# gsΛ Continuous Function in Topological Space

Authors: S. Pious Missier; Vijilius Helena Raj
Journal: ISRN Geometry
 (2013)

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## Abstract

We introduce the different notions of a new class of continuous functions called generalized semi Lambda (gsΛ) continuous function in topological spaces. Its properties and characterization are also discussed.


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## Body

## 1. Introduction

In 1986, Maki [1] continued the work of Levine and Dunham on generalized closed sets and closure operators by introducing the notion of Λ-sets in topological spaces. A Λ-set is a set A which is equal to its kernel (= saturated set), that is, to the intersection of all open supersets of A. Arenas et al. [2] introduced and investigated the notion of λ-closed sets and λ-open sets by involving Λ-sets.In 2008 Caldas et al. [3] introduced Λ generalized closed sets (Λg, Λ-g, gΛ) and their properties. They also studied the concept of λ closed maps. In 2007, Caldas et al. [4] introduced the concept of λ irresolute maps.In this paper, we establish a new class of maps called gsΛ continuous function and study its properties and characteristics.Throughout this paper, (X,τ), (Y,σ), and (Z,η) (or simply X,Y, and Z will always denote topological spaces on which no separation axioms are assumed unless explicitly stated. Int(A), Cl(A), IntλA, ClλA, gsΛCl(A), and gsΛInt(A) denote the interior of A, closure of A, lambda interior of A, lambda closure of A, gs Lambda closure of A and gs Lambda interior of A, respectively.


## 2. Preliminary Definitions

Definition 1.
A subsetA of a space (X,τ) is called (1)
a semiopen set [5] if A⊂Cl(Int(A)),(2)
a preopen set [6] if A⊂Int(Cl(A)),(3)
a regular open set [6] if A=Int(Cl(A)).
The complement sets of semi open (resp., preopen and regular open) are called semi closed sets (resp., preclosed and regular closed). The semiclosure (resp., preclosure) of a subsetA of X denoted by sCl(A), (pCl(A)) is the intersection of all semi closed sets (pre closed sets) containing A.A topological space (X,τ) is said to be (1)
a generalized closed [7] if Cl(A) ⊂U,  whenever A⊂U and U is open in (X,τ),(2)
ag* closed [8] if Cl(A) ⊂U, whenever A⊂U and U is g-open in (X,τ),(3)
semigeneralized closed (denoted by sg-closed) [9] if sCl(A)⊂U, whenever A⊂U and U is semi open in (X,τ),(4)
generalized semiclosed (denoted by gs-closed) [9] if sCl(A)⊂U, whenever A⊂U and U is open in (X,τ),(5)
a subsetA of a space (X,τ) is called λ-closed [2] if A=B∩C, where B is a Λ-set and C is a closed set,(6)
a subsetA of (X,τ) is said to be a Λg closed set [3] if Cl(A)⊂U whenever A⊂U, where U is λ open in (X,τ),(7)
a subsetA of (X,τ) is said to be a Λ-g closed set [3] if Clλ(A)⊂U  whenever A⊂U, where U is λ open in (X,τ),(8)
a subsetA of (X,τ) is said to be a gΛ closed set [3] if Clλ(A)⊂U whenever A⊂U, where U is open in (X,τ),(9)
a subsetA of (X,τ) is said to be a gsΛ closed set [10] if Clλ(A)⊂U  whenever A⊂U, where U is semi open in X.The complement of the above closed sets are called its respective open sets.Definition 2.
