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International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 ON IR*-CONTINUITY IN IDEAL TOPOLOGICAL SPACES RENU THOMAS*, C. JANAKI** & SHEEBA K S*** * Sree Narayana Guru college, K.G. Chavadi, Coimbatore-105, Tamil Nadu, India [email protected] **L.R.G Govt. Arts college for women, Tirupur-4, Tamil Nadu, India janakicsekar@ yahoo.com ***Sree Narayana Guru college, K.G. Chavadi, Coimbatore-105, Tamil Nadu, India [email protected] Abstract: In this paper, we introduce IR*- continuity in ideal topological spaces. Its relationship with continuous functions, R*-continuous functions, *-continuous functions, contra R*-continuous functions are compared and its basic properties are studied. The concept of IR*-irresolute in ideal topological space is also defined. Some new continuous functions namely strongly IR*-continuous functions, maximal IR*-continuous functions, IR*-totally continuous functions are defined. Key words: IR*-continuous functions, IR*-irresolute functions, Strongly IR*-continuous functions, Maximal IR*-continuous functions and IR*-totally continuous functions. 1. Introduction In 1930 Kuratowski and Vaidyanathaswamy [15] introduce the concept of ideals in topological spaces. After several decades in 1990, Jancovic and Hamlett [14] investigated the topological ideals which is generalization of general topology. Very recently C. Janaki and Renu Thomas introduced IR*-closed sets [6] in Ideal topological spaces. The purpose of the present paper is to study the new classes of functions called IR*-continuous functions, IR*-irresolute, Strongly IR*-continuous functions, Maximal IR*-continuous functions and IR*-totally continuous functions along with some of properties, characterizations and relationships are obtained. 2. Preliminaries A subset A of a topological space (X, τ ) is called (i) a regular open [4] if A = int(cl (A)) and regular closed [4] if A = cl(int(A)) (ii) a regular semi-open set [4], if there is a regular open set U such that U ⊂ A ⊂ cl (U) (iii) a R*- closed set [4], if rcl (A) ⊂ U, whenever A ⊂ U and U is regular semi-open in (X, τ). The intersection of all regular closed subset of (X, τ) ©2015 RS Publication, [email protected] Page 99 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 containing A is called the regular closure of A and is denoted by rcl (A). Let 𝑋𝛼 𝛼∈𝐽 be an indexed family of topological spaces. Let us take as a basis for a topology on the product space 𝑋𝛼 𝛼∈𝐽 the collection of all sets of the form 𝑈𝛼 𝛼∈𝐽 Where 𝑈𝛼 is open in 𝑋𝛼 for each 𝛼 ∈ 𝐽. The topology generated by this basis is called the box topology [7]. An ideal [1] I on a topological space (X, τ) is a nonempty collection of subsets of X which satisfies (i) A ∈ I and B ⊂ A implies B ∈ I and (ii) A ∈ I and B ∈ I implies A ∪ B∈ I. Given a topological space (X, τ) with an ideal I on X and if 𝜌 (X) is the set of all subsets of X, a set operator (.)