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ON SPACES VIA DENSE SETS and SMPC FUNCTIONS 1. Definitions and Preliminaries

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ON SPACES VIA DENSE SETS and SMPC FUNCTIONS 1. Definitions and Preliminaries 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 interior 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 topology 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 topologies on X having the same semiregularizalion as T. A topological property 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 open set 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 dense set 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.
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