Mixed multiset topological space and Separation axioms

Abstract

Very recently the concept of separation axioms in multiset topological space and the concept of mixed multiset topological space were developed. In this article we redefine the definition of mixed multiset topological space.We introduce the separation axioms in it and investigate some of their properties.We also show that the properties of being $T_{0} , T_{1} ,T_{2}$ spaces are hereditary.

1 Introduction

The theory of multiset (Bag or mset) as a generalization of set theory was developed by R.R. Yager [19] and later it was studied by W. D. Blizard [26, 27] and many others [1, 4,5,6, 11, 14, 22, 28]. After the introduction of fuzzy set by L.A. Jadeh [15] in 1965, the concepts of fuzzy multiset and fuzzy topology were developed and studied by many people [3, 7, 14, 21, 23, 24]. In the year 2012 Girish and John [10] developed the idea of multiset topology (M-topology). Many results of general topology were investigated in multiset topology [9, 12, 18]. Specially the separation axioms in multiset topological space were studied by Sheikh,Omar and Raafat [25]. Mixed topology is a technique of mixing two topologies on a set to get a third topology. The works on mixed topology were done by Buck [20], Cooper [8] and many more. In 1995 Das and Baishya [17] introduced the idea of mixed fuzzy topological space. This definition of Mixed fuzzy topological space was generalised by Tripathy and Ray [2]. Separation axioms in mixed fuzzy topological space were introduced and studied by M.H. Rashid and D.M. Ali [16] in 2008. In the year 2019, Shravan and Tripathy in [13] developed the concept of Mixed topological space on multisets (Mixed M-topological space) by using quasi-coincidence between a multipoint and a multiset. In this article we redefine definition of mixed M-topological space introduced by Shravan and Tripathy. We introduce the separation axioms in it and investigate some of their properties. We also show that the properties of being $T_{0} , T_{1} ,T_{2}$ spaces are hereditary.

2 Preliminaries

In classical set theory, a set is a well-defined collection of distinct objects. If repeated occurrences of any object is allowed in a set then that is known as multiset (mset or bag, for short).

Definition 2.1

A multiset M drawn from the set X is represented by a function Count M or $C_{M}$: $X \rightarrow N$ , where N represents the set of non negative integers.

The mset M drawn from the set $X = \{ x_{1},x_{2}, ... ,x_{n} \}$ is denoted by M = $\{ k_{1}/x_{1}, k_{2}/x_{2}, ... , k_{n}/x_{n}\}$ , where M is an mset with $x_{1}$ appearing $k_{1}$ times, $x_{2}$ appearing $k_{2}$ times and so on. In definition 2.1 , $C_{M}(x)$ is the number of occurrences of the element x in the mset M. However those elements which are not included in the mset M have zero count.

Clearly, a set is a special case of an mset.

Example 2.2.

Let $X = \lbrace a, b, c, d, e\rbrace$ be any set. Then $M = \lbrace a, a, b, b, d, d, b, d, d, e, d \rbrace$ is a multiset drawn from X. The multiset M is also denoted by $M =\lbrace 2/a, 3/b, 5/d, 1/e \rbrace$. Here $C_{M}(a) = 2, C_{M}(b) = 3, C_{M}(c) = 0, C_{M}(d) = 5, C_{M}(e) = 1$.

Definition 2.3.

Let M be an mset drawn from a set X. The support set of M, denoted by $M^{*}$, is a subset of X and $M^{*} =\lbrace x \in X : C_{M}(x) > 0 \rbrace$ i.e., $M^{*}$ is an ordinary set and it is also called the root set.

An mset M is said to be an empty mset if for all $x \in X, C_{M}(x) = 0$.

Definition 2.4.

A domain X is defined as a set of elements from which msets are constructed. The mset space $[X]^{m}$ is the set of all msets whose elements are in X such that no element in the mset occurs more than m times.

If $X = \lbrace x_{1},x_{2}, ... ,x_{k} \rbrace$ then $[X]^{m} = \lbrace \lbrace {m_{1}/x_{1}, m_{2}/x_{2},... ,m_{k}/x_{k}}\rbrace$ : for $i = 1, 2,...,k \, ; \, m_{i} \in \lbrace 0, 1, 2,... ,m \rbrace \rbrace$.

Let $M, N \in [X]^{m}$. Then the following are defined :

• 1. $M = N$ if $C_{M}(x) = C_{N}(x)\ \forall x \in X$.

• 2. $M \subseteq N$ if $C_{M}(x) \le C_{N}(x) \, \forall \, x \in X.$

• 3. $M \subset N$ if $C_{M}(x) \le C_{N}(x) \, \forall \, x \in X$ and there exists an element $x \in X$ such that $C_{M}(x)< C_{N}(x)$

• 4. $P = M \cup N$ if $C_{P}(x) = \max \lbrace C_{M}(x), C_{N}(x) \rbrace \, \forall \, x \in X$.

• 5. $P = M \cap N$ if $C_{P}(x)= \min \lbrace C_{M}(x) , C_{N}(x) \rbrace \, \forall \, x \in X$.

• 6. $P = M -N$ if $C_{P}(x) = \max \lbrace C_{M}(x)- C_{N}(x) , \, 0 \rbrace \, \forall \, x \in X$.

Let $[X]^{m}$ be am mset space and $\{ M_{1}, M_{2}, M_{3}, ... \}$ be a collection of msets drawn from $[X]^{m}$. The following operations are possible under an arbitrary collection of msets.