* : 𝜌(X) 𝜌(X), called a local function [1] of A with respect to τ and I is defined as follows: A ⊆ X, A* (I, τ) ={x ∈ X | U ∩ A ∉ I for every U τ (x)} where τ (x) ={U ∈ τ | x ∈ U} . A Kuratowski closure operator [1] cl * (.) for a topology τ* (X, τ), called the *- topology, finer than τ is defined by cl *(A) = A ∪ A* (I, τ ). Given a topological space(X,τ) with an ideal I on X, cl*(A) and int*(A) will denote the closure and interior of A in (X, τ*).When there is no chance for confusion, A* is substituted for A*(I, τ). A subset A of an ideal topological space (X, τ, I) is said to be (.)* - dense in itself (resp. τ* - closed, (.)* - perfect) if A ⊂ A* (resp. A* ⊂ A, A=A*). If (X, τ, I) is an ideal topological space and A is a subset of X, then (A, τA, IA) is an ideal topological space , were τA is the related topology on A and IA = {A ∩ J : J∈ I }. A subset A of an ideal topological space (X, τ, I) is (i) *- closed [4] if A* ⊆ A (ii) IR-closed [6] if A = cl*(int (A)) and is denoted by ri∗∗ C(X) (iii) IR*-closed [6] if ri∗∗ cl(A) ⊂ U, whenever A ⊂ U and U is regular semi-open (iv) IR*-open [4] if Ac is IR*-closed in X. (v) is T-dense[8] if every subset of X is *-dense in itself. Definition 2.1: A function f : (X, τ) → (Y, σ) is called 1. R*-continuous [4] if every f-1(V) is R* closed in (X, τ) for every closed set V in (Y, σ) . 2. Contra R*-continuous [5] if every f-1(V) is R*- closed in (X, τ) for every open set V in (Y,σ). 3. Strongly continuous [10] if f-1(V) is clopen in X for each subset V in Y. 4. Totally continuous [3] if f-1(V) is clopen set in X for each open set V of Y. ©2015 RS Publication, [email protected] Page 100 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Definition 2.2: A function f: (X, τ, I) → (Y, σ) is said to be * - continuous [10] if f-1(A) is *- closed set in Y for every closed A in Y. Definition 2.3: For a function f: (X, τ, I) → (Y, σ) , the subset {(X, f(x)): x ε X} ⊂ X × Y is called the graph of f [11]. Theorem 2.4: [6] Let (X, τ, I) be an Ideal topological space and U ⊂ A ⊂ X. If A is R.S.O set in X and U is IR*-closed set in X, then U is IR*-closed set in A. 3. IR*-Continuous in Ideal Topological space Definition 3.1: A function f: (X, τ, I) → (Y, σ) is called IR*-continuous if f-1(V) is IR*-closed in X for every closed V of Y. Example 3.2: Let X=Y={a,b,c,d}, τ = {X,,{a},{b},{a,b},{a,b,c}}, σ = {Y, ,{b}, {d}, {b,d}}, I={, {a}}, f : (X, τ, I) → (Y, σ) be the function defined by f(a) = c, f(b) = a, f(c) = d, f(d) = b, then f is IR*- continuous . Remark 3.3: IR*- continuity and continuity are independent concepts. Example 3.4: Let X = {a,b,c,d},τ = {X,,{a},{b},{a,b},{a,b,c}},Y = X, σ = {Y,,{a,b,c}}, I={, {a}}, f : (X, τ, I) → (Y, σ) be the identity mapping. Here f is continuous function but not IR*-continuous . Example 3.5: Let X = {a,b,c,d},τ = {X,,{a},{b},{a,b},{a,b,c}},Y = X, σ = {Y, ,{b},{d},{b,d}}, I={, {a}}, f : (X, τ, I) → (Y, σ) be the function defined by f(a) = c, f(b) = a, f(c) = d, f(d) = b. Here f is IR*- continuous function but not continuous . Remark 3.6: R*-continuous and IR*- continuous are independent concepts. Example 3.7 : Let X = {a,b,c,d}, τ = {X, ,{a},{c,d},{a,c,d}},Y = X, σ = {Y, , {a}, {d}, {a,d},{a,b}, {a,b,d}}, I={, {a}}, f: (X, τ, I ) → (Y, σ ) be the function defined by f(a) = c , f(b)=a, f(c) = d, f(d) = b. Then