Continuous Mapping of Covering Approximation Spaces and Topologies Induced by Arbitrary Covering Relations

Abstract

In rough set theory, there are many covering approximation spaces, so how to classify covering approximation spaces has become a hot issue. In this paper, we propose the concepts of a covering approximation $T_1$-space, F-symmetry, covering rough continuous mapping, and covering rough homeomorphism mapping, and we obtain some interesting results. We have used the above definitions and results to classify covering approximation spaces. Finally, we find a new method for constructing topologies, obtain some properties, and provide an example to illustrate our method’s similarities and differences with other construction methods.

1. Introduction

With developments in science and technology, the task of data classification has become more important in many fields, including medical diagnosis, decision-making, data mining, and others. Data exist in various spaces, so information processing must rely on the relevant properties of that space. Therefore, how to obtain the properties of a space has become an urgent problem. To address this issue, fuzzy set theory and granular computing were proposed by L. A. Zadeh in 1965 and 1996 [1,2]. They have been used in various fields, including machine learning, approximation reasoning, and knowledge discovery. Although they can be used to deal with uncertainties and incompleteness, if we want to use them, we need to add human factors, which can lead to inaccurate results. Classical rough set theory is a suitable tool for dealing with incomplete and imprecise information systems and is also an effective tool for dealing with uncertain knowledge. Therefore, it has been successfully applied to many fields [3,4,5,6,7]. Lower and upper approximation operators are defined by classical rough sets based on the equivalence relation [8,9,10,11,12], which exhibits many limitations because the equivalence relation is too strong. In order to remove these limitations, Zakowski used the covering relation to replace the equivalence relation and defined the concept of a covering rough set [13]. Many scholars have carried out a lot of work in this field. For example, Zhang Y.L., et al. studied the invariance of covering rough sets using compatible mappings [14], and Li J.J. studied covering generalized approximation spaces using topological methods [15,16]. Wang P., et al. studied the necessary and sufficient conditions for the covering upper approximation operator to become a topological closure operator and investigated its membership functions [17,18]. Zhang W.X., et al. studied the rough set of general relations, and so on [19,20,21,22]. There are many ways to generate approximation operators from the above. Due to the large coverage space, it is unrealistic to study its properties individually. Whether we can classify the covering approximation spaces and study the nature of the class is a question worth thinking about.

A topological structure is one of the most crucial structures in mathematics, offering mathematical tools to deal with information systems and rough sets [22]. Using the topological method to study covering approximation spaces is a highly effective approach. For example, Yang L.Y., et al. defined topologies using a lower approximation operator. Z. Zhao proposed topologies induced by coverings [23] .Yu H., et al. obtained topologies using lower and upper covering approximation operators [24], and so on. Any topological concepts and symbols that we did not mention in this paper can be found in [25].

Continuous mapping plays a key role in general topology and other fields. Therefore, we can investigate the relationship between two topological spaces using continuous mapping, such as homeomorphism, separation, connectedness, and so on. Covering approximation spaces are an important part of generalized approximation spaces. To study the properties of covering approximation spaces, many scholars have studied the properties of covering approximation spaces by means of various relations. The important topological concept of continuity is not used. We define covering rough continuous mapping and covering rough homeomorphism mapping, obtain a covering approximation $T_1$-space, and provide a classification method. In addition to this, we also propose a method to construct topologies. With the help of this method, we can provide a unified case for using a covering to construct a topology as well as relations to construct topologies. Compared with the topology we constructed, this topology becomes a special case. In other words, we generalize the method of constructing a topology.

The organization of this paper is as follows. In Section 1, the background is introduced. In Section 2, several preliminary concepts in rough approximation space are briefly recalled. In Section 3, we propose some new concepts regarding covering rough continuous mapping F-symmetry and covering rough homeomorphism mapping, and we obtain many interesting results. We also establish a classification method for covering approximation spaces using these definitions. In Section 4, by analyzing the properties of covering spaces using covering rough continuous mapping and covering rough homeomorphism mapping, we define the separation property of covering approximation spaces, obtain the $T_1$ covering approximation space, distinguish covering approximation spaces using $T_1$ separation, and then establish another classification method for covering approximation spaces. In Section 5, we successfully apply a rough set to construct a new topological space, obtaining a new topology induced by ∗. We also compare it with other topologies, find that this topology is finer than others, and research its properties. Section 6 concludes this paper.

2. Preliminaries

Definition 1 

(Covering [13]). Let U be a universe of discourse and $\mathcal{C}$ be a family of subsets of U; no subsets in $\mathcal{C}$ are empty. If $\cup \mathcal{C}=U$, then $\mathcal{C}$ is called a covering of U.

Let U be a non-empty set and ∗ be an arbitrary relation on U. Then, $(U,\ast )$ is called an approximation space. For any $X\subseteq U$, the lower and upper approximations of X are defined as follows:

$$\begin{array}{c}{\displaystyle \underline{R}}\left(X\right)=\{x\in U:\ast \left(x\right)\subseteq X\}\hfill \\ \overline{R}\left(X\right)=\{x\in U:\ast \left(x\right)\cap X\ne Ø\}.\hfill \end{array}$$

If ∗ is an equivalence relation on U, then $\ast \left(x\right)={\left[x\right]}_{\ast }$; if ∗ is a binary relation on U, then $\ast \left(x\right)={\ast }_s\left(x\right)$, where ${\ast }_s\left(x\right)=\{y\in U:(x,y)\in \ast \}$; and if ∗ is a covering relation on U, then $\ast \left(x\right)=\{C\in \mathcal{C}:x\in C\}$, where $\mathcal{C}$ is a cover of U. We have the following:

$$\begin{array}{c}{\displaystyle \underline{D}}\left(X\right)=\cup \{{\ast }_s\left(x\right):{\ast }_s\left(x\right)\subseteq X\}\hfill \\ \overline{D}\left(X\right)=\cup \{{\ast }_s\left(x\right):{\ast }_s\left(x\right)\cap X\ne Ø\}.\hfill \end{array}$$

A binary relation ∗ on U is said to be reflexive if for any $x\in U$, $(x,x)\in \ast$; ∗ is said to be symmetric if for any $x,y\in U$, $(x,y)\in \ast$ implies $(y,x)\in \ast$; and ∗ is called a Euclidean relation if for any $x,y,z\in U$, $(x,y)\in \ast$ and $(x,z)\in \ast$ imply $(y,z)\in \ast$.

Definition 2 

(Covering approximation space [13]). Let U be a universe of discourse and $\mathcal{C}$ be a covering of U. Then, we call U together with covering $\mathcal{C}$ a covering approximation space, denoted by $(U,\mathcal{C})$.

Definition 3. 