• 1. The union $\cup _{i \in \vartriangle } M_{i}$ = $\{ C_{\cup M_{i}}(x)/x : C_{\cup M_{i}}(x)$ = $Max \{ C_{M_{i}} (x) : x \in X \} \}$.

• 2. The intersection $\cap _{i \in \vartriangle } M_{i}$ = $\{ C_{\cap M_{i}}(x)/x : C_{\cap M_{i}}(x)$ = $Min \{ C_{M_{i}} (x) : x \in X \} \}$.

Definition 2.5.

Let X be a support set and $[X]^{m}$ be the mset space defined over X. Then for any mset $M \in [X]^{m}$, the complement $M^{c}$ of M in $[X]^{m}$ is an element of $[X]^{m}$ such that $C_{M^{c}} (x) = m - C_{M}(x)$.

Definition 2.6.

A submset N of M is a whole submset of M with each element in N having full multiplicity as in M, i.e., $C_{N} (x) = C_{M} (x)$ for every $x \in N$.

Definition 2.7.

Let $M \in [X]^{m}$. The power whole mset of M, denoted by PW(M), is defined as the set of all whole submsets of M.

Definition 2.8.

Let $M \in [X]^{m}$. The power mset P(M) of M is the set of all submsets of M .

Definition 2.9.

Let $M \in [X]^{m}$. The power set of M is the support set of the power mset P(M) and is denoted by $P^{*}(M)$.

Definition 2.10.

Let $M \in [X]^{m}$ and $\tau \subseteq P^{*}(M)$. Then $\tau$ is called a multiset topology on M if $\tau$ satisfies the following properties.

• 1. $\phi$ and M are in $\tau$.

• 2. The union of the elements of any subcollection of $\tau$ is in $\tau$.

• 3. The intersection of any two elements of $\tau$ is in $\tau$.

Mathematically, a multiset topological space is an ordered pair $( M, \tau )$ consisting of an mset $M \in [X]^{m}$ and a multiset topology $\tau \subseteq P^{*}(M)$ on M. The multiset topology is abbreviated as an M-topology.

Definition 2.11.

Let $[X]^{m}$ be a space of multisets. A multipoint is a multiset M in $[X]^{m}$ such that

$$\begin{aligned} C_{M}(x) = {\left\{ \begin{array}{ll} k, &{} \text{ for } \; x \in X ; \\ 0,&{} \text{ for } \; y \ne x , \, y \in X \end{array}\right. } \end{aligned}$$

Remark 2.12.

A multipoint $\{k/x \}$ is a subset of a multiset M or $\{k/x \} \in M$ if $k \le C_{M}(x)$.

Definition 2.13.

Let M be a multiset in the space $[X]^{m}$. Let $k/x \in M$. Then k/x is said to be quasi-coincident with $j/y \in M$ if $k + j > m$.

Definition 2.14.

Let M be a multiset in the space $[X]^{m}$. Let $N \subseteq M$. Then k/x is said to be quasi-coincident with N if $k > C_{N^{c}}(x)$.

Definition 2.15.

A multiset M is said to be quasi-coincident with N, i.e., MqN at x iff $C_{M}(x) > C_{N^{c}}(x)$.

Remark 2.16.

If M and N are quasi-coincident at x then both $C_{M}(x)$ and $C_{N}(x)$ are non-zero and so M and N intersect at x.

Definition 2.17.

A multiset N in an M-topological space $(M,\tau _{1})$ is said to be Q-neighbourhood (Q-nbd)of k/x if and only if there exists an open mset P such that $k/x q P \subseteq N$.

Proposition 2.18.

Let $[X]^{m}$ be a space of multisets. Let $M ,N \in [X]^{m}$. Then $M \subseteq N$ iff M and $N^{c}$ are not quasi-coincident , i.e., $k/x \in M$ iff k/x is not quasi-coincident with $M^{c}$.

In the year 2019, Shravan and Tripathy developed the concept of Mixed multiset topological space. By mixing two M-topologies they obtained a new M-topology in the following way.

Let $[X]^{m}$ be an mset space and $M \in [X]^{m}$. Suppose for each multipoint $k/x \in M$ , there exists a collection ${\mathcal {U}}_{k/x}$ of submsets of M such that

• 1. k/xqU for all $U \in {\mathcal {U}}_{k/x}$.

• 2. $U,V \in {\mathcal {U}}_{k/x} \Rightarrow U \cap V \in {\mathcal {U}}_{k/x}$.

• 3. $U \in {\mathcal {U}}_{k/x}$ and $U \subseteq V \Rightarrow V \in {\mathcal {U}}_{k/x}$.

Then a new topology on M was defined in the following manner.

Definition 2.19.

Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces in $[X]^{m}$. Define $\tau _{1}(\tau _{2}) = \{A \subseteq [X]^{m}$ : For every multipoint k/x with k/xqA, there exists a $\tau _{2}$-Q-nbd B of k/x such that k/xqB and $\tau _{1}$-closure ${\overline{B}} \subseteq A \}$. This collection $\tau _{1}(\tau _{2})$ is an M-topology on M known as mixed multiset topology (mixed M-topology) on M and the pair $(M, \tau _{1}(\tau _{2}))$ as mixed M-topological space.

Definition 2.20.

Let $(M,\tau _{1})$ be an M-topological space and $A \subseteq M$. The closure of A is defined as the mset intersection of all the closed msets containing A and is denoted as Cl(A) or ${\overline{A}}$, i.e., $Cl(A)= \min \{C_{K}(x) : A \subseteq K, K \text {is a closed mset} \}$.

