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HOUSTON JOURNAL Volume MINIMAL TOTALLY DISCONNECTED OF MATHEMATICS 20, No. 4, 1994 SPACES Y. BDEIR AND R.M. STEPHENSON,JR. ABSTRACT. 7)-minimal and 7)-closed spaces are studied for the cases: 7) is totally disconnected and Hausdorff; and 7) is totally separated. Several characterization, embedding and product theorems are obtained, and some examples are given. 1. Introduction and terminology. For terms not definedhere, see [8], [9] or [16]. As is the casein [16],all hypothesized topologies and spacesare assumed to be Hausdorff (exceptin Theorem5.13,whereit is explicitlystatedotherwise).Thusthe word "space"means"Hausdorffspace."Occasionallywe shall usethe word "Hausdorff" for emphasis. Recallthat, for a property79 of topologies, a 79-space (X, T) is called: 79-closed if (X, T) is a closedsubsetof every79-space in whichit can be embedded as a subspace;79-completeif there does not exist an extension space(Y,U) of (X,T) suchthat (Y,U)is a 79-space and Y • X; and 79minimal if there does not exist a topology S on X strictly weaker than 7' for which (X,S) is a 79-space.For most properties79 studiedin the past, compact 79 implies 79-minimal,79-minimalimplies 79-closed,and 79closedand 79-completeare equivalent. For someproperties79, 79-closedalso implies79-minimalor compact.In [1],B. Banaschewski studied79-minimal and 79-completespacesfor a number of properties 79. One of the results he obtained was the following. 1991 Mathematics Subject Classification. Primary 54D25; Secondary 54D35, 54B10, or 54A10. The first author's contribution is a portion of a doctoral dissertation at the University of South Carolina. 721 722 Y. BDEIR AND R.M. STEPHENSON, JR. Theorem 1.1. ([1]). Let X be a zero-dimensional space.Then the œollowing are equivalent. (a) (b) (c) (d) X X X X is zero-dimensional-minimal. is zero-dimensional-complete. is zero-dimensional-closed. is compact. In this note we consider•P-minimaland •P-closedspacesfor the properties defined below. A set C is called nondegenerate if it containsmore than one point. A nondegeneratesubsetC of a topologicalspaceX is said to be separatedby U andV if U and V are subsets of X suchthat {C NU, C NV} is a partition of C into nonemptyopensubsetsof the spaceC. A topologicalspaceX will be called: totally disconnected if no nondegeneratesubsetof X is connected; absolutelytotally disconnectedif whenever C is a nondegeneratesubset of X there exist disjoint opensubsetsU and V of X suchthat C is separated by U and V; and totallyseparated if the closed-and-open (clopen)subsets of X separatethe points of X. Each of the last two propertiesimplies the precedingone, but it is well known that if X is totally disconnectedand locally compact, then it is also zero-dimensional.A number of new, as well as old, examplesillustrating distinctionsamongspaceshaving thesepropertiesare consideredin Section5. We notethat someauthorsusethesetermsdifferently([9]), and othersuse the terms "hereditarilydisconnected" ([9]) and "ful_•.y disconnected"([16])or "ultra-HausdortT' ([15])insteadof "totallydisconnected" and "totally separated." Givena filter baseY ona space,the setof adherentpointsof Y, •{clF: F ß Y}, will be denotedadY, and Y will be calledfixed (free)if adY • A filter base )c on a topological spaceX is called an openfilter base (openfilter) if eachmemberof )c is an openset (and wheneverG is an openset whichcontainssomememberof )c then G ß )c). An openfilter base )c on a spaceX will be called: totally disconnected (absolutelytotally disconnected) if for everynondegenerate setC C X • ad)c thereexist (disjoint)opensubsetsU and V of X suchthat C is separatedby U and V, and V containssome member of )c; totally separatedif for every point x ß X • ad)c there existsa clopenneighborhoodof x in X which misses MINIMAL TOTALLY DISCONNECTED SPACES 723 someset in .Y; zero-dimensionalif .Y has a baseconsistingof clopensubsets of X; and semiregularif .Y has a base of regular open sets,i.e., sets of the form F = intclF. The letter "Z)" will denote one of the propertiestotally disconnected, absolutelytotally disconnected, and totally separated.If "Z)" (or another letter denotingone of severalproperties)appearsmore than oncein the statement of a result, it denotesthe sameproperty throughout the result. Givena property79, a spaceX will saidto satisfythe condition:79(i) if everyP-filter baseon X is fixed;or 79(ii) if everyP-filter baseon X which has a unique adherent point is convergent. One characterization theorem we shall use involvessuchconditionsand the property semiregular. Recall that given a topologicalspace(X, T), the family of regular opensetsof (X, T) is a basefor a topologysT on X, and (X, T) is called semiregular if 7' = sT. It is known([8], [16])that for eachT • T, clsTT = cl•-T, and hence(X, sT) and (X, T) havethe sameregularopensets. Theorem 1.2. ([7],[1]) Let (X, T) be a topologicaJ space.The œoJlowing hold. (a) (X, T) isHausdorff-closed if andonlyif (X, sT) isHausdorff-closed if and only if (X, T) satisfiesthe conditionopen(i) (or semiregular(i)) if and only if (X, sT) is Hausdorff-minimal. (b) (X, T) is Hausdorff-minimal if and only if (X, T) satisfiesthe conditionopen(ii) (or semiregular (ii)) if and on/yif (X, T) is Hausdorffclosedand semiregular. Some notation concerningtotally separatedspaceswill be needed. Given a totally separatedtopologyT on a set X, bT will denotethe topology on X whichhas as a basethe family of all clopensubsetsof (X, T). If the topology on X has not been named, then the correspondingzerodimensionalor semiregularspaceswill be denotedbX or sX. The following is immediate. Lemma 1.3. Let T be a totally separatedtopologyon a set X. (•) The spaces(X, :r) and (X, b•) have the s•e clopenset,, T, (X, bT) is zero-dimensional, and (X, T) is zero-dimensional if T = bT. (b) For any topologyS on X, if T C S then bT C bS, and if T C • then bT= 724 2. Y. BDEIR Characterization AND R.M. STEPHENSON, JR. theorems. As indicated above, in the past severalauthorshave usedconditionsof the form 7•(ii) or 7•(i) to