Let $(U,\mathcal{C})$ be a covering approximation space. For any $X\subseteq U$, the operators are defined as follows:

$$\begin{array}{c}{\displaystyle \underline{apr}}\left(X\right)=\{y\in U:N\left(y\right)\subseteq X,N\left(y\right)=\cap C_i,\forall C_i\in \mathcal{C},y\in C_i\}\hfill \\ \overline{apr}\left(X\right)=\{y\in U:N\left(y\right)\cap X\ne Ø,N\left(y\right)=\cap C_i,\forall C_i\in \mathcal{C},y\in C_i\}\hfill \end{array}$$

Lemma 1. 

Let X and Y be two sets and $f:X\to Y$ be a mapping; then, we have the following:

(1) 

For any $A\subseteq X$, we have $A\subseteq f^{-1}\left(f\left(A\right)\right)$;

(2) 

For any $B\subseteq Y$, we have $f\left(f^{-1}\left(B\right)\right)\subseteq B$; if f is a surjection, then there is $f\left(f^{-1}\left(B\right)\right)=B$;

(3) 

For any $A\subseteq X$ and $B\subseteq Y$, there is $f\left(A\right)\subseteq B$ if and only if $A\subseteq f^{-1}\left(B\right)$.

3. Covering Rough Continuous Mapping and Covering Rough Homeomorphism Mapping

It is well known that a covering approximation space is the most important generalized approximation space. Many scholars have defined many covering approximation operators. It is very important to distinguish these operators. In this section, we propose a covering rough continuous mapping and a covering rough homeomorphism mapping using the upper operator $\overline{apr}$ and obtain some properties of the covering rough continuous mapping and the covering rough homeomorphism mapping.

Lemma 2. 

Let $(U,\mathcal{C})$ be a covering approximation space. For any $X\subseteq U$ and $X_i\subseteq U(i\in I)$, we have the following:

(1) 

$\overline{apr}\left(X^c\right)={\left({\displaystyle \underline{apr}}\left(X\right)\right)}^c$;

(2) 

${\displaystyle \underline{apr}}\left({\bigcap }_{i\in I}{X}_i\right)={\bigcap }_{i\in I}{\displaystyle \underline{apr}}\left(X_i\right)$;

(3) 

$\overline{apr}\left({\bigcup }_{i\in I}{X}_i\right)={\bigcup }_{i\in I}\overline{apr}\left(X_i\right)$;

(4) 

$\overline{apr}\left(X\right)={\bigcup }_{x\in X}\overline{apr}\left(\left\{x\right\}\right)$, where $X^c$ represents the complement of X in U.

Proof. 

(1) For any $y\in \overline{apr}\left(X^c\right)$, by Definition 3 , we have $N\left(y\right)\bigcap (U\backslash X)\ne Ø$; then, $N\left(y\right)⊈X$. Hence, $y\notin {\displaystyle \underline{apr}}\left(X\right)$ by Definition 3, and it follows that $y\in {\left({\displaystyle \underline{apr}}\left(X\right)\right)}^c$, which implies $\overline{apr}\left(X^c\right)\subseteq {\left(\overline{apr}\left(X\right)\right)}^c$.

We only need to prove the converse. For any $x\in {\left({\displaystyle \underline{apr}}\left(X\right)\right)}^c$, we have $x\notin {\displaystyle \underline{apr}}\left(X\right)$. By Definition 3, we can obtain $N\left(x\right)⊈X$ and claim that $N\left(x\right)\bigcap (U\backslash X)\ne Ø$. If $N\left(x\right)\bigcap (U\backslash X)=Ø$, then $N\left(x\right)\subseteq X$. It is a contradiction to $N\left(x\right)⊈X$. In other words, $N\left(x\right)\bigcap X^c\ne Ø$. From Definition 3, we have $x\in \overline{apr}\left(X^c\right)$. It follows that $\overline{apr}\left(X^c\right)={\left({\displaystyle \underline{apr}}\left(X\right)\right)}^c$.

(2) For any $i\in I$, since ${\bigcap }_{i\in I}{X}_i\subseteq X_i$, then ${\displaystyle \underline{apr}}\left({\bigcap }_{i\in I}{X}_i\right)$$\subseteq {\displaystyle \underline{apr}}\left(X_i\right)$; thus, ${\displaystyle \underline{apr}}\left({\bigcap }_{i\in I}{X}_i\right)\subseteq {\bigcap }_{i\in I}{\displaystyle \underline{apr}}\left(X_i\right)$, and we shall prove the converse.

For any $x\in {\bigcap }_{i\in I}{\displaystyle \underline{apr}}\left(X_i\right)$, we have $x\in {\displaystyle \underline{apr}}\left(X_i\right)$ for any $i\in I$. According to Definition 3, we can obtain $N\left(x\right)\subseteq X_i$. By the arbitrariness of i, we have $N\left(x\right)\subseteq {\bigcap }_{i\in I}{X}_i$. Thus, we have $x\in {\displaystyle \underline{apr}}\left({\bigcap }_{i\in I}{X}_i\right)$. From the above, we know that ${\displaystyle \underline{apr}}\left({\bigcap }_{i\in I}{X}_i\right)={\bigcap }_{i\in I}{\displaystyle \underline{apr}}\left(X_i\right)$.

(3) The proof is similar to (2).

(4) It is easy to prove that ${\bigcup }_{x\in X}\overline{apr}\left(\left\{x\right\}\right)\subseteq \overline{apr}\left(X\right)$ by (2); we only need to prove $\overline{apr}\left(X\right)\subseteq {\bigcup }_{x\in X}\overline{apr}\left(\left\{x\right\}\right)$. For any $y\in \overline{apr}\left(X\right)$, according to Definition 3, we obtain $N\left(y\right)\bigcap X\ne Ø$. Thus, there exists $x_0\in N\left(y\right)\bigcap X$ such that $N\left(y\right)\bigcap \left\{x_0\right\}\ne Ø$. Therefore, $y\in \overline{apr}\left(\left\{x_0\right\}\right)$. By the arbitrariness of y, we can obtain $\overline{apr}\left(X\right)\subseteq {\cup }_{x\in X}\overline{apr}\left(\left\{x\right\}\right)$. From the above, it follows that $\overline{apr}\left(X\right)={\bigcup }_{x\in X}\overline{apr}\left(\left\{x\right\}\right)$. □

Definition 4. 

Let $(U_1,{\mathcal{C}}_1)$ and $(U_2,{\mathcal{C}}_2)$ be two covering approximation spaces and $f:U_1\to U_2$ be a mapping. If $f\left(\overline{apr}\left(X\right)\right)\subseteq \overline{apr}\left(f\left(X\right)\right)$ for any $X\subseteq U_1$, then we call f a covering rough continuous mapping from $(U_1,{\mathcal{C}}_1)$ to $(U_2,{\mathcal{C}}_2)$. If f is a bijective, f and $f^{-1}$ are covering rough continuous mappings; then, f is called a covering rough homeomorphism mapping from $(U_1,{\mathcal{C}}_1)$ to $(U_2,{\mathcal{C}}_2)$, where $f\left(X\right)=\left\{f\right(x):x\in X\}$.

From Definition 4, we can obtain the properties of a covering rough continuous mapping as follows:

Proposition 1. 

