Kyungpook Malhemalical J ou m꾀 Vol.34, No. l, 109- 11 6, June 1994

ON SPACES VIA DENSE SETS AND SMPC FUNCTIONS

D. A. Rose a nd R. A. Mabmoud

Density is one of t he basic propcrt ies in topological spaces. So, some spac않 sucb as hyperconnected space, submaximal spacc and all resolvabil­ ity spaces have been defined via lhis properly. Therefore, we devoled this paper for the study and investigated of new properties of these types of spaces. Also, a strong M-precontinuous function (abbrivaled as: SMPC) was characterized by using some previolls spaces. Finall y, some effects of SMP C on olher spaces are studied.

1. Definitions and preliminaries Topological spaces used here will not include any separation properties wh ich are assumed unless they arc otherwise needed in which case lhey will be explicilly staled. Given a lopological space (X, T), for any A 드 X , we denote the and the c1 0sure of A with respecl lo T , by IntA and CIA, respectively. A is called preopen [1] if A IntCIA, and PO(X,T) means the collection of all preopen sets in (X, r). For any space (X, r) let 감 be the smallest tO]lolgy on X containing PO(X, T). The topolgy r O [2] is PO(X, r ) n SO(X, T) where A E SO(X, T) iff A is semi-open [3] . i.e. O A 드 CllntA. Thus, for any space (X, T), T 드 T 드 PO(X ,T) 드 Tp ' II is also known thal PO(X, T''')'= PO(X, r) (Coroll ary 1 of [4]). For any T on X , the semi l'eg ulal'ization of r is t he topol ogy 감 having fo r a basis lhe set of regular open subsets of (X , r ). The semiregular c1 ass of T is the set [T] , of a ll on X having the same semiregularizalion as T. A P is a semiregular property if il is shared by a ll members of [r]s when possessed by any one

Received June 9, 1992 Rev ised March 20 , 1994

109 110 D. A. Rose and R. A. Mahmoud member. This is equivalent to saying that (X,T ) and (X, T.) both have P whenever either does [5]. An a -topological property is any topologi­ cal property shared by aU members of the a-class when possessed by any one member of the a-class. In particular, it is any topological property possessed by both (X, T) and (X, T") when possessed by either [5] . A space (X, T) is called hyperconnected [6] if each nonempty is dense. (X , T) is resolvable [7] if there is a dense subset D 드 X for which X - D is also dense. A space whi ch is not resolvable is called irresolvable A subset of X is resolvable (irresolvable) if it is resolvable (irresolvable) as a subspace. A space is hered itaril y irresolvable if each of its nonempty subsets is irresolvahle. (X,T) is submaximal [8] if each of its dense subsets are open. Al so, (X, T) is strongly ∞ mpact (strongly Lindelöf) [9] if for each preopen cover of X , there is a finite (countible) subcover. Spaces pre낀 (i 0,1,2) are defined likewise the spacés T; (i = 0,1,2) except that the open sets are replaced by preopen sets [10] A function f : (X, T) • (Y, a) is said to be precontinuous [1 ], preirres­ 이 ute [11] and strongly M-precontinuous [12] (SMP C) if the inverse image 。 f each open, preopen and preopen in (Y,a) is preopen , preopen and open in (X , T), respectively

Resolvable spaces An important res ult related to the resolvable spaces which would be called the Hewitt-representation of (X, T) in [13] Theorem 1 [13]. Every space (X,T) can be represented uniquelly as a disjoint union X = F U G where F is c/osed and π so/va ble and G is open and hereditari/ν 11γ'e so/vab/e. Hence (X ,T) 상 reso/vable iJJ G = 0 and (X, T) is heπ ditaπly iπ'eso/v­ able 퍼 F= 0. x Lemma 1. (Corollary 5 of [7]) If (X, T) is resolvable th en Tp = 2 .

Proof Let D) and D 2 be disjoint dense subsets of X and let x E X. Then D) U {x} and D 2 U {x} are dense and hence preopen. Thus {x} = (C) U {x}) n (D2 U {x}) E 강 · Other properties of resolvable spaces have studied in [6] by using an anti-operation due to Bankston Now, we give the condition under which that is property of being a resolvable space is hereditrary. On spaces via dense sets and SMPC functions 111

Theorem 2. Each semi-open subset of a resolvable spaée is resolvable.

Proof Let A E SO(X,T) , i.e. A 드 ClIntA 드 X and X be resolvable Then IntA is resolvable and A-IntA is nowhere dense in (A, TjA). Thus, if Dl U Dz is a disjoint union of dense subsets of IntA then Do = D1 U (A-IntA) and Dz are disjoint and also are dense in A.

It is obvious that every subspace of a hereditarily irresolvable space is heredi tarily irresolvable, whereas the converse follows nextly.