f is IR*- continuous function but not R*- continuous. Example 3.8 : Let X = {a,b,c,d},τ = {X, ,{a},{c,d},{a,c,d}},Y = X, σ = {Y, ,{a},{b}, {a,b,c}}, I={, {a}}, f : (X, τ, I) → (Y, σ ) be the function defined by f(a) = a, f(b) = d, f(c) = b, f(d) = c. Then f is R*-continuous function but not IR*- continuous . Remark 3.9: Contra R*- continuity and IR*-continuity are independent concepts. ©2015 RS Publication, [email protected] Page 101 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Example 3.10: Let X = {a,b,c,d}, τ = {X, ,{a},{b},{a,b},{a,b,c}},Y = X, σ = {Y,, {b}, {d}, {b,d}}, I={, {a}}, f: (X, τ, I) → (Y, σ) be the function defined by f(a) = c, f(b) = a, f(c) = d, f(d) = b. Here f is IR*- continuous function but not contra R*- continuous. Example 3.11 : Let X = {a,b,c}, τ = {X,,{a},{b},{a,b}}, Y = X, σ = {Y,,{a},{a,b},{a,c}}, I={, {a}}, f: (X, τ, I) → (Y, σ) be the function defined by f(a) = b, f(b) = c, f(c) = a. Here f is contra IR*- continuous function but not IR*- continuous. Theorem 3.12: * - continuous and IR*- continuous are independent. Example 3.13: Let X = {a,b,c,d} = Y, τ = {X,,{a},{b},{a,b},{a,b,c}} and σ = {Y,,{b},{d}, {b,d}}, I={, {a}}, f : (X, τ, I) → (Y, σ) be the function defined by f(a) = c , f(b) = a , f(c) = d , f(d) = b. Here f is IR*- continuous but not *-continuous. Example 3.14: Let X = {a,b,c,d} = Y, τ = {X,,{a},{b},{a,b},{a,b,c}}, σ = {Y, ,{a},{c},{a,c}, {b,c}, {a,b,c},{b,c,d}}, I={, {a}}, f : (X, τ, I) → (Y, σ) be the function defined by f(a) = a, f(b) = c, f(c) = b, f(d) = d. Here f is *- continuous but not IR*- continuous. Remark 3.15: The above results are diagrammatically represented below: Remark 3.16: The composition of two IR*- continuous function need not be IR*- continuous. Example 3.17 : Let X = {a,b,c,d} = Y = Z, τ = {X,,{a},{c,d},{a,c,d}}, σ = {Y,,{b}, {d},{b,d}}, η = {Z,,{a,b}} I=J={, {a}}, f : (X, τ, I) → (Y, σ) be the function defined by f(a) = c, f(b) = a, f(c) = d, f(d) = b and g : (Y, σ, J) → (Z, η) defined by g(a) = a, g(b) = d, g(c) = b, g(d) = c. Here both f and g are IR*- continuous but g о f is not IR*- continuous. ©2015 RS Publication, [email protected] Page 102 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Theorem 3.18: A map f: (X, τ, I) → (Y, σ) is IR*-continuous if the inverse image of every closed set in (Y, σ) is IR*-closed in (X, τ, I). Proof: Necessity: Let V be open set in (Y, σ). Since f is IR*-continuous, f-1(VC) is IR*-closed in (X, τ, I). But f-1(VC) = X - f-1(V). Hence f-1(V) is IR*- closed in (X, τ, I) Sufficiency: Assume that inverse image of every closed set in (Y, σ) is IR*-closed in (X, τ, I). Let V be the closed set in (Y, σ). By our assumption f-1(VC) = X-f-1(V) is IR*-closed in (X, τ, I) which implies that f-1(V) is IR* closed in (X, τ, I). Hence f is IR*- continuous. Theorem 3.19: Let f: (X, τ, I) → (Y, σ) be a map. Then the following statements are equivalent 1) f is IR*-continuous 2) the inverse image of each open set in Y is IR*- open in X Proof: Assume that f: (X, τ, I) → (Y, σ) is IR*-continuous. Let G be open set in Y .Then GC is closed in Y. Since f is IR*-continuous, f-1(GC) is