ON LOCALLY s-CLOSED SPACES

in the prent papz, the concepts of s-closed sub-spaces, loCd(ly s- closed spaces have been lt(oduced and chardCt(Ized. We have seen that local s- c{osednss is seml-rgulat propetty; the concept ot s-8-closed mapping has been introduced here and the following impo(t=nt ptopertles are establlshed Let X --) be an s-8-closed sur3ectlon wlth s-set (Malo and Nolrl 8i) polnt Inve(ses. Then (d) it iS completely continuous (Ary and Gupta |i)) ad Y is locally compact q -space, then, X Is locally s-closed. g (b) It f is -conttnuous IGanguly and BdSU {5) and X s a locally compact T-

Throughout the present paper, by (X,T) or slmply by X we shall mean a topologlcal space.A subset A of a topological space s said to be regular open (resp.regular closed) if int(cl(A))=A (resp.cl(Int(A))=A), where cl(A) (resp.int(A)) denotes the closure (resp.nterior) of A. A subset A of a space X is sald to be semi-open [7] if there exlsts an open set 0 such that O Acl(O).The complelnent of a seml-open set is called seml-closed (Crossley and Hildebrand 13]).
The semi-closure of a subset A of a space, denoted by sclA, s the ntersectlon of all semi-closed sets containlng A (Crossley and Hildebrand [3]).A set A whlch is both semi-open as well as seml-closed is called a seml-regular set (Malo and Nolrl [8]).The collecton of all seml-open (resp.semi-regular, regular open) sets contalnlng a pont x of X wll be denoted by SO(x) (resp.SR(x), RO(x)) and for the whole space X these wlll be denoted by SO(X) (resp.SR(X), RO(X)).A polnt x of X sad to be s-0-cluster [8] polnt of a subset A of X If for every U (SO(x), sclU A#.Snce, for a seml-open set U, sclU is a semi-regular set [8], a polnt x of X Is said to be an s-0-cluster pont of A ff R A, for all R SR(x).The collection of all s-0-cluster points of A will be denoted by s-O-clA ([A] for s-O short).A set A is s-0-closed if A=[A]s_ 0 A complement of an s-0-closed set called an s-0-open set.For a space (X,T), RO(X,T) is a base for a topology T on X S coarser than T and (X,Ts) is called the semi-regularizatlon space of (X,T).A topologlcal property P xs said to be semi-regular property if whenever a space (X,T) possesses that property P so does its seml-regularzation space (X,Ts).A subset A of X Is s-closed [8] (resp.S-closed (Nolrl [II])) relatlve to X or slmply an s-set (resp.S-set) if every cover of A by sets of SO(X) admits a finite subfamlly such that A sclU (resp.A C clU).In case A X and A Is an s-set (resp.S- set), then X is called s-closed [8] (resp.S-closed [14]).A subset A is called Nearl compact (NC-set (Carnahan [2]), for short) if every cover of A by means of open sets of X has a finite subfamily U 1 U (say) such that AU intclU.n Clearly every s-set (resp.compact) set, is an NC-set, but not conversely.A subset A of a space X is said to be an O-set (Noiri [10]) if Aint(cl(int(A))).
At the very outset, an example is glven to assert that, every set, s-closed relative to X, is not necessarily an s-closed subspace of X.
EXAMPLE I. Every countable set in an uncountable set X with co-countable topology T s s-closed relative to (X,T), but is not even an S-closed subspace.
DE'INITION i.A subset A of X s said to be pre-open (Mashour et al. [9]) A intclA.This collection includes all open sets and, even more, all -open sets.
LEMMA i. (See Dorsett [4]) Let (X,T) be a topological space and let A be pre- open set in (X,T), then SR(A,TA)=SR(X,T) A, where T A is the subspace topology on A.
THEOREM i.A pre-open set A of X is s-closed as a subspace iff it is s-closed relative to X.
PROOF.Let A be s-closed relative to X and also let IVm I } be a cover of A by semi-regular sets of the subspace A. Then by Lemma I, there exists a semi- regular set U m in X, for each I, such that V U A. Therefore, AU.Since A is s-closed relative to X, there exlsts a finite subset I of I such that o A U, which shows that A (U A) i.e., A V m Therefore, A is s-closed m%Iub_space I 0 I 0 as Conversely, let A be s-closed as a subspace.Let Vm I be a cover of A by semi-regular sets o X.Then A U (V A).Since A as s-closed as a subspace, there exists a flnlte subset I of I such that A U (Vm A), which shows that o ( I A V a. Therefore A is s-closed relative to X. 0 THEOREM 2. Let B be a pre-open set tzl (X,T).Then a subset A of B as s-closed relative to the subspace B iff A as s-closed relatlve to X.
PROOF.The proof follows by Lemma I.
COROLLARY I. Let A and B be open sets of a space X such that A B. Then A as an s-cloed subspace ot B ff A as an s-closed subspace of X.
PROOF.Applyng Theorem 1 and Theorem 2, we get the result.
DEFINITION 2. Let (X,T) be a topological space, then SR(X,T} forms a sub-base for a topology called T -topology on X. SR Let A be s-closed relatlve to (X,T}.Then every cover of A by sets of SR(X,T} has a finite subcover.But SR(X,T) forms a sub-base for (X,' ).So every

