summary:The aim of the paper is to study the preservation and the reflection of acc and hacc spaces under various kinds of mappings. In particular, we show that acc and hacc are not preserved by perfect mappings and that acc is not reflected by closed (nor perfect) mappings while hacc is reflected by perfect mappings

Bonanzinga, M.

Institute of Mathematics AS CR, v. v. i.

Commentationes Mathematicae Universitatis CarolinaeMaddalena BonanzingaPreservation and reflection of properties acc and haccCommentationes Mathematicae Universitatis Carolinae, Vol. 37 (1996), No. 1, 147--153Persistent URL: http://dml.cz/dmlcz/118818Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 1996Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.czComment.Math.Univ.Carolin. 37,1 (1996)147–153 147Preservation and reflection of properties acc and haccM. BonanzingaAbstract. The aim of the paper is to study the preservation and the reflection of accand hacc spaces under various kinds of mappings. In particular, we show that acc andhacc are not preserved by perfect mappings and that acc is not reflected by closed (norperfect) mappings while hacc is reflected by perfect mappings.Keywords: acc space, hacc space, countably compact space, ordinal space, cofinalityClassification: 54D20, 54D30Introduction and preliminaryRecently Matveev in ([5], [6]) introduced a new property acc (absolutely count-ably compact) which is stronger than countably compact and the property hacc(hereditarily absolutely countably compact); Matveev proved that the continuousimage of an acc space need not to be acc and that the continuous pseudoopenimage of an acc space is acc. Note in the Matveev’s terminology a mappingf : X → Z is pseudoopen provided Int(f(O)) 6= ∅ for every nonempty open setO ⊂ X ; this property is stronger than the classical definition of “pseudoopenmapping” as given in [1], and [2]. In this paper we refer to Matveev’s property asstrongly pseudoopen.We want to give answers concerning the preservation of acc and hacc spacesunder various kinds of mappings; further we will show that the properties acc andhacc are not reflected by perfect mappings. Our counterexamples use product ofordinal spaces; the key of this problem is to know when a product of two ordinalspaces is acc or hacc. In the first part of this paper we give a solution to thisproblem by a characterization.We will use the following definitions: X means a topological space; A, Int(A)the closure and the interior of A respectively, where A is a subset of X ; if Int(A) =A (Int(A) = A) we say that A is a regular open (regular closed) subset of thespace. A clopen subset of a topological space X is a set that is both open andclosed subspace of X . X1⊕X2 denotes the discrete sum of disjoint spaces X1 andX2. Further a mapping f : X → Y is said to be closed (open) if for every closed(open) set A ⊂ X , the image f(A) is closed (open); a mapping p : X → Y isThis research was supported by a grant from the C.N.R. (G.N.S.A.G.A.) and M.U.R.S.T.through “Fondi 40 %” Italy148 M.Bonanzingacalled perfect if it is a continuous closed surjection and each fiber p−1(y), (y ∈ Y )is compact. All mappings are assumed to be continuous.Recall the following definitions:Definition 1 [4]. A space X is called countably compact provided every count-able open cover has a finite