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 Total Downloads : 621
 Authors : A. Pushpalatha, K. Kavithamani
 Paper ID : IJERTV1IS10447
 Volume & Issue : Volume 01, Issue 10 (December 2012)
 Published (First Online): 28122012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Strongly C*G Continuous Maps In Topological Space
A. Pushpalatha
Professor Department of Mathematics, Government Arts College,
Udumalpet642 126, Tirupur District, Tamil Nadu, India.
Abstract
In this paper, we have introduced the concept of strongly c*gcontinuous, perfectly c*g continuous, c*g locally closed, c*g locally continuous in Topological space.
Key words: Strongly c*g continuous, perfectly c*g – continuous, c*glocally closed, c*g locally continuous.

Kavithamani
Research Scholar Karpagam University Coimbatore District Tamil Nadu, India

INTRODUCTION
Levine [3] introduced and investigated the concept of strong continuity in topological spaces. Sundaram [12] introduced strongly g continuous maps and perfectly g continuous maps in topological spaces. Pushapalatha [8] introduced strongly g* continuous and perfectly g* continuous maps in topological spaces. In this section we have introduced two strong forms of continuous maps in topological spaces, namely strongly c*g continuous maps, perfectly c*g continuous maps and study some of their properties .
The notation of a locally closed set in a topological space was introduced by Kurotowski & Sierpinski [13]. According to Bourbaki [20], a subset of a topological space X is locally closed in X if it is the intersection of an open set in X and closed set in X .Stone [14 ] has used the term FG for a locally closed subset . Locally closed sets are of some interest in the setting of local compactness,StoneCech Compactifications (or) Cech complete Spaces [15 ] . Sundaram [12 ] introduced the concept of generalized locally continuous function in topological space and investigated some of their properties.
Pushpalatha [8] introduced strongly generalized locally continuous functions & some of their properties in topological spaces. In the chapter, we have introduced the concept of c*g locally continuous functions and study some of their properties.

PRELIMINARIES DEFINITION: 2.1
A map f: XY from a topological space X into a topological space Y is called

Strongly continuous if f1(V) is both open and closed in X for each subset V in Y [3].

Perfectly continuous if f1(V) is both open and closed in X for each open subset V in Y [10].

generalized continuous(g continuous) if f1(V) is gopen in X for each open set V in Y [12].

Strongly g continuous if f 1(V) is both open in X for each g open set V in Y [12].

Perfectly g continuous if f 1(V) is both open and closed in X for each gopen set V in Y [12].

Strongly g* continuous if f
Example 3.3: Let X = Y = {a,b,c} with the
1(V) is both open in X for each g*
topologies
1 = { ,X,{a}{a,b}} & 2 =
open set V in Y [8].

Perfectly g* continuous if f 1(V) is both open and closed in X for each g*open set V in Y [8].


