# On μ -Continuous Functions In Topological Spaces

DOI : 10.17577/IJERTV2IS4400

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#### On μ -Continuous Functions In Topological Spaces

Tuticorin 628 008

Department of Mathematics G.Venkataswamy Naidu College Kovilpatti.

In this paper, we introduce -continuous map and their relations with some generalized continuous maps. Various properties and characterizations of – continuous map are discussed by using -closure and -interior under certain conditions.

-continuous map, -closure, -interior.

2000 Mathematics Subject Code Classification : 54C05, 54C08.

1. Many others ([4] [5] [6] [13]) working in the field of general topology have shown more interest in studying the concepts of generalizations of continuous map. A weak form of continuous map called gcontinuous map was introduced by Balachandran, Sundaram and Maki [2].

M.K.R.S.Veerakumar has introduced several generalized closed sets namely, g -closed sets, g*-closed sets, g*pclosed sets, *gclosed sets, *gclosed sets, *gs-closed sets, -closed sets, p-closed sets and s-closed sets and their continuity. The concept of,

-closed sets was introduced by S.Pious Missier and E.Sucila [12]. In this paper we

introduce the concept of -continuous map in topological spaces.

Throughout this paper, we consider spaces on which no separation axioms are assumed unless explicity stated . For A X, the closure and interior of A is denoted by cl(A) and int(A) respectively. The complement of A is denoted by AC.

A subset A of a topological space (X, ) is called

1. a preopen set [9] if A int(cl(A)) and preclosed if cl(int(A)) A.

2. a semiopen set [6] if A cl(int(A)) and a semiclosed set if int(cl(A)) A.

3. an -open set [11] if A int(cl(int(A)) and -closed set if cl(int(cl(A))) A.

4. a semi preopen set [1] if A cl(int(cl(A))) and a semipreclosed set if int (cl(int(A))) A.

The intersection of all semiclosed (resp. preclosed, semipreclosed, -closed) sets containing a subset A of X is called semiclosure (resp. preclosure, semipreclosure, -closure) of A is denoted by scl(A) (resp. pcl(A), spcl(A), cl(A)).

A subset A of a topological space (X, ) is called

1. a generalized closed set (briefly gclosed [7] if cl(A) U whenever A U and U is open in (X, ).

2. an -generalized closed set (briefly gclosed ) [8] if cl(A) U whenever A U and U is open in (X, ).

3. a g -closed set [15] if cl(A) U whenever A U and U is semiopen in (X, ).

4. a *gclosed set [16] if cl(A) U whenever A U and U is g -open in (X, ).

5. a g*-closed set [16] if cl(A) U whenever A U and U is g-open in (X, ).

6. a g*-preclosed set (briefly g*pclosed ) [17] if pcl(A) U whenever A U and U is gopen in (X, ).

7. a *g- semiclosed set [20] (briefly *gs-closed ) if scl(A) U whenever A U and U is g -open in (X, ).

8. a *g-closed set [22] if cl(A) U whenever A U, and U is g -open in (X, ).

9. a g*-closed set [8] if cl(A) int (U) whenever A U and U is -open in (X, ).

10. a -closed set [19] if scl(A) U whenever A U and U is sgopen in (X, ).

11. a g* -closed set [19] if cl(A) U whenever A U and U is gopen in (X, ).

12. a -closed set [21] if cl(A) U whenever A U and U is g*-open in (X, ).

13. a -preclosed set (briefly pclosed ) [22] if pcl(A) U whenever A U and U is g*- open in (X, ) .

14. a -semiclosed set (briefly sclosed) [23] if scl(A) U whenever A U and U is g*-open in (X, ).

15. a -closed set [12] if scl(A) U whenever A U and U is -open in (X, ). The complement of -closed set is called – open set. The class of all -open (resp.

-closed) subsets of X is denoted by o(X, ), (resp. c(X, )).

A map f : (X, ) (Y, ) is called

1. semicontinuous [6] if f-1(V) is semiclosed in (X, ) for every closed set V in (Y, ).

2. g-continuous [2] if f-1(V) is g-closed in (X, ) for every closed set V in (Y, ).

3. -continuous [10] if f -1(V) is -closed in (X,) for every closed set V in (Y, ).

4. gcontinuous [8] if f -1(V) is g-closed in (X, ) for every closed set V in (Y, ).

5. g -continuous [15] if f -1(V) is g -closed in (X, ) for every closed set V in (Y, ).

6. *g-continuous [15] if f -1(V) is *g-closed in (X, ) for every closed set V in (Y, ).

7. *g- continuous [15] if f -1(V) is *g-closed in (X, ) for every closed set V in (Y, ).

8. g*- continuous [16] if f -1(V) is g*-closed in (X, ) for every closed set V in (Y, ).