subcover.Note that a characterization of countable compactness (see [4, 3.12.22 (d)])states that a T2 space is countably compact iff for every open cover U of X thereexists a finite set F ⊂ X such that St(F,U) =⋃{U ∈ U : U ∩ F 6= ∅} = X .Definition 2 [5]. A space X is said to be absolutely countably compact (acc)provided for every open cover U of X and every dense D ⊂ X , there exists a finiteset F ⊂ D such that St(F,U) =⋃{U ∈ U : U ∩ F 6= ∅} = X .Matveev ([5]) noted that:compact =⇒ accT2=⇒ countably compactand he proved that every countably compact space X with countable tightness(see Definition 1.1) is acc. Further Vaughan ([10]) proved that every countablycompact, orthocompact space (see Definition 1.2) is acc.Matveev ([5]) proved that a regular closed subspace of an acc space need not beacc. Since every regular closed subspace of a topological space is a closed subspace,absolute countable compactness is a not hereditary with respect to closed subsetsproperty. Then he introduced the following definition:Definition 3 [5]. A space X is hereditarily absolutely countably compact (hacc)if all closed subspaces of X are acc.1. Acc and hacc spaces as products of ordinal spacesDefinition 1.1 [4]. If p is an element of a space X , the tightness of the point pin X is t(p, X) = min{κ : for all Y ⊆ X with p ∈ Y , there is A ⊆ Y with |A| ≤ κand p ∈ A}; the tightness of the space X is t(X) = Sup{t(p, X) : p ∈ X}.The following lemma is a basic result for the proof of the Theorem 1.2 thatrepresent the key in the solution of the original problem.Lemma 1.1. Let α and β be ordinals. If ω ≤ cf(β) < α, then α × β is not acc.Proof: If ω = cf(β) we have that α× β is not countably compact and then it isnot acc. Suppose ω < cf(β) and consider the clopen subspace (cf(β) + 1)× β ⊂α × β. Since all clopen subsets of an acc space are acc, we want to show that(cf(β) + 1) × β is not acc. Put cf(β) = κ and let {xδ : δ < κ} be a sequencesuch that {xδ : δ < κ} is cofinal in β and δ < γ ⇔ xδ < xγ . Consider the spaceX = (κ+1)×β. Put T = {κ}×β and D = X −T . T is a closed subset of X andD is dense in X . For every δ < κ define Uδ = (δ, κ]× [0, xδ]. Since {xδ : δ < κ}Preservation and reflection of properties acc and hacc 149is cofinal in β, the family {Uδ : δ < κ} is an open cover of T and then the familyU = {X − T } ∪ {Uδ : δ < κ} is an open cover of X . Suppose F ⊂ D is finite.Denote ξ = max{η ∈ (κ+1) : (η, γ) ∈ F} and consider the point (κ, xξ) ∈ T . Forevery U ∈ U such that (κ, xξ) ∈ U , we have that U = Uσ, where σ ≥ ξ. Sinceevery point (η, γ) ∈ F is such that η ≤ ξ, we have that F ∩ Uσ = ∅, for everyσ ≥ ξ. So (κ, xξ) /∈ St(F,U). By this, St(F,U) 6= X for every finite F ⊂ D whichshows that X is not acc. Definition 1.2 [7]. A space X is called orthocompact provided for every opencover U there exists an open refinement V such that for every V | ⊂ V , we have⋂{V ∈ V | : x ∈ V } is open for each x ∈ X .We prove a close analog of the following theorem of Brian Scott which willanswer the question of when α × β is acc or hacc.Theorem 1.1 [7]. Let α and β be ordinals. If α ≤ β, then the following areequivalent:1. α × β is orthocompact;2. α × β is normal;3. one of the following holds:(a) α = β and cf(α) ≤ ω;(b) α = β and cf(α) = α;(c) cf(α) ≤ ω and α ≤ cf(β); and(d) cf(α) ≤ ω and cf(β) ≤ ω.The following is the main result