Strongly C*G- Continuous Maps In Topological Space

Strongly C*G- Continuous Maps In Topological Space

A. Pushpalatha

Professor Department of Mathematics, Government Arts College,

Udumalpet-642 126, Tirupur District, Tamil Nadu, India.

Abstract

In this paper, we have introduced the concept of strongly c*g-continuous, 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*g-locally closed, c*g locally continuous.

1. Kavithamani

   Research Scholar Karpagam University Coimbatore District Tamil Nadu, India

   1. 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,Stone-Cech 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.

   2. PRELIMINARIES DEFINITION: 2.1

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

      1. Strongly continuous if f-1(V) is both open and closed in X for each subset V in Y [3].

      2. Perfectly continuous if f-1(V) is both open and closed in X for each open subset V in Y [10].

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

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

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

      1. 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].

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

   3. 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, f-1(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, f-1 (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, f-1 (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 f-1 ({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, f-1 (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 f-1(Gc) open in

      X. But f-1 (Gc) =X- f-1 (G) and so f-1 (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, f-1 (Gc) is closed in X. But f-1 (Gc) = X – f-1 (Gc) and so f-1 (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 , g-1 (G) is c*g- open in

      Y. Since f is strongly c*g- continuous, f -1

      [g-1 (G) ] is open in X. But (g f)-1(G) = f – 1[g-1 (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,f-1(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, f-1 (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 f-1 {(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 f-1(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, f-1 (Fc ) is

      both open & closed in X . But f-1 (Fc ) = X- f-1 (F ) and also f-1 (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 f-1 (Gc ) is both open and closed in Y. But f-1 (Gc ) = X- f-1 (G) and so f-1 (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*g-continuity

      Strongly c*g-continuity

      Strongly g*- continuous

      Continuity.

   4. 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:

      1. 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.

      2. If A is glc **- set in X and B is closed then A B is c*glc**.

      Proof :-

      1. 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.

      2. 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*g-open 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

1. LC- irresolute [2] if f-1 (V) LC(X) for each V in LC(Y).

2. S*GLC- irresolute [17] if f-1 (V)

   S*GLC(X) for each V S*GLC(X).

3. LC-continuous [2] if f-1(V)LC(X) for each open set V in Y .

4. S*GLC – continuous [17] if f-1 (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

1. C*GLC-irresolute if f-1(V) C*GLC(X) for each V C*GLC(X).

2. C*GLC-continuous if f-1 (V)

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

3. C*GLC* irresolute (respectively C*GLC**- irresolute) if f-1(V) C*GLC*(X) (respectively f-1(V)

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

4. C*GLC* – continuous (respectively C*GLC**- continuous) if f-1(V) C*GLC*(X) ( respectively f-1 (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 f-1(v) is locally closed in Y. But locally closed sets are c*g- locally closed sets. Therefore f-1(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 LC-continuous. Because {b} is open in Y but f-1{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*GLC-irresoluteness

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.

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

2. 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 f-1(V) ic c*glc in X. Hence f :

XB is C*GLC- continuous.

1. Let V be c*glc set in Z. Since g is C*GLC- irresolute, g-1 (V) is c*glc in Y. Since f is C*GLC irresolute,

   f-1( g-1(V)) is c*glc in X .But f-1( g-1(V)) = (g f)-1(V) and so g f is C*GLC- irresolute.

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

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