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s
-POINT FINITE REFINABLE SPACES
SHELDON W. DAVIS, ELISE M. GRABNER, and GARY C. GRABNER
(Received 4 October 1996 and in revised form 28October 1996)
Abstract.A spaceXis calleds-point finite refinable (ds-point finite refinable) provided every open coverᐁofXhas an open refinementᐂsuch that, for some (closed discrete) C⊆X,
(i) for all nonemptyV∈ᐂ, V∩C≠∅and
(ii) for alla∈Cthe set(ᐂ)a= {V∈ᐂ:a∈V}is finite.
In this paper we distinguish these spaces, study their basic properties and raise several interesting questions. Ifλis an ordinal withcf (λ)=λ > ωandSis a stationary subset of λthenSis nots-point finite refinable. Countably compactds-point finite refinable spaces are compact. A spaceXis irreducible of orderωif and only if it isds-point finite refinable. IfXis a strongly collectionwise Hausdorffds-point finite refinable space without isolated points thenXis irreducible.
Keywords and phrases.s-point finite refinable, metacompact, irreducible space.
1991 Mathematics Subject Classification. 54D20.
Suppose thatᐁ is a cover of a set X and C⊆X. We say thatᐁ is C-point finite provided
(i) for all nonemptyU∈ᐁ, U∩C≠∅,
(ii) for alla∈Cthe set(ᐁ)a= {U∈ᐁ:a∈U}is finite.
We will call a spaceX s-point finite refinable (ds-point finite refinable)provided every open coverᐁofXhas aCᐁ-point finite open refinementᐂfor some (closed discrete) Cᐁ⊆X. Notice that ifᐁis a C-point finite cover of a spaceX for someC⊆X and
∅∉ᐁthen condition (i) above can be restated asᐁ= ∪{(ᐁ)a:a∈C}.In [10] Chaber observed that ifᐁis an open cover of a countably compact spaceXthen{st(x,ᐁ):
x∈X}has a finite subcover. That is, there is a finiteC⊆Xsuch that the collection
∪{(ᐁ)a :a∈C}is a subcover ofᐁ. AT1-space X is compact (aT3-space Lindelöf) if and only if every open coverᐁ ofX has an open refinementᐂ which isC-point finite for some finite (countable)C ⊆X. More generally, in [2, 1] it is noted that if ᐁis an open cover of aT1-spaceX then there is a closed discreteA⊆X such that ᐁA = ∪{(ᐁ)a :a∈A}is a subcover of ᐁ. It is readily seen that a (T3-) spaceX is metacompact (paracompact) if and only if for every open coverᐁ ofX there is an open refinementᐂsuch that for some closed discrete setC⊆X,ᐂisC-point finite and
(iii) the collection{st(a,ᐂ):a∈C}is point finite (locally finite).
A collection of subsetsᏴ of a set X is calledminimal provided that for allH∈
shown that every point finite cover of a set has a minimal subcover. It follows from [6, Thm. 1.1], that a spaceX is irreducible if and only if every open coverᐁ has a
C-point finite open refinementᐂfor some closed discrete setC⊆Xsuch that for all
a∈C,|ᐂa| =1.We show that every spaceXcan be embedded as a closed subset of an irreducible space, Theorem 1.3. We do not know if everyds-point finite refinable space is irreducible. However countably compactds-point finite refinable spaces are compact and we show thatds-point finite refinable spaces are irreducible of orderω, Theorem 2.5. Also strongly collectionwise Hausdorffds-point finite refinable spaces without isolated points are irreducible, Corollary 2.3.
