Beyda S. Abdullah, Amir A. Mohammed
Department of Mathematics, College of Education for pure Sciences, University of Mosul, Mosul, Iraq.
DOI: 10.4236/oalib.1105604   PDF    HTML   XML   281 Downloads   1,389 Views   Citations

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 T_0ii, T_1ii.

Keywords

α-Open Set, ii-Open Set, Separation Axioms

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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 $\tau$ of X, such that: $G\ne \varphi ,X$, $A\subseteq CL\left(A\cap G\right)$ and $Int\left(A\right)=G$, the complement of an ii-open set is an ii-closed set. We denote the family of ii-open sets in $\left(X,\tau \right)$ by ${\tau }^{ii}$. It is shown in [4] that each of $\tau \subset {\tau }^{ii}$ and ${\tau }^{ii}$ 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 $T_{0ii}$, $T_{1ii}$. We prove similar properties and characterizations of $T_0$ and $T_1$.

2. Preliminaries

Throughout this paper, $\left(X,\tau \right)$ and $\left(Y,\sigma \right)$ (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 $A\subseteq CL\left(Int\left(A\right)\right)$.

2) α-open set [2] if $A\subseteq Int\left(CL\left(Int\left(A\right)\right)\right)$.

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

i) $G\ne \varphi ,X$

ii) $A\subseteq CL\left(A\cap G\right)$

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 $\tau$ of X, such that

i) $G\ne \varphi ,X$

ii) $A\subseteq CL\left(A\cap G\right)$

iii) $Int\left(A\right)=G$

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 $\tau$ of X and $G\ne \varphi ,X$ such that $Int\left(A\right)=G$. 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) ${\tau }^i$, ${\tau }^{ii}$ denote the family of all i-open sets and ii-open sets respectively.

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

1) $T_{0i}$ if a, b are to distinct points in X, there exist an i-open set U such that either $a\in U$ and $b\notin U$, or $b\in U$ and $a\notin U$.

2) $T_{1i}$ if $a,b\in X$ and $a\ne b$, there exist i-open sets U, V containing a, b respectively, such that $b\notin U$ and $a\notin V$.

Definition 2.3. A function $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ is called

1) Continuous [6] , if $f^{-1}\left(G\right)$ is open in $\left(X,\tau \right)$ for every open set G of $\left(Y,\sigma \right)$.

2) α-continuous [6] , if $f^{-1}\left(G\right)$ is α-open in $\left(X,\tau \right)$ for every open set G of $\left(Y,\sigma \right)$.

3) Semi-Continuous [5] , if $f^{-1}\left(G\right)$ is semi-open in $\left(X,\tau \right)$ for every open set G of $\left(Y,\sigma \right)$.

4) i-Continuous [3] , if $f^{-1}\left(G\right)$ is i-open in $\left(X,\tau \right)$ for every open set G of $\left(Y,\sigma \right)$.

3. Applications of ii-Open Sets

Definition 3.1. Let A be a subset of a topological space $\left(X,\tau \right)$. A derived set of A denoted by D(A) is defined as follows: $D\left(A\right)=\left\{x\in X:\left(G\cap A\right)\backslash \left\{x\right\}\ne \varphi ,\forall x\in G\right\}$. A point $x\in X$ is said to be ii-limit point of A if it satisfies the following assertion: $\left(\forall G\in {\tau }^{ii}\right)\left(x\in G\Rightarrow \left(G\cap A\right)\backslash \left\{x\right\}\ne \varphi \right)$. The set of all ii-limit points of A is called the ii-derived set of A and is denoted by $D_{ii}\left(A\right)$. Note that $x\in X$ is not ii-limit point of A if and only if there exist an ii-open set G in X such that ( $x\in G$ and $\left(G\cap A\right)\backslash \left\{x\right\}=\varphi$ ).

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

1) $D_{ii}\left(A\right)\subset D(A)$

2) If $A\subseteq B$, then $D_{ii}\left(A\right)\subseteq D_{ii}(B)$

3) $D_{ii}\left(A\right)\cup D_{ii}\left(B\right)\subset D_{ii}\left(A\cup B\right)$ and $D_{ii}\left(A\cap B\right)\subset D_{ii}\left(A\right)\cap D_{ii}(B)$

4) $D_{ii}\left(D_{ii}\left(A\right)\right)\backslash A\subset D_{ii}(A)$

5) $D_{ii}\left(A\cup D_{ii}\left(A\right)\right)\subset A\cup D_{ii}(A)$

Proof. 1) Since every open set is ii-open [4] , it follows that $D_{ii}\left(A\right)\subset D\left(A\right)$.

2) Let $x\in D_{ii}\left(A\right)$. Then G is ii-open set containing x such that

$\left(A\cap G\right)\backslash \left\{x\right\}\ne \varphi$ (3.1)

Since $A\subseteq B$ we get $\left(A\cap G\right)\subseteq \left(B\cap G\right)$, it implies that $\left(A\cap G\right)\backslash \left\{X\right\}\subseteq \left(B\cap G\right)\backslash \left\{X\right\}\ne \varphi$, from (3.1) we get $\left(B\cap G\right)\backslash \left\{X\right\}\ne \varphi$.

Hence, $x\in D_{ii}\left(B\right)$. Therefore $D_{ii}\left(A\right)\subseteq D_{ii}\left(B\right)$.

3) Since $A\subseteq A\cup B$ and $B\subseteq A\cup B$, from (2) we get $D_{ii}\left(A\right)\subseteq D_{ii}\left(A\cup B\right)$, $D_{ii}\left(B\right)\subseteq D_{ii}\left(A\cup B\right)$.

This implies to $D_{ii}\left(A\right)\cup D_{ii}\left(B\right)\subset D_{ii}\left(A\cup B\right)$.