3 Main Results

Definition 3.1.

Let $M , N \in [X]^{m}$. Then M is said to be quasi-coincident with N, i.e., MqN iff there exists an element x in X such that $C_{M}(x) > C_{N^{c}}(x)$.

Clearly MqN iff $C_{M}(x) + C_{N}(x)> m$.

Shravan and Tripathy in [13] developed the concept of Mixed topological space on multisets (Mixed M-topological space) by using quasi-coincidence between a multipoint and a multiset. Here we redefine the mixed multiset topological space by using quasi-coincidence between two multisets. So our definition will generalise the existing definition.

Theorem 3.2.

Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces on M. Let $\tau _{1}(\tau _{2}) = \{A \subseteq M$ : For any submset B of M with BqA, there exists a $\tau _{2}$-open mset C such that BqC and $\tau _{1}$-closure ${\overline{C}} \subseteq A\}$. This collection $\tau _{1}(\tau _{2})$ will form an M-topology on M. We shall call it mixed multiset topology (Mixed M-topology) on M and the pair $(M, \tau _{1}(\tau _{2})$ as mixed multiset topological space (Mixed M-topological space).

Proof.

• (1) Clearly $\phi , M \in \tau _{1}(\tau _{2})$.

• (2) Let $A_{1},A_{2} \in \tau _{1}(\tau _{2})$. Let A be a submset of M such that $A q (A_{1} \cap A_{2})$. Clearly $A q (A_{1} \cap A_{2})$ implies that $A q A_{1}$ and $A q A_{2}$. Since $A_{1} \in \tau _{1}(\tau _{2})$, so, there exists a $\tau _{2}$-open mset $B_{1}$ such that $A q B_{1}$ and $\tau _{1}$-closure $\overline{B_{1}} \subseteq A_{1}$. Again since $A_{2} \in \tau _{1}(\tau _{2})$ so, there exists a $\tau _{2}$-open mset $B_{2}$ such that $A q B_{2}$ and $\tau _{1}$-closure $\overline{B_{2}} \subseteq A_{2}$. Obviously $B_{1} \cap B_{2} \in \tau _{2}$ and $\overline{B_{1} \cap B_{2} } \subseteq \overline{B_{1}} \cap \overline{B_{2}} \subseteq A_{1} \cap A_{2}$. Also $A q B_{1}$ , $A q B_{2}$ $\Rightarrow A q (B_{1} \cap B_{2})$. Thus for any submset A of M with $A q (A_{1} \cap A_{2})$, there exists a $\tau _{2}$-open mset $B_{1} \cap B_{2}$ such that $A q (B_{1} \cap B_{2})$ and $\tau _{1}$-closure $\overline{B_{1} \cap B_{2}} \subseteq A_{1} \cap A_{2}$. Therefore $A_{1} \cap A_{2} \in \tau _{1}(\tau _{2})$.

• (3) Let $\{A_{i} : i \in \vartriangle \}$ be a collection of elements of $\tau _{1}(\tau _{2})$. Let A be a submset of M such that $A q (\bigcup _{i \in \vartriangle } A_{i})$. Now $A q (\bigcup _{i \in \vartriangle } A_{i}) \Rightarrow C_{A}(x)+ C_{\bigcup _{i \in \vartriangle } A_{i}}(x) > m$ for some $x \Rightarrow C_{A}(x)+ \max _{i \in \vartriangle } \{C_{A_{i}} (x) \} > m \Rightarrow$ there exists a $A_{i_{0}} \in \{A_{i} : i \in \vartriangle \} , i_{0} \in \vartriangle$ such that $C_{A}(x)+ C_{A_{i_{0}}} (x) > m \Rightarrow Aq A_{i_{0}}$. Since $A_{i _{0}} \in \tau _{1}(\tau _{2})$, so there exists a $\tau _{2}$-open mset $B_{i_{0}}$ such that $A q B_{i_{0}}$ and $\tau _{1}$-closure $\overline{B_{i_{0}}} \subseteq A_{i_{0}}$. But $A_{i_{0}} \subseteq \bigcup _{i \in \vartriangle } A_{i}$. Therefore $\overline{B_{i_{0}}} \subseteq \bigcup _{i \in \vartriangle } A_{i}$. Thus for any submset A of M with $A q (\bigcup _{i \in \vartriangle } A_{i})$, there exists a $\tau _{2}$-open mset $B_{i_{0}}$ such that $A q B_{i_{0}}$ and $\tau _{1}$-closure $\overline{B_{i_{0}}}$ $\subseteq \bigcup _{i \in \vartriangle } A_{i}$. Therefore $\bigcup _{i \in \vartriangle } A_{i} \in \tau _{1}(\tau _{2})$.

Hence $\tau _{1}(\tau _{2})$ is an M-topology on M, i.e. $(M, \tau _{1}(\tau _{2}))$ is an M-topological space. $\square$

Example 3.3.

Let $X = \{a,b,c \}$ and $M = \{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M, \{2/a\},\{3/b\},\{2/a, 3/b\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{1/c\}, \{2/a, 1/c\}\}$. Clearly $\tau _{1}$ and $\tau _{2}$ are two M-topologies on M. From theorem 3.2, we obtain $\tau _{1}(\tau _{2})= \{\phi ,M,\{1/c\},\{2/a,1/c\}\}$ , which is clearly a mixed M-topology on M, i.e.,$(M,\tau _{1}(\tau _{2}))$ is a mixed M-topological space.

Definition 3.4.