Many covering properties have been characterized in terms of monotone or directed open covers. For example, a space is metacompact if and only if every monotone open cover has a point finite open refinement, [17]. Fords-point finite refinability we cannot restrict ourselves to monotone open covers, Example 3. However the question for directed open covers remains open. We have several results that demonstrate the central importance of this question. Suppose that ᐁ is a directed open cover of a
T1-space X and thatC is a closed discrete subset of X. If ᐁ has an open C-point finite refinement then it has a minimal openC-point finite refinement, Theorem 2.1. Supposef:X→Y is a perfect map from a spaceX onto ads-point finite refinable spaceY. Every directed open cover ofXhas a minimal open refinement, Theorem 3.2. Throughout this paper all spaces areT1. When we topologize an ordinal space it will be with the order topology and subsets of topological spaces will have the subspace topology. IfXis a set,Ᏼa collection of subsets ofXandC⊆XthenᏴis said to be point finite onCprovidedC⊆ ∪Ᏼand everyx∈Cis in only finitely many members ofᏴ. For any collectionᏭof subsets of a setXand anyx∈X,(Ꮽ)x= {A∈Ꮽ:x∈A} andst(x,Ꮽ)= ∪(Ꮽx).
1. Examples and basics
Theorem1.1. Ifλis an ordinal with cf(λ)=λ > ωandS is a stationary subset of λthenSis nots-point finite refinable.
Proof. Suppose thatᐂis an open refinement of the open cover{[0,α]∩S:α∈S}
ofS. For every limit ordinalα∈SletVα∈(ᐂ)αandβ(α) < αsuch that[β(α),α]∩S⊆ Vα.By the “Pressing Down Lemma” there is aβ∗< λsuch that the setS= {α∈S:α is a limit ordinal andβ∗=β(α)}is a stationary subset ofλ.Notice that ifα∈S and (ᐂ)a is finite thenα < β∗.Thus ifD⊆S andᐂis point finite onDthenD⊆[0,β∗] and therefore|D|< λ.
SupposeD⊆S andᐂisD-point finite. Then, since|D|< λand|S| =λ,there is an η∈Dsuch that the setS= {α∈S:η∈V
a}has cardinalityλ.Since(ᐂ)η is finite there is aV∈ᐂsuch thatS∗= {α∈S:V=V
α}has cardinalityλ.But this is not possible sinceV⊆[0,γ]someγ < λand therefore has cardinality less thanλ.Hence ᐂis notD-point finite for anyD⊆S.
The following clearly holds.
Example1. A countably compact, noncompacts-point finite refinable LOTS which is notds-point finite refinable.
LetX=ω1×(ω1+1)be given the lexicographic order. That is, define a linear or-dering<∗onXby letting(α,β) <∗(γ,δ)provided
α < γ, or α=γ and β < δ for allα,γ < ω1andβ,δ≤ω1. (1) We topologizeXby giving it the order topology. SinceXis countably compact but not compact it is notds-point finite refinable.
Notice that for a limit ordinalβ < ω1,if x∈X and x <∗(β,0),then there is an ordinalα < βsuch thatx <∗(α,0).For ordinalsα < β < ω1, define
(α,β]∗=x∈X:(α,0) <∗x≤∗(β,0)
= ∪(γ,δ):δ≤ω1:α≤γ < β\(α,0)∪(β,0). (2) Ifβis a limit ordinal such sets form an open local base at the point(β,0).Clearly,
{(α,0):α < ω1}is a closed subset ofXhomeomorphic toω1with the usual order topology.
Claim. Xiss-point finite refinable.
Proof. Letᐂ be an open cover ofX. For each limit ordinal β < ω1, letVβ∈ᐂ such that(β,0)∈Vβand letα(β) < βsuch that(α(β),β]∗⊆Vβ. By the Pressing Down Lemma, there is an uncountable setB⊆ω1and an ordinalγ < ω1such thatγ=α(β) for allβ∈B.For allβ∈BletUβ= {(γ,β+1)} ∪(γ+1,β]∗and letᐁ= {Uβ:β∈B}. Note that
(1) {x∈X:(γ+1,0) <∗x} ⊆ ∪ᐁ.