We shall prove that $D_{ii}\left(A\cap B\right)\subset D_{ii}\left(A\right)\cap D_{ii}\left(B\right)$. Since $A\cap B\subseteq A$, $A\cap B\subseteq B$, from (2) we get $D_{ii}\left(A\cap B\right)\subset D_{ii}\left(A\right)$ and $D_{ii}\left(A\cap B\right)\subset D_{ii}\left(B\right)$. Therefore $D_{ii}\left(A\cap B\right)\subset D_{ii}\left(A\right)\cap D_{ii}\left(B\right)$.

4) If $x\in D_{ii}\left(D_{ii}\left(A\right)\right)\backslash A$ and G is an ii-open set containing x, then $G\cap \left(D_{ii}\left(A\right)\backslash \left\{X\right\}\right)\ne \varphi$. Let $y\in G\cap \left(D_{ii}\left(A\right)\backslash \left\{x\right\}\right)$. Then, since $y\in D_{ii}\left(A\right)$ and $y\in G$, $G\cap \left(A\backslash \left\{Y\right\}\right)\ne \varphi$. Let $z\in G\cap \left(A\backslash \left\{Y\right\}\right)$. Then, $z\ne x$ for $z\in A$ and $x\notin A$. Hence, $G\cap \left(A\backslash \left\{X\right\}\right)\ne \varphi$. Therefore, $x\in D_{ii}\left(A\right)$.

5) Let $x\in D_{ii}\left(A\cup D_{ii}\left(A\right)\right)$. If $x\in A$, the result is obvious. So, let, $x\in D_{ii}\left(A\cup D_{ii}\left(A\right)\right)\backslash A$ then for ii-open set G containing x, $\left(G\cap \left(A\cup D_{ii}\left(A\right)\backslash \left\{x\right\}\right)\right)\ne \varphi$. Thus, $G\cap \left(A\backslash \left\{X\right\}\right)\ne \varphi$ or $G\cap \left(D_{ii}\left(A\right)\backslash \left\{x\right\}\right)\ne \varphi$.

Now, it follows similarly from (4) that $G\cap \left(A\backslash \left\{X\right\}\right)\ne \varphi$. Hence, $x\in D_{ii}\left(A\right)$.

Therefore, in any case, $D_{ii}\left(A\cup D_{ii}\left(A\right)\right)\subset A\cup D_{ii}\left(A\right)$.

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 $X=\left\{a,b,c\right\}$ and $\tau =\left\{\varphi ,X,\left\{b\right\}\right\}$. Thus, $iio\left(x\right)=\left\{\varphi ,X,\left\{b\right\},\left\{a,b\right\},\left\{b,c\right\}\right\}$. Take the following:

1) $A=\left\{c\right\}$. Then, $D\left(A\right)=\left\{a,b\right\}$ and $D_{ii}\left(A\right)=\varphi$. Hence, $D\left(A\right)\not\subset D_{ii}\left(A\right)$ ;

2) $C=\left\{a,b\right\}$ and $E=\left\{c\right\}$. Then $D_{ii}\left(c\right)=\left\{a,c\right\}$ and $D_{ii}\left(E\right)=\varphi$. Hence $D_{ii}\left(C\cup E\right)\ne D_{ii}\left(C\right)\cup D_{ii}\left(E\right)$.

Theorem 3.4. For any subset A of a space X, $C{L}_{ii}\left(A\right)=A\cup D_{ii}\left(A\right)$.

Proof. Since $D_{ii}\left(A\right)\subset C{L}_{ii}\left(A\right)$, $A\cup D_{ii}\left(A\right)\subset C{L}_{ii}\left(A\right)$. On the other hand, Let $x\in C{L}_{ii}\left(A\right)$. If $x\in A$, then the proof is complete. If $x\notin A$, each ii-open set G containing x intersects A at a point distinct from x; so $x\in D_{ii}\left(A\right)$. Thus, $C{L}_{ii}\left(A\right)\subset A\cup D_{ii}\left(A\right)$, which completes the proof.

Definition 3.5. A point $x\in X$ is said to be ii-interior point of A if there exist an ii-open set G containing x such that $G\subset A$. The set of all ii-interior points of A is said to be ii-interior of A and denoted by $In{t}_{ii}\left(A\right)$.

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

1) $In{t}_{ii}\left(A\right)$ is the union of all ii-open subset of A

2) A is ii-open if and only if $A=In{t}_{ii}(A)$

3) $In{t}_{ii}\left(In{t}_{ii}\left(A\right)\right)=In{t}_{ii}(A)$

4) $In{t}_{ii}\left(A\right)=A\backslash D_{ii}\left(X\backslash A\right)$

5) $X\backslash In{t}_{ii}\left(A\right)=C{L}_{ii}\left(X\backslash A\right)$

6) $X\backslash C{L}_{ii}\left(A\right)=In{t}_{ii}\left(X\backslash A\right)$

7) If $A\subseteq B$, then $In{t}_{ii}\left(A\right)\subseteq In{t}_{ii}(B)$

8) $In{t}_{ii}\left(A\right)\cup In{t}_{ii}\left(B\right)\subseteq In{t}_{ii}\left(A\cup B\right)$

9) $In{t}_{ii}\left(A\right)\cap In{t}_{ii}\left(B\right)\supseteq In{t}_{ii}\left(A\cap B\right)$

Proof. 1) Let $\left\{G_{ii}\backslash ii\in \wedge \right\}$ be a collection of all ii-open subsets of A. If $x\in In{t}_{ii}\left(A\right)$, then there exist $j\in \wedge$ such that $x\in G_j\subseteq A$. Hence $x\in {\displaystyle {\cup }_{ii\in \wedge }{G}_{ii}}$, and so $In{t}_{ii}\left(A\right)\subseteq {\displaystyle {\cup }_{ii\in \wedge }{G}_{ii}}$. On the other hand, if $y\in {\displaystyle {\cup }_{ii\in \wedge }{G}_{ii}}$, then $y\in G_k\subseteq A$ for some $k\in \wedge$. Thus $y\in In{t}_{ii}\left(A\right)$, and ${\displaystyle {\cup }_{ii\in \wedge }{G}_{ii}}\subseteq In{t}_{ii}\left(A\right)$. Accordingly, ${\displaystyle {\cup }_{ii\in \wedge }{G}_{ii}}\subseteq In{t}_{ii}\left(A\right)$.