A functionf:(X,τ)→(Y,σ) is called (1)
semicontinuous [5] if f-1(V) is semi open in (X,τ) for every open set V in (Y,σ),(2)
semi open (semiclosed) [9] if f(F) is semi open (semi closed) in (Y,σ) for any open (closed) set F in (X,τ),(3)
pre-semiopen (pre-semiclosed) [9] if f(F) is semi open (semi closed) in (Y, σ) for every semi open (semi closed) set in (X,τ),(4)
irresolute [9] if for any semi open set S of (Y,σ), f-1(S) is semi open in (X,τ),(5)
g continuous [11] if f-1(V) is g open in (X,τ) for every open set V in (Y,σ),(6)
g^ continuous [12] if f-1(V) is g^ open in (X,τ) for every open set V in (Y,σ),(7)
contracontinuous [13] if f-1(V) is closed in (X,τ) for every open set V in (Y,σ),(8)
λ continuous [2, 14] if f-1(V) is λ open (λ closed) in (X,τ) for every open (closed) set V in (Y,σ),(9)
λ closed [3] if f(F) is λ closed in (Y, σ) for every λ closed set F of (X,τ),(10)
λ irresolute [4] if the inverse image of λ open sets in Y are λ open in (X,τ),(11)
sg continuous [15] if f-1(V) is sg open (sg closed) in (X,τ) for every open (closed) set in (Y,σ),(12)
gsΛ closed map (gsΛ  open map) [16]  if the image of each closed (open) set in X  is gsΛ  closed (gsΛ open) in Y.(13)
M.gsΛ closed map (gsΛ open map) [16]  if the image of each gsΛ closed (gsΛ  open) set in X is (gsΛ  open) gsΛ closed in Y.Lemma 3 (see [7]).
Iff:X→Y is continuous and closed and if B is g closed (or g open) subset of Y, then f-1(B) is g closed (or g open) in X.Definition 4.
A space(X,τ) is called(i)
[7] a T1/2 space if every g closed subset of X is closed in X,(ii)
[12] a Tg^ space if every g^ closed subset of X is closed in X,(iii)
[12] a Tb space if every gs closed subset of X is closed in X.Proposition 5 (see [10]).
In a topological space (X, τ), the following properties hold. (1)
Every closed set is gsΛ closed.(2)
Every open set is gsΛ closed.(3)
Everyλ closed set is gsΛ closed.(4)
InT1 space every gsΛ closed set is λ closed.(5)
InT1 space every Λg closed set is gsΛ closed.(6)
In partition space every gsΛ closed set is g closed and g^ closed.(7)
In a door space every subset is gsΛ closed.(8)
InT1/2 space every subset is gsΛ closed.Definition 6.
Let (X,τ) be the topological space and A⊆X. We define gsΛ closure of A (briefly gsΛCl(A)) to be the intersection of all gsΛ closed sets containing A, gsΛ interior of A (briefly gsΛInt(A)) to be the union of all gsΛ open sets contained in A.Lemma 7 (see [10]).
LetA and B be subsets of a topological space (X,τ). The following properties hold. (1)
gsΛCl(A) is the smallest gsΛ closed set containing A.(2)
IfA is gsΛ closed then A = gsΛCl(A). Converse need not be true.(3)
A⊂ gsΛCl(A)⊂Clλ(A)⊂Cl(A).(4)
IfA⊂B, then gsΛCl(A)⊂ gsΛCl(B).(5)
gsΛCl(gsΛCl(A)) = gsΛCl(A).(6)
gsΛInt(A) is the largest gsΛ open set contained in A.(7)
IfA is gsΛ open then A= gsΛint(A). Converse need not be true.Proofs are obvious from the definition and properties of gsΛ closed sets and gsΛ open sets.


## 3. gsΛ Continuous Function

Definition 8.
A mapf:(X,τ)→(Y,σ) is called gsΛ continuous function if the inverse image f-1(V) of each closed set V in (Y,σ) is gsΛ closed in (X,τ).Theorem 9.
Every continuous function is gsΛ continuous function.Proof.
LetF be a closed set in (Y,σ) and a function f:(X,τ)→(Y,σ) be a continuous function. Hence, f-1(F) is closed in (X,τ). As every closed set is gsΛ closed set by Proposition 5, we have f-1(F)   is gsΛ closed in X. Thus, f is a gsΛ continuous function.Converse need not be true as seen from the following example.Example 10.