provideusefulcharacterizations of 7•-minimalor P-closedspaces, e.g.,see[1]-[8]or [11]-[18].We shallusesimilartechniques for the case 7• is 79. Theorem 2.1. Let X be a 79-space.The followingare equivalent. (a) X is a 79-closed space. x v(i). Proof. A proof is given for the case79 is totally disconnected.The other cases are similar. Suppose(a) is false.Then thereexista totally disconnected spaceY containingX as a subspace and a point p • clX • X. Let ow= {T n X: T is an openneighborhood of p in Y}. Then owis a free openfilter on the spaceX. Considerany nondegenerate subsetC of X. SinceC U {p} is not a connectedsubsetof the totally disconnectedspaceY, there exist open subsetsL and M of Y suchthat C U {p} is separatedby L and M, where, sayp CM. If M ClC • 0, then C is separatedin the spaceX by the open setsU = L f• X and V = M ClX, and V Cow.If M ClC = 0, then there exist open subsetsJ and K of Y which separateC, and C is separatedin the spaceX by the opensetsU = J fflX and V - (K U M) • X, with V • ow. Henceowis a free totally disconnected filter on X, and (b) mustbe false. Conversely,supposethere is a free totally disconnectedfilter base ow on X. Choosea pointp not in X anddefineY = X U {p}, andcall a subset V of Y openprovidedthat (a) V nX is openin X, and (b) if p 6 V thenV contains a member of ow.The spaceY is a totally disconnectedextension spaceof X with Y • X, and so X is not totally connected-closed.[] Corollary 2.2. SupposeX is totally separated. Tt•e followingare equivalent. (a) X is totally separated-closed. (b) Every zero-dimensional filter baseon X is fixed. (c) bX is compact. Proof. It follows from the definition of a totally separated filter base that (b) holdsif and only if X satisfiesthe conditiontotally separated(i), so (a) and (b) are equivalent.If (b) holdsand 6 is any zero-dimensional filter base on the zero-dimensionalspacebX, then 6 is a zero-dimensionalfilter MINIMAL TOTALLY DISCONNECTED SPACES 725 baseon X and henceis fixed. Thus bX is totally separated-closed by (a) and hencecompactby Theorem1.1. Finally, suppose(c) holdsand Y is any zero-dimensionalfilter base on X. Then Y has an adherent point p on the compact space bX. Since Y has a base of clopen subsetsof X, and X and bX havethe sameclopensets,thenp • Cq•'.Thus (b) is true. [] Corollary 2.3. Let 72 be totally disconnectedor absolutelytotally disconnected, and let X be a P-space. The following are equivalent. (a) X is P-closed. (b) X is Hausdorff-closed. Proof. A proof is given for the case P is totally disconnected. Obviously (b) implies(a). Suppose(b) is false. By 1.2 thereis a freeopenfilter on X. Then by Zorn's lemma, there exists a free maximal open filter Y on X. Considerany nondegenerate subsetC of X. SinceX is totally disconnected, there exist open subsetsL and M of X such that C is separatedby L and M. Supposefirst that F f• C • 0 for all F • f'. SinceY is a maximal open filter, L •YifF•(L•C) 5• 0forallF•Y, but ifFf•(L•C) =0for someF C .T, then F VI(M CIC) 5• 0 for all F • .T, in whichcaseM Supposenext that there is a set G • 5r suchthat G VIC = 0. Then one can define U = L and V = M U G, and note that C is separated by the open sets U and V, and V C .T. Thus 5r is a free totally disconnectedfilter on X, and (a) must be falseby Theorem2.1. [] The next three theoremsrelate D-minimal and D-closed spaces. We sketch proofs of them. Theorem 2.4. Let (X, T) bea topological space.The followingare equivalent. (a) (X, T) is totallyseparated-minimal. (b) (X, is zro-dimensional-minimal. (c) (X, T) is compactand totally disconnected. Proof. Suppose(a) holds. It then followsfrom 1.3 (a) that bT = 7' and (X, T) is zero-dimensional. Thus (X, T) is zero-dimensional-minimal. If (b) holdsthen (X, T) is compactby 1.1. If (c) holdsthen (X, T) is both zero-dimensionaland Hausdorff-minimal,and henceit is totally separatedminimal. [] 726 Y. BDEIR AND R.M. STEPHENSON, JR. Theorem 2.5. Let P be totally disconnectedor absolutely totally discon- nected,and let (X, T) be a P-space.The œollowing are equivalent. (a) (X, T) is P-minimal. (b) (X, P Proof. Supposethere existsa topologyb/on X strictly weakerthan T such that (X,L/) is a P-space. There must exist a point p 6 X and an open neighborhood of p in (X, T) whichcontainsno/,/-neighborhood of p. Then .• = {U • b/: p • U} is a P-filter baseon the space(X, T) whichhas a unique adherent point but is not convergent.Conversely,supposethere is a P-filter base.Y on (X, •r) whichhasa uniqueadherentpointp but is not convergent. Defineb/-- {T E (r: if p E T, thenT containssomememberof Then b/is a P-topology on X which is strictly weakerthan (r. [] Theorem 2.6. Let 72 be totally disconnectedor absolutely totally disconnected, and let X be a topological space. If X is 72-minimal, then it is 72-closed. Proof. By Theorems 2.1 and 2.5, it sufficesto show that if each P-filter base with a unique adherent point is convergent,then each P-filter base is fixed. We prove the contrapositive. If p E X and .Y is a free P-filter baseon X, let • = {T: T is open,p E T, andT contains somememberof •). It is easyto verify that G is a filter basewith a uniqueadherentpoint (namely p). The fact that .Y is free impliesthat G is nonconvergent, and that p is its onlyadherentpoint. We showthat • is a P-filter base.Let C C X • {p} be any nondegenerate set. There exist (disjoint)opensetsL and M of X which separateC, where M containssomemember of .Y. If L n C consistsof one point, there exist disjoint open sets I and J with L n C C [ and p E J. If L n C is nondegenerate, there exist (disjoint)open setsI and J which separate L • C, where p • J. In either case, U = L • I and V = M t3 J are (disjoint)opensetswhichseparateC, and V containsa memberof •. [] As a semiregularHausdorff-closedspaceis Hausdorff-minimalby 1.2, one obtains the followingconsequence of 2.3 and 2.6. Corollary 2.7. Let X be a semiregular totally disconnectedspace. The œollowing are equivalent. (a) X is totally disconnected-closed. (b) X is totally disconnected-minimal. MINIMAL TOTALLY DISCONNECTED SPACES 727 (c) X is Hausdorff-minimal. The authors do not know if every totally disconnected-minimalspace is Hausdorff-minimal, but one can use 1.2, 2.6 and the next lemma to provethat everyabsolutelytotally disconnected-minimal spaceis Hausdorffminimal. Lemma 2.8. If a spaceX is absolutely totally disconnected,then so is its associatedsemiregularspacesX. Proof. If U and V are disjoint open sets of X, then intclU and intclV are disjoint open sets of sX which contain U and V. [] Corollary 2.9. Let X be an absolutely totally disconnectedspace. Then X is absolutley totally disconnected-minimal if and only if it is Hausdorffminimal. 