Download Minimal Totally Disconnected Spaces

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
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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
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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.
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(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.
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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
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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.
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Received April 14, 1992
Revised Version Received October 28, 1993
THE UNIVERSITY OF SOUTH CAROLINA, COLUMBIA, SC 29208
E-mail address: [email protected]