summary:We give a characterization of normal and countably paracompact spaces via continuous set-avoiding selections

Barov, Stoyu

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

Commentationes Mathematicae Universitatis CarolinaeStoyu BarovOn a characterization of normal and countably paracompact spaces viaset-avoiding selectionsCommentationes Mathematicae Universitatis Carolinae, Vol. 49 (2008), No. 1, 45--52Persistent URL: http://dml.cz/dmlcz/119700Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 2008Institute 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. 49,1 (2008)45–52 45On a characterization of normal and countablyparacompact spaces via set-avoiding selectionsStoyu BarovAbstract. We give a characterization of normal and countably paracompact spaces viacontinuous set-avoiding selections.Keywords: normal and countably paracompact, set-avoiding selectionsClassification: 54C60, 54C651. IntroductionThroughout this note all spaces are assumed to be T1-spaces and B stands fora Banach space. We often denote by (B, ρ) the Banach space B with the metric ρgenerated by the given norm on B. In order to state our main result let us set upsome of the terminology. If Y is a space, then 2Y denotes the set of all non-emptysubsets of Y . For a space B we denoteE(B) = {A ∈ 2B : A is convex and dimA <∞}.Let φ : X → 2Y be a map. Then φ is lower semi-continuous (LSC for short)if {x ∈ X :φ(x) ∩ U 6= ∅} is open in X for every open U in Y . A map f : X → Yis called a selection for φ if f(x) ∈ φ(x) for every x ∈ X . An open ball with aradius ε > 0 and a center x in a given metric space will be denoted by B(x, ε).All other definitions and concepts are introduced at the end of this section.Now, we formulate our main theorem.Theorem 1. For a T1-space X the following are equivalent.(i) X is normal and countably paracompact.(ii) For every separable Banach space B and for every LSC map φ : X → E(B)such that dimφ(x) = dimφ(y) for every x, y ∈ X there exists a continuousselection f for φ such that f(x) ∈ [φ(x)]◦.(iii) For every separable Banach space B and for every LSC map φ : X → E(B)such that dimφ(x) = dimφ(y) for every x, y ∈ X there exists a continuousselection f for φ such that for every x ∈ X there exist a neighborhood Vxof x and an εx > 0 with B(f(y), εx) ∩ φ(y) ⊂ [φ(y)]◦ for every y ∈ Vx.46 S.BarovIn the case when B is the real line the equivalence (i)⇔(ii) in the above theoremwas virtually shown by Dowker [4] and Katětov [6]. By [7, Theorem 3.1′′′], wehave that (ii) with the dimension requirement being removed is equivalent for Xto be perfectly normal. The following example also shows that we cannot removethe dimension requirement for φ in the above theorem.Example 1. Let X be normal and countably paracompact which is not perfectlynormal and let F ⊂ X be closed but not a Gδ-set in X . Define a LSC mapφ : X → 2[0,1] as follows:φ(x) = {0} if x ∈ F, andφ(x) = (0, 1) if x /∈ F.Clearly, dimφ(x) 6= dimφ(y) if x ∈ F and y ∈ X \ F . On the other hand, φdoes not have a continuous selection since otherwise, we would have that F is aGδ-set in X .Let A be a subset of B. Let aff A denote the affine hull of A, 〈A〉 the convexhull and A the closure of A in B. ∂A means the relative boundary of A, thatis, the boundary with respect to aff A, and we define A◦ = A \ ∂A. Recallthat if A is convex in Rn, n ∈ N, then A◦ 6= ∅. A set {a0, a1, . . . , ak} in