9. g*p- continuous [17] if f -1(V) is g*p-closed in (X, ) for every closed set V in (Y, ).

10. *gs- continuous [20] if f -1(V) is *gs-closed in (X, ) for every closed set V in (Y, ).

11. g*- continuous [19] if f -1(V) is g*-closed in (X, ) for every closed set V in

(Y, ).

12. -continuous [21] if f -1(V) is -closed in (X, ) for every closed set V in (Y, ).

13. p- continuous [22] if f -1(V) is p-closed in (X, ) for every closed set V in (Y, ).

14. s- continuous [23] if f -1(V) is s-closed in (X, ) for every closed set V in

(Y, ).

A topological space (X, ) is called a

1. T -space [12] if every -closed set is closed.

2. T -space [12] if every -closed set is -closed.

3. sT -space [12] if every -closed set is semiclosed.

4. pT -space [12] if every -closed set is preclosed.

5. spT -space [12] if every -closed set is semipreclosed.

6. T -space [12] if every -closed set is -closed.

1. -CONTINUITY

We introduce the following definition.

A function f : (X, ) (Y, ) is called -continuous if f-1(V) is -closed subset of (X, ) for every closed subset V of (Y, ).

Every continuous (resp. semicontinuous) map is -continuous but not conversely.

The proof follows from the fact that every closed (resp. semiclosed) set is -closed.

Let X = Y = {a, b, c}, = {X, , {a}, {b, c}} and = {Y, , {b}}. Define a map f : X Y by f(a) = a, f(b) = b and f(c) = c is -continuous. However f is neither continuous nor semicontinuous, since for the closed set U = {a, c} in Y, f-1(U) = {a, c} which is neither closed nor semiclosed in X.

Every – continuous map is -continuous but not conversely.

Proof : The proof follows from the fact that every -closed set is -closed.

Let X = Y = {a, b, c}, = {X, , {b}, {c, a}} and = {Y, , {c}, {b, c}}.Define a map f : X Y by f(a) = a, f(b) = b and f(c) = c. This map is -continuous but not

-continuous, since for the closed set U = {a} in Y, f-1 (U) = {a} is not -closed in X.

Thus the class of all -continuous maps properly contains the classes of continuous maps, semicontinuous maps and -continuous maps.

The following examples shows that -continuity is independent of -continuity.

Let X = Y= {a, b, c}, = {X, , {b}, {a, c}} and = {Y, , {b, c}}. Define f : X Y by f(a) = a, f(b) = c, f(c) = b is -continuous but not -continuous, since for the closed set U = {a} in Y, f-1(U) = {a} is not -closed in X.

Let X = Y = {a, b, c}, = {X, , {b}, {a, b}} and = {Y, , {a}}. Let f : X Y be an identity map. Here the map f is -continuous but not -continuous, since for the closed set U ={b, c} in Y, f-1(U) = {b, c} is not -closed in X.

The following examples shows that -continuous is independent of p continuous and s-continuous.

Let X = Y = {a, b, c}, = {X, , {a}, {a, b}} and = {Y, , {c}}. Let f : X Y be an identity map. Then f is p continuous and s continuous but not -continuous, since for the closed set U = {a, b} in Y, f-1(U) = {a, b} is not -closed in X.

Let X = Y = {a, b, c}, = {X, , {a}, {c}, {a, c}} and = {Y, , {b, c}}. Define a map f : X Y by f(a) = a, f(b) = c and f(c) = b. Here the map f is -continuous but not

p-continuous, since for the closed set U = {a} in Y, f-1(U) = {a} is not p-closed in X.

Let X = Y = {a, b, c}, = {X, , {a}, {b, c}} and = {Y, , {b}}. Let f : X Y be an identity map. Then the map f is -continuous but not s continuous, since for the closed set U = {a, c} in Y, f-1(U) = {a, c} is not s closed in X.

The following examples shows that -continuous is independent of *g-continuous,

*g-continuous and *gs-continuous.

Let X = Y = {a, b, c}, = {X, , {a}, {b, c}} and = {Y, , {c}}. Define a map f : X Y by f(a) = a, f(b) = c and f(c) = b. Here the map f is -continuous but it is not

*g-continuous, *gcontinuous and *gscontinuous. Since for the closed set U = {a, b} in Y, f-1(U) = {a,c} which is not *g-closed, *g-closed and *gs-closed in X.

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

*g-continuous and *gs-continuous but not -continuous, since for the closed set U = {a, b} in Y, f-1(U) = {a, c} which is not -closed in X.

The following examples shows that -continuous is independent of g*-continuous, g-continuous, g continuous and g*p-continuous.

