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On Standard Concepts Using ii-Open Sets

DOI: 10.4236/oalib.1105604    28 Downloads   78 Views  

ABSTRACT

Following Caldas in [1] we introduce and study topological properties of ii-derived, ii-border, ii-frontier, and ii-exterior of a set using the concept of ii-open sets. Moreover, we prove some further properties of the well-known notions of ii-closure and ii-interior. We also study a new decomposition of ii-continuous functions. Finally, we introduce and study some of the separation axioms specifically T0ii, T1ii.

1. Introduction

The notion of α-open set was introduced by Njastad in [2] . Caldas in [1] introduced and studied topological properties of α-derived, α-border, α-frontier, α-exterior of a set by using the concept of α-open sets. In this paper, we introduce and study the same above concepts by using ii-open sets. A subset A of X is called ii-open set [3] if there exists an open set G in the topology τ of X, such that: G ϕ , X , A C L ( A G ) and I n t ( A ) = G , the complement of an ii-open set is an ii-closed set. We denote the family of ii-open sets in ( X , τ ) by τ i i . It is shown in [4] that each of τ τ i i and τ i i is a topology on X. This property allows us to prove similar properties of α-open set. Also, we define ii-continuous functions and we study the relation between this type of function and continuous, semi-continuous, α-continuous and i-continuous functions. Finally, we introduce a new type of separation axioms namely T 0 i i , T 1 i i . We prove similar properties and characterizations of T 0 and T 1 .

2. Preliminaries

Throughout this paper, ( X , τ ) and ( Y , σ ) (simply X and Y) always mean topological spaces. For a subset A of a space X, Cl (A) and Int (A) denote the closure of A and the interior of A respectively. We recall the following definitions, which are useful in the sequel.

Definition 2.1. A subset A of a space X is called

1) Semi-open set [5] if A C L ( I n t ( A ) ) .

2) α-open set [2] if A I n t ( C L ( I n t ( A ) ) ) .

3) i-open set [3] if there exist an open set G in the topology τ of X, such that

i) G ϕ , X

ii) A C L ( A G )

The complement of an i-open set is an i-closed set.

4) ii-open set [4] if there exist an open set G in the topology τ of X, such that

i) G ϕ , X

ii) A C L ( A G )

iii) I n t ( A ) = G

The complement of an ii-open set is an ii-closed set.

5) int-open set [4] if there exist an open set G in the topology τ of X and G ϕ , X such that I n t ( A ) = G . The complement of int-open set is int-closed set.

6) αo (X), So (X), io (X), iio (X), into (X) are family of α-open, semi-open, i-open, ii-open, int-open sets respectively.

7) τ i , τ i i denote the family of all i-open sets and ii-open sets respectively.

Definition 2.2. [3] A topological space X is called

1) T 0 i if a, b are to distinct points in X, there exist an i-open set U such that either a U and b U , or b U and a U .

2) T 1 i if a , b X and a b , there exist i-open sets U, V containing a, b respectively, such that b U and a V .

Definition 2.3. A function f : ( X , τ ) ( Y , σ ) is called

1) Continuous [6] , if f 1 ( G ) is open in ( X , τ ) for every open set G of ( Y , σ ) .

2) α-continuous [6] , if f 1 ( G ) is α-open in ( X , τ ) for every open set G of ( Y , σ ) .

3) Semi-Continuous [5] , if f 1 ( G ) is semi-open in ( X , τ ) for every open set G of ( Y , σ ) .

4) i-Continuous [3] , if f 1 ( G ) is i-open in ( X , τ ) for every open set G of ( Y , σ ) .

3. Applications of ii-Open Sets

Definition 3.1. Let A be a subset of a topological space ( X , τ ) . A derived set of A denoted by D(A) is defined as follows: D ( A ) = { x X : ( G A ) \ { x } ϕ , x G } . A point x X is said to be ii-limit point of A if it satisfies the following assertion: ( G τ i i ) ( x G ( G A ) \ { x } ϕ ) . The set of all ii-limit points of A is called the ii-derived set of A and is denoted by D i i ( A ) . Note that x X is not ii-limit point of A if and only if there exist an ii-open set G in X such that ( x G and ( G A ) \ { x } = ϕ ).