2) Straightforward.

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

4) If $x\in A\backslash D_{ii}\left(X\backslash A\right)$, then $x\notin D_{ii}\left(X\backslash A\right)$ and so there exist an ii-open set G containing x such that $G\cap \left(X\backslash A\right)=\varphi$. Thus, $x\in G\subset A$ and hence $x\in In{t}_{ii}\left(A\right)$. This shows that $A\backslash D_{ii}\left(X\backslash A\right)\subset In{t}_{ii}\left(A\right)$. Now let $x\in In{t}_{ii}\left(A\right)$. Since $In{t}_{ii}\left(A\right)\in {\tau }^{ii}$ and $In{t}_{ii}\left(A\right)\cap \left(X\backslash A\right)=\varphi$. We have $x\notin D_{ii}\left(X\backslash A\right)$. Therefore, $In{t}_{ii}\left(A\right)=A\backslash D_{ii}\left(X\backslash A\right)$.

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

$X\backslash In{t}_{ii}\left(A\right)=X\backslash \left(A\backslash D_{ii}\left(X\backslash A\right)\right)=\left(X\backslash A\right)\cup D_{ii}\left(X\backslash A\right)=C{L}_{ii}\left(X\backslash A\right)$.

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

$In{t}_{ii}\left(X\backslash A\right)=\left(X\backslash A\right)\backslash D_{ii}\left(A\right)=X\backslash \left(A\cup D_{ii}\left(A\right)\right)=X\backslash C{L}_{ii}(A)$

7) Since $A\subseteq B$ and $In{t}_{ii}\left(A\right)\subseteq A$, $In{t}_{ii}\left(B\right)\subseteq B$, we get $In{t}_{ii}\left(A\right)\subseteq In{t}_{ii}\left(B\right)$.

8) Since $A\subseteq \left(A\cup B\right)$ and $B\subseteq \left(A\cup B\right)$, from (7) we get $In{t}_{ii}\left(A\right)\subseteq In{t}_{ii}\left(A\cup B\right)$, $In{t}_{ii}\left(B\right)\subseteq In{t}_{ii}\left(A\cup B\right)$. Therefore $In{t}_{ii}\left(A\right)\cup In{t}_{ii}\left(B\right)\subseteq In{t}_{ii}\left(A\cup B\right)$.

9) Since $A\cap B\subseteq A$ and $A\cap B\subseteq B$, from (7) we get $In{t}_{ii}\left(A\cap B\right)\subseteq In{t}_{ii}\left(A\right)$, $In{t}_{ii}\left(A\cap B\right)\subseteq In{t}_{ii}\left(B\right)$. Therefore $In{t}_{ii}\left(A\cap B\right)\subseteq In{t}_{ii}\left(A\right)\cap \left(B\right)$.

Definition 3.7. $b_{ii}\left(A\right)=A\backslash In{t}_{ii}\left(A\right)$ 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) $b_{ii}\left(A\right)\subset b\left(A\right)$ where $b\left(A\right)$ denotes the border of A

2) $In{t}_{ii}\left(A\right)\cup b_{ii}\left(A\right)=A$

3) $In{t}_{ii}\left(A\right)\cap b_{ii}\left(A\right)=\varphi$

4) $b_{ii}\left(A\right)=\varphi$ if and only if A is ii-open set

5) $b_{ii}\left(In{t}_{ii}\left(A\right)\right)=\varphi$

6) $In{t}_{ii}\left(b_{ii}\left(A\right)\right)=\varphi$

7) $b_{ii}\left(b_{ii}\left(A\right)\right)=b_{ii}(A)$

8) $b_{ii}\left(A\right)=A\cap C{L}_{ii}\left(X\backslash A\right)$

9) $b_{ii}\left(A\right)=A\cap D_{ii}\left(X\backslash A\right)$

Proof.

1) Since $Int\left(A\right)\subset In{t}_{ii}\left(A\right)$, we have $b_{ii}\left(A\right)=A\backslash In{t}_{ii}\left(A\right)\subseteq A\backslash Int\left(A\right)=b\left(A\right)$.

2) and (3). Straightforward.

4) Since $In{t}_{ii}\left(A\right)\subseteq A$, it follows from Theorem 3.6 (2). That A is ii-open $\iff$ $A=In{t}_{ii}\left(A\right)$ $\iff$ $b_{ii}\left(A\right)=A\backslash In{t}_{ii}\left(A\right)=\varphi$.

5) Since $In{t}_{ii}\left(A\right)$ is ii-open, it follows from (4) that $b_{ii}\left(In{t}_{ii}\left(A\right)\right)=\varphi$.

6) If $x\in In{t}_{ii}\left(b_{ii}\left(A\right)\right)$, then $x\in b_{ii}\left(A\right)$. On the other hand, since $b_{ii}\left(A\right)\subset A$, $x\in In{t}_{ii}\left(b_{ii}\left(A\right)\right)\subset In{t}_{ii}\left(A\right)$. Hence, $x\in In{t}_{ii}\left(A\right)\cap b_{ii}\left(A\right)$.

Which contradicts (3). Thus $In{t}_{ii}\left(b_{ii}\left(A\right)\right)=\varphi$.

7) Using (6), we get $b_{ii}\left(b_{ii}\left(A\right)\right)=b_{ii}\left(A\right)\backslash In{t}_{ii}\left(b_{ii}\left(A\right)\right)=b_{ii}\left(A\right)$.

8) Using Theorem 3.6 (6), we have $b_{ii}\left(A\right)=A\backslash In{t}_{ii}\left(A\right)=A\backslash \left(X\backslash C{L}_{ii}\left(X\backslash A\right)\right)=A\cap C{L}_{ii}\left(X\backslash A\right)$

9) Applying (8) and the Theorem (3.4), we have $b_{ii}\left(A\right)=A\cap C{L}_{ii}\left(X\backslash A\right)=A\cap \left(\left(X\backslash A\right)\cup D_{ii}\left(X\backslash A\right)\right)=A\cap D_{ii}\left(X\backslash A\right)$.

