On faintly πg-continuous functions

A new class of functions, called faintly πg-continuous functions, has been defined and studied. The relationships among faintly πg-continuous functions and πg-connected spaces, strongly πg-normal spaces and πg-compact spaces are investigated. Furthermore, the relationships between faintly πg-continuous functions and graphs are investigated.


Introduction
Generalized open sets play a very important role in General Topology and they are now the research topics of many topologists worldwide.Indeed a significant theme in General Topology and Real analysis concerns the variously modified forms of continuity, separation axioms etc. by utiliaing generalized closed sets.In 1970, Levine [9] initiated the study of so-called g-closed sets, that is, a subset A of a topological space (X, τ ) is g-closed if the closure of A included in every open superset of A and defined a T 1/2 space to be one in which the closed sets and the g-closed sets coincide.Zaitsev [17] defined the concept of π-closed sets and a class of topological spaces called quasi-normal spaces.Recently, Dontchev and Noiri [1] defined the notion of πg-closed sets and used this notion to obtain a characterization and some preservation theorems for quasi-normal spaces.In this paper, faintly πg-continuity is introduced and studied.Moreover, basic properties and preservation theorems of faintly πg-continuous functions are investigated and relationships between faintly πg-continuous functions and graphs are investigated.

Preliminaries
In the present paper, (X, τ ) and (Y, σ) represent topological spaces on which no separation axioms are assumed unless otherwise mentioned.For a subset A of a space (X, τ ), Cl(A), Int(A) and A c denote the closure of A, the interior of A and 2000 Mathematics Subject Classification: 54C10, 54D10 10 N. Rajesh the complement of A in X, respectively.A subset A of a space (X, τ ) is said to be regular open [15] (resp.semiopen [8], preopen [11]) if A = Int(Cl(A)) (resp.A ⊂ Cl(Int(A)), A ⊂ Int(Cl(A))).The complement of a regular open (resp.preopen) set is called a regular closed (resp.preclosed) set.The finite union of regular open sets is said to be π-open [1].The complement of π-open set is said to be π-closed.A subset A of a topological space (X, τ ) is said to be πg-closed [17] if The family of all πg-open (resp.πg-closed, πg-clopen) sets of (X, τ ) is denoted by πGO(X) (resp.πGC(X), πGCO(X)).Assume throughout this paper πGO(X) is closed under arbitrary unions.The intersection (resp.union) of all πgclosed (resp.πg-open) sets of X containing (resp.contained in) A ⊂ X is called the πg-closure [6] (resp.πg-interior [6]) of A and is denoted by πg-Cl(A) (resp. The union of all θ-open sets contained in a subset A is called the θ-interior of A and is denoted by Int θ (A).It follows from [16] that the collection of θ-open sets in a topological space (X, τ ) forms a topology τ θ on X.A subset A ⊂ X is said to be δ-open [16] if it is the union of regular open sets of X.The complement of a δ-open set is called a δ-closed set.The intersection of all δ-closed sets containing A is called the δ-closure [16] of A and is denoted by Cl δ (A).

Definition 2.1
The intersection of all preclosed sets containing the set A in a space X is called the preclosure of A and is denoted by p Cl(A)) [11].Definition 2.2 A subset A of a topological space (X, τ ) is said to be πgp-closed [12] ) is said to be slightly πg-continuous [13] if for each x ∈ X and each clopen set V of Y containing f (x), there exists U ∈ πGO(X, x) such that f (U ) ⊂ V .

Faintly πg-continuous functions
, there exists U ∈ πGO(X, x) such that f (U ) ⊂ V .If f has this property at each point of X, then it is said to be faintly πg-continuous.Theorem 3.2 For a function f : (X, τ ) → (Y, σ), the following statements are equivalent: (i) f is faintly πg-continuous; On faintly πg-continuous functions 11 The other equivalances are Obvious. 2 Then f is faintly πgp-continuous but not faintly πg-continuous.
Remark 3.7 Every almost contra-super-continuous function, every contra πg-continuous function and every contra-R-map is (πg, s)-continuous function.
Theorem 3.8 Suppose that the collection of πg-closed sets of X is closed under arbitrary intersections.Then if f : (X, τ ) → (Y, σ) is (πg, s)-continuous, then f is faintly πg-continuous.
Proof: Clear. 2 Remark 3.10 The converse of Theorem 3.9 is not true in general as can be seen from the following example.
Proof: Follows from the Definition 3.12. 2 Theorem 3.14 If a function f : (X, τ ) → (Y, σ) is faintly πg-continuous and (Y, σ) is a regular space, then f is πg-continuous.Proof: Let x ∈ X and V be any clopen subset of (Y, σ) Since f is faintly πg-continuous, there exists U ∈ πGO(X, x) containing x such that f (U ) ⊂ V .This shows that f is slightly πg-continuous. 2 Definition 3.19 Let (X, τ ) be a topological space.Since the intersection of two clopen sets of (X, τ ) is clopen, the clopen sets of (X, τ ) may be use as a base for a topology for X.This topology is called the ultra-regularization of τ [8] and is denoted by τ u .A topological space (X, τ ) is said to be ultra-regular if τ = τ u .