Theorem 3. If(X , T) is a space, X = X,U Xz and X,n xz = 0 and X 1 is c/osed th en ifboth (X1 ,TjX1) and (XZ ,TjXZ) are hereditarily irresolvable th en (X , T) is h e re d i ta π Iy i1"1'esolvable. Proof Suppose that ø ￥ A 드 X and (A , T / A) is reso lvahle. Then there exist di sjoint, dense in A, subsets Dl and Dz with A = D, U D2 . Suppose that D1 n X z # 0 and Dz n x 2 # 0. Then since Xz is open in X , D1 n X z and Dz n Xz are di sjoint and dense in A n X 2 . For if x E D 2 n X z and V is open wi th x E V , since D 1 is dense in A, V n A n D, 폼 0. lf U is open in X and x E U then, for V = U n X z, V E T and x E V so that U n A n (D, n X z) ￥ 0. Thus, D, n Xz and similarly Dz n X 2 are dense in A n X 2 and di sjoint. T hus, A n X 2 is a resolvable subspace of X 2 which contradicts X 2 being hereditarily irresolvable. Apparently, either Dl n X 2 = ø or D2 n X z = 0. But in either case A n x 1 contains a in A. Thus, CIA(A n X t} = A 드 An X, 드 X, sin ce X! is closed. Thus A is a resolvable subspace of X1 whi ch cannot be sin ce X 1 is hereditarily irresolvable. This fin al contradiction proves that (X , T) is heredi tarily irresolvable. Theorem 4 [5J. Let f : (X,T) • (Y, 0") be a bijection . Th en f is a se mihomeomorphism ifJ f is an a-homeomorphism Where a bijection f (X, T) • (Y,O") is said to be a semihomeo morphism [14J if both f and f -' preserve semiopen sets. Any property transmitted by semihomeomorphisms is called semitopological [1 4J By a previous result, a property P is semitopological if and only if P is an a -topological property and it is clear that the laUer holds if and only if X and X " both have P whenever either does [5J . Since it is obviously that spaces (X , T) and (X , T"' ) share the same family of dense subsets, also resolvability is one of the a-topologi cal and hence semitopological properties. This illustrates our belief that gener all y the best way to demonstrate that a property P is semitopological 112 D. A. Rose and R. A. Mahmoud is to show that it is a-topological. Whereas serniregular properties are a-topological [15J. Hence as the following example shows that semitopo logical properties are not semiregular. In particular, resolvability space.

Example 1. Let (X, T) be the two-point Sierpinski space. Then (X, T) is not resolvable whereas the indiscrete semiregularization (X, Ts ) is resolv able.

3. Hyperconnected and submaximal spaces

Lemma 2. (Proposition 1 of [7]) A E PO(X, T) iff A = U n D f0 1" some U E T and de η se D 드 X

Proof A E PO(X,T) • A 드 IntCIA = U E T. L.et D = X - (U - A) = (X -U)UA. Then D is dense since X = CIAU(X - CIA) 드 CIAU(X - U) = ClD. Also. A = U n D. Conversely, if A = U n D with U E T and D dense, A 드 U ç IntClU = IntCl(A) so that A E PO(X,T)

Theorem 5. A space (X, T) is hyperconnecled iff the class of dense sets

Of(X,T) is Tp - {Ø}.

Proof Follow directly by using previous lemma and the fact that T 드 Tp •

Lemma 3. If(X ， 꺼T ) is submaxiη7πmí

Proof Clearly T 드 PO(X ， 끼).T N‘ o아、w A E PO(X ， 서)T • A = U n D for some U E T and dense D 드 X. Therefore, if (X, T) is submaximal, D E T • A E T.

Theorem 6. For a space (X, T) the fo l/ owi때 are equivalent (i) (X, T) is submaximal (ii) Theπ exists a η open, dense and heredilarily irresolν able subspace

D c X and T = Tp (iii) PO(X, T) 드 SO(X,T) a 뼈 T= 감，

(iv) A 드 X is η0ψ here de η se iff InlA = ø and T = T p

(v) (X, T p ) is s띠 maximal

Proof By (i) and Lemma 3 we get T = PO(X, T) = Tp and hence (X, Tp ) is submaximal. Therefore, the equivalents of (i) with each of other statments follow from Theorems 2 and 4 of [7J Since submaximal spaces are hereditarily irresolvable. Which sub spaces of submaximal spaces are submaximal? Since each open and hence On spaces via dense sets and SMPC functions 113 preopen subsets of a submaximal space are submaximal and we show a bit more. We first note the following useful known lemma

Lemma 4. If A E S O(X ,r ) then r "' /A = (r / A )" Theorem 7. If(X,r ) is submaximal and A E SO(X ,r) then (A,r/A) is submaximal Pro of Since (X,r) is submaximal and A E SO(X,r). Then r r '" and there is an open, dense, hereditarily irresolvable subset D 드 X. lf A ￥ 0, t hen D n IntA is a dense, open, herediiarily irresolvable su bspace of (A,r/A ), and also CIA( Dn Int A) = An Cl(D n IntA) = An ClInt A = A 5ince r/A = r" /A = (r/A)"' , we have that (A ,r/A) is s 배 maximal ‘

Theorem 8. Sμ bmax i malit ν is preserved b ν open surjeclions

Pro of If f: (X,r) • (Y , σ) is an open surjection and (X , r) is submaxi­ mal and if D 드 Y is dense, f-l(D) is dense and hence open in X so that D = f (J-l(D )) is open.

