<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">OALibJ</journal-id><journal-title-group><journal-title>Open Access Library Journal</journal-title></journal-title-group><issn pub-type="epub">2333-9705</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/oalib.1105604</article-id><article-id pub-id-type="publisher-id">OALibJ-95072</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Biomedical&amp;Life Sciences</subject><subject> Business&amp;Economics</subject><subject> Chemistry&amp;Materials Science</subject><subject> Computer Science&amp;Communications</subject><subject> Earth&amp;Environmental Sciences</subject><subject> Engineering</subject><subject> Medicine&amp;Healthcare</subject><subject> Physics&amp;Mathematics</subject><subject> Social Sciences&amp;Humanities</subject></subj-group></article-categories><title-group><article-title>
 
 
  On Standard Concepts Using &lt;i&gt;ii&lt;/i&gt;-Open Sets
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Beyda</surname><given-names>S. Abdullah</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Amir</surname><given-names>A. Mohammed</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Department of Mathematics, College of Education for pure Sciences, University of Mosul, Mosul, Iraq</addr-line></aff><pub-date pub-type="epub"><day>03</day><month>09</month><year>2019</year></pub-date><volume>06</volume><issue>09</issue><fpage>1</fpage><lpage>13</lpage><history><date date-type="received"><day>15,</day>	<month>July</month>	<year>2019</year></date><date date-type="rev-recd"><day>14,</day>	<month>September</month>	<year>2019</year>	</date><date date-type="accepted"><day>17,</day>	<month>September</month>	<year>2019</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  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
  <sub style="text-align:justify;white-space:normal;">0ii</sub>
  , T
  <sub style="text-align:justify;white-space:normal;">1ii</sub>
  .
 
</p></abstract><kwd-group><kwd>&lt;i&gt;α&lt;/i&gt;-Open Set</kwd><kwd> &lt;i&gt;ii&lt;/i&gt;-Open Set</kwd><kwd> Separation Axioms</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>The notion of α-open set was introduced by Njastad in [<xref ref-type="bibr" rid="scirp.95072-ref2">2</xref>] . Caldas in [<xref ref-type="bibr" rid="scirp.95072-ref1">1</xref>] 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 [<xref ref-type="bibr" rid="scirp.95072-ref3">3</xref>] 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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] 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 .</p></sec><sec id="s2"><title>2. Preliminaries</title><p>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.</p><p>Definition 2.1. A subset A of a space X is called</p><p>1) Semi-open set [<xref ref-type="bibr" rid="scirp.95072-ref5">5</xref>] if A ⊆ C L ( I n t ( A ) ) .</p><p>2) α-open set [<xref ref-type="bibr" rid="scirp.95072-ref2">2</xref>] if A ⊆ I n t ( C L ( I n t ( A ) ) ) .</p><p>3) i-open set [<xref ref-type="bibr" rid="scirp.95072-ref3">3</xref>] if there exist an open set G in the topology τ of X, such that</p><p>i) G ≠ ϕ , X</p><p>ii) A ⊆ C L ( A ∩ G )</p><p>The complement of an i-open set is an i-closed set.</p><p>4) ii-open set [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] if there exist an open set G in the topology τ of X, such that</p><p>i) G ≠ ϕ , X</p><p>ii) A ⊆ C L ( A ∩ G )</p><p>iii) I n t ( A ) = G</p><p>The complement of an ii-open set is an ii-closed set.</p><p>5) int-open set [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] 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.</p><p>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.</p><p>7) τ i , τ i i denote the family of all i-open sets and ii-open sets respectively.</p><p>Definition 2.2. [<xref ref-type="bibr" rid="scirp.95072-ref3">3</xref>] A topological space X is called</p><p>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 .</p><p>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 .</p><p>Definition 2.3. A function f : ( X , τ ) → ( Y , σ ) is called</p><p>1) Continuous [<xref ref-type="bibr" rid="scirp.95072-ref6">6</xref>] , if f − 1 ( G ) is open in ( X , τ ) for every open set G of ( Y , σ ) .</p><p>2) α-continuous [<xref ref-type="bibr" rid="scirp.95072-ref6">6</xref>] , if f − 1 ( G ) is α-open in ( X , τ ) for every open set G of ( Y , σ ) .</p><p>3) Semi-Continuous [<xref ref-type="bibr" rid="scirp.95072-ref5">5</xref>] , if f − 1 ( G ) is semi-open in ( X , τ ) for every open set G of ( Y , σ ) .