#### Abstract

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

#### 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 -sets in topological spaces. A -set is a set which is equal to its kernel (= saturated set), that is, to the intersection of all open supersets of . Arenas et al. [2] introduced and investigated the notion of -closed sets and -open sets by involving -sets.

In 2008 Caldas et al. [3] introduced generalized closed sets (g, -g, g) and their properties. They also studied the concept of closed maps. In 2007, Caldas et al. [4] introduced the concept of irresolute maps.

In this paper, we establish a new class of maps called gs continuous function and study its properties and characteristics.

Throughout this paper, (), (), and () (or simply , and will always denote topological spaces on which no separation axioms are assumed unless explicitly stated. , , , , gs, and gs denote the interior of , closure of , lambda interior of , lambda closure of , gs Lambda closure of and gs Lambda interior of , respectively.

#### 2. Preliminary Definitions

*Definition 1. *A subset of a space () is called (1)a semiopen set [5] if , (2)a preopen set [6] if ,(3)a regular open set [6] if .

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 subset of denoted by sCl(), (pCl()) is the intersection of all semi closed sets (pre closed sets) containing .

A topological space () is said to be (1)a generalized closed [7] if Cl() , whenever and is open in (),(2)a closed [8] if Cl() , whenever and is *g*-open in (,), (3)semigeneralized closed (denoted by sg-closed) [9] if sCl(, whenever and is semi open in (), (4)generalized semiclosed (denoted by gs-closed) [9] if sCl(, whenever and is open in (), (5)a subset of a space () is called -closed [2] if , where is a -set and is a closed set, (6)a subset of () is said to be a g closed set [3] if Cl( whenever , where is open in (), (7)a subset of () is said to be a -g closed set [3] if whenever , where is open in (), (8)a subset of () is said to be a g closed set [3] if whenever , where is open in (), (9)a subset of () is said to be a gs closed set [10] if whenever , where is semi open in .

The complement of the above closed sets are called its respective open sets.

*Definition 2. *A function ) is called (1)semicontinuous [5] if is semi open in () for every open set in (), (2)semi open (semiclosed) [9] if is semi open (semi closed) in () for any open (closed) set in (),(3)pre-semiopen (pre-semiclosed) [9] if is semi open (semi closed) in (, ) for every semi open (semi closed) set in (), (4)irresolute [9] if for any semi open set of (), is semi open in (), (5) continuous [11] if () is open in () for every open set in (), (6) continuous [12] if () is open in () for every open set in (), (7)contracontinuous [13] if () is closed in () for every open set in (), (8) continuous [2, 14] if () is open ( closed) in () for every open (closed) set in , (9) closed [3] if is closed in (, ) for every closed set of (), (10) irresolute [4] if the inverse image of open sets in are open in (), (11)sg continuous [15] if () is sg open (sg closed) in for every open (closed) set in (),(12)gs closed map (gs open map) [16] if the image of each closed (open) set in is gs closed (gs open) in .(13)M.gs closed map (gs open map) [16] if the image of each gs closed (gs open) set in is (gs open) gs closed in .

Lemma 3 (see [7]). *If is continuous and closed and if B is closed (or open) subset of , then is closed (or open) in . *

*Definition 4. *A space is called(i) [7] a space if every closed subset of is closed in , (ii) [12] a space if every closed subset of is closed in ,(iii) [12] a space if every closed subset of is closed in .

Proposition 5 (see [10]). *In a topological space (, ), the following properties hold. *(1)Every closed set is gs closed. (2)Every open set is gs closed. (3)Every closed set is gs closed. (4)In space every gs closed set is closed. (5)In space every g closed set is gs closed. (6)In partition space every gs closed set is closed and closed. (7)In a door space every subset is gs closed. (8)In space every subset is gs closed.

*Definition 6. *Let () be the topological space and . We define gs closure of (briefly gs ) to be the intersection of all gs closed sets containing , gs interior of (briefly gs ) to be the union of all gs open sets contained in .

Lemma 7 (see [10]). *Let A and B be subsets of a topological space (). The following properties hold. *(1)

*gs*(2)

*Cl*(*A*) is the smallest gs closed set containing .*If*(3)

*A*is gs closed then*A*= gs. Converse need not be true.*(4)*

*A*gs.*If , then gs gs*(5)

*gs gs) = gs.*(6)

*gs is the largest gs open set contained in .*(7)

*If is gs open then gs. Converse need not be true.*

Proofs are obvious from the definition and properties of gs closed sets and gs open sets.