(2) for allβ∈B , Uβis the only member ofᐁcontaining the isolated point(γ,β+1). The set[0,γ+1]∗ is compact, since[0,γ+1]∗ is a closed subset of the compact subspace[0,γ+1]×(ω1+1)ofX. Thus, there is a finite subsetᐃ= {W1,W2,...,Wn} ofᐂ that covers[0,γ+1]∗. For each k∈ {1,2,...,n}let x
k∈Wk. Note that if k∈
{1,2,...,n} and β∈B with xk∈Uβ then xk =(γ,β+1). Thus,xk is in at most 1 member ofᐁfor eachk∈ {1,2,...,n}.LetC= {(γ,β+1):β∈B} ∪ {x1,x2,...,xn}. The collectionᐃ∪ᐁis an open refinement ofᐂand
(1) G∩C≠∅for allG∈ᐃ∪ᐁ,
(2) each member ofCis in at mostn+1 members ofᐃ∪ᐁ.
Henceᐃ∪ᐁis aC-point finite open refinement ofᐂ. Thus,Xiss-point finite refinable.
Metacompact spaces (among many others) are isocompact. That is every closed countably compact subset of a metacompact space is compact [3].
Example2. Irreducible GO-spaceY which is not isocompact.
Closed subspaces ofs-point finite refinable,ds-point finite refinable and even irre-ducible spaces need not bes-point finite refinable. In fact, Examples 1 and 2 both have a closed subset homeomorphic toω1. We can say more.
Theorem1.3. Every spaceX can be embedded as a closed subspace of an irre-ducible spaceY.
Proof. LetYbe the space obtained from the product spaceX×(ω+1)by isolating the points ofX×ω.Supposeᐁis an open cover ofY .For eachx∈X, chooseUxopen inX containingxandnx∈ωsuch that there is aU∈ᐁwithUx×[nx,ω]⊆U.Let Xm= {x∈X:nx=m}for allm∈ω.Let ᐁ0= {(Ux×[1,ω]∪ {(x,0)}:x∈X0}. Forn >0,letᐁn= {(Ux×[n+1,ω])∪ {(x,n)}:x∈Xn and x∈ ∪k<n∪y∈XkUy}.
Finally, letᐁω= {{x}:x∈Y andx∈ ∪ ∪n<ωᐁn}. Letᐂ= ∪n≤ωᐁn.Clearly,ᐂis an open refinement ofᐁ. SupposeV∈ᐂ. IfV∈ᐁω,thenV= {x}andx is not in any other element ofᐂ.IfV∈ᐁn,forn < ω,then choosex∈Xsuch thatV=(Ux×[n+ 1,ω])∪{(x,n)}.We show that(x,n)∉Wfor anyW∈ᐂ\{V}. IfW∈ᐁω,(x,n)∉W
since(x,n)∈V. IfW∈ᐁkfor somek, then eitherk=n,k > n, ork < n. Ifk < n, then
(x,n)∉W by the definition ofᐁn. Ifk > n, then W⊆X×[n+1,ω]so(x,n)∉W. Ifk=n, and(x,n)∈W, thenW=V. Henceᐂ is minimal. ThusY is an irreducible space, and clearlyXis homeomorphic to the closed subspaceYo.
Every space contains a left-separated dense subspace [16, 2.6(ii)]. M. Ismail has ob-served that in [14, Lem. 2.2] where it is shown that every left-separatedT2-space is
C-closed (countably compact subsets are closed), they in fact prove that such spaces are hereditarily irreducible.
Theorem1.4(Ismail). EveryT2-spaceX has a dense hereditarily irreducible sub-space.
In [13, Thm. 2.3] they characterize paracompact GO-spaces as those GO-spaces which do not have a closed subspace homeomorphic to a stationary subset of a reg-ular uncountable cardinal. GO-spaces are monotonically normal and in [4] they show that this elegant characterization of paracompactness in GO-spaces remarkably holds for the broader class of monotonically normal spaces.
Theorem1.5(Balogh and Rudin). A monotonically normal space is paracompact if and only if it does not have a closed subspace homeomorphic to a stationary subset of a regular uncountable cardinal.
Example 1 shows thats-point finite refinable LOTS and Example 2 irreducible GO-spaces can have closed subsets homeomorphic to ω1 and therefore, in spite of Theorem 1.1, need not be paracompact. However, H. Bennett has proved the following surprising result.