LetX=Y={a,b,c,d,e} and (X,τ)={∅,X,{a},{b,c,d},{a,b,c,d}}, (Y,σ)={∅,Y,{a},{b,c},{a,b,c}}. The identity function f:(X,τ)→(Y,σ) is a gsΛ continuous function but not continuous function as A={a,d,e} is closed in (Y,σ) but f-1(A)={a,d,e}  is not closed in (X,τ).Theorem 11.
Everyλ continuous function is gsΛ continuous function.Proof.
LetF be a closed set in (Y,σ) and a function f:(X,τ)→(Y,σ) be a λ continuous function. Hence, f-1(F) is λ closed in (X,τ). As every λ closed set is gsΛ closed set by Proposition 5, we have f-1(F) is gsΛ closed in X. Thus, f is a gsΛ continuous function.Converse need not be true as seen from the following example.Example 12.
LetX=Y={a,b,c,d,e} and (X,τ)={∅,X,{a},{b,c,d},{a,b,c,d}}, (Y,σ)={∅,Y,{a},{b,c},{a,b,c}}. The identity function f:(X,τ)→(Y,σ) is a gsΛ continuous function but not λ continuous function as A={a,d,e} is closed in (Y,σ) but f-1(A)={a,d,e} is not λ closed in (X,τ).Theorem 13.
If (X,τ) is a T1 space then every gsΛ continuous function f:(X,τ)→(Y,σ) is λ continuous function.Proof.
The proof is simple to the readers as inT1 space every gsΛ closed set is λ closed.Theorem 14.
Everyg^ continuous function is gsΛ continuous function.Proof.
As everyg^ closed set is gsΛ closed set [10]  f is a gsΛ continuous function.Converse need not be true as seen from the following example.Example 15.
LetX=Y={a,b,c,d,e} and (X,τ)={∅,X,{a},{b},{a,b},{a,b,c},{a,b,c,d},{a,b,c,e}, {a,b,d,e},{a,b,d},{a,b,e}}, (Y,σ) = {∅,Y,{a},{b,c,d},{a,b,c,d}}. The identity function f:(X,τ)→(Y,σ) is a gsΛ continuous function but not g^ continuous function as A={a,e} is closed in (Y,σ) but f-1(A)={a,e} is not g^ closed in (X,τ).Theorem 16.
If (X,τ) is a partition space, then every gsΛ continuous function f:(X,τ)→(Y,σ) is g^ continuous function.Proof.
The proof is simple as in a partition space every gsΛ closed set is g^ closed [10].Theorem 17.
Every contra continuous function is gsΛ continuous function.Proof.
LetF be a closed set in (Y,σ) and a function f:(X,τ)→(Y,σ) be a contra continuous function. Hence, f-1(F) is open in (X,τ). As every open set is gsΛ closed set by Proposition 5, we have f-1(F) is gsΛ closed in X. Thus,  f is a gsΛ continuous function.Converse need not be true as seen from the following example.Example 18.
LetX=Y={a,b,c,d,e} and (X,τ)={∅,X,{a},{b},{a,b},{a,b,c},  {a,b,c,d},{a,b,c,e},{a,b,d,e},{a,b,d},{a,b,e}}, (Y,σ) = {∅,Y,{a},{b,c,d},{a,b,c,d}}. The identity function f:(X,τ)→(Y,σ) is a gsΛ continuous function but not contra continuous function as A={e} is closed in (Y,σ) but f-1(A)={e} is not open in (X,τ).Theorem 19.
Everyλ irresolute function is gsΛ continuous function.Proof.
