Ratio Mathematica                                                                            Volume 44, 2022 
 

  

 

Further Diversification of  

Nano Binary Open Sets 

 

 

J. Jasmine Elizabeth1  

G. Hari Siva Annam2 

 

 

Abstract 

The purpose of this paper is to introduce and study the nano binary exterior, nano binary 

border and nano binary derived in nano binary topological spaces. Also studied their 

characterizations.  

 

Keywords: 𝑁𝐵- Derived, 𝑁𝐵- Exterior, 𝑁𝐵- Border. 
 

2010 AMS Subject Classification: 54A05, 54A993. 

 

 

 

 

 

 

 

 

 

 

 

 

 
1Assistant Professor, Kamaraj College, Thoothukudi, (Part Time Research Scholar [Reg no. 

19122102092008], Manonmaniam Sundaranar University, Tirunelveli-627012) Tamil Nadu, India. 

jasmineelizabeth89@gmail.com 
2Assistant Professor, PG and Research Department of Mathematics, Kamaraj College, Thoothukudi-

628003, Tamil Nadu, India. hsannam@yahoo.com. Affiliated to Manonmaniam Sundaranar University, 

Tirunelveli-627012, Tamil Nadu, India. 
3Received on June 4th, 2022. Accepted on Sep 1st, 2022. Published on Nov 30th, 2022. doi: 

10.23755/rm.v44i0.884. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper is published 

under the CC-BY licence agreement 

15

mailto:hsannam@yahoo.com


J. Jasmine Elizabeth, G. Hari Siva Annam 
 

 
 

1. Introduction 

M. Lellis Thivagar [1] introduced the concept of nano topological space with 

respect to a subset X of a universe U. S. Nithyanantha Jothi and P. Thangavelu [2] 

introduced the concept of binary topological spaces. By combining these two concepts 

Dr. G. Hari Siva Annam and J. Jasmine Elizabeth [3] introduced nano binary 

topological spaces. In this paper we have introduced the nano binary border, nano 

binary derived and nano binary exterior in nano binary topological spaces. Also studied 

their properties and characterizations with suitable examples. 

 

2. Preliminaries 
Definition 2.1: [3] Let (𝑈1, 𝑈2) be a non-empty finite set of objects called the universe 
and R be an equivalence relation on (𝑈1, 𝑈2)named as the indiscernibility relation. 
Elements belonging to the same equivalence class are said to be indiscernible with one 

another. The pair (𝑈1, 𝑈2, 𝑅)is said to be the approximation space. Let (𝑋1, 𝑋2) ⊆
(𝑈1, 𝑈2). 
 

1. The lower approximation of (𝑋1, 𝑋2) with respect to R is the set of all objects, which 
can be for certain classified as (𝑋1, 𝑋2)with respect to R and it is denoted by 𝐿𝑅 (𝑋1, 𝑋2). 
That is,𝐿𝑅 (𝑋1, 𝑋2) = ⋃ {𝑅(𝑥1,   𝑥2)𝜖(𝑈1,𝑈2) (𝑥1, 𝑥2): 𝑅(𝑥1, 𝑥2) ⊆ (𝑋1, 𝑋2)} 

Where 𝑅(𝑥1, 𝑥2)denotes the equivalence class determined by (𝑥1, 𝑥2). 
 

2. The upper approximation of (𝑋1, 𝑋2) with respect to R is the set of all objects, which 
can be possibly classified as (𝑋1, 𝑋2)with respect to R and it is denoted by 𝑈𝑅 (𝑋1, 𝑋2). 
That is,𝑈𝑅 (𝑋1, 𝑋2) = ⋃ {𝑅(𝑥1,𝑥2)𝜖(𝑈1,𝑈2) (𝑥1, 𝑥2): 𝑅(𝑥1, 𝑥2)⋂(𝑋1, 𝑋2) ≠ ϕ}. 

 

3. The boundary region of (𝑋1, 𝑋2) with respect to R is the set of all objects, which can 
be classified neither as (𝑋1, 𝑋2) nor as not −(𝑋1, 𝑋2) with respect to R and it is denoted 
by 𝐵𝑅 (𝑋1, 𝑋2). 
That is, 𝐵𝑅 (𝑋1, 𝑋2) = 𝑈𝑅 (𝑋1, 𝑋2) − 𝐿𝑅 (𝑋1, 𝑋2). 
 

