PII. S0161171204404050 http://ijmms.hindawi.com © Hindawi Publishing Corp.
T(α,β)
-SPACES AND THE WALLMAN COMPACTIFICATION
KARIM BELAID, OTHMAN ECHI, and SAMI LAZAAR
Received 4 April 2004
Some new separation axioms are introduced and studied. We also deal with maps having an extension to a homeomorphism between the Wallman compactifications of their domains and ranges.
2000 Mathematics Subject Classification: 54D10, 54D35, 54F65, 18A40.
1. Introduction. Among the oldest separation axioms in topology there are three famous ones,T0,T1, andT2.
TheT0-axiom is usually credited to Kolmogoroff and theT1-axiom to Fréchet or Riesz
(and spaces satisfying the axioms are sometimes called Kolmogoroff spaces, Fréchet spaces, or Riesz spaces, accordingly). TheT2-axiom is included in the original list of
axioms for a topology given by Hausdorff [10].
We denote by Top the category of topological spaces with continuous maps as mor-phisms, and by Topithe full subcategory of Top whose object isTi-spaces. It is a part of
the folklore of topology that Topi+1is a reflective subcategory of Topi, fori= −1,0,1,
with Top−1=Top. Thus Topiis reflective in Top, for eachi=0,1,2 (see MacLane [17]).
In other words, there is a universalTi-space for every topological spaceX; we denote it
byTi(X). The assignmentXTi(X)defines a functorTifrom Top onto Topi, which
is a left adjoint functor of the inclusion functor TopiTop.
The first section of this paper is devoted to the characterization of morphisms in Top rendered invertible by the functorT0.
LetX be a topological space. ThenTi(X)is aTi-space; moreover,Ti(X)may be a
Ti+1-space. The second section deals with spaceXsuch thatTi(X)is aTi+1-space.
Definition1.1. Leti,jbe two integers such that 0≤i < j≤2. A topological space Xis said to be aT(i,j)-space ifTi(X)is aTj-space (thus there are three new types of
separation axioms; namely,T(0,1),T(0,2), andT(1,2)).
More generally, one may introduce the following categorical concept.
Definition1.2. LetC be a category andF,Gtwo (covariant) functors from Cto
itself.
(1) An objectXofCis said to be aT(F,G)-object ifG(F(X))is isomorphic withF(X).
(2) LetPbe a topological property. An objectXofC is said to be aT(F,P)-object if F(X)satisfies the propertyP.
Recall that a topological spaceX is said to be aTD-space [1] if every one-point set
the following implications:
T2 ⇒T1 ⇒TD ⇒T0. (1.1)
Following Definition 1.2, one may define another new separation axiom; namely, T(0,D). Unfortunately, we have no intrinsic topological characterization ofT(1,2)-spaces.
However,T(0,D)-,T(0,1)-, andT(0,2)-spaces are completely characterized inSection 3.
Section 4deals with the separation axiomsT(0,S),T(S,D),T(S,1), andT(S,2), whereS is
the functor of soberification from Top to itself (followingDefinition 1.2, a spaceX is said to beT(S,D)ifS(X)is aTD-space).
One of the two anonymous referees of this paper has notified that theTDproperty is
not reflective in Top; the second author has asked Professor H. P. Kunzi (University of Cape Town) for an explanation of this fact. We give this explanation as communicated by Kunzi.
In [5, Remark 4.2, page 408], Brümmer has proved that the countable product of the Sierpinski space is not aTD-space. On the other hand, according to Herrlich and
Strecker [12], if a subcategoryAis reflective in a categoryB, then for each categoryI,
Ais closed under the formation ofI-limits inB(see [12, Theorem 36.13]). (TakingIa discrete category, you see that in particularAis closed under products inB.) Therefore the full subcategory TopDof Top whose objects areTD-spaces is not reflective in Top.
The importance and usefulness of compactness properties in topology and func-tional analysis is universally recognized. Compactifications of topological spaces have been studied extensively, as well as the associated Stone-ˇCech compactification.
In [11], Herrlich has stated that it is of interest to determine if the Wallman com-pactification may be regarded as a functor, especially as an epireflection functor, on a suitable category of spaces. This problem was solved affirmatively by Harris in [9].
LetX,Ybe twoT1-topological spaces andf:X→Ya continuous map. Aw-extension
off is a continuous mapw(f ):wX→wY such thatw(f )◦ωX=ωY◦f, wherewX
is the Wallman compactification ofXandωX:X→wXis the canonical embedding of
Xinto its Wallman compactificationwX.
InSection 5, we attempt to characterize when Wallman extensions of maps are home-omorphisms.
2. Topologically onto quasihomeomorphisms. Recall that a continuous map q : Y→Zis said to be aquasihomeomorphismifU→q−1(U)defines a bijectionᏻ(Z)→ ᏻ(Y ) [8], whereᏻ(Y ) is the set of all open subsets of the space Y. A subsetS of a topological spaceXis said to bestrongly denseinXifS meets every nonempty locally closed subset ofX[8]. Thus a subsetSofXis strongly dense if and only if the canoni-cal injectionSXis a quasihomeomorphism. It is well known that a continuous map q:X→Y is a quasihomeomorphism if and only if the topology of X is the inverse image byqof that ofY and the subsetq(X)is strongly dense inY [8]. The notion of quasihomeomorphism is used in algebraic geometry and it has recently been shown that this notion arises naturally in the theory of some foliations associated to closed connected manifolds (see [3,4]).
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3719
Remark2.1. (1) Iff:X→Y,g:Y →Z are continuous maps and two of the three
mapsf,g,g◦fare quasihomeomorphisms, then so is the third one.
(2) Letq:X→Y be a continuous onto map. Then the following statements are equiv-alent (see [7, Lemma 1.1]):
(i) qis a quasihomeomorphism;
(ii) qis open and for each open subsetUofX, we haveq−1(q(U))=U;
(iii) qis closed and for each closed subsetCofX, we haveq−1(q(C))=C.
