Irresolute Maps in Topological Spaces

International Journal of Applied Engineering Research, ISSN 0973-4562 Vol. 11 No.1 (2016) © Research India Publications; http/www.ripublication.com/ijaer.htm

∆*- Irresolute Maps in Topological Spaces K.Meena1

K.Sivakamasundari2

Department of Mathematics, Kumaraguru College of Technology, Coimbatore, TamilNadu, India1 e-mail : [email protected]

Department of Mathematics, Avinashilingam Institute for Home Science for Women, Coimbatore, TamilNadu, India2 e-mail : [email protected]

Abstract−In this paper a new class of irresolute map called

iii) δ-open set if it is the union of regular open sets. [13]

∆*-irresolute map via ∆*-open set is introduced and some of its

The complement of the above sets i), ii) and iii) are called Semi closed, -closed and δ-closed sets respectively.

characteristics are proved. Also the independent relationship of this map with various irresolute maps are investigated. Moreover the

Definition 2.2 A subset A of a topological space (X, ) is called δg*-closed set if δcl(A) ⊆ U whenever A ⊆ U where U is g-open in (X, ). [11]

association of ∆*-irresolute map with δ-open and δg*-open surjective maps under separation axioms are analysed. Furthermore some of its properties under composition mapping are derived in

Definition 2.3 A subset A of a topological space (X, ) is called ∆*-closed set if δcl(A) ⊆ U whenever A ⊆ U where U is δg-open in (X, ). [6] The class of all ∆*closed sets of (X, ) is denoted by * Δ C(X, ). The complement of ∆*-closed set is called ∆*open set.

this paper. Key words− ∆*-closed set, ∆*-continuous map, and

T

* g*

T -space * 

-space.

Mathematics Subject Classification − 54A05, 54C10

I.

Definition 2.4 The ∆*-closure operator of a topological space (X, ) is denoted by ∆*cl(A) and it is defined as follows. ∆*cl(A) = ∩{ F ⊆ X : A ⊆ F and F is ∆*-closed in (X, )}. [7]

INTRODUCTION

The concept of irresolute functions was introduced and investigated by S.G.Crossley and S.K.Hildebrand [1] in the year 1972. He proved that the irresolute functions are stronger than semi continuous but are independent of continuous functions. Many researchers have continued his work and introduced various strong and weak forms of irresolute functions. In 1980, S.N.Maheshwari et.al., [5] introduced α-irresolute maps. Further the idea of δgirresolute maps, αg-irresolute and δg*-irresolute maps were explained by J.Dontchev et.al.[3], R.Devi et.al.[2] and R.Sudha[12] respectively. Throughout this paper (X, ), (Y, σ) and (Z, η) represent non empty topological spaces on which no separation axioms are mentioned unless otherwise specified. Remark : In [6], the ∆*-closed sets were introduced and initially denoted by δ(δg)*-closed sets by the author. II.

Definition 2.5 A space (X, ) is said to be a i)

* *T -space if every ∆ -closed subset of (X, ) is δ-

closed in (X, ). [8] ii)

T

* g*

-space if every ∆*-closed subset

of (X, ) is δg*-closed in (X, ). [8] Definition 2.6 A mapping f : (X, )→(Y, σ) is said to be ∆*-continuous if the inverse image of every closed set in (Y, σ) is ∆*-closed in (X, τ). [9] Definition 2.7 A map f : (X, τ)→(Y, σ) is called i) irresolute if f−1(V) is a semi open set of (X, τ) for every semi open set V of (Y, σ). [1]

PRELIMINARIES

Definition 2.1 A subset A of (X, ) is called a

ii) α-irresolute if f−1(V) is a α-open set of (X, τ) for every α-open set V of (Y, σ). [5]

i) Semi open set if A  cl[int(A)]. [4] ii) -open set if A  int[cl{int(A)}]. [10]

