CHAPTER 2STUDY OF SUBSETS USING GENERALIZED OPEN SETS 2.1 INTRODUCTION Topology is an important object of study in Mathematics with open sets as well as closed sets being the most fundamental concepts in topological spaces. Open sets and closed sets have been generalized by several mathematicians. The concepts of preclosure and preinterior of a set are due to A.

S.Mashhour et al 38. In 2002, Navalagi 66 has defined preneighbourhoods, pre-interior point, prelimit point, prederived set and prefrontier of set. The present chapter deals with some properties of gp-open and gp-closed sets This chapter contains three sections. In section 2 of this chapter, we introduce a generalized pre interior of sets, called gp-interior (briefly gp-interior) sets by using g-open sets in topological spaces. We obtain several characterizations of this class and study its properties and investigate the relationship with existing ones..

In section 3, we introduce gp-neighbourhoods, gp-limit-points and gp-derived sets in topological spaces and obtained some of their basic properties. In section 4, we introduce gp-frontier and gp-exterior of a set and obtained the some of the characterizations and properties. 2.2 Some Results On gp-Interior Of a Set . We define the followingDEFINITION 2.2.1: For a subset A of a topological space ( x, ?), generalized preinterior of A, is denoted by gpInt(A) and is defined as gpInt(A) = ? { G / G ? A and G is gp open in X}. That is gpInt(A) is the union of all gp – open set contained in A.

DEFINITION 2.2.2: Let x ? X. A set U ? X is called generalized preneighbourhood of x if there exists A ? GPO(X) such that A ? U. THEOREM 2.2.

3: Let A be a subset of X, then sgpInt(A) is the largest sgp-open subset of X contained in A if A is sgp-open.PROOF: Let A X be gp-open. Then gpInt(A) = {G: G A and G is gp-open in (X, )}. Since A A and A is gp-open, A = gpInt(A) is the largest gp-open subset of X contained in A.

The converse of the above theorem need not be true as seen from the following example.EXAMPLE 2.2.4: Let X= {a, b, c} and = {X, , {a}, {b}, {a, b}}. Let A = {b, c}, gpInt(A) ={b} is gp-open in (X, ). But A is not gp-open set in (X, ).REMARK 2.

2.5: For any subset A of X, Int(A) gpInt(A) A.REMARK 2.2.6: For a subset A of X, gpInt(A) Int(A) as seen from the following example.

EXAMPLE 2.2.6: Let X= {a, b, c} and = {X, , {a}, {a, b}}. Let A = {a, c}. Then gpint(A) = {a, c} and int(A) = {a}. Hence gpInt(A) Int(A).THEOREM 2.

2.7: If A B, then gpInt(A) gpInt(B).PROOF: Suppose A B, we know that gpInt(A) A. Also we have A B. which implies gpInt(A) B, gpInt(A) is an open set which is contained in B. But gpInt(B) is the largest open set contained in B.

Therefore gpInt(B) is larger that gpInt(A). That is gpInt(A) gpInt(B).REMARK 2.2.8: gpInt(A) = gpInt(B) does not imply that A = B. This shown by the following example.

EXAMPLE 2.2.9: X = {a, b, c} and ={X, , {a}, {b}, {a, b}, {a, c}}. Then (X, ) be a topological space. Here gp-open sets are X, , {a}, {b}, {a, b}, {a, c}.

Let A = {a} and B = {a, c}, then gpInt(A) = gpInt(B) but A B.THEOREM 2.2.10: For any subset A of X, following holds:gpInt() = gpInt(X) = XIf A B then gpInt(A) gpInt(B)gpInt(A) is the largest gp-open set contained in AgpInt(AB) = gpInt(A) gpInt(B)gpInt(AB) gpInt(A) gpInt(B)gpIntgpInt(A) = gpInt(A)PROOF: Proof follows from the Definition 2.2.1.THEOREM 2.2.

11: If A is gp-open if and only if gpInt(A) = A. REMARK 2.2.12: For any subset A of X, int(A) gpInt(A) A. 2.