Definition 2.2: [3] Let (𝑈1, 𝑈2) be the universe, R be an equivalence on (𝑈1, 𝑈2) and 
𝜏𝑅 (𝑋1, 𝑋2) = {(𝑈1, 𝑈2), (𝜙, 𝜙), 𝐿𝑅 (𝑋1, 𝑋2), 𝑈𝑅 (𝑋1, 𝑋2), 𝐵𝑅 (𝑋1, 𝑋2)} where (𝑋1, 𝑋2) ⊆
(𝑈1, 𝑈2). Then by the property R(𝑋1, 𝑋2)satisfies the following axioms 
 

1. (𝑈1, 𝑈2)and (𝜙, 𝜙) 𝜖 𝜏𝑅 (𝑋1, 𝑋2). 
2. The union of the elements of any sub collection of 𝜏𝑅 (𝑋1, 𝑋2) is in 𝜏𝑅 (𝑋1, 𝑋2). 
3. The intersection of the elements of any finite sub collection of 𝜏𝑅 (𝑋1, 𝑋2) is in 
𝜏𝑅 (𝑋1, 𝑋2). 
 

That is,𝜏𝑅 (𝑋1, 𝑋2) is a topology on (𝑈1, 𝑈2) called the nano binary topology on 
(𝑈1, 𝑈2) with respect to(𝑋1, 𝑋2). 

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Further Diversification of Nano Binary Open Sets 

We call (𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2))as the nano binary topological spaces. The elements of 
𝜏𝑅 (𝑋1, 𝑋2)are called as nano binary open sets and it is denoted by 𝑁𝐵 open sets. Their 
complement is called 𝑁𝐵 closed sets. 
 

Definition 2.3: [3] If (𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2))is a nano binary topological spaces with 

respect to (𝑋1, 𝑋2) and if (𝐻1, 𝐻2) ⊆ (𝑈1, 𝑈2), then the nano binary interior of (𝐻1, 𝐻2) 
is defined as the union of all 𝑁𝐵open subsets of (𝐴1, 𝐴2) and it is defined by 
𝑁𝐵

∘(𝐻1, 𝐻2). 
 

That is, 𝑁𝐵
∘(𝐻1, 𝐻2) is the largest 𝑁𝐵open subset of(𝐻1, 𝐻2). The nano binary 

closure of (𝐻1, 𝐻2) is defined as the intersection of all 𝑁𝐵closed sets containing 

(𝐻1, 𝐻2) and it is denoted by 𝑁𝐵 (𝐻1, 𝐻2). 
 

That is, 𝑁𝐵 (𝐻1, 𝐻2) is the smallest 𝑁𝐵closed set containing(𝐻1, 𝐻2). 
 

3. Nano Binary Derived 

Definition 3.1: A point (𝑥1, 𝑥2) ∈ (𝑈1, 𝑈2) is said to be a 𝑁𝐵 limit point of (𝐴1, 𝐴2) if 

for each 𝑁𝐵-open set (𝐾1, 𝐾2) containing (𝑥1, 𝑥2) satisfies (𝐾1, 𝐾2) ∩ ((𝐴1, 𝐴2) −

(𝑥1, 𝑥2)) ≠ (∅, ∅). 

 

Definition 3.2: The set of all 𝑁𝐵 limit points of (𝐴1, 𝐴2) is said to be nano binary 
derived set and is denoted by 𝑁𝐵 _𝐷(𝐴1, 𝐴2). 
 

Theorem 3.3: In (𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)), let (𝐴1, 𝐴2) and (𝐵1, 𝐵2) be two subsets of 
(𝑈1, 𝑈2). Then the following holds:  
1) 𝑁𝐵 _𝐷(∅, ∅) = (∅, ∅). 
2) If (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) then (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷((𝐴1, 𝐴2) − (𝑥1, 𝑥2)). 
3) If  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2), then 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). 
4) 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐵1, 𝐵2) = 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)). 
Proof: 1) Let (𝑥1, 𝑥2) ∈ (𝑈1, 𝑈2) and (𝐺1, 𝐺2) be a 𝑁𝐵-open set containing (𝑥1, 𝑥2). 

Then ((𝐺1, 𝐺2) − (𝑥1, 𝑥2)) ∩ (∅, ∅) = (∅, ∅) ⇒ (𝑥1, 𝑥2) ∉ 𝑁𝐵 _𝐷(∅, ∅). Therefore, for 

any (𝑥1, 𝑥2) ∈ (𝑈1, 𝑈2), (𝑥1, 𝑥2) is not a𝑁𝐵 limit point of (∅, ∅). Hence 𝑁𝐵 _𝐷(∅, ∅) =
(∅, ∅). 