STRONGLY c*g CONTINUOUS MAPS IN TOPOLOGICAL SPACE
Definition: 3.1
A map f : X Y from a topological space X into a topological space Y is said to be strongly c*g continuous if the inverse image of every c*g open set in Y is open in X.
Theorem 3.2
If a map : X Y from a topological space X into a topological space Y in strongly c*g continuous, then it is continuous but not conversely.
Proof:Assume that f in strongly c*g continuous. Let G be any open set in Y. Since, every open set in c*g open, G is c*g open in Y. Since f is strongly c*g continuous, f1(G) is open in X. Therefore f is continuous.
The converse need not be true as seen from the following example.
{ ,Y,{a,b}}. Define a map f(X, 1 ) (Y, 2 ) be the identity. Then f is continuous. But f is not strongly c*g continuous since, for the c*g open set G={b} in Y, f1 (G) = {G} is not open in X. Theorem 3.4: If f :XY from a topological space X into a topological space Y is strongly continuous then it is strongly c*g continuous but not conversely.
Proof: Assume that f is strongly continuous. Let G be any c*g open set in Y. Since f is strongly continuous, f1 (G) open in X by the definition of strongly continuous. Therefore f is strongly c*g continuous.
The converse need not be true as seen from the following example.
Example 3.5: Let X =Y={a,b,c) with topologies ={,x,{a},{b},{a,b}} and =
{ ,y {a}}.Consider a map : (x, ) (y ,
) is defined by f(a)=f(c)=c & f(b)=b. Then f is strongly c*g continuous. But not strongly continuous. For the subset {a} of Y f1 ({a}) = {a} is open in X, but is not closed in X.
Theorem 3.6: If : X Y is strongly c*g continuous, then it is strongly g*continuous but not conversely.
Proof: Assume that f is strongly c*g continuous. Let G be any strongly g open set in Y. Since every strongly g open set is c*g open, G is c*g open in Y. Since f is strongly c*g continuous, f1 (G) is open in
. Therefore f is strongly g*continuous.
The converse need not be true as seen from the following example.
Example 3.7: Let X =Y={a,b,c} be topological spaces with the topologies =
{, X, {a}, {c}, {a,b}, {a,c}} and = { ,y {a,c}. Let f: (X, ) (y , ) be the identity map. Then f is strongly g* continuous, but not strongly c*g continuous. For, {b} is a c*g open in Y, but f ({b} = {b } is not open in X .
Therorem 3.8: A map f: (X Y ) from a topological spaces X into a topological space Y is strongly c*g continuous if and if only if the inverse image of every c*g closed set in Y is closed in X.
Proof: Assume that f is strongly c*g continuous. Let G be any c*g closed set in
Y. Then Gc is c*g open in Y. Since f is strongly c*g continuous f1(Gc) open in
X. But f1 (Gc) =X f1 (G) and so f1 (G) is closed in X.
Conversely assume that the inverse image of every c*g closed set in Y is closed in x. Let G be any c*g open set in Y. Then Gc is c*g closed in Y. By assumption, f1 (Gc) is closed in X. But f1 (Gc) = X – f1 (Gc) and so f1 (G) is open in X. Therefore f is strongly c*g continuous.
Remarks3.9: From the above observation we get the following diagram.
Strongly continuity
Strongly c*g continuous
Continuity.
In the above diagram none of the implications can be reversed.
Theorem 3.10: If a map f : X Y is strongly c*g continuous and a map g: Y
Z is c*g continuous , then the composition g f: X Z is continuous
Proof: Let G be any open set in Z . Since g is c*g continuous , g1 (G) is c*g open in
Y. Since f is strongly c*g continuous, f 1
[g1 (G) ] is open in X. But (g f)1(G) = f – 1[g1 (G) ]. Therefore g f is continuous.Definition 3.10 A map f : X Y is said to be perfectly c*g continuous if the inverse image of every c*g open set in Y is both open and closed in X.
Theorem 3.11: A map f : X Y from a topological space X into a topological space Y is perfectly c*g continuous, then it is strongly c*g continuous but not conversely.
Proof: Assume that f is perfectly c*g continuous. Let G be any c*g open set in
Y. Since f is perfectly c*g continuous,f1(G) is open in X. Therefore f is strongly c*g continuous.
The converse need not be true as seen from the following example.
Example 3.12: Let X =Y= {a,b,c} ,with topologies 1 ={,x,{a},b {a,b}} and 2 =
{ 1,y, {a,b}. Define f: (X, 1 ) (y , 2 ) as the identity map. Then f is strongly c*g contnuous but not perfectly c*g continuous. Since, for the c*g open set G={a} in Y, 1 (G) = {G} is open but not closed in X.
Theorem 3.13: If a map f : X Y is perfectly c*g continuous then it is perfectly g* continuous but not conversely.
Proof: Assume that f is perfectly c*g continuous. Let G be a a c*g open set in
Y. Then G is c*g open in Y. Since f is perfectly c*g continuous, f1 (G) is both open and closed in X . Therefore f is perfectly g* continuous.
The converse need not be true as seen from the following example.
Example 3.14:Let X =Y={a,b,c) with topologies ={,X,{a},{b,c}} and =
{ ,Y {a}}.Define a map f: (X, ) (Y ,
) as the identify function. Then f is perfectly g* continuous, but not perfectly c*g continuous, since for the c*g open set
{b} in Y f1 {(b) ={b} is not both open and closed in X.
Theorem 3.15: If a map f : X Y from a topological space X into a topological space Y is perfectly c*g continuous if and only if f1(G) is both open and closed set in X for every c*g closed set G in Y.
Proof: Assume that f is perfectly c*g continuous. Let F be any c*g closed set in
Y. Then Fc is c*g open set in Y. Since f is perfectly c*g continuous, f1 (Fc ) is
both open & closed in X . But f1 (Fc ) = X f1 (F ) and also f1 (F ) is both open and closed in X.
Conversely assume that the inverse image of every c*g closed set in Y is both open and closed in X. Let G be any c*g open set in Y
. Then Gc is c*g closed in Y. By assumption f1 (Gc ) is both open and closed in Y. But f1 (Gc ) = X f1 (G) and so f1 (G) is both open and closed in Y. Therefore f is perfectly c*g continuous.
Remark 3.16: From the above observations we have the following implications and none of them are revereable.
Perfectly c*gcontinuity
Strongly c*gcontinuity
Strongly g* continuous
Continuity.