Bis called a geometrically independent set (see, for example, [1], [8]) if the set{a1 − a0, a2 − a0, . . . , ak − a0} is a linearly independent set.If δ > 0 and A ⊂ (B, ρ) then B(A, δ) is the open ball around A with radius δ,i.e.B(A, δ) = {x : ρ(x,A) < δ}.If x ∈ (B, ρ) then we always write B(x, δ) instead of B({x}, δ).Further, if A,C are nonempty compact subsets of a normed linear space (Y, ‖·‖)then dH(A,C) stands for the Hausdorff metric between A and C (see [5] for moredetails).We conclude this section with one more definition. For a map φ : X → 2Y wedefine the following condition.(*) for every x ∈ X there is a neighborhood Ux of x such that dimφ(x) ≥dimφ(y) for every y ∈ Ux.We see later on that Theorem 1 remains valid if we replace the dimensionrequirement for φ everywhere in the theorem with the requirement for φ to satis-fy (*).The paper is arranged as follows. In the next section we establish some lemmasthat we need in the sequel. In Section 3 we prove our main results.Characterization of normal and countably paracompact spaces 472. Some lemmasIn this section we state and prove several lemmas that are needed for provingour main results. Let us point out that some of the lemmas are of independentinterest.The proof of the next lemma directly follows from [3, Lemma 2]. For reader’sconvenience we give the proof.Lemma 2. Let k ∈ N, and let {a1, . . . , ak} be a linearly (geometrically) inde-pendent set in a normed linear space (Y, ‖ · ‖). Then there is a δ > 0 such thatevery subset {b1, . . . , bk} of Y with ‖ai − bi‖ < δ for every 1 ≤ i ≤ k is also alinearly (geometrically) independent set.Proof: LetM ={(t1, . . . , tk) ∈ Rk :∥∥∥∥k∑i=1tiai∥∥∥∥≤ 2}.Since {a1, . . . , ak} is a linearly independent set we have that M is compact.Indeed, let us consider a map f : Rk → span{a1, . . . , ak}, defined by the formulaf((t1, . . . , tk)) =k∑i=1tiai for (t1, . . . , tk) ∈ Rk.Then f is bijective because {a1, . . . , ak} is a linearly independent set. In addi-tion, f is a continuous linear function between two finite dimensional Banachspaces and hence f is a linear homeomorphism (see, for example, [8, Corol-lary 1.1.14]). Therefore, M is compact as an inverse image of a compact setunder a homeomorphism. Thus, we can take an α to be an upper bound for theset {∑ki=1 |ti|: (t1, . . . , tk) ∈M}. Further, we can choose a δ such that 0 < δ <12α .Let {b1, . . . , bk} ⊂ Y with ‖ai − bi‖ < δ for every 1 ≤ i ≤ k. For (t1, . . . , tk) ∈Mwe have that∥∥∥∥k∑i=1ti(ai − bi)∥∥∥∥≤k∑i=1|ti|‖(ai − bi)‖ <k∑i=1|ti|δ ≤ δα ≤12.DefineD ={(t1, . . . , tk) ∈ Rk :∥∥∥∥k∑i=1tibi∥∥∥∥≤ 1}.Now, we are going to show that D ⊂ M . Indeed, let (t1, . . . , tk) ∈ D. Seta′ =∑ki=1 tibi and notice that ‖a′‖ ≤ 1. Consider a′′ =∑ki=1 tiai. We need48 S.Barovto show that ‖a′′‖ ≤ 2. Clearly, we may assume that ‖a′′‖ 6= 0, so we considera′′/‖a′′‖. Obviously, (t1, . . . , tk)/‖a′′‖ ∈M and therefore,‖a′ − a′′‖/‖a′′‖ =∥∥∥∥k∑i=1ti‖a′′‖(ai − bi)∥∥∥∥≤12.So, ‖a′′‖ − ‖a′‖ ≤ ‖a′ − a′′‖ ≤ ‖a′′‖/2 and therefore, ‖a′′‖ ≤ 2‖a′‖ ≤ 2. Hence(t1, . . . , tk) ∈M .Next, being a subset of the compact set M , D is bounded. That implies that{b1, . . . , bk} is a linearly independent set. Indeed, if we assume that {b1, . . . , bk}is a linearly dependent set then there is (t1, . . . , tk) ∈ Rk, not all entries equal tozero, such that∑ki=1 tibi = 0. Thus, (βt1, . . . , βtk) ∈ D for every real β. Thatmeans