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 Authors : S. Pious Missier, E.Sucila
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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.

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, *gsclosed sets, closed sets, pclosed sets and sclosed 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

semicontinuous [6] if f1(V) is semiclosed in (X, ) for every closed set V in (Y, ).

gcontinuous [2] if f1(V) is gclosed in (X, ) for every closed set V in (Y, ).

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

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

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

*gcontinuous [15] if f 1(V) is *gclosed in (X, ) for every closed set V in (Y, ).

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

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

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

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

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

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

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

s continuous [23] if f 1(V) is sclosed in (X, ) for every closed set V in
(Y, ).
A topological space (X, ) is called a

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

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

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

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

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

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

CONTINUITY
We introduce the following definition.
A function f : (X, ) (Y, ) is called continuous if f1(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, f1(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, f1 (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, f1(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, f1(U) = {b, c} is not closed in X.
The following examples shows that continuous is independent of p continuous and scontinuous.
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, f1(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
pcontinuous, since for the closed set U = {a} in Y, f1(U) = {a} is not pclosed 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, f1(U) = {a, c} is not s closed in X.
The following examples shows that continuous is independent of *gcontinuous,
*gcontinuous and *gscontinuous.
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
*gcontinuous, *gcontinuous and *gscontinuous. Since for the closed set U = {a, b} in Y, f1(U) = {a,c} which is not *gclosed, *gclosed and *gsclosed 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 *gcontinuous,
*gcontinuous and *gscontinuous but not continuous, since for the closed set U = {a, b} in Y, f1(U) = {a, c} which is not closed in X.
The following examples shows that continuous is independent of g*continuous, gcontinuous, g continuous and g*pcontinuous.
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, gcontinuous, gcontinuous and g*pcontinuous. Since for the closed set U = {c} in Y, f1(U) = {c} which is not g*closed, gclosed, gclosed and g*pclosed 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, gcontinuous, gcontinuous and g*pcontinuous but not continuous. Since for the closed set U = {a, b} in Y, f1(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, f1(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, f1(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.
scontinuous continuous pcontinuous  g*pcontinuous g*continuous
continuous gcontinuous
continuous continuous gcontinuous semicontinuous g*continuous
*gcontinuous *gcontinuous *gscontinuous
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) = f1 (g1(U)) = f1({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 f1(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 f1(Uc) is closed in X. But f1(Uc) = (f1(U))c and f1(U) is open in X. Converse is similar.

– 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,

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.

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

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

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


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

Follows by taking complements in (i)

Follows by replacing A by Ac in (i)
For a subset A of a topological space X, the following conditions are equivalent :

o (X, ) is closed under any union,

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

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

(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.

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

(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 :

The map f is continuous;

The inverse of each open set is open;

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;

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

For each subset B of Y, cl(f1(B) f1(cl(B));

For each subset B of Y, f1(int(B)) int(f1(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 f1(V).
Then f(x) V and thus there exist an open set Ux such that x Ux and f(Ux) V. Now
x Ux f1(V) and f1(V) =
f is continuous.
U Ux. By assumption f1(V) is open in X and therefore
x f 1 (V)
Conversely, suppose that (i) holds and let f(x) V. Then x f1(V) which is open in X, since f is continuous. Let U = f1(V). Then x U and f(U) V.
(iv) (i) Suppose that (i) holds and A be a subset of X. Since A f1(f(A)). We have A f1(cl(f(A))). Since cl(f(A)) is a closed set in Y, by assumption f1(cl(f(A))) is an
closed set containing A. Consequently, cl(A) f1(cl(f(A))). Thus f( cl(A)) f(f1cl(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(f1(F))) cl(f(f1(F))) cl(F) = F. Thus cl(f1(F)) f1 (F) and so f1(F) is closed.

(v) : Suppose that (iv) holds and B be any subset of Y. Then replacing A by f1(B) in (iv) we get f( cl(f1(B))) cl(f(f1(B))) cl(B). Thus cl(f1(B)) f1(cl(B)). Conversely, suppose that (v) holds. Let B = f(A) where A is a subset of X. Then we have
cl(A) cl(f1(B)) f1(cl(f(A))) and so f( cl(A)) cl(f(A)).

(vi) : Let B be any subset of Y. Then by (v) we have cl(f1(Bc)) f1(cl(Bc)) and hence ( int f1(B))c (f1 int (B))c. Therefore we obtain f1(int(B)) int(f1(B)).

(i) : Suppose that (vi) holds. Let F be any closed subset of Y. We have f1(Fc) = f1(int(Fc)) int(f1(Fc)) = ( cl(f1(F)))c and hence cl(f1(F)) f1(F). By proposition 4.10 f1(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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