3 gp-Neighbourhoods and gp-Limit Points In this section we define the notions of gp-neighbourhoods, gp-limit points and gp-derived sets in topological spaces and obtained some of their properties. DEFINITION 2.3.

1: A subset A of a topological space (X, ) is called generalized pre-neighbourhood (briefly gp-nbd) of a point x of X if there exists a gp-open set U such that x U A.Definition 2.3.2: Let (X, ) be a topological space and A be a subset of X. A subset N of X is said to be gp-neighbourhood of A if there exists a gp-open set G such that A G N.THEOREM 2.3.

3: Let A be a subset of a topological space (X, ). Then A is gp-open if and only if A contains a gp-nbd of each of its points.PROOF: Let A be a gp-open set in (X, ). Let x A, which implies x A A.

Thus A is gp-nbd of x. Hence A contains a gp-nbd of each of its points. Conversely, A contains a gp-nbd of each of its points. For every x A there exists a neighbourhood Nx of x such that x Nx A. By the definition of gp-nbd of x, there exists a gp-open set Gx such that x Gx Nx A. Now we shall prove that A = {Gx: x A}. Let x A. Then there exist gp-open set Gx such that x Gx.

Therefore, x {Gx: x A} which implies A {Gx: x A}. Now let y{Gx: x A} so that y some Gx for some x A and hence y A. Therefore, {Gx: x A} A. Hence A = {Gx: x A}. Also each Gx is a gp-open set. And hence A is a gp-open set.

THEOREM 2.3.4: If A is a gp-closed subset of X and x X – A, then there exists a gp-nbd N of x such that N A =.

PROOF: If A is a gp-closed set in X, then X – A is a gp-open set. By the Theorem 2.3.

3, X – A contains a gp-nbd of each of its points. Which implies that, there exist a sgp-nbd N of x such that N X-A. That is, no point of N belongs to A and hence N A = .THEOREM 2.3.

5: Let (X, ) be a topological space. If A is a gp-closed subset of X and x X – A then there exists a gp-neighbourhood N of x such that A N .PROOF: Since A is gp-closed, X – A is gp-open set in (X, ).

By the Theorem 2.3.3, X – A contains a gp-neighbourhood of each of its points. Hence there exists a gp-neighbourhood of N of x such that N X-A. Thus N A . DEFINITION 2.3.

6: Let (X,) be a topological space and A be a subset of X. Then a point xX is called a generalized pre-limit point of A if and only if every gp-nbd of x contains a point of A distinct from x. That is N – {x} A , for every gp-nbd N of x. Also equivalently if and only if every gp-open set G containing x contains a point of A other than x. The set of all gp-limit points of A is called gp-derived set of A and is denoted by gpd(A).THEOREM 2.3.

7: Let A and B be subsets of X and A B implies gpd(A) gpd(B).PROOF: Let x gpd(A) implies x is a gp-limit point of A that is every gp-nbd of x contains a point of A other than x. Since A B, every gp-nbd of x contains a point of B other than x.

Consequently x is a gp-limit point of B. That is xgpd(B). Therefore gpd(A) gpd(B).

THEOREM 2.3.8: Let (X,) topological space and A be subset of X. Then A is gp-closed if and only if gpd(A) A.PROOF: If A is gp-closed set.

That is X-A is gp-open set. Now we prove that gpd(A) A. Let x gpd(A) implies x is a gp-limit point of A, that is every gpnbd of x contains a point of A different from x. Now suppose xA so that x X-A, which is gp-open and by definition of gp-open sets, there exists a gp-nbd N of x such that N X-A.

From this we conclude that N contains no point of A, which is a contradiction. Therefore xA and hence gpd(A) A. Conversely assume that gp d(A) A and we will prove that A is a gp-closed set in X or X-A is gp-open set. Let x X-A. Let x be an arbitrary point of X-A, so that x A which implies that x gpd(A). That is there exists a gp-nbd N of x which consists of only points of X – A.

This means that X-A is gp-open. And hence A is gp-closed set in X.THEOREM 2.3.