3. Product spaces, subspaces, and images. Dependingon the particular property 72 , there is considerablevariation in the behaviorof the properties72-closed and 72-minimal. For example, if 72is regular,the twopropertiesarenot productive([14])or closedhereditary [5], but it is knownthat if 72 is Hausdorff,the property72-minimal (72-closed) is productive([11], [17]), and 72-closed is inheritedby the closuresof opensetsand preserved undercontinuous maps([8, p.146]). As a consequence of the latter knownresultsand 2.3, 2.7 and 2.9, we immediately obtain analogoustheoremsconcerningthe case72is totally disconnectedor absolutelytotally disconnected.In this sectionsuchresultsare stated without proof, and resultsare given concerningthe case72is totally separated. Theorem 3.1. Supposethat X is a topologicalspace. (a) If X is totally separated-closed and U is any opensubsetof X whoseboundary B is a totally separated-closed space, then clU is totally separated-closed. (b) Iœ72is totally disconnected or absolutelytotally disconnected, X is 72-closed,and U is any open subsetof X, then clU is 72-closed. The proofof 3.1 (a) is straightforward:If .• is a zero-dimensional filter baseon clU, then either F D B • 0 for all F E .•, sothat • = {F D B: F E .•} is a zero-dimensional filter baseon the spaceB, or there existsG E .• with G D B -- 0, in whichcase• = {F D G: F E .•} is a zero-dimensional filter base on the spaceX. In either case,• and hence.• must be fixed. [] 728 Y. BDEIR AND R.M. STEPHENSON, JR. It is notedin ([8, p. 147])that everycountable Hausdorff-closed space has a denseset of isolated points. For D-closedspaces,a similar r•sult can be obtained. Theorem 3.2. Let X = {xn} be a countablyinfinitetotally disconnected space. (a) If X satisfiesthe conditionzero-dimensional (i), thenX hasinfinitely many isolated points. (b) If X is D-dosed,thenit hasa denseset of isolatedpoints. Proof. (a). If every suchspacehas at least one isolatedpoint, then the set I of isolatedpointsof X must be infinite, for, otherwise,X • I would be a clopenset satisfyingzero-dimensional (i) and wouldhavean isolated point p, but thenp wouldbe an isolatedpoint of X in X • I. Supposethen that X has no isolated points. Since X is not connected it has a nonempty clopen subsetK• not containingxl. Since/t'• is infinite and not connected it contains a nonempty clopen set N2 not containing x2. Proceeding by induction,onecanconstructa freezero-dimensional filter base{Kn} on X, whichwouldcontradictzero-dimensional (i). Thus (a) holdsby Corollary 2.2. (b). Suppose X is totally separated-closed. By 2.2, everyzerodimensionalfilter baseon X mustbe fixed. Supposealsothe openset Y = X • clI is nonempty, where ! is the set of isolated points of X. Then the space Y contains no isolated points since any suchpoint would be isolated in X, so Y is infiniteand thereexisty 6 Y • {Xl} and a clopenset L• containingy but not x•. Now L1 fflI/' is infinite sinceit is nonempty and doesnot contain an isolatedpoint, sothereexistz 6 L• fflY • {x2) and a clopenset L2 containing z but not x2. Continuing,one obtainsa sequenceof clopensetsthat generatesa free zero-dimensionalfilter baseon X, which is a contradiction. Thus X =clI. [] Theorem 3.3. Suppose that X and Y are D-spaces and there exists a continuousmap f of X onto Y. (a) If X is D-closed,thensois Y. (b) If 72 is totally disconnected or absolutelytotally disconnected, ¾ is 72-closed,and f: X -* Y is a perfect open map, then X is 72-closed. (c) If Y is totallyseparated-closed, f-1 (p) is totallyseparated-closed for eachp • ¾, and f(N) is clopenin Y wheneverK is clopenin X, then X is totally separated-closed. MINIMAL TOTALLY DISCONNECTED SPACES 729 Proof. It is straightforward to check that if • is a free/p-filter base on Y thenf-•(•r) = {f-•(F): F E •r} is alsoa free/P-filterbaseon X. Thus (a) holdsby 2.1. The proofsof (b) and (c) aresimilar.We prove(b). Suppose that the hypothesis of (b) holdsand •r is an openfilter baseon X. Then f(5r) = {/(F): F E 5r} is an openfilter baseon Y and hasan adherentpointp. Thusp • clf(F): f(clF) for everyF • •r, so{f-•(p) r3clF: F • •r} is a filterbase,and•r musthavean adherent pointin the compact setf-• (p). Hence by 2.1, X is P-closed. [] Now, let us considernext the productivity of these properties. It is not difficultto show(andis knownin two of the cases([9, 6.2])) that/P is productiveand hereditary, and it is knownthat semiregularityis productive, sousingthe remarksat the beginningof this section,aswellas 1.1, 3.3 (a), and related results, one can easily obtain the following product theorem. Theorem 3.4. Let X be a nonempty product space. (a) If X is/P-closed(/P-minimal),thensois eachof its factorspaces. (b) If eachfactorspaceof X is totaJly-separated minimaJ,sois X. (c) If eachfactor spaceof X is totally totaJlydisconnected-closed (disconnected-minimM and semiregular), thensois X. (d) If eachfactorspaceof X is absoluteJy totalJydisconnected-closed (absolutelytotally disconnected-minimal), thensois X. Using 3.3 (c) and the characterizations 3.5 and 3.7 below, we can provethat if certain additional restrictionsare satisfied,the property totally separated-closedis productive. Theorem 3.5. Let X and Y be totMly separated-closed spaces. The following are equivalent. (a) For eachclopenset K of X x Y, prx K is a closedsubsetof X. (b) X x Y is totally separated-dosed. (c) b(X x V)= bX x Proof. Suppose(a) holds.Applying3.3 (c) to the map f = prx, oneobtains (b). If (b) holds,then b(X x Y) is