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International Journal of Applied Engineering Research, ISSN 0973-4562 Vol. 11 No.1 (2016) © Research India Publications; http/www.ripublication.com/ijaer.htm

iii) αg-irresolute if f−1(V) is a αg-open set of (X, τ) for every αg-open set V of (Y, σ). [3]

Definition 3.5 A map f : (X, τ)→(Y, σ) is said to be ∆ *open map if the image of each open set in (X, τ) is ∆*-open in (Y, σ).

iv) δg-irresolute if f−1(V) is a δg-open set of (X, τ) for every δg-open set V of (Y, σ). [2]

Proposition 3.6 Let f : (X, τ)→(Y, σ) be any map and g : (Y, σ)→(Z, η) be injective and ∆*-irresolute map. If their composition (g ∘ f) : (X, τ)→(Z, η) is ∆*-open then f is ∆*open in (Y, σ). Proof: Let V be any open set in (X, τ). Since (g ∘ f) is ∆*-open, (g ∘ f)(V) is ∆*-open in (Z, η). Since g is ∆*irresolute and injective, f−1[(g ∘ f)(V)] = f(V) is ∆*-open in (Y, σ). Hence f is ∆*-open in (Y, σ).

v) δg*-irresolute if f−1(V) is a δg*-open set of (X, τ) for every δg*-open set V of (Y, σ). [12] Definition 2.8 Let f : (X, τ)→(Y, σ) be a map. Let A1 and A2 be subsets of (X, τ). Let B1 and B2 be subsets of (Y, σ). Then the following results are always true. i) If A1 ⊆ A2 then f(A1) ⊆ f(A2)

Proposition 3.7 Let f : (X, τ)→(Y, σ) be δ-open

ii) If B1 ⊆ B2 then f −1(B1) ⊆ f −1(B2). iii) A1 ⊆ f [f(A1)]. If f is injective then A1 = f [f(A1)].

surjective and ∆*-irresolute map. If (X, τ) is a

iv) f [f−1(B1)] ⊆ B1. If f is surjective then B1 = f [f−1(B1)].

then (Y, σ) is a

−1

−1

c

c

v) If f is surjective then [f(A)] ⊆ f( A ).

III.

T -space * 

T -space. *  *

Proof: Let V be any ∆ -open in (Y, σ). Since f is ∆*irresolute map, f−1(V) is ∆*-open in (X, τ). Since (X, τ) is a

T -space, f−1(V) is δ-open in (X, τ). Since f is δ-open *  −1

*

∆ -IRRESOLUTE MAPS

surjective map, f [f (V)] = V is δ-open in (X, τ). Hence (Y,

Definition 3.1 A map f : (X, τ)→(Y, σ) is called ∆*irresolute if f−1(V) is ∆*-open set in (X, τ) for every ∆*open set V in (Y, σ).

σ) is a

T -space. * 

Proposition 3.8 Let f : (X, τ)→(Y, σ) be δg*-open

*

Proposition 3.2 A map f : (X, τ)→(Y, σ) is ∆ -irresolute if and only if for every ∆*-closed set V in (Y, σ), f−1(V) is ∆*-closed set in (X, τ). Proof : (Necessary) : Let f : (X, τ)→(Y, σ) be ∆*irresolute map. Let V be any ∆*-closed set in (Y, σ). Then Y−V is ∆*-open in (Y, σ). Since f is ∆*-irresolute map, f −1(Y−V) = X− f−1(V) is ∆*-open in (X, τ) and hence f−1(V) is ∆*-closed in (X, τ). (Sufficiency) : Assume that f−1(V) is ∆*-closed in (X, τ) for each ∆*-closed set V in (Y, σ). Let U be any ∆*-open set in (Y, σ). Then Y−U is ∆*-closed set in (Y, σ). By the assumption f−1(Y−U) is ∆*-closed set in (X, τ). That is f−1(Y−U) = X−f−1(U) is ∆*-closed set in (X, τ) which implies that f−1(U) is ∆*-open set in (X, τ). Hence f is irresolute map.

surjective ∆*-irresolute map. If (X, τ) is a then (Y, σ) is a *T



g*

T

* g*

-space

-space.