Let X = Y = {a, b, c}, = {X, , {b}, {c}, {b, c}} and = {Y, , {a, b}}. Let f : X Y be an identity map. Here the map f is -continuous but not g*-continuous, g-continuous, g-continuous and g*p-continuous. Since for the closed set U = {c} in Y, f-1(U) = {c} which is not g*-closed, g-closed, g-closed and g*p-closed in X.

Let X = Y = {a, b, c}, = {X, , {a}, {a, c}} and = {Y, , {c}}. Let f : X Y be an identity map. Then the map f is g*-continuous, g-continuous, g-continuous and g*p-continuous but not -continuous. Since for the closed set U = {a, b} in Y, f-1(U) = {a, b} is not -closed in X.

The following examples shows that -continuous is independent of g* -continuous.

Let X = Y = {a, b, c}, = {X, , {a}, {b, c}} and = {Y, , {b, c}}. Define a map f : X Y by f(a) = c, f(b) = a and f(c) = b. Here f is -continuous but not g* – continuous, since for the closed set U = {a} in Y, f-1(U) = {b} which is not g*-closed in X.

Let X = Y = {a, b, c}, = {X, , {b}, {a, b}} and = {Y, , {a}}. Let f : X Y be an identity map. Then f is g* – continuous but not -continuous, since for the closed set U = {b, c} in Y, f-1(U) = {b, c} is not -closed in X.

The following diagram shows the relationship established between -continuous function and some other continuous functions. A B (resp.A B) represents A implies B but not conversely (resp. A and B are independent of each other).

From the above Propositions and Examples, we have the following diagram.

s-continuous -continuous p-continuous | g*p-continuous g*-continuous

continuous g-continuous

-continuous -continuous g-continuous semicontinuous g*-continuous

*g-continuous *g-continuous *gs-continuous

The composition of two -continuous maps need not be -continuous.

Let X = Y = Z = {a, b, c}, = {X, , {a}, {a, c}}, = {Y, , {b, c}} and

= {Z, , {c}}. Define a map f : X Y by f(a) = b, f(b) = c and f(c) = a. Let g : Y Z be an identity map. Then both f and g are -continuous, but g f is not -continuous. Since for

the closed set U = {a, b} in Z, (g f )-1 (U) = f-1 (g-1(U)) = f-1({a, b}) = {c, a} which is not

-closed in X.

If f : X Y is -continuous and g : Y Z is continuous then their composition

g f : X Z is – continuous.

Clearly follows from the definitions.

A map f : (X, ) (Y, ) is -continuous if and only if f-1(U) is -open in (X, ), for every open set U in (Y, ).

Let f : X Y be -continuous and U be an open set in Y. Then f-1(Uc) is -closed in X. But f-1(Uc) = (f-1(U))c and f-1(U) is -open in X. Converse is similar.

2. – CLOSURE AND – INTERIOR

For every set E X we define the -closure of E to be the intersection of all

-closed sets containing E. In symbols, cl(E) = {A : E A, A C(X, )}.

For any E X, E cl(E) cl(E).

Since every closed set is -closed but not conversely.

If A B, then cl(A) cl(B)

Clearly follows from Definition 4.1.

If E is -closed, then cl(E) = E.

Clearly follows from Definition 4.1.

Let A be a subset of a topological space X. For any x X, x cl(A) if and only if U A for every -open set U containing x.

Necessity : Suppose that x cl(A). Let U be an -open set containing x such that

U A = and so A Uc. But Uc is -closed and hence cl(A) Uc. Since x Uc we obtain x cl (A) which is contrary to the hypothesis.

Sufficiency: Suppose that every -open set of X containing x meets A. If x cl(A), then there exists an -closed F of X such that A F and x F. Therefore, x Fc and Fc is an -open set containing x. But Fc A = . This is contrary to the hypothesis.

For any A X, int(A) is defined as the union of all -open sets contained in A. That is, int(A) = U{U : U A and U o(X, )}

For any set A X, int(A) int(A).

For any two subsets A1 and A2 of X,

1. If A1 A2, then int(A1) int(A2). (ii) int (A1 A2) int(A1) int(A2).

If A is -open, then A = int(A).

Clearly follows from the Definition 4.6.

Let A be a subset of a space X, then the following are true.

1. ( int(A))c = cl(Ac)

2. int(A) = ( cl(Ac))c

3. cl(A) = ( int(Ac))c

1. Let x ( int(A))c. Then x int(A). That is, every -open set U containing

x such that U A. Thus every -open set U containing x is such that U Ac . By Proposition 4.5, x cl(Ac) and therefore, ( int(A))c cl(Ac).