</p><p>4) i-Continuous [<xref ref-type="bibr" rid="scirp.95072-ref3">3</xref>] , if f − 1 ( G ) is i-open in ( X , τ ) for every open set G of ( Y , σ ) .</p></sec><sec id="s3"><title>3. Applications of ii-Open Sets</title><p>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 ).</p><p>Theorem 3.2. For subsets A, B of a space X, the following statements hold:</p><p>1) D i i ( A ) ⊂ D (A)</p><p>2) If A ⊆ B , then D i i ( A ) ⊆ D i i (B)</p><p>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)</p><p>4) D i i ( D i i ( A ) ) \ A ⊂ D i i (A)</p><p>5) D i i ( A ∪ D i i ( A ) ) ⊂ A ∪ D i i (A)</p><p>Proof. 1) Since every open set is ii-open [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] , it follows that D i i ( A ) ⊂ D ( A ) .</p><p>2) Let x ∈ D i i ( A ) . Then G is ii-open set containing x such that</p><p>( A ∩ G ) \ { x } ≠ ϕ (3.1)</p><p>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 .</p><p>Hence, x ∈ D i i ( B ) . Therefore D i i ( A ) ⊆ D i i ( B ) .</p><p>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 ) .</p><p>This implies to D i i ( A ) ∪ D i i ( B ) ⊂ D i i ( A ∪ B ) .</p><p>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 ) .</p><p>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 ) .</p><p>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 .</p><p>Now, it follows similarly from (4) that G ∩ ( A \ { X } ) ≠ ϕ . Hence, x ∈ D i i ( A ) .</p><p>Therefore, in any case, D i i ( A ∪ D i i ( A ) ) ⊂ A ∪ D i i ( A ) .</p><p>In general, the converse of (1) may not true and the equality does not hold in (3) of theorem 3.2.</p><p>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:</p><p>1) A = { c } . Then, D ( A ) = { a , b } and D i i ( A ) = ϕ . Hence, D ( A ) ⊄ D i i ( A ) ;</p><p>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 ) .</p><p>Theorem 3.4. For any subset A of a space X, C L i i ( A ) = A ∪ D i i ( A ) .</p><p>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.</p><p>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 ) .</p><p>Theorem 3.6. For subset A, B of a space X, the following statements are true:</p><p>1) I n t i i ( A ) is the union of all ii-open subset of A</p><p>2) A is ii-open if and only if A = I n t i i (A)</p><p>3) I n t i i ( I n t i i ( A ) ) = I n t i i (A)</p><p>4) I n t i i ( A ) = A \ D i i ( X \ A )</p><p>5) X \ I n t i i ( A ) = C L i i ( X \ A )</p><p>6) X \ C L i i ( A ) = I n t i i ( X \ A )</p><p>7) If A ⊆ B , then I n t i i ( A ) ⊆ I n t i i (B)</p><p>8) I n t i i ( A ) ∪ I n t i i ( B ) ⊆ I n t i i ( A ∪ B )</p><p>9) I n t i i ( A ) ∩ I n t i i ( B ) ⊇ I n t i i ( A ∩ B )</p><p>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 ) .</p><p>2) Straightforward.</p><p>3) It follows from (1) and (2).</p><p>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 ) .</p><p>5) Using (4) and Theorem (3.4), we have</p><p>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 ) .</p><p>6) Using (4) and Theorem (3.4), we get.</p><p>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)</p><p>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 ) .</p><p>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 ) .</p><p>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 ) .</p><p>Definition 3.7. b i i ( A ) = A \ I n t i i ( A ) is said to be the ii-border of A.</p><p>Theorem 3.8. For a subset A of a space X, the following statements hold:</p><p>1) b i i ( A ) ⊂ b ( A ) where b ( A ) denotes the border of A</p><p>2) I n t i i ( A ) ∪ b i i ( A ) = A</p><p>3) I n t i i ( A ) ∩ b i i ( A ) = ϕ</p><p>4) b i i ( A ) = ϕ if and only if A is ii-open set</p><p>5) b i i ( I n t i i ( A ) ) = ϕ</p><p>6) I n t i i ( b i i ( A ) ) = ϕ</p><p>7) b i i ( b i i ( A ) ) = b i i (A)</p><p>8) b i i ( A ) = A ∩ C L i i ( X \ A )</p><p>9) b i i ( A ) = A ∩ D i i ( X \ A )</p><p>Proof.</p><p>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 ) .</p><p>2) and (3). Straightforward.</p><p>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 .</p><p>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 .</p><p>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 ) .</p><p>Which contradicts (3). Thus I n t i i ( b i i ( A ) ) = ϕ .</p><p>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 ) .</p><p>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 )</p><p>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 ) .</p><p>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.</p><p>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.