#### 3. gs Continuous Function

*Definition 8. *A map is called gs continuous function if the inverse image of each closed set in () is gs closed in ().

Theorem 9. *Every continuous function is gs continuous function. *

*Proof. * Let be a closed set in () and a function be a continuous function. Hence, is closed in (). As every closed set is gs closed set by Proposition 5, we have is gs closed in . Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

*Example 10. *Let and , . The identity function is a gs continuous function but not continuous function as is closed in () but is not closed in ().

Theorem 11. *Every continuous function is gs continuous function. *

*Proof. * Let be a closed set in () and a function be a continuous function. Hence, is closed in (). As every closed set is gs closed set by Proposition 5, we have is gs closed in *X*. Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

*Example 12. *Let and , . The identity function is a gs continuous function but not continuous function as is closed in () but is not closed in ().

Theorem 13. *If () is a space then every gs continuous function is continuous function. *

*Proof. *The proof is simple to the readers as in space every gs closed set is closed.

Theorem 14. *Every continuous function is gs continuous function. *

*Proof. * As every closed set is gs closed set [10] is a gs continuous function.

Converse need not be true as seen from the following example.

*Example 15. *Let and ,,,, ,,, = ,,,,,,,,,. The identity function is a gs continuous function but not continuous function as is closed in () but is not closed in ().

Theorem 16. *If () is a partition space, then every gs continuous function is continuous function. *

*Proof. * The proof is simple as in a partition space every gs closed set is closed [10].

Theorem 17. *Every contra continuous function is gs continuous function. *

*Proof. * Let be a closed set in () and a function be a contra continuous function. Hence, () is open in (). As every open set is gs closed set by Proposition 5, we have () is gs closed in . Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

*Example 18. *Let and , , = ,,,,,,,,,. The identity function is a gs continuous function but not contra continuous function as is closed in () but is not open in .

Theorem 19. *Every irresolute function is gs continuous function. *

* Proof. *Let be a closed set in (). As every closed set is closed set, is a closed set in . Since is a irresolute function, () is closed in (). As every closed set is gs closed set by Proposition 5, we have () is gs closed in (). Thus, is a gs continuous function.

Converse need not be true as seen from the following example.

*Example 20. *Let and ,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,. The identity function is a gs continuous function but not irresolute function as is closed in but is not closed in .

Theorem 21. *If () is a partition space, then every gs continuous function is continuous function. *

*Proof. * Let be a closed set in . Since a function is a gs continuous function is gs closed in . Since is a partition space by Proposition 5, we have is closed in . Thus, is a continuous function.

Theorem 22. *Any function , where the domain is a (door) space is a gs continuous function. *

*Proof. * Since in (door) space every subset is gs closed set [10], the proof is clear to the readers.

Theorem 23. *Let a bijective map is a irresolute and pre-semiopen function. If is gs closed in then () is gs closed in . *

*Proof. *Let be a gs closed set of (). Let be a semi open set of containing . Then , where is semi open in (), as is a pre semi open function. Since is a gs closed set of , we have . Thus, we have (, where is closed in (). Thus, we have (. Hence, is gs closed in .

Lemma 24 (see [10]). *If is semi open and gs closed then is closed. *

Lemma 25 (see [17]). * A function is contra irresolute if and only if the inverse image of each semi closed set in is semi open in . *

Theorem 26. *If a function is a gs continuous function and contra irresolute function then is a continuous function. *

*Proof. *Let be a closed set of . As every closed set is semi closed, is a semi closed set in . Since is a gs continuous function and contra irresolute function, is gs closed and semi open in . Now, by Lemma 24 the proof is very clear.

Lemma 27 (see [10]). *Let , where is open in . If is gs closed in , then is gs closed in . *

Theorem 28. *If a function is a continuous function and is an open subset of , then the restriction is also gs continuous. *

*Proof. * Let be a closed set of and be an open subset of . As every open set is closed, is closed in and since is continuous function is closed in . Hence, we have is closed in , which is also gs closed in . Since where is an open subset of , by Lemma 27 is gs closed in . Thus, the restriction is also gs continuous.