Theorem 3.22
The set of all points x ∈ X in which a function f : (X, τ ) → (Y, σ) is not faintly πg-continuous is the union of πg-frontier of the inverse images of θopen sets containing f (x).
Proof: Suppose that f is not faintly πg-continuous at x ∈ X.Then there exists Theorem 3.23 Let f : (X, τ ) → (Y, σ) be a function and g : (X, τ ) → (X × Y, τ × σ) the graph function of f , defined by g(x) = (x, f (x)) for every x ∈ X.If g is faintly πg-continuous, then f is faintly πg-continuous.
. This shows that f is faintly πg-continuous. 2

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N. Rajesh Definition 3.24 A space (X, τ ) is said to be πg-connected [ [7], [5]] if X cannot be written as a disjoint union of two nonempty πg-open sets.
Proof: Assume that (Y, σ) is not connected.Then there exist nonempty open sets Proof: The proof is similar to Theorem 3.27.

Separation Axioms
Definition 4.1 A topological space (X, τ ) is said to be: (i) πg-T 1 [7] (resp.θ-T 1 ) if for each pair of distinct points x and y of X, there exists πg-open (resp.θ-open) sets U and V containing x and y, respectively such that y / ∈ U and x / ∈ V .
(ii) πg-T 2 [7] (resp.θ-T 2 [14]) if for each pair of distinct points x and y in X, there exists disjoint πg-open (resp.θ-open) sets U and V in X such that x ∈ U and y ∈ V .
Proof: Suppose that Y is θ-T 1 .For any distinct points x and y in X, there exist Proof: Suppose that Y is θ-T 2 .For any pair of distinct points x and y in X, there exist disjoint θ-open sets U and V in Y such that f (x) ∈ U and f (y) ∈ V .Since f is faintly πg-continuous, f −1 (U ) and f −1 (V ) are πg-open in X containing x and y, respectively.Therefore, Definition 4.4 A space (X, τ ) is called strongly θ-regular (resp.strongly πgregular) if for each θ-closed (resp.πg-closed) set F and each point x / ∈ F , there exist disjoint θ-open (resp.πg-open) sets U and V such that F ⊂ U and x ∈ V .Definition 4.5 A space (X, τ ) is said to be strongly θ-normal (resp.strongly πgnormal) if for any pair of disjoint θ-closed (resp.πg-closed) subsets F 1 and Theorem 4.7 If f is faintly πg-continuous πg-θ-open injective function from a strongly πg-regular space (X, τ ) onto a space (Y, σ), then (Y, σ) is strongly θ-regular.

N. Rajesh
Proof: Let F be an θ-closed subset of Y and y / ∈ F .Take Proof: Let F 1 and F 2 be disjoint θ-closed subsets of Y .Since f is faintly πgcontinuous, f −1 (F 1 ) and f −1 (F 2 ) are πg-closed sets.Take U = f −1 (F 1 ) and V = f −1 (F 2 ).We have U ∩ V = ∅.Since X is strongly πg-normal, there exist disjoint πg-open sets A and B such that U ⊂ A and V ⊂ B. We obtain that Recall that for a function f : (X, τ ) → (Y, σ), the subset {(x, f (x)) : x ∈ X} ⊂ X × Y is called the graph of f and is denoted by G(f ).Definition 4.9 A graph G(f ) of a function f : (X, τ ) → (Y, σ) is said to be θ-πgclosed if for each (x, y) ∈ (X × Y )\G(f ), there exist U ∈ πGO(X, x) and V ∈ σ θ containing y such that Proof: It is an immediate consequence of Definition 4.9. 2 Proof: Let x and y be any two distinct points of X.Then since f is injective, we have f (x) = f (y).Then, we have (x, f (y)) ∈ (X × Y )\G(f ).By Lemma 4.10, U ∈ πGO(X) and

N. Rajesh
Proof: Suppose that y / ∈ f (K).Then (x, y) / ∈ G(f ) for each x ∈ K. Since G(f ) is θ-πg-closed, there exist U x ∈ πGO(X, x) and a θ-open set V x of Y containing y such that f (U x ) ∩ V x = ∅ by Lemma 4.10.The family {U x : x ∈ K} is a cover of K by πg-open sets.Since K is πg-compact relative to (X, τ ), there exists a finite subset K 0 of K such that K ⊂ ∪{U x : x ∈ K 0 }.Set V = ∩{V x : x ∈ K 0 }.Then V is a θ-open set in Y containing y.Therefore, we have