compact,henceHausdorff-minimal, and sothe topologyon bX x bY (whichis weakerby 1.3 (b)) cannotbe strictly weakerthan that on b(X x Y). Finally,suppose (c) holds.Then for every clopenset K of X x Y, K is a compactsubsetof b(X x Y) = bX x bY, and henceprx(K) is a closedsubsetof bX and thereforeof X. [] 730 Y. BDEIR AND R.M. STEPHENSON, JR. Using 3.5 and a techniquesimilar to one due to H. Tamano, one can obtain an analogueof his theorem that the product of two pseudocompact Tychonoffspacesis pseudocompact if oneof the spacesis a k-space([19]). Corollary 3.6. Let X and Y be totally separated-dosedspaces,and suppose X is a k-space. Then X x Y is totally separated-closed. Proof. Let K by an clopensubsetof X x Y. We wishto provethat prx (K) is a closedsubsetof X. SinceX is a k-space,it is enoughto prove that for everycompactsubsetC of X, C n prx (K) is a closedsubsetof C. Applying a.a (c) to the spacesC x Y and Y and the perfectmap f = pry, where pry: C x Y -• Y, oneseesthat the spaceC x Y is totally separated-closed. Let !/= pro, wherepro: C x Y -, C. By 3.5, the image9((C x Y)n K) of the clopensubset(C x Y) n K of C x Y is a closedsubsetof C. But g((C x Y) n K) = C n prx(K), andsoC n prx(K) is a closedsubsetof C. Thusprx(K) is a closedsubset of X. By 3.5, X x Y is totally separated-closed.[] By 3.6, the product of finitely many totally separated-closedspaces is totally separated-closed if at most one of the factor spacesfails to be compactor first countable.For infinitely many spaces,an analogueof 3.5 can be obtained. Theorem 3.7. Let X = I•{Xa : a • A} be a productof totally separatedclosedspaces. The following are equivalent. (a) For eachclopensetK ofX andsubsetR of A, prR(K) is a clopen subsetofXR = I-I{X,.:r G R}. (b) For eachclopensetK of X andfinitesubsetR of A, pr•(K) is a clopensubsetof Xn = I•{Xr : r • R}. (c) X is totally separated-closed. (d) bX = I•{bXa : a • A}. Proof.Proofsthat (c) implies(d) and (d) implies(a) are similarto proofs givenof the corresponding implicationsfor 3.5. Suppose(b) is true. Let us prove(c) holds. Note first that for any finite subset_Rof A and clopen subsetK of Xn, the projection of K onto anyfactorspaceXr is a closedsubsetof X,--just apply(b) to the clopen set K x I-i{X• :a 5• r} and map pr, X -• X, Next onecan apply3.5 a MINIMAL TOTALLY DISCONNECTED SPACES 731 finite number of times to the factor spacesof Xn and concludethat each XR is totally separated-closed. Consider any zero-dimensionalfilter 6 on X. We wish to prove that 6 is fixed. Let )r be a maximal zero-dimensional filter on X such that 6 C )r. It followsfrom (b) that for eacha G A, pra()r) is a maximal zero-dimensionalfilter on the totally separated-closedspaceXa, and hence there existsexactly one point p• • 91pr•()•). Definep to be the point {Pa : a • A}. Similarly,sincefor eachfinite subsetR of A, prR()r) is a maximal zero-dimensionalfilter on the totally separated-closedspace X•, there is a uniquepoint q • (•pr•()r), sofor eachsuchR, we haveq = {p,.: r • R} • (•pr•()r). Finally,notethat by the preceding if N is any basic open neighborhoodof p, and if R is the set of restricted coordinatesof then for each F • )r there exists x • F with x,• = p7-for each r • R, and so N fflF • •. Thus p is an adherentpoint of )r in X, and p HenceX is totally separated-closed, and (c) is true. [] Beforestating one applicationof 3.7, we recall that a topologicalspace is calledfeeblycompact ([16, 1.1], [17])if everycountableopenfilter base on the space is fixed. Corollary 3.8. Let X = ri{x• ßa • A} bea productof totally separatedclosed spaces which are first countable and feebly compact. Then X is totally separated-closed. Proof. By [17,Theorem4.4], the spaceX is feeblycompact.Sincecontinuousimagesand clopcnsubsetsof feebly compactspacesare feebly compact, and feebly compactsubsetsof first countablespacesare closedsubsets,then for everyclopcnsubsetK of X and countablesubsetR of A, pr•(I•) is a clopcn subsetof XR. Hence X is totally separated-closed by 3.7. [] In order to obtain a secondconsequenceon 3.7, we briefly consider the following two concepts.Given topologies$ and T on a set X, if $ C T, then we shall write •q _<T provided that for every S • •q,cl$S = clzS. If (X, T) is a totally separatedspace,then (X, T) will be calledclopen Hausdorff-closed providedthat every open filter base on (X, bT) has an adherentpoint in (X, T). Note that for a totally separatedspace,clopcn Hausdorff-closcdis implied by Hausdorff-closcd,and as indicated below, it implies totally separated-closed.Three lcmmas will be needed. 732 Y. BDEIR AND R.M. STEPHENSON, JR. Lemma 3.9. ([20,Theorem2]) Let S and7- be topologies ona setX such that $ C T. Then $ < T if and only if for every set T C T, T C intsclsT. Lemma 3.10. Let (X,T) = l-I{(Xa,•) ßa C A} be a nonemptyproduct of totally separated spacessuch that /'or each a G A, bT• < T•, and let (x,s) = ßa e A}. Thenb7,= S, and b7,< 7,. Proof. By 1.3, $ C b7'. To show that b7' C $, it sufficesto prove that every7,-continuous mapof X into J = {0, 1} is alsoS-continuous. Suppose f: (X, 7') -* J is 7,-continuous. Letp • X andB = 91{pr71(B•):r• R} be a 7'-basicopenneighborhood of p, whereR C A is finite, and f I B is constant,say, f(B) = 0. Fix s C R and q • B. Let g : X• • J be definedby the rule g(x) - f(t) if t • X satisfies t• = x and ta = q• for any a e A • {s}. Clearly,g is continuous on (X•, T•) and henceon (X•, bT•), and g(B•) = 0, sog(B•*)= 0, whereB•* denotesthe interiorof the closure of B• in (X•,bT•). Likewise,f(t) = 0 for any point t suchthat t• e Bj and t• = q• for any a C A • {s}. Sinceq wasan arbitrary point of B, the function f vanishes throughout pr71(B• *) 91(91{pr71(B•): r e R • {s}}). If oneappliessimilarreasoningto the latter set and someu • R • {s}, then doesso again to the new set that results, --1 * pr•1(B•)91pr• (B•)fl(ffl(pr71(B•)'r GR•{s,u})), and so on, then after a finite numberof stepsit will follow that f vanishes throughout the set B* = 91{pr71(B•):r • R}. Now by 3.9, eachB• C B•*, sop C B C B* • $. Hencef is S-continuousat p. Sincep was arbitrary, f is •q-continuous.Therefore,b7' = •q. To verify the last statement,considerany set S • b7,. We wish to show that clb•-S= clzS. Sinceb7, C 7,, clbzS D clzS. Supposep • X • clzS and B (denotedas above)is a basic7,-openneighborhood of p missingS. Since S•bT, S91B=13, andbT=$,wehave S 91B* = S 91intsclsB = S 91int6crc16zB= 13. Becauseeach b7,• _< T•, B C B* by 3.9. Thusp • B* • b7,, andp • X \ cl•S. [] MINIMAL TOTALLY DISCONNECTED SPACES 733 Lemma 3.11. Let (X, T) be a totally separatedspace.The followingare equivalent. (a) (X, T) is clopenmusdor-closed. (b) (X, T) is totallyseparated-closed and bT _<T. Proof. Suppose(b) holds.Let )r be any openfilter baseon (X, bT). Since (X, bT) iscompact,)r hasan adherentpointp in (X, bT), andsincebT _<T, then p • clbxrF = clxrF for each F • )r, which shows)r has an adherent point in the space(X, T). Thus (a) is true. Conversely, assume(a) holds.Obviously,(X, bT) mustthen be zerodimensional-closed, hencecompact,and so (X,T) is totally separatedclosed. Consider any set V • bT. We wish to show that clxrV = clbxrV. SincebT C T, cl•rV C cl•:rV. Let x • cl•rV and A/' be the family of open neighborhoods of x in (X, bT). ThenA/'I V is an openfilterbaseon and must have a T-adherent point q. As q is also a bT-adherent point of A/'] V, q = x. Thenx • clzV. [] Theorem 3.12. The property clopenHausdorff-closedis productive. Theorem 3.12 follows from 3.7 and 3.9-3.11. 4. Extension spaces. In this section it will be shown that every/)-space has a/)-closed ex- tension space,and every semiregularabsolutelytotally disconnectedspace has an absolutely totally disconnected-minimalextension space. The fol- lowingterminologybasedon [3] will be used. Let X be a topologicalspaceand A/[ a family of free open filters on X such that whenever jr, • 6 A/[ are distinct then there exist disjoint sets F • jr and G • •. Let E(X) = X UA/[ be their disjointunion.Then E(X) is called the simple extensionof X with filter trace A/I if it is topologized sothat a subsetV of E(X) is openif and onlyif (a) V Cq X is openin X, and (b) if Jr • V•A/[ then F C V for someF • •. E(X) is called the strict extensionof X with filter traceA/I if it is endowedwith the topology whichhasas a base{V*: V is openin X}, wherefor any opensubsetV of x, v*=vu{•Aa:v•}. Theorem 4.1. Let X be any totally separatedspace which is not totally separated-closed,and let ./Vl be the set oœall œreemaximal zero-dimensional tilters on X. Then the simpleextensionE(X) oœX with filter trace Ad 734 Y. BDEIR AND R.M. STEPHENSON, JR. is totally separated-closed,and, for every compactzero-dimensionalspace Y and continuous map f: X -• Y, thereis a continuous extensionfs : s(x) Y of f. Proof. Sinceeach •r • j• is a free maximal zero-dimensionalfilter on X, it is easy to show that for each clopen subset K of X, elk = IV* is clopen in E(X), and to usethe latter to showthat E(X) is a totally separated extensionspaceof X. Let 6 be any zero-dimensional filter on E(X) and considerthe zero-dimensional filter T/= 6 I X of the spaceX. If NT/ • 0 then 6 is fixed. If NT/= 0 then there exists•r G j• with T/C •r, and hence the "point" •r is an adherent point of 6, for given any basicneighborhood K U {•r} of •r, whereK is a clopensubsetof the spaceX, onehasK G •r and therefore for all G • 6. Thus E(X) is totally separated-closed. To verify the last statementof 4.1, first note that sinceX and bX (E(X)and bE(X)) have the sameclopensets,then the compactzero-dimensional spacebE(X) is an extensionspaceof bX whosefilter traceis {•r: •r is a freemaximalzerodimensionalfilter on bX}. Now let f bc asin the hypothesisof 4.1. Because Y has a base of clopen sets and X and bX have the same clopen sets, f: bX -• Y is alsocontinuous. Thusthereis a continuous extensionfE: bE(X) -• Y ([3],[16,4.7]).ThenfE: E(X) -• Y is alsocontinuous.[] Theorem 4.2. Let 7p be totally disconnectedor absolutely totally disconnected, let X be any 7P-spacewhich is not 7P-closed,and let A/t be the set of all free maximal openfilterson X. Then the simpleextensionE(X) of X with filter trace A/t is a 7P-closedextensionspaceof X. Proof. A proof is given for the case7p is totally disconnected.It is known that E(X) is a Hausdorff-closed extensionspaceof X ([3], [16,4.8]). Considerany nondegenerate subsetC of E(X). We wishto provethat C is not a connectedsubspace of E(X). SinceX is totally disconnected and E(X) • X is a discretespace,C is not connected if C C X or C C E(X)•X. Let usassumethe setsD = C•X and R = C • X are nonempty. If D consistsof only one point, then the setsS = E(X) • D and T = X are open,and C is separatedby S and T. Supposethen that D contains more than one point. MINIMAL TOTALLY DISCONNECTED SPACES 735 Since D is not connectedthere exist open subsetsU and V of X which separate D in X. Define S= Ut• {.? e R :.? e cl(U C3D)} and T - V t• {• e R: • e cl(V V3D)} U (E(X) • ciD). Then C C S U T and one has S ClC • (3 • T N C. Considerany point .• • S • X. As )r • cl(U ClD), U ClF • 13for all F • )r, and henceU since)r is a maximalopenfilter. Thus U U {)r} is an openset contained in S, and S is a neighborhoodof .•. Since )r was an arbitrary element of S • X, S is a neighborhood of eachpoint of S • X. In addition,S ClX = U is open, so $ is an open set. Similarly, one can show that T is a union of open sets and hence is open. Clearly,S OlT OlD = U r3(V u (X \ clD)) r3D = 13. OnealsohasSr3Tr3R= 13,for consider anypoint)r •R. If )r • E(X) •clD, then )r ½ S. If .• • cl(U ClD) and )r • cl(V ClD), then U ClF • 13for all F 6 )r, which implies U • )r, and similarly also V • .•, so that U But fromthe latter it wouldfollowthat {St} U (U r3V) is a neighborhood of )r, whereasDCl ({)r} U (S ClV))= 13while.• • clD. Thusif )r • R ClclD, then .