Theorem 3.2. For subsets A, B of a space X, the following statements hold:

1) D i i ( A ) D (A)

2) If A B , then D i i ( A ) D i i (B)

3) D i i ( A ) D i i ( B ) D i i ( A B ) and D i i ( A B ) D i i ( A ) D i i (B)

4) D i i ( D i i ( A ) ) \ A D i i (A)

5) D i i ( A D i i ( A ) ) A D i i (A)

Proof. 1) Since every open set is ii-open [4] , it follows that D i i ( A ) D ( A ) .

2) Let x D i i ( A ) . Then G is ii-open set containing x such that

( A G ) \ { x } ϕ (3.1)

Since A B we get ( A G ) ( B G ) , it implies that ( A G ) \ { X } ( B G ) \ { X } ϕ , from (3.1) we get ( B G ) \ { X } ϕ .

Hence, x D i i ( B ) . Therefore D i i ( A ) D i i ( B ) .

3) Since A A B and B A B , from (2) we get D i i ( A ) D i i ( A B ) , D i i ( B ) D i i ( A B ) .

This implies to D i i ( A ) D i i ( B ) D i i ( A B ) .

We shall prove that D i i ( A B ) D i i ( A ) D i i ( B ) . Since A B A , A B B , from (2) we get D i i ( A B ) D i i ( A ) and D i i ( A B ) D i i ( B ) . Therefore D i i ( A B ) D i i ( A ) D i i ( B ) .

4) If x D i i ( D i i ( A ) ) \ A and G is an ii-open set containing x, then G ( D i i ( A ) \ { X } ) ϕ . Let y G ( D i i ( A ) \ { x } ) . Then, since y D i i ( A ) and y G , G ( A \ { Y } ) ϕ . Let z G ( A \ { Y } ) . Then, z x for z A and x A . Hence, G ( A \ { X } ) ϕ . Therefore, x D i i ( A ) .

5) Let x D i i ( A D i i ( A ) ) . If x A , the result is obvious. So, let, x D i i ( A D i i ( A ) ) \ A then for ii-open set G containing x, ( G ( A D i i ( A ) \ { x } ) ) ϕ . Thus, G ( A \ { X } ) ϕ or G ( D i i ( A ) \ { x } ) ϕ .

Now, it follows similarly from (4) that G ( A \ { X } ) ϕ . Hence, x D i i ( A ) .

Therefore, in any case, D i i ( A D i i ( A ) ) A D i i ( A ) .

In general, the converse of (1) may not true and the equality does not hold in (3) of theorem 3.2.

Example 3.3. Let X = { a , b , c } and τ = { ϕ , X , { b } } . Thus, i i o ( x ) = { ϕ , X , { b } , { a , b } , { b , c } } . Take the following:

1) A = { c } . Then, D ( A ) = { a , b } and D i i ( A ) = ϕ . Hence, D ( A ) D i i ( A ) ;

2) C = { a , b } and E = { c } . Then D i i ( c ) = { a , c } and D i i ( E ) = ϕ . Hence D i i ( C E ) D i i ( C ) D i i ( E ) .

Theorem 3.4. For any subset A of a space X, C L i i ( A ) = A D i i ( A ) .

Proof. Since D i i ( A ) C L i i ( A ) , A D i i ( A ) C L i i ( A ) . On the other hand, Let x C L i i ( A ) . If x A , then the proof is complete. If x A , each ii-open set G containing x intersects A at a point distinct from x; so x D i i ( A ) . Thus, C L i i ( A ) A D i i ( A ) , which completes the proof.

Definition 3.5. A point x X is said to be ii-interior point of A if there exist an ii-open set G containing x such that G A . The set of all ii-interior points of A is said to be ii-interior of A and denoted by I n t i i ( A ) .

Theorem 3.6. For subset A, B of a space X, the following statements are true:

1) I n t i i ( A ) is the union of all ii-open subset of A

2) A is ii-open if and only if A = I n t i i (A)

3) I n t i i ( I n t i i ( A ) ) = I n t i i (A)

4) I n t i i ( A ) = A \ D i i ( X \ A )

5) X \ I n t i i ( A ) = C L i i ( X \ A )

6) X \ C L i i ( A ) = I n t i i ( X \ A )

7) If A B , then I n t i i ( A ) I n t i i (B)

8) I n t i i ( A ) I n t i i ( B ) I n t i i ( A B )

9) I n t i i ( A ) I n t i i ( B ) I n t i i ( A B )

Proof. 1) Let { G i i \ i i } be a collection of all ii-open subsets of A. If x I n t i i ( A ) , then there exist j such that x G j A . Hence x i i G i i , and so I n t i i ( A ) i i G i i . On the other hand, if y i i G i i , then y G k A for some k . Thus y I n t i i ( A ) , and i i G i i I n t i i ( A ) . Accordingly, i i G i i I n t i i ( A ) .