Proof: Let V be any ∆*-open set in (Y, σ). Since f is ∆*irresolute map, f−1(V) is ∆*-open in (X, τ). Since (X, τ) is a

T

* g*

-space, f−1(V) is δg*-open in (X, τ). Since f is δg*-

open surjective f [f−1(V)] = V is δg* -open in (X, τ). Hence (Y, σ) is a

T

* g*

-space.

Proposition 3.9 Let f : (X, τ)→(Y, σ) be ∆*-irresolute map and g : (Y, σ)→(Z, η) be ∆*-continuous map. Then their composition mapping (g ∘ f) : (X, τ)→(Z, η) is ∆*continuous. Proof: Let V be any closed set in (Z, η). Since g is ∆*continuous map f−1(V) is ∆*-closed in (Y, σ). Since f is ∆*irresolute, f−1 [g−1(V)] = [g ∘ f]−1(V) is ∆*-closed in (X, τ). Hence (g ∘ f) is ∆*-continuous.

Proposition 3.3 If a map f : (X, τ)→(Y, σ) is ∆*irresolute then for every subset A of (X, τ), f [∆*cl(A)] ⊆ δcl[f(A)]. Proof : Let A ⊆ X. Then δcl[f(A)] is δ-closed in(Y, σ) and hence it is ∆*-closed in (Y, σ). Since f is ∆*-irresolute, f−1{δcl[f(A)]} is ∆*-closed in (X, τ). Since f(A) ⊆ δcl[f(A)], we have A ⊆ f−1{δcl[f(A)]}. By the definition of ∆*-closure, ∆*cl(A) ⊆ ∆*cl{f−1[δcl{f(A)}] = f−1[δcl{f(A)}. Hence f[∆*cl(A)] ⊆ δcl[f(A)].

Remark 3.10 The following examples show that ∆*irresolute map and irresolute map are independent. Counter Example 3.11 Let X = {a, b, c} = Y with τ = {, X, {a, b}} and σ = {, Y, {a}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = a, f(b) = c, f(c) = c. Then f is ∆*irresolute but not irresolute since for the semi open set {a} in (Y, σ), f−1{a} = {a} is not semi open set in (X, τ). Counter Example 3.12 Let X = {a, b, c} = Y with τ = {, X, {a}} and σ = {, Y, {a}, {a, b}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is irresolute but not ∆*-

Proposition 3.4 If a map f : (X, τ)→(Y, σ) is ∆*irresolute then for any subset B ⊆ Y, ∆*cl [(f−1(B)] ⊆ f−1 [δcl(B)]. Proof : Let B ⊆ Y. Then δcl(B) is δ-closed in (Y, σ). Since every δ-closed set is ∆*-closed set (Proposition 3.2 of [6]), δcl(B) is ∆*-closed in (Y, σ). By hypothesis, f−1 [δcl(B)] is ∆*-closed in (X, τ). Since B⊆δcl(B), f−1(B) ⊆f−1 [δcl(B)]. By the definition of ∆*-closure, ∆*cl [f−1(B)] ⊆ f−1 [δcl(B)].

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International Journal of Applied Engineering Research, ISSN 0973-4562 Vol. 11 No.1 (2016) © Research India Publications; http/www.ripublication.com/ijaer.htm

irresolute since for the ∆*-open set {b} in (Y, σ), f−1{b} = {b} is not ∆*-open set in (X, τ).

Counter Example 3.24 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}, {a, c}} and σ = {, Y, {a}, {b}, {a, b}, {a, c}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is ∆*- irresolute but not δg*-irresolute map and αg-irresolute since for the δg*-open set {a, c} in (Y, σ), f−1{a, c} = {a, c} is not δg*-open in (X, τ) and for the αg-open set {b} in (Y, σ), f−1{b} = {b} is not αg-open in (X, τ).