IR*-closed in X. But f-1(GC) = X-f-1(G). Thus f-1(G) is IR* -open in Y. Conversely assume that the inverse image of each open in Y is IR*- open in X. Let F be any closed in Y. By assumption f-1(FC) is IR*-open in X. But f-1(FC) = X – f-1(F). Thus X- f-1(F) is IR* open in X and so f-1(F) is IR*- closed in X. Therefore f is IR*- continuous. Hence they are equivalent. Theorem 3.20: Let f: (X, τ, I) → (Y, σ) be a function and g: X → X × Y is the graph of f. If g is IR* continuous then f is IR*- continuous. Proof: Let g : X → X×Y be the graph function of f. Suppose g is IR*-continuous and V be closed set in Y, then X × V is closed set in X × Y. Since g is IR*-continuous, g-1(X× 𝑉) = f-1(V) is IR*- closed set in X. Thus the inverse image of every closed set in Y is IR*- closed set in X. Therefore f is IR*- continuous. Theorem 3.21: Let {Xα } αJ be any family of topological space . If f : (X, τ, I) →П αJ Xα is an IR*- contiuous function , then Pα ∘ f : X → Xα is IR*- continuous for each α J where Pα is the projection of П α J Xα onto Xα. Proof: We will consider a fixed α0 J . Let Gα0 be a closed set in Xα0. Then (Pα0 )-1(Gα0 ) is closed in П Xα, since f is IR*- continuous , f-1((Pα0 )-1(Gα0 )) = ((Pα0 ∘ f )-1(Gα0 ) is IR*- closed in X. Thus Pα ∘ f is IR*- continuous. ©2015 RS Publication, [email protected] Page 103 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Corollary 3.22: For any bijective function f: (X, τ) → (Y, σ, J) , the following are equivalent 1) f-1 : (Y, σ, J)→(X, τ) is IR*- continuous 2) f(U) is IR*- open in Y for every open set U in X 3) f(U) is IR*- closed in Y for every closed set in X Proof: It is trivial. Theorem 3.23: Let f: (X, τ, I) → (Y, σ, J) be IR*-continuous, and U ∈ RSO(X). Then the restriction f/U: (U, τU, IU) → (Y, σ, J) is IR*- continuous. Proof: Let V be any closed set of (Y, σ), since f is IR*- continuous, f-1(V) is IR*- closed in X. By theorem 2.4 f-1(V) ∩ U is IR* closed in X. Thus (f/U)-1(V) = f-1(V) ∩ U is IR* -closed in U because U is regular semi open in X. This proves that f/U: (U, τU, IU) → (Y, σ, J) is IR*- continuous. Theorem 3.24: If X = A ∪ B and A⊂B⊂X, A is IR*-closed in B and B is IR*-closed in X. Then A is IR*-closed in X. Proof: Suppose A is IR*-closed in B. Then ri∗∗ cl(A)⊂U, whenever U is regular semi open in B. Also U is regular semi open in X since X = A ∪ B. Therefore A is IR*-closed in X. Theorem 3.25: Let X = A ∪ B be a topology space with topology τ and Y be a topological space with topology σ .Let f: (A, τA, IA) → (Y, σ) and g: (B, τB, JB) → (Y, σ) be IR*-continuous map such that f(x) = g(x) for every x ∈ A ∩ B. Suppose that A and B are IR*- closed sets in X. Then the combination α: (X, τ, I) → (Y, σ) is IR*-continuous. Proof: Let F be any closed set in Y. Clearly α-1(F) = f-1(F) ∪ g-1(F) = C ∪ D were C = f-1(F) and D = g-1(F). But C is IR*-closed in A and A is IR*-closed in X and so C is IR*-closed in X. Since we have proved if D ⊆ B ⊆ X, D is IR*-closed in B and B is IR*- closed in X then D is IR*-closed in X. Also C ∪ D is IR*-closed in X. Therefore α-1(F) is IR*-closed in X. Hence α is IR*-continuous. Definition 