subset in the following two cases:

• 1 If for some $O_0 \in g_\delta ,$ A intersects $O_0$ and A intersects each $O \in g_\delta ,$ intersecting $O_0.$ Moreover, $A \subset \bigcup _{U\cap A\ne \emptyset ,U\in g_\delta }(X-U).$

• 2 If A intersects every $O \in g_\delta .$

$\square$

Remark 41

The nonempty subset A is $\beta -$open subset, if A intersects every $O \in g_\delta .$ It follows that X is $\beta -$open set.

$\beta -$generalized topological spaces which are defined by special generalized topological spaces $(X, g_{i\delta }), i\in \{1,2,3\}$

By using Theorem 40, the $\beta -$generalized topological spaces can be constructed as follows:

• 1 The $\beta -$generalized topology $g_\beta$ on $g_{1\delta }$ is defined as follows:

  A is $\beta -$open subset:

  • (i) If A intersects each of the subsets $\{O_1,O_2,O_3\}$ and X is $1\delta -$open set.

  • (ii) If A intersects each of the subsets $\{O_1,O_2,O_3\}$ and X is not $1\delta -$open set, then $A\subset O_1\cup O_2\cup O_3.$

• 2 The $\beta -$generalized topology $g_\beta$ on $g_{2\delta }$ is defined as follows:

  A is $\beta -$open subset in the following cases:

  • (i) $A \cap O_1 \ne \emptyset , A \cap O_2 =\emptyset .$ Therefore, each subset of $O_1$ is $\beta -$open subset.

  • (ii) $A \cap O_2 \ne \emptyset , A \cap O_1 =\emptyset .$ Therefore, each subset of $O_2$ is $\beta -$open subset.

  • (iii) A intersects $O_1 \cup O_2.$

• 3 The $\beta -$generalized topology $g_\beta$ on $g_{3\delta }$ is defined as follows:

  A is $\beta -$open subset in the following cases:

  • (i) If it intersects three subsets only of $\{O_1,O_2,O_3,O_4\},$ and $A \subset X- O_i$ or $A \cap O_i=\emptyset ,$ for only one element i in the family $\{1,2,3,4\}.$

  • (ii) If it intersects all the elements of $g_{3\delta }.$

Properties of the composition binary operation on the monotonic functions $int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta$

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generated by $\delta \in \Gamma _\delta .$ Then, for any $A \subset X,$ the definitions of the monotonic maps $\sigma ,\alpha ,\pi$ and $\beta ,$ implies the following relations:

• 1 $\alpha (A)=int_\delta (\sigma (A)).$

• 2 $\beta (A)=cl_\delta (\pi (A)).$

• 3 $int_\delta (A)\subset int_\delta (\sigma (A)),\quad cl_\delta (\pi (A))\subset cl_\delta (A).$

• 4 $\sigma (A)=\sigma (int_\delta (A)),\quad \pi (A)=\pi (cl_\delta (A)).$

Theorem 42

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generated by $\delta \in \Gamma _\delta .$ Then, for any $A \subset X,$ the following conditions are satisfied:

• 1 $\pi (X-A)=X-\sigma (A).$

• 2 $\sigma (X-A)=X-\pi (A).$

• 3 $\alpha (X-A)=X-\beta (A).$

• 4 $\beta (X-A)=X-\alpha (A).$

Proof

The proof is straightforward, using the relations: $int_\delta (X - A) = X - cl_\delta (A)$ and $cl_\delta (X - A) = X - int_\delta (A).$ $\square$

Theorem 43

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generating by $\delta \in \Gamma _\delta .$ Then, for any $A \subset X,$ and $\gamma \in \{\sigma ,\alpha ,\pi ,\beta \},$ it follows that $\gamma (A)=\gamma ^2(A).$

Proof

• 1
  $$\begin{aligned} \sigma (A)= & {} \bigcap _{V\in g_\delta }\left\{ X-V: int_\delta (A)\cap V=\emptyset \right\} ,\\ \sigma ^2(A)= & {} \bigcap _{V\in g_\delta }\left\{ X-V: int_\delta (\sigma (A))\cap V=\emptyset \right\} . \end{aligned}$$

  Since $int_\delta (A)\subset \sigma (A),$ then $int_\delta (A)\subset int_\delta (\sigma (A)),$ which implies that for any $V \in g_\delta ,$ if $int_\delta (\sigma (A))\cap V = \emptyset ,$ then $int_\delta (A)\cap V = \emptyset .$ Therefore $\sigma ^2(A)\subset \sigma (A).$

  Conversely, let $x \in \sigma (A)$ and $x \not \in \sigma ^2(A),$ then there exists $V \in g_\delta$ such that $x \in V$ and $int_\delta (\sigma (A))\cap V = \emptyset .$ Since $int_\delta (A)\subset \sigma (A),$ then $int_\delta (A)\cap V = \emptyset ,$ which contradicts $x \in \sigma (A).$ Hence $x \in \sigma ^2(A),$ and $\sigma ^2(A)\supset \sigma (A),$ which implies that $\sigma ^2(A)=\sigma (A).$

• 2 $\pi ^2(A) = \pi (\pi (A)) = \pi (X-\sigma (X-A)) = X-\sigma (\sigma (X-A)) = X-\sigma ^2(X-A) = X -\sigma (X - A) = \pi (A).$