We introduce the concept of “topologically onto (resp., one-to-one) maps” as follows.
Definition2.2. Letq:X→Y be a continuous map.
(1) It is said thatqis topologically onto if, for eachy∈Y, there existsx∈X such that{y} = {q(x)}.
(2)qis said to be topologically one-to-one if, for eachy,x∈Xsuch thatq(x)=q(y), {y} = {x}.
(3)qis said to be topologically bijective if it is topologically onto and topologically one-to-one.
We recall theT0-identification of a topological space which is done by Stone [18].
LetXbe a topological space and define∼onXbyx∼y if and only if{x} = {y}. Then∼is an equivalence relation onXand the resulting quotient spaceX/∼is aT0
-space. This procedure and the space it produces are referred to as theT0-identification
ofX. Clearly,T0(X)=X/∼. The canonical onto map fromXonto itsT0-identification T0(X)will be denoted byµX. Of course,µXis an onto quasihomeomorphism.
As recalled in the introduction,T0defines a (covariant) functor from Top to itself. If q:X→Y is a continuous map, then the diagram
X
µX
q
Y
µY
T0(X) T0(q)
T0(Y )
(2.1)
is commutative.
Example2.3. (1) Every one-to-one continuous map is topologically one-to-one.
(2) Every onto continuous map is topologically onto. (3) A topologically bijective map need not be one-to-one.
LetXbe a topological space which is notT0. Of course,µX is topologically bijective
andµXis not one-to-one.
For any functorF :C→Dbetween two given categories, the set of all arrows inC
Theorem2.4. Letq:X→Y be a continuous map. Then the following statements are
equivalent:
(i) qis a topologically onto quasihomeomorphism;
(ii) T0(q)is a homeomorphism.
Proof. (i)⇒(ii). The mapT0(q)◦µX is a quasihomeomorphism, sinceT0(q)◦µX= µY◦q. HenceT0(q)is a quasihomeomorphism, byRemark 2.1(1).
T0(q) is onto. LetµY(y)∈T0(Y ). Then there exists anx ∈X such that {y} = {q(x)}. HenceµY(y)=µY(q(x)). ThusµY(y)=T0(q)(µX(x)).
T0(q) is one-to-one. Let µX(x),µX(x) ∈ T0(X) be such that T0(q)(µX(x)) =
T0(q)(µX(x)). ThenµY(q(x))=µY(q(x)). Hence{q(x)} = {q(x)}. We are aiming
to prove that{x} = {x}; it is sufficient to show that{x} ⊆ {x}. Indeed, letUbe an
open subset ofXcontainingxandVan open subset ofY such thatU=q−1(V ). Then
q(x)∈V. It follows thatq(x)∈V; that is,x∈U.
Therefore,T0(q)is a bijective quasihomeomorphism. But one may check easily that bijective quasihomeomorphisms are homeomorphisms.
(ii)⇒(i). The equalityT0(q)◦µX=µY◦q forcesq to be a quasihomeomorphism, by
Remark 2.1(1). It remains to prove that q is topologically onto. To do so, lety ∈Y. Then there existsx ∈X such that T0(q)(µX(x))=µY(y). Thus µY(q(x))=µY(y).
Therefore,{y} = {q(x)}, completing the proof.
As a direct consequence of [6, Lemma 1.2, Theorem 1.3] andTheorem 2.4, one may give an external characterization ofT0-spaces.
Theorem2.5. (1)For any topological spaceX, the following statements are
equiva-lent:
(i) Xis aT0-space;
(ii) for each topologically onto quasihomeomorphismq:Y→Zand each continuous mapf:Y→X, there is a unique continuous mapf˜:Z→Xsuch thatf˜◦q=f.
(2)Letq:Y→Zbe a continuous map. Then the following statements are equivalent:
(i) qis a topologically onto quasihomeomorphism;
(ii) for eachT0-spaceXand each continuous mapf:Y→X, there is a unique
con-tinuous mapf˜:Z→Xsuch thatf˜◦q=f.
Question2.6. Give an intrinsic topological characterization of morphisms in Top
rendered invertible by the functorF, whereF∈ {T1,T2}.
3. T(0,D)-,T(0,1)-, andT(0,2)-spaces. We begin by recalling theT1-reflection. LetXbe
a topological space andRthe intersection of all closed equivalence relations onX(an equivalence relation onXis said to be closed if its equivalence classes are closed inX). The quotient spaceX/Ris homeomorphic to theT1-reflection ofX.
We begin with some straightforward examples and remarks.
Remark3.1. (1) Since
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3721
then we get
T(0,2) ⇒T(0,1) ⇒T(0,D). (3.2)
(2) EachT(0,2)-space isT(1,2).
(3) There is aT(1,2)-space which is not T(0,1). Let X be the Sierpinski space. Then T1(X)is a one-point set, thusXis aT(1,2)-space. However,T0(X)=X; henceXis not a
T(0,1)-space.
(4) There is aT(0,1)-space which is notT(1,2): it suffices to consider aT1-space which
is notT2.
(5) There is aT(0,1)-space which is notT(0,2): take aT1-space which is notT2.
(6) There is aT(1,2)-space which is notT(0,2): the Sierpinski space does the job.
(7) There is aT(0,D)-space which is notT(0,1): it suffices to consider aTD-space which
is notT1.
(8) There is aT(0,D)-space which is notT(1,2). The example in(4)does the job.
(9) There is aT(1,2)-space which is notT(0,D): letY be an infinite set. Letw∉Y and
X=Y∪{w}. EquipXwith the topology whose closed sets areXand all finite subsets ofY. Clearly,T1(X)is a one-point set; thenXis aT(1,2)-space. ButXis notT(0,D)-space,
sinceXisT0and{w}is not locally closed.
Remark3.2. LetXbe a topological space. ThenT0(X)is aT1-space if and only if,
for eachx∈X,{x} =µ−1
X (µX({x})), whereµX:X→T0(X)=X/∼is the canonical onto
map.