Let $(U_1,{\mathcal{C}}_1)$ and $(U_2,{\mathcal{C}}_2)$ be two covering approximation spaces and $f:U_1\to U_2$ be a mapping. Then, the followings are equivalent:

(1) 

f is a covering rough continuous mapping from $(U_1,{\mathcal{C}}_1)$ to $(U_2,{\mathcal{C}}_2)$;

(2) 

For any $Y\subseteq U_2$, $\overline{apr}\left(f^{-1}\left(Y\right)\right)\subseteq f^{-1}\left(\overline{apr}\left(Y\right)\right)$;

(3) 

For any $Y\subseteq U_2$, $f^{-1}\left({\displaystyle \underline{apr}}\left(Y\right)\right)\subseteq {\displaystyle \underline{apr}}\left(f^{-1}\left(Y\right)\right)$;

(4) 

For any $x\in U_1$, $f\left(\overline{apr}\left\{x\right\}\right)\subseteq \overline{apr}\left(f\left(x\right)\right)$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$. For any $Y\subseteq U_2$, by Definition 4 and Lemma 2 (2), it is not difficult to prove $f\left(\overline{apr}\left(f^{-1}\left(Y\right)\right)\right)\subseteq \overline{apr}\left(f\left(f^{-1}\left(Y\right)\right)\right)\subseteq \overline{apr}\left(X\right)$. It follows that $\overline{apr}\left(f^{-1}\left(Y\right)\right)\subseteq f^{-1}\left(\overline{apr}\left(Y\right)\right)$ according to Lemma 2 (3).

$\left(2\right)\Rightarrow \left(1\right)$. For any $X\subseteq U_1$, we have $f^{-1}\left(\overline{apr}\left(f\left(X\right)\right)\right)\supseteq \overline{apr}\left(f^{-1}\left(f\left(X\right)\right)\right)\supseteq \overline{apr}\left(X\right)$ by $\left(2\right)$. So, $f\left(\overline{apr}\left(X\right)\right)\subseteq \overline{apr}\left(f\left(X\right)\right)$ by Lemma 2 (3). Hence, f is a covering rough continuous mapping from $(U_1,C_1)$ to $(U_2,C_2)$.

$\left(2\right)\Rightarrow \left(3\right)$. For any $Y\subseteq U_2$, $f^{-1}\left(\overline{apr}\left(Y^c\right)\right)\supseteq \overline{apr}\left(f^{-1}\right)\left(Y^c\right))$ by $\left(2\right)$. From Lemma 2 (1), we have $f^{-1}\left(\overline{apr}\left(Y^c\right)\right)=f^{-1}\left({\left({\displaystyle \underline{apr}}\left(Y\right)\right)}^c\right)={\left(f^{-1}\left({\displaystyle \underline{apr}}\left(Y\right)\right)\right)}^c$, which implies that $f^{-1}\left({\displaystyle \underline{apr}}\left(Y\right)\right)\subseteq {\displaystyle \underline{apr}}\left(f^{-1}\left(Y\right)\right)$ for any $Y\subseteq U_2$.

$\left(3\right)\Rightarrow \left(2\right)$. It is easy to prove, so we omit the proof.

$\left(1\right)\Rightarrow \left(4\right)$. It is obvious by Definition 4.

$\left(4\right)\Rightarrow \left(1\right)$. For any $X\subseteq U_1$, $X={\cup }_{x\in X}\left\{x\right\}$. We have $f\left(\overline{apr}\left\{x\right\}\right)\subseteq \overline{apr}\left(f\left(x\right)\right)$ by (4). From Lemma 2 (4), f preserves the union operations; thus, $f\left(\overline{apr}\left(X\right)\right)=f\left(\overline{apr}\left(U_{x\in X}\left\{x\right\}\right)\right)=U_{x\in X}f\left(\overline{apr}\left(\left\{x\right\}\right)\right)\subseteq U_{x\in X}\overline{apr}\left(f\left(x\right)\right)=\overline{apr}\left(f\left(x\right)\right)$. Therefore, $\left(1\right)\iff \left(4\right)$. □

Theorem 1. 

Let $(U_1,{\mathcal{C}}_1)$, $(U_2,{\mathcal{C}}_2)$, and $(U_3,{\mathcal{C}}_3)$ be covering approximation spaces. If $f:(U_1,{\mathcal{C}}_1)\to (U_2,{\mathcal{C}}_2)$ and $g:(U_2,{\mathcal{C}}_2)\to (U_3,{\mathcal{C}}_3)$ are both covering rough continuous mappings, then $g\circ f:(U_1,C_1)\to (U_3,C_3)$ is also a covering rough continuous mapping.

Proof. 

For any $X\subseteq U_1$, f is a covering rough continuous mapping. According to Definition 4, we have $f\left(\overline{apr}\left(X\right)\right)\subseteq \overline{apr}\left(f\left(X\right)\right)$. Since $f\left(X\right)\subseteq U_2$ and g is a covering rough continuous mapping, it is easy to obtain that $g\left(\overline{apr}\left(f\left(X\right)\right)\right)\subseteq \overline{apr}\left(g\left(f\left(X\right)\right)\right)$. Therefore, $g\left(f\left(\overline{apr}\left(X\right)\right)\right)\subseteq \overline{apr}\left(g\left(f\left(X\right)\right)\right)$, and thus, $g\circ f\left(\overline{apr}\left(X\right)\right)\subseteq \overline{apr}(g\circ f\left(X\right))$. From Definition 4, we have that $g\circ f$ is a covering rough continuous mapping. □

By Definition 4, Proposition 1, and Theorem 1, it is easy to obtain theorems of covering rough homeomorphism mappings:

Theorem 2. 

Let $(U_1,{\mathcal{C}}_1)$ and $(U_2,{\mathcal{C}}_2)$ be two covering approximation spaces and f be a bijective mapping. Then, the following cases are equivalent:

(1) 

If f is a covering rough homeomorphism mapping from $(U_1,{\mathcal{C}}_1)$ to $(U_2,{\mathcal{C}}_2)$, then $f^{-1}$ is a covering rough homeomorphism mapping from $(U_2,{\mathcal{C}}_2)$ to $(U_1,{\mathcal{C}}_1)$;

(2) 

For any $X\subseteq U_1$, $f\left(\overline{apr}\left(X\right)\right)=\overline{apr}\left(f\left(X\right)\right)$;

(3) 

For any $Y\subseteq U_2$, $f^{-1}\left({\displaystyle \underline{apr}}\left(Y\right)\right)={\displaystyle \underline{apr}}\left(f^{-1}\left(Y\right)\right)$;

(4) 

For any $Y\subseteq U_2$, $f^{-1}\left(\overline{apr}\left(Y\right)\right)=\overline{apr}\left(f^{-1}\left(Y\right)\right)$.

Theorem 3. 