• 3 $\alpha ^2(A) = \alpha (\alpha (A)) = \alpha (int_\delta (\sigma (A)))=int_\delta (\sigma (int_\delta (\sigma (A))))=int_\delta (\sigma (\sigma (A)))= int_\delta (\sigma (A))=\alpha (A).$

• 4 $\beta ^2(A) = \beta (\beta (A)) = \beta (cl_\delta (\pi (A)))=cl_\delta (\pi (cl_\delta (\pi (A))))=cl_\delta (\pi (\pi (A)))= cl_\delta (\pi (A))=\beta (A).$

$\square$

Theorem 44

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generated by $\delta \in \Gamma _\delta .$

The composition operation $\circ$ on the set of all functions is a binary operation on the family of monotonic maps $\{int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta \}.$

Proof

One can construct the following table easily.

$\circ$	$int_\delta$	$cl_\delta$	$\sigma$	$\pi$	$\alpha$	$\beta$
$int_\delta$	$int_\delta$	$\pi$	$\alpha$	$\pi$	$\alpha$	$\pi$
$cl_\delta$	$\sigma$	$cl_\delta$	$\sigma$	$\beta$	$\sigma$	$\beta$
$\sigma$	$\sigma$	$\beta$	$\sigma$	$\beta$	$\sigma$	$\beta$
$\pi$	$\alpha$	$\pi$	$\alpha$	$\pi$	$\alpha$	$\pi$
$\alpha$	$\alpha$	$\pi$	$\alpha$	$\pi$	$\alpha$	$\pi$
$\beta$	$\sigma$	$\beta$	$\sigma$	$\beta$	$\sigma$	$\beta$

Therefore, the composition operation $\circ$ is a binary operation on the family of monotonic maps $\{int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta \}.$ $\square$

Corollary 45

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generated by $\delta \in \Gamma _\delta .$

For any $\gamma \in \{int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta \},$ it follows that:

• 1 $\beta \circ \gamma =\sigma \circ \gamma .$

• 2 $\pi \circ \gamma =\alpha \circ \gamma .$

• 3 $\gamma \circ \sigma =\gamma \circ \alpha .$

• 4 $\gamma \circ \pi =\gamma \circ \beta .$

• 5 $int_\delta \circ \gamma \circ int_\delta =\alpha ; \gamma \ne int_\delta .$

• 6 $cl_\delta \circ \gamma \circ cl_\delta =\beta ; \gamma \ne cl_\delta .$

• 7 $\sigma \circ \gamma \circ \sigma =\sigma .$

• 8 $\pi \circ \gamma \circ \pi =\pi .$

• 9 $\alpha \circ \gamma \circ \alpha =\alpha .$

• 10 $\beta \circ \gamma \circ \beta =\beta .$

Proof

The proof can be constructed from the above table in Theorem 44. $\square$

The relation between the $\gamma -$generalized topological spaces, where $\gamma \in \{int_\delta , cl_\delta , \sigma ,\alpha ,\pi ,\beta \},$ can be studied in the following theorem.

Theorem 46

Let $(X, g_\delta )$ be any $\delta -$generalized topological space, generated by $\delta \in \Gamma _\delta ,$ and $(X,D_X)$ be the discrete space, then:

• 1 $g_{int_\delta }=g_\delta \subset g_\alpha \subset g_\sigma \subset g_\beta \subset g_{cl_\delta }=D_X.$

• 2 $g_\alpha \subset g_\pi \subset g_\beta .$

Proof

The proof is easy, since for all $A\subset X,$ it follows that:

(a):

$int_\delta (A)\subset A\subset cl_\delta (A).$

(b):

$int_\delta (A)\subset \alpha (A)\subset \sigma (A)\subset \beta (A)\subset cl_\delta (A).$

(c):

$\alpha (A)\subset \pi (A)\subset \beta (A).$

$\square$

Conclusion

In this paper, the family of monotonic functions $\Gamma (X)$ have the following properties:

• 1 The monotonic map $int_{\delta }\in \Gamma (X)$ is the smallest monotonic map in the equivalence class of all monotonic maps $\Gamma _{\delta },$ which is defined by the same generalized topology $\delta .$ Moreover, the monotonic map $cl_{\delta }\in \Gamma (X)$ is the largest monotonic map in the associated equivalence class $\Gamma ^{\delta }$ to the class $\Gamma _{\delta }.$

• 2 Using the invariant systems $\prec G,{\mathcal {H}}_G\succ ,$ it is shown that there exists a one-to-one correspondence between the family of $G-$continuous functions and the family of $f(G)-$continuous functions: $h\leftrightarrow \gamma _{f^{-1}} \circ h \circ \gamma _f; h\in {\mathcal {H}}_G;$ for any one-to-one correspondence $f\in X^X.$

• 3 The family of monotonic maps $\{int_{\delta },cl_{\delta },\sigma ,\alpha , \pi ,\beta \},$ for every $\delta -$Cs$\acute{a}$zs$\acute{a}$r generalized topological space $(X, g_{\delta })$ is closed under the composition operation and has interesting relations (see article 4).