Remark 3.13 The following examples show that ∆*irresolute map and α-irresolute map are independent. Counter Example 3.14 Let X = {a, b, c} = Y with τ = {, X, {a, b}} and σ = {, Y, {a}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = a, f(b) = c, f(c) = c. Then f is ∆*irresolute but not α-irresolute since for the α-open set {a, b} in (Y, σ), f−1{a, b} = {a} is not α-open in (X, τ).

Definition 3.25 A map f : (X, τ)→(Y, σ) is said to be Contra ∆*-irresolute if f−1 (V) is ∆*-open in (X, τ) for every ∆*-closed set V in (Y, σ). Theorem 3.26 Let f : (X, τ)→(Y, σ) and g : (Y, σ)→(Z, η) be any two maps. Then the following two statements hold. a) If f is ∆*-irresolute and g is Contra ∆*-irresolute maps then the composition mapping (g ∘ f) : (X, τ)→(Z, η) is Contra ∆*-irresolute map. b) If f is Contra ∆*-irresolute and g is ∆*-irresolute then (g ∘ f) : (X, τ)→(Z, η) is Contra ∆*- irresolute map. Proof: a) Let f : (X, τ)→(Y, σ) be ∆*-irresolute and g : (Y, σ)→(Z, η) be Contra ∆*-irresolute map. Let U be any ∆*-closed set in (Z, η). Since g is Contra ∆*-irresolute, f−1(U) is ∆*-open in (Y, σ). Since f is ∆*-irresolute, f−1[g−1(U)] = (g ∘ f)−1(U) is ∆*-open in (X, τ). Hence the composition mapping (g ∘ f) is Contra ∆*-irresolute map. b) Let f be Contra ∆*-irresolute and g be ∆*-irresolute map. Let V be any ∆*-open set in (Z, η). Since g is ∆*irresolute, f−1(V) is ∆*-open in (Y, σ). Also f is Contra ∆*irresolute. Therefore f−1[g−1(V)] = (g ∘ f)−1(U) is ∆*-closed in (X, τ). Hence the composition mapping (g ∘ f) is Contra ∆*-irresolute map.

Counter Example 3.15 Let X = {a, b, c} = Y with τ = {, X, {a}} and σ = {, Y, {a}, {a, b}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is α-irresolute but not ∆*irresolute map since for the ∆*-open set {a, b} in (Y, σ), f−1{a, b} = {a, b} is not ∆*-open in (X, τ). Remark 3.16 The following examples show that ∆*irresolute map is independent with δg*-irresolute map. Counter Example 3.17 Let X = {a, b, c} = Y with τ = {, X, {a}} and σ = {, Y, {a}, {a, b},{a, c}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is δg*-irresolute but not ∆*-irresolute since for the ∆*-open set {c} in (Y, σ), f−1{c} = {c} is not ∆*-open in (X, τ). Counter Example 3.18 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}, {a, c}} and σ = {, Y, {a}, {b}, {a, b}, {a, c}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is ∆*-irresolute but not δg*-irresolute map since for the δg*open set {a, c} in (Y, σ), f−1{a, c} = {a, c} is not δg*-open in (X, τ).

Proposition 3.27 Let f : (X, τ)→(Y, σ) and g : (Y, σ)→(Z, η) be any two Contra ∆*-irresolute maps. Then their composition mapping (g ∘ f) : (X, τ)→(Z, η) is ∆*-irresolute map. Proof: Let U be any ∆*-closed set in (Z, η). Since g is Contra ∆*-irresolute, f−1(U) is ∆*-open in (Y, σ). Then by the assumption, f−1[g−1(U)] is ∆*-closed in (X, τ). Hence f−1[g−1(U)] = (g ∘ f)−1(U) is ∆*-closed set in (X, τ). Thus (g ∘ f) : (X, τ)→(Z, η) is a ∆*-irresolute map.

Remark 3.19 The following examples show that ∆*irresolute map is independent with ∆*-continuous map. Counter Example 3.20 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}, {a, c}} and σ ={, Y, {a, b}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is ∆*-irresolute but not ∆*-continuous since for the closed set {a} in (Y, σ), f−1{a} = {a} is not ∆*-closed in (X, τ).