Let $(U_1,{\mathcal{C}}_1)$ ,$(U_2,{\mathcal{C}}_2)$, and $(U_3,{\mathcal{C}}_3)$ be covering approximation spaces. Then, we have the following:

(1) 

Identity mapping $i:(U_1,{\mathcal{C}}_1)\to (U_1,{\mathcal{C}}_1)$ is a covering rough homeomorphism mapping.

(2) 

If $f:(U_1,{\mathcal{C}}_1)\to (U_2,{\mathcal{C}}_2)$ is a covering rough homeomorphic mapping, then $f^{-1}:(U_2,{\mathcal{C}}_2)\to (U_1,{\mathcal{C}}_1)$ is also a covering rough homeomorphism mapping.

(3) 

If $f:(U_1,{\mathcal{C}}_1)\to (U_2,{\mathcal{C}}_2)$ and $g:(U_2,{\mathcal{C}}_2)\to (U_3,{\mathcal{C}}_3)$ are covering rough homeomorphism mappings, then $g\circ f:(U_1,{\mathcal{C}}_1)\to (U_3,{\mathcal{C}}_3)$ is also a covering rough homeomorphism mapping.

4. Separation Properties of Covering Approximation Spaces

Separation properties play an important role in many spaces. In this section, we will define a $T_1$-space and discuss its properties. In this segment, we provide a new method for the classification of covering approximation spaces.

Definition 5. 

Let $(U,\mathcal{C})$ be a covering approximation space. For any $x\in U$, if $\overline{apr}\left(\left\{x\right\}\right)=\left\{x\right\}$, covering approximation space $(U,\mathcal{C})$ is called a $T_1-$space.

Example 1. 

Let U be an arbitrary non-empty universe and $\mathcal{C}=\left\{\right\{x\}:x\in U\}$. By Definition 3 and Definition 5, it is easy to check that $(U,\mathcal{C})$ is a covering approximation space and a $T_1-$space.

Example 2. 

Let U be an arbitrary non-empty universe, taking $\mathcal{C}=\left\{U\right\}$. By Definition 5, we can easily obtain that $(U,\mathcal{C})$ is a covering approximation space but not a $T_1-$space.

Let U be an arbitrary non-empty universe and $\mathcal{C}$ be a covering of U. Then, $\mathcal{C}$ can induce a topology $\tau$. Since $\tau$ is related to $\mathcal{C}$, we use ${\tau }_{\mathcal{C}}$ instead of $\tau$. We say that topological space $(U,{\tau }_{\mathcal{C}})$ is induced by covering approximation space $(U,\mathcal{C})$.

Lemma 3. 

Topological space $(U,{\tau }_{\mathcal{C}})$ is a topological $T_1-$space if and only if every single point set in $(U,{\tau }_{\mathcal{C}})$ is a closed set.

Theorem 4. 

Let $(U,\mathcal{C})$ be a covering approximation space and $(U,{\tau }_{\mathcal{C}})$ be a topological space induced by $(U,\mathcal{C})$. Then, covering approximation space $(U,\mathcal{C})$ is a $T_1-$space if and only if $(U,{\tau }_{\mathcal{C}})$ is a topological $T_1-$space.

Proof. 

(⇒) Let covering approximation space $(U,\mathcal{C})$ be a $T_1-$space; by Definition 5, we claim that $\left\{x\right\}=N\left(x\right)$ for any $x\in U$. Otherwise, there exists $a\in N\left(x\right)$ such that $x\ne a$. From Definition 3, we have $x\in \overline{apr}\left(\left\{x\right\}\right)$. Since $\overline{apr}\left(\left\{x\right\}\right)$ has more than two elements, it contradicts that $(U,\mathcal{C})$ is a $T_1-$space.

We shall prove that $\overline{\left\{x\right\}}$ is a closed set. $\overline{\left\{x\right\}}=\bigcap \{C:C\in \mathcal{C}\}=N\left(x\right)$ = $\left\{x\right\}$. By Lemma 3, we obtain that $(U,{\tau }_{\mathcal{C}})$ is a topological $T_1-$space.

(⇐) Let $(U,{\tau }_{\mathcal{C}})$ be a topological $T_1-$space. Then, for any $x\in U$ , by Definition 3, $\overline{apr}\left(\left\{x\right\}\right)=\{y:N\left(y\right)\bigcap \left\{x\right\}\ne Ø\}=\{y:x\in N\left(y\right)\}=\{y:x\in \left\{y\right\}\}=$$\left\{x\right\}$; therefore, covering approximation space $(U,\mathcal{C})$ is a $T_1-$space. □

Theorem 5. 

Let $(U_1,{\mathcal{C}}_1)$ and $(U_2,{\mathcal{C}}_2)$ be two covering approximation spaces and f be a covering rough homeomorphism mapping from covering approximation space $(U_1,{\mathcal{C}}_1)$ to covering approximation space $(U_2,{\mathcal{C}}_2)$; then, covering approximation space $(U_1,{\mathcal{C}}_1)$ is a $T_1-$space if and only if covering approximation space $(U_2,{\mathcal{C}}_2)$ is a $T_1-$space.

Proof. 

(⇒) Let covering approximation space $(U_1,{\mathcal{C}}_1)$ be a $T_1-$space and f be a mapping from the covering approximation space $(U_1,{\mathcal{C}}_1)$ to the covering approximation space $(U_2,{\mathcal{C}}_2)$; then, for any $y\in U_2$, there exists a unique $x\in U_1$ such that $x=f^{-1}\left(y\right)$. Since the covering approximation space $(U_1,{\mathcal{C}}_1)$ is a $T_1-$space, $\overline{apr}\left(\left\{x\right\}\right)=\left\{x\right\}=$$f^{-1}\left(y\right)$; therefore, $f\left(\overline{apr}\left(\left\{x\right\}\right)\right)=\left\{y\right\}$. By Theorem 2, we obtain $\overline{apr}\left(f\left(\left\{x\right\}\right)\right)=\overline{apr}\left(\left\{y\right\}\right)=\left\{y\right\}$; thus, $(U_2,{\mathcal{C}}_2)$ is a $T_1-$space.

(⇐) We can use a similar method to prove the converse; therefore, we omit it here.

The topological operator plays an important role in general topology and rough sets and provides a method for exchanging information systems [22]. We discuss the necessary and sufficient conditions for $\overline{D}$ to be a topological closure operator. □

Theorem 6. 

Let U be a universe and ∗ be a preorder and Euclidean relation on U; then, $\overline{D}$ is a topological closure operator on U.

Proof. 