Theorem1.6(Bennett). A LOTS is paracompact if and only if it isds-point finite refinable.
need not be irreducible.
Question1. Is everyds-point finite refinable space irreducible? The following is readily verified.
Theorem2.1. A spaceXis irreducible, ifXo, the set of all nonisolated points ofX, is irreducible.
Notice thatX being irreducible does not necessarily imply that the subspaceXo is irreducible or evens-point finite refinable, see Example 1.4. In fact, Theorem 1.3, shows that any spaceXcan be embedded as a closed subspace of an irreducible space
Y withX=Yo.
A spaceXis said to be(strongly) collectionwise Hausdorff if for any closed discrete setAinX, the points ofAcan be separated by a (discrete) disjoint collection of open sets.
Theorem2.2. Suppose thatXis a strongly collectionwise Hausdorff space. IfXois ds-point finite refinable thenXis irreducible.
Proof. Suppose thatᐁis an open cover ofX. Letᐁo= {U∩Xo:U∈ᐁ}andᐂo be an open (inXo)D-point finite refinement ofᐁowhereDis a closed discrete subset ofXo. For all∅≠V∈ᐂoletW (V )be an open subset ofXsuch thatV=W (V )∩Xo andW (V )⊆U for someU∈ᐁ. Notice thatᐃ= {W (V ):V∈ᐂo}is an open partial refinement ofᐁcoveringXosuch that
(1) W∩D≠∅,for allW∈ᐃ, (2) for allx∈D,(ᐃ)xis finite.
SinceD is closed discrete andXo which is closed inX, D is closed discrete inX. SinceXis strongly CWH, for eachx∈D, letG(x)be an open neighborhood ofxsuch thatᏳ= {G(x):x∈D}is discrete. For allx∈D, letH(x)=G(x)∩(∩(W )x)and notice, by (2),H(x)is open. SinceᏳis discrete so isᏴ= {H(x):x∈D}. Also, since no point ofD is isolated inX, for allx∈D, H(x)\{x}is infinite. For eachx∈D, letnx = |(W )x|< ωand fori=1,2,...,nx, letyix∈H(x)\{x} such that ifi,k∈
{1,2,...,nx}withi≠kthenyix≠ykx. SinceᏴis discrete so isD∗= {yix:x∈Dandi∈
{1,2,...,nx}}. For eachx∈D,list(ᐃ)x= {V1x,...,Vnxx}and, for eachi∈ {1,...,nx},
letWx
i =Vix\(D∗\{yix}). The collectionᐃ∗= {Wix:x∈D,i∈ {1,2,...,nx}}is clearly a minimal open partial refinement ofᐃand henceᐁcoveringXo. Thenᐂ=ᐃ∗∪{{x}: x∈X\∪ᐃ∗}is a minimal open refinement ofᐁ.
Corollary2.3. IfXis a strongly collectionwise Hausdorffds-point finite refinable space without isolated points thenXis irreducible.
Clearly normal collectionwise Hausdorff spaces are strongly collectionwise Haus-dorff. AT2-space is said to be of point-countable typeif every point is contained in a compact set of countable character. For example, first countable spaces, locally compact space, and evenp-spaces [12].V=Limplies that normal spaces of point-countable type are collectionwise Hausdorff [19].
A collectionof sets is said to bemonotoneprovided for allS,S∈eitherS⊆S orS⊆S. More generally, collectionof sets is said to bedirectedprovided for allS, S∈there is aT∈such thatS∪S⊆T. A spaceXis metacompact if and only if every monotone open cover ofXhas a point-finite open refinement [18].
Example3. A nonds-point finite refinable space in which every monotone open cover has a minimal open refinement.
Let Y = ∞i=1(ωi+1). For each natural number k, let Xk = (ki=1(ωi+1))× (∞i=k+1(ωi))and let X = ∪∞k=1Xk with the usual subspace topology. The space X is ℵ1-compact and every monotone open cover of X has a countable subcover but
Xis not Lindelöf, [19]. Since a countable open cover of any space has an irreducible open refinement, every monotone open cover ofXhas an irreducible open refinement. However, sinceXisℵ1-compact but not Lindelöf,Xis notds-point finite refinable.
