 Open Access
 Authors : R. Rajeswari , S. Darathi , D. Deva Margaret Helen
 Paper ID : IJERTV9IS020255
 Volume & Issue : Volume 09, Issue 02 (February 2020)
 Published (First Online): 02032020
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Regular Strongly Compactness and Regular Strongly Connectedness in Topological Space
1R. Rajeswari, 2S. Darathi, 3D. Deva Margaret Helen
1Assistant professor in Mathematics,
2,3Masters in Mathematics
1,2,3Thassim Beevi Abdul Kader College for Women Kilakarai, Ramanathapuram.
Abstract:In this paper we proved that the regular strongly separated sets in the topological space.Also we extend our work in regular strongly compactness and connectness in the topological space.

INTRODUCTION
M.Karpagadevi and A.Pushpalatha already proved the RWClosed maps and RW Open maps in topological spaces later they worked on RWcontinuous map and RWirresolute maps in the topological space. M.Karpagadevi and A.Pushpalatha already proved that regular weakly separated sets in the topological space.We introduce the structure of regular strongly separated sets.Also we give the concept of regular strongly compactness and regular strongly connectedness in the topological space.

PRELIMINARIES

Definition:
In a topological space X is called regular space if every closed subset C of X and the point p not contained in C admit non overlapping open neighourhoods.Thus p and C can be separated by neighbourhoods this is known as regular space.

Definition:
A topological space X is called separated if given any two distinct points x and y are disjoint by neighourhoods.

Definition:
A topological space X is called compact if every open cover has a finite subcover that covers X.

Definition:
A topological space X is called complete if every Cauchy sequence has converges in X.

Definition:
The weak topology on X is the initial topology with respect to dual space family X*.It is also known as Coarsest topology such that each element of X* remains the continuous function and this is differ from ordinary topology that ordinary topology is called strong topology.

Definition:
Let (X,) be a topological space, two non empty subsets A and B are said to be weakly separated if A cl(B)= and rw cl(A) =

Analogy:
Two regular weakly separated sets are always disjoint.

Definition
A collection {Ai : iI} of regular weakly open sets in a topological space X is called a rwopen cover of a subset B in X if B iI
{Ai : iI}

Definition:
A topological space X is regular weakly compact if every rwopen cover of X has finite sub cover of X.

Definition:
A subset B of a topological space X is called rwcompact relative to X if for every collection {Ai : iI} of rwopen subsets of x such that B iI {Ai iI} there exist a finite subset I0 of I such that B iIo {Ai : iI0}.

Definition:
A subset B of a topological space X is called rwcompact if B is rwcompact as the subspace of X.

Definition:
A topological space X is regular weakly connected if X cannot be written as a union of two disjoint non empty rwopen sets.


MAIN RESULT

Definition:
Let (X,) be a topological space, two non empty subsets A and B are said to be strongly separated if there exist an open sets U and V such that AU;BV and UV=.

Theorem:
Two regular strongly separated sets are always disjoint.
Proof:
Let A and B be regular strongly sets,by the 3.1 definition there exist an openset U and V such that AU;BV and UV=.This implies that AB=.Hence A and B are disjoint.

Example:
If X={a,b,c,d,e,f,g,h}, ={X,,{c,d},{d,f},{a,e},{a,g},{b,c,d,e},{a,c,d}},U={c,d,f} then the interior is U0={c,d,f} so U is an open set,let A={c,d}[A ] and V0={a,e,g} so V is an open set,let B={a,e} [BV] if UV= then AB=.

Theorem:
Let A and B be two regular strongly separated sets of (X,).If CA and DB then C and D are regular strongly separated.
Proof:
Let A and B be regular strongly separated sets of a topological space (X,) then by the definition there exist the open sets U,V such that AU,BV
UV=.let CA and DB then we have two open set U,V then CU,D , UV= thus C and D are regular strongly separated.

Example:
We have to follow 3.3 example let C={d} ,D={e} it follows that CU,D , UV= thus C and D are regular strongly separated.