2)Let (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐴1, 𝐴2). Then (𝐺1, 𝐺2) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) ≠ (∅, ∅), for 

every 𝑁𝐵-open set (𝐺1, 𝐺2) containing (𝑥1, 𝑥2) implies every 𝑁𝐵-open set (𝐺1, 𝐺2) of 
(𝑥1, 𝑥2), contains at least one point other than (𝑥1, 𝑥2) of  (𝐴1, 𝐴2). Therefore (𝑥1, 𝑥2) ∈
𝑁𝐵 _𝐷((𝐴1, 𝐴2) − (𝑥1, 𝑥2)). 
3)Let (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐴1, 𝐴2). Then (𝐺1, 𝐺2) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) ≠ (∅, ∅), for 

every 𝑁𝐵-open set (𝐺1, 𝐺2) containing (𝑥1, 𝑥2). Since  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2) implies 
(𝐺1, 𝐺2) ∩ ((𝐵1, 𝐵2) − (𝑥1, 𝑥2)) ≠ (∅, ∅) ⇒ (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). Thus (𝑥1, 𝑥2) ∈

𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⇒ (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). Therefore 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). 
4)Since (𝐴1, 𝐴2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2) and (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2). By (3), 
𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) and 𝑁𝐵 _𝐷(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪

17



J. Jasmine Elizabeth, G. Hari Siva Annam 
 

 
 

(𝐵1, 𝐵2)). Therefore, 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)).… 
(1). Let (𝑥1, 𝑥2) ∉ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). Then (𝑥1, 𝑥2) ∉ 𝑁𝐵 𝐷(𝐴1,𝐴2)

and 

(𝑥1, 𝑥2) ∉ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). Therefore, there exists 𝑁𝐵-open sets (𝐺1, 𝐺2) and (𝐻1, 𝐻2) 
containing (𝑥1, 𝑥2) such that (𝐺1, 𝐺2) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) = (∅, ∅) and (𝐻1, 𝐻2) ∩

((𝐵1, 𝐵2) − (𝑥1, 𝑥2)) = (∅, ∅). Since (𝐺1, 𝐺2) ∩ (𝐻1, 𝐻2) ⊆ (𝐺1, 𝐺2) and (𝐻1, 𝐻2), 

((𝐺1, 𝐺2) ∩ (𝐻1, 𝐻2)) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) = (∅, ∅) and ((𝐺1, 𝐺2) ∩ (𝐻1, 𝐻2)) ∩

((𝐵1, 𝐵2) − (𝑥1, 𝑥2)) = (∅, ∅). Also (𝐺1, 𝐺2) ∩ (𝐻1, 𝐻2) is a 𝑁𝐵-open set 

containing(𝑥1, 𝑥2). Therefore, ((𝐺1, 𝐺2) ∩ (𝐻1, 𝐻2)) ∩ (((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) −

(𝑥1, 𝑥2)) = (∅, ∅). That is, (𝑥1, 𝑥2) is not a 𝑁𝐵 limit point of (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2). Hence 
(𝑥1, 𝑥2) ∉ 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)). Therefore, 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆
𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐵1, 𝐵2)…. (2). From (1) and (2),𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∪
𝑁𝐵 _𝐷(𝐵1, 𝐵2) = 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)). 
 

Theorem 3.4: Let (𝐴1, 𝐴2)and (𝐵1, 𝐵2) be two subsets of 𝑁𝐵 topological space 
(𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)).Then 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). 
 

Proof: Since (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2)and (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐵1, 𝐵2). By the 
previous theorem,𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐷(𝐴1, 𝐴2)and 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩
(𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐷(𝐵1, 𝐵2). Therefore,𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∩
𝑁𝐵 _𝐷(𝐵1, 𝐵2). 
 

Remark 3.5: The reverse inclusion may not true as shown in the following example. 

 

Example 3.6: U1 = {𝑎,  𝑏,  𝑐},  U2 = {1,  2}   with
(U1, U2)

R
⁄ = {({a, b}, {2}), ({c}, {1})}. 

Let (X1, X2) =({b}, {2}). Then τR(X1, X2) ={(Φ, Φ), (U1, U2), ({a, b}, {2})}. Here 
(𝐴1, 𝐴2) = ({a, b}, {1})and (𝐵1, 𝐵2) = ({b, c}, {1,2}), (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)=({b}, {1}) 
and hence 𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2))={({a}, {1}), ({a}, {2}), ({b}, {2}), 
({c}, {1}), ({c}, {2})}. Also 𝑁𝐵 _𝐷(𝐴1, 𝐴2)={({a}, {1}), ({a}, {2}), ({b}, {1}), 
({b}, {2}), ({c}, {1}), ({c}, {2})} and 𝑁𝐵 _𝐷(𝐵1, 𝐵2)={({a}, {1}), ({a}, {2}), ({b}, {1}), 
({c}, {1}), ({c}, {2})}. But 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐷(𝐵1, 𝐵2)={({a}, {1}), ({a}, {2}), 
({b}, {1}), ({c}, {1}), ({c}, {2})}.  
Thus, 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐷(𝐵1, 𝐵2)⊈𝑁𝐵 _𝐷((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 
 