2) Straightforward.

3) It follows from (1) and (2).

4) If x A \ D i i ( X \ A ) , then x D i i ( X \ A ) and so there exist an ii-open set G containing x such that G ( X \ A ) = ϕ . Thus, x G A and hence x I n t i i ( A ) . This shows that A \ D i i ( X \ A ) I n t i i ( A ) . Now let x I n t i i ( A ) . Since I n t i i ( A ) τ i i and I n t i i ( A ) ( X \ A ) = ϕ . We have x D i i ( X \ A ) . Therefore, I n t i i ( A ) = A \ D i i ( X \ A ) .

5) Using (4) and Theorem (3.4), we have

X \ I n t i i ( A ) = X \ ( A \ D i i ( X \ A ) ) = ( X \ A ) D i i ( X \ A ) = C L i i ( X \ A ) .

6) Using (4) and Theorem (3.4), we get.

I n t i i ( X \ A ) = ( X \ A ) \ D i i ( A ) = X \ ( A D i i ( A ) ) = X \ C L i i (A)

7) Since A B and I n t i i ( A ) A , I n t i i ( B ) B , we get I n t i i ( A ) I n t i i ( B ) .

8) Since A ( A B ) and B ( A B ) , from (7) we get I n t i i ( A ) I n t i i ( A B ) , I n t i i ( B ) I n t i i ( A B ) . Therefore I n t i i ( A ) I n t i i ( B ) I n t i i ( A B ) .

9) Since A B A and A B B , from (7) we get I n t i i ( A B ) I n t i i ( A ) , I n t i i ( A B ) I n t i i ( B ) . Therefore I n t i i ( A B ) I n t i i ( A ) ( B ) .

Definition 3.7. b i i ( A ) = A \ I n t i i ( A ) is said to be the ii-border of A.

Theorem 3.8. For a subset A of a space X, the following statements hold:

1) b i i ( A ) b ( A ) where b ( A ) denotes the border of A

2) I n t i i ( A ) b i i ( A ) = A

3) I n t i i ( A ) b i i ( A ) = ϕ

4) b i i ( A ) = ϕ if and only if A is ii-open set

5) b i i ( I n t i i ( A ) ) = ϕ

6) I n t i i ( b i i ( A ) ) = ϕ

7) b i i ( b i i ( A ) ) = b i i (A)

8) b i i ( A ) = A C L i i ( X \ A )

9) b i i ( A ) = A D i i ( X \ A )

Proof.

1) Since I n t ( A ) I n t i i ( A ) , we have b i i ( A ) = A \ I n t i i ( A ) A \ I n t ( A ) = b ( A ) .

2) and (3). Straightforward.

4) Since I n t i i ( A ) A , it follows from Theorem 3.6 (2). That A is ii-open A = I n t i i ( A ) b i i ( A ) = A \ I n t i i ( A ) = ϕ .

5) Since I n t i i ( A ) is ii-open, it follows from (4) that b i i ( I n t i i ( A ) ) = ϕ .

6) If x I n t i i ( b i i ( A ) ) , then x b i i ( A ) . On the other hand, since b i i ( A ) A , x I n t i i ( b i i ( A ) ) I n t i i ( A ) . Hence, x I n t i i ( A ) b i i ( A ) .

Which contradicts (3). Thus I n t i i ( b i i ( A ) ) = ϕ .

7) Using (6), we get b i i ( b i i ( A ) ) = b i i ( A ) \ I n t i i ( b i i ( A ) ) = b i i ( A ) .

8) Using Theorem 3.6 (6), we have b i i ( A ) = A \ I n t i i ( A ) = A \ ( X \ C L i i ( X \ A ) ) = A C L i i ( X \ A )

9) Applying (8) and the Theorem (3.4), we have b i i ( A ) = A C L i i ( X \ A ) = A ( ( X \ A ) D i i ( X \ A ) ) = A D i i ( X \ A ) .