For each pointxof a spaceX, we denote byγ(x)the set
{x}\y∈X:{y} = {x}. (3.3)
With this notation, we have the following.
Theorem3.3. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(0,D)-space;
(ii) for eachx∈X,γ(x)is a closed subset ofX.
The proof needs the following lemma.
Lemma3.4. Letq:X→Ybe an onto quasihomeomorphism andCa subset ofY. Then
the following statements are equivalent:
(i) Cis locally closed inY;
(ii) q−1(C)is locally closed inX.
Proof. LetᏸᏯ(X),ᏸᏯ(Y )be the sets of all locally closed subsets ofXandY,
re-spectively.
It is well known [8] that the mapᏸᏯ(Y )→ᏸᏯ(X)defined byFq−1(F)is bijective.
It is sufficient to show (ii)⇒(i).
Indeed, ifq−1(C)is locally closed inX, then there is anL∈ᏸᏯ(Y )such thatq−1(C)=
Proof ofTheorem3.3. Let µX :X→T0(X)be the canonical map fromX to its
T0-reflectionT0(X).
According toLemma 3.4, X is aT(0,D)-space if and only ifµX−1(µX({x})) is locally
closed inX, for eachx∈X. One may check easily thatµ−1
X (µX({x}))\µ−1X (µX({x}))=γ(x). But it is well known
that a subsetS of a spaceX is locally closed if and only ifS\S is closed, completing the proof.
The following result gives a characterization ofT(0,1)-spaces.
Theorem3.5. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(0,1)-space;
(ii) for eachx,y∈Xsuch that{x}≠{y}, there is a neighborhood ofxnot contain-ingy;
(iii) for eachx∈Xand each closed subsetCofXsuch that{x}∩C≠∅,x∈C;
(iv) for each open subsetUofXand eachx∈U,{x} ⊆U;
(v) for eachx∈X,{U:U∈ᏻ(x)} ={U:U∈ᐂ(x)} = {x}, whereᏻ(x)is the set of all open subsets ofXcontainingxandᐂ(x)is the set of all neighborhoods of
X.
Proof. Of course, for any topological spaceX and anyx∈X, we have{U:U∈
ᏻ(x)} ={U:U∈ᐂ(x)}.
We show the implications (i)(ii), (i)⇒(iii)⇒(iv)⇒(i), and (i)⇒(v)⇒(iv).
(i)⇒(ii). Letx,y∈X be such that{x}≠{y}; thenµX(x)≠µX(y). SinceT0(X)is a
T1-space, there exists a neighborhoodUofµX(x)not containingµX(y). Henceµ−1X (U)
is a neighborhood ofxnot containingy.
(ii)⇒(i). LetµX(x),µX(y)be two distinct points inT0(X). Then {x}≠{y}. Hence
there is a neighborhoodUofxnot containingy. SinceµXis a quasihomeomorphism,
there exists a neighborhoodUinT0(X)such thatU=µ−1
X (U)and thusUis a
neigh-borhood ofµX(x)not containingµX(y). ThereforeT0(X)is aT1-space.
(i)⇒(iii). LetCbe a closed subset ofXandx∈Xsuch that{x}∩C≠∅. SinceµXis a
quasihomeomorphism, there exists a closed subsetC0ofT0(X)such thatC=µX−1(C0).
Henceµ−1
X (C0∩{µX(x)})≠∅; and thusC0∩{µX(x)}≠∅. Therefore,µX(x)∈C0; that
is,x∈C.
(iii)⇒(iv). Let U be an open subset of X and, x ∈U. Then {x} ∩X\U = ∅, and consequently,{x} ⊆U.
(iv)⇒(i). It is easily seen thatµ−1
X (µX({x}))⊆ {x}. Conversely, lety∈ {x}. For each
open subsetU ofXcontainingx, we have {x} ⊆U. Thusy∈U; so that{x} ⊆ {y}. Hence {x} = {y}; that is, µX(x) = µX(y). This yields {x} ⊆ µ−X1(µX({x})). Thus
µ−1
X (µX({x}))= {x}. Therefore,T0(X)is aT1-space, byRemark 3.2.
(i)⇒(v). SinceT0(X)is aT1-space, we have{x} =µX−1({µX(x)})and{µX(x)} = ∩{V:
V∈ᏻ(µX(x))}, for eachx∈X. On the other hand, sinceµXis a continuous surjective
open map, we haveᏻ(µX(x))= {µX(U):U∈ᏻ(x)}. Therefore,
{x} =µ−1 X
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3723
Moreover, sinceµXis a quasihomeomorphism, we haveµX−1(µX(U))=U, for each open
subsetUofX(seeRemark 2.1). It follows that{x} ={U:U∈ᏻ(x)}. (v)⇒(iv). Straightforward.
Corollary 3.6. The T(0,1)-property is a productive and hereditary property: any
product ofT(0,1)-spaces isT(0,1)and any subspace of aT(0,1)-space isT(0,1).
A subspaceY ofXis calledirreducibleif each nonempty open subset ofY is dense inY (equivalently, ifC1andC2are two closed subsets ofXsuch thatY⊆C1∪C2, then
Y ⊆C1orY⊆C2). LetCbe a closed subset of a spaceX. We say thatChas a generic
pointif there existsx∈Csuch thatC= {x}.
Recall that a topological spaceXis said to be quasisober [14] (resp., sober [8]) if any nonempty irreducible closed subset ofXhas a generic point (resp., a unique generic point).
Lemma3.7. Letq:X→Y be a quasihomeomorphism. (1) IfXis aT0-space, thenqis one-to-one.
(2) IfY is aTD-space, thenqis onto.
(3) IfY is aTD-space andXis aT0-space, thenqis a homeomorphism.
(4) IfXis sober andY is aT0-space, thenqis a homeomorphism.
Proof. (1) Letx1,x2be two points ofXwithq(x1)=q(x2). Suppose thatx1≠x2. Then there exists an open subsetU ofX such that, for example,x1∈U andx2∉U.