• is in exactly one of the sets cl(U ClD) and cl(V ClD), that is, )r • (SUT) • (S ClT). Therefore,$ ClTClC = 13,whichcompletes the proof that C is not a connected subsetof E(X). [] Remark•.3. It is known([8, p.148])that the spaceE(X) in Theorem4.2 satisfies the following property: If ¾ is any Hausdorff-closedspace and f : X -• Y is a continuous open mapping, then there is a continuous extension f•; :E(X) -• Y. Theorem 4.4. Let X be a semiregular absolutely totally disconnected spacewhich is not absolutely totally disconnected-closed, and let 2M be the set of all free maximal semiregularfilters on X. Then the strict extension E(X) ofX withfilter trace./Misan absolutely totallydisconnected-minimaJ extensionspace of X. Proof. Banaschewski proved([2]) that for any semiregularspaceX, the spaceE(X) in the statementof 4.4 is a Hausdorff-minimal extensionspace of X. Considerany nondegenerate subsetC of E(X). We provethat there exist disjointopensubsetsS and T of E(X) suchthat C is separatedby S and T. 736 Y. BDEIR AND R.M. STEPHENSON, JR. SupposeC C E(X) • X. Sinceeverymemberof .A4is a freemaximal semiregularfilter, if •' and 6 are distinct membersof .A4 in C, there exists a regularopen set F of X suchthat F E •' and X • clxF • 6. Hence $ = F* andT = (X • clxF)* aredisjointopensetswhichseparate C (and alsoseparateE(X) • X sinceX • clxF is a regularopenset of X whose union with F is a densesubsetof X). If C F•X consistsof exactlyonepointp and if •' • C • X, then there existsa regularopenneighborhood R of p in X suchthat X • clxR • •'. We may then take $ = R* andT = (X • clxR)*. Assume C F•X is nondegenerate.Since X is absolutely totally disconnected, there exist disjoint open sets U and V of X such that C F•X is separatedby U and V in X. Then $ = (intxclxU)* and T -- (intxclxV)* U (X • clx(U U V))* have the desiredproperties. [] Lemma 4.5. Let X be a completely norma/ totally disconnectedspace. Then X is absolutely totally disconnected. Proof. SupposeC is a nondegeneratesubsetof X. Then there are nonempty subsetsA and B of C whichseparateC. HenceA 91clB = (3= B 91clA, so by the completenormality of X, there are disjoint open sets U and V such that A C U and B C V ([9, 2.1.7]). [] Corollary 4.6. Every completelynorma/ totally disconnectedspacehas an absolutely totally disconnected-minimalextensionspace. 5. Examples and related results. Severalexamplesand resultswill be given which illustrate distinctions among these spaces,as well as ways in which someof the previousresults cannot be extended. Example 5.1. Let X be the well knowncountableminimal Hausdorffspace of N. Bourbakiwhichis not compact.X is describedin [4], [7] and [16]. It is not difficult to prove that X is absolutelytotally disconnected.Hence X is absolutely totally disconnected-minimal. It is not, however, totally separated since there exist two points in the space X which do not have disjointclosedneighborhoods.If Y is oneof the Hausdorff-closed subspaces of X which is not minimal Hausdorff, then Y is totally separated but not zero-dimensional. MINIMAL TOTALLY DISCONNECTED SPACES 737 Example 5.2. In [12],H. Herrlichmodifiedthe unit intervalto showthat not every minimal Hausdorff space is of the secondcategory. Applying his method to the Cantor set C, one can construct an absolutely totally disconnectedminimal spaceX which is not of the secondcategory. Choose disjoint dense subsetsL and R of the Cantor set C whose union is C, and let X = L x {1, 2} U R x {3}, wherethe topologyon X hasas a base {x n (T x {1, 2,3)) ßT is openin {X t• (T x {i)): i E {1,2} and T is openin C}. It is straightforwardto checkthat: X is Hausdorff-closed and semiregular; and for any point (c,i) E X, the quasi-component of (c,i) in X is either {(c, i)} if/= 3, or {(c, 1), (c,2)} if/5k 3. If L is chosen to be countablethen X is not of the secondcategory. If one definesM to be the equivalence relation on X which collapseseach quasi-componentto a point, then the quotientmap q: X -• X/M is a perfectopenmap, but the totally separated spaceX/M is not totally disconnected-minimal. Example 5.3. Applyinganothermethodof Herrlich[11]to the Cantorset C, let T be the usual topology on C, choosethree disjoint densesubsets {L,M,R} of (C,T) whoseunionis C, and let Then the space(C,L/) is clopenHausdorff-closed, for onecan easilyshow that (C,b/d) = (C, T), which is compact,and b/d _</d, and then apply 3.11. Note that althoughthe closureof everyopensubsetof (C,/d) with compact boundary is totally separated-closedby 3.1, the closure of the open set L, clL = LUR, is not, sinceifp • M, {TtqL 'p • T • T} is a free zero-dimensional filter baseon (clL,/d clL). Hence(C,/d) is not Hausdorff-closed. Example 5.4. There are a number of non-compact,regular spaceswhich have appeared in articles concerningregular-closedspaceswhich one can easily verify are totally separated-closedspaces. We identify several: it is shownin [11] that a certainsubspace Y of the Berriozdtbal-Sorgenfrey regular-minimalspacein [5] is not regular-minimal,but is regular-closed and hasthe propertythat C(Y) is pointseparating--one canprovethat Y 738 Y. BDEIR AND R.M. STEPHENSON, JR. and the corresponding generalizations of Y in [6]are alsototally separatedclosed;and the regular,but not regular-closed, completelyHausdorffspace constructed in Theorem2 of [13]is totally separated-closed. One can use Lemma 3.11 to provethat Y is not clopenHausdorff-closed by taking V to be the interior of one "plank" in Y, and noting that bY - the one-point compactification of Y • {p), andp 6 (clot'V)• clyV. If IV is Hausdorffor absolutely totally disconnected, a IV-space(X, 7,) is IV-closedif and only if its associated semiregularspace(X, sT) is IVminimal, but the constructionbelow showsthat there is no totally disconnected analogueof this result. First we state a result due to Banaschewski that will be needed(seealso[16,p. 88]for a proof). Lemma 5.5. ([2]) If (Y,];) is a densesubspace of (E, T), then (Y, s];) = (Y, (s7,) I Y). In particular,everydensesubspace oœ a semiregular spaceis semiregular. Example 5.6. Thereexistsa totallydisconnected-closed space(X, 7,) such that there existsno topology/d on X weakerthan 7' for which (X,/d) is a semiregulartotally disconnectedspace. In particular, the Hausdorffminimalassociated semiregular space(X, s7') of (X, 7') fails to be totally disconnected, and so (X, 7') is not absolutelytotally disconnected. Proof. Let P be any nondegenerateconnectedsubspaceof