Theorem 3.7: 𝑁𝐵 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2), where (𝐴1, 𝐴2) ⊆ (U1, U2). 
Proof: If (𝑥1, 𝑥2) ∈ (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2), Then (𝑥1, 𝑥2) ∈ (𝐴1, 𝐴2) or  (𝑥1, 𝑥2) ∈
𝑁𝐵 _𝐷(𝐴1, 𝐴2). Let (𝑥1, 𝑥2) ∉ (𝐴1, 𝐴2). Then (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐴1, 𝐴2). Therefore, for 
every 𝑁𝐵-open set (𝐺1, 𝐺2)containing (𝑥1, 𝑥2), (𝐺1, 𝐺2) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) ≠

(∅, ∅). Since (𝑥1, 𝑥2) ∉ (𝐴1, 𝐴2), (𝐺1, 𝐺2) ∩ (𝐴1, 𝐴2) ≠ (∅, ∅). Therefore, (𝑥1, 𝑥2) ∈

𝑁𝐵 (𝐴1, 𝐴2). Therefore, (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⊆ 𝑁𝐵 (𝐴1, 𝐴2)…. (1). Let (𝑥1, 𝑥2) ∈

𝑁𝐵 (𝐴1, 𝐴2)and (𝑥1, 𝑥2) ∈ (𝐴1, 𝐴2). Then the result is obvious. If (𝑥1, 𝑥2) ∈

𝑁𝐵 (𝐴1, 𝐴2)and (𝑥1, 𝑥2) ∉ (𝐴1, 𝐴2). Therefore, (𝐺1, 𝐺2) ∩ (𝐴1, 𝐴2) ≠ (∅, ∅) for every 
𝑁𝐵-open set (𝐺1, 𝐺2) containing (𝑥1, 𝑥2) and hence (𝐺1, 𝐺2) ∩ ((𝐴1, 𝐴2) − (𝑥1, 𝑥2)) ≠

18



Further Diversification of Nano Binary Open Sets 

(∅, ∅). Therefore, (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐷(𝐴1, 𝐴2) and hence (𝑥1, 𝑥2) ∈ (𝐴1, 𝐴2) ∪

𝑁𝐵 _𝐷(𝐴1, 𝐴2). Therefore, 𝑁𝐵 (𝐴1, 𝐴2) ⊆ (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2)… (2). From (1) and 

(2), 𝑁𝐵 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2). 
 

Result 3.8: 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) − 𝑁𝐵 _𝐷[(U1, U2) − (𝐴1, 𝐴2)], where (𝐴1, 𝐴2) ⊆

(U1, U2). 

Proof: By the previous theorem, 𝑁𝐵 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐷(𝐴1, 𝐴2)⇒(U1, U2) −

𝑁𝐵 (𝐴1, 𝐴2) = ((U1, U2) − (𝐴1, 𝐴2)) ∩ ((U1, U2) − 𝑁𝐵 _𝐷(𝐴1, 𝐴2)) ⇒ (U1, U2) −

𝑁𝐵 (𝐴1, 𝐴2) = ((U1, U2) − (𝐴1, 𝐴2)) − 𝑁𝐵 _𝐷(𝐴1, 𝐴2) ⇒ 𝑁𝐵
𝑜

((U1, U2) − (𝐴1, 𝐴2)) =

((U1, U2) − (𝐴1, 𝐴2)) − 𝑁𝐵 _𝐷(𝐴1, 𝐴2). By replacing (U1, U2) − (𝐴1, 𝐴2) by (𝐴1, 𝐴2) 

and (𝐴1, 𝐴2)by (U1, U2) − (𝐴1, 𝐴2), 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) − 𝑁𝐵 _𝐷[(U1, U2) −

(𝐴1, 𝐴2)]. 
 

4. Nano Binary Exterior 

Definition 4.1: For a subset(𝐴1, 𝐴2) ⊆ (𝑈1, 𝑈2), the nano binary exterior of (𝐴1, 𝐴2) is 
defined as𝑁𝐵

𝑜
((𝑈1, 𝑈2) − (𝐴1, 𝐴2)). It is denoted by 𝑁𝐵 _𝐸(𝐴1, 𝐴2). 