Example 3.9. Consider the topological space $\left(X,\tau \right)$ given in Example (3.3). If $A=\left\{a,b\right\}$, then $b_{ii}\left(A\right)=\varphi$ and $b\left(A\right)=\left\{a\right\}$. Hence, $b\left(A\right)\not\subset b_{ii}\left(A\right)$, that is, in general, the converse Theorem 3.9 (1) may not be true.

Definition 3.10. $F{r}_{ii}\left(A\right)=C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)$ 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) $F{r}_{ii}\left(A\right)\subset Fr\left(A\right)$ where $Fr\left(A\right)$ denotes the frontier of A

2) $C{L}_{ii}\left(A\right)=In{t}_{ii}\left(A\right)\cup F{r}_{ii}(A)$

3) $In{t}_{ii}\left(A\right)\cap F{r}_{ii}\left(A\right)=\varphi$

4) $b_{ii}\left(A\right)\subset F{r}_{ii}(A)$

5) $F{r}_{ii}\left(A\right)=b_{ii}\left(A\right)\cup D_{ii}(A)$

6) $F{r}_{ii}\left(A\right)=D_{ii}\left(A\right)$ if and only if A is ii-open set

7) $F{r}_{ii}\left(A\right)=C{L}_{ii}\left(A\right)\cap C{L}_{ii}\left(X\backslash A\right)$

8) $F{r}_{ii}\left(A\right)=F{r}_{ii}\left(X\backslash A\right)$

9) $F{r}_{ii}\left(A\right)$ is ii-closed

10) $F{r}_{ii}\left(F{r}_{ii}\left(A\right)\right)\subset F{r}_{ii}(A)$

11) $F{r}_{ii}\left(In{t}_{ii}\left(A\right)\right)\subset F{r}_{ii}(A)$

12) $F{r}_{ii}\left(C{L}_{ii}\left(A\right)\right)\subset F{r}_{ii}(A)$

13) $In{t}_{ii}\left(A\right)=A\backslash F{r}_{ii}(A)$

Proof.

1) Since $C{L}_{ii}\left(A\right)\subseteq CL\left(A\right)$ and $Int\left(A\right)\subseteq In{t}_{ii}\left(A\right)$, it follows that $F{r}_{ii}\left(A\right)=C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)\subseteq CL\left(A\right)\backslash In{t}_{ii}\left(A\right)\subseteq CL\left(A\right)\backslash Int\left(A\right)\subseteq Fr\left(A\right)$.

2) $In{t}_{ii}\left(A\right)\cup F{r}_{ii}\left(A\right)=In{t}_{ii}\left(A\right)\cup \left(C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)\right)=C{L}_{ii}\left(A\right)$.

3) $In{t}_{ii}\left(A\right)\cap F{r}_{ii}\left(A\right)=In{t}_{ii}\left(A\right)\cap \left(C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)\right)=\varphi$.

4) Since $A\subseteq C{L}_{ii}\left(A\right)$, we have $b_{ii}\left(A\right)=A\backslash In{t}_{ii}\left(A\right)\subseteq C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)=F{r}_{ii}\left(A\right)$.

5) Since $In{t}_{ii}\left(A\right)\cup F{r}_{ii}\left(A\right)=In{t}_{ii}\left(A\right)\cup b_{ii}\left(A\right)\cup D_{ii}\left(A\right)$, $F{r}_{ii}\left(A\right)=b_{ii}\left(A\right)\cup D_{ii}\left(A\right)$.

6) Assume that A is ii-open. Then $F{r}_{ii}\left(A\right)=b_{ii}\left(A\right)\cup D_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)=\varphi \cup \left(D_{ii}\left(A\right)\backslash A\right)=D_{ii}\left(A\right)\backslash A=b_{ii}\left(X\backslash A\right)$, by using (5), Theorem 3.6 (2), Theorem 3.8 (4) and Theorem 3.8 (9).

Conversely, suppose that $F{r}_{ii}\left(A\right)=b_{ii}\left(X\backslash A\right)$. Then $\varphi =F{r}_{ii}\left(A\right)\backslash b_{ii}\left(X\backslash A\right)=\left(C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)\right)\backslash \left(X\backslash A\right)\backslash In{t}_{ii}\left(X\backslash A\right)=A\backslash In{t}_{ii}\left(A\right)$. by using (4) and (5) of Theorem 3.6, and so $A\subseteq In{t}_{ii}\left(A\right)$. Since $In{t}_{ii}\left(A\right)\subseteq A$ in general, it follows that $In{t}_{ii}\left(A\right)=A$ so from Theorem 3.6 (2) that A is ii-open set.

7) $F{r}_{ii}\left(A\right)=C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)=C{L}_{ii}\left(A\right)\cap \left(C{L}_{ii}\left(X\backslash A\right)\right)$.

8) It follows from (7).

9) $\begin{array}{l}C{L}_{ii}\left(F{r}_{ii}\left(A\right)\right)=C{L}_{ii}\left(C{L}_{ii}\left(A\right)\right)\cap \left(C{L}_{ii}\left(X\backslash A\right)\right)\\ \subset C{L}_{ii}\left(C{L}_{ii}\left(A\right)\right)\cap C{L}_{ii}\left(C{L}_{ii}\left(X\backslash A\right)\right)=F{r}_{ii}\left(A\right).\end{array}$ Hence, $F{r}_{ii}\left(A\right)$ is ii-closed.

10) $\begin{array}{l}F{r}_{ii}\left(F{r}_{ii}\left(A\right)\right)\\ =C{L}_{ii}\left(F{r}_{ii}\left(A\right)\right)\cap C{L}_{ii}\left(X\backslash F{r}_{ii}\left(A\right)\right)\subset C{L}_{ii}\left(F{r}_{ii}\left(A\right)\right)=F{r}_{ii}\left(A\right)\end{array}$.