Since there exists an open subsetVofY satisfyingq−1(V )=U, we getq(x
1)∈V and
q(x2)∉V, which is impossible. It follows thatqis one-to-one.
(2) Lety∈Y. Then{y}is a locally closed subset ofY. Hence{y}∩q(X)≠∅, since q(X)is strongly dense inY. Thusy∈q(X), proving thatqis onto.
(3) One may check easily that bijective quasihomeomorphisms are homeomorphisms. (4) By(1),qis one-to-one.
Now, observe that ifSis a closed subset ofY, thenSis irreducible if and only if so isq−1(S).
We prove thatqis onto. To this end, lety∈Y. According to the above observation, q−1({y})is a nonempty irreducible closed subset ofX. Henceq−1({y})has a generic
pointx. Thus we have the containments
{x} ⊆q−1{q(x)}⊆q−1{y}= {x}. (3.5)
So thatq−1({q(x)})=q−1({y}). It follows from the fact thatqis a
quasihomeomor-phism that{q(x)} = {y}. SinceY is aT0-space, we getq(x)=y. This proves that q
is onto, and thusqis bijective. But a bijective quasihomeomorphism is a homeomor-phism.
Proposition 3.8. Letq:X→Y be a quasihomeomorphism. If Y is aT(0,1)-space,
then so isX.
Proof. Clearly, T0(q): T0(X)→ T0(Y ) is a quasihomeomorphism. Hence, since
T0(X)is aT0-space andT0(Y )is aT1-space,T0(q)is a homeomorphism, byLemma 3.7.
Example3.9. A quasihomeomorphismq:Y→Xsuch thatY is aT(0,1)-space butX
is not.
TakeY andXas inRemark 3.1(9). Then each nonempty locally closed subset ofX meetsY. Hence, the canonical embeddingq:Y →Xis a quasihomeomorphism.
Of course,Y is aT(0,1)-space. However,Xis not aT(0,1)-space. Indeed, for eachx∈X \{w},{x} = {x}and{w} =X; henceXis aT0-space which is notT1. Therefore,X is
notT(0,1).
The following proposition follows immediately fromTheorem 2.4.
Proposition3.10. Letq:X→Ybe a topologically onto quasihomeomorphism. Then
the following statements are equivalent:
(i) Xis aT(0,1)-space;
(ii) Y is aT(0,1)-space.
It is well known that a spaceXis aT2-space if and only if, for eachx∈X,{U:U∈ ᐂ(x)} = {x}, whereᐂ(x)is the set of all neighborhoods ofx.
Before giving a characterization ofT(0,2)-spaces, we need a technical lemma.
Lemma 3.11. Let q:X→Y be an onto quasihomeomorphism. Then the following
properties hold.
(1)For each subsetBofY,q−1(B)=q−1(B).
(2)For each x∈X,{q−1(V ):V∈ᐂ(q(x))} =q−1({V :V ∈ᐂ(q(x))})={U:
U∈ᐂ(x)}.
Proof. (1) We observe that a continuous mapq:X→Y is open if and only if, for
each subset B of Y, we have q−1(B)=q−1(B)(see [8, Chapter 0, (2.10.1)]). Now by
Remark 2.1(2) an onto quasihomeomorphism is open, so that (1) follows immediately. (2) Straightforward.
Theorem3.12. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(0,2)-space;
(ii) for eachx,y∈Xsuch that{x}≠{y}, there are two disjoint open setsUandV
inXwithx∈Uandy∈V;
(iii) for eachx∈X,{U:U∈ᐂ(x)} = {x}.
Proof. (i)⇒(ii). Letx,y∈Xsuch that{x}≠{y}, thenµX(x)≠µX(y). SinceT0(X) is aT2-space, there exist two disjoint open setsUandVinT0(X)withµX(x)∈U
andµX(y)∈V. Therefore,U=µX−1(U)andV=µ−X1(V)are two disjoint open sets in
Xwithx∈Uandy∈V.
(ii)⇒(i). Let µX(x),µX(y)be two distinct points inT0(X). Then{x}≠{y}and, by
(ii), there are disjoint open setsU andV ofX with x∈U and y∈V. SinceµX is a
quasihomeomorphism, there exist two disjoint open setsU,VofT0(X)withµX(x)∈
U and µX(y)∈V and such thatU =µ−1
X (U)and V =µX−1(V), so thatT0(X)is a
T2-space.
(i)⇒(iii). LetµX be the canonical map from X onto itsT0-reflection. Then for each
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3725
According toRemark 3.2,µ−1
X ({µX(x)})= {x}. Hence
{x} =µ−1
X (V ):V∈ᐂ
µX(x). (3.6)
Thus, for eachx∈X, we have{U:U∈ᐂ(x)} = {x}, byLemma 3.11(2). (iii)⇒(i). Letx∈X. First, we prove that
U
:U∈ᐂ(x)=U:U∈ᐂ(x)= {x}. (3.7)
Clearly,{U:U∈ᐂ(x)} ⊆{U:U∈ᐂ(x)}. Conversely, lety∈ {x}. Thenx∈V, for eachV∈ᐂ(y). Hence
x∈V:V∈ᐂ(y)= {y}. (3.8)
Thusy∈{U:U∈ᐂ(x)}; so that (3.7) holds for eachx∈X. Therefore,Xis aT(0,1)
-space, byTheorem 3.5. Thus, according toRemark 3.2,{x} =µ−1
X ({µX(x)}). Applying
Lemma 3.11, we have
µX(x)=V:V∈ᐂµX(x), (3.9)
proving thatT0(X)is aT2-space and thusXis aT(0,2)-space.
Corollary3.13. It is clear that theT(0,2)-property is a productive and hereditary
property.
Proposition 3.14. Letq:X →Y be a quasihomeomorphism. Then the following
statements are equivalent:
(i) Xis aT(0,2)-space;
(ii) Y is aT(0,2)-space.