that D is unbounded — a contradiction. Consequently, {b1, . . . , bk} is alinearly independent set. The result for the geometrically independent sets followseasily from that of linearly independent sets. That completes the proof. Lemma 3. Let A = {a1, . . . , ak} and C = {c1, . . . , ck} be two sets of points in anormed linear space (Y, ‖ · ‖) such that ‖ai − ci‖ < δ for every 1 ≤ i ≤ k. ThendH(〈A〉, 〈C〉) < δ.Proof: Let m = max{‖ai − ci‖: 1 ≤ i ≤ k} and notice that m < δ. Then, by theassumptions, we have that C ⊂ 〈A〉+B(0, δ). Next, since 〈A〉+B(0, δ) is convexwe get that 〈C〉 ⊂ 〈A〉 + B(0, δ). Hence 〈C〉 ⊂ B(〈A〉, δ). Finally, by symmetrywe have that 〈A〉 ⊂ B(〈C〉, δ) and that completes the proof. Lemma 4. Let X be a topological space and (Y, ‖ · ‖) be a normed linear space.Let φ : X → E(B) be LSC satisfying (*). Then Pn = {x ∈ X : dimφ(x) = n− 1}is clopen for every n ∈ N.Proof: Clearly, we have that X =⋃∞n=1 Pn with Pi ∩ Pj = ∅ if i 6= j. We aregoing to show that each Pn is open. Let n ∈ N and x ∈ Pn. Since φ satisfies(*) it suffices to show that there is a neighborhood Wx of x such that dimφ(y) ≥dimφ(x) for every y ∈Wx. Indeed, if n = 1 then it is a triviality. So, we assumethat n ≥ 2. We find a geometrically independent set {a1, . . . , an} ⊂ φ(x). ByLemma 2, we find a δ > 0 such that if ‖bi − ai‖ < δ for each 1 ≤ i ≤ n thenthe subset {b1, . . . , bn} of Y is also a geometrically independent set. For every1 ≤ i ≤ n letUi = {y ∈ X : φ(y) ∩B(ai, δ) 6= ∅}.Since φ is LSC we have that each Ui is open in X . Set Wx =⋂ni=1 Ui. Takey ∈Wx. Observe that φ(y) ∩B(ai, δ) 6= ∅ for every 1 ≤ i ≤ n. By the choice of δwe get that φ(y) contains at least n geometrically independent vectors and hencedimφ(y) ≥ dimφ(x) = n− 1. Consequently, each Pn is open. Furthermore, eachPn is closed since the complement of each Pn is open as a union of open sets.That completes the proof. Characterization of normal and countably paracompact spaces 49Remark 1. Observe that, in the proof of Lemma 4, we show that if φ : X → E(B)is LSC then for every x ∈ X there is a neighborhood U of x such that dimφ(y) ≥dimφ(x) for every y ∈ U .Lemma 5. Let X be a topological space and (B, ρ) be a Banach space. Letφ : X → E(B) be LSC such that dimφ(x) = dimφ(y) for every x, y ∈ X . Thenfor every x ∈ X there are a neighborhood Wx of x, a µx > 0 and a LSC mapψx : Wx → 2B such that each ψx(y) is convex and compact and B(ψx(y), µx) ∩φ(y) ⊂ [φ(y)]◦ for every y ∈Wx.Proof: Let x ∈ X . If dimφ(x) = 0 then the lemma is trivially satisfied bysetting Wx = X , ψx = φ and taking µx to be any positive number. So, we mayassume that n− 1 = dimφ(x) > 0.We find a geometrically independent set {a1, . . . , an} ⊂ [φ(x)]◦ and setF = 〈{a1, . . . , an}〉.Pick p ∈ F ◦. By Lemma 2, we can find a δ such that if ‖bi − ai‖ < δ then{b1, . . . , bn} is also a geometrically independent set. Letδ∗ = ρ(p, ∂F ),µx = min{13δ∗, δ}, andUi = {y ∈ X : φ(y) ∩B(ai, µx) 6= ∅}.Notice that every Ui is open in X since φ is LSC. SetWx =n⋂i=1Ui.Further, we are going to construct a LSC map ψx : Wx → 2B such that eachimage is compact and convex andψx(x) = {p} andB(ψx(y), µx) ∩ φ(y) ⊂ [φ(y)]◦ for every y ∈ Wx.Let y ∈ Wx be arbitrary. Take {b1, . . . , bn} such that bi ∈ φ(y) and ρ(ai, bi) < µxfor every 1 ≤ i ≤ n. Next, observe that {b1, . . . , bn} is a geometrically independentset. SetG = 〈{b1, . . . , bn}〉 and Gy = B(p, µx) ∩ φ(y).Now, by Lemma 3, we have thatρ(p,G) < µx ≤13δ∗.In addition, we have the following claim.50 S.BarovClaim 1. ρ(p, ∂G) ≥ 23δ∗.Proof: Let z ∈ ∂G. Then z =∑ni=1 tibi, where the right-hand side is a convexcombination of the bi’s with at least one coefficient zero. Consider z∗ =∑ni=1 tiai.Then we have that z∗ ∈ ∂F since at least one of the ti’s is zero and {a1, . . . , an}is a geometrically independent set. Moreover, ρ(z, z∗) = ‖∑ni=1 ti(bi − ai)‖ ≤∑ni=1 ti‖bi − ai‖ <∑ni=1 tiµx = µx. Consequently, ρ(p, z) ≥ ρ(p, z∗)− ρ(z, z∗) >δ∗ − µx ≥23δ∗. Furthermore, we prove the following claim.Claim 2. B(Gy , µx) ∩ φ(y) ⊂ G◦ ⊂ [φ(y)]◦.Proof: Obviously, we have that dimφ(y) = n − 1 and dimG = n − 1. Con-sequently, aff G = aff φ(y) and G◦ ⊂ [φ(y)]◦ since both G and φ(y) are convexand finite-dimensional. Next, having in mind that ρ(p,G) < µx and B(Gy , µx) ⊂B(p, 2µx) ⊂ B(p,23δ∗) we have that∅ 6= Gy ∩G ⊂ Gy ⊂ B(Gy , µx) ∩ aff φ(y) ⊂⊂ B(Gy , µx) ∩ aff G ⊂ B(p,23δ∗) ∩ aff G.Further, applying Claim 1, we obtain thatB(p,23δ∗) ∩ aff G ⊂ G◦ ⊂ [φ(y)]◦.Hence the claim results. Now, define ψx : Wx → 2B, as follows: ψx(x) = {p} and ψx(y) = Gy fory ∈ Wx \ {x}. Let us verify that ψx is LSC. Indeed, let U be open in B and letO = {z ∈ Wx:U ∩ ψx(z) 6= ∅}. We need to show that O is open. Observe that,by the definition of ψx, we have thatO ={z ∈ Wx : U ∩ (B(p, µx) ∩ φ(z)) 6= ∅}={z ∈ Wx:U ∩B(p, µx) ∩ φ(z) 6= ∅}.Now, O is open since φ is LSC and U ∩B(p, µx) is open in B.All other required properties of ψx are obvious. We are done. Remark 2. Considering Lemma 5, let us suppose that (B, ρ) is separable. Then,without loss of generality, we can assume that the set Q = {Wx:x ∈ X} iscountable. Indeed, if dimφ(x) = 0 then we can take Q = {X}. If n − 1 =dimφ(x) > 0 then we consider a countable base O for B. Then, observe that,for every 1 ≤ i ≤ n, B(ai, µx) can be replaced by Ox,i ∈ O such that ai ∈Ox,i ⊂ B(ai, µx). Further, we redefine Ui’s, i.e. Ui = {y ∈ X :φ(y) ∩ Ox,i 6= ∅}.Thus, Wx is completely determined by (Ox,1, . . . , Ox,n). Since the set of all finitesubcollections of O is countable we may assume that Q is countable as well.Characterization of normal and countably paracompact spaces 513. Main resultsFirst, we are going to prove Theorem 1.Proof of Theorem 1: (i)⇒(iii). For every x ∈ X we can find a neighborhoodWx of x, a positive number µx and a LSC map ψx : Wx → 2B as in Lemma 5.Next, by Remark 2, without loss of generality, we can assume that the set Q ={Wx:x ∈ X} is countable. Further, for each W ∈ Q we fix an xW ∈W such thatW =WxW . Then, we can arrange the countable set {(W,µxW , ψxW ):W ∈ Q} ina sequence {(Wi, µi, ψi)}∞i=1. Since X is normal and countably paracompact wecan find an open locally finite refinement (Oi) of (Wi) such that X =⋃∞n=1Oiand Oi ⊂ Wi for each i ∈ N. Moreover, by Michael [7, Theorem 3.1′′] (or, by [7,Theorem 3.1′]), we find a continuous selection fOifor ψi↾Oi. Further, for x ∈ Xwe defineAx = {k ∈ N:x ∈ Ok} and ε∗x = min{µk: k ∈ Ax}.Now, we define a function θ : X → 2B as follows:θ(x) = 〈{fOk(x) : k ∈ Ax}〉 for x ∈ X.Let us verify that θ is a LSC function. Let x ∈ X . Consider B(p, δ) with δ > 0and p ∈ θ(x). For every k ∈ Ax we find a neighborhood Uk of x in Ok such thatfOk(y) ∈ B(fOk(x), δ) for every y ∈ Uk. Set U =⋂k∈AxUk. Clearly, U is aneighborhood of x in X . Observe that, for each y ∈ U and k ∈ Ax we have thatB(fOk(x), δ) ∩ θ(y) 6= ∅. Now, by