Let R be a binary on U; ∗ is a preorder and Euclidean relation. We shall prove that $\overline{D}$ is a topological closure operator on U. Since ∗ is a preorder, then for any $X\subseteq U$, we can obtain ${\displaystyle \underline{D}}\left(X\right)\subseteq X$, $X\subseteq \overline{D}\left(X\right)$, and $\overline{D}\left(\overline{D}\left(X\right)\right)\subseteq \overline{D}\left(X\right)$, and we have $X\subseteq \overline{D}\left(X\right)$. We only need to prove that $\overline{D}$ satisfies $\overline{D}(X\cup Y)=\overline{D}\left(X\right)\cup \overline{D}\left(Y\right)$ and $\overline{D}(Ø)=Ø$ for any $X,Y\subseteq U$. By definition $\overline{D}$, we can obtain $\overline{D}(Ø)=Ø$. We shall prove $\overline{D}(X\cup Y)=\overline{D}\left(X\right)\cup \overline{D}\left(Y\right)$. For any $X,Y\subseteq U$, we know that $\overline{D}\left(X\right)\subseteq \overline{D}(X\cup Y)$ and $\overline{D}\left(Y\right)\subseteq \overline{D}(X\cup Y)$ by $X\subseteq X\cup Y$; thus, $\overline{D}\left(X\right)\cup \overline{D}\left(Y\right)\subseteq \overline{D}(X\cup Y)$. We prove the converse.

For any $y\in \overline{D}(X\cup Y)$, by definition $\overline{D}$, there exists $x_0\in U$ such that $y\in {\ast }_s\left(x_0\right)$ and ${\ast }_s\left(x_0\right)\cap (X\cup Y)\ne Ø$. Therefore $({\ast }_s\left(x_0\right)\cap X)\cup ({\ast }_s\left(x_0\right)\cap Y)\ne Ø$; thus, ${\ast }_s\left(x_0\right)\cap X\ne Ø$ or ${\ast }_s\left(x_0\right)\cap Y\ne Ø$. By definition $\overline{D}$, we have $y\in \overline{D}\left(X\right)$ or $y\in \overline{D}\left(Y\right)$. □

In Theorem 6, the converse may not true.

Example 3. 

Let $U=\{a,b\}$, $\ast =\left\{\right(a,a),(a,b),(b,b\left)\right\}$. By definition $\overline{D}$, we have ${\ast }_s\left(\left\{a\right\}\right)=\{a,b\},{\ast }_s\left(\left\{b\right\}\right)=\left\{b\right\}$:

(1) 

$\overline{D}(Ø)=Ø$;

(2) 

$\overline{D}\left(\overline{D}(Ø)\right)=Ø$$=\overline{D}(Ø)$;

$\overline{D}\left(\overline{D}\left(\left\{a\right\}\right)\right)=\{a,b\}$$=\overline{D}\left(\left\{a\right\}\right)$;

$\overline{D}\left(\overline{D}\left(\left\{b\right\}\right)\right)=\{a,b\}$$=\overline{D}\left(\left\{b\right\}\right)$;

$\overline{D}\left(\overline{D}\left(\{a,b\}\right)\right)=\{a,b\}$$=\overline{D}\left(\{a,b\}\right)$;

(3) 

$Ø\subseteq \overline{D}(Ø)$;

$\left\{a\right\}\subseteq \overline{D}\left(\left\{a\right\}\right)$,

$\left\{b\right\}\subseteq \overline{D}\left(\left\{b\right\}\right)$;

$\{a,b\}\subseteq \overline{D}\left(\{a,b\}\right)$;

(4) 

$\overline{D}(Ø)\cup \overline{D}(Ø)=Ø=\overline{D}(Ø\cup Ø)$;

$\overline{D}(Ø)\cup \overline{D}\left(\left\{a\right\}\right)=\{a,b\}=\overline{D}(Ø\cup \left\{a\right\})$;

$\overline{D}(Ø)\cup \overline{D}\left(\left\{b\right\}\right)=\{a,b\}=\overline{D}(Ø\cup \left\{b\right\})$;

$\overline{D}(Ø)\cup \overline{D}\left(\{a,b\}\right)=\{a,b\}=\overline{D}(Ø\cup \{a,b\})$;

$\overline{D}\left(\left\{a\right\}\right)\cup \overline{D}\left(\left\{b\right\}\right)=\{a,b\}=\overline{D}(\left\{a\right\}\cup \left\{b\right\})$,

$\overline{D}\left(\left\{a\right\}\right)\cup \overline{D}\left(\{a,b\}\right)=\{a,b\}=\overline{D}(\left\{a\right\}\cup \{a,b\})$,

$\overline{D}\left(\left\{b\right\}\right)\cup \overline{D}\left(\{a,b\}\right)=\{a,b\}=\overline{D}(\left\{b\right\}\cup \{a,b\})$,

$\overline{D}\left(\{a,b\}\right)\cup \overline{D}\left(\{a,b\}\right)=\{a,b\}=\overline{D}(\{a,b\}\cup \{a,b\})$,

$b\in {\ast }_s\left(\left\{a\right\}\right)$ and $a\in {\ast }_s\left(\left\{a\right\}\right)$, but $a\notin {\ast }_s\left(\left\{b\right\}\right)$.

From the above, we know that $\overline{D}$ is a topological closure operator on U, but ∗ is not a Euclidean relation.

It is natural to inquire about the necessary and sufficient conditions for it to be a topological closure operator.

Definition 6 

([23]). Let $\ast =R$ be an arbitrary binary relation on U. The smallest transitive relation on U containing the relation R is called the transitive closure of R. They denote the transitive closure of R by $t\left(R\right)$.

Yu H. et al. obtained that if R is a reflexive relation, then $t\left(R\right)$ is also a reflexive relation. D, Pei, et al. showed that R is a binary relation on U; then, $t\left(R\right)=R\cup R^2\cup \dots R^n\dots$. It is then natural to ask what other properties it has. We will discuss this question.

Definition 7. 

Let U be a non-empty universe and R a binary relation on U. R is called $F-$symmetry if for any $x,y\in U$, we have $(x,y)\in R$ and $(y,x)\in R$.

Theorem 7. 

Let U be a universe, R be an $F-$symmetry relation on U, and $n\ge 2$; then, $t\left(R\right)$ is a reflexive and $F-$symmetry relation on U.

Proof. 

Since $R\subseteq t\left(R\right)$, we have that $t\left(R\right)$ is an $F-$symmetry relation. We only need to prove that $t\left(R\right)$ is a reflexive relation. For any $x,y\in U$, we have $(x,y)\in R$ and $(y,x)\in R$. Thus, $(x,x)\in R^2$ and $(y,y)\in R^2$ by $n\ge 2$ and the composition of $R^2$. □

5. Topologies Induced by Relation ∗

Zhao defined the topology of coverings in [23] using only covering approximation spaces. L. Yang et al. defined an open set with ${\displaystyle \underline{R}}\left(X\right)=A^0$ and constructed a topology. We shall use relation ∗ and a subset of $\mathcal{P}\left(U\right)$ to construct a topology. If ∗ is a covering relation, the topology induced by ∗ is finer than Zhao’s; if ∗ is a relation R, the topology induced by R is finer than Yang’s. In other words, we propose a new method of constructing topologies using ∗ and take the topology that they constructed as our special case. That is to say, we generalize their methods of constructing topologies.

Lemma 4 

([25]). Let X be a topological space. For every $A\subseteq X$, the following conditions are equivalent:

(1) 

Point x belongs to $\overline{A}$;

(2) 

For every neighborhood U of x, we have $U\cap A\ne Ø$.