Example 3.9. Consider the topological space ( X , τ ) given in Example (3.3). If A = { a , b } , then b i i ( A ) = ϕ and b ( A ) = { a } . Hence, b ( A ) b i i ( A ) , that is, in general, the converse Theorem 3.9 (1) may not be true.

Definition 3.10. F r i i ( A ) = C L i i ( A ) \ I n t i i ( A ) is said to be the ii-frontier of A.

Theorem 3.11. For a subset A of a space X, the following statements hold:

1) F r i i ( A ) F r ( A ) where F r ( A ) denotes the frontier of A

2) C L i i ( A ) = I n t i i ( A ) F r i i (A)

3) I n t i i ( A ) F r i i ( A ) = ϕ

4) b i i ( A ) F r i i (A)

5) F r i i ( A ) = b i i ( A ) D i i (A)

6) F r i i ( A ) = D i i ( A ) if and only if A is ii-open set

7) F r i i ( A ) = C L i i ( A ) C L i i ( X \ A )

8) F r i i ( A ) = F r i i ( X \ A )

9) F r i i ( A ) is ii-closed

10) F r i i ( F r i i ( A ) ) F r i i (A)

11) F r i i ( I n t i i ( A ) ) F r i i (A)

12) F r i i ( C L i i ( A ) ) F r i i (A)

13) I n t i i ( A ) = A \ F r i i (A)

Proof.

1) Since C L i i ( A ) C L ( A ) and I n t ( A ) I n t i i ( A ) , it follows that F r i i ( A ) = C L i i ( A ) \ I n t i i ( A ) C L ( A ) \ I n t i i ( A ) C L ( A ) \ I n t ( A ) F r ( A ) .

2) I n t i i ( A ) F r i i ( A ) = I n t i i ( A ) ( C L i i ( A ) \ I n t i i ( A ) ) = C L i i ( A ) .

3) I n t i i ( A ) F r i i ( A ) = I n t i i ( A ) ( C L i i ( A ) \ I n t i i ( A ) ) = ϕ .

4) Since A C L i i ( A ) , we have b i i ( A ) = A \ I n t i i ( A ) C L i i ( A ) \ I n t i i ( A ) = F r i i ( A ) .

5) Since I n t i i ( A ) F r i i ( A ) = I n t i i ( A ) b i i ( A ) D i i ( A ) , F r i i ( A ) = b i i ( A ) D i i ( A ) .

6) Assume that A is ii-open. Then F r i i ( A ) = b i i ( A ) D i i ( A ) \ I n t i i ( A ) = ϕ ( D i i ( A ) \ A ) = D i i ( A ) \ A = b i i ( X \ A ) , by using (5), Theorem 3.6 (2), Theorem 3.8 (4) and Theorem 3.8 (9).

Conversely, suppose that F r i i ( A ) = b i i ( X \ A ) . Then ϕ = F r i i ( A ) \ b i i ( X \ A ) = ( C L i i ( A ) \ I n t i i ( A ) ) \ ( X \ A ) \ I n t i i ( X \ A ) = A \ I n t i i ( A ) . by using (4) and (5) of Theorem 3.6, and so A I n t i i ( A ) . Since I n t i i ( A ) A in general, it follows that I n t i i ( A ) = A so from Theorem 3.6 (2) that A is ii-open set.

7) F r i i ( A ) = C L i i ( A ) \ I n t i i ( A ) = C L i i ( A ) ( C L i i ( X \ A ) ) .

8) It follows from (7).

9) C L i i ( F r i i ( A ) ) = C L i i ( C L i i ( A ) ) ( C L i i ( X \ A ) ) C L i i ( C L i i ( A ) ) C L i i ( C L i i ( X \ A ) ) = F r i i ( A ) . Hence, F r i i ( A ) is ii-closed.

10) F r i i ( F r i i ( A ) ) = C L i i ( F r i i ( A ) ) C L i i ( X \ F r i i ( A ) ) C L i i ( F r i i ( A ) ) = F r i i ( A ) .

11) Using Theorem 3.6 (3), we get F r i i ( I n t i i ( A ) ) = C L i i ( I n t i i ( A ) ) \ I n t i i ( I n t i i ( A ) ) C L i i ( A ) \ I n t i i ( A ) = F r i i ( A ) .