Proof. (i)⇒(ii). Clearly,T0(q):T0(X)→T0(Y )is a quasihomeomorphism. On the
other hand, sinceT0(X)is aT2-space, it is a sober space; and since in additionT0(Y )
is aT0-space, thenT0(q)is a homeomorphism, byLemma 3.7. Therefore, T0(Y ) is a
T2-space. This means thatY is aT(0,2)-space.
(ii)⇒(i). AgainT0(q):T0(X)→T0(Y )is a quasihomeomorphism. Now, sinceT0(X) is aT0-space andT0(Y )is aT1-space, thenT0(q)is a homeomorphism, byLemma 3.7.
Therefore,T0(X)is aT2-space. This means thatXis aT(0,2)-space.
Example3.15. A quasihomeomorphismq:X→Y such thatXis aT2-space andY
is not aT2-space.
LetY= {0,1,2}equipped with the topology{∅,Y ,{1,2},{0}}and letX= {0,1}be provided with a discrete topology. ThenXis aT2-space andY is not aT2-space. The
4. T(0,S)-,T(S,D)-,T(S,1)-, andT(S,2)-spaces. LetXbe a topological space andS(X)the
set of all nonempty irreducible closed subset ofX[8]. LetUbe an open subset ofX; set U= {C∈S(X):U∩C≠∅}; then the collection{U:Uis an open subset ofX}provides a topology onS(X)and the following properties hold [8].
(i) The mapηX:X→S(X)which carriesx∈XtoηX(x)= {x}is a
quasihomeomor-phism.
(ii)S(X)is a sober space.
(iii) The topological spaceS(X)is called the soberification ofX, and the assignment
S(X)defines a functor from the category Top to itself. (iv) Letq:X→Y be a continuous map, then the diagram
X
ηX
q
Y
ηY
S(X) S(q) S(Y )
(4.1)
is commutative.
In [14, Proposition 2.2], Hong has proved that a space is quasisober if and only if its T0-reflection is sober. The following result makes [14, Proposition 2.2] more precise.
Theorem4.1. Letq:X→Y be a quasihomeomorphism. Then the following
proper-ties hold.
(1) IfXis aT(0,S)-space, then so isY.
(2) Suppose thatY is aT(0,S)-space. Then the following statements are equivalent:
(i) Xis aT(0,S)-space;
(ii) qis topologically onto.
(3) Ifqis topologically onto, then the following statements are equivalent:
(i) Xis aT(0,S)-space;
(ii) Y is aT(0,S)-space.
Proof. (1) SinceT0(q)is a quasihomeomorphism andT0(X)is sober, we deduce
thatT0(q)is a homeomorphism, byLemma 3.7(4). HenceT0(Y )is sober, proving that Y is aT(0,S)-space.
(2) (i)⇒(ii). According toLemma 3.7(4),T0(q)is a homeomorphism. Thusqis topo-logically onto, byTheorem 2.4.
(ii)⇒(i). Again, according toTheorem 2.4,T0(q)is a homeomorphism. HenceT0(X) is sober, sinceT0(Y )is.
(3) Combine(1)and(2).
Proposition4.2[14, Proposition 2.2]. A topological space is quasisober if and only
if itsT0-reflection is sober.
Proof. The canonical mapµX:X→T0(X)is an onto quasihomeomorphism. Then,
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3727
Now, the result follows from the following simple facts: ifq:X→Y is a quasihome-omorphism andXis quasisober, thenY is quasisober; ifqis an onto quasihomeomor-phism andY is quasisober, thenXis quasisober.
Remark4.3. (1) EachT(S,2)-space is aT(S,1)-space.
(2) EachT(S,1)-space is aT(S,D)-space.
(3) EachT(0,2)-space is aT(0,S)-space.
(4) There is aT(0,1)-space which is not aT(0,S)-space. It suffices to consider aT1-space
which is not sober.
(5) There is aT(S,1)-space which is not aT(S,2)-space. It suffices to consider a sober
T1-space which is notT2(see [19, pages 675-676] and [13, pages 12-13]).
(6) There is aT(0,S)-space which is notT(S,1)-space.
TakeXas inRemark 3.1(9). Note that a nonempty closed subsetCofXis irreducible if and only ifC=XorC= {y}, wherey∈Y, and thusXis a sober space. ThereforeX is aT(0,S)-space which is not aT(S,1)-space.
Example4.4. A quasihomeomorphismq:Y →Xsuch thatXis aT(0,S)-space, butY
is not.
TakeXandY as inRemark 3.1(9). ThenT0(X)=Xis a sober space, butT0(Y )=Y is not becauseY is a nonempty irreducible closed subset without a generic point. Thus the canonical embeddingYXdoes the job.
We next give a characterization ofT(S,D)-spaces.
Theorem4.5. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(S,D)-space;
(ii) Xis a quasisoberT(0,D)-space.
Proof. (i)⇒(ii). LetµX:X→T0(X)(resp.,ηX:X→S(X)) be the canonical map from Xonto itsT0-reflectionT0(X)(resp., soberificationS(X)). Then the diagram
X
µX
ηX
S(X)
µS(X)
T0(X) T0(ηX ) T0S(X)
(4.2)
is commutative.
HenceT0(ηX)is a quasihomeomorphism. ThusT0(X)is homeomorphic to the
sub-spaceT0(ηX)(T0(X)), sinceT0(X)is aT0-space.
According to [13, Theorem 2.2], every subspace of a soberTD-space is sober. Thus T0(X)is sober and, consequently,T0(ηX)is a homeomorphism, byLemma 3.7.
There-fore,T0(X)is a soberTD-space; this means thatXis aT(0,S)- and aT(0,D)-space; but the
(ii)⇒(i). ByProposition 4.2,T0(X)is a sober space. On the other hand, X is quasi-homeomorphic toT0(X); thenS(X)is homeomorphic toS(T0(X)), by [2, Theorem 2.2]. ThusS(X)is homeomorphic toT0(X)so thatS(X)is aTD-space, proving thatXis a
T(S,D)-space.
Now, we give a characterization ofT(S,1)-spaces.