SR
sub-basic open cover of (X,TsR) has a flnte subcover.Therefore by Alexander sub- base theorem A is comDact n (X,TsR)-Coverse[y, if A is compact in (X,TsR) then every sub-basic open cover has a finite subcover.So every cover by sets of SR(X,T} has a finite subcover.Therefore A is s- closed relative to (X,T).

THEOREM 3.
Let B be a T -closed set in X and let A be any subset of X which SR s s-closed relative to (X,T).Then AB is s-closed relative to (X,T}. PROOF.Let { U 6I be a TsR-Open cover of AnB.Then clearly { Ua I} {X-B) as a TsR-Open cover of A. By Lemma 2, A as compact relative to (X,TsR}; and so, there exists a flnlte subset I of 1 such that A e Ua} U (X--B) which o o imples that ABC Ua Therefore AB s compact in (X,T }.Then by Lemma 2, SR I 0 A s s-closed relative to (X,T).COROLLARY 2. If B is regular open or regular closed and A as any subset of X which s s-closed relative to X, then A 5 as s-closed relative to X.

PROOf.
Since every regular closed or regular open set is sem-reular, the corollary follows from Theorem 2. COROLLARY 3. If X is an s-closed space and A is a regular open set of X, then A is an s-closed subspce of X.
PROOF.The proof follows from Theorem 1 and Theorem 3. PROOf.The proof follows from Corollary 2 and Theorem 1 and second prt follows from Corollary 1. n THORFq 4.
If A i i 1,2,...,n are s-sets i.e., s-closed relative to X. then A. is s-closed relative to X. i=l PROOF.Straightforward.
THEOREM 5. Let X be an s-closed spce and let A be a closed set of X and let frontier of A, denoted by Fr(A), be s-closed relative to X. Then A is s-closed relative to X.
C.K. BASU PROOF.Since X is s-closed, by Corollary 3 and Theorem I, intA is s-closed relatave to X whenever A as closed set.Sance A=intA UFr(A), by Theorem 4, A as s- closed relative to X.

3.
LOCALLY s-CLOSED SPACES DEFINITION 3. A space X as saad to be locally s-closed aff each polnt belongs to a regular open neaghbourhood (nbd.for short) which is an s-closed subspace of X.
REMARK I. Clearly every s-closed space is locally s-closed space.However, the converse is not true, ingeneral, because any uncountable set with discrete topology is locally s-closed but not s-closed.THEOREM 6.A topologacal space (S,T) is locally s-closed iff for each point x { X, there exists a regular open set U containing x such that U is locally s-closed.