12) F r i i ( C L i i ( A ) ) = C L i i ( C L i i ( A ) ) \ I n t i i ( C L i i ( A ) ) = C L i i ( A ) \ I n t i i ( C L i i ( A ) ) = C L i i ( A ) \ I n t i i ( A ) = F r i i ( A ) .

13) A \ F r i i ( A ) = ( A \ C L i i ( A ) ) \ I n t i i ( A ) = I n t i i ( A ) .

The converses of (1) and (4) of Theorem 3.11 are not true in general, as shown by Example

Example 3.12. Consider the topological space ( X , τ ) given in Example 3.3. If A = { c } , then F r ( A ) = { a , c } { c } = F r i i ( A ) , and if B = { a , b } , then F r i i ( B ) = { c } b i i ( B ) .

Definition 3.13. E x t i i ( A ) = I n t i i ( X \ A ) is said to be an ii-exterior of A.

Theorem 3.14. For a subset A of a space X, the following statements hold:

1) E x t ( A ) E x t i i ( A ) where E x t ( A ) denotes the exterior of A

2) E x t i i ( A ) is ii-open

3) E x t i i ( A ) = I n t i i ( X \ A ) = X \ C L i i (A)

4) E x t i i ( E x t i i ( A ) ) = I n t i i ( C L i i (A))

5) If A B , then E x t i i ( A ) E x t i i (B)

6) E x t i i ( A B ) E x t i i ( A ) E x t i i (B)

7) E x t i i ( A B ) E x t i i ( A ) E x t i i (B)

8) E x t i i ( X ) = ϕ

9) E x t i i ( ϕ ) = X

10) E x t i i ( A ) = E x t i i ( X \ E x t i i (A))

11) I n t i i ( A ) E x t i i ( E x t i i (A))

12) X = E x t i i ( A ) E x t i i ( A ) F r i i (A)

Proof. 1) It follows from Theorem 3.6 (1).

2) It is straightforward by Theorem 3.6 (6).

3) E x t i i ( E x t i i ( A ) ) = E x t i i ( X \ C L i i ( A ) ) = I n t i i ( X \ X \ C L i i ( A ) ) = I n t i i ( C L i i ( A ) ) .

4) Assume that A B . Then E x t i i ( B ) = E x t i i ( X \ B ) E x t i i ( X \ A ) = E x t i i ( A ) , by using Theorem 3.6 (7).

5) Applying Theorem 3.6 (8), we get

E x t i i ( A B ) = I n t i i ( X \ ( A B ) ) = I n t i i ( ( X \ A ) ( X \ B ) ) I n t i i ( X \ A ) I n t i i ( X \ B ) = E x t i i ( A ) E x t i i ( B ) .

6) Applying Theorem 3.6 (9), we obtain

E x t i i ( A B ) = I n t i i ( X \ ( A B ) ) = I n t i i ( ( X \ A ) ( X \ B ) ) I n t i i ( X \ A ) I n t i i ( X \ B ) = E x t i i ( A ) E x t i i ( B ) . .

7) Straightforward.

8) Straightforward.

9) E x t i i ( X \ E x t i i ( A ) ) = E x t i i ( X \ I n t i i ( X \ A ) ) = I n t i i ( X \ ( X \ I n t i i ( X \ A ) ) ) = I n t i i ( I n t i i ( X \ A ) ) = I n t i i ( X \ A ) = E x t i i ( A ) .

10) I n t i i ( A ) I n t i i ( C L i i ( A ) ) = I n t i i ( X \ I n t i i ( X \ A ) ) = I n t i i ( X \ E x t i i ( A ) ) = E x t i i ( E x t i i ( A ) ) .

Example 3.15. Let X = { a , b , c , d } and τ = { ϕ , X , { c , d } } . Thus, i i o ( x ) = { ϕ , X , { c , d } , { b , c , d } , { a , c , d } } . If A = { a } and B = { b } . Then E x t i i ( A ) E x t ( A ) . E x t i i ( A B ) E x t i i ( A ) E x t i i ( B ) and E x t i i ( A B ) E x t i i ( A ) E x t i i ( B ) .

4. A New Decomposition of ii-Continuity

We begin by the following definition:

Definition 4.1. A function f : ( X , τ ) ( Y , σ ) is called ii-continuous if f 1 ( G ) is ii-open set in ( X , τ ) for any open set G of ( Y , σ ) .