Notice that ifᐂ is an open cover of Example 3 having no countable subcover and we letᐂ∗be the collection of all finite unions of members ofᐂthenᐂ∗is a directed open cover ofXwhich does not have aC-point finite open refinement for any closed discreteC⊆X.
Theorem2.5. Suppose thatᐁis a directed open cover of aT1-spaceXand thatC is a closed discrete subset ofX. If ᐁhas an openC-point finite refinement then it has a minimal openC-point finite refinement.
Proof. Suppose thatᐂis an openC-point finite refinement ofᐁ. For eachV∈ᐂ and eachx∈C∩V, letW (V ,x). Since the setCis closed discrete for eachx∈Cthe setC\{x}is closed. Hence for eachV∈ᐂand eachx∈C∩Vthe setW (V ,x)is open. Letᐃ= {W (V ,x):V∈ᐂandx∈C∩V}andᐃ= {st(x,ᐃ):x∈C}. Sinceᐂis an open cover ofXand for eachV∈ᐂand eachx∈C∩V,W (V ,x)=(V\C)∪ {x} ⊆V
the collectionᐃis an open refinement ofᐁ. Sinceᐁis directed and elements ofᐃ are finite unions of elements ofᐃ,ᐃis an open refinement ofᐁ. To see thatᐃis minimal we need only note that for everyx∈Cthe setst(x,ᐃ)is the only member ofᐃcontainingx.
Question2. If every directed open coverᐁof a spaceXhas aCᐁ-point finite open refinement for some closed discreteCᐁ⊆X, isXirreducible?
It follows from Theorem 2.5 that the answer to Question 2 being yes for irreducible spaces would imply an affirmative answer to Question 1.
In [7] the concept of irreducible of orderκis introduced for any infinite cardinalκ, generalizing irreducible. In this paper, we are interested only in the case whereκ=ω. A spaceXisirreducible of orderωprovided for every open coverᐁofXthere is an open refinementᐂofᐁsuch thatᐂ=ᐂ1∪ ··· ∪ᐂn and a family of discrete closed collections{Υ1,Υ2,...,Υn}such that
(1) For allk∈ {1,2,...,n}and for allT∈Υk,ᐂkT= {V∈ᐂk:T⊆V}≠∅but finite. (2) {V:V∈ᐂk
T,T∈Υk,k∈ {1,2,...,n}}coversX.
Theorem2.6. A spaceXis irreducible of orderωif and only if it isds-point finite refinable.
(⇒)Suppose X is irreducible of order ωand ᐁ is an open cover of X. Let ᐂ=
ᐂ1∪···∪ᐂnbe an open refinement ofᐁand supposeΥ1,Υ2,...,Υnare discrete closed collections such that
(1) For allk∈ {1,2,...,n}and for allT∈Υk,ᐂkT= {V∈ᐂk:T⊆V}≠∅but finite. (2) {V:V∈ᐂk
T,T∈Υk,k∈ {1,2,...,n}}coversX.
For eachk∈ {1,2,...,n}andT ∈Υkletx(k,T )∈T and note thatAk= {x(k,T ): T∈Υk}is a discrete collection sinceΥkis discrete andA=A1∪···∪Anis also closed discrete. For eachk∈ {1,2,...,n},T∈ΥkandV∈ᐂkTletG(k,T ,V )=V\(A\{x(k,T )}) andᏳ= {G(k,T ,V):k∈ {1,2,...,n},T ∈Υk,V∈ᐂkT}. SinceAis closed discrete,Ᏻis a collection of open sets. Lety∈A. There is ak∈ {1,2,...,n}and aT∈Υksuch that y=x(k,T )∈T. By (1),ᐂk
T≠∅so letV∈ᐂkT andy∈G(k,T ,V ).