Let $(M, \tau _{1}(\tau _{2}))$ be a mixed M-topological space. The complement of a submset N of M is the set $M - N$. The complement of N is denoted by $N^{c}$.

Definition 3.5.

Let $(M, \tau _{1}(\tau _{2}))$ be a mixed M-topological space. A submset N of M is called a $\tau _{1}(\tau _{2})$-open mset if $N \in \tau _{1}(\tau _{2})$. A submset N of M is called a $\tau _{1}(\tau _{2})$-closed mset if $M - N$ is open, i.e., if $N^{c} \in \tau _{1}(\tau _{2})$.

Definition 3.6.

Let $(M, \tau _{1}(\tau _{2}))$ be a mixed M-topological space and let $N \subseteq M$. Then the collection $\tau _{1}(\tau _{2})_{N}$ = $\{ N \cap G$ : $G \in \tau _{1}(\tau _{2})\}$ is a mixed M-topology on N. We shall call $\tau _{1}(\tau _{2})_{N}$ a subspace mixed M-topology and $(N, \tau _{1}(\tau _{2})_{N})$ a subspace of $(M, \tau _{1}(\tau _{2}))$.

Example 3.7.

Let $X = \{a,b,c \}$ and $M = \{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M, \{2/a\},\{3/b\},\{2/a, 3/b\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{1/c\}, \{2/a, 1/c\}\}$. Then $\tau _{1}(\tau _{2})= \{\phi ,M,\{1/c\},\{2/a, 1/c\}\}$. Let $N = \{1/a,1/c \}$. Then $\tau _{1}(\tau _{2})_{N} = \{\phi ,N, \{1/a\},\{1/c\} \}$ is a subspace mixed M-topology.

Remark 3.8.

Throughout this article a multipoint $\{ k/x \}$ of M in a mixed M-topological space $(M, \tau _{1}(\tau _{2}))$ will simply be denoted by k/x.

Theorem 3.9.

Let $M \in [X]^{m}$. Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces on M. If every $\tau _{1}$-Q-neighbourhood of k/x is a $\tau _{2}$-Q-neighbourhood of k/x for all multipoints k/x then $\tau _{1}$ is coarser than $\tau _{2}$.

Proof.

Let $A \in \tau _{1}$. We show that $A \in \tau _{2}$.

Now $A \in \tau _{1}$

$\Rightarrow A$ is a $\tau _{1}$-Q-neighbourhood of every multipoint k/x quasi-coincident with A.

$\Rightarrow A$ is a $\tau _{2}$-Q-neighbourhood of every multipoint k/x quasi-coincident with A.

$\Rightarrow$ there exists a $B_{k/x} \in \tau _{2}$ such that $k/x q B_{k/x}$ and $B_{k/x} \subseteq A$.

We now show that $A=\bigcup _{k/x} \, B_{k/x}$, union being taken over all such $B_{k/x} \subseteq A$.

We have $C_{A}(x)+ m - C_{A}(x)+1>m$

$\Rightarrow m-C_{A}(x)+1> m-C_{A}(x)$

$\Rightarrow m-C_{A}(x)+1>C_{A^{c}}(x)$

$\Rightarrow [m-C_{A}(x)+1]/x q A$

$\Rightarrow [m+1-C_{A}(x)]/x q D_{k/x}$ for some $D_{k/x} \in \tau _{2}$ with $D_{k/x} \subseteq A$

$\Rightarrow m+1-C_{A}(x)+C_{D_{k/x}}(x)>m$

$\Rightarrow C_{D_{k/x}}(x)>C_{A}(x)-1$

$\therefore C_{A}(x)= max_{k/x} \, \{C_{B_{k/x}}(x)\}$, i.e., $A=\bigcup _{k/x} \, B_{k/x}$.

Since every $B_{k/x}$ is a $\tau _{2}$-open mset, so A is a $\tau _{2}$-open mset. Therefore $\tau _{1}\subseteq \tau _{2}$. Hence proved. $\square$

Theorem 3.10.

Let $M \in [X]^{m}$. Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces on M. Then the mixed M-topology $\tau _{1}(\tau _{2})$ is coarser than $\tau _{2}$.

Proof.

Let A be a $\tau _{1}(\tau _{2})$-Q-neighbourhood of k/x. Then there exists $B \in \tau _{1}(\tau _{2})$ such that k/xqB and $B \subseteq A$. So, there exists a $\tau _{2}$-open mset $B_{2}$ such that $k/x q B_{2}$ and $\tau _{1}$-closure $\overline{B_{2}} \subseteq B$. Therefore there exists a $\tau _{2}$-open mset $B_{2}$ such that $k/x q B_{2}$ and $B_{2} \subseteq A$.[Because $B_{2}\subseteq \overline{B_{2}} \subseteq B \subseteq A$] and so A is a $\tau _{2}$-Q-neighbourhood of k/x. Therefore, by theorem 3.9 , we have $\tau _{1}(\tau _{2}) \subseteq \tau _{2}$. Hence proved. $\square$

Definition 3.11.

Let $(M,\tau _{1}(\tau _{2}))$ be a mixed M-topological space. Then M is said to be a $T_{0}$-space iff for any pair of multipoints $k_{1}/x_{1}$, $k_{2}/x_{2}$ in M such that $x_{1} \ne x_{2}$, there exists a $\tau _{1}(\tau _{2})$-open mset G such that $k_{1}/x_{1} \in G$ , $k_{2}/x_{2}$ $\notin$ G or there exists a $\tau _{1}(\tau _{2})$-open mset H such that $k_{1}/x_{1} \notin H$ , $k_{2}/x_{2} \in H$.