3.26: Let N be a subset of (X, τ, I) and x ∈ X.Then N is called an IR*-open neighbourhood of x[9] if there exists an IR*-open set U containing x such that U⊂ N. Theorem 3.27: Let (X, τ, I) be T-dense. Then, for a function f: (X, τ, I) → (Y, σ) the following statements are equivalent: (1) f is IR*-continuous. (2) For each x ∈ X and each open set V in Y with f (x) ∈ V, there exists an IR*-open set U containing x such that f (U) ⊂ V. ©2015 RS Publication, [email protected] Page 104 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 (3) For each x ∈ X and each open set V in Y with f (x) ∈ V, f -1(V) is an IR*-open neighborhood of x. Proof: (1) ⇒ (2) Let x ∈ X and let V be an open set in Y such that f (x) ∈ V. Since f is IR*-continuous, f -1 (V) is IR*-open in X. By putting U = f -1 (V), we have x ∈ U and f (U) ⊂ V. (2) ⇒ (3) Let V be an open set in Y and let f (x) ∈ V. Then by (2), there exists an IR*-open set U containing x such that f (U) ⊂ V. So x ∈ U ⊂ f -1(V). Hence f -1(V) is an IR*-open neighborhood of x. (3) ⇒ (1) Let V be an open set in Y and let f (x) ∈ V. Then by (3), f -1(V) is an IR*-open neighborhood of x. Thus for each x ∈ f -1(V), there exists an IR*-open set UX containing x such that x ∈ UX ⊂ f -1(V). Hence f -1(V) = 𝑥∈f −1 (v) 𝑈𝑥 , f -1(V) is IR*-open in X. 4. IR*- irresolute in ideal topological space Definition 4.1: A function f: (X, τ, I) → (Y, σ, J) is called IR*- irresolute if f-1(V) is IR*- closed in (X, τ) for every IR*- closed set V in (Y, σ). Example 4.2 : Let X={a,b,c,d}, τ = {X,,{b},{c},{d},{b,c},{b,d},{c,d},{b,c,d}},Y=X, σ = {Y,,{a},{b},{a,b},{a,b,c}}, I={, {a}}, f:(X, τ, I) → (Y, σ, J) be the identity map, then f is IR*- irresolute. Remark 4.3: Every IR*- irresolute is IR*- continuous but not conversely. Example 4.4 : Let X = {a,b,c,d} = Y, τ = {X,,{a},{b},{a,b},{a,b,c}}, σ = {Y, , {b},{d},{b,d}}, I={, {a}}, f:(X, τ, I) → (Y, σ, J) be the function defined by f(a) = c, f(b) = a, f(c) = d, f(d) = b. Here f is IR*-continuous but not IR*- irresolute. Theorem 4.5: A function f: (X, τ, I) → (Y, σ, J) is IR*- irresolute iff the inverse image of every IR*-open in (Y, σ, J) is IR*- open in (X, τ, I). Proof: Obvious. Theorem 4.6: Let f: (X, τ, I) → (Y, σ, J) and g:(Y, σ, J) → (Z, η, K) be any two function then, 1) g о f is IR*- continuous if g is continuous and f is IR*- continuous. 2) g о f is IR* -irresolute if g is IR*- irresolute and f is IR*- irresolute. 3) g о f is IR*- continuous if g is IR*- continuous and f is IR*- irresolute. ©2015 RS Publication, [email protected] Page 105 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Proof : 1) Let V be closed set in (Z, η).Then g-1(V) is closed in (Y, σ), since g is continuous and IR*- continuity of f implies f-1(g-1(V)) = (g о f)-1(V) is IR*- closed in (X, τ, I).That is (g о f)-1(V) is IR*- closed in (X, τ, I).Hence g о f is IR*- continuous. 2) Let V be IR*- closed in closed set in (Z, η, K). Since g is IR*- irresolute, g-1(V) is IR*-closed set in (Y, σ, J). As f is IR*- irresolute f-1(g-1(V)) = (g о f)-1(V) is IR*- closed in (X, τ, I). Hence g о f is IR*- irresolute. 