 

Definition 4.2: For a subset(𝐴1, 𝐴2) ⊆ (𝑈1, 𝑈2), the nano binary border of (𝐴1, 𝐴2) is 
defined as (𝐴1, 𝐴2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2). It is denoted by 𝑁𝐵 _𝐵(𝐴1, 𝐴2). 
 

Theorem 4.3: Let (𝐴1, 𝐴2)and (𝐵1, 𝐵2) be two subsets of 𝑁𝐵 topological space 
(𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)). Then the following holds:  
1) If  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2), then 𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐸(𝐴1, 𝐴2). 

2) 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐸(𝐵1, 𝐵2). 

3) 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 

Proof: 1) If  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2)then (𝑈1, 𝑈2) − (𝐵1, 𝐵2) ⊆ (𝑈1, 𝑈2) − (𝐴1, 𝐴2) ⇒
𝑁𝐵

𝑜 ((𝑈1, 𝑈2) − (𝐵1, 𝐵2)) ⊆ 𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) ⇒ 𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊆

𝑁𝐵 _𝐸(𝐴1, 𝐴2). 
2) Since (𝐴1, 𝐴2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)and (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2). By (1), 
𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐸(𝐴1, 𝐴2)and 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆

𝑁𝐵 _𝐸(𝐵1, 𝐵2). Therefore, 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐸(𝐵1, 𝐵2). 

3) Since (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2)and (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐵1, 𝐵2). By (1) 
𝑁𝐵 _𝐸(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2))and 𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∩

(𝐵1, 𝐵2)). Therefore, 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 

 

Remark 4.4: The inclusion may be strict. We can see in the following example. 

 

Example 4.5: Let U1 = {𝑎, 𝑏, 𝑐}, U2 = {1, 2}   with 
(U1, U2)

R
⁄ {({a, b}, {2}), ({c}, {1})}.   

Let (X1, X2) =({b}, {2}). Then τR(X1, X2) ={(Φ, Φ), (U1, U2), ({a, b}, {2})}. 
2) Take (𝐴1, 𝐴2) = ({𝑎, 𝑏}, {2}) 𝑎𝑛𝑑 (𝐵1, 𝐵2) = ({c}, {1}). 𝑁𝐵 _𝐸({𝑎, 𝑏}, {2}) ∪
𝑁𝐵 _𝐸({c}, {1}) = 𝑁𝐵

𝑜 ({c}, {1}) − 𝑁𝐵
𝑜 ({𝑎, 𝑏}, {2}) = (Φ, Φ) − ({𝑎, 𝑏}, {2}) =

({𝑎,   𝑏}, {2}). Also, 𝑁𝐵 _𝐸(({𝑎, 𝑏}, {2}) ∪ ({c}, {1})) = 𝑁𝐵 _𝐸(U1, U2) = 𝑁𝐵
𝑜 (Φ, Φ) =

19



J. Jasmine Elizabeth, G. Hari Siva Annam 
 

 
 

(Φ, Φ). Therefore, ({a, b}, {2}) ⊈ (Φ, Φ) and hence 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊂

𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐸(𝐵1, 𝐵2). 
 

3) Take (𝐴1, 𝐴2) = ({𝑐}, {1,2}) 𝑎𝑛𝑑 (𝐵1, 𝐵2) = ({a, c}, {1}). 𝑁𝐵 _𝐸(({𝑐}, {1,2}) ∩
 ({a, c}, {1})) = 𝑁𝐵 _𝐸({c}, {1}) = 𝑁𝐵

𝑜 ({𝑎, 𝑏}, {2}) = ({𝑎, 𝑏}, {2})and 
𝑁𝐵 _𝐸({𝑐}, {1,2}) ∩ 𝑁𝐵 _𝐸({a, c}, {1}) = 𝑁𝐵

𝑜 ({a, b}, {∅}) ∩ 𝑁𝐵
𝑜 ({b}, {2}) = (Φ, Φ) ∩

(Φ, Φ) = (Φ, Φ). Therefore, ({a, b}, {2}) ⊈ (Φ, Φ) and hence 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩
𝑁𝐵 _𝐸(𝐵1, 𝐵2) ⊂ 𝑁𝐵 _𝐸((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 
 

Theorem 4.6: Let (𝐴1, 𝐴2) and (𝐵1, 𝐵2) be two subsets of 𝑁𝐵 topological space 
(𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)). Then the following holds:  

1) 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = (𝑈1, 𝑈2) − 𝑁𝐵 (𝐴1, 𝐴2) 

2)𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2)) = 𝑁𝐵
𝑜 (𝑁𝐵 (𝐴1, 𝐴2)) 