Theorem 4.2. Let f : ( X , τ ) ( Y , σ ) be a function then:

1) Every continuous function is an ii-continuous,

2) Every ii-continuous function is an i-continuous,

3) Every α-continuous function is an ii-continuous.

Proof. 1) Let G be open set in ( Y , σ ) . Since f is continuous, it follows that f 1 ( G ) is open set in ( X , τ ) . But every open set is ii-open set [4] . Hence f 1 ( G ) is ii-open set in ( X , τ ) . Thus f is ii-continuous.

2) Let G be open set in ( Y , σ ) . Since f is an ii-continuous, it follows that f 1 ( G ) is an ii-open set in ( X , τ ) . But every ii-open set is i-open set [4] . Hence f 1 ( G ) is i-open set in ( X , τ ) . Thus f is i-continuous.

3) Let G be open set in ( Υ , ϭ). Since f is α-continuous, it follows that f 1 ( G ) is α-open set in ( X , τ ) . But every α-open set is ii-open set [4] . Hence f 1 ( G ) is ii-open set in ( X , τ ) . Thus f is an ii-continuous.

The converse need not be true by the following example.

Example 4.3. Let

X = { a , b , c , d } , τ = { ϕ , X , { a } , { b } , { a , b } }

and

Y = { a , b , c , d } , σ = { ϕ , Y , { a } , { b } , { a , b } , { a , d } , { a , b , d } }

and

i i o ( x ) = { ϕ , X , { a } , { b } , { a , b } , { a , d } , { a , b , c } , { a , b , d } } ,

i o ( x ) = { ϕ , X , { a } , { b } , { a , b } , { a , c } , { a , d } , { b , c } , { b , d } , { a , b , c } , { a , b , d } , { a , c , d } , { b , c , d } } ,

α o ( x ) = { ϕ , X , { a } , { b } , { a , b } , { a , b , c } , { a , b , d } } .

Let f : ( X , τ ) ( Y , σ ) be the identity function then f 1 ( { a } ) = { a } , f 1 ( { b } ) = { b } , f 1 ( { c } ) = { c } , f 1 ( { d } ) = { d } . Then f is ii-continuous, but f is not α-continuous, since for the open set { a , d } in ( Y , σ ) , f 1 ( { a , d } ) = { a , d } is not α-open in ( X , τ ) and f is not continuous, since for the open set { a , d } in ( Y , σ ) , f 1 ( { a , d } ) = { a , d } is not open in ( X , τ ) . Now when f : ( X , τ ) ( Y , σ ) be defined by f 1 ( { a } ) = { b } , f 1 ( { b } ) = { a } , f 1 ( { c } ) = { c } , f 1 ( { d } ) = { d } we get f is i-continuous, but f is not ii-continuous, since for the open set { a , d } in ( Y , σ ) , f 1 ( { a , d } ) = { b , d } is not ii-open in ( X , τ ) .

Theorem 4.4. Let f : ( X , τ ) ( Y , σ ) be a function then every semi-continuous function is an ii-continuous.

Proof. Let G be open set in ( Y , σ ) . Since f is semi-continuous, it follows that f 1 ( G ) is semi-open set in ( X , τ ) . But every semi-open set is ii-open set [4] . Hence f 1 ( G ) is ii-open set in ( X , τ ) . Thus f is an ii-continuous.

Definition 4.5. A function f : ( X , τ ) ( Y , σ ) is called int-continuous if f 1 ( G ) is int-open set in ( X , τ ) for any open set G in ( Y , σ ) .

Theorem 4.6. Let f : ( X , τ ) ( Y , σ ) be a function then:

1) Every continuous function is int-continuous,

2) Every ii-continuous function is int-continuous,

3) Every α-continuous function is int-continuous.

Proof. 1) Let G be open set in ( Y , σ ) . Since f is continuous, it follows that f 1 ( G ) is open set in ( X , τ ) . But every open set is int-open set [4] . Hence f 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.

2) Let G be open set in ( Y , σ ) . Since f is ii-continuous, it follows that f 1 ( G ) is an ii-open set in ( X , τ ) . But every ii-open set is int-open set [4] . Hence f 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.

3) Let G be open set in ( Y , σ ) . Since f is α-continuous, it follows that f 1 ( G ) is α-open set in ( X , τ ) . But every α-open set is int-open set [4] . Hence f 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.

The converse need not be true by the following example.