Lemma 3 and by the definition of θ, we havethat B(p, δ) ∩ θ(y) 6= ∅ for every y ∈ U . Hence θ is LSC.Further, obviously, each image θ(x) is compact and is a subset of [φ(x)]◦.Moreover, for each x ∈ X , we have thatB(θ(x), ε∗x) ∩ φ(x) ⊂ [φ(x)]◦.Now, again by [7, Theorem 3.1′′], we can find a continuous selection f forθ. Let x ∈ X be arbitrary. Find a neighborhood Vx of x such that the set{k:Ok ∩ Vx 6= ∅} is finite. SetCx = {k : Vx ∩Ok 6= ∅} and εx = min{µk : k ∈ Cx}.Let us prove that Vx, εx and f are as required. Indeed, pick an y ∈ Vx. ThenAy ⊂ Cx and therefore ε∗y ≥ εx. Thus,B(θ(y), εx) ∩ φ(y) ⊂ B(θ(y), ε∗y) ∩ φ(y) ⊂ [φ(y)]◦.Now, since f(y) ∈ θ(y), we get that B(f(y), εx) ∩ φ(y) ⊂ [φ(y)]◦. Hence theimplication (i)⇒(ii) results.52 S.BarovThe implication (iii)⇒(ii) is obvious; we need to show that (ii)⇒(i). To showthat X is normal we use a very standard idea. Let F1, F2 be closed in X withempty intersection. Define φ : X → E([0, 1]) as follows: φ(x) = [0, 13 ] for x ∈ F1,φ(x) = [23 , 1] for x ∈ F2 and φ(x) = [0, 1] for x ∈ X \ (F1∪F2). Then, clearly, φ isLSC. By the hypotheses, we can find a continuous selection f : X → [0, 1] for φ.Then the open sets f−1([0, 12 )) and f−1((12 , 1]), that have no common points,contain F1 and F2 respectively. The fact, that X is countably paracompact,follows from the proof of [2, Theorem 4] ((2)⇒(1)). That completes the proof ofthe theorem. Now, we are in a position to state and prove the following variation of Theo-rem 1.Theorem 6. For a T1-space X the following are equivalent.(i) X is normal and countably paracompact.(ii) For every separable Banach space B and for every LSC map φ : X → E(B)satisfying (*) there exists a continuous selection f for φ such that forevery x ∈ X there exist a neighborhood Vx of x and an εx > 0 withB(f(y), εx) ∩ φ(y) ⊂ [φ(y)]◦ for every y ∈ Vx.Proof: It follows directly from Lemma 4 and Theorem 1. Acknowledgment. The author would like to thank the referee for their manyvaluable remarks.References[1] Alexandroff P., Pasynkov B.A., Introduction to Dimension Theory (Russian), Nauka, Mos-cow, 1973.[2] Barov S., Dijkstra J.J., On boundary avoiding selections and some extension theorems,Pacific J. Math. 203 (2002), no. 1, 79–87.[3] Barov S., Dijkstra J.J., On closed sets with convex projections under narrow sets of direc-tions, Trans. Amer. Math. Soc., to appear.[4] Dowker C.H., On countably paracompact spaces, Canad. J. Math. 3 (1951), 219–224.[5] Engelking R.,General Topology, translated from Polish by the author, 2nd ed., Sigma Seriesin Pure Mathematics 6, Heldermann Verlag, Berlin, 1989.[6] Katětov M., On real valued functions in topological spaces, Fund. Math. 38 (1951), 85–91.[7] Michael E., Continuous selections I, Ann. of Math. (2) 63 (1956), 361–382.[8] van Mill J., The infinite dimensional topology of function spaces, North-Holland Mathe-matical Library, 64, Elsevier Science B.V., Amsterdam, 2001.Institute of Mathematics, Bulgarian Academy of Sciences, 8 Acad. G. BonchevStr., 1113 Sofia, BulgariaE-mail : stoyu@yahoo.com(Received May 28, 2007, revised December 12, 2007)

On a characterization of normal and countably paracompact spaces via set-avoiding selections

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On a characterization of normal and countably paracompact spaces via set-avoiding selections

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