the plane con- taining a point p suchthat P • {p} is totally disconnected. An exampleof such a spaceconstructedby B. Knaster and C. Kuratowski can be found in varioustexts (e.g., [9, 6.323]). The spaceP • {p} is absolutelytotally disconnectedby Lemma 4.5, but it can be usedto constructa spacehavingthe statedproperty.Let Y = P x {1,2}, andlet 1; = the familyof all subsets V of Y suchthat: if x E P \ {p} and (x, 2) E V, thenfor someneighborhood U of x in P\ {p},V D (U\ {x}) x {1,2}; andif (p,2) • V, thenfor some openneighborhood U of p in P, V D (U • {p}) x {1}. Then it is easyto showthat the space(Y,1;) is totallydisconnected, for the pointsin P x {1} forma (dense)setof isolatedpints,andsincePx {2} ishomeomorphic with the directsumof P \ {p} and a singletonset,the subspace P x {2} is totally disconnected.Note, however,that in the associatedsemiregularspace (Y, s];) the sameset,P x {2}, whenconsidered asa subspace of (Y, s];), is homeomorphic with P and henceis connected.Thus (Y, s];) is not totally disconnected. MINIMAL TOTALLY DISCONNECTED SPACES 739 Let (E, 5r) bethe totallydisconnected-closed extension spacefor (¾, obtainedin 4.2. By 5.5, (¾,slY)= (¾,(s(T) lY). Hencethe space is not totally disconnected. Now consider anytopologyW on E suchthat (E, W) is a semiregular space and W C •. We shall prove that W = sZ, which will show that (E, W) is not totally disconnected. If J is anyregularopensetof the space (E, T), thenits complement E• J = cl(E• clJ) is the closureof an openset and sois a Hausdorff-closed subspace of (E, Z). Thusthe continuous image of E • J underthe identitymap i: (E, • (E, W) is Hausdorff-closed, and E • J is closedsubsetof (E, W), i.e., J • W. Sincethe regularopen setsof (E, T) form a basefor sT, the precedingshowsthat sT C W. Then by the semiregularity of (E, W), oneobtainsW = sT. The following example shows that the property totally separatedclosedis not productive. Example 5.7. Let X be the spaceconstructed in [18].The proof(d) given in [18] that the continuous real valuedfunctionsseparatethe pointsof X showsthat the clopensubsetsof X separateits points. It is also immediate from page77 and the proof(e) in [18]that X satisfiesthe conditiontotally separated(i). ThusX is totally separated-closed. Finally,it wasnotedthat X x X contains an infinite closedsubsetC such that every point of C is an isolated point of X x X. Since the complementsof the finite subsetsof C form a base for a free zero-dimensionalfilter on X x X, the spaceX x X is not totally separated-closed. Analogousto RemarkI in [18],we note that by 2.2 this example also showsthat there is totally separated spaceX for whichbX is compact,but b(X x X) • bX x bX. The next two results show that in Theorem 4.1 the requirement that the space Y be zero-dimensionaland compact cannot be replaced by the requirement that Y be totally separated-closed.Before stating it, we recall that by an almost compactspaceis meant any completelyregular spaceX suchthat I•X • X I • 1. Theorem 5.8. Let X be any almost compactzero-dimensionalspacewhich contains a regular open set R whoseboundary is not compact. Then, [or every totally separated-closedextensionspaceE of X, there existsa totally separated-closedextension space Y of X such that the identity map f X • Y hasno continuous extension fs: E • Y. 740 Y. BDEIR AND R.M. STEPHENSON, JR. Proof. Let X be as in the hypothesis.Let )r = {T :T is openand T D K for somenon-compact,clopensubsetK of X}. By [10, 6J], of any two disjoint zero-setsof X, at least one is compact. Hence for non-compact, clopensetsK and L,X \ (K n L) = (X \ K) U (X \ L) is compact,and so K n L C )r. Thus )r is a zero-dimensionalfilter on X, and since X is locally compactand zero-dimensional,)r is free. In addition, )r is maximal zero-dimensional, for givenany clopensubsetK of X, oneof K and X • K is non-compact,and so either I• C )r or X • K G )r. )r is also the only free maximal zero-dimensionalfilter on X, for given distinct maximal zerodimensionalfilters M and B on any space,there exist disjoint sets A G M and B G B, and no free filter on X could contain a compact set. Let E be an arbitrary totally separated-closed extensionspaceof X. Let • = {T: T is openin E and for someclopenset K of E, T D K and K O X G )r}. Then • is a zero-dimensional filter on E, so there existsa point p C N•. Now for eachclopenset K of X in )r, X • K is compact and thusa clopensubsetof E, andhencethe setG = E • (X • K) • • and satisfies O fqX = K. Thus6 [ X = • andp • E \ X. Also,E \ X = {p}, for givenany point x • E \ X, the restrictionto X of the family P of all clopen neighborhoodsof x would be a filter base of non-compact,clopen subsetsof X, and soonewouldhaveP I X C .T, whichwouldimply x and p have the same quasi-componentin E. Let R be as in the hypothesis.Note that for every clopen set K of X in .T, K•R y• O,for if somesuchK satisfiesKOR = 0, then K•clxR = 0 and henceclxR C X \ K, whereasX \ K is compactand clxR is not. Furthermore,sinceclx(X \ clxR) = X\ intxclxR = X \ R, the setsR and X • clxR havethe sameboundaryin the spaceX, and soa similarargument showsthat for everyclopenset K of X in •, K (q(X \ clxR) y• 0. Let C and 7• be maximal open filters on X strongerthan .T suchthat R • 7• and X \ clx R C C. Let Af be the familyof all openneighborhoods of p in E. Then A/' I X •: C or AfIX •: •. Suppose AfIX •: •. Because T• is a maximalopen filter on X, there must exist a set T G T• suchthat T containsno member ofN' I X. Let Y = E, andcalla subsetV of Y openin Y providedthat V • {p} is openin X andifp E V thenV•{p} • 'R. Since5gis a freeopenfilter on the spaceX, ¾ is a Hausdorffspace,and sinceX = ¾ • {p} is locallycompact and zero-dimensional,¾ is totally separated. If N is any zero-dimensional MINIMAL TOTALLY DISCONNECTED SPACES 741 filter on Y whichhas no adherentpoint in X, then 7-/ [ X C ,T C 7•, and so p is an adherent point of 7-/. Thus ¾ is totally separated-closed. Let f : X -• Y be the identity map of X. Then f is continuous. Suppose next that thereis a continuous extension fE : E -• Y of f. As f' C .A/[X, andf' andfE(A/'[ X) = A/'[ X arefreefilterbases onthespace X, wehaverE(p) •kx foranyx in X. ThusrE(p) _p. But theset isanopenneighborhood ofp in thespace Y, andrE(N) = (N AX)[J {p} T [J{p} for anyneighborhood N ofp in E. ThusfE cannotbe continuous at p. Therefore,thereis no continuous extension fE: E -• Y of f. [] Example 5.9. It can be shownthat if X is the spaceof countableordinals, or X is the Tychonoff plank(or a generalized versionof it as in [6]),then X satisfies the hypothesisof Theorem 5.8. It is known that for each such space,[fix • X[ = 1. If R is the set of isolatedpointsof X which(or both of whosecoordinates)are "even"non-limitordinals(i.e., of the form 2i or c•+ 2i, wherei is finite and c• is a limit ordinal),then R is a regularopen set of X whoseboundary is not compact. Recall that if X is a topologicalspaceand œ is a family of extension spacesof X then a spaceE C œis calleda projectire maximum in E provided that for everyspaceY C œ,thereis a continuous extension fE: E -• Y of the identity map f on X. If X is a zero-dimensionalspacewhich is not compact, 2•/tis definedas in Theorem 4.1, and E is definedto be the strict extensionspaceof X with filter trace.A/t,then it is well known([3]) that E is a projectivemaximum in the family E of all compact zero-dimensionalextensionspacesof X. In contrast, the results 5.8 and 5.9 establishthe following. Example 5.10. Let X be one of the spacesin Example 5.9. Then there is no projective maximum for X in the family œof all totally separated-closed extension spacesof X. Example 5.11. Although every totally separated space X has a totally separated-closed extensionspaceE(X), and every(absolutely)totally disconnected spaceX hasa (an absolutely)totally disconnected-closed extensionspaceE(X), the results1.1and5.5 (of Banaschewski) or Theorem5.12 (a) below(dueto J.R. Porterand R.G. Woods[15])showthat in general E(X) cannotbe constructedso as to be both totally disconnected-closed and totally separated. For, if that were the casefor somespaceX, then the 742 Y. BDEIR AND R.M. STEPHENSON, JR. associated semiregular spacesE(X) of E(X) wouldbe totally separated and Hausdorfff-minimal,hencezero-dimensional,and so the subspacesX of sE(X) wouldalsohaveto be zero-dimensional. Thus if X weresemiregular, i.e., if X = sX, then X would also have to be zero-dimensionalin orderfor E(X) to haveboth properties.In particular,if X is oneof the spacesin 5.3 or 5.4, or as notedin [15],if X is the totally separated,but not zero-dimensional, separablemetrizablespaceof ErdSs(describedin [9, 6.2.19],[10, 16L],and [16, 1V]), then X doesnot havean extension space which is both Hausdorff-closedand totally separated. After this articlewassubmitted,we learnedof [15]in whichPorter and Woodsstudiedtotally separatedHausdorff-closed extensionspaces(or equivalentlyin their terminology,"ultra-Hausdorff H-closedextensions") of certaintotally separatedspaces.One of the theoremsthey provedcharacterizes spaceswhich have totally separatedHausdorff-closedextensionspaces, and another providesfurther information about the relationship between Theorems 4.1 and 4.2. These theoremscan be re-stated in our terminology as follows. Theorem 5.12. ([15])Let X be a topological space. (a) If X has a totaI1yseparatedHausdorff-closcd extensionspace,then sX is a zero-dimensionalspace. (b) Supposethat sX is a zero-dimensional space.Thenthe spaceE(X) in 4.1 is a totally separatedHausdorff-cIoscdextensionspace of X which is a projectire maximum in the farnily œ of all totally separated Hausdorff-cIosedextension spaces of X. Furthermore, the extension spacesin 4.1 and 4.2 are equivalent extension spacesof X if and only if in the spacesX eachregular open set has compact boundary. The variety of examplescited in this sectionillustrate that there are many different kinds of/P-minimal and/P-closed spaces. A natural question to consider,nevertheless,is if one can obtain additional examplesof interest by not followingthe often standard practice of requiring spacesto be Hausdorff. Our last result showsthat the oppositeis the case. Theorem 5.13. ff the definitions of "totally disconnccted-minimaI,""totally disconnected-closed," and related terms are modified by removing the MINIMAL TOTALLY DISCONNECTED SPACES 743 requirement that spacesbe Hausdorff,then the followinghold. (a) No infinite topological spaceis totally disconnected-minimal or totally disconnected-closed. (b) A compacttotallydisconnected spaceX neednot bezero-dimensional. (c) Only an indiscretespaceis zero-dimensional-minimal, and onlythe empty spaceis zero-dimensional-closed. Proof. (a). Let (X, 7') be infiniteandtotallydisconnected. Then it contains a countably infinite pairwisedisjoint family/C of nonemptyclopensubsets. Choosea set It' C /C and a point p C /t', and define/d = {T: T is open in (X, 7'), and if p E T then T containsall but finitely many membersof/C). Then/2 is a topologyon X strictlyweakerthan 7'. Consider any nondegenerate subsetC of X. Since(X, 7') is totally disconnected, (C, 7'1 C) isnotconnected. If Cf•If = 0 or if Cf•L = 0 forall L e/C• {If}, then L/ I C = 7'1 C, and henceC is not a connectedsubsetof (X,L/). On the otherhand, if C f• K • 0 and C f• L • 0 for someL e/C • {It'}, then C is not a connectedsubsetof (X,/J) sinceL is a clopensubsetof (X,b/). Thus (X,/2) is totally disconnected. To showthat (X, 7') is not totally disconnected-closed, choosea point q not in X, set Y = X U {q}, and let )2 = the set of all subsetsV of Y such that V f• X is open in X, and if q • V then V containsall but finitely many of the setsin /C. Then (Y,)2) is a totally disconnected extensionspaceof (X, 7') with X • Y. Statement(b) canbe verifiedby addingtwo newpointsto an infinite discrete space ¾ and topologizingthe resulting set X so that the neighborhood system of each new point is the set of all cofinite subsetsof X containing that point. [] We would like to concludeby thanking the refereefor a numberof very goodsuggestions concerningstyle, expositionand mathematicalcontent. REFERENCES 1. 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