Lemma 5 

([23]). Let $(U,\mathcal{C})$ be a covering approximation space. The topology $\mathcal{T}$ induced by the covering $\mathcal{C}$ is defined as follows: a subset G of U is said to be open in U if for each $x\in G$, there are finite elements $C_1,C_2,\dots ,C_n$ of $\mathcal{C}$ such that $x\in {\bigcap }_{i=1}^n{C}_i\subseteq G$.

Theorem 8. 

Let U be an arbitrary non-empty universe and ∗ a reflexive relation on U. We construct topology τ using ∗ as follows: $V\subseteq U$ is called an open set in U if for any $x\in V$, there exists a finite family $\{X_i:i\in I\}\subseteq \mathcal{P}\left(U\right)$ such that $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq V$.

Proof. 

We prove that $\tau$ is a topology.

(1)

It is obvious that Ø and U in $\tau$ by the definition condition of an open set.

(2)

For any $V_1,V_2\in \tau$, we show that $V_1\bigcap V_2\in \tau$. For any $x\in V_1\bigcap V_2$, there exist finite families $\{X_{i_k}:k\in I\}$ and $\{X_{j_m}:m\in L\}$ such that $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in I}}{X}_{i_k})\subseteq V_1$ and $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{m\in L}}{X}_{i_m})\subseteq V_2$. Thus, $x\in ({\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in I}}{X}_{i_k}))\bigcap ({\displaystyle \underline{\ast }}({\displaystyle \bigcap _{m\in L}}{X}_{i_m}))$ = ${\displaystyle \underline{\ast }}(({\displaystyle \bigcap _{m\in L}}{X}_{i_m})\bigcap ({\displaystyle \bigcap _{k\in I}}{X}_{i_k}))\subseteq V_1\bigcap V_2$. This prove that any finite intersections of elements of $\tau$ are still in $\tau$.

(3)

Let $\{V_{\alpha }:\alpha \in J\}$ be a family of elements of $\tau$. We need to prove that $V={\displaystyle \bigcup _{\alpha \in G}}{V}_{\alpha }\in \tau$. For any $x\in V$, an indexed ${\alpha }_0\in J$ must exist such that $x\in V_{{\alpha }_0}$. Since $V_{{\alpha }_0}$ is open and ∗ is a reflexive relation on U, we can find a finite family $\{X_i:i\in I\}\subseteq \mathcal{P}\left(U\right)$ such that $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq V_{{\alpha }_0}\subseteq V$, so V is open. Therefore, the desired result is proved.

□

Example 4. 

Let $U=\{a,b,c\}$ be an arbitrary non-empty universe, taking $\mathcal{C}=\left\{\right\{a,c\},\{b,c\left\}\right\}$, $\mathcal{A}=\left\{\right\{b\},\{a,b\left\}\right\}\subseteq \mathcal{P}\left(U\right)$, and $X=\left\{b\right\}\subseteq U$. By Definition 2, we have $(U,\mathcal{C})$ as a covering approximation space, but X is not an open set, by Lemma 5. We also know that $\mathcal{A}=\left\{\right\{b\},\{a,b\left\}\right\}$ is not a covering of U, but we can obtain $b\in {\displaystyle \underline{\ast }}(\bigcap \mathcal{A})\subseteq X$; thus, X is an open set in U by Theorem 8.

From Example 4 and Lemma 5, we obtain a topology, which is $\{Ø,U,\{b,c\left\}\right\}$. This topology no longer contains more elements. From this angle, we can say that the method of defining the topology in Lemma 5 is limited by coverings. We can construct a finer topology, which is $\mathcal{P}\left(U\right)$, using Example 4 and Theorem 8. Our method solves the limitation of coverings.

Remark 1. 

(1) For any $X\subseteq U$, if $x\in X$ and ∗ is a reflexive relation on U, we have ${\displaystyle \underline{\ast }}\left(X\right)\in \tau$ by Theorem 8.

(2) If ∗ is a reflexive relation and ${\displaystyle \underline{\ast }}\left(X\right)=X^0$, we have ${\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq {\displaystyle \underline{\ast }}\left(X\right)=X^0$ by Theorem 8 and $X^0\in \tau$. Therefore, our topology τ induced by ∗ is finer than Yang’s. We generalize Yang’s results.

(3) If ∗ is a reflexive relation and $\{{\displaystyle \underline{\ast }}\left(X\right):X\subseteq U\}=\mathcal{C}$ is a covering of U, then $\mathcal{T}=\tau$ by Lemma 5 and Theorem 8. We can conclude that our topology is much finer than Zhao’s because ${\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq {\displaystyle \underline{\ast }}\left(X\right)\subseteq X$. We also generalize their definition, and their definitions become a special case of our definition.

Theorem 9. 

Let U be an arbitrary non-empty universe, ∗ be a reflexive relation on U, and τ be a topology induced by ∗. Then, we have the following properties:

(1) 

For each $x\in U$, $\{X_i:i\in I\}\subseteq \mathcal{P}\left(U\right)$ is all the subsets of U which contain x and ∗ is a reflexive relation on U. If $V\in \tau$ is a subset of U and $x\in V$, then ${\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)$ is the smallest subset of U and $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq V$, denoted by $C\left(x\right)=\bigcap \{{\displaystyle \underline{\ast }}\left(X_i\right):x\in {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$.

(2) 

If $V\in \tau$ is a subset of U, then $V=$ ${\displaystyle \bigcup _{x\in V}}C\left(x\right)$.

(3) 

Let $\{{\displaystyle \underline{\ast }}\left(X_i\right):x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$; then, $\overline{\left\{x\right\}}$=$U\backslash {\displaystyle \bigcup _{x\notin \underline{\ast }\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right)$;

(4) 

Let ∗ be a reflexive relation on U. For every $X\subseteq U$, we have $int\left({\displaystyle \underline{\ast }}\left(X\right)\right)={\displaystyle \underline{\ast }}\left(X\right)$, where $int\left({\displaystyle \underline{\ast }}\left(X\right)\right)$ represents the interior of ${\displaystyle \underline{\ast }}\left(X\right)$.

(5) 

Let $\{{\displaystyle \underline{\ast }}\left(X_i\right):x\in {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$ be a family subsets of U; then, $\{{\displaystyle \underline{\ast }}\left(X_i\right):x\in {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$ is a base for $(U,\tau )$ at point x.

Proof. 

(1) Let $V\in \tau$ be a subset of U and $x\in V$. There is a finite subset family $\{X_k:k\in K\}\subseteq \mathcal{P}\left(U\right)$ such that $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{i\in I}}{X}_i)\subseteq V$ by Theorem 8. $\{X_i:i\in I\}\subseteq \mathcal{P}\left(U\right)$ is all the subsets of U which contains x; then, $C\left(x\right)$ is the smallest subset of U containing x and $C\left(x\right)\subseteq {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in K}}{X}_k)$. Thus, $x\in C\left(x\right)\subseteq {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in K}}{X}_k)\subseteq V$.