Theorem 3.12.

Let $(M,\tau _{1}(\tau _{2}))$ be a $T_{0}$-space. Then every subspace of it is a $T_{0}$-space and hence the property is hereditary.

Proof.

Let $N \subseteq M$. Then $(N,\tau _{1}(\tau _{2})_{N})$ is a subspace mixed M-topology, where $\tau _{1}(\tau _{2})_{N}$= $\lbrace N \cap G$ : $G \in \tau _{1}(\tau _{2})\rbrace$. We want to show $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{0}$-space. Let $k_{1}/x$ and $k_{2}/y$ be any two multipoints in N such that $x \ne y$ . Then $k_{1}/x, k_{2}/y \in M$ , $x\ne y$. Since $(M,\tau _{1}(\tau _{2}))$ is $T_{0}$-space, so there exists a $\tau _{1}(\tau _{2})$-open mset H such that $(k_{1}/x \in H$ , $k_{2}/y \notin H)$ or there exists a $\tau _{1}(\tau _{2})$-open mset K such that $(k_{1}/x \notin K$ , $k_{2}/y \in K)$. Therefore $(k_{1}/x \in N \cap H$ and $k_{2}/y \notin N \cap H)$ or $(k_{1}/x \notin N \cap K$ and $k_{2}/y \in N \cap K)$. Also $N \cap H, N \cap K \in \tau _{1}(\tau _{2})_{N}$. Thus for any two multipoints $k_{1}/x$ and $k_{2}/y$ in N ,there exists a $\tau _{1}(\tau _{2})_{N}$-open mset $N \cap H$ such that $(k_{1}/x \in N \cap H$ , $k_{2}/y \notin N \cap H)$ or there exists a $\tau _{1}(\tau _{2})_{N}$-open mset $N \cap K$ such that $(k_{1}/x \notin N \cap K$ , $k_{2}/y \in N \cap K)$. Hence $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{0}$-space and the property is hereditary. $\square$

Theorem 3.13.

Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1}(\tau _{2}))$ is $T_{0}$-space then $(M,\tau _{2})$ is also $T_{0}$-space.

Proof.

Let r/x and s/y, $x \ne y$ be two multipoints of M. Since $(M,\tau _{1}(\tau _{2}))$ is $T_{0}$, so there exists a $\tau _{1}(\tau _{2})$-open mset G such that $(r/x \in G$ and $s/y \notin G)$ or there exists a $\tau _{1}(\tau _{2})$-open mset H such that $(r/x \notin H$ and $s/y \in H)$. Since $\tau _{1}(\tau _{2})\subseteq \tau _{2}$, so $G,H \in \tau _{2}$. Thus there exists a $\tau _{2}$-open mset G such that $(r/x \in G$ and $s/y \notin G)$ or there exists a $\tau _{2}$-open mset H such that $(r/x \notin H$ and $s/y \in H)$. Therefore $(M,\tau _{2})$ is $T_{0}$-space. $\square$

Remark 3.14.

The converse of the theorem 3.13 is not true. The following counter example establishes that.

Example 3.15.

Let $X = \{a,b,c \}$ and $M = \{2/a,3/b \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M, \{2/a\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{3/b\}\}$. It is Clear that $\tau _{1}$ and $\tau _{2}$ are two M-topologies on M. Also from the definition of mixed M-topology we obtain that $\tau _{1}(\tau _{2})= \{\phi ,M,\{1/a\}\}$. We can observe that $(M,\tau _{2})$ is a $T_{0}$-space but $(M,\tau _{1}(\tau _{2}))$ is not a $T_{0}$-space.

Theorem 3.16.

Let $M \in [X]^{m}$. Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1})$ is $T_{0}$-space and $\tau _{1} \subseteq \tau _{2}$ then $(M,\tau _{1}(\tau _{2}))$ is a $T_{0}$-space.

Proof.

Let $k_{1}/x_{1}$ and $k_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1})$ is $T_{0}$, so there exists a $\tau _{1}$-open mset G such that $k_{1}/x_{1} \in G$ and $k_{2}/x_{2} \notin G$ or there exists a $\tau _{1}$-open mset H such that $k_{1}/x_{1} \notin H$ and $k_{2}/x_{2} \in H$. Suppose that $G \in \tau _{1}$ exists such that $k_{1}/x_{1} \in G$ and $k_{2}/x_{2} \notin G$. Let U be a $\tau _{1}$-Q-nbd of $k_{1}/x_{1}$ such that $k_{2}/x_{2} \notin U$ . Then for $G \in \tau _{1}$ , we have $k_{1}/x_{1} q G \subseteq U$. So, $k_{1} > C_{G^{c}}(x_{1})$ $\Rightarrow$ $k_{1} + C_{G}(x_{1}) > m$ $\Rightarrow$ $k_{1} + C_{U}(x_{1}) > m$ $\Rightarrow$ $k_{1}/x_{1} q U$. Since $\tau _{1} \subseteq \tau _{2}$ and $G \in \tau _{1}$, so G a is $\tau _{2}$-open mset such that $k_{1}/x_{1} q G$ and $\tau _{1}$-closure ${\overline{G}} \subseteq U$. Therefore $U \in \tau _{1} (\tau _{2})$. Also $k_{1}/x_{1} \in G \Rightarrow$ $k_{1}/x_{1} \in U$. So $U \in \tau _{1} (\tau _{2})$ such that $k_{1}/x_{1} \in U$ and $k_{2}/x_{2} \notin U$. Again if $H \in \tau _{1}$ exists such that $k_{1}/x_{1} \notin H$ and $k_{2}/x_{2} \in H$ and if V is a $\tau _{1}$-Q-nbd of $k_{2}/x_{2}$ such that $k_{1}/x_{1} \notin V$ then exactly in the same way we can show that $V \in \tau _{1} (\tau _{2})$ such that $k_{1}/x_{1} \notin V$ and $k_{2}/x_{2} \in V$. Thus there exists a $\tau _{1} (\tau _{2})$-open mset U such that $k_{1}/x_{1} \in U$ and $k_{2}/x_{2} \notin U$ or there exists a $\tau _{1} (\tau _{2})$-open mset V such that $k_{1}/x_{1} \notin V$ and $k_{2}/x_{2} \in V$. Hence $(M,\tau _{1}(\tau _{2}))$ is a $T_{0}$-space. $\square$