Theorem4.6. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(S,1)-space;
(ii) wheneverF andGare distinct nonempty irreducible closed subsets ofX, there is an open subsetUofXsuch thatU∩F≠∅andU∩G= ∅;
(iii) for each nonempty irreducible closed subsetsF andGofX,F⊆Gif and only if
F=G.
Proof. (i)⇒(ii). LetF andGbe two distinct nonempty irreducible closed subsets of X. SinceS(X)is aT1-space andF ≠G, there exists an open setU ofS(X)such that
F∈UandG∉U. ThusUis an open subset ofXwithU∩F≠∅andU∩G= ∅. (ii)⇒(iii). Straightforward.
(iii)⇒(i). First, note that ifH is a closed subset ofX, then{G∈S(X):G⊆H}is a closed subset ofS(X). Now, letF be a nonempty irreducible closed subset ofX. Then, by (iii),{G∈S(X):G⊆F} = {F}is a closed subset ofS(X). Hence each point ofS(X)is closed. Therefore,Xis aT(S,1)-space.
We finish this section by characterizingT(S,2)-spaces.
Theorem4.7. LetXbe a topological space. Then the following statements are
equiv-alent:
(i) Xis aT(S,2)-space;
(ii) wheneverF andG are distinct nonempty irreducible closed subsets ofX, there are disjoint open subsetsUandVofXsuch thatF∩U≠∅andG∩V≠∅.
Proof. (i)⇒(ii). LetF andGbe two distinct nonempty irreducible closed subsets of X. SinceS(X)is aT2-space andF≠G, then there exist two disjoint open subsetsUand
VinS(X)such thatF∈UandG∈V, so that the two open setsUandVsatisfy (ii). (ii)⇒(i). LetF andGbe two distinct points inS(X)and letUandVbe as in (ii). Then it is clear thatF ∈U and G∈V. Clearly U∩V =U∩V =∅ = ∅; therefore,X is a T(S,2)-space.
Proposition4.8. Letq:X→Ybe a quasihomeomorphism. Then the following
state-ments are equivalent:
(i) Xis aT(S,1)- (resp.,T(S,D)-, resp.,T(S,2)-)space;
(ii) Y is aT(S,1)- (resp.,T(S,D)-, resp.,T(S,2)-)space.
Proof. This follows easily from the fact thatS(q):S(X)→S(Y )is a
homeomor-phism, by [2, Theorem 2.2].
5. The Wallman compactification. The Wallman compactification of aT1-space is
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3729
LetXbe aT1-space, and letwXbe the collection of all closed ultrafilters onX. For
each closed setD⊆X, define D∗ to be the set D∗ = {ᐁ∈wX :D∈ᐁ}if D≠∅
and ∅∗ = ∅. Then {D∗ :Dis a closed subset ofX}is a base for the closed sets of
a topology on wX. LetU be an open subset ofX; we defineU∗⊆wX to be the set
U∗= {ᐁ∈wX:A⊆Ufor someAinᐁ}. The class{U∗:Uis an open subset ofX}is
a base for the open sets of the topology ofwX.
The following properties are well known and may be found in any standard textbook on general topology (see, e.g., Kelley [16]).
Properties5.1. LetXbe aT1-space. Consider the mapωX:X→wXwhich takes x∈XtoωX(x)= {A:Ais a closed subset ofXandx∈A}. Then the following
prop-erties hold.
(1) IfDis closed inX, thenωX(D)=D∗. In particular,ωX(X)is dense inwX.
(2) ωXis continuous and it is an embedding ofXinwXif and only ifXis aT1-space.
(3) IfAandBare closed subsets ofX, thenωX(A∩B)=ωX(A)∩ωX(B).
(4) wXis a compactT1-space.
(5) Every continuous map onXto a compact Hausdorff spaceKcan be extended to wX.
For aT(0,1)-spaceX, we defineW X=w(T0(X))and we call it the Wallman
compacti-fication ofX. The notationwXis reserved only forT1-spaces so that it is better to use
some other notation forT(0,1)-spaces; the same forωX:ωX is reserved forT1-spaces;
forT(0,1)-spaces, we definewX=ωT0(X)◦µX.
SinceµX is an onto quasihomeomorphism, one obtains immediately thatW X can
be described exactly as wX is for T1-spaces. Properties 5.1 are also true for T(0,1)
-spaces.
Remark5.2. LetXbe aT(0,1)-space. Then the following properties hold:
(1) For each open subsetUofX, we havewX(U)⊆U∗.
(2) For each closed subsetDofX, we havewX(D)⊆D∗.
(3) Let U be open and Dclosed in a T(0,1)-space. Then U∩D≠∅if and only if
U∗∩D∗≠∅.
Now we give some new observations about Wallman compactifications.
Proposition5.3. LetXbe aT(0,1)-space andU an open or closed subset ofX. IfU
is compact, thenU∗=wX(U).
Proof. Suppose thatU is open inX. Let ᐁ∈U∗. Then there exists F ∈ᐁ such
thatF ⊆U.F is compact, by the compactness ofU. Thus{HF:H∈ᐁ}≠∅. Let x∈{H∩F:H∈ᐁ}; thenᐁ=wX(x). Hence,ᐁ∈wX(U).
On the other hand, according toRemark 5.2,wX(U)⊆U∗. Therefore,wX(U)=U∗.
Now suppose thatUis closed inX. Letᐁ∈U∗; thenU∈ᐁ. Since{H:H∈ᐁ}≠∅,
pick anx∈{H:H∈ᐁ}. It is easily seen thatᐁ=wX(x). Therefore, according to
Remark 5.2,wX(U)=U∗.
Remark5.5. The converse ofCorollary 5.4does not hold. LetX be a noncompact T(0,1)-space. ThusX∗=W Xis compact; however,Xis not compact.
Question5.6. LetUbe an open subset of aT(0,1)-space withU≠∅andU≠X. If we suppose thatU∗is compact, isUcompact?