Example 4.7. Let X = { a , b , c } , τ = { ϕ , X , { a } , { b , c } } and Y = { a , b , c } , σ = { ϕ , Y , { a } , { a , c } } and i n t o ( x ) = { ϕ , X , { a } , { a , b } , { b , c } , { a , c } } , i i o ( x ) = α o ( x ) = { ϕ , X , { a } , { b , c } } . Let f : ( X , τ ) ( Y , σ ) be the identity function then f 1 ( { a } ) = { a } , f 1 ( { b } ) = { b } , f 1 ( { c } ) = { c } . Then f is int-continuous, but f is not ii-continuous, since for the open set { a , c } in ( Y , σ ) , f 1 ( { a , c } ) = { a , c } is not ii-open in ( X , τ ) and f is not continuous, since for the open set { a , c } in ( Y , σ ) , f 1 ( { a , c } ) = { a , c } is not open in ( X , τ ) and f is not α-continuous, since for the open set { a , c } in ( Y , σ ) , f 1 ( { a , c } ) = { a , c } is not α-open.

Definition 4.8. A subset A of X is called weakly ii-open set if A is ii-open set and A C L ( I n t ( A ) A ) .

Theorem 4.9. A subset A of a space X is α-open set if and only if A is weakly ii-open.

Proof. Let A be α-open set. Since A I n t ( C L ( I n t ( A ) ) ) and A C L ( A ) . Therefore A C L ( I n t ( A ) ) C L ( A ) , this implies that A C L ( I n t ( A ) A ) . Now, put G = I n t ( A ) where G ϕ , X , then A is ii-open set. Therefore, A is weakly ii-open set.

Conversely, Let A be weakly ii-open set, then there exist an open set G ϕ , X , such that G = I n t ( A ) satisfying A C L ( I n t ( A ) A ) and A is ii-open set. Since A C L ( I n t ( A ) A ) , this implies that A C L ( I n t ( A ) ) and I n t ( A ) I n t ( C L ( I n t ( A ) ) ) . Since A is ii-open set, using (2) from Theorem (3.6), we get A = I n t ( A ) . Therefore A I n t ( C L ( I n t ( A ) ) ) . Thus A is α -open set.

As a summary the following Figure 1 shows the relations among semi-continuous, ii-continuous, i-continuous, int-continuous, α-continuous and continuous.

Figure 1. Relations among semi-continuous, ii-continuous, i-continuous, int-continuous, α-continuous and continuous.

Corollary 4.10. A function f : ( X , τ ) ( Y , σ ) is α-continuous if and only if it is weakly ii-continuous.

Proof. Clear from Theorem 4.9.

5. ii-Separating Axioms

In this section we define T 0 i i and T 1 i i spaces for ii-open sets and we determine them by giving many examples. Specially, we define T 1 , T 1 α and T 1 i spaces to compare them with T 1 i i space.

Definition 5.1. A topological space X is called

1) T 0 i i if a, b are to distinct points in X, there exists ii-open set U such that either a U and b U , and b U and a U .

2) T 1 i i if a , b X and a b , there exist ii-open sets U, V containing a, b respectively, such that b U and a V .

Example 5.2. Let X = { a , b } , τ i i = τ = { ϕ , X , { a } , { b } } ( X , τ ) and ( X , τ i i ) are topological spaces.

1) a , b X ( a b ) there exists { a } τ i i such that a { a } , b { a } . Therefore ( X , τ ) is T 0 i i .

2) a , b X ( a b ) there exists { a } , { b } τ i i such that a { a } , b { b } . Therefore ( X , τ ) is T 1 i i .

Theorem 5.3.

1) Every T 0 -space is T 0 i i -space,

2) Every T 1 -space is T 0 i i -space,

3) Every T 1 -space is T 1 i i -space,

4) Every T 1 i i -space is T 0 i i -space.

Proof. (1), (2), (3) and (4) follow using the fact that every open set is ii-open [4] .

The converse needs not to be true by the following example.

Example 5.4. Let

X = { a , b , c } , τ = { ϕ , X , { a } } and τ i i = { ϕ , X , { a } , { a , b } , { a , c } } .

( X , τ ) and ( X , τ i i ) are topological spaces.

( X , τ ) is not T 0 -space because, b , c X ( b c ) there is no open set G such that b G , c G .