LetF be a closed set in (Y,σ). As every closed set is λ closed set, F is a λ closed set in (Y,σ). Since f:(X,τ)→(Y,σ) is a λ irresolute function, f-1(F) is λ closed in (X,τ). As every λ closed set is gsΛ closed set by Proposition 5, we have f-1(F) is gsΛ closed in (X,τ). Thus, f is a gsΛ continuous function.Converse need not be true as seen from the following example.Example 20.
LetX=Y={a,b,c,d,e} and (X,τ)={∅,X,{a},{a,b},{b,e},{c,d},{a,c,d},{a,b,e},{a,b,c,d},{b,c,d,e}}, (Y,σ)={∅,Y,{a},{b,c},{a,b,c}}. The identity function f:(X,τ)→(Y,σ) is a gsΛ continuous function but not λ irresolute function as A={d,e} is λ closed in (Y,σ) but f-1(A)={d,e} is not λ closed in (X,τ).Theorem 21.
If (X,τ) is a partition space, then every gsΛ continuous function f:(X,τ)→(Y,σ) is g continuous function.Proof.
LetF be a closed set in (Y,σ). Since a function f:(X,τ)→(Y,σ) is a gsΛ continuous function f-1(F) is gsΛ closed in (X,τ). Since (X,τ) is a partition space by Proposition 5, we have f-1(F)  is g closed in X. Thus, f is a g continuous function.Theorem 22.
Any functionf:(X,τ)→(Y,σ), where the domain (X,τ) is a T1/2 (door) space is a gsΛ continuous function.Proof.
Since inT1/2 (door) space every subset is gsΛ closed set [10], the proof is clear to the readers.Theorem 23.
Let a bijective mapf:(X,τ)→(Y,σ) is a λ irresolute and pre-semiopen function. If F is gsΛ closed in (Y,σ) then f-1(F) is gsΛ closed in  (X,τ).Proof.
LetA be a gsΛ closed set of (Y,σ). Let U be a semi open set of (X,τ) containing f-1(A). Then A⊆f(U), where f(U) is semi open in (Y,σ), as f is a pre semi open function. Since A is a gsΛ closed set of (Y,σ), we have Clλ(A)⊆f(U). Thus, we have f-1(Clλ(A))⊆U, where f-1(Clλ(A)) is λ closed in (X,τ). Thus, we have Clλ(f-1(A))⊆Clλ(f-1(Clλ(A))=f-1(Clλ(A))⊆U. Hence,  f-1(A) is gsΛ closed in (X,τ).Lemma 24 (see [10]).
IfA is semi open and gsΛ closed then A is λ closed.Lemma 25 (see [17]).
A functionf:(X,τ)→(Y,σ) is contra irresolute if and only if the inverse image of each semi closed set in Y is semi open in X.Theorem 26.
If a functionf:(X,τ)→(Y,σ) is a gsΛ continuous function and contra irresolute function then f  is a λ continuous function.Proof.
LetV be a closed set of Y. As every closed set is semi closed, V is a semi closed set in Y. Since fis a gsΛ continuous function and contra irresolute function, f-1(V) is gsΛ closed and semi open in (Y,σ). Now, by Lemma 24 the proof is very clear.Lemma 27 (see [10]).
LetF⊆A⊆X, where A is open in X. If F is gsΛ closed in X, then F is gsΛ closed in A.Theorem 28.
If a functionf:(X,τ)→(Y,σ) is a λ continuous function and A is an open subset of X, then the restriction fA:A→Y is also gsΛ continuous.Proof.
LetV be a closed set of Y and A be an open subset of X. As every open set is λ closed, A is λ closed in X and since f is λ continuous function f-1(V) is λ closed in X. Hence, we have f-1(V)∩A is λ closed in X, which is also gsΛ closed in X. Since f-1(V)∩A⊆A⊆X where A is an open subset of  X, by Lemma 27f-1(V)∩A=(fA-1(V)) is gsΛ closed in A. Thus, the restriction fA:A→Y is also gsΛ continuous.Definition 29.