3) 𝑁𝐵 _𝐸(𝑈1, 𝑈2) = (∅, ∅)and𝑁𝐵 _𝐸(∅, ∅) = (𝑈1, 𝑈2) 
4) 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = 𝑁𝐵 _𝐸[(𝑈1, 𝑈2) − 𝑁𝐵 _𝐸(𝐴1, 𝐴2)] 
5) 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2) 
6) 𝑁𝐵

𝑜 (𝐴1, 𝐴2), 𝑁𝐵 _𝐸(𝐴1, 𝐴2), 𝑁𝐵 _𝐹(𝐴1, 𝐴2)  are mutually disjoint and (𝑈1, 𝑈2) =
𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐹(𝐴1, 𝐴2). 
7) (𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = (∅, ∅) 
8) 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ⊆ (𝑈1, 𝑈2) − (𝐴1, 𝐴2) 

Proof: 1)𝑁𝐵 _𝐸(𝐴1, 𝐴2) = 𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) = (𝑈1, 𝑈2) − 𝑁𝐵 (𝐴1, 𝐴2). Hence 

(1) is proved. 

 

2)𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2)) = 𝑁𝐵 _𝐸[𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2))] = 𝑁𝐵 _𝐸[(𝑈1, 𝑈2) −

𝑁𝐵 (𝐴1, 𝐴2)] = 𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − [(𝑈1, 𝑈2) − 𝑁𝐵 (𝐴1, 𝐴2)]) =

𝑁𝐵
𝑜

(𝑁𝐵 (𝐴1, 𝐴2)).Therefore, 𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2)) = 𝑁𝐵
𝑜

(𝑁𝐵 (𝐴1, 𝐴2)). 

 

3)𝑁𝐵 _𝐸(𝑈1, 𝑈2) = 𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (𝑈1, 𝑈2)) = 𝑁𝐵
𝑜 (∅, ∅) = (∅, ∅)and 𝑁𝐵 _𝐸(∅, ∅) =

𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (∅, ∅)) = 𝑁𝐵
𝑜 (𝑈1, 𝑈2) = (𝑈1, 𝑈2). Therefore, 𝑁𝐵 _𝐸(𝑈1, 𝑈2) =

(∅, ∅)and 𝑁𝐵 _𝐸(∅, ∅) = (𝑈1, 𝑈2). 
 

4)𝑁𝐵 _𝐸[(𝑈1, 𝑈2) − 𝑁𝐵 _𝐸(𝐴1, 𝐴2)] = 𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − [(𝑈1, 𝑈2) − 𝑁𝐵 _𝐸(𝐴1, 𝐴2)]) =

𝑁𝐵
𝑜

(𝑁𝐵 _𝐸(𝐴1, 𝐴2)) = 𝑁𝐵
𝑜

[𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (𝐴1, 𝐴2))] = 𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) =

𝑁𝐵 _𝐸(𝐴1, 𝐴2). Therefore, 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = 𝑁𝐵 _𝐸[(𝑈1, 𝑈2) − 𝑁𝐵 _𝐸(𝐴1, 𝐴2)]. 
 

5)Since (𝐴1, 𝐴2) ⊆ 𝑁𝐵 (𝐴1, 𝐴2) ⇒ 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ⊆ 𝑁𝐵

𝑜
(𝑁𝐵 (𝐴1, 𝐴2)) = 𝑁𝐵

𝑜
((𝑈1, 𝑈2) −

𝑁𝐵
𝑜 [(𝑈1, 𝑈2) − (𝐴1, 𝐴2)]) = 𝑁𝐵 _𝐸(𝑁𝐵

𝑜 [(𝑈1, 𝑈2) − (𝐴1, 𝐴2)]) =

𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2)). Therefore, 𝑁𝐵
𝑜

(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐸(𝑁𝐵 _𝐸(𝐴1, 𝐴2)). 
 

6)Assume that 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩ 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ≠ (∅, ∅). Then there exists (𝑥1, 𝑥2) ∈

𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩ 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ⇒ (𝑥1, 𝑥2) ∈ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) and (𝑥1, 𝑥2) ∈ 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ⇒
(𝑥1, 𝑥2) ∈ (𝑈1, 𝑈2) − (𝐴1, 𝐴2)and  (𝑥1, 𝑥2) ∈ (𝐴1, 𝐴2), which is not possible. Hence our 

20



Further Diversification of Nano Binary Open Sets 

assumption is wrong. Therefore, 𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∩ 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (∅, ∅). In the same 

way we can prove the others. 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = (𝑈1, 𝑈2) − 𝑁𝐵 (𝐴1, 𝐴2) = (𝑈1, 𝑈2) −

𝑁𝐵 (𝐴1, 𝐴2) = (𝑈1, 𝑈2) − [𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐹(𝐴1, 𝐴2)]. Therefore, (𝑈1, 𝑈2) =

𝑁𝐵 _𝐸(𝐴1, 𝐴2) ∪ 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐹(𝐴1, 𝐴2). 