11) Using Theorem 3.6 (3), we get $F{r}_{ii}\left(In{t}_{ii}\left(A\right)\right)=C{L}_{ii}\left(In{t}_{ii}\left(A\right)\right)\backslash In{t}_{ii}\left(In{t}_{ii}\left(A\right)\right)\subseteq C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)=F{r}_{ii}\left(A\right)$.

12) $\begin{array}{l}F{r}_{ii}\left(C{L}_{ii}\left(A\right)\right)=C{L}_{ii}\left(C{L}_{ii}\left(A\right)\right)\backslash In{t}_{ii}\left(C{L}_{ii}\left(A\right)\right)=C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(C{L}_{ii}\left(A\right)\right)\\ =C{L}_{ii}\left(A\right)\backslash In{t}_{ii}\left(A\right)=F{r}_{ii}\left(A\right).\end{array}$

13) $A\backslash F{r}_{ii}\left(A\right)=\left(A\backslash C{L}_{ii}\left(A\right)\right)\backslash In{t}_{ii}\left(A\right)=In{t}_{ii}\left(A\right)$.

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 $\left(X,\tau \right)$ given in Example 3.3. If $A=\left\{c\right\}$, then $Fr\left(A\right)=\left\{a,c\right\}\not\subset \left\{c\right\}=F{r}_{ii}\left(A\right)$, and if $B=\left\{a,b\right\}$, then $F{r}_{ii}\left(B\right)=\left\{c\right\}\not\subset b_{ii}\left(B\right)$.

Definition 3.13. $Ex{t}_{ii}\left(A\right)=In{t}_{ii}\left(X\backslash A\right)$ 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) $Ext\left(A\right)\subset Ex{t}_{ii}\left(A\right)$ where $Ext\left(A\right)$ denotes the exterior of A

2) $Ex{t}_{ii}\left(A\right)$ is ii-open

3) $Ex{t}_{ii}\left(A\right)=In{t}_{ii}\left(X\backslash A\right)=X\backslash C{L}_{ii}(A)$

4) $Ex{t}_{ii}\left(Ex{t}_{ii}\left(A\right)\right)=In{t}_{ii}(C{L}_{ii}(A))$

5) If $A\subseteq B$, then $Ex{t}_{ii}\left(A\right)\supset Ex{t}_{ii}(B)$

6) $Ex{t}_{ii}\left(A\cup B\right)\subset Ex{t}_{ii}\left(A\right)\cup Ex{t}_{ii}(B)$

7) $Ex{t}_{ii}\left(A\cap B\right)\supset Ex{t}_{ii}\left(A\right)\cap Ex{t}_{ii}(B)$

8) $Ex{t}_{ii}\left(X\right)=\varphi$

9) $Ex{t}_{ii}\left(\varphi \right)=X$

10) $Ex{t}_{ii}\left(A\right)=Ex{t}_{ii}(X\backslash Ex{t}_{ii}(A))$

11) $In{t}_{ii}\left(A\right)\subset Ex{t}_{ii}(Ex{t}_{ii}(A))$

12) $X=Ex{t}_{ii}\left(A\right)\cup Ex{t}_{ii}\left(A\right)\cup F{r}_{ii}(A)$

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

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

3) $\begin{array}{l}Ex{t}_{ii}\left(Ex{t}_{ii}\left(A\right)\right)=Ex{t}_{ii}\left(X\backslash C{L}_{ii}\left(A\right)\right)\\ =In{t}_{ii}\left(X\backslash X\backslash C{L}_{ii}\left(A\right)\right)=In{t}_{ii}\left(C{L}_{ii}\left(A\right)\right)\end{array}$.

4) Assume that $A\subset B$. Then $Ex{t}_{ii}\left(B\right)=Ex{t}_{ii}\left(X\backslash B\right)\subseteq Ex{t}_{ii}\left(X\backslash A\right)=Ex{t}_{ii}\left(A\right)$, by using Theorem 3.6 (7).

5) Applying Theorem 3.6 (8), we get

$\begin{array}{l}Ex{t}_{ii}\left(A\cup B\right)=In{t}_{ii}\left(X\backslash \left(A\cup B\right)\right)=In{t}_{ii}\left(\left(X\backslash A\right)\cup \left(X\backslash B\right)\right)\\ \subseteq In{t}_{ii}\left(X\backslash A\right)\cup In{t}_{ii}\left(X\backslash B\right)=Ex{t}_{ii}\left(A\right)\cup Ex{t}_{ii}\left(B\right).\end{array}$

6) Applying Theorem 3.6 (9), we obtain

$\begin{array}{l}Ex{t}_{ii}\left(A\cap B\right)=In{t}_{ii}\left(X\backslash \left(A\cap B\right)\right)=In{t}_{ii}\left(\left(X\backslash A\right)\cap \left(X\backslash B\right)\right)\\ \supset In{t}_{ii}\left(X\backslash A\right)\cap In{t}_{ii}\left(X\backslash B\right)=Ex{t}_{ii}\left(A\right)\cap Ex{t}_{ii}\left(B\right).\end{array}$.

7) Straightforward.

8) Straightforward.