Definition 4.1:
A subset S of X is called c*g locally closed set [c*glc–set] if S= AB, Where A is c*g open in X and B is c*g closed in X
.C*GLC(X) denotes the class of all c*g sets in X.
Theorem 4.2: If a subset S of X is locally closed then it is c*g locally closed but not conversely.
Proof :Let S= PQ, Where P is open in X and Q is closed in X . Since every open set is c*g open and every closed, S is c*g locally closed in X.
The converse need not be true as seen from the following examples.
Example 4.3: Consider the topological space x = {a ,b, c} with topology =
{ ,X,{a}}. Then the set {a,c} c*g – locally closed but is not locally closed.
Theorem 4.4: If a subset S of X is strongly generalized locally closed in X then S is c*g locally closed but not conversely.
Proof :Let S= PQ, where P is strongly g open and q is strongly g closed in X. Since strongly g open implies c*g open and strongly g closed implies c*g closed , S is c*g locally closed set in X.
Example 4.4: Consider the topological space X = {a ,b, c} with topology =
{ ,X,{b}}. Then the set {a,b} c*g – locally closed but is not strongly generalized locally closed.
Theorem 4.5: If a subset S of X is c*g locally closed in X, then S is regular generalized locally closed but not conversely.
Proof : Let S= PQ, Where P is c*g locally closed and Q is c*g locally closed in X.Since c*g locally closed implies rg closed and c*g locally open implies rg open. Therefore S is regular generalized locally closed.
Example 4.6: Let X={a,b,c,d}, ={,X,{a}{b},{a,b}}.Then {d} is rg locally closed but is not c*g locally closed set in X.
Theorem 4.7: If A is c*g locally closed in X and B is c*g open (respectively closed) in X, then A B is c*g locally closed in X. Proof :There exist a c*g – open set P and a c*g closed set Q such that A =PQ. Now, AB= (PQ)B = (PB)Q [
respectively AB = P(QB)]. Since PB is c*g open [respectively QB is c*g closed], AB is c*g – locally closed in X.
Definition 4.8:
A subset S of a topological space X is called c*glc* set if S= PQ where P is c*g open in X and Q is closed in X.
Definition 4.9:
A subset S of a topological space X is called c*glc** set if S= PQ where P is open in X and Q is c*g closed in X.
Theorem 4.10:

If A is c*glc* set in X and B is c*g open (or closed), then A B is c*glc* set in X.

If A is glc ** set in X and B is closed then A B is c*glc**.
Proof :

Since A is c*glc* set ,there exist a c*g open set P and a closed set Q .Such that A = PQ. Now AB = (PQ) B = ( PB)
Q. Since PQ is c*g open and Q is closed, AB is c*glc* set. In the case of B being a closed set, we have
AB = (PQ) B = P (QB). Since P is c*g open and QB is closed, AB is c*glc* set.

Since A is c*glc**, there exist an open set P and a c*g closed set Q such that A= (PQ). Now AB = (PQ) B = P (Q
B). Since Q is c*g closed and B is closed , QB is c*g closed. Therefore , AB is c*glc** set .
Theorem 4.11: A subset A of a topological space X is c*glc*set if and only if there exists a c*g open set P such that A= Pcl(A).
Proof :Assume that A is c*glc*set. There exists a c*g open set P and a closed set Q such that A = PQ. Since A Q and Q is closed, A cl(A) Q. Then A P and A cl(A), and hence A Pcl(A). To prove the reverse inclusion let X
Pcl(A).Then X P and X cl(A) Q and so X PQ =A. Hence Pcl(A) A. Therefore A=Pcl(A).
Conversely assume that there exist a c*g open set P such that A= Pcl(A). Now P is c*g open set and cl(A) is closed. Therefore A is c*glc* – set.
Theorem 4.12: If a subset A of a topological space X is c*glc** – set then there exists an open set P such that A = Pcl*(A), where cl*(A) is the closure of A as defined by Dunham [19].
Proof : By definition there exist an open set P and a c*g closed set Q such that A= PQ. Then, since A cl* (A) Q, We have A Pcl*(A). Conversely , if X Pcl*(A),then X Q and X P. Then, X QP = A and hence Pcl*(A) A. Therefore A = Pcl*(A).
Theorem 4.13: If A and B are c*glc* set in a topological space X then AB is c*glc* set in X.
Proof :From the assumptions there exist c*g open sets P and Q such that
A = Pcl(A) and B = Qcl(B).Then AB
= ( PQ) [( cl (A) cl(B)]. Since PQ is c*gopen and cl(A)cl(B) is closed , AB is c*glc* – set.