Theorem:
Two regular strongly closed subset of a topological space (X,) are rsseparated iff they are disjoint. Proof:
If rsseparatedd sets are disjoint then rs closed separated sets are disjoint. Conversely,
Let A and B be two disjoint rs closed sets,since AU,BVthen we have rscl(A)=A,rscl(B)=B and AB= this implies that U = .consequently A cl(B)= and rscl(A) = hence A and B are rsseparated.

lemma:
If the union of two rsseparated sets in a rsclosed set then the individual sets are rsclosures are rsclosures of themselves.

Theorem:
Two disjoint sets A and B are rsseparated in (X,) iff they are both rsopen and rsclosed in A . Proof:
Two disjoint sets A and B are rsseparated in (X,) then A cl(B)= and rscl(A) = .let R=A then rsclR(A)=rs cl(A) = () ( ) = therefore A is rsclosed in R similarly B is rsclosed inR.now A = = so A and B are complements of each other in R thus each one of A and B is rsopen in R.
Conversely,let A and B be disjoint sets which are both rsopen and rsclosed in R=A = rsclR(A)=rscl(A) =
() ( )=A [ () ] but A = this implies A[rscl(A) ] = . Now A and [rscl(A) ] are disjoint and their union is R.So,rscl(A) = cl(B)=.
Hence A and B are rsseparated.

Definition:
A collection {Ai : iI} of regular strongly open sets in a topological space X is called a rsopen cover of a subset B in X if B iI
{Ai : iI}

Definition:
A topological space X is regular strongly compact if every rsopen cover of X has finite sub cover of X.

Definition:
A topological space is said to be strongly compact if every cover of pre open sets admits a finite sub cover.

Theorem:
A rsclosed subset of rscompact space is rscompact relative to X.
Proof :
Let A be a rsclosed subset of a rscompact space X. Then Ac is rsopen in X. Let S be a rsopen cover of A in X. Then, S along with Ac forms a rsopen cover of X. Since X is rscompact, it has a finite sub cover say {G1, G2,……Gn}. If this subcover contains Ac, we discard it. Otherwise leave the subcover as it is. Thus we have obtained a finite subcover of A and so A is rscompact relative to X.

Theorem:
A closed subset of rscompact space is rscompact relative to X.
Proof :
Let A be a closed subset of a rscompact space X. Then A is rsclosed. Therefore by Theorem 3.5,every rsclosed subset of a rs compact space is rscompact relative to X. Hence A is rscompact space related to X.

Definition :
A topological space (X, ) is said to be regular strongly connected (briefly rsconnected) if X cannot be written as a union of two disjoint nonempty rsopen sets.

Theorem:
Every rwconnected space is connected .
Proof :
Let (X, ) be an rsconnected space. Suppose that X is not connected. Then X = A B, where A and B are disjoint nonempty open sets in (X, ) Since every open set is rsopen, A and B are rsopen. Therefore, X = A B, where A and B are disjoint non empty rsopen sets in (X, ). This contradicts the fact that X is rsconnected and so X is connected.

Theorem:
A topological space (X, ) is rsdisconnected if there exists a nonempty proper subset of X which is both rsopen and rsclosed.
Proof :
Let A be a nonempty proper subset of X which is both rsopen and rsclosed. Then clearly Ac is a nonempty proper subset of X which is both rsopen and rsclosed. Thus A Ac = and therefore A rscl(Ac) = and rscl(Ac) A= . Also, X = A Ac . Thus X is the union of two nonempty rsseparated sets. Hence X is rsdisconnected.

Theorem:
If the sets A and B form an rsseparation of (X, ) and if (Y, ) is an rsconnected subspace of (X, ) the Y lies entirely within either A or B.
Proof :
Since A and B are both rsopen in X, the sets A Y and B Yare rsopen in Y. These two sets are disjoint and their union is Y. If they were both nonempty, they would constitute an rsseparation of Y. Therefore one of them is empty. Hence Y must lie entirely in A or in B.

Theorem:
If every two points of a set A are contained in some rsconnected subset of A, then A is rsconnected.
Proof :
Suppose that A is not rsconnected. Then A is the union of two nonempty disjoint rsopen sets G and H. Let g G and h H. Then g and h are two distinct points of A. By hypothesis, there exists a rsconnected subset B of A such that g,h B. But B is an rsconnected subset of a rsdisconnected set A, we have B G or B H. Since G and H are disjoint and B contains atleast one point of G and one of that of H. This is a contradiction. Hence A is rsconnected.


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