Conversely, let x cl(Ac). Then by Proposition 4.5, every -open set U containing x is such that U Ac . By definition 4.6, x int(A), hence x ( int(A))c and so

cl(Ac) ( int(A))c. Thus cl(Ac) = ( int (A))c

2. Follows by taking complements in (i)

3. Follows by replacing A by Ac in (i)

For a subset A of a topological space X, the following conditions are equivalent :

1. o (X, ) is closed under any union,

2. A is -closed if and only if cl(A) = A.

3. A is -open if and only if int(A) = A.

1. (ii) : Let A be an -closed set. Then by definition of -closure, we get

cl(A) = A. Conversely, assume that cl(A) = A. For each x Ac, x cl(A), by Proposition 4.5 there exist an -openset Gx such that Gx A = and hence x Gx Ac.

Therefore, we obtain Ac =

xAc

Gx. By (i) Ac is -open and hence A is -closed.

2. (iii) = Follows by (ii) and proposition 4.9.

3. (i) = Let {U/ } be a family of -open sets of X. Put U = U U . For

each x U, there exist (x) such that x U(x) U. Since U(x) is -open, x int(U) and so U = int(U). By (iii), U is -open. Thus o (X, ) is closed under any union.

In a topologial space X, assume that o (X, ) is closed under any union. Then

cl(A) is an -closed set for every subset A of X.

set.

Since cl(A) = cl ( cl(A)) and by Proposition 4.10, we get cl(A) is an -closed

Let f : X Y be a map. Assume that o(X, ) is closed under any union. Then the following are equivalent :

1. The map f is -continuous;

2. The inverse of each open set is -open;

3. For each point x in X and each open set V in Y with f(x) V, there is an

-open set U in X such that x U, f(U) V;

4. For each subset A of X, f( cl(A)) cl(f(A));

5. For each subset B of Y, cl(f-1(B) f-1(cl(B));

6. For each subset B of Y, f-1(int(B)) int(f-1(B)).

(i) (ii) By theorem 3.26

(i) (iii) : Suppose that (iii) holds and let V be an open set in Y and let x f-1(V).

Then f(x) V and thus there exist an -open set Ux such that x Ux and f(Ux) V. Now

x Ux f-1(V) and f-1(V) =

f is -continuous.

U Ux. By assumption f-1(V) is -open in X and therefore

x f 1 (V)

Conversely, suppose that (i) holds and let f(x) V. Then x f-1(V) which is -open in X, since f is -continuous. Let U = f-1(V). Then x U and f(U) V.

(iv) (i) Suppose that (i) holds and A be a subset of X. Since A f-1(f(A)). We have A f-1(cl(f(A))). Since cl(f(A)) is a closed set in Y, by assumption f-1(cl(f(A))) is an

-closed set containing A. Consequently, cl(A) f-1(cl(f(A))). Thus f( cl(A)) f(f-1cl(f(A))) cl(f(A)).

Conversely, suppose that (iv) holds for any subset A of X. Let F be a closed subset of Y. Then by assumption, f( cl(f-1(F))) cl(f(f-1(F))) cl(F) = F. Thus cl(f-1(F)) f-1 (F) and so f-1(F) is -closed.

1. (v) : Suppose that (iv) holds and B be any subset of Y. Then replacing A by f-1(B) in (iv) we get f( cl(f-1(B))) cl(f(f-1(B))) cl(B). Thus cl(f-1(B)) f-1(cl(B)). Conversely, suppose that (v) holds. Let B = f(A) where A is a subset of X. Then we have

cl(A) cl(f-1(B)) f-1(cl(f(A))) and so f( cl(A)) cl(f(A)).

2. (vi) : Let B be any subset of Y. Then by (v) we have cl(f-1(Bc)) f-1(cl(Bc)) and hence ( int f-1(B))c (f-1 int (B))c. Therefore we obtain f-1(int(B)) int(f-1(B)).

3. (i) : Suppose that (vi) holds. Let F be any closed subset of Y. We have f-1(Fc) = f-1(int(Fc)) int(f-1(Fc)) = ( cl(f-1(F)))c and hence cl(f-1(F)) f-1(F). By proposition 4.10 f-1(F) is -closed in X. Hence f is -continuous.

Let f : (X, ) (Y, ) be a -continuous map. If (X, ) is a T space, then f is continuous.

The proof follows from definition.

Let f : (X, ) (Y, ) be a -continuous map. If (X, ) is a T space, then f is

-continuous.

The proof follows from definition.

Let f : (X, ) (Y, ) is -continuous map. If (X, ) is a sT space, then f is semicontinuous.

The proof follows from definition.

If f : (X, ) (Y, ) is -continuous map. If (X, ) is a pT space, then f is precontinuous.

The proof follows from definition.

If f : (X, ) (Y, ) is -continuous map. If (X, ) is a spT space, then f is semiprecontinuous.

The proof follows from definition.

If f : (X, ) (Y, ) is -continuous map. If (X, ) is a T space, then f is

-continuous.

The proof follows from definition.

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