9) $\begin{array}{l}Ex{t}_{ii}\left(X\backslash Ex{t}_{ii}\left(A\right)\right)=Ex{t}_{ii}\left(X\backslash In{t}_{ii}\left(X\backslash A\right)\right)=In{t}_{ii}\left(X\backslash \left(X\backslash In{t}_{ii}\left(X\backslash A\right)\right)\right)\\ =In{t}_{ii}\left(In{t}_{ii}\left(X\backslash A\right)\right)=In{t}_{ii}\left(X\backslash A\right)=Ex{t}_{ii}\left(A\right).\end{array}$

10) $\begin{array}{l}In{t}_{ii}\left(A\right)\subset In{t}_{ii}\left(C{L}_{ii}\left(A\right)\right)=In{t}_{ii}\left(X\backslash In{t}_{ii}\left(X\backslash A\right)\right)\\ =In{t}_{ii}\left(X\backslash Ex{t}_{ii}\left(A\right)\right)=Ex{t}_{ii}\left(Ex{t}_{ii}\left(A\right)\right).\end{array}$

Example 3.15. Let $X=\left\{a,b,c,d\right\}$ and $\tau =\left\{\varphi ,X,\left\{c,d\right\}\right\}$. Thus, $iio\left(x\right)=\left\{\varphi ,X,\left\{c,d\right\},\left\{b,c,d\right\},\left\{a,c,d\right\}\right\}$. If $A=\left\{a\right\}$ and $B=\left\{b\right\}$. Then $Ex{t}_{ii}\left(A\right)\not\subset Ext\left(A\right)$. $Ex{t}_{ii}\left(A\cap B\right)\ne Ex{t}_{ii}\left(A\right)\cap Ex{t}_{ii}\left(B\right)$ and $Ex{t}_{ii}\left(A\cup B\right)\ne Ex{t}_{ii}\left(A\right)\cup Ex{t}_{ii}\left(B\right)$.

4. A New Decomposition of ii-Continuity

We begin by the following definition:

Definition 4.1. A function $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ is called ii-continuous if $f^{-1}\left(G\right)$ is ii-open set in $\left(X,\tau \right)$ for any open set G of $\left(Y,\sigma \right)$.

Theorem 4.2. Let $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ 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 $\left(Y,\sigma \right)$. Since f is continuous, it follows that $f^{-1}\left(G\right)$ is open set in $\left(X,\tau \right)$. But every open set is ii-open set [4] . Hence $f^{-1}\left(G\right)$ is ii-open set in $\left(X,\tau \right)$. Thus f is ii-continuous.

2) Let G be open set in $\left(Y,\sigma \right)$. Since f is an ii-continuous, it follows that $f^{-1}\left(G\right)$ is an ii-open set in $\left(X,\tau \right)$. But every ii-open set is i-open set [4] . Hence $f^{-1}\left(G\right)$ is i-open set in $\left(X,\tau \right)$. Thus f is i-continuous.

3) Let G be open set in ( ${\rm Y},$ ϭ). Since f is α-continuous, it follows that $f^{-1}\left(G\right)$ is α-open set in $\left(X,\tau \right)$. But every α-open set is ii-open set [4] . Hence $f^{-1}\left(G\right)$ is ii-open set in $\left(X,\tau \right)$. Thus f is an ii-continuous.

The converse need not be true by the following example.

Example 4.3. Let

$X=\left\{a,b,c,d\right\}$, $\tau =\left\{\varphi ,X,\left\{a\right\},\left\{b\right\},\left\{a,b\right\}\right\}$

and

$Y=\left\{a,b,c,d\right\}$, $\sigma =\left\{\varphi ,Y,\left\{a\right\},\left\{b\right\},\left\{a,b\right\},\left\{a,d\right\},\left\{a,b,d\right\}\right\}$

and

$iio\left(x\right)=\left\{\varphi ,X,\left\{a\right\},\left\{b\right\},\left\{a,b\right\},\left\{a,d\right\},\left\{a,b,c\right\},\left\{a,b,d\right\}\right\}$,

$\begin{array}{l}io\left(x\right)=\{\varphi ,X,\left\{a\right\},\left\{b\right\},\left\{a,b\right\},\left\{a,c\right\},\left\{a,d\right\},\left\{b,c\right\},\left\{b,d\right\},\\ \left\{a,b,c\right\},\left\{a,b,d\right\},\left\{a,c,d\right\},\left\{b,c,d\right\}\},\end{array}$

$\alpha o\left(x\right)=\left\{\varphi ,X,\left\{a\right\},\left\{b\right\},\left\{a,b\right\},\left\{a,b,c\right\},\left\{a,b,d\right\}\right\}$.

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

Theorem 4.4. Let $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ be a function then every semi-continuous function is an ii-continuous.

Proof. Let G be open set in $\left(Y,\sigma \right)$. Since f is semi-continuous, it follows that $f^{-1}\left(G\right)$ is semi-open set in $\left(X,\tau \right)$. But every semi-open set is ii-open set [4] . Hence $f^{-1}\left(G\right)$ is ii-open set in $\left(X,\tau \right)$. Thus f is an ii-continuous.

Definition 4.5. A function $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ is called int-continuous if $f^{-1}\left(G\right)$ is int-open set in $\left(X,\tau \right)$ for any open set G in $\left(Y,\sigma \right)$.

Theorem 4.6. Let $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ 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 $\left(Y,\sigma \right)$. Since f is continuous, it follows that $f^{-1}\left(G\right)$ is open set in $\left(X,\tau \right)$. But every open set is int-open set [4] . Hence $f^{-1}\left(G\right)$ is int-open set in $\left(X,\tau \right)$. Thus f is int-continuous.

2) Let G be open set in $\left(Y,\sigma \right)$. Since f is ii-continuous, it follows that $f^{-1}\left(G\right)$ is an ii-open set in $\left(X,\tau \right)$. But every ii-open set is int-open set [4] . Hence $f^{-1}\left(G\right)$ is int-open set in $\left(X,\tau \right)$. Thus f is int-continuous.

3) Let G be open set in $\left(Y,\sigma \right)$. Since f is α-continuous, it follows that $f^{-1}\left(G\right)$ is α-open set in $\left(X,\tau \right)$. But every α-open set is int-open set [4] . Hence $f^{-1}\left(G\right)$ is int-open set in $\left(X,\tau \right)$. Thus f is int-continuous.

The converse need not be true by the following example.