Supposey∈X\A. By (2) there is ak∈ {1,2,...,n},T ∈Υk andV∈ᐂkT such that y ∈V. Since G(k,T ,V)\A= V\A, y ∈G(k,T ,V ). HenceᏳ is an open cover ofX. Fork∈ {1,2,...,n}, T∈ΥkandV ∈ᐂkT,G(k,T ,V)⊆V. Hence, sinceᐂrefinesᐁ,Ᏻ refinesᐁ.
Finally supposey∈A. Ifk∈ {1,2,...,n},T∈ΥkandV∈ᐂkTsuch thaty∈G(k,T ,V) theny=x(k,T ). Since for eachk∈ {1,2,...,n}there is at most oneT∈Υksuch that y=x(x,T )(Υkdiscrete (pwd)) and for eachT∈ΥkᐂkT is finite, the pointycan be in only finitely many members ofᏳ. ThusᏳis anA-point finite refinement ofᐁ. SinceA
is closed discrete, we conclude thatXisds-point finite refinable.
Question 1 can be restated as “Are irreducible and irreducible of orderω equiva-lent?”.
3. Mappings. The properties of the image and pre-image of spaces having various covering properties under a variety of mappings have been well studied (see [8, 9]). In particular, the perfect pre-image of a metacompact space is metacompact [12] and the closed image of a metacompact space is metacompact [18]. Fors-point finite refinable spaces little positive is known. In the next two theorems the possible significance of the question of characterizing these spaces in terms of directed open covers (Ques-tion 2) is seen. Notice that the proof of Theorem 3.1 is a modifica(Ques-tion of the proof of [12, Thm. 5.1.35].
Theorem3.1. Supposef:X→Y is a perfect map from a spaceXonto ans-point finite refinable spaceY. Every directed open cover ofXhas aC-point finite open refine-ment for some setC⊆X.
Proof. Letᐁbe a directed open cover ofX. For eachy∈Y, letU(y)∈ᐁsuch thatf−1(y)⊆U(y)(f−1(y)is compact). For eachy∈Y, letV (y)be an open neigh-borhood ofy inY such thatf−1(y)⊆f−1(V(y))⊆U(y)(f is closed, continuous). Nowᐂ= {V(y):y∈Y}is an open cover ofY so letA⊆Y andᐃan open refinement ofᐂsuch that
(1) W∩A≠∅for allW∈ᐃ.
(2) {W∈ᐃ:a∈W}is finite for alla∈A.
leta∈A and supposexa∈f−1(W ). Then a=f (xa)∈W. However, the collection
{W∈ᐃ:a∈W}is finite so{f−1(W ):W∈ᐃandx
a∈f−1(W )}is also finite. Notice in the above proof if the setAis closed discrete then the setA∗is also closed discrete. The following result is a direct consequence of this and Theorem 2.5.
Theorem3.2. Supposef:X→Y is a perfect map from a spaceXonto ads-point finite refinable spaceY. Every directed open cover ofXhas a minimal open refinement. The following examples show that even irreducible spaces are not very well behaved under closed mappings.
Example4. The perfect image of ans-point finite refinable space need not bes -point finite refinable.
LetXbe the space from Example 1. Define a functionf:X→ωas follows
f (α,β)=
α, ifβ=0
α+1, otherwise for allα < ω1andβ≤ω1. (3)
Example5. The closed image of an irreducible space need not bes-point finite refinable.
The restriction of the function in Example 4 to the subspaceY ofXin Example 2 is a closed mapping ontoω1.
Question3. Is the perfect pre-image of ans-point finite refinable (ds-point finite refinable) [irreducible] space s-point finite refinable (ds-point finite refinable) [irre-ducible]?
Notice that if the answer to Question 1 is yes for irreducible spaces then Theorem 3.2 implies that the perfect pre-image of an irreducible space is irreducible.
Question4. Is the perfect image of ads-point finite refinable (irreducible) space
ds-point finite refinable (irreducible)?
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Davis, Grabner, and Grabner:229Vincent Science Hall, Slippery Rock, PA16057-1326, USA