PROOF. Sufficaency
At farst we prove that if A is a regular-open set in (X,T) then every regular-open set in the subspace (A,T A) is also regular-open in (X,T).
Let VCA be regular-open an the subspace (A,TA).Then V antAClAV intA(AClxV) . Therefore V is regular open an (X,T).Now let x be any X X X X poant of X.Then, by hypothesis, there exists a regular-open set U of (X,T) containlng x such that U as locally s-closed.Then there exists a regular open set V in U such that x E V and V is an s-closed subspace of U. Therefore V as a regular- open set in (X,T) and by Corollary I, V is s-closed subspace of X. Therefore (X,T) Is locally s-closed.

Necessity
The proof as straightforward.
THEOREM 7. Let (X,T) be a topological space.The following are equivalent (i) X is locally s-closed; (ii) every polnt has a regular-open set whach is s-closed relative to X; (iii) every point x of X has an open nbd U such that int cl U is s-closed X X relative to X; (iv) every point x of X has an open nbd U such that sclU is s-closed relative to X; (v) for every point x of X, there exists an -open set V contaanang x such that sclV is s-closed relatave to X; (vi) for every point x of X, there exists an -open set V containing x such that int cl V is s-closed relative to X; X X (vii) for each x X, there exists a pre-open set V containing x such that sclV is s-closed relative to X; (viii) for every x of X, there exists a pre-open set V containing x such that int cl V is s-closed relative to X; X X (ix) for every x X, there exists a pre-open set V containing x such that intxclxV is an s-closed subspace of X.
PROOF.(i) --(ii) Follows from Theorem 1 and from the fact that every regular-open set is pre-open set. (ii) --) (iii) is obvious.
PROOF.Let (X,T) be locally s-closed.Dorsett [4] proved that SR(X,T)=SR(X,T S) and hence a subset A of X is s-closed relatlve to (X,T) iff A Is s-closed relative to (X,Ts).We know that if U Is an open and V a closed subset of (X,T), then ClTU cl T U and IntTV int T V. Using these results we can easily prove that foz a S S ClTsU.Therefore every regular-open set regular-open set U of (X T) IntTclTU intTs n (X,T) is regular open in (X,T S) and vice-versa.So (X,T) and (X,T S) have the sdme collection of regular-open sets.Therefore by deflniton, (X,T) s locally s-closed Iff (X,T S) is locally s-closed.
for any subset A of X.

PROOF Let
be s-8-closed and A be any subset of X.Then f( ). s-8 Conversely let A be an arbitrary s-8-closed set in X.By hypothesls f PROOF.Sufficiency Suppose that the glven hypothesis holds.Let A be any s- 8-closed set in X.Let y be an arbitrary point n Y--f(A); then X--A s an s-8-open -I set containing f (y); by hypothesis, there exists an open set v contalning y such LEMMA 3. A subset A of a space X is an s-set iff every cover of A by s--open sets has a flnite subfamlly which covers A.
PROOF.Sufflciency part is straightforward.

Necessity
Let A be an s-set.Let THEOREM Ii.Let f X ---Y be an s-8-closed surjectlon with s-set point -I inverses; if A Is any compact set in Y then f (A) is an s-set in X.