 

7)(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∩ 𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) ⊆ (𝐴1, 𝐴2) ∩
((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) = (∅, ∅). Therefore, (𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐸(𝐴1, 𝐴2) = (∅, ∅). 
 

8)𝑁𝐵 _𝐸(𝐴1, 𝐴2) = 𝑁𝐵
𝑜

((𝑈1, 𝑈2) − (𝐴1, 𝐴2)) ⊆ (𝑈1, 𝑈2) − (𝐴1, 𝐴2). 
 

Note 4.7: If (𝐴1, 𝐴2) is 𝑁𝐵 closed, then equality holds in (5).  
 

Theorem 4.8: In (𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)), (𝐴1, 𝐴2)and (𝐵1, 𝐵2) be two subsets of (𝑈1, 𝑈2). 
Then the following holds:  

1) 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (∅, ∅) 

2) (𝐴1, 𝐴2) is 𝑁𝐵-open if and only if 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (∅, ∅) 
3) 𝑁𝐵

𝑜 (𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = (∅, ∅) 
4) 𝑁𝐵 _𝐵(𝑁𝐵

𝑜
(𝐴1, 𝐴2)) = (∅, ∅) 

5) 𝑁𝐵 _𝐵(𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = 𝑁𝐵 _𝐵(𝐴1, 𝐴2) 

6) 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∩ 𝑁𝐵 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2)). 

7) If  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2), then 𝑁𝐵 _𝐵(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐵(𝐴1, 𝐴2). 
8) 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐵(𝐵1, 𝐵2). 
9) 𝑁𝐵 _𝐵(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐵(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 
10) 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = 𝑁𝐵 _𝐷((𝑈1, 𝑈2) − (𝐴1, 𝐴2))and 𝑁𝐵 _𝐷(𝐴1, 𝐴2) = 𝑁𝐵 _𝐵((𝑈1, 𝑈2) −
(𝐴1, 𝐴2)). 
11) (𝐴1, 𝐴2) = 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐵(𝐴1, 𝐴2). 
Proof: 1) 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∩ [(𝐴1, 𝐴2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2)] =
𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∩ [(𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2))] = 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) −
𝑁𝐵

𝑜 (𝐴1, 𝐴2)) ∩ (𝐴1, 𝐴2) = (∅, ∅) ∩ (𝐴1, 𝐴2) = (∅, ∅).Therefore, 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∩

𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (∅, ∅) 
 

2)Any subset (𝐴1, 𝐴2) of 𝑁𝐵 topological space (𝑈1, 𝑈2, 𝜏𝑅 (𝑋1, 𝑋2)) is 𝑁𝐵-open 
⇔(𝐴1, 𝐴2) = 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ⇔ (𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (∅, ∅) ⇔ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) =

(∅, ∅). 
 

3)𝑁𝐵
𝑜

(𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = 𝑁𝐵
𝑜

((𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2)) = 𝑁𝐵

𝑜
[(𝐴1, 𝐴2) ∩

((𝑈1, 𝑈2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2))] ⊆ 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∩ 𝑁𝐵
𝑜 ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2)) ⊆

𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2)) = (∅, ∅). Therefore, 𝑁𝐵
𝑜 (𝑁𝐵 _𝐵(𝐴1, 𝐴2)) =

(∅, ∅). 
 

4)𝑁𝐵 _𝐵(𝑁𝐵
𝑜

(𝐴1, 𝐴2)) = 𝑁𝐵
𝑜 (𝐴1, 𝐴2) − 𝑁𝐵

𝑜 (𝑁𝐵
𝑜(𝐴1, 𝐴2)) = 

𝑁𝐵
𝑜 (𝐴1, 𝐴2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2) = (∅, ∅). Therefore, 𝑁𝐵 _𝐵(𝑁𝐵
𝑜

(𝐴1, 𝐴2)) = (∅, ∅). 
 

5)𝑁𝐵 _𝐵(𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = 𝑁𝐵 _𝐵(𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = 𝑁𝐵 _𝐵(𝐴1, 𝐴2) −

(∅, ∅) (By (3)) = 𝑁𝐵 _𝐵(𝐴1, 𝐴2). Therefore, 𝑁𝐵 _𝐵(𝑁𝐵 _𝐵(𝐴1, 𝐴2)) = 𝑁𝐵 _𝐵(𝐴1, 𝐴2). 