of this section.Theorem 1.2. Let α and β be ordinals. If α ≤ β and α × β is countablycompact, then the following are equivalent:1. α × β is acc;2. α × β is hacc;3. α × β is orthocompact;4. α × β is normal;5. one of the following holds:(a) α = β and cf(α) < ω;(b) α = β and cf(α) = α;(c) cf(α) < ω and α ≤ cf(β); and(d) cf(α) < ω and cf(β) < ω.Proof: Clearly (5) in Theorem 1.2 implies (3) in Theorem 1.1; so for Theorem 1.2we have (5)⇒ (4)⇒ (3). Since every countably compact, orthocompact space isacc ([10]) and orthocompactness and countable compactness are hereditary withrespect to closed subsets, we have that (3)⇒ (2). As (2)⇒ (1), we have to showthat (1)⇒ (5). The proof is by contradiction, and we note that by the hypothesis“α×β is countably compact”, we have cf(α) 6= ω and cf(β) 6= ω, however, α andβ can be successor ordinals. Assume α = β. If (a) and (b) do not hold, we havethat cf(α) > ω and cf(α) 6= α. Then ω < cf(α) < α and, by Lemma 1.1, α × α150 M.Bonanzingais not acc, a contradiction. Assume α 6= β; therefore α < β, by hypothesis. If (c)and (d) do not hold, we have that (cf(α) > ω or cf(β) < α) and (cf(α) > ω orcf(β) > ω) hold. This is equivalent to say that one of the following holds:(i) cf(α) > ω and cf(α) > ω, i.e. cf(α) > ω;(ii) cf(α) > ω and cf(β) > ω;(iii) cf(β) < α and cf(α) > ω;(iv) cf(β) < α and cf(β) > ω.Since (ii) ⇒ (i) and (iii) ⇒ (i), we have that cf(α) > ω or ω < cf(β) < αholds.If cf(α) > ω, then ω < cf(α) ≤ α < β. So, by Lemma 1.1, β × α is not acc.But β × α is homeomorphic to α × β, a contradiction. If ω < cf(β) < α, byLemma 1.1, α × β is not acc, a contradiction. 2. Preservation of property hacc under open and closed mappingsRecall the following proposition:Proposition 2.1 [4]. If f : X → Y is an open mapping, then for any subspaceL ⊂ Y the restriction f|f−1(L) : f−1(L)→ L is open.Proposition 2.2. The hacc property is preserved by open mappings.Proof: Suppose f : X → Y is an open onto mapping and F a closed subsetof Y . Since X is hacc and f−1(F ) is closed in X , we have that f−1(F ) is acc.Further the restriction f|f−1(F ) is open hence strongly pseudoopen; so the imageF = f−1f(F ) is acc. Proposition 2.3. The hacc property is preserved by closed mappings.Proof: Suppose f : X → Y is a closed onto mapping, F is a closed subset of Y ,U is an open cover of F and D a dense subset of F . Then U0 = {f−1(U) : U ∈ U}is an open cover of f−1(F ). Since f−1(D) is a closed subset of X and X is hacc,f−1(D) is acc. U0 is an open cover of f−1(D) because f−1(D) ⊂ f−1(F ); asf−1(D) is a dense subset of f−1(D), there exists A0, a finite subset of f−1(D)such that St(A0,U0) ⊃ f−1(D). Then f(St(A0,U0)) ⊃ f(f−1(D)). Since f is aclosed surjection, F is a closed subset of Y and D is a dense subset of F , we havef(f−1(D)) = F and hence that St(f(A0),U) = f(St(A0,U0)) ⊃ F . This provesthat Y is hacc. Corollary 2.1. The hacc property is preserved by perfect mappings.3. Non-preservation and non-reflection of acc and hacc under variouskinds of mappingsIn this section we will give some examples to prove that the acc property isnot preserved or reflected by perfect mappings; further we will show that thePreservation and reflection of properties acc and hacc 151hacc property is not preserved by strongly pseudoopen mappings nor reflected byperfect mappings.Recall the following