9: Let (X,) topological space, every gp-derived set in (X,) is gp-closed set.PROOF: Let A be a subset of X and gpd(A) is gp-derived set of A. By Theorem 2.4.8, A is gp-closed if and only of d(A) A. Hence gpd(A) is gp-closed if and only if gpd(gpd(A)) gpd(A) . That is every gp-limit point of gpd(A) belongs to gpd(A).Now let x be a gp-limit point of gpd(A).

That is xgpd(gpd(A)). So that there exist a gp-open set G containing x such that {G – {x}} gpd(A) which implies {G – {x}} A , because every gpnbd of an element of gpd(A) has at least one point of A. Hence x is a gp-limit point of A. That is x belongs to gpd(A).

Thus x gpd(gpd(A)) implies x gpd(A). Therefore gpd(A) is gp-closed set in (X,).THEOREM 2.3.10: For subsets A and B of X we have gp(AB) gpInt(A) – gpInt(B).PROOF: Let xgpInt(A-B) which implies that there exists a gp-neighbourhood Ux of x such that Ux A – B A.

This shows that Ux B = and hence x gpInt(B).The equality in the Theorem 2.2.1 does not hold in general as illustrated by the following:EXAMPLE 2.

3.11: Let X = { a, b, c, d} with topology = { , {b}, {d}, {b,d}, X }. Then PO(X) = { , {b}, {d}, {b,d}, {a,b,d}, {b,c,d}, X }. We take A = { a, b, d } and B = { a, b}. Then, gpInt(A) = {a, b, d}, gpInt(B) = {b} and gpInt(A-B) = gpInt({d}) = {d}.

Therefore, gpInt(AB) ? gpInt(A) gpInt(B)LEMMA 2.3.12: A and B are gp – open subsets of topological spaces X and Y respectively iff then A B is gp – open in XY.PROOF: If A is gp – open in X and B is gp – open in Y then M pint (A) whenever M A and M is closed in X and N pint (B) whenever N B and N is closed in Y.

Now M x N ? pInt (A) pInt(B) = pInt (AB) Hence, AB is gp-open in XY whenever M x N ? A x B , for M x N closed in X x Y.Conversely, if AB is gp-open in X Y then, M x N ? pInt (AB) whenever M x N ? A x B , for M x N closed in X x Y.Now pInt (A) pInt(B) = pInt (AB) ? AxB It follows that, M pint (A), whenever M A and M is closed in X and N pint (B) whenever N B and N is closed in YThus A and B are gp – open.THEOREM 2.

3.13: For subset A of X and subset B of Y we have gpCl(AB) = gpCl(A) gpCl(B) and gpInt(AB) = gpInt(A) gpInt(B).PROOF: Let (a,b)gpCl(AB). We will show that agpCl(A) and bgpCl(B). Let aU be gp-open in X. Since (a,b) UY, which is gp-open in XY, we have ? (UY) (AB) = (UA) (YB) = (UA) B.

This implies that UA ? and agpCl(A). Similarly we can show that, bgpCl(B). Hence gpCl(AB) gpCl(A) gpCl(B).

Now let (a,b) gpCl(A) gpCl(B), that is agpCl(A) and bgpCl(B). If possible assume that (a,b)gpCl(AB). Then there exists a gp-open set G1G2 in XY containing (a,b) such that (G1G2) (AB) = . That is (G1A) ( G2B) = . This implies that either (G1A) = or (G2B) = . But then by Lemma 2.2 G1 is preopen in X and G2 preopen in Y and so either apCl(A) or bpCl(B) which is a contradiction to the hypothesis.

Hence (a,b)pCl(AB). Therefore pCl(A) pCl(B) pCl(AB). By similar arguments we can show that pInt(AB) = pInt(A) pInt(B). THEOREM 2.3.14:A subset G of a topological space is preopen iff pCl(GpCl(A)) = pCl(GA) for every AX.

PROOF: (if part) Let G be a preopen subset of X and A X. Let xpCl(GpCl(A)). This implies that every preopen set U containing x intersects GpCl(A). That is (UG)(pCl(A)) = U(GpCl(A)) ? ()Now if xpCl(GA), then there exists a preopen set V containing x such that V(GA) = , or (VG)A= which shows that no point of VG belongs to pCl(A) and thus (VG) pCl(A) = a contradiction to () above. Thus xpCl(GA) and hence pCl(GpCl(A))pCl(GA). But already pCl(GA)pCl(GpCl(A)) and hence the equality.

