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Generalized topology and the family of monotonic maps \(\Gamma (X)\)
Journal of the Egyptian Mathematical Society volume 31, Article number: 3 (2023)
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 \(A\subset X:\)
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
Definition 5
To each equivalence class \(\Gamma _{\delta },\) there exists an associated class
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 \(B\subset X:\)
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 \(A \in P(X)\)
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:
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
\(\square\)
For each \(G\subset P(X),\) define the family \(\mathbf {\Omega }_G\subset P(\Phi )\) as:
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:
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:
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 \(f\in X^X,\)
\(\square\)
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
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:
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
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
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
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
The four steps (shown above) which constructs the map \(\gamma\) will be denoted by the following notation:
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,
Therefore,
The map \(\gamma ^2\) can be constructed using the following symbols:
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,
Consequently,
and
and
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:
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,
Then,
And
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:
For each \(O\in \triangle _A,\) we define
And
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
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:
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).
Availibility of data and materials
None.
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Kamel, G.A., Dib, K.A. Generalized topology and the family of monotonic maps \(\Gamma (X)\). J Egypt Math Soc 31, 3 (2023). https://doi.org/10.1186/s42787-023-00162-5
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DOI: https://doi.org/10.1186/s42787-023-00162-5
Keywords
- Family of monotonic maps \(\Gamma (X)\)
- Cs\(\acute{a}\)zs\(\acute{a}\)r generalized topological space
- Interesting monotonic maps \(\{int_{\delta }, cl_{\delta }, \sigma , \alpha , \pi , \beta \}\)
Mathematics Subject Classification
- Primary 54C50
- Secondary 54A05