(i) Letx be a point of (X,τ) and V be a subset of X. Then V is called a gsΛ neighbourhood of x in (X,τ), if there exist a gsΛ open set U of (X,τ), such that x∈U⊆V.
(ii) The intersection of all gsΛ closed sets each containing a set A in a topological space X is called the gsΛ closure of A and is denoted by gsΛCl(A).Theorem 30.
LetA be a subset of (X,τ). Then x∈gsΛCl(A)  if and only if for any gsΛ neighbourhood Nx of x in (X,τ),A∩Nx≠ϕ.Proof.
Necessity. Assume that x∈ gsΛCl(A). Suppose that there exist a gsΛ neighbourhood Nx of x such that A∩Nx=ϕ. Since Nx is a gsΛ neighbourhood of x in (X,τ) by definition, there exists a gsΛ open set V such that x∈V⊆Nx. Since Nx is such that A∩Nx=ϕ, we have A∩V=ϕ we have A⊆(V)C. Since (V)C is a gsΛ closed set containing A, we have gsΛCl(A) ⊆(V)C and therefore x∉ gsΛ Cl(A), which is a contradiction.
Sufficiency. Assume that for each gsΛ neighbourhood Nx of x in (X,τ),  A∩Nx≠ϕ. Suppose x∉ gsΛCl(A). Then there exist a gsΛ closed set F of (X,τ), such that A⊆F and x∉F. Thus, x∈FC and FC is a gsΛ open set in (X,τ). But A∩FC=ϕ which is a contradiction. Thus,  x∈ gsΛCl(A).Theorem 31.
Letf:(X,τ)→(Y,σ) be a function.Then the following are equivalent. (i)
The function is gsΛ continuous.(ii)
The inverse of each open set in(Y,σ) is gsΛ open in (X,τ).(iii)
For eachx in (X,τ), the inverse of every neighbourhood of f(x) is a gsΛ neighbourhood of x.(iv)
For eachx in (X,τ), and each neighbourhood N of f(x), there is a gsΛ neighbourhood W of x, such that f(W)⊆N.(v)
For each subsetA of (X,τ),  f(gsΛCl(A))⊆Cl(f(A)).(vi)
For each subsetB of (Y,σ),  gsΛCl(f-1(B))⊆f-1Cl(B).Proof.
(i)⇔(ii). Let  f:(X,τ)→(Y,σ) be a gsΛ continuous function and U be an open set in (Y,σ). Then UC is closed set in Y. Hence, by the definition of gsΛ continuous function f-1(UC)=(f-1(U))C is gsΛ closed set in X. Thus, f-1(U) is a gsΛ open set in X. The converse is analogous.
(ii)⇔(iii). For x∈(X,τ), let N be the neighbourhood of f(x). Then there exist an open set V in (Y,σ) such that f(x)∈V⊆N. Consequently, f-1(V) is gsΛ open set in (X,τ) and x∈f-1(V)⊆f-1(N). Thus, f-1(N) is a gsΛ neighbourhood in (X,τ). Converse is very simple to the reader.
(iii)⇔(iv). Let x∈X and N be the neighbourhood of f(x). Then by assumption W=f-1(N) is a gsΛ neighbourhood of x and f(W)=f(f-1(N))⊆N. Converse is clear.
(iv)⇔(v). Suppose that (iv) holds and let  y∈f(gsΛCl(A)) and let N be the neighbourhood of y. Then there exist x∈X and a gsΛ neighbourhood W of x, such that f(x)=y,x∈W,  x∈ gsΛCl(A) and f(W)⊆N. Since x∈ gsΛCl(A), by Theorem 30  W∩A≠ϕ and hence f(A)∩N≠ϕ. Hence, y=f(x)∈Cl(f(A)).