(2) Let $V\in \tau$ be a subset of U and $x\in V$; then, we can obtain $C\left(x\right)\subseteq V$ by (1). Since $\left\{C\right(x):x\in V\}$ is a cover of V, $V\subseteq {\displaystyle \bigcup _{x\in G}}C\left(x\right)$; thus, we have $V=$${\displaystyle \bigcup _{x\in V}}C\left(x\right)$.

(3) Picking $M\left(x\right)={\displaystyle \bigcup _{x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right)$, ${\displaystyle \underline{\ast }}\left(X_i\right)$ is an open subset of U for very $i\in I$ by Remark 1 (1), and the union ${\displaystyle \bigcup _{x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right)$ is an open one by $\left(O_3\right)$. It follows that the set $U\backslash {\displaystyle \bigcup _{x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right)$ is closed, and $x\in (U\backslash {\displaystyle \bigcup _{x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right))$. It is easily seen that $\overline{\left\{x\right\}}$ is the smallest closed set containing x. Obviously, $\overline{\left\{x\right\}}\subseteq U\backslash {\displaystyle \bigcup _{x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I}}{\displaystyle \underline{\ast }}\left(X_i\right)$. We shall prove the converse.

For any $y\in U\backslash \overline{\left\{x\right\}}$. Since $U\backslash \overline{\left\{x\right\}}$ is an open set containing y, $C\left(y\right)\subseteq U\backslash \overline{\left\{x\right\}}$ by (1); therefore, $C\left(y\right)\cap \overline{\left\{x\right\}}=Ø$ and $x\notin C\left(y\right)$. There exists a ${\displaystyle \underline{\ast }}\left(X_{i_0}\right)$ of $\{{\displaystyle \underline{\ast }}\left(X_i\right):x\notin {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$ such that $y\in {\displaystyle \underline{\ast }}\left(X_{i_0}\right)\wedge x\notin {\displaystyle \underline{\ast }}\left(X_{i_0}\right)$. Thus, $y\in M\left(x\right)$ and $U\backslash \overline{\left\{x\right\}}\subseteq M\left(x\right)$; therefore, $U\backslash M\left(x\right)\subseteq \overline{\left\{x\right\}}$.

(4) Let ∗ be a reflexive relation on U, for every $X\subseteq U$, from Remark 1 (1), it follows that ${\displaystyle \underline{\ast }}\left(X\right)$ is an open set. Then, we obtain ${\displaystyle \underline{\ast }}\left(X\right)=int\left({\displaystyle \underline{\ast }}\left(X\right)\right)$ directly by definition of an open set.

(5) Let $\mathcal{B}\left(x\right)$=$\{{\displaystyle \underline{\ast }}\left(X_i\right):x\in {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$. For any $V_1,V_2\in \mathcal{B}\left(x\right)$ and every $x\in V_1\cap V_2$, there exist finite collections $\{X_{i_k}:k\in I\}$ and $\{X_{j_m}:m\in L\}$ such that $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in I}}{X}_{i_k})\subseteq V_1$ and $x\in {\displaystyle \underline{\ast }}({\displaystyle \bigcap _{m\in L}}{X}_{i_m})\subseteq V_2$. Thus, $x\in ({\displaystyle \underline{\ast }}({\displaystyle \bigcap _{k\in I}}{X}_{i_k}))\bigcap ({\displaystyle \underline{\ast }}({\displaystyle \bigcap _{m\in L}}{X}_{i_m}))$ = ${\displaystyle \underline{\ast }}(({\displaystyle \bigcap _{m\in L}}{X}_{i_m})\bigcap ({\displaystyle \bigcap _{k\in I}}{X}_{i_k}))\subseteq V_1\bigcap V_2$. Picking $X_0=({\displaystyle \bigcap _{m\in L}}{X}_{i_m})\bigcap ({\displaystyle \bigcap _{k\in I}}{X}_{i_k})$, ${\displaystyle \underline{\ast }}\left(X_0\right)\in \mathcal{B}\left(x\right)$, which satisfies $\left(B_1\right)$ of Definition 2.6. $\mathcal{B}\left(x\right)$ obviously satisfies $\left(B_2\right)$ of Definition 2.6. We have $\{{\displaystyle \underline{\ast }}\left(X_i\right):x\in {\displaystyle \underline{\ast }}\left(X_i\right),i\in I\}$ as a base for $(U,\tau )$ at point x. □

Theorem 10. 

Let U be an arbitrary non-empty universe, ∗ be a reflexive relation on U, and τ be a topology induced by ∗; then, we have the following properties:

(1) 

For any $A\subseteq U$ and $x\in U$, we have $x\in \overline{A}$ if and only if $C\left(x\right)\cap A\ne Ø$ if and only if there exists a base $\mathcal{B}\left(x\right)$ at x such that for every $V\in \mathcal{B}\left(x\right)$, $V\cap A\ne Ø$.

(2) 

For any $x,y\in U$ and $x\ne y$, $C\left(x\right)=C\left(y\right)$ if and only if $\overline{\left\{x\right\}}=\overline{\left\{y\right\}}$.

(3) 

Let ${\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$=$\{y\in U:\forall i\in I,{\displaystyle \underline{\ast }}\left(X_i\right)\in \{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}(x\in {\displaystyle \underline{\ast }}\left(X_i\right)⟷y\in {\displaystyle \underline{\ast }}\left(X_i\right))\}$. For any $x\in U$, ${\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$=$N\left(x\right)\cap \overline{\left\{x\right\}}$.

(4) 

Let F be a closed subset of U; then, $F={\displaystyle \bigcup _{x\in F}}\overline{\left\{x\right\}}$= ${\displaystyle \bigcup _{x\in F}}{\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$.

Proof. 

(1) For any $x\in \overline{A}$, from Lemma 4, $x\in \overline{A}$ if and only if $C\left(x\right)\cap A\ne Ø$ can be proved in a similar way.

If $C\left(x\right)\cap A\ne Ø$, then $V\cap A\ne Ø$ is obvious. It remains to be shown that if there exists a base $\mathcal{B}\left(x\right)$ at x such that for every $V\in \mathcal{B}\left(x\right)$, $V\cap A\ne Ø$, then $x\in \overline{A}$. Suppose that $x\in \overline{A}$ does not hold, i.e., $x\notin \overline{A}$. There exists a closed set F such that $x\notin F$. For the open set $V=U\backslash F$, we have $x\in V$ and $V\cap A=Ø$. For every base $\mathcal{B}\left(x\right)$ at x, there exists a $V^{\prime }\in \mathcal{B}\left(x\right)$ such that $x\in V^{\prime }\subseteq V$, and from $V\cap A=Ø$, it follows that $V^{\prime }\cap A=Ø$, i.e., for every $V\in \mathcal{B}\left(x\right)$, $V\cap A\ne Ø$ does not hold.

