International Journal of Scientific & Engineering Research Volume 4, Issue 1, January-2013 1

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J3* *- ***Continuous Maps and Pasting Lemma in Topological Spaces**

P.G.Palanimani, R.Parimelazhagan

Abstract— In this paper, the authors introduce a new class of maps called �* continuous maps and �* irresolute maps in topological spaces and study some of its basic properties and relations among

Index Terms— g-closed, g-continuous,J3 * -closed,J3 * continuous, J3 * irresolute.

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1.Introduction **2. Preliminaries**

Biswas[3], Husain[10], Ganster and Reilly[9], Levine[11,13], Marcus[15], Mashour [16]et al,Noiri[18], Noiri and Ahmed[19] and Tong[15,16,17] have introduced and investigated simple continuous, almost continuous, LC-continuity, weak continui- ty, semi-continuity,quasi-continuity, a-continuity, strong

semi-continuity, semi-weak continuity, weak almost continui-

ty, A-continuity and B-continuity respectively. Balachandran et al have introduced and studied generalized semi- continuous maps, semi locally continuous maps, semigenera- lized locally con-tinuous maps and generalized locally conti- nuous maps. Maki and Noiri studied the pasting lemma for a- continuous mappings. El Etik[7] also introduced the concept

of gb-continuous functions with the aid of b-open sets,Omari

and Noorani introduced and studied the concept of genera- lized g-closed sets and g -continuous maps in topological spaces. Palanimani and Parimelazhagan[26] introduced and studied theproperties of *-closed set in topological spaces. Crossley and Hildebrand[5] introducedand investigated irre- solute functions which are stronger than semi continuous maps but are independent of continuous maps. Since then several researchers have introduced several strong and weak forms of irresolute functions. Di Maio and Noiri[6], Faro[8], Cammaroto and Noiri [4], Maheswari and Prasad[16] and Sundaram [21] have introduced and studied quasi-irresolute and strongly irresolute maps strongly a-irresolute maps, al-

most irresolute maps, a -irresolute maps and gc-irresolute

maps are respectively.The aim of this paper is to introduce and study the concepts of new class of maps namely *- continuous maps and *-irresolute maps.Throughout this pa-

per (X, r) and (Y, 6)(or simply X and Y) represents the non-

-empty topological spaces on which no separation axiom are

assumed, un-less otherwise mentioned. For a subset A of X, cl(A) and int(A) represents the closure of A and interior of A respectively.

We recall the following definitions.

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• *R.Parimalazhagan is currently working as Professor and Head , Depart-*

ment of Science and Humanities in Karpagam College of Engineering ,

Definition 2.1[13] : A subset A of a topological space (X, r) is called generalized closed set (briefly g-closed) if cl(A)� U,

whenever A � U and U is open in X .

Definition 2.2 : A subset A of a topological space (X, r) is

called * closed set. If cl(int(A))� U, whenever A � U and U

is g-open in X .

Definition 2.3[25] : A subset A of a topological space (X, r) is called g*-closed if cl(A)� U,whenever A � U and U is g-open

in X .

Definition 2.4: A map f : (X, r) �(Y, a) from a topological

space X into a topological space Y is called g-continuous if

f-1 (V) is g-closed in X for every closed set V of Y.

Definition 2.5: A map f : (X, r) �(Y, a) from a topological

space X into a topological space Y is called *-continuous if

f-1 (V) is *- closed in X for every closed set V of Y.

Definition 2.6: A map f : (X, r) �(Y, a) from a topological

space X into a topological space Y is called irresolute if f-1 (V)

is semi-closed in X for every semi-closed set V of Y.

Definition 2.7[5]: A map f : (X, r) �(Y, a) from a topological

space X into a topological space Y is called semi-generalized

continuous(briefly sg continuous) if f-1 (V) is sg-closed

in X for every closed set V of Y.

3. J3***- Continuous Maps**

In this section we introduce the concept of *-Continuous maps in topological spaces.

topological space Y is called *-continuous if the inverse im- age of every closed set in Y is *-closed in X.

a topological space Y is continuous, then it is *-continuous but not conversely.

Y. Then the inverse image f -1 (F) is closed in X. Since every closed set is *-closed, f -1 (F) is *-closed in X. Therefore f is

*-continuous.

Example 3.4 : Let X = Y={a, b, c} with topologies " = {X, q, {a},

{a, b}} and a = {Y, q, {a},{b,c}}. Let f : X� Y be a map defined by f (a) = c, f (b) = b, f (c) = a. Here f is *-continuous. But f is

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not continuous since for the closed set F={b,c} in Y,f-1 (F)={b,

a} is not closed in X.