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J. Jasmine Elizabeth, G. Hari Siva Annam 
 

 
 

 

6)𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2)) =

(𝐴1, 𝐴2) ∩ 𝑁𝐵 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2)). Therefore, 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (𝐴1, 𝐴2) ∩

𝑁𝐵 ((𝑈1, 𝑈2) − (𝐴1, 𝐴2)).  
 

7)If  (𝐴1, 𝐴2) ⊆ (𝐵1, 𝐵2), then  𝑁𝐵
𝑜 (𝐴1, 𝐴2) ⊆ 𝑁𝐵

𝑜 (𝐵1, 𝐵2) ⇒ (𝑈1, 𝑈2) −
𝑁𝐵

𝑜 (𝐵1, 𝐵2) ⊆ (𝑈1, 𝑈2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ⇒ (𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐵1, 𝐵2)) ⊆
(𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2)) ⇒ (𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2) −

𝑁𝐵
𝑜 (𝐴1, 𝐴2) ⇒ 𝑁𝐵 _𝐵(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) (By (6)) 

 

8)Since (𝐴1, 𝐴2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)and (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2). By (4) 
𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐵(𝐴1, 𝐴2)and 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆
𝑁𝐵 _𝐵(𝐵1, 𝐵2). Therefore, 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∪ (𝐵1, 𝐵2)) ⊆ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐵(𝐵1, 𝐵2). 
 

9)Since (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐴1, 𝐴2)and (𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2) ⊆ (𝐵1, 𝐵2). By (4), 
𝑁𝐵 _𝐵(𝐴1, 𝐴2) ⊆ 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2))and 𝑁𝐵 _𝐵(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∩
(𝐵1, 𝐵2)). Therefore, 𝑁𝐵 _𝐵(𝐴1, 𝐴2) ∩ 𝑁𝐵 _𝐵(𝐵1, 𝐵2) ⊆ 𝑁𝐵 _𝐵((𝐴1, 𝐴2) ∩ (𝐵1, 𝐵2)). 
 

10)𝑁𝐵 _𝐵(𝐴1, 𝐴2) = (𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2). By result 3.8, (𝐴1, 𝐴2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2) =
(𝐴1, 𝐴2) − [(𝐴1, 𝐴2) − 𝑁𝐵 _𝐷((𝑈1, 𝑈2) − (𝐴1, 𝐴2))] = 𝑁𝐵 _𝐷((𝑈1, 𝑈2) − (𝐴1, 𝐴2))By 
replacing (𝐴1, 𝐴2)by (𝑈1, 𝑈2) − (𝐴1, 𝐴2),  𝑁𝐵 _𝐷(𝐴1, 𝐴2) = 𝑁𝐵 _𝐵((𝑈1, 𝑈2) − (𝐴1, 𝐴2)). 
 

11)𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐵(𝐴1, 𝐴2) = 𝑁𝐵

𝑜 (𝐴1, 𝐴2) ∪ ((𝐴1, 𝐴2) − 𝑁𝐵
𝑜 (𝐴1, 𝐴2)) =

𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ ((𝐴1, 𝐴2) ∩ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2))) = (𝑁𝐵
𝑜(𝐴1, 𝐴2) ∪ (𝐴1, 𝐴2)) ∩

(𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ ((𝑈1, 𝑈2) − 𝑁𝐵

𝑜 (𝐴1, 𝐴2))) = (𝐴1, 𝐴2) ∩ (𝑈1, 𝑈2) = (𝐴1, 𝐴2). 

Therefore, (𝐴1, 𝐴2) = 𝑁𝐵
𝑜 (𝐴1, 𝐴2) ∪ 𝑁𝐵 _𝐵(𝐴1, 𝐴2). 

 

5. Conclusion 

Nano binary derived, nano binary border and nano binary exterior in nano binary 

topological spaces were introduced and their properties were discussed. In future we 

will discuss generalized closed sets in nano binary topological spaces. 

 

References 
 

[1] Lellis Thivagar. M and Carmel Richard, On Nano forms of weakly open sets, 

International Journal of Mathematics and Statistics Invention, 1(1) 2013, 31-37. 

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[3] Hari Siva Annam. G and J. Jasmine Elizabeth, Cognition of Nano Binary 

Topological Spaces, Global Journal of Pure and Applied Mathematics, 6(2019), 1045-

1054. 

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Further Diversification of Nano Binary Open Sets 

[4] Jasmine Elizabeth. J and G. Hari Siva Annam, Some notions on nano binary 

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[5] Jasmine Elizabeth. J and G. Hari Siva Annam, Nano binary contra continuous 

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[6] Lellis Thivagar. M and Sutha Devi. V, On Multi-granular Nano Topology, Accepted 

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