sc*g-Homeomorphisms in Topological Spaces

sc*g-Homeomorphisms in Topological Spaces

A. Pushpalatha

Department of Mathematics, Government Arts College

Udumalpet-642 126, Tirupur District Tamil Nadu, India

R. Nithyakala

Department of Mathematics Vidyasagar College of Arts and Science Udumalpet-642 126, Tirupur District Tamil Nadu, India

Abstract

In this paper, we introduced a new class sc*g- Homeomorphisms in Topological space.

Key words: Homeomorphism – strongly g-homeomorphism- sc*g – homeomorphism.

1. Introduction

Several mathematicians have generalized homeomorphisms in topological spaces. Biswas[18], Crossely and Hildebrand[19], Gentry and Hoyle[20] and Umehara and Maki[21] have introduced and investigated semi-homeomorphisms somewhat homeomorphisms and g-A- homeomorphisms Crossely and Hildebrand defined yet another semi-homeomorphism which is also a generalization of homeomorphisms. Sundaram[6] introduced g-homeomorphisms and gc- homeomorphisms in topological spaces.

In the section, we introduce the concept of sc*g- homeomorphisms and study some of their properties.

Definition: 2.7.1 A bijection f : (X, ) (Y, ) from a topological space X into a topological space Y is called a sc*generalized homeomorphism (sc*g- homeomorphism) if f is both sc*g-open and sc*g-continuous.

Theorem: 2.7.2 Every homeomorphism is a sc*g-homeomorphism.

Proof : Since every continuous function is sc*g- continuous and every open map is sc*g-open, the proof follows.

The converse of the above theorem need not be true as seen from the following example.

Example: 2.7.3 Consider the topological spaces X = Y= {a, b, c} with topologies ={ ,X,{a, b}} and = { ,Y,{a}, {a, b}}. Then the identity

map f : (X, ) (Y, ) is a sc*g- homeomorphism but not a homeomorphism. Since for the open set {a} in Y is not open in X.

Theorem:2.7.4 Everystrongly g- homeomorphism is a sc*g-homeomorphism but not conversely.

Proof : Let f : X Y be a strongly g- homeomorphism. Then f is strongly g -continuous and strongly g -open. Since every strongly g – continuous function is sc*g-continuous and every strongly g -open map is sc*g-open, f is sc*g- continuous and sc*g-open. Hence f is a sc*g- homeomorphism.

The converse of the above theorem need not be true as seen from the following example.

Example : 2.7.5 Let X=Y={a, b, c} with

={ ,X,{a}}and = { ,Y,{a, b}} respectively.

Then the identity map f : (X, ) (Y, ) is sc*g-homeomorphism but not a strongly g- homeomorphism. Since for the open set {a, b} in Y is not a strongly g-open set in X.

Next we shall characterize the sc*g- homeomorphism and sc*g-open maps.

Theorem : 2.7.6 For any bijection f : X Y the following statements are equivalent.

(a)f -1: Y X is sc*g-continuous. (b)f is a sc*g-open map.

(c)f is a sc*g-closed map.

Proof : (a) (b) Let G be any open set in X. Since f -1 is sc*g-continuous, the inverse image of G under f -1, namely f(G) is sc*g-open in Y and so f

is sc*g-open map.

1. (c) Let f be any closed set in X. Then Fc is open in X. Since f is sc*g-open, f(Fc) is

   sc*g-open in Y. But f(Fc) = Y- f(F) and so f(F) is

   sc*g-closed in Y. Therefore f is a sc*g-closed map.

2. (a) Let F be any closed set in X. Then

(f -1)-1(F) = f(F) is sc*g-closed in Y . Since f is a sc*g-closed map. Therefore f -1 is sc*g-continuous.

Theorem : 2.7.7 Let f : (X, ) (Y, ) be a bijective and sc*g-continuous map, the following statements are equivalent.

(a)f is a sc*g-open map.

1. f is a sc*g-homeomorphism.

2. f is a sc*g-closed map.

Proof : (a)(b) By assumption, f is bijective and sc*g-continuous and sc*g-open. Then by definition, f is sc*g- homeomorphism.

(b)(c) By assumption, f is sc*g-open and bijective. By theorem 2.7.4 f is sc*g-closed map. (c)(a) By assumption, f is sc*g-closed and bijective. By theorem 2.7.4 f is sc*g-open map.

The following example shows that the composition of two sc*g-homeomorphisms need not be sc*g- homeomorphism.

Example : 2.7.8 Consider the topological spaces X = Y= Z= {a, b, c} with topologies

1={ , X, {a}, {a, b}}, 2={ , Y, {a}}and

3={ , Z, {b, c}} respectively. Let f and g identity maps such that f : X Y and g : Y Z. Then f and g are sc*g-homeomorphisms but their

composition g f : X Z is not a sc*g-

homeomorphism. For the open set {a, b} in X, g(f{a, b}) = {a, b} is not sc*g-open set in Z. Definition :2.7.9 A bijection f : (X, ) (Y, ) is said to be a (sc*g)*-homeomorphism if f and its inverse f -1are sc*g-irresolute maps. Notation : Let family of all (sc*g)*- homeomorphisms from (X, ) onto itself be denoted by (sc*g)*h(X, ) and family of all sc*g- homeomorphisms from (X, ) onto itself be

denoted by sc*g h(X, ). The family of all homeomorphisms form from (X, ) onto itself be denoted by h(X, ).

Theorem :2.7.10 Let X be a Topological space.

Then

1. The set (sc*g)*h(X) is a group under composition of maps.

   (ii)h(X) is a subgroup of (sc*g)*h(X). (iii)(sc*g)*h(X) sc*g h(X).

   Proof: (i) Let f, g (sc*g)*h(X). Then g h (sc*g)*h(X) and so (sc*g)*h(X) is closed under the composition of maps. The composition of maps is associative. The identity map i : xx is a (sc*g)*- homeomorphism and so i (sc*g)*h(X). Also f . i =

   i. f = f for every f (sc*g)*h(X). If f (sc*g)*h(X),

   then f -1 (sc*g)*h(X) and f. f -1 = f -1. f = i. Hence (sc*g)*h(X) is a group under the

   composition of maps.

2. Let f : X Y be a homeomorphism. Then by theorem 2.6.5 both f and f -1 are

   (sc*g)*-irresolute and so f is a (sc*g)*- homeomorphism. Therefore every homeomorphism is a (sc*g)*-homeomorphism and so h(X) is a subset of (sc*g)*h(X). Also h(X) is a group under the composition of maps. Therefore h(X) is a subgroup of the group (sc*g)*h(X).

3. Since every (sc*g)*-irresolute map is sc*g- continuous, (sc*g)*h(X) is a subset of sc*gh(X).

From the above observations we get the following diagram:

Homeomorphism strongly g-homeomorphism sc*g – homeomorphism

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