(only if part) Let G be any subset of X satisfying the condition pCl(GpCl(A)) = pCl(GA) for every AX. Thus in particular, pCl(GpCl(X-G)) = pCl(G(X-G)) = . Therefore,GpCl(X-G) = and hence pCl(X-G) X-G. This shows that pCl(X-G) = X-G which implies that X-G is preclosed. Therefore G is preopen.Navalagi in 4 has shown that for subsets A, B X, pCl (A) pCl (B) pCl (AB) and that the equality does not hold in general. By imposing a stronger condition the following theorem shows that the equality holds even for an arbitrary family of subsets of X.THEOREM 2.

3.15: Let {A| I} be any family of subsets of X. If ??IpCl(A?) is preclosed then ??IpCl(A?)=pCl(??IA?) .PROOF: As A? ??IA?, therefore pCl(A? )pCl(??IA?) and hence ??IpCl(A?) pCl(??IA?). We will show that pCl??IA???IpCl(A?).

Let x pCl??IA?. Now if possible let x??IpCl(A?). Since ??IpCl(A?) is preclosed there it contains all its prelimit points and therefore x is not a prelimit point of ??IpCl(A?) and therefore there exists a preneighbourhood Ux of x such that Ux(??IpCl(A?)) = . This implies that UxpCl(A?) = for every I and hence UxA?= for every I, a contradiction to x pCl??IA?. Therefore pCl??IA???IpCl(A?) and hence the result.THEOREM 2.3.

16: If for subsets A and B of X, pfr(A)pfr(B)= then pInt (A)pInt (B)=pInt(AB) PROOF: We already have pInt (A) pInt (B) pInt (AB). Now let xpInt(AB). This implies that there exists preopen set Ux containing x such that Ux AB.Case(I) If Ux A or Ux B then xpInt (A) pInt (B). Case(II) Without loss of generality we can assume that all preopen sets Ux containing x is such that Ux AB, for if there is some preopen set Vx containing x which is not contained in AB then ? Vx( pInt(AB)) is a nonempty preopen set containing x and contained in AB.

Therefore, if for all preopen sets Ux of x such that Ux AB we have UxA and UxB, it will imply that xpInt(A) and xpInt(B) and that Ux(B-A) ? and Ux(A-B) ? . This shows that xpCl(A) and xpCl(B). Therefore xpfr(A) and xpfr(B) and hence pfr(A)pfr(B) ? a contradiction.THEOREM 2.3.17: In general for any subset A of X we have pfr(pfr(A))pfr(A).

PROOF. pfr(pfr(A)) = pCl(pfr(A)) pCl(Xpfr(A)) pCl(pfr(A)) = pfr(A), as pfr(A) is preclosed.THEOREM 2.3.

18: For a subset A of X we have pfr(pInt(A)) pfr(A) and pfr(pCl(A))pfr(A).PROOF: pfr(pInt(A)) = pCl(pInt(A)) – pInt(pInt(A)) = pCl(pInt(A)) – pInt(A) pCl(A)–pInt(A)= pfr(A) . pfr(pCl(A))=pCl(pCl(A))–pInt(pCl(A))=pCl(A)–pInt(pCl(A)) pCl(A) – pInt(A) = pfr(A)THEOREM 2.3.

19: A subset A of X is preopen iff pfr(A) = pd(A). PROOF. Let A be preopen. Then pInt(A) = A. Now, pfr(A) = pCl (A) pInt(A) = pCl (A) A. As, pCl(A) = Apd(A) sopfr(A) = pCl (A) A = (Apd(A)) A = pd(A).

Conversly, let pfr(A) = pd(A), that is pd(A)= pCl (A) pInt(A) = (Apd(A)) pInt(A). Hence, ApInt(A) = . Therefore A pInt(A). But pInt(A) A and hence A= pInt(A) which shows that A is preopen.2.4 Results on gp-frontier and gp-exterior of a set.DEFINITION 2.