</p><p>Theorem 3.11. For a subset A of a space X, the following statements hold:</p><p>1) F r i i ( A ) ⊂ F r ( A ) where F r ( A ) denotes the frontier of A</p><p>2) C L i i ( A ) = I n t i i ( A ) ∪ F r i i (A)</p><p>3) I n t i i ( A ) ∩ F r i i ( A ) = ϕ</p><p>4) b i i ( A ) ⊂ F r i i (A)</p><p>5) F r i i ( A ) = b i i ( A ) ∪ D i i (A)</p><p>6) F r i i ( A ) = D i i ( A ) if and only if A is ii-open set</p><p>7) F r i i ( A ) = C L i i ( A ) ∩ C L i i ( X \ A )</p><p>8) F r i i ( A ) = F r i i ( X \ A )</p><p>9) F r i i ( A ) is ii-closed</p><p>10) F r i i ( F r i i ( A ) ) ⊂ F r i i (A)</p><p>11) F r i i ( I n t i i ( A ) ) ⊂ F r i i (A)</p><p>12) F r i i ( C L i i ( A ) ) ⊂ F r i i (A)</p><p>13) I n t i i ( A ) = A \ F r i i (A)</p><p>Proof.</p><p>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 ) .</p><p>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 ) .</p><p>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 .</p><p>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 ) .</p><p>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 ) .</p><p>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).</p><p>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.</p><p>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 ) ) .</p><p>8) It follows from (7).</p><p>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.</p><p>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 ) .</p><p>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 ) .</p><p>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 ) .</p><p>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 ) .</p><p>The converses of (1) and (4) of Theorem 3.11 are not true in general, as shown by Example</p><p>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 ) .</p><p>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.</p><p>Theorem 3.14. For a subset A of a space X, the following statements hold:</p><p>1) E x t ( A ) ⊂ E x t i i ( A ) where E x t ( A ) denotes the exterior of A</p><p>2) E x t i i ( A ) is ii-open</p><p>3) E x t i i ( A ) = I n t i i ( X \ A ) = X \ C L i i (A)</p><p>4) E x t i i ( E x t i i ( A ) ) = I n t i i ( C L i i (A))</p><p>5) If A ⊆ B , then E x t i i ( A ) ⊃ E x t i i (B)</p><p>6) E x t i i ( A ∪ B ) ⊂ E x t i i ( A ) ∪ E x t i i (B)</p><p>7) E x t i i ( A ∩ B ) ⊃ E x t i i ( A ) ∩ E x t i i (B)</p><p>8) E x t i i ( X ) = ϕ</p><p>9) E x t i i ( ϕ ) = X</p><p>10) E x t i i ( A ) = E x t i i ( X \ E x t i i (A))</p><p>11) I n t i i ( A ) ⊂ E x t i i ( E x t i i (A))</p><p>12) X = E x t i i ( A ) ∪ E x t i i ( A ) ∪ F r i i (A)</p><p>Proof. 1) It follows from Theorem 3.6 (1).</p><p>2) It is straightforward by Theorem 3.6 (6).</p><p>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 ) ) .</p><p>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).</p><p>5) Applying Theorem 3.6 (8), we get</p><p>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 ) .</p><p>6) Applying Theorem 3.6 (9), we obtain</p><p>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 ) . .</p><p>7) Straightforward.</p><p>8) Straightforward.</p><p>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 ) .</p><p>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 ) ) .</p><p>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 ) .</p></sec><sec id="s4"><title>4. A New Decomposition of ii-Continuity</title><p>We begin by the following definition:</p><p>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 ) .</p><p>Theorem 4.2. Let f : ( X , τ ) → ( Y , σ ) be a function then:</p><p>1) Every continuous function is an ii-continuous,</p><p>2) Every ii-continuous function is an i-continuous,</p><p>3) Every α-continuous function is an ii-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is ii-open set in ( X , τ ) . Thus f is ii-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is i-open set in ( X , τ ) . Thus f is i-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is ii-open set in ( X , τ ) . Thus f is an ii-continuous.</p><p>The converse need not be true by the following example.</p><p>Example 4.3. Let</p><p>X = { a , b , c , d } , τ = { ϕ , X , { a } , { b } , { a , b } }</p><p>and</p><p>Y = { a , b , c , d } , σ = { ϕ , Y , { a } , { b } , { a , b } , { a , d } , { a , b , d } }</p><p>and</p><p>i i o ( x ) = { ϕ , X , { a } , { b } , { a , b } , { a , d } , { a , b , c } , { a , b , d } } ,</p><p>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 } } ,</p><p>α o ( x ) = { ϕ , X , { a } , { b } , { a , b } , { a , b , c } , { a , b , d } } .</p><p>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 ) .</p><p>Theorem 4.4. Let f : ( X , τ ) → ( Y , σ ) be a function then every semi-continuous function is an ii-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is ii-open set in ( X , τ ) . Thus f is an ii-continuous.