Conversely suppose that (v) holds and letx∈X, N be the neighbourhood of f(x). Let A=f-1(Y∖N). Since f(gsΛCl(A))⊆Cl(f(A))⊆Y∖N, we have gsΛCl(A)=A. Since x∉ gsΛ Cl(A), there exist a gsΛ neighbourhood W of x such that W∩A=ϕ and hence f(W)⊆f(X∖A)⊆N.
(v)⇔(vi). Suppose that (v) holds and B be any subset of (Y,σ). Then replacing A by f-1(B) in (v), we obtain f(gsΛCl(f-1(B)))⊆Cl(f(f-1(B))⊆Cl(B). That is gsΛCl(f-1(B)))⊆f-1(Cl(B).
Conversely, suppose that (vi) holds, and letB=f(A), where A is the subset of (X,τ). Then we have gsΛCl(A) ⊆ gsΛCl(f-1(B))⊆f-1Cl(f(A)) and so f(gsΛCl(A))⊆Cl(f(A)).
This completes the proof of the theorem.Theorem 32.
Let{Xi/i∈I} be any family of topological spaces. If f:X→ΠXj is a gsΛ continuous function, then Pri of :X→Xi is gsΛ continuous for each i∈I, where Pri is the projection of ΠXj on Xi.Proof.
We will consider a fixedi∈I. Suppose Ui is an arbitrary open set in Xi. Then Pri-1(Ui) is open in ΠXj. Since f is a gsΛ continuous function, we have f-1(Pri-1(Ui))=(Pri∘f)-1(Ui) is gsΛ open in X. Hence,  Pri of is a gsΛ continuous function.Definition 33.
LetA be a subset of X. A mapping r:(X,τ)→A is called a gsΛ continuous retraction if r is gsΛ continuous and the restriction rA is the identity mapping on A.Definition 34.
A topological space (X,τ) is called a gsΛ Hausdroff if for each pair x,y of distinct points of X, there exist gsΛ neighbourhoods U1 and U2 of x and y, respectively, that are disjoint.Theorem 35.
LetA be a subset of X and r:X→A be a gsΛ continuous retraction. If X is gsΛ Hausdroff, then A is a gsΛ closed set of X.Proof.
Suppose thatA is not gsΛ closed. Then there exist a point x in X such that x∈ gsΛCl(A) but x∉A. It follows that r(x)≠x because r is gsΛ continuous retraction. Since X is gsΛ Hausdroff there exist disjoint gsΛ open sets U1 and U2 in X such that x∈U1 and r(x)∈U2. Now let W be an arbitrary gsΛ neighbourhood of x. Then W∩U1 is a gsΛ neighbourhood of x. Since x∈ gsΛCl(A), by Theorem 30, we have (W∩U1)∩A≠∅. Therefore, there exist a point y∈W∩U1∩A. Since y∈A, we have r(y)=y∈U1 and hence r(y)∉U2. This implies that r(W)⊈U2 because y∈W. This is a contrary to gsΛ continuity of r. Consequently, A is a gsΛ closed set of X.


## 4. On Composite Functions

Theorem 36.
Composition of continuous functions is a gsΛ continuous function.Proof.
Since every closed set is a gsΛ closed set, the proof is simple to prove.Theorem 37.
Composition ofλ irresolute functions is a gsΛ continuous function.Proof.
Letf:(X,τ)→(Y,σ) and g:(Y,σ)→(Z,ς) are λ irresolute functions and let F be a closed set of (Z,ς), which is also λ closed set by [2]. Then g-1(F) is a λ closed set in (Y,σ) and f-1(g-1(F))=(g∘f)-1(F) is a λ closed set in (X,τ). Since every λ closed set is gsΛ closed set by Proposition 5, we get g∘f is a gsΛ continuous function.Theorem 38.
Composition of contra continuous functions is a gsΛ continuous function.Proof.
Proof is clear by definition.Remark 39.
But composition of gsΛ continuous function is not a gsΛ continuous function.Example 40.