Definition 3.17.

Let $(M,\tau _{1}(\tau _{2}))$ be a mixed M-topological space. Then M is said to be a $T_{1}$-space iff for any pair of multipoints $k_{1}/x_{1}$, $k_{2}/x_{2}$ in M such that $x_{1} \ne x_{2}$, there exist $\tau _{1}(\tau _{2})$-open msets G , H such that $(k_{1}/x_{1} \in G$ , $k_{2}/x_{2}$ $\notin$ G) and $(k_{1}/x_{1} \notin H$ , $k_{2}/x_{2} \in H)$.

Theorem 3.18.

Let $(M,\tau _{1}(\tau _{2}))$ be a $T_{1}$-space. Then every subspace of it is a $T_{1}$-space and hence the property is hereditary.

Proof.

Let $N \subseteq M$. Then $(N,\tau _{1}(\tau _{2})_{N})$ is subspace mixed M-topology, where $\tau _{1}(\tau _{2})_{N}= \{N \cap G : G \in \tau _{1}(\tau _{2})\}$. We want to show $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{1}$-space. Let $k_{1}/x$ and $k_{2}/y$ be any two multipoints in N such that $x\ne y$. Then $k_{1}/x, k_{2}/y \in M$ such that $x \ne y$. Since $(M,\tau _{1}(\tau _{2}))$ is $T_{1}$-space, so there exist two $\tau _{1}(\tau _{2})$-open msets H and K such that $(k_{1}/x \in H$ and $k_{2}/y \notin H)$ and $(k_{1}/x \notin K$ and $k_{2}/y \in K)$. Therefore $(k_{1}/x \in N \cap H$ and $k_{2}/y \notin N \cap H)$ and $(k_{1}/x \notin N \cap K$ and $k_{2}/y \in N \cap K)$. Also $N\cap H, N \cap K \in \tau _{1}(\tau _{2})_{N}$. Therefore $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{1}$-space and hence the property is hereditary. $\square$

Theorem 3.19.

Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1}(\tau _{2}))$ is $T_{1}$-space then $(M,\tau _{2})$ is also $T_{1}$-space.

Proof.

Let r/x and s/y, $x \ne y$ be two multipoints of M. Since $(M,\tau _{1}(\tau _{2}))$ is $T_{1}$, so there exist two $\tau _{1}(\tau _{2})$-open msets G and H such that $(r/x \in G$ , $s/y \notin G)$ and $(r/x \notin H$ , $s/y \in H)$. Since $\tau _{1}(\tau _{2})\subseteq \tau _{2}$, so $G,H \in \tau _{2}$. Thus there exist two $\tau _{2}$-open msets G and H such that $(r/x \in G$ , $s/y \notin G)$ and $(r/x \notin H$ and $s/y \in H)$. Therefore $(M,\tau _{2})$ is $T_{1}$-space. $\square$

Remark 3.20.

The converse of the theorem 3.19 is not true. We show this in the following example.

Example 3.21.

Let $X=\{a,b,c \}$ and $M=\{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M,\{2/a\},\{3/b\},\{2/a,3/b\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{3/b\}\ , \{1/c\},\{2/a, 3/b\}, \{1/c,3/b\},\{2/a,1/c\}\}$. Clearly $\tau _{1}$ and $\tau _{2}$ are two M-topologies on M. Also from the definition of mixed M-topology we obtain that $\tau _{1}(\tau _{2})= \{\phi ,M,\{1/c\},\{1/c, 3/b\},\{2/a, 1/c\}\}$. We can observe that $(M,\tau _{2})$ is a $T_{1}$-space but $(M,\tau _{1}(\tau _{2}))$ is not a $T_{1}$-space.

Theorem 3.22.

Let $M \in [X]^{m}$. Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1})$ is $T_{1}$-space and $\tau _{1} \subseteq \tau _{2}$ then $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space.

Proof.