</p><p>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 ) .</p><p>Theorem 4.6. Let f : ( X , τ ) → ( Y , σ ) be a function then:</p><p>1) Every continuous function is int-continuous,</p><p>2) Every ii-continuous function is int-continuous,</p><p>3) Every α-continuous function is int-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] . Hence f − 1 ( G ) is int-open set in ( X , τ ) . Thus f is int-continuous.</p><p>The converse need not be true by the following example.</p><p>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.</p><p>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 ) .</p><p>Theorem 4.9. A subset A of a space X is α-open set if and only if A is weakly ii-open.</p><p>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.</p><p>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.</p><p>As a summary the following <xref ref-type="fig" rid="fig1">Figure 1</xref> shows the relations among semi-continuous, ii-continuous, i-continuous, int-continuous, α-continuous and continuous.</p><p>Corollary 4.10. A function f : ( X , τ ) → ( Y , σ ) is α-continuous if and only if it is weakly ii-continuous.</p><p>Proof. Clear from Theorem 4.9.</p></sec><sec id="s5"><title>5. ii-Separating Axioms</title><p>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.</p><p>Definition 5.1. A topological space X is called</p><p>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 .</p><p>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 .</p><p>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.</p><p>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 .</p><p>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 .</p><p>Theorem 5.3.</p><p>1) Every T 0 -space is T 0 i i -space,</p><p>2) Every T 1 -space is T 0 i i -space,</p><p>3) Every T 1 -space is T 1 i i -space,</p><p>4) Every T 1 i i -space is T 0 i i -space.</p><p>Proof. (1), (2), (3) and (4) follow using the fact that every open set is ii-open [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] .</p><p>The converse needs not to be true by the following example.</p><p>Example 5.4. Let</p><p>X = { a , b , c } , τ = { ϕ , X , { a } } and τ i i = { ϕ , X , { a } , { a , b } , { a , c } } .</p><p>( X , τ ) and ( X , τ i i ) are topological spaces.</p><p>( 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 .</p><p>( 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 } .</p><p>a , c ∈ X ( a ≠ c ) there exists { a } ∈ τ i i such that a ∈ { a } , c ∉ { a } .</p><p>b , c ∈ X ( b ≠ c ) there exists { a , b } ∈ τ i i such that b ∈ { a , b } , c ∉ { a , b } .</p><p>( X , τ ) is not T 1 -space because, a , b ∈ X ( a ≠ b ) there exists X ∈ τ such that a ∈ X , b ∈ X .</p><p>( 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 } .</p><p>Theorem 5.5. Every T 1 α -space is T 1 i i -space.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] , U, V is an ii-open set in X. Hence X is T 1 i i -space.</p><p>Theorem 5.6. Every T 1 i i -space is T 1 i -space.</p><p>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 [<xref ref-type="bibr" rid="scirp.95072-ref4">4</xref>] , U, V is an i-open set in X. Hence X is T 1 i -space.</p><p>The converse needed not to be true by the following example.</p><p>Example 5.7. Let X = { a , b , c } , τ = { ϕ , X , { a } , { b , c } } and.</p><p>τ i = { ϕ , X , { a } , { b } , { c } , { b , c } } . τ i i = { ϕ , X , { a } , { b , c } } .</p><p>( X , τ ) and ( X , τ i i ) are topological spaces.</p><p>( 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 } .</p><p>a , c ∈ X ( a ≠ c ) there exists { a } , { c } ∈ τ i such that a ∈ { a } , c ∉ { a } and c ∈ { c } , a ∉ { c } .</p><p>b , c ∈ X ( b ≠ c ) there exists { c } , { b } ∈ τ i such that c ∈ { c } , b ∉ { c } and b ∈ { b } , c ∉ { b } .</p><p>( 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 } .</p><p>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.</p><p>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 } ) .</p><p>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.</p><p>Theorem 5.9. A space ( X , τ ) is T 1 i i -space if and only if the singletons are ii-closed sets.</p><p>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.</p><p>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.</p><p>As a consequence the following <xref ref-type="fig" rid="fig2">Figure 2</xref> shows the relations among T 0 , T 0 i i , T 1 , T 1 i i and T 1 α .</p></sec><sec id="s6"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s7"><title>Cite this paper</title><p>Abdullah, B.S. and Mohammed, A.A. 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