Y. Then the inverse image f -1 (F) is g-closed in X. Since every

g-closed set is * -closed in X, f-1 (F) is * -closed in

X. Therefore f is * -continuous.

Example 3.7 Let X = Y = {a, b, c} with topologies " = {X, q, {a},

{a,b}} and a = {Y,q,{a},{b}, {a,b}}.Let f : X � Y be the identity map. Here f is * -continuous. But f is not g-continuous since for the closed set F={ a,c} f is not * -closed set in X.

into a topological space Y is sg-continuous, then it is *- continuous but not conversely.

set in Y. Then the inverse image f -1 (V ) is * -closed in X.

Since every * -closed set is sg-closed, f-1 (V ) is * -closed in

X. Therefore f is * -continuous.

Example 3.10 Let X = Y = {a, b, c} with topologies " = {X, q, {a},

{a,b}} and a = {Y, q,{b}, {a,c}}. Let f : X � Y be a map defined

by f (a) = b, f (b) = a, f (c) =c. Here f is * -continuous. But f is not sg-continuous since for the closed set F={ a,c} f is not * - closed set in X.

into a topological space Y is gs-continuous, then it is * - continuous but not conversely.

(a) For each point x E X and each open set V in Y with f ( x ) E

V,there is a * -open set U in X such that x E U , f (U ) c V .

(b) For every subset A of X , f ( * (A)) ccl(int (f(A))) holds.

(c) For each subset B of Y, * (f-1(B)) c f-1(cl(int ((A)). Proof: (i) Assume that f : X � Y be *-continuous. Let G be open in Y.Then G c is closed in Y . Since f is *-continuous ,

f-1 (G c ) is *-closed in X. But f-1 (Gc )= X - f-1 (G) .Thus X -

f-1 (G) is *- closed in X and so f-1 (G) is *-open in X. There-

fore

(a) implies (b).

Conversely assume that the inverse image of each open set in

Y is *-open in X. Let F be any closed set in Y. The Fc is open

in Y. By assumption, f-1 (Fc ) is *-open in X. But f-1 (Fc ) = X

- f-1 (F). Thus X - f-1 (F) is *-open in X and so f-1 (F) is *-

closed in X. Therefore f is *-continuous.

Hence (b) implies (a). Thus (a) and (b) are equivalent.

(ii) Assume that f is *-continuous. Let A be any subset of X. Then cl(int(f (A)))is closed in Y. Since f is *-continuous ,

f-1 cl int(f(A)) is *-closed in X and it contains A. Therefore *(A)c f-1 cl int(f(A)) and so f(f3* (A) c

cl int(f(A)) .

(a) � (b).

Let y E *(A) and xE X , f (x) E V . Let V be any neighborhood of y. Then there exists a point x E X and a f3* -open set U such that f (x) = y , x E U , x E *(A),f (U) E V .Since x E f3* (A),UnA:;t q; holds and hence f (A)nV= q; .Therefore we have y = f(x) E cl(int(A).

(b) � (a).

Let x E X and V be a open set containing f (x) .Let A = f -1 (Vc ), then x ft. A.Sincef(f3* (A)) c cl(f(A)) c Vc .Then x ft. f3* (A) ,

there exist a f3* (A) open set U containing x such that U nA E q

and hence f (U ) c f (Ac ) c V .

set in Y. Then the inverse image f -1 (V ) is * -closed in X.

(b) �

(c)

Since every * -closed set is gs-closed,f-1 (V ) is *-closed in

X. Therefore f is * -continuous.

Let B be any subset of Y. Replacing A by f-1

(B) we get from

Example 3.13 Let X = Y = {a, b, c} with topologies " = {X, q, {a},

{a,b}} and a = {Y,q,{a},{c},{a,c}}. Let f : X � Y be a map defined by f (a) = b, f (b) = c, f (c) =b. Here f is *-continuous. But f is not sg-continuous since for the closed set F={ a,b} f is not *- closed set in X.

into a topological space Y

(i) The following statements are equivalent. (a) f is *-continuous.

(b) The inverse image of each open set in Y is *-open in X. (ii) If f : X � Y is *-continuous, then f ( *(A)) c cl(int (f(A))) for every subset A of X.

(iii) The following statements are equivalent.

(b),f f3* (f-1(B)) � cl int(f-1 (B)) � B. Hence

f3* (f-1 (B)) c f-1 cl(int(B)) .

(c) � (b)

Let B = f (A) where A is a subset of X. Then f3* (A) c f3* (f-1 (B)) c f-1 cl(int(A)) Therefore f(f3* (A)) c

cl int(f(A))

true as seen from the following example.