Proposition5.7. LetDbe a closed subset of aT(0,1)-space. ThenDis irreducible if
and only ifD∗is irreducible inW X.
Proof. LetDbe an irreducible closed subset ofX. SinceD∗=wX(D), D∗is
irre-ducible.
Conversely, suppose thatD∗is irreducible. LetU,Vbe two open subsets ofXsuch
thatU∩D≠∅andV∩D≠∅. HenceU∗∩D∗≠∅andV∗∩D∗≠∅, byRemark 5.2. Thus U∗∩V∗∩D∗≠∅, so that (U∩V )∗∩D∗ ≠∅. Therefore, U∩V∩D≠∅, by
Remark 5.2.
We need to introduce new concepts.
Definition5.8. (1) A subsetCof a topological spaceXis said to be closedly dense
ifCmeets each nonempty closed subset ofX.
(2) A subsetSof a topological spaceXis said to be sufficiently dense ifSmeets each nonempty closed subset and each nonempty open subset ofX.
(3) By an almost-homeomorphism (a-homeomorphism, for short), we mean a contin-uous mapq:X→Y such thatq(X)is sufficiently dense inY and the topology ofXis the inverse image of that ofY byq.
(4) By a Wallman morphism (W-morphism, for short), we mean a continuous map q:X→Y such thatq(X)is closedly dense inY and the topology ofXis the inverse image of that ofY byq.
Thus we have the following implications:
Strongly dense Sufficiently dense closedly dense
Dense
Homeomorphism quasihomeomorphism a-homeomorphism W-morphism. (5.1)
The converses fail, as shown by the following examples.
Example5.9. ConsiderX=[0,ω]to be the set of all ordinal numbers less than or
equal to the first limit ordinalω. We equipXwith the natural order≤.
The discrete Alexandroff topology onX associated to the reverse order is ᏻ(X)= {∅,X,[0,ω[} ∪ {(↓x):x∈X}, where (↓x)= {y ∈X:y≤x}. SetD=[0,ω[,C = X−{0}andK= {0,ω}.
Note that usually the Alexandroff topology is defined by the upper sets (see Johnstone [15]); here the topology used is associated with the reverse order.
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3731
(b) Since{0}is open and{0} ∩C= ∅,C is not dense inX. However,C is closedly dense inX.
(c) The subsetKis sufficiently dense but not strongly dense inX.
(d) It is easily seen that the canonical embeddingiC:C→Xis aW-morphism which
is not ana-homeomorphism. The canonical embeddingiK:K→Xis an
a-homeomor-phism which is not a quasihomeomora-homeomor-phism.
(e) It is well known that a quasihomeomorphism need not be a homeomorphism.
Remark5.10. An ontoW-morphismq:X→Y is a quasihomeomorphism. Indeed it
suffices to show that ifFandGare two closed subsets ofY such thatq−1(F)=q−1(G),
thenF=Gwhich is clear sinceqis an onto map.
Proposition5.11. (1)The composite of twoa-homeomorphisms (resp.,W
-morph-isms) is ana-homeomorphism (resp.,W-morphism).
(2)Ifq:X→Y is aW-morphism andXisT0, thenqis one-to-one.
(3)Ifq:X→Y is aW-morphism andY isT1, thenqis an onto quasihomeomorphism.
(4)Ifqis aW-morphism and ifXisT0andY isT1, thenqis a homeomorphism.
Proof. (1) We show (1) for twoa-homeomorphisms. Letp:X→Y andq:Y→Zbe
twoa-homeomorphisms. Clearly, the topology ofXis the inverse image of that ofZ byq◦p.
LetAbe a closed (resp., an open) subset ofZ. Sinceq−1(A)is closed (resp., open) inY,
thenp(X)∩q−1(A)≠∅, so thatA∩q(p(X))≠∅. Henceq◦pis ana-homeomorphism.
(2) and (3) have the same proof asLemma 3.7.
(4) follows immediately from (2), (3), andRemark 5.10.
Letf:X→Y andg:Y→Z be two continuous maps. One may check easily that if two among the three mapsg◦f,f,gare quasihomeomorphisms, then so is the third one (seeRemark 2.1).
Fora-homeomorphisms andW-morphisms, we get the following result which has a straightforward proof.
Proposition5.12. Letf:X→Y andg:Y→Zbe two continuous maps.
(1)Suppose thatg◦f andgarea-homeomorphisms (resp.,W-morphisms). Thenf is ana-homeomorphism (resp., aW-morphism).
(2)Suppose thatg◦fis ana-homeomorphism (resp., aW-morphism) andfis a quasi-homeomorphism. Thengis ana-homeomorphism (resp., aW-morphism).
The following example shows thatRemark 2.1(1) fails to be true for a-homeomor-phisms orW-morphisms.
Example 5.13. Let f : X→Y and g: Y →Z be two continuous maps such that g◦fandfarea-homeomorphisms (resp.,W-morphisms). Thengis not necessarily an a-homeomorphism (resp., aW-morphism).
Letf :{0,ω} → {0,1,ω}be the canonical embedding and let g:{0,1,ω} →X be defined byg(0)=0,g(1)=0,g(ω)=ω. Clearly,g◦f andf area-homeomorphisms. However, since the closed subset{1,ω}of{0,1,ω}is not an inverse image bygof a closed subset ofX,gis not aW-morphism.
Recall from [9] that a continuous mapq:X→Y betweenT1-spaces is said to be aw
-extensionif there is a continuous mapw(q):wX→wY such thatωY◦q=w(q)◦ωX.
In an analogous manner, one may defineW-extensionsforT(0,1)-spaces.
The following gives a class of morphismsq:X→Ywhich yield aW-extensionW (q): W X→W Y that is a homeomorphism.
Proposition5.14. LetX,Y be twoT(0,1)-spaces andq:X→Y aW-morphism. Then qhas aW-extension which is a homeomorphism.
Proof. We first remark that diagram (2.1) commutes. HenceT0(q)◦µX =µY◦q.