PROOF. Let
{ U mI } be any cover of f-l(A) by s-8-open sets of X.For ach point yA, f-l(y) U By hypothesis f-l(y) is an s-set, by Lemma 3,  there exists a finite subfamily I of 1 such that f-l(y)CU{ u El 'Yn of A such that A C )V which shows that f (A) is covered i=l Yl by a finlte number of s-8-open sets from and hence f-l(A) Is an s-set.
COROLLARY 5. Let f X ---Y be an s--closed surjection with s-set point inverses; If X is T and Y is compact then f is continuous. 2 PROOF.Let A be a closed set in Y. Therefore A is also compact; by Theorem iI, -i f (A) is an s-set n X.Since every s-set is an NC-set and X is T2, by Theorem 2.1 -I of T. Noirl [12], f (A) is closed and hence f is continuous.THEOREM 12. Let f X ---Y be a completely-continuous s-8-closed surjection with s-set point inverses.If Y is locally compact T2, X is locally s-closed.PROOF.Since Y is locally compact T2, for each point x X, there exists a -I closed compact nbd.U of f(x).Since is completely continuous, f (int U) is a regular open set containing x.But it is easy to see that every regular-open set is semi-regular and hence an s-8-closed set (see Malo and Noiri [8]).Slnce U is -I compact and f is an s-@-closed function, by Theorem Ii, f (U) Is an s-set in X and -I -i -I xf (nt U)Cf (U).Hence, by Corollary 2, f (int U) is an s-set in X. Therefore X is locally s-closed.DEFINITION 6.A function f X ---Y is said to be -continuous (Ganguly and   Basu [5]) if for each xX and each W SO(f(x)), there is an open set V containlng x such that f(V) sclW.Equivalently f is -continuous iff the inverse image of each semi-regular set is clopen.
LEMMA 4. If f X --) Y is -continuous and KX is compact; then f(K) is an s-set in Y.
PROOF.Let { U m I } be a cover of f(K) by semi-regular sets of Y. Then { f-l(u EI } is a cover of K by open sets of X.Since K is compact, there -I exists a finite subset I of I such that K C [12]) Let X be a T2-space.Then for any disjoint NC-sets A and B, there exist dlsjoint regular open sets U and V such that AU and BV.THEOREM 13.If f X --Y is an s-8-closed, -continuous surjection with s- set point inverses and if X is locally compact T2, then Y is locally s-closed.
PROOF.We shall first prove that Y is T 2.
Let Yl and Y2 be two distinct points -1 of Y. Then f (yl) and f (y2) are disjoint s-sets and hence disjoint NC-sets.By ACKNOWLEDGEMENT.The author gratefully acknowtedges the kind guidance of Prof. S.
Ganguiy of the Department of Pure Mathematics, Calcutta Unuerslty, and is also thankful to the referee for certain constructlve suggestion towards the Improvement of the paper.He is also grateful to the University Grants Commission, New Delhi, for sponsoring ths research work.

COROLLARY 4 .
If A is s-closed open subspace of X and is a regular open set of X, hen AO is an s-closed subspace of X and (hence of A and B).
set there exists a semi-open setXv such that x V sclV U Since we o o know that Union of any collectlon s-8-open sets is s-@-open and since Is an s-8- closed function, by Theorem 10, there exists an open set V of Y containing y such Y that f-l(v CU { Vy y E A } is a cover of a compact set A and hence there

DEFINITION 5 .
A function f X ---) Y is said to be completely continuous (Arya and Gupta [I]) if inverse image of each open set in Y is regular-open in X.

2 .
But every regular-open set is an s-8-open set and so, by Theorem I0, -I there exist open sets V., 1,2 containing yj in Y such that f (V.)U.where 3 Ii j=l,2.Thus Y is T 2.Let X be locally compact T 2 for each point x of f (y), there exists a compact closed nbd.U of x in X.Since interior of a closed nbd. is a x regular-open set, it is semi-regular as well.Therefore the family { intU -I -I xx f (y)} is a cover of an s-set f (y) by semi-regular sets.By Proposition 4.1 -iof Ma].o and Noirl[8], there exist polnts x s-8-open set contalnng (y) and since, IS an s-@-closed tunctlon by -I Theorem I0, there xists an open set 9 conta[nlng y such that f (9 )xntU i.e., Y Y y6 (intU)Cf(U).But f bng -contlnuous, (U) xs an s-set by Le,a 4is an s-set.Hence Y Is locally s-closed.Y