Notations: – LC(X) denotes the class of all locally closed sets in a topological space X and C*GLC(X) denotes the class of all c*glc sets in X.
Similarly, C*GLC*(X) [respectively C*GLC**(X) ] denotes the class of all c*glc*sets [ respectively c*glc** – sets ]
Ganster and Reilly [2] have proved that
Continuity
LC irresolute
LC continuity Pushpalatha [17] has proved that
LC continuity
S*GLC irresoluteness
S*GLC continuity
GLC continuous.
But none of these implications can be reversed. Also they observed that the composition of two S*GLC irresolute functions is S*GLC irresolute and the composition of a S*GLC continuous function is S*GLC continuous.
Definition 5.1.
A function f : XY from a space X into a space Y is called

LC irresolute [2] if f1 (V) LC(X) for each V in LC(Y).

S*GLC irresolute [17] if f1 (V)
S*GLC(X) for each V S*GLC(X).

LCcontinuous [2] if f1(V)LC(X) for each open set V in Y .

S*GLC – continuous [17] if f1 (V)
LC(X) for each open set V in Y.
Definition 5.2.
A function f: xy from a space X into a space Y is called

C*GLCirresolute if f1(V) C*GLC(X) for each V C*GLC(X).

C*GLCcontinuous if f1 (V)
C*GLC(X) for each open et V in Y.

C*GLC* irresolute (respectively C*GLC** irresolute) if f1(V) C*GLC*(X) (respectively f1(V)
C*GLC**(X)) for each V C*GLC*(Y) (respectively V C*GLC**(X)).

C*GLC* – continuous (respectively C*GLC** continuous) if f1(V) C*GLC*(X) ( respectively f1 (V) C*GLC**(X) ) for each open set V inY.
Theorem 5.3: If a function f : XY from a space X into a space Y is LC continuous then it is C*GLC continuous but not conversely .
Proof: – Assume that f is LC continuous. Let V be an open set in Y. Then f1(v) is locally closed in Y. But locally closed sets are c*g locally closed sets. Therefore f1(V)
C*GLC(X) and so f is C*GLC continuous. The converse need not be true as seen from the following example.
Example 5.4: Let X = Y = {a,b,c},
= { ,x,{a}} and be the discrete topology. Define f : (X,) (Y,) as the identity function .Then f is not LCcontinuous. Because {b} is open in Y but f1{b} = {b} is not locally closed in y, clearly f is C*GLC continuous.
Theorem 5.5: If function f: XY from a space X into a space Y is C*GLC irresolute then it is C*GLC continuous.
Proof: – Let V be open in Y. Since every open set is c*g open set and every c*g set open set is c*g – locally closed , V C*GLC(Y). Since f is C*GLC irresolute, f 1(V)C*GLC(X). Therefore f is C*GLC continuous. Thus we have the following implications
i) Continuity
LC irresolute
LC continuity
C*GLC continuous
ii)
C*GLCirresoluteness
C*GLC continuity
However none of the above implications can be reversed.
Theorem 5.6: If function f: XY from a space X into a space Y be C*GLC continuous and A be a c*g open subset of X (respectively closed). Then the restriction f/A: AY is C*GLC continuous.
Proof: – Let V be open in Y.Let f 1(V) = W. Then W is c*glc in X.Since f is C*GLC continuous .Let W= PQ where P is c*g – open in X and Q is c*g – closed in
X. Now (f/A)1 (V) = WA = (PQ) A = (PA )Q.
But PA [respectively AG] is c*g closed by [18] is c*g open in X and so the restriction f/A is C*GLC continuous.
Theorem 5.7: (i) Let f : XY be C*GLC continuous and B be an open subset of Y containing f(X). Then f : XB is C*GLC continuous.

If f: XYand g : YZ are both C*GLC irresolute then the composition g f: XZ is C*GLC irresolute.

If f : XY is C*GLC continuous and g : YZ is continuous then the composition g f : XZ is C*GLC continuous.
Proof : – (i) Let V be open in B.Since B is open in Y, the set V is open in Y .Therefore f1(V) ic c*glc in X. Hence f :
XB is C*GLC continuous.

Let V be c*glc set in Z. Since g is C*GLC irresolute, g1 (V) is c*glc in Y. Since f is C*GLC irresolute,
f1( g1(V)) is c*glc in X .But f1( g1(V)) = (g f)1(V) and so g f is C*GLC irresolute.

Let V be open in Z. Since g is continuous g1(V) is open in Y. Since f is C*GLC continuous, f1( g1(V)) is c*glc in X.But f1( g1(V)) =(g f)1(V) and so g f is C*GLC continuous.

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