(2) If $C\left(x\right)=C\left(y\right)$, we have $x\in \overline{\left\{y\right\}}$ and $y\in \overline{\left\{x\right\}}$ from (1). It is well known that $\overline{\left\{x\right\}}$ is the smallest closed set containing x; thus, $\overline{\left\{x\right\}}\subseteq \overline{\left\{y\right\}}$. Similarly, $\overline{\left\{y\right\}}\subseteq \overline{\left\{x\right\}}$; therefore, we have $\overline{\left\{x\right\}}=\overline{\left\{y\right\}}$. If $\overline{\left\{x\right\}}=\overline{\left\{y\right\}}$, we shall prove $C\left(x\right)=C\left(y\right)$. Since $x\in \overline{\left\{y\right\}}$, we have $\left\{y\right\}\cap C\left(x\right)\ne Ø$. Thus, $y\in C\left(x\right)$, and therefore, $C\left(y\right)\subseteq C\left(x\right)$. Similarly, we have $C\left(x\right)\subseteq C\left(y\right)$; thus, $C\left(x\right)=C\left(y\right)$.

(3) It is not difficult to prove that ${\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$=$\{y\in U:\forall i\in I,{\displaystyle \underline{\ast }}\left(X_i\right)\in \{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}(x\in {\displaystyle \underline{\ast }}\left(X_i\right)⟷y\in {\displaystyle \underline{\ast }}\left(X_i\right))\}$ = $\{y\in U:C(x)=C(y\left)\right\}$. Suppose that y is an element of ${\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$; then, $x\in C\left(y\right)$. Thus, $y\in \overline{\left\{x\right\}}$, and then, $y\in C\left(x\right)\cap \overline{\left\{x\right\}}$. Therefore, ${\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}\subseteq C\left(x\right)\cap \overline{\left\{x\right\}}$. We shall prove the converse. For any $y\in C\left(x\right)\cap \overline{\left\{x\right\}}$. We have $y\in C\left(x\right)$ and $x\in C\left(y\right)$ by (1); therefore, $C\left(x\right)=C\left(y\right)$. Thus, we have $y\in {\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}$.

(4) For any $x\in F$, we have $\overline{\left\{x\right\}}\subseteq F$, and then, ${\displaystyle \bigcup _{x\in F}}\overline{\left\{x\right\}}\subseteq F$. $\{\overline{\left\{x\right\}}:x\in F\}$ is a closed covering of F; thus, we have $F\subseteq {\displaystyle \bigcup _{x\in F}}\overline{\left\{x\right\}}$. Therefore, $F={\displaystyle \bigcup _{x\in F}}\overline{\left\{x\right\}}$. $F\subseteq {\displaystyle \bigcup _{x\in F}}{\mathcal{P}}_x^{\{{\displaystyle \underline{\ast }}\left(X_i\right):i\in I\}}\subseteq {\displaystyle \bigcup _{x\in F}}\overline{\left\{x\right\}}=F$ by (3). □

Zhao defined the upper approximation operator COM${}^{+}\left(X\right)$ in [23] by $U/\sim$ as follows: COM${}^{+}\left(X\right)=\bigcup \{Y_i\in U/\sim :Y_i\cap X\ne Ø\}$. The following property gives its characterizations.

Proposition 2. 

COM${}^{+}\left(X\right)=\bigcup \{Y_i\in U/\sim :Y_i\cap X\ne Ø\}$ is a topological closure operator on U.

Proof. 

COM${}^{+}(Ø)=\bigcup \{Y_i\in U/\sim :Y_i\cap Ø\ne Ø\}$ = ∅ satisfies ($C_3$) by the definition of closure operator ; for any $X\subseteq U$, we claim that $X\subseteq$ COM${}^{+}\left(X\right)=\bigcup \{Y_i\in U/\sim :Y_i\cap X\ne Ø\}$. $\forall x\in X$, since $U/\sim$ is a partition of U, there exists $Y_{i_0}\in U/\sim$ such that $x\in Y_{i_0}$; thus, $Y_{i_0}\cap \left\{x\right\}\ne Ø$, so we have $x\in$ COM${}^{+}\left(\left\{x\right\}\right)$. Therefore, $X\subseteq$ COM${}^{+}\left(X\right)$ satisfies ($C_2$) by the definition of closure operator. We shall prove that COM${}^{+}$ satisfies ($C_1$). For any $X,Y\subseteq U$, COM${}^{+}\left(X\right)$∪ COM${}^{+}\left(Y\right)$⊆ COM${}^{+}(X\cup Y)$ is obvious. We only need to prove the converse.

For any $x\in$ COM${}^{+}(X\cup Y)$, there exists a $Y_{i_0}\in U/\sim$ such that $x\in Y_{i_0}$ and $Y_{i_0}\cap (X\cup Y)\ne Ø$. Thus, $(Y_{i_0}\cap X)\cup (Y_{i_0}\cap Y)\ne Ø$, and then, at least one of $(Y_{i_0}\cap X)\ne Ø$ and $(Y_{i_0}\cap Y)\ne Ø$ is satisfied. Therefore, $x\in$ COM${}^{+}\left(X\right)$ or $x\in$ COM${}^{+}\left(Y\right)$. We have COM${}^{+}(X\cup Y)\subseteq$ COM${}^{+}\left(X\right)$∪ COM${}^{+}\left(Y\right)$, which satisfies ($C_1$) by the definition of closure operator.

Finally, we prove that COM${}^{+}$ satisfies ($C_4$) by the definition of closure operator. For any $X\subseteq U$, COM${}^{+}\left(X\right)$⊆ COM${}^{+}$(COM${}^{+}\left(X\right)$) is obvious. We need to prove the converse. For any $x\in$ COM${}^{+}$(COM${}^{+}\left(X\right)$), there exists $Y_{i_0}\in U/\sim$ such that $x\in Y_{i_0}$ and $Y_{i_0}\cap$ COM${}^{+}\left(X\right)\ne Ø$. Then, there exists an $x_0$ such that $x_0\in Y_{i_0}\cap$ COM${}^{+}\left(X\right)$; therefore, there exists a $Y_{i_1}\in U/\sim$ such that $x_0\in Y_{i_1}$ and $Y_{i_1}\cap X\ne Ø$. Since $U/\sim$ is a partition of U and $x_0\in Y_{i_0}$, $x_0\in Y_{i_1}$; then, $Y_{i_0}=Y_{i_1}$. Thus, $x\in Y_{i_1}$ and $Y_{i_1}\cap X\ne Ø$, and then, $x\in$ COM${}^{+}\left(X\right)$. We can obtain COM${}^{+}$(COM${}^{+}\left(X\right)$)⊆ COM${}^{+}\left(X\right)$. □

6. Conclusions

In this paper, we have proposed new concepts and investigated the relationship among covering approximation spaces. We have obtained some interesting properties of covering spaces by covering rough continuous mapping and covering rough homeomorphism mapping. We can provide a method for the classification of covering approximation spaces. In future work, it is worth investigating the properties of the other approximation spaces and obtaining the method for general approximation spaces.