Proposition 3.28 Let f : (X, τ)→(Y, σ) and g : (Y, σ)→(Z, η) be any two Contra ∆*-irresolute maps. Then their composition mapping (g ∘ f) : (X, τ)→(Z, η) need not be a Contra ∆*-irresolute map as seen from the following example.

Counter Example 3.21 Let X = {a, b, c} = Y with τ = {, X, {a}, {b}, {a, b}, {a, c}} and σ ={, Y, {a, b}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = a, f(b) = c, f(c) = b. Then f is ∆*-continuous but not ∆*-irresolute since for the ∆*-open set {b} in (Y, σ), f−1{b} = {c} is not ∆*-open in (X, τ).

Counter Example 3.29 Let X = {a, b, c} = Y = Z with τ = {, X, {a, b}}, σ = {, Y, {a}} and η = {, Z, {a}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = c, f(b) = b, f(c) = a. Then f is Contra ∆*-irresolute. Let g : (Y, σ)→(Z, η) be a map such that g(a) = b, g(b) = a, g(c) = a. Then g is also Contra ∆*-irresolute map. But the composition mapping (g ∘ f) : (X, τ)→(Z, η) is not Contra ∆*-irresolute map since for the ∆*-closed set {b, c} in (Z, η), (g ∘ f)−1[{b, c}] ={c} is not ∆*-open in (X, τ). Hence (g ∘ f) is not a Contra ∆*-irresolute map.

Remark 3.22 The following examples show that ∆*irresolute map is independent with δg-irresolute map and αg-irresolute map. Counter Example 3.23 Let X = {a, b, c} = Y with τ = {, X, {a}} and σ = {, Y, {a}, {a, b}, {a, c}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is δg-irresolute and αg-irresolute but not ∆*-irresolute since for the ∆*-open set {c} in (Y, σ), f−1{c} = {c} is not ∆*-open (X, τ).

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International Journal of Applied Engineering Research, ISSN 0973-4562 Vol. 11 No.1 (2016) © Research India Publications; http/www.ripublication.com/ijaer.htm

Remark 3.30 The following examples show that the notions of Contra ∆*-irresolute and ∆*-irresolute maps are independent.

References [1]

Counter Example 3.31 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}, {a, c}} and σ = {, Y, {a, b}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is ∆*-irresolute map but not Contra ∆*-irresolute map. Since for the ∆*-closed set {b, c} in (Y, σ), f−1[{b, c}] = {b, c} is not ∆*-open in (X, τ).

[2]

[3]

Counter Example 3.32 Let X = {a, b, c} = Y with τ = {, X, {a, b}} and σ = {, Y, {a}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = c, f(b) = b, f(c) = a. Then f is Contra ∆*-irresolute but not ∆*-irresolute since for the ∆*-open set {a} in(Y, σ), f−1[{a}] = {c} is not ∆*-open in (X, τ).

[4] [5] [6]

Remark 3.33 The Contra ∆*-irresolute and ∆*continuous maps are independent as seen from the following examples.

[7]

Counter Example 3.34 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}} and σ ={, Y, {a}}. Let f : (X, τ)→(Y, σ) be an identity map. Then f is ∆*-continuous but not Contra ∆*-irresolute since for the ∆*-closed set {b, c} in (Y, σ), f−1[{b, c}] = {b, c} is not ∆*-open in (X, τ).

[8]

[9]

Counter Example 3.35 Let X = {a, b, c} = Y with τ = {, X, {a}, {a, b}} and σ = {, Y, {a}}. Let f : (X, τ)→(Y, σ) be a map such that f(a) = c, f(b) = b, f(c) = a. Then f is Contra ∆*-irresolute but not ∆*-continuous map since for the closed set {b, c} in (Y, σ), f−1[{b, c}] = {a, b} is not ∆*closed in (X, τ).

[10] [11]

[12]

Remark 3.26 From the above results we have the following diagram.

[13]

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