*Definition 29. *(i) Let be a point of and be a subset of . Then is called a gs neighbourhood of in , if there exist a gs open set of (), such that .

(ii) The intersection of all gs closed sets each containing a set in a topological space is called the gs closure of and is denoted by gs.

Theorem 30. *Let be a subset of . Then if and only if for any gs neighbourhood of in . *

*Proof. **Necessity.* Assume that gs. Suppose that there exist a gs neighbourhood of such that . Since is a gs neighbourhood of in by definition, there exists a gs open set such that . Since is such that , we have we have . Since is a gs closed set containing , we have gs) and therefore gs Cl(), which is a contradiction.*Sufficiency.* Assume that for each gs neighbourhood of in , . Suppose gs). Then there exist a gs closed set of (), such that and . Thus, and is a gs open set in . But which is a contradiction. Thus, gs.

Theorem 31. *Let be a function.Then the following are equivalent. *(i)*The function is gs continuous. *(ii)*The inverse of each open set in is gs open in . *(iii)*For each in , the inverse of every neighbourhood of is a gs neighbourhood of .*(iv)*For each in , and each neighbourhood of , there is a gs neighbourhood of , such that .*(v)*For each subset of , . *(vi)*For each subset of , gs. *

*Proof. *(i)(ii). Let be a gs continuous function and be an open set in . Then is closed set in . Hence, by the definition of gs continuous function is gs closed set in . Thus, is a gs open set in . The converse is analogous.

(ii)(iii). For , let be the neighbourhood of . Then there exist an open set in such that . Consequently, is gs open set in and . Thus, is a gs neighbourhood in . Converse is very simple to the reader.

(iii)(iv). Let and be the neighbourhood of . Then by assumption is a gs neighbourhood of and . Converse is clear.

(iv)(v). Suppose that (iv) holds and let gs) and let be the neighbourhood of . Then there exist and a gs neighbourhood of , such that , gs and . Since gs, by Theorem 30 and hence . Hence, .

Conversely suppose that (v) holds and let , be the neighbourhood of . Let . Since , we have gs. Since gs Cl(), there exist a gs neighbourhood of such that and hence .

(v)(vi). Suppose that (v) holds and be any subset of (). Then replacing by () in (), we obtain gs. That is gs.

Conversely, suppose that (vi) holds, and let , where is the subset of . Then we have gs) gs and so (gs.

This completes the proof of the theorem.

Theorem 32. *Let be any family of topological spaces. If is a gs continuous function, then of is gs continuous for each , where is the projection of on . *

*Proof. * We will consider a fixed . Suppose is an arbitrary open set in . Then is open in . Since is a gs continuous function, we have is gs open in . Hence, of is a gs continuous function.

*Definition 33. *Let be a subset of . mapping is called a gs continuous retraction if is gs continuous and the restriction is the identity mapping on .

*Definition 34. *A topological space () is called a gs Hausdroff if for each pair of distinct points of , there exist gs neighbourhoods and of and , respectively, that are disjoint.

Theorem 35. *Let be a subset of and be a gs continuous retraction. If is gs Hausdroff, then is a gs closed set of . *

*Proof. *Suppose that is not gs closed. Then there exist a point in such that gsCl() but . It follows that because is gs continuous retraction. Since is gs Hausdroff there exist disjoint gs open sets and in such that and . Now let be an arbitrary gs neighbourhood of . Then is a gs neighbourhood of . Since gs, by Theorem 30, we have . Therefore, there exist a point . Since , we have and hence . This implies that because . This is a contrary to gs continuity of . Consequently, is a gs closed set of .

#### 4. On Composite Functions

Theorem 36. *Composition of continuous functions is a gs continuous function.*

*Proof. * Since every closed set is a gs closed set, the proof is simple to prove.

Theorem 37. *Composition of irresolute functions is a gs continuous function. *

*Proof. * Let and are irresolute functions and let be a closed set of (), which is also closed set by [2]. Then is a closed set in () and is a closed set in . Since every closed set is gs closed set by Proposition 5, we get is a gs continuous function.

Theorem 38. *Composition of contra continuous functions is a gs continuous function. *

*Proof. *Proof is clear by definition.