Thus T0(q)◦µX is a W-morphism. Now, since µX is a quasihomeomorphism, T0(q)
is aW-morphism, byProposition 5.12(2). Therefore, T0(q) is a homeomorphism, by Proposition 5.11(4). It follows thatT0(q)has a canonicalw-extensionw(T0(q))which is a homeomorphism. Thus the diagram
T0(X)
ωT0(X)
T0(q)
T0(Y )
ωT0(Y )
wT0(X)=W Y ω(T0(q)) wT0(Y )=W Y
(5.2)
commutes.
If we denoteW (q)=w(T0(q)), then the above diagrams indicate clearly thatW (q) is aW-extension ofqwhich is a homeomorphism.
It is well known that the Wallman compactification of aT1-spaceX is Hausdorff if
and only ifXis normal and in this casewX=β(X)(the Stone-ˇCech compactification ofX) (see, e.g., Wallman [20]).
Corollary5.15. W X is Hausdorff if and only if T0(X)is a normal space. In this
caseW X=β(T0(X)).
Remark5.16. If a continuous mapq:X→Y has aW-extension which is a
homeo-morphism, then q need not be a homeomorphism. To see this it suffices to take a noncompactT1-spaceX. Of course, 1wXis aw-extension ofωX; however,ωXis not a
homeomorphism.
Definition5.17. LetXbe aT(0,1)-space andY a subspace ofX.
(α,β)-SPACES AND THE WALLMAN COMPACTIFICATION 3733
(2)Yis called astrong Wallman generator(sW-generator, for short) ofXif the canon-ical embeddingi:YXhas aW-extensionW (i)which is a homeomorphism.
Clearly, eachsW-generator is aW-generator.
As an immediate consequence ofProposition 5.14, we get the following examples.
Example5.18. LetXbe aT(0,1)-space.
(1) wX(X)is ansW-generator ofW X.
(2) Each closedly dense subset ofXis ansW-generator ofX.
We think that it is of interest to answer the following questions.
Questions5.19. LetXbe aT(0,1)-space.
(1) CharacterizeW-generators ofX. (2) CharacterizesW-generators ofX.
(3) Is there aW-generator which is not ansW-generator?
Now we are in a position to give a characterization of continuous maps betweenT(0,1)
-spaces having an extension to Wallman compactifications that is a homeomorphism.
Theorem5.20. LetX,Y be twoT(0,1)-spaces andq:X→Y a continuous map. Then
the following statements are equivalent:
(i) qhas aW-extension which is a homeomorphism;
(ii) q(X)is ansW-generator of Y and the topology of Xis the inverse image of that ofY byq.
Proof. (i)⇒(ii). (a)The topology ofXis the inverse image of that ofY byq.LetCbe a closed subset ofX. SinceW (q)is a homeomorphism,W (q)(C∗)=Kis a closed subset
ofW Y. SetH=w−1
Y (K). We prove thatC=q−1(H).
(1) Let x∈C. Then wX(x)∈wX(C)⊆C∗. HenceW (q)(wX(x))∈W (q)(C∗)=K,
which giveswY(q(x))∈K. It follows thatq(x)∈wY−1(K)=H. Therefore,x∈q−1(H).
(2) Conversely, letx ∈q−1(H). Thenq(x)∈H =w−1
X (K); this means that (wY◦
q)(x)∈ K, so that W (q)(wX(x))∈ K= W (q)(C∗). Since W (q)is bijective, wX(x) ∈C∗. Hencex∈w−1
X (C∗)=C. We have thus proved thatC=q−1(H). In other words,
the topology ofXis the inverse image of that ofY byq.
(b)q(X)is ansW-generator of Y.According to(a), the induced mapq1:X→q(X)
byqis aW-morphism. Henceq1has aW-extensionW (q1)which is a homeomorphism,
byProposition 5.14. Thus the diagrams
X
wX
q1
q(X)
wq(X)
W X W (q1) W q(X)
X
wX
q
Y
wY
W X W (q) W Y
(5.3)
Letj:q(X)Y be the canonical embedding. Clearly, the diagram
q(X)
wq(X)
j
Y
wY
wq(X) W (q)◦(W (q1))−1 W Y
(5.4)
commutes. Therefore,jhasW (q)◦(W (q1))−1as aW-extension which is a
homeomor-phism. This means thatq(X)is ansW-generator ofY.
(ii)⇒(i). Under the assumptions of (ii), the induced mapq1:X→q(X)byqis aW
-morphism. Thus, according toProposition 5.14,q1has aW-extensionW (q1)which is
a homeomorphism. On the other hand, the canonical embeddingj :q(X)Y has a W-extension which is a homeomorphism, byProposition 5.14. It follows that the two diagrams
X q1
wX
q(X)
wq(X) j
Y
wY
W X W (q1) W q(X) W (j) W Y
(5.5)
commute. Therefore,W (j)◦W (q1)is aW-extension ofq:X→Y which is a
homeomor-phism.
Question5.21. Is it possible to replace the word “sW-generator” inTheorem 5.20 by “W-generator”?
Acknowledgments. The authors gratefully acknowledge helpful corrections,
com-ments, and suggestions of the two anonymous referees. Many thanks to one of the two referees who gave us pertinent remarks aboutTD-spaces. We also thank Professors H.
P. Kunzi and G. Brümmer from the University of Cape Town.
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Karim Belaid: Département de Mathématiques, Ecole Supérieure des Sciences et Techniques de Tunis, 5 Avenue Taha Hussein, BP 56 Bab Manara, 1008 Tunis, Tunisia
E-mail address:belaid412@yahoo.fr
Othman Echi: Department of Mathematics, Faculty of Sciences of Tunis, University of Tunis El Manar, 2092 Tunis, Tunisia
E-mail address:othechi@yahoo.com
Sami Lazaar: Département de Mathématiques, Institut Préparatoire aux Etudes d’Ingénieurs de Nabeul, Université du 7 Novembre à Carthage, 8000 Nabeul, Tunisia