LetX = Y = Z = {a,b,c,d,e} and (X,τ) = {∅,X,{a},{b},{a,b},{b,c},{a,b,c},{b,c,d},{a,b,c,d},{b,c,d,e}}, (Y,σ) = {∅,Y,{a},{b,c,d},{a,b,c,d}}, and (Z,ς) = {∅,Z,{a},{b},{a,b},{b,e},{c,d},{b,c,d},{a,b,e},{a,c,d},{a,b,c,d},{b,c,d,e}}. The identity function f:(X,τ)→(Y,σ) and g:(Y,σ)→(Z,ς) is a gsΛ continuous functions but (g∘f) is not gsΛ continuous function as A={b,e} is closed in (Z,ς) but (g∘f)-1(A)={b,e} is not gsΛ closed in (X,τ).Theorem 41.
(i) Iff:(X,τ)→(Y,σ) is a gsΛ continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(ii) Iff:(X,τ)→(Y,σ) is a λ continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(iii) Iff:(X,τ)→(Y,σ) is a λ irresolute function and g:(Y,σ)→(Z,ς) is a λ continuous function, then g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(iv) Iff:(X,τ)→(Y,σ) is a g^ continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(v) Iff:(X,τ)→(Y,σ) is a contra continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.Proof.
(i) Letf:(X,τ)→(Y,σ) is a gsΛ continuous function and g:(Y,σ)→(Z,ς) is a continuous function and let F be a closed set of (Z,ς). Then g-1(F) is a closed set in (Y,σ) and f-1(g-1(F))=(g∘f)-1(F) is a gsΛ closed set in (X,τ) as f:(X,τ)→(Y,σ) is a gsΛ continuous function. Thus g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(ii) It is clear thatg∘f:(X,τ)→(Z,ς) is a λ continuous function and hence gsΛ continuous function by Theorem 11.
(iii) Since every closed set isλ closed and every λ closed set is gsΛ closed set (Proposition 5),  g∘f:(X,τ)→(Z,ς) is a gsΛ continuous function.
(iv) Letf:(X,τ)→(Y,σ) is a g^ continuous function and g:(Y,σ)→(Z,ς) is a continuous function and let F be a closed set of (Z,ς). Then g-1(F) is a closed set in (Y,σ) and f-1(g-1(F))=(g∘f)-1(F) is a g^ closed set in (X,τ) as f:(X,τ)→(Y,σ) is a g^ continuous function. Since every g^ continuous function is gsΛ continuous function by Theorem 14, we get g∘f is a gsΛ continuous function.
(v) As every closed set is gsΛ open by Proposition 5  the proof is very clear.Theorem 42.
(i) Letf:(X,τ)→(Y,σ) is a gsΛ continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς) is a g continuous (g^ continuous) function if (X,τ) is a partition space.
(ii) Letf:(X,τ)→(Y,σ) is a gsΛ continuous function and g:(Y,σ)→(Z,ς) is a continuous function, then g∘f:(X,τ)→(Z,ς)  is a λ continuous function if (X,τ) is a T1 space.Proof.
(i) In partition space every gsΛ closed set is g closed (g^ closed) set by Proposition 5, the theorem is proved.
(ii) The proof is clear as in aT1 space every gsΛ closed set is λ closed by Proposition 5.Lemma 43 (see [10]).
If a mapf:(X,τ)→(Y,σ) is irresolute and a λ closed map then for every gsΛ closed set B of (X,τ),f(B)  is gsΛ closed set in (Y,σ).Theorem 44.
Letf:(X,τ)→(Y,σ) and g:(Y,σ)→(Z,ς) are bijective.