Corollary 1. [f fI X " is submaximal th en each X " is submaximal

4. SMPC functions Now, we characterize the SMPC-function based on the concepts : r. a nd some previously spaces via dense sets

Theorem 9 . f: (X ,r) • (Y, O" ) is SMPC iff f : (X ,r ) • (Y ,O".) is continuous. Proof A basic open set in 0". has the form V = n k=1 Bk where each B k E PO(Y, 0") ‘ 50 if f ‘ (X,r) • (Y,O") is SMPC, and V is a basic open set in 0"., f-l (V) = n k=lf-1 (B k) E r so that f: (X ,r ) • (Y , O".) is continuous. The converse is clear since PO(Y, 0" ) ζ 0" •. We offer the following consequences Theorem 10. If (Y, 0") is resolvable, th e following are equivalent (i ) f: (X,r) • (Y,O") is SMPC. (ii ) f - l (B ) is clopen (closed and open) for each B 드 Y (i ii ) f - l(y) is clopen for each y E Y . (iv) f-l (y ) is op e η for each y E Y . Y (v) f: (X,r) • (Y,2 ) is coη ti때 114 D. A. Rose and R. A. Mahmoud

Proof By Lemma 1 and Theorem 9 Corollary 2. If (X , T) is connected and (Y,a) is resolvable then f (X,T) • (Y ,a) 성 SMPC iff f is a constant function. For example if R is the usual space of real numbers, every non-constant fun ction f : R • R is not SMPC.

Corollary 3.’ If (X πT， 샤) is 값d ense- 따'l17 (Y， 이a ) is a nonempty resolvable space, then there is π o SMPC injection f:(X,T) • (Y,a).

Lemma 5. (Theroem 5 of [7]) For a space (X, 'T) let X = F u G d e η o t e the Hewitt-representation of (X, T). Then PO(X, T) is a topology on X iff CI G is open and {x } E PO(X,'T) for each x E Iπ t F.

Corollary 4. If a space (Y,a) as (X,'T) in Lemma 5, th eη f: (X ,'T) • (Y,a ) 엉 SMPC iff f : (X, T) • (Y , PO(Y, (7)) is continuous

Theorem 1 1. For f : (X ,'T) • (Y, (7 ), the follo wing holds (i) If (X , T) is submaximal, then f is SMPC 펴 it is preirresolute. (ii) SMPC coincides with continu 때 if (Y, (7) is submaximal (i페li prπ'C continu“i“ty and continuity m'C equivalent.

Theorem 12. Let f : (Xπ) • (Y, (7 ) be a surjective SMPC function and (X,T) is compact. Then (Y,a) is st1'Ongly compact. Proof Let {끼 : i E I} be a preopen cover of Y. By SMPC of f , f-l( V; ) E 'T for each i E 1. Hence {f -l (κ) : i E I} is open cover of X which is compact‘ then there is 10 (finite) of 1 such that X = U {f -l (κ):iE 1o}. So, Y = u{ 끼 : i E Io} . Hence (Y,a) is strongly compact.

Corollary 5. St1'Ongly compactness is prese π ed uπ d er a SMP C surjec­ tíon.

Theorem 13. Jf f : (X, T) • (Y,a) is SMPC su η e ctive and (X, T) is LindelöJ, th en (Y, (7) is st1'O ngly Lindelöf Proof Let {κ : i E I} be a preopen cover of Y , then {f - l(κ) : i E I} is open cover of X . Since X is Lindelöf, there ex ists a countable subcover with X = U{f- I(V;) : i E 10 (countable) }. Hence Y = u{κ ‘ i E 10}, this gives the result.

Theorem 14. The inverse image under' an SMPC injection of a pr'C -T, On spaces via den se sets and S MPC fun ct ions 11 5 space, is T; f or i = (0 , 1, 2) P roof We prove this result in one case (say i = 0). So, let f : (X , T) • (Y, (7 ) be SMP C injective, (Y, (7) be pre-To and Xl , X2 be two distinct points of X. T hen f(Xl) 켜 f(X2) io Y. Hence for each preopen set V 드 Y containing one of f(xj), (j = 1,2). So, f - '(V) is open a od contains a corresponding point Xj , (j = 1, 2). Then (X, T) is a 낀o -s pace. The proof of t he others are similarly

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DEPARTMENT OF MATHEMATICS , EAST CENTRAL UNIVERSJTY, ODA , OKLAHOMA , 74820, U.S.A

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