3) Let V be closed in (Z, η). Since g is IR*- continuous g-1(V) is IR*- closed in (Y, σ, J). As f is IR*- irresolute f-1(g-1(V)) = (g о f)-1(V) is IR*- closed in (X, τ, I). Therefore g о f is IR*- continuous. 5. Strongly IR*-continuous in ideal topological space Definition 5.1: A function f : (X, τ) → (Y, σ, J) is said to strongly IR*- continuous if for every IR*- closed set U in Y , f-1(U) is closed in X. Example 5.2: Let X = {a,b,c,d},τ = {X,,{a},{d},{c},{a,d},{a,c},{b,c},{c,d},{a,b,c},{a,c,d}, {b,c,d}} Y = X , σ = {Y, ,{a},{c},{a,c},{a,b,c},{a,c,d}}, I={, {a}}, f:(X, τ) → (Y, σ, J) be defined as f(a) = b, f(b) = c, f(c) = d, f(d) = a, then f is strongly IR*- continuous. Remark 5.3: Every strongly IR*-continuous is continuous but not converse. Example 5.4 : Let X = {a,b,c,d},τ = {X,,{a},{b},{a,b},{a,b,c},},Y = X, σ = {Y, ,{b},{d},{b,d}}, I={, {a}}, f:(X, τ, I) → (Y, σ, J) be defined as f(a) = c, f(b) = a, f(c) = d, f(d) = b, then f is IR*-continuous but not strongly IR*-continuous. Remark 5.5: Strongly IR*-continuous and IR*-continuous are independent. Example 5.6: Let X = {a,b,c,d},τ = {X,,{a},{b},{a,b},{a,b,c}}, Y = X, σ = {Y,,{a,b,c}}, I={, {a}}, f:(X, τ, I) → (Y, σ, J) be the identity map, then f is continuous but not strongly IR*-continuous. Example 5.7 : Let X = {a,b,c,d},τ ={X,,{a},{c},{d},{a,c},{a,d},{b,c},{c,d},{a,b,c},{a,c,d}, {b,c,d}} and Y = X , σ = {Y,,{a},{b},{a,b},{a,b,c}}, I={, {a}}, f : (X, τ) → (Y, σ, J) be the identity map, then f is strongly IR*-continuous but not continuous. Remark 5.8: The above results are diagrammatically represented below: ©2015 RS Publication, [email protected] Page 106 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Theorem 5.9: A function f : (X, τ) → (Y, σ, J) is strongly IR*-continuous iff the inverse image of every IR*-closed set in Y is closed in X. Proof: Obvious. Theorem 5.10: Let f: (X, τ) → (Y, σ, J) is strongly IR*-continuous and g: (Y, σ, J)→(Z,η,K) be IR*-continuous , then g о f is continuous. Proof: obvious. 6. Maximal IR*-continuous in ideal topological spaces Definition 6.1: A proper non empty IR*-closed subset U of an ideal space (X, τ, I) is said to be maximal IR*-closed if any IR*-closed set containing U is either X or U. Example 6.2 : Let X = {a,b,c,d} , τ = {X,,{b},{d},{b,d}}, I = {,{a}}, the IR*-closed sets are {a,c}, {b,d}, {a,b,c},{a,b,d},{a,c,d},{b,c,d} , then the sets {a, b, c},{a, c, d},{a, b, d} and {b,c,d} are maximal IR*-closed sets. Remark 6.3: Every maximal IR*-closed sets is IR*-closed set. But IR*-closed set need not be a maximal IR*-closed set. In Example 6.2 {a,c} is IR*-closed but not maximal IR*-closed. Theorem 6.4: The following statements hold true for any ideal space (X, τ, I). (i) Let F be a maximal IR*-closed set and G be a IR*-closed set. Then F∪G = X or G⊂ F. (ii) If F and G are maximal IR*-closed sets then F ∪ G = X or F = G. Proof: (i) Let F be a maximal IR*-closed set and G be IR*- closed set. If F ∪ G = X then there is nothing to prove. Assume that F∪ G ≠ X. F ⊆ F ∪ G. F∪ G is IR*-closed. Since F is a maximal IR*-closed set F ∪ G = X or F ∪ G = F. Hence F ∪ G = X or G ⊂ F. (ii) Let F and G are maximal IR*-closed sets. If