Example 4.7. Let $X=\left\{a,b,c\right\}$, $\tau =\left\{\varphi ,X,\left\{a\right\},\left\{b,c\right\}\right\}$ and $Y=\left\{a,b,c\right\}$, $\sigma =\left\{\varphi ,Y,\left\{a\right\},\left\{a,c\right\}\right\}$ and $into\left(x\right)=\left\{\varphi ,X,\left\{a\right\},\left\{a,b\right\},\left\{b,c\right\},\left\{a,c\right\}\right\}$, $iio\left(x\right)=\alpha o\left(x\right)=\left\{\varphi ,X,\left\{a\right\},\left\{b,c\right\}\right\}$. Let $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ be the identity function then $f^{-1}\left(\left\{a\right\}\right)=\left\{a\right\}$, $f^{-1}\left(\left\{b\right\}\right)=\left\{b\right\}$, $f^{-1}\left(\left\{c\right\}\right)=\left\{c\right\}$. Then f is int-continuous, but f is not ii-continuous, since for the open set $\left\{a,c\right\}$ in $\left(Y,\sigma \right)$, $f^{-1}\left(\left\{a,c\right\}\right)=\left\{a,c\right\}$ is not ii-open in $\left(X,\tau \right)$ and f is not continuous, since for the open set $\left\{a,c\right\}$ in $\left(Y,\sigma \right)$, $f^{-1}\left(\left\{a,c\right\}\right)=\left\{a,c\right\}$ is not open in $\left(X,\tau \right)$ and f is not α-continuous, since for the open set $\left\{a,c\right\}$ in $\left(Y,\sigma \right)$, $f^{-1}\left(\left\{a,c\right\}\right)=\left\{a,c\right\}$ 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 $A\subseteq CL\left(Int\left(A\right)\cap A\right)$.

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 $A\subseteq Int\left(CL\left(Int\left(A\right)\right)\right)$ and $A\subseteq CL\left(A\right)$. Therefore $A\subseteq CL\left(Int\left(A\right)\right)\cap CL\left(A\right)$, this implies that $A\subseteq CL\left(Int\left(A\right)\cap A\right)$. Now, put $G=Int\left(A\right)$ where $G\ne \varphi ,X$, 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 $G\ne \varphi ,X$, such that $G=Int\left(A\right)$ satisfying $A\subseteq CL\left(Int\left(A\right)\cap A\right)$ and A is ii-open set. Since $A\subseteq CL\left(Int\left(A\right)\cap A\right)$, this implies that $A\subseteq CL\left(Int\left(A\right)\right)$ and $Int\left(A\right)\subseteq Int\left(CL\left(Int\left(A\right)\right)\right)$. Since A is ii-open set, using (2) from Theorem (3.6), we get $A=Int\left(A\right)$. Therefore $A\subseteq Int\left(CL\left(Int\left(A\right)\right)\right)$. 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.

Corollary 4.10. A function $f:\left(X,\tau \right)\to \left(Y,\sigma \right)$ 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 $T_{0ii}$ and $T_{1ii}$ spaces for ii-open sets and we determine them by giving many examples. Specially, we define $T_1$, $T_{1\alpha }$ and $T_{1i}$ spaces to compare them with $T_{1ii}$ space.

Definition 5.1. A topological space X is called

1) $T_{0ii}$ if a, b are to distinct points in X, there exists ii-open set U such that either $a\in U$ and $b\notin U$, and $b\in U$ and $a\notin U$.

2) $T_{1ii}$ if $a,b\in X$ and $a\ne b$, there exist ii-open sets U, V containing a, b respectively, such that $b\notin U$ and $a\notin V$.

Example 5.2. Let $X=\left\{a,b\right\}$, ${\tau }^{ii}=\tau =\left\{\varphi ,X,\left\{a\right\},\left\{b\right\}\right\}$ $\left(X,\tau \right)$ and $\left(X,{\tau }^{ii}\right)$ are topological spaces.

1) $a,b\in X$ ( $a\ne b$ ) there exists $\left\{a\right\}\in {\tau }^{ii}$ such that $a\in \left\{a\right\}$, $b\notin \left\{a\right\}$. Therefore $\left(X,\tau \right)$ is $T_{0ii}$.

2) $a,b\in X$ ( $a\ne b$ ) there exists $\left\{a\right\},\left\{b\right\}\in {\tau }^{ii}$ such that $a\in \left\{a\right\}$, $b\in \left\{b\right\}$. Therefore $\left(X,\tau \right)$ is $T_{1ii}$.

Theorem 5.3.

1) Every $T_0$ -space is $T_{0ii}$ -space,

2) Every $T_1$ -space is $T_{0ii}$ -space,

3) Every $T_1$ -space is $T_{1ii}$ -space,

4) Every $T_{1ii}$ -space is $T_{0ii}$ -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

$X=\left\{a,b,c\right\}$, $\tau =\left\{\varphi ,X,\left\{a\right\}\right\}$ and ${\tau }^{ii}=\left\{\varphi ,X,\left\{a\right\},\left\{a,b\right\},\left\{a,c\right\}\right\}$.

$\left(X,\tau \right)$ and $\left(X,{\tau }^{ii}\right)$ are topological spaces.

$\left(X,\tau \right)$ is not $T_0$ -space because, $b,c\in X$ ( $b\ne c$ ) there is no open set G such that $b\in G$, $c\notin G$.

$\left(X,\tau \right)$ is $T_{0ii}$ -space because, $a,b\in X$ ( $a\ne b$ ) there exists $\left\{a\right\}\in {\tau }^{ii}$ such that $a\in \left\{a\right\}$, $b\notin \left\{a\right\}$.

$a,c\in X$ ( $a\ne c$ ) there exists $\left\{a\right\}\in {\tau }^{ii}$ such that $a\in \left\{a\right\}$, $c\notin \left\{a\right\}$.

$b,c\in X$ ( $b\ne c$ ) there exists $\left\{a,b\right\}\in {\tau }^{ii}$ such that $b\in \left\{a,b\right\}$, $c\notin \left\{a,b\right\}$.

$\left(X,\tau \right)$ is not $T_1$ -space because, $a,b\in X$ ( $a\ne b$ ) there exists $X\in \tau$ such that $a\in X$, $b\in X$.

$\left(X,\tau \right)$ is not $T_{1ii}$ -space because, $b,a\in X$ ( $a\ne b$ ) there exists $\left\{a,b\right\}\in {\tau }^{ii}$ such that $a\in \left\{a,b\right\}$, $b\in \left\{a,b\right\}$.