proposition:Proposition 3.1 [5]. If X = X1 ∪ . . . ∪ Xn, Xi is acc for 1 ≤ i ≤ n, andXi ⊂ Int(Xi) for 1 ≤ i ≤ n then X is acc.Now we show the following:Example 3.1. The acc property is not preserved by perfect mappings.Let us consider the space H = H1 ⊕ H2, where H1 = ω2 × (ω1 + 1) andH2 = ω2× (ω2+1). Now consider the factor mapping h : H → X which identifiesthe points of ω2 × {ω1} ⊂ H1 with the corresponding points of ω2 × {ω2} ⊂ H2.Now we show that the resulting factor space X is acc. Let X1 = h(H1) andX2 = h(ω2 × ω2) with ω2 × ω2 ⊂ H2. Note that the restriction of h to H1 andof h to ω2 × ω2 ⊂ H2 are homeomorphisms, and that X = X1 ∪ X2. Since,by Theorem 1.2, H1 and ω2 × ω2 are acc, X1 and X2 are acc. Let T be theimage of the top-line of H1 and H2, i.e. T = h(ω2 × {ω1}) = h(ω2 × {ω2}); thenT ⊂ X1 and T ∩ X2 = ∅. X1 = h(H1) ⊃ h(H1) − T = X − (X2 ∪ T ) that isopen in X because X2 ∪ T is closed in X (in fact h−1(X2 ∪ T ) is closed in H).So Int(h(H1)) ⊃ Int(h(H1) − T ) = h(H1) − T ; then Int(X1) = Int(h(H1)) ⊃h(H1)− T = h(H1) − Int(T ) = h(H1) = X1. Since h−1(X2) is open in H , X2is open in X and then X2 ⊂ Int(X2). So, by Proposition 3.1 X is acc. In orderto define a mapping with domain X , we identify X1 = ω2 × (ω1 + 1) = H1 andX2 = h(ω2 × ω2) with ω2 × ω2 ⊂ H2. Let Z = ω2 × (ω2 + 1). Then Z is not accby Theorem 1.2. Define the mappingϕ : X → Zby ϕ(α, β) = (α, β) ∈ (ω2 × ω2), for all (α, β) ∈ X2 and ϕ(α, x) = (α, ω2) ∈(ω2 × {ω2}), for all (α, x) ∈ X1. Note that the restriction ϕ|X2 is an homeo-morphism (thus a continuous mapping) onto ω2 ×ω2, and the restriction ϕ|X1 isthe projection (thus a continuous mapping) of X1 onto ω2 ×{ω1} identified withω2×{ω2}. Since X = X1 ∪X2, it is easy to check that ϕ is continuous, and sincethe projection X × Y → X where Y is compact, is a closed mapping, it followsthat ϕ is a closed mapping. Further each fiber ϕ−1(y), (y ∈ ω2 × (ω2 + 1)) iscompact and, by Theorem 1.2, Z is a non acc space; then ϕ is a perfect mappingfrom an acc space onto a non acc space and the proof is complete. Example 3.2. The acc property is not reflected by perfect mappings.Consider the factor mapping h : H → X from Example 3.1; h is a continuoussurjection such that each fiber h−1(x), (x ∈ X) is compact (in fact, |h−1(x)| ≤ 2,for each x ∈ X). Further, as H1 is homeomorphic to h(H1), H2 is homeomorphicto X2 ∪ T and h(H1) and X2 ∪ T are closed subsets of X , every closed subset Fof H is closed in X ; then h is a closed mapping. So h is a perfect mapping onto152 M.Bonanzingaan acc space such that h−1(X) = H is a non acc space (in fact, H2 is a non acc,clopen subset of H). Example 3.3. The hacc property is not preserved by strongly pseudoopen map-ping.Consider the factor mapping h : H → X from Example 3.1. As the restrictionof h to H2 is an homeomorphism, h(H2) is a closed, non acc subspace of X ; thenX is a non hacc space. Let ϕ be the restriction of h to [ω2 × (ω1 + 1)] ⊕ [ω2 ×ω2] ⊂ H . By Theorem 1.2, ω2 × (ω1 + 1) and ω2 × ω2 are hacc spaces, then[ω2× (ω1+1)]⊕ [ω2×ω2] is hacc. Since h is a strongly pseudoopen mapping andthe restriction of any strongly pseudoopen mapping to an open subset is stronglypseudoopen, we have that ϕ is a strongly pseudoopen (non open) mapping froman hacc space onto a non hacc space. Example 3.4. The hacc property is not reflected by perfect mappings.Consider