F ∪ G = X, then there is nothing to prove. Assume that F ∪ G ≠ X. Then by (i) F ⊂ G, G ⊂ F which implies that F = G. ©2015 RS Publication, [email protected] Page 107 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Definition 6.5: A function f: (X, τ, I) → (Y, σ) is said to be maximal IR*-continuous if f-1 (V) is maximal IR*-closed in X for every closed set V in Y. Theorem 6.6: Every surjective maximal IR*-continuous function is IR*-continuous. Proof: Let f: (X, τ, I) → (Y, σ) be a surjective maximal IR*- continuous map. Let V be a closed set in Y then f-1 (V) is a maximal IR*-closed set in X which is a IR*-closed set in X. Hence f is IR*-continuous. Remark 6.7: The converse of the theorem 6.6 is not true. Example 6.8 : Let X = {a, b, c} = Y, τ = {X, , {a},{b},{a, b}}, I = { ,{c}}, σ = {Y, , {a, b}}. The identity function from X to Y is IR*-continuous but not maximal IR*-continuous. Remark 6.9: Composition of two maximal IR*-continuous functions need not be maximal IR*-continuous. Example 6.10 : Let X = {a, b, c, d}, Y = Z = {a, b, c}, τ = {X,,{b},{d},{b, d}}, I = {,{b}}, σ = {Y,,{a}}, J = {,{c}}. The function f is defined as f(a) = b, f(b) = a, f(c) = d, f(d) = c. Here f is maximal IR*-continuous where f: (X, τ, I) → (Y, σ). The identity function g: (Y, σ, J) → (Z, η) with η = {Z,,{b}}. g is also maximal IR*-continuous. But their composition g о f is not IR*-continuous. Since for the closed set {a, c} in Z (g о f)-1 ({a, c}) = f -1({a, c}) = {b, d} is not maximal IR*-closed in X. Hence g о f is not IR*-continuous. Theorem 6.11: Let f: (X, τ, I) → (Y, σ) be a maximal IR*- continuous function and g: (Y, σ) → (Z, η) be surjective continuous function then (g о f): (X, τ, I) → (Z, η) is a maximal IR*-continuous function. Proof: Let V be a nonempty proper closed set in Z. Since g is continuous g-1(V) is a proper nonempty closed set in Y. Since f is maximal IR*-continuous f -1 (g-1 (V)) is a maximal IR*- closed set in X. 7. IR*-totally continuous functions in ideal topological spaces Definition 7.1: A function f: (X, τ) → (Y, σ, J) is said to be IR*-totally continuous if the inverse image of every IR*-open set of Y is clopen in X. Example7.2: Let X = {a,b,c} = Y, τ = {X, , {a}, {b}, {c}, {a,b}, {a,c}, {b,c} }, σ = {Y, , {a}, {b}, {a,b} }, I={, {a}}. Let f: (X, τ) → (Y, σ, J) be an identity map. Then f is IR*-totally continuous. ©2015 RS Publication, [email protected] Page 108 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 Theorem 7.3: A bijective function f: (X, τ) → (Y, σ, J) is a IR*-totally continuous function if and only if the inverse image of every IR*-closed subset of Y is clopen in X. Proof: Let F be any IR*-closed set in Y. Then Y\F is a IR*-open set in Y. By definition f -1(Y\ F) is clopen in X. That is, X\f -1(F) is clopen in X. This implies f -1(F) is clopen in X. Conversely if V is IR*-open in Y, then Y\V is IR*-closed in Y. By assumption, f -1 (Y\V) = X\f -1 (V) is clopen in X, which implies f-1(V) is clopen in X. Therefore f is IR*-totally continuous function. Theorem 7.4: (i) Every IR*-totally continuous function is IR*-continuous. (ii) Every totally