Theorem 5.5. Every $T_{1\alpha }$ -space is $T_{1ii}$ -space.

Proof. Let X be $T_{1\alpha }$ -space. Let a, b be two distinct points in X. Since X is $T_{1\alpha }$ -space there exist two α-open sets U, V in X such that $a\in U$, $b\notin U$, $a\notin V$, $b\in V$. Since every α-open set is ii-open set [4] , U, V is an ii-open set in X. Hence X is $T_{1ii}$ -space.

Theorem 5.6. Every $T_{1ii}$ -space is $T_{1i}$ -space.

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

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

Example 5.7. Let $X=\left\{a,b,c\right\}$, $\tau =\left\{\varphi ,X,\left\{a\right\},\left\{b,c\right\}\right\}$ and.

${\tau }^i=\left\{\varphi ,X,\left\{a\right\},\left\{b\right\},\left\{c\right\},\left\{b,c\right\}\right\}$. ${\tau }^{ii}=\left\{\varphi ,X,\left\{a\right\},\left\{b,c\right\}\right\}$.

$\left(X,\tau \right)$ and $\left(X,{\tau }^{ii}\right)$ are topological spaces.

$\left(X,\tau \right)$ is $T_{1i}$ -space because, $a,b\in X$ ( $a\ne b$ ) there exists $\left\{a\right\},\left\{b\right\}\in {\tau }^i$ such that $a\in \left\{a\right\}$, $b\notin \left\{a\right\}$ and $b\in \left\{b\right\}$, $a\notin \left\{b\right\}$.

$a,c\in X$ ( $a\ne c$ ) there exists $\left\{a\right\},\left\{c\right\}\in {\tau }^i$ such that $a\in \left\{a\right\}$, $c\notin \left\{a\right\}$ and $c\in \left\{c\right\}$, $a\notin \left\{c\right\}$.

$b,c\in X$ ( $b\ne c$ ) there exists $\left\{c\right\},\left\{b\right\}\in {\tau }^i$ such that $c\in \left\{c\right\}$, $b\notin \left\{c\right\}$ and $b\in \left\{b\right\}$, $c\notin \left\{b\right\}$.

$\left(X,\tau \right)$ is not $T_{1ii}$ -space because, $b,c\in X$ ( $c\ne b$ ) there exists $\left\{b\right\},\left\{c\right\}\in {\tau }^{ii}$ such that $c\in \left\{b,c\right\}$, $b\in \left\{b,c\right\}$.

Theorem 5.8. A space X is $T_{0ii}$ if and only if $C{L}_{ii}\left(\left\{x\right\}\right)\ne C{L}_{ii}\left(\left\{y\right\}\right)$ for every pair of distinct points x, y of X.

Proof. Let X be a $T_{0ii}$ -space. Let $x,y\in X$ such that $x\ne y$, then there exists an ii-open set U containing one of the points but not the other, then $x\in U$ and $y\notin U$. Then $X\backslash U$ is ii-closed set containing y but not x. But $C{L}_{ii}\left(\left\{y\right\}\right)$ is the smallest ii-closed set containing y. Therefore $C{L}_{ii}\left(\left\{y\right\}\right)\subset X\backslash U$ and hence $x\notin C{L}_{ii}\left(\left\{y\right\}\right)$. Thus $C{L}_{ii}\left(\left\{x\right\}\right)\ne C{L}_{ii}\left(\left\{y\right\}\right)$.

Conversely, Suppose for any $x,y\in X$ with $x\ne y$, $C{L}_{ii}\left(\left\{x\right\}\right)\ne C{L}_{ii}\left(\left\{y\right\}\right)$. Let $z\in X$ such that $z\in C{L}_{ii}\left(\left\{x\right\}\right)$ but $z\notin C{L}_{ii}\left(\left\{y\right\}\right)$. If $x\in C{L}_{ii}\left(\left\{y\right\}\right)$ then $C{L}_{ii}\left(\left\{x\right\}\right)\subset C{L}_{ii}\left(\left\{y\right\}\right)$ and hence $z\in C{L}_{ii}\left(\left\{y\right\}\right)$. This is contradiction. Therefore $x\notin C{L}_{ii}\left(\left\{y\right\}\right)$. That is $x\in X\backslash C{L}_{ii}\left(\left\{y\right\}\right)$. Therefore $X\backslash C{L}_{ii}\left(\left\{y\right\}\right)$ is ii-open set containing x but not y. Hence X is an $T_{0ii}$ -space.

Theorem 5.9. A space $\left(X,\tau \right)$ is $T_{1ii}$ -space if and only if the singletons are ii-closed sets.

Proof. Let X be $T_{1ii}$ -space and let $x\in X$, to prove that $\left\{x\right\}$ is ii-closed set. We will prove $X\backslash \left\{x\right\}$ is ii-open set in X. Let $y\in X\backslash \left\{x\right\}$, implies $x\ne y$ and since X is $T_{1ii}$ -space then their exist two ii-open sets U, V such that $x\notin U$, $y\in V\subset X\backslash \left\{x\right\}$. Since $y\in V\subset X\backslash \left\{x\right\}$, then $X\backslash \left\{x\right\}$ is ii-open set. Hence $\left\{x\right\}$ is ii-closed set.

Conversely, Let $x\ne y\in X$ then $\left\{x\right\},\left\{y\right\}$ are ii-closed sets. That is $X\backslash \left\{x\right\}$ is ii-open set clearly, $x\notin X\backslash \left\{x\right\}$ and $y\in X\backslash \left\{x\right\}$. Similarly $X\backslash \left\{y\right\}$ is ii-open set, $y\notin X\backslash \left\{y\right\}$ and $x\in X\backslash \left\{y\right\}$. Hence X is an $T_{1ii}$ -space.

As a consequence the following Figure 2 shows the relations among $T_0$, $T_{0ii}$, $T_1$, $T_{1ii}$ and $T_{1\alpha }$.