(i) Ifg∘f:(X,τ)→(Z,ς) is a continuous function and f:(X,τ)→(Y,σ) is gsΛ closed map then g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(ii) Ifg∘f:(X,τ)→(Z,ς) is a λ continuous function and f:(X,τ)→(Y,σ) is λ closed map then g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(iii) Ifg∘f:(X,τ)→(Z,ς) is a λ irresolute function and f:(X,τ)→(Y,σ) is λ closed map then  g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(iv) Ifg∘f:(X,τ)→(Z,ς) is a continuous function and f:(X,τ)→(Y,σ) is λ closed map then g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(v) Ifg∘f:(X,τ)→(Z,ς) is a gsΛ continuous function and f:(X,τ)→(Y,σ) is irresolute and a λ closed map then g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(vi) Ifg∘f:(X,τ)→(Z,ς) is a gsΛ continuous function and f:(X,τ)→(Y,σ) is M.gsΛ closed map then g:(Y,σ)→(Z,ς) is a gsΛ continuous function.Proof.
(i) LetF be a closed set in (Z,ς). Since g∘f is a continuous function (g∘f)-1(F)=f-1(g-1(F)) is closed in (X,τ). Now since f is a gsΛ closed map f(f-1(g-1(F))=g-1(F)  is gsΛ closed set in (Y,σ). Thus g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(ii) LetF be a closed set in (Z,ς). Since g∘f is a λ continuous function (g∘f)-1(F) = f-1(g-1(F)) is λ closed in (X,τ). Now since f  is a λ closed map f(f-1(g-1(F))=g-1(F) is λ closed which is also gsΛ closed set in (Y,σ) [10]. Thus, g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(iii) LetF be a closed set in (Z,ς) which is also λ closed in Z [10]. Since g∘f is a λ irresolute function (g∘f)-1(F)=f-1(g-1(F)) is λ closed in (X,τ). Now, since f is a λ closed map f(f-1(g-1(F))=g-1(F) is λ closed which is also gsΛ closed set in (Y,σ) [10]. Thus, g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(iv) LetF be a closed set in (Z, ς). Since g∘f is a continuous function (g∘f)-1(F)=f-1(g-1(F)) is closed in (X,τ), which is also λ closed in X [10]. Now, since f is a λ closed map f(f-1(g-1(F))=g-1(F) is λ closed which is also gsΛ closed set in (Y,σ) [10]. Thus, g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(v) LetF be a closed set in (Z,ς). Since g∘f is a gsΛ continuous function (g∘f)-1(F)=f-1(g-1(F)) is gsΛ closed in (X,τ). By Lemma 43  f(f-1(g-1(F))=g-1(F) is gsΛ closed set in (Y,σ). Thus g:(Y,σ)→(Z,ς) is a gsΛ continuous function.
(vi) LetF be a closed set in (Z,ς). Since g∘f is a gsΛ continuous function (g∘f)-1(F)=f-1(g-1(F)) is gsΛ closed in (X,τ). Now, since fis a M.gsΛ closed map f(f-1(g-1(F))=g-1(F) is gsΛ closed. Thus, g:(Y,σ)→(Z,ς) is a gsΛ continuous function.Theorem 45.
For any bijectionf:(X,τ)→(Y,σ), the following are equivalent:(i)
f-1:(Y,σ)→(X,τ) is gsΛ continuous,(ii)
f:(X,τ)→(Y,σ) is gsΛ open,(iii)
f:(X,τ)→(Y,σ) is gsΛ closed.Proof.
(i)→(ii) Let U be an open set in (X,τ). By assumption (f-1)-1(U))=f(U) is gsΛ open in (Y,σ) and hence f  is gsΛ open map.
(ii)→(iii) Let F be a closed set in (X,τ). Then FC is open in (X,τ). By assumption f(FC)=(f(F))C is gsΛ open in (Y,σ) and therefore f(F) is gsΛ closed in (Y,σ). Hence, f is gsΛ closed.
(iii)→(i) Let F be a closed set in (X,τ). By assumptionf(F) is gsΛ closed in (Y,σ). But f(F)=(f-1)-1(F)) and therefore f-1 is is gsΛ continuous.



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*Source: 787014-2013-02-26.xml*