continuous function is IR*-continuous. Proof: Obvious. Remark 7.5: The converse of Theorem 7.4 need not be true, which can be verified from the following examples. Example 7.6: (i) Let X = Y = {a,b,c}, τ = {X,,{a,b}} and σ = {Y,,{c},{a,b}}, I={, {a}}. Let f: (X, τ, I) → (Y, σ, J) be defined as f (a) = c, f (b) = b, f(c) = a. Then f is IR*-continuous, but f is not IR*-totally continuous. (ii) Let X = Y = {a,b,c,d}, τ = {X,,{a},{b},{a,b},{a,b,c}} and σ = {Y, ,{b},{d},{b,d}}, I={, {a}}. Let f: (X, τ, I) → (Y, σ, J) be defined as f (a) = c, f (b) = a, f(c) = d, f (d) = b. Then f is IR*-continuous but not totally continuous. Theorem 7.7: Let f: (X, τ) → (Y, σ, J) be a function, where X and Y are topological spaces. Then the following are equivalent: 1. f is IR*-totally continuous. 2. For each x ∈ X and each IR*-open set V in Y with f(x) ∈ V, there is a clopen set U in X such that x ∈ U and f(U) ⊂ V. Proof: (1) ⇒ (2): Suppose f is IR*-totally continuous and V be any IR*-open set in Y containing f(x) such that x ∈ f -1(V). Since f is IR*-totally continuous, f -1(V) is clopen in X. Let U = f-1(V) then U is a clopen set in X and x ∈ U. Also f(U) = f(f -1(V)) ⊂ V. This implies f(U) ⊂ V. (2) ⇒ (1): Let V be IR*-open set in Y. Let x∈ f -1(V) be any arbitrary point. This implies f(x) ∈ V. Therefore by (2) there is a clopen set GX containing x such that f(GX) ⊂ V , which implies GX ⊂ f-1(V) is a clopen neighbourhood of x. Since x is arbitrary, it implies f -1(V) is a ©2015 RS Publication, [email protected] Page 109 International Journal of Advanced Scientific and Technical Research Available online on http://www.rspublication.com/ijst/index.html Issue 6 volume 1, Jan. –Feb. 2016 ISSN 2249-9954 clopen neighbourhood of each of its points. Hence it is a clopen set in X. Therefore f is IR*-totally continuous. Reference: 1. Erdal Ekici, Sena Ozen, A generalized class of 𝜏* in ideal spaces, Filomat 27:4(2013), 529-535. 2. K.Indirani, V.Rajendran, P.Sathishmohan and L.Chinnapparaj, On g*α-I-continuous in topological space Journal of global research in Mathematical Archives, Vol.1,No.12,Dec 2013. 3. R.C.Jain, The role of regularly open sets in general topology, Ph.D, Thesis, Meeraut University, Institute of Advanced studies, Meerut,India,1980. 4. C.Janaki and Renu Thomas, R*-closed sets in topological spaces, International Journal of Mathematical Archive – 3[8], 2012, 3067-3074. 5. 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Nirmala Rebecca Paul, RgI-closed sets in ideal topological spaces, International Journal of computer applications (0975-8887), Vol.69, No.4, May 2013. 13. G.Sindu and K.Indirani,New class of rg*b-continuous functions in topological spaces, Intern.J.Fuzzy Mathematical Archive, Vol.5, No.2, 103-111. 14. R.Santi and M.Rameshkumar, A decomposition of continuity in ideal by using semi local functions, Asian Journal of Mathematics and applications, vol.2014. 15. A. Vadivel, Mohanarao Navuluri, Regular weakly closed sets in ideal topological spaces, International journal of pure and applied mathematics, Volume 86 No. 4 2013, 607-619. ©2015 RS Publication, [email protected] Page 110