Let $k_{1}/x_{1}$ and $k_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1})$ is $T_{1}$, so there exist $\tau _{1}$-open msets G , H such that $(k_{1}/x_{1} \in G$ , $k_{2}/x_{2}$ $\notin$ G) and $(k_{1}/x_{1} \notin H$ , $k_{2}/x_{2} \in H)$ Let U be a $\tau _{1}$-Q-nbd of $k_{1}/x_{1}$ such that $k_{2}/x_{2} \notin U$ . Then for $G \in \tau _{1}$ , we have $k_{1}/x_{1} q G \subseteq U$. So, $k_{1} > C_{G^{c}}(x_{1})$ $\Rightarrow$ $k_{1} + C_{G}(x_{1}) > m$ $\Rightarrow$ $k_{1} + C_{U}(x_{1}) > m$ $\Rightarrow$ $k_{1}/x_{1} q U$. Since $\tau _{1} \subseteq \tau _{2}$ and $G \in \tau _{1}$, so G a is $\tau _{2}$-open mset such that $k_{1}/x_{1} q G$ and $\tau _{1}$-closure ${\overline{G}} \subseteq U$. Therefore $U \in \tau _{1} (\tau _{2})$. Also $k_{1}/x_{1} \in G \Rightarrow$ $k_{1}/x_{1} \in U$. So $U \in \tau _{1} (\tau _{2})$ such that $k_{1}/x_{1} \in U$ and $k_{2}/x_{2} \notin U$. Again if V is a $\tau _{1}$-Q-nbd of $k_{2}/x_{2}$ such that $k_{1}/x_{1} \notin V$ then exactly in the same way we can show that $V \in \tau _{1} (\tau _{2})$ such that $k_{1}/x_{1} \notin V$ and $k_{2}/x_{2} \in V$. Thus there exist $\tau _{1} (\tau _{2})$-open msets U , V such that $(k_{1}/x_{1} \in U$ and $k_{2}/x_{2} \notin U)$ and $(k_{1}/x_{1} \notin V$ and $k_{2}/x_{2} \in V)$. Hence $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space. $\square$

Theorem 3.23.

If $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space then it is a $T_{0}$-space.

Proof.

Let $m_{1}/x_{1}$ and $m_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space, so there exist $\tau _{1}(\tau _{2})$-open msets H and K such that $(m_{1}/x_{1} \in H$ , $m_{2}/x_{2} \notin H)$ and $(m_{1}/x_{1} \notin K$ , $m_{2}/x_{2} \in K)$. Hence $(M,\tau _{1}(\tau _{2}))$ is a $T_{0}$-space. $\square$

Remark 3.24.

The converse of the theorem 3.23 is not true. The following counter example reveals that.

Example 3.25.

Let $X = \{a,b,c \}$ and $M = \{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M,\{2/a\},\{3/b\},\{2/a, 3/b\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{1/c\}, \{2/a, 1/c\}\}$. Clearly $\tau _{1}$ and $\tau _{2}$ are two M-topologies on M. Also we obtain $\tau _{1}(\tau _{2})= \{\phi ,M,\{1/c\},\{2/a, 1/c\}\}$. We see that $(M,\tau _{1}(\tau _{2}))$ is a $T_{0}$-space but not $T_{1}$-space.

Theorem 3.26.

Let $(M,\tau _{1}(\tau _{2}))$ be a mixed M-topological space. If k/x is $\tau _{1}(\tau _{2})$-closed mset for all $x \in M^{*}$, $k = C_{M}(x)$ then $(M,\tau _{1}(\tau _{2}))$ is $T_{1}$-space.

Proof.

Let $m_{1}/x_{1}$ and $m_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Let $C_{M}(x_{1}) = k_{1}$ and $C_{M}(x_{2}) = k_{2}$. Then by hypothesis, $k_{1}/x_{1}$ and $k_{2}/x_{2}$ are $\tau _{1}(\tau _{2})$-closed msets of M. Let $k_{1}/x_{1} = r$ and $k_{2}/x_{2} = s$. Then $r^{c} , s^{c} \in \tau _{1}(\tau _{2})$ and so $m_{1}/x_{1} \in s^{c} ,m_{2}/x_{2} \notin s^{c}$ and $m_{1}/x_{1} \notin r^{c} ,m_{2}/x_{2} \in r^{c}$. Hence $(M,\tau _{1}(\tau _{2}))$ is $T_{1}$-space. $\square$

Definition 3.27.

Let $(M,\tau _{1}(\tau _{2}))$ be a mixed M-topological space. Then M is said to be a $T_{2}$-space iff for any pair of multipoints $k_{1}/x_{1}$, $k_{2}/x_{2}$ in M such that $x_{1} \ne x_{2}$, there exist $\tau _{1}(\tau _{2})$-open msets G , H such that $k_{1}/x_{1} \in G , k_{2}/x_{2} \in$ H and $G \cap H = \phi$.

Theorem 3.28.

Let $(M,\tau _{1}(\tau _{2}))$ be a $T_{2}$-space. Then every subspace of it is a $T_{2}$-space and hence the property is hereditary.

Proof.

Let $N \subseteq M$. Then $(N,\tau _{1}(\tau _{2})_{N})$ is subspace mixed M-topology, where $\tau _{1}(\tau _{2})_{N}= \{N \cap G : G \in \tau _{1}(\tau _{2})\}$. We want to show $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{2}$-space. Let $k_{1}/x$ and $k_{2}/y$ be any two multipoints in N such that $x\ne y$. Then $k_{1}/x, k_{2}/y \in M$ such that $x \ne y$. Since $(M,\tau _{1}(\tau _{2}))$ is $T_{2}$-space, so there exist two $\tau _{1}(\tau _{2})$-open msets H and K such that $k_{1}/x \in H$ , $k_{2}/y \in K$ and $H \cap K = \phi$. Therefore $k_{1}/x \in N \cap H$ , $k_{2}/y \in N \cap K$ and $(N \cap H) \cap (N \cap K) = \phi$. Also $N\cap H, N \cap K \in \tau _{1}(\tau _{2})_{N}$. Therefore $(N,\tau _{1}(\tau _{2})_{N})$ is $T_{2}$-space and hence the property is hereditary. $\square$

Theorem 3.29.

Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1}(\tau _{2}))$ is $T_{2}$-space then $(M,\tau _{2})$ is also a $T_{2}$-space.

Proof.

Let $m_{1}/x_{1}$ and $m_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1}(\tau _{2}))$ is a $T_{2}$-space, so there exist $\tau _{1}(\tau _{2})$-open msets H and K such that $m_{1}/x_{1} \in H$ , $m_{2}/x_{2} \in K$ and $H\cap K = \phi$. Since $\tau _{1}(\tau _{2})\subseteq \tau _{2}$, so $H,K \in \tau _{2}$. Therefore $(M,\tau _{2})$ is $T_{2}$-space. $\square$

Remark 3.30.

The converse of the theorem 3.29 is not true. We establish that in the following example.

Example 3.31.

Let $X=\{a,b,c \}$ and $M=\{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M,\{2/a\},\{3/b\},\{2/a,3/b\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a\},\{3/b\}\ , \{1/c\},\{2/a, 3/b\}, \{1/c,3/b\},\{2/a,1/c\}\}$. Clearly $\tau _{1}$ and $\tau _{2}$ are two M-topologies on M. Also from the definition of mixed M-topology we obtain that $\tau _{1}(\tau _{2}))= \{\phi ,M,\{1/c\},\{1/c, 3/b\},\{2/a, 1/c\}\}$. We see that $(M,\tau _{2})$ is a $T_{2}$-space but $(M,\tau _{1}(\tau _{2}))$ is not a $T_{2}$-space.

Theorem 3.32.

Let $M \in [X]^{m}$. Let $(M,\tau _{1})$ and $(M,\tau _{2})$ be two M-topological spaces. If $(M,\tau _{1})$ is $T_{2}$-space and $\tau _{1} \subseteq \tau _{2}$ then $(M,\tau _{1}(\tau _{2}))$ is a $T_{2}$-space.

Proof.

Let $k_{1}/x_{1}$ and $k_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1})$ is $T_{2}$, so there exist $\tau _{1}$-open msets G , H such that $k_{1}/x_{1} \in G$ , $k_{2}/x_{2}$ $\in H$ and $G \cap H = \phi$. Let U be a $\tau _{1}$-Q-nbd of $k_{1}/x_{1}$. Then for $G \in \tau _{1}$ , we have $k_{1}/x_{1} q G \subseteq U$. So, $k_{1} > C_{G^{c}}(x_{1})$ $\Rightarrow$ $k_{1} + C_{G}(x_{1}) > m$ $\Rightarrow$ $k_{1} + C_{U}(x_{1}) > m$ $\Rightarrow$ $k_{1}/x_{1} q U$. Since $\tau _{1} \subseteq \tau _{2}$ and $G \in \tau _{1}$, so G a is $\tau _{2}$-open mset such that $k_{1}/x_{1} q G$ and $\tau _{1}$-closure ${\overline{G}} \subseteq U$. Therefore $U \in \tau _{1} (\tau _{2})$. Also $k_{1}/x_{1} \in G \Rightarrow$ $k_{1}/x_{1} \in U$. Again if V is a $\tau _{1}$-Q-nbd of $k_{2}/x_{2}$ then exactly in the same way we can show that $V \in \tau _{1} (\tau _{2})$ such that $k_{2}/x_{2} \in V$. Also $G \cap H = \phi \Rightarrow U \cap V = \phi$. Thus there exist $\tau _{1} (\tau _{2})$-open msets U , V such that $k_{1}/x_{1} \in U$ , $k_{2}/x_{2} \in V$ and $U \cap V = \phi$.. Hence $(M,\tau _{1}(\tau _{2}))$ is a $T_{2}$-space. $\square$

Theorem 3.33.

If $(M,\tau _{1}(\tau _{2}))$ is a $T_{2}$-space then it is a $T_{1}$-space.

Proof.

Let $m_{1}/x_{1}$ and $m_{2}/x_{2}$ be any two multipoints of M such that $x_{1} \ne x_{2}$. Since $(M,\tau _{1}(\tau _{2}))$ is a $T_{2}$-space, so there exist $\tau _{1}(\tau _{2})$-open msets H and K such that $m_{1}/x_{1} \in H$ , $m_{2}/x_{2} \in K$ and $H \cap K = \phi$. Since $H \cap K = \phi$, so $m_{1}/x_{1} \notin K$ and $m_{2}/x_{2} \notin H)$ Thus there are $\tau _{1}(\tau _{2})$-open msets H and K such that $(m_{1}/x_{1} \in H$ , $m_{2}/x_{2} \notin H)$ and $(m_{1}/x_{1} \notin K$ , $m_{2}/x_{2} \in K)$. Hence $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space. $\square$

Remark 3.34.

The converse of the theorem 3.33 is not true. The following example confirms that.

Example 3.35.

Let $X=\{a,b,c \}$ and $M=\{2/a,3/b,1/c \} \in [X]^{3}$. Let $\tau _{1} = \{\phi ,M,\{2/a\},\{3/b\},\{1/c\}\}$ and $\tau _{2} = \{\phi ,M,\{2/a, 3/b\}, \{1/c,3/b\},\{2/a,1/c\}\}$. Then $\tau _{1}(\tau _{2})= \{\phi ,M,\{2/a, 3/b\}, \{1/c,3/b\},\{2/a,1/c\}\}$. We see that $(M,\tau _{1}(\tau _{2}))$ is a $T_{1}$-space but not a $T_{2}$-space.