*Remark 39. *But composition of gs continuous function is not a gs continuous function.

*Example 40. *Let *X* = *Y* = *Z* = {*a,b,c,d,e*} and (*X*,) = ,*X*,{*a*},{*b*}*,*{*a,b*}*,*{*b,c*}*,*{*a,b,c*}*,*{*b,c,d*}*,*{*a,b,c,d*},{*b,c,d,e *, , = ,*Y,*{*a*}*,*{*b,c,d*}*,*{*a,b,c,d *, and , = ,*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 and is a gs continuous functions but () is not gs continuous function as is closed in but is not gs closed in .

Theorem 41. *(i) If is a gs continuous function and is a continuous function, then is a gs continuous function.**(ii) If is a continuous function and is a continuous function, then is a gs continuous function. **(iii) If is a irresolute function and is a continuous function, then is a gs continuous function.**(iv) If is a continuous function and is a continuous function, then is a gs continuous function.**(v) If is a contra continuous function and is a continuous function, then is a gs continuous function.*

*Proof. *(i) Let is a gs continuous function and is a continuous function and let be a closed set of (). Then () is a closed set in () and (() is a gs closed set in as is a gs continuous function. Thus is a gs continuous function.

(ii) It is clear that is a continuous function and hence gs continuous function by Theorem 11.

(iii) Since every closed set is closed and every closed set is gs closed set (Proposition 5), is a gs continuous function.

(iv) Let is a continuous function and is a continuous function and let be a closed set of . Then () is a closed set in and is a closed set in as is a continuous function. Since every continuous function is gs continuous function by Theorem 14, we get is a gs continuous function.

(v) As every closed set is gs open by Proposition 5 the proof is very clear.

Theorem 42. *(i) Let is a gs continuous function and is a continuous function, then is a g continuous ( continuous) function if is a partition space. *

*(ii) Let is a gs continuous function and is a continuous function, then is a continuous function if is a space.*

*Proof. *(i) In partition space every gs closed set is closed ( closed) set by Proposition 5, the theorem is proved.

(ii) The proof is clear as in a space every gs closed set is closed by Proposition 5.

Lemma 43 (see [10]). * If a map is irresolute and a closed map then for every gs closed set of is gs closed set in . *

Theorem 44. *Let and are bijective. **(i) If is a continuous function and is gs closed map then is a gs continuous function. **(ii) If is a continuous function and is closed map then is a gs continuous function.**(iii) If is a irresolute function and is closed map then is a gs continuous function.**(iv) If is a continuous function and is closed map then is a gs continuous function.**(v) If is a gs continuous function and is irresolute and a closed map then is a gs continuous function.**(vi) If is a gs continuous function and is M.gs closed map then is a gs continuous function.*

*Proof. *(i) Let be a closed set in . Since is a continuous function ( is closed in (). Now since is a gs closed map is gs closed set in (). Thus is a gs continuous function.

(ii) Let be a closed set in (). Since is a continuous function () = (()) is closed in (). Now since is a closed map is closed which is also gs closed set in [10]. Thus, is a gs continuous function.

(iii) Let be a closed set in which is also closed in [10]. Since is a irresolute function is closed in . Now, since is a closed map is closed which is also gs closed set in () [10]. Thus, is a gs continuous function.

(iv) Let be a closed set in (, ). Since is a continuous function is closed in , which is also closed in [10]. Now, since is a closed map is closed which is also gs closed set in [10]. Thus, is a gs continuous function.

(v) Let be a closed set in . Since is a gs continuous function is gs closed in . By Lemma 43 is gs closed set in . Thus is a gs continuous function.

(vi) Let be a closed set in . Since is a gs continuous function is gs closed in . Now, since is a M.gs closed map is gs closed. Thus, is a gs continuous function.

Theorem 45. *For any bijection , the following are equivalent:*(i)* is gs continuous,*(ii)* is gs open, *(iii)* is gs closed. *

*Proof. *(i)(ii) Let be an open set in (). By assumption is gs open in and hence is gs open map.

(ii)(iii) Let be a closed set in . Then is open in . By assumption is gs open in and therefore is gs closed in . Hence, is gs closed.

(iii)(i) Let be a closed set in (). By assumption is gs closed in . But and therefore is is gs continuous.