( X , τ ) is T 0 i i -space because, a , b X ( a b ) there exists { a } τ i i such that a { a } , b { a } .

a , c X ( a c ) there exists { a } τ i i such that a { a } , c { a } .

b , c X ( b c ) there exists { a , b } τ i i such that b { a , b } , c { a , b } .

( X , τ ) is not T 1 -space because, a , b X ( a b ) there exists X τ such that a X , b X .

( X , τ ) is not T 1 i i -space because, b , a X ( a b ) there exists { a , b } τ i i such that a { a , b } , b { a , b } .

Theorem 5.5. Every T 1 α -space is T 1 i i -space.

Proof. Let X be T 1 α -space. Let a, b be two distinct points in X. Since X is T 1 α -space there exist two α-open sets U, V in X such that a U , b U , a V , b V . Since every α-open set is ii-open set [4] , U, V is an ii-open set in X. Hence X is T 1 i i -space.

Theorem 5.6. Every T 1 i i -space is T 1 i -space.

Proof. Let X be a T 1 i i -space. Let a, b be two distinct points in X. Since X is T 1 i i -space there exist two ii-open sets U, V in X such that a U , b U , a V , b V . Since every ii-open set is i-open set [4] , U, V is an i-open set in X. Hence X is T 1 i -space.

The converse needed not to be true by the following example.

Example 5.7. Let X = { a , b , c } , τ = { ϕ , X , { a } , { b , c } } and.

τ i = { ϕ , X , { a } , { b } , { c } , { b , c } } . τ i i = { ϕ , X , { a } , { b , c } } .

( X , τ ) and ( X , τ i i ) are topological spaces.

( X , τ ) is T 1 i -space because, a , b X ( a b ) there exists { a } , { b } τ i such that a { a } , b { a } and b { b } , a { b } .

a , c X ( a c ) there exists { a } , { c } τ i such that a { a } , c { a } and c { c } , a { c } .

b , c X ( b c ) there exists { c } , { b } τ i such that c { c } , b { c } and b { b } , c { b } .

( X , τ ) is not T 1 i i -space because, b , c X ( c b ) there exists { b } , { c } τ i i such that c { b , c } , b { b , c } .

Theorem 5.8. A space X is T 0 i i if and only if C L i i ( { x } ) C L i i ( { y } ) for every pair of distinct points x, y of X.

Proof. Let X be a T 0 i i -space. Let x , y X such that x y , then there exists an ii-open set U containing one of the points but not the other, then x U and y U . Then X \ U is ii-closed set containing y but not x. But C L i i ( { y } ) is the smallest ii-closed set containing y. Therefore C L i i ( { y } ) X \ U and hence x C L i i ( { y } ) . Thus C L i i ( { x } ) C L i i ( { y } ) .

Conversely, Suppose for any x , y X with x y , C L i i ( { x } ) C L i i ( { y } ) . Let z X such that z C L i i ( { x } ) but z C L i i ( { y } ) . If x C L i i ( { y } ) then C L i i ( { x } ) C L i i ( { y } ) and hence z C L i i ( { y } ) . This is contradiction. Therefore x C L i i ( { y } ) . That is x X \ C L i i ( { y } ) . Therefore X \ C L i i ( { y } ) is ii-open set containing x but not y. Hence X is an T 0 i i -space.

Theorem 5.9. A space ( X , τ ) is T 1 i i -space if and only if the singletons are ii-closed sets.

Proof. Let X be T 1 i i -space and let x X , to prove that { x } is ii-closed set. We will prove X \ { x } is ii-open set in X. Let y X \ { x } , implies x y and since X is T 1 i i -space then their exist two ii-open sets U, V such that x U , y V X \ { x } . Since y V X \ { x } , then X \ { x } is ii-open set. Hence { x } is ii-closed set.

Conversely, Let x y X then { x } , { y } are ii-closed sets. That is X \ { x } is ii-open set clearly, x X \ { x } and y X \ { x } . Similarly X \ { y } is ii-open set, y X \ { y } and x X \ { y } . Hence X is an T 1 i i -space.

As a consequence the following Figure 2 shows the relations among T 0 , T 0 i i , T 1 , T 1 i i and T 1 α .

Figure 2. Relations among T 0 , T 0 i i , T 1 , T 1 i i and T 1 α .

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Abdullah, B. and Mohammed, A. (2019) On Standard Concepts Using ii-Open Sets. Open Access Library Journal, 6, 1-13. doi: 10.4236/oalib.1105604.

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