4.1: A point xX is called a pre-exterior point of a subset A of X if x is a pre-interior point of XA. The set of all pre-exterior points of X is called the pre-exterior of A and is denoted by pExt(A). Thus by definition, pExt(A) = pInt(XA).DEFINITION 2.

4.2 : A point xX is called pre-frontier point of a subset A of X if it is neither pre-interior point of A nor pre-exterior point of A.THEOREM 2.4.

3:If X is a topological space and A, B X, then the following properties hold for the pre-exterior( pExt) operator but the converses are not true in general:(i) If A B then pExt(B)pExt(A)(ii) pExt(AB)pExt(A) pExt(B)(iii) pExt(A)pExt(B)pExt(AB)PROOF :(i) If A B, then XB XA and hence pInt(XB)pInt(XA) which implies that pExt(B)pExt(A).(ii) Since AAB and BAB, so by (i) above we have pExt(AB) pExt(A) and pExt(AB) pExt(B) and therefore pExt(AB)pExt(A)pExt(B).(iii) Since ABA and ABB, again by (i) above we have pExt(A) pExt(AB) and pExt(B) pExt(AB) and hence pExt(A)pExt(B) pExt(AB).EXAMPLE 2.4.4: Let X={a,b,c,d} with topology = {,{b},{d},{b,d},X}. Then PO(X) ={ ,{b},{d},{b,d},{a,b,d},{b,c,d}, X}. Take A={d}and B={b,c}.

Hence, pExt(A) = pInt({a,b,c}) = {b},pExt(B) = pInt({a,d}) = {d} and pExt(AB) = pExt({b,c,d}) = pInt({a}) = and pExt(AB) = pExt() = pInt(X)=X. Hence AB butpExt(A)pExt(B).Also,pExt(A)pExt(B)pExt(AB)and pExt(AB)pExt(A)pExt(B).THEOREM 2.

4.5: For a subsets A and B of a topological space X the following properties hold for the pre-exterior ( pExt) operator:(i) Ext(A)pExt(A)(ii) pExt(X) = and pExt() = X(iii) pExt(A) is preopen(iv) pExt(A) = XpCl(A)(v) pExt(pExt(A)) = pInt(pCl(A))(vi) pExt(AB) = pExt(A) pExt(B)(vii) pExt(A) = pExt(XpExt(A))(viii) pInt(A) pExt(pExt(A))(ix) A pExt(A) = (x) pInt(A), pExt(A) and pfr(A) are mutually disjoint (xi) X = pInt(A)pExt(A)pfr(A) PROOF:(i) Let x Ext(A) which implies that xInt(XA) pInt(XA) = pExt(A)(ii) pExt(X)=pInt(XX)=pInt()= and pExt()=pInt(X)=pInt(X)=X(iii) As pExt(A) = pInt(XA), therefore pExt(A) is preopen.(iv)Again pExt(A) = pInt(XA) = XpCl(A).(v) By (iv) above we have,pExt(pExt(A)) = pExt(XpCl(A)) = pInt(X(XpCl(A))) = pInt(pCl(A))(vi) pExt(A) pExt(B) = pInt(XA) pInt(XB)pInt((XA)(XB)) = pInt(X(AB)) = pExt(AB). Again, since AAB and B AB, so pExt(AB)pExt(A) and pExt(AB)pExt(B) therefore Ext(AB)pExt(A)pExt(B). Thus pExt(AB) = pExt(A) pExt(B)(vii)pExt(XpExt(A)) = pInt(X(XpExt(A))) = pInt(pExt(A)) = pInt(pInt(XA)) = pInt(XA) = pExt(A).(viii) By definition pExt(A) = pInt(XA) XA and hence from (i) of the previous theorem we have pExt(XA)pExt(pExt(A)) which implies that pInt(A)pExt(pExt(A)).(ix)p Ext(A) = pInt(XA) XA, which implies that A pExt(A) = .(x) pExt(A) = pInt(XA) XA and pInt(A) A shows that pExt(A)pInt(A) = .(xi) By (iv) we have pExt(A) = XpCl(A) =X(pInt(A)pfr(A)) which implies that X = pInt(A)pExt(A)pfr(A).