the factor space X from Example 3.1 and define the mappingϕ : X → ω2 × (ω1 + 1)by ϕ((α, β)) = (α, β) ∈ ω2×ω1, for all (α, β) ∈ X1−T and ϕ((α, x)) = (α, ω1) ∈ω2×{ω1}, for all (α, x) ∈ X2∪T . ϕ is a continuous mapping onto the hacc spaceω2 × (ω1 + 1), by Theorem 1.2. Further each fiber ϕ−1(y), (y ∈ ω2 × (ω1 + 1)) iscompact because if y ∈ ω2 × ω1, |ϕ−1(y)| = 1 while if y = (α, ω1) ∈ ω2 × {ω1},ϕ−1(y) = h({(α, ω1)} ∪ {α} × (ω2 + 1)), where {(α, ω1)} ⊂ H1 and {α} × (ω2 +1) ⊂ H2, i.e. ϕ−1(y) is a continuous image of a compact space, then it is acompact space. Further the mapping ϕ is closed (the proof is similar to theproof of Example 3.1). So ϕ is a perfect mapping onto an hacc space such thatϕ−1(ω2 × (ω1 + 1)) = X is a non hacc space. 4. Open questionsIt is now natural to pose a question: is hacc preserved by strongly pseudoopenmappings with compact fibers?Another natural question is the following: are acc and hacc reflected by per-fect mappings (a) with first-countable fibers, (b) with sequential fibers, (c) withcountably tight fibers.Conclude the paper with the following diagram summing up all information wegive concerning preservation and reflection of acc and hacc under various kindsof mappings.Preservation and reflection of properties acc and hacc 153TablePRESERVATION AND REFLECTION OF ACC AND HACCTYPES OF MAPPINGSTYPES OFPROPERTIES open closed perfect strongly pseudoopenpreservation of acc + − − +∗preservation of hacc + + + −reflection of acc ? − − ?reflection of hacc ? − − ?(∗) See ([5]).References[1] Arhangel’skii A.V.,Mappings and spaces (in Russian), Uspehi Mat. Nauk 21 (1966), No. 4,133–184; English Translation: Russian Math. Sur. 21 (1966), No. 4, 115–162.[2] Arhangel’skii A.V., Ponomarev V.I., Foundations of General Topology in Problems andExercises (in Russian), Nauka, Moscow, 1974.[3] Dugundji J., Topology, Allyn and Bacon, Inc., Boston London Sydney Toronto.[4] Engelking R., General Topology, PWN, Warszawa, 1977.[5] Matveev M.V., Absolutely countably compact spaces, Topology and Appl. 58 (1994), 81–92.[6] Matveev M.V., A countably compact topological group which is not absolutely countablycompact, Question and Answers 11 (1993), 173–176.[7] Scott B.M., Towards a product theory for orthocompactness, Studies in Topology, eds. M.Stravrakas and K.R.Allen, Academic Press, Inc., New York, 1975, pp. 517–537.[8] van Douwen E.K., Reed G.M., Roscow A.W., Tree I.J., Star covering properties, Topologyand Appl. 39 (1991), 71–103.[9] Vaughan J.E., Countably compact and sequentially compact spaces, in Handbook ofSet-theoretic Topology, K.Kunen, J.E. Vaughan, eds., North-Holland, Amsterdam, 1984,pp. 569–602.[10] Vaughan J.E., On the product of a compact space with an absolutely countably compactspace, Commenationes Math. Univ. Carolinae, to appear.Dipartimento di Matematica, Università di Messina, Contrada Papardo, SalitaSperone, 98168 Messina, Italy(Received February 21, 1995, revised September 11, 1995)

English

Preservation and reflection of properties acc and hacc

Czech digital mathematics library

http://dml.cz/bitstream/handle/10338.dmlcz/118818/CommentatMathUnivCarolRetro_37-1996-1_9.pdf

Preservation and reflection of properties acc and hacc

Abstract

Similar works

Full text

Available Versions

Institute of Mathematics AS CR, v. v. i.

Czech digital mathematics library