Example 3.16 Let X = Y = {a, b, c} with topologies " = {X, q, {a},

{a,b}} and a = {Y,q,{a},{b}, {a,b}}.Let f : X � Y be the identity

map. Here f is *-continuous. But f is not g-continuous since for the closed set F={ a,c} f is not * -closed set in X.Let X = Y =

{a, b, c} with topologies " = {X, q, {a}, {a,b}} and a = {Y,q,{a},{c},

{a,c}}. f : X � Y be a map defined by f (a) = a, f (b) = b, f (c) = c.

then for every subset A of X f(f3* (A)) c cl(int(A)) holds but it

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is not *-continuous since for a closed set {a,c} in Y

f-1 ({a, c}) = {a, b} is not *-closed in X .

Then g o f: X � z is *-continuous if g is continuous and f is

*-continuous

Proof: Let V be a closed set in Z, Since g is continuous, g-1(V) is closed in V. and since f is * continuous.f-1(g-1(V) is *-closed

in X . thus g@ f is * continuous.

not be *- continuous. Let us prove the remark by the follow- ing example.

Example 3.18 Let X = Y = Z = {a, b, c} with topologies " = {X, q,

{a}, {a, b}}, a = {Y, q,{a},{c}, {a, c}} and fJ = {Z, q, {a, c}}. Let g : (X, " ) �(Y, a) be a map defined by g(a) = a, g(b) = c and g(c) = b ,let f : (Z, fJ) � (X, " ) be a map defined by f (a) = a, f (b) = c and f(c) = b. Both f and g are *-continuous.

closed in X. But f-1(g-1(F))=(g o f )-1(F). Therefore g o f : X --+ Z is

*-continuous.

Example 4.8 Let X = Y = {a, b, c} with topologies "= {X, q, {a},

{b}, {a, b}} and a = {Y, q, {b}, {b, c}}. Then the identity map

f : (X, ") � (Y, a) is irresolute, but it is not *-irresolute. Since

G = {a,c} is *-closed in (Y, a),where f-1(G) = {a,c} is not *-clos ed in (X, ").

Example 4.9 Let X = Y = {a, b, c} with topologies " = {X, q, {a, b}} and a = {Y, q, {a}}. Then the identity map f : (X, ") � (Y, a)

is *-irresolute, but it is not irresolute. Since G = {a, c}

is semi-open in (Y, a), where f-1(G) = {a,c} is not semi-open in

(X, " ).

There fore it is evident that

Define g o f : (Z, fJ) � (Y, a). Here {b} is a closed set of (Y, a). Therefore (g o f) -1({b})= {c} is not a *-closed set of (Z, fJ). Hence g o f is not *-continuous.

�

irresolute

�

* irresolute

�

* continous

called *-irresolute if the inverse image of every *-closed set in Y is *-closed in X.

the inverse image of every * -open set in Y is * -open in X. Proof: Assume that f is *-irresolute. Let A be any *-open set in Y. Then Ac is *-closed set in Y. Since f is *-irresolute,

f-1(Ac) is *-closed in X.But f-1(Ac) = X -f-1(A) and so f-1(A) is * -

open in X. Hence the inverse image of every *-open set in Y is

*-open in X. Conversely assume that the inverse image of every * -open set in Y is *-open in X. Let A be any *-closed set in Y. Then Ac is *-open in Y. By assumption, f-1(Ac) is *-

open in X. But f-1(Ac) = X - f-1(A) and so f-1(A) is

*-closed in X. Therefore f is *-irresolute.

Y. Since every closed set is *-closed, F is *-closed in Y. Since f is *-irresolute, f-1(F) is *-closed in X. Therefore f is *-conti nuous.

as seen from the following example.

Example 4.5 Let X = Y = {a, b, c} with topologies " = {X, q, {a},

{a,b}} and a = {Y,q,{a},{c}, {a,c}} Let f : (X, ") � (Y, a) be map

defined by f (a) = a, f (b) = c and f (c) = b. . Here f is *- conti- nuous. But {a, c} is *-closed in Y but f-1({a, c}) = {a, b} is not

*-closed in X. Therefore f is not *-irresolute.

*- irresolute map f : X � Y and any *-continuous map g : Y

� Z, the composition gof :(X, ") � (Z, fJ) is *-continuous.

, g-1(F) is *-closed in Y. Since f is *-irresolute, f-1(g-1(F)) is *-

gy " and Y be a topological space with topology a.

Let f : (A, "/A) � (Y, a) and g :(B, "/B) � (Y, a) be *- conti-

nuous maps such that f(x) = g(x) for every x @A n B. Suppose that A and B are *-closed sets in X. Then the combination

a : (X, ") � (Y, a) is *-continuos.

Proof: Let F be any closed set in Y. Clearly a-1(F) = f-1(F) u

g-1(F)= C u D where C = f-1(F) and D = g-1(F). But C is *-closed

in A and A is *-closed in X and so C is *-closed in X. Since we have proved that if B@A@ X, B is *-closed in A and A is

*-closed in X then B is *-closed in X. Also C u D is *-closed

in X. Therefore a-1(F) is *-closed in X. Hence a is *- continous.

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