Received August 13, 1998; revised May 6, 1999

Transcription

1 Scientiae Mathematicae Vol. 2, No. 3(1999), THE STRONG PARACOMPACTNESS OF σ-products KEIKO CHIBA Dedicated to the memory of Professor Amer Be slagić Received August 13, 1998; revised May 6, 1999 Abstract. In this paper we shall prove that the following questions are negative for strong paracompactness, orthocompactness, star-lindelöfness and normality : Does a σ-product satisfiy a topological property P if every finite subproduct of the σ-product satisfies P? 1. Introduction. We later define σ-products which were introduced by H. H. Corson [6]. Throughout this paper we assume that each space is a Hausdorff space having at least two points. A. P. Kombarov [10] proved that, if every finite subproduct of a σ-product is paracompact (resp. Lindelöf), then the σ-product is paracompact (rep. Lindelöf). The following question arises naturally. Question. Suppose every finite subproduct of a σ-product satisfies a topological property P. Does the σ-product satisfy P? Here Π{X α α F } is said to be a finite subproduct of σ for a finite subset F of A where S = {X α α A} and σ is a σ-product of S. Concerning many covering properties, the above questions are affirmative. In fact, it is known for the following properties : metacompactness, hereditarily metacompactness, weak θ-refinability, weak δθ-refinability, hereditarily weak θ(δθ)-refinability; (under the assumption of σ is subnormal) subparacompactness; (under the assumption of σ is normal)submetacompactness, submeta-lindelöfness, B-property, D-property, shrinkability, screenability, collectionwise normality and ultraparacompactness [metacompactness and subparacompactness are due to Teng [14] and others due to Chiba [2,3,4,5]]. In this paper we shall prove that the above question is no for strong paracompactness, orthocompactness, star-lindelöfness and normality. 2. Preliminaries. Definition 1. σ = σ(s) isaσ-product of S means there is a point x X =Π{X α α A} (called the base point of σ) such that σ is the subspace of X consisting of {x X Q(x) is finite}. Here x α denotes the α-coordinate of x and Q(x) ={α α A, x α x α}. Let A n = {a A : a = n} each n ω and put A <ω = {A n n ω}. Here a denotes the cardinal number of a. For each a A <ω, let Y a =Π{X α α a} Π{{x α } α A a}. For each α A, let p α : σ X α be the projection map and we denote Y {α} by Y α and for each α, β A, Y {α,β} by Y αβ Mathematics Subject Classification. Primary 54D20,54B10;Secondary 54B05,54D18. Key words and phrases. σ-product, strongly paracompact, orthocompact, star-lindelöf, normal.

2 286 KEIKO CHIBA Let X be a space and τ an infinite cardinal such that A = τ. In case X α = X for each α A, let us denote σ(s) byσ(x, τ). The following facts concerning σ-products are known. Fact 1. Let σ = σ(s) and σ n = {x σ : Q(x) n} for each n ω. Then σ n is closed in σ. Fact 2. If S = {X α α A} is a family of metacompact spaces such that each finite subproduct of S is metacompact, then σ = σ(s) is metacompact. A space X is strongly paracompact if for every open over G of X, there is a star finite open cover H of X such that H G. Here H Gmeans H is a refinement of G. A cover H is called star finite (resp. star countable if each H H, {H H H,H H φ} is finite (resp. countable). Concerning strong paracompactness, the following facts are well known. Fact 3. Let X be a strongly paracompact space and Y a compact space. Then X Y is strongly paracompact. Fact 4. A space X is strongly paracompact iff for every open cover G of X, there is a star countable open cover H of X such that H G. By Fact 4, every Lindelöf space is strongly paracompact. A space X is called orthocompact if for every open over G of X, there is an interior preserving open cover H of X such that H G. Here H Gmeans H is a refinement of G. A cover H is called interior preserving if each H H, H is an open set in X. Concerning orthocompactness, the following facts are well known. Fact 5. The following statement for a space X is equivalent. (1) X is orthocompact. (2) For every open cover G of X, there exists an interior preserving open cover H of X such that H is an one to one refinement of G. Here H is an one to one refinement of G means if G = {G λ λ<τ}, then H = {H λ λ<τ} where H λ G λ for each λ. Fact 6. The product of finitely many locally compact GO-space is orthocompact iff it is normal. A space X is called star-lindelöf ([7, 9]) if for every open cover U, there is a countable subset A X such that st(a, U) =X. Here st(a, U) = {U U U,A U }. Fact 7. Every separable space is star-lindelöf. Definition 2. Let λ be an infinite ordinal and denote by T (λ) the space {α α is an ordinal, 0 α λ} with the topology as follows: {α} is open for each α<λ, {(α, λ] α <λ} is a neighborhood base of λ. Here (α, λ] ={β β is an ordinal, α<β λ}. Refer to [8] for undefined definitions of covering properties.

3 THE STRONG PARACOMPACTNESS OF σ-products Strong paracompactness. In this section, we shall prove the following. Example 1. Let S = {X n n ω} where X n =[0, 1] for each n<ωand X ω = T (ω ω ). Here [0, 1] = {x x R, 0 x 1} is the closed unit interval with the usual subspace topology of the real line R. Let(x n ) n ω be defined by x n =0for each n<ωand x ω = ω ω and (x n ) n ω be the base point of σ(s). Then (i) every finite subproduct of σ(s) is strongly paracompact, (ii) σ(s) is not strongly paracompact. (i) is obvious by Fact 3. Proof of (ii). Let S 1 = {X n n<ω}. Then σ(s) is homeomorphic to σ(s 1 ) T (ω ω ) with the product topology where x =(x n ) n<ω is the base point of σ(s 1 ). Therefore, it is sufficient to prove (ii) σ(s 1 ) T (ω ω ) is not strongly paracompact. To prove (ii), we first prove the following. Lemma 1. 1 R + can be embedded as a closed subspace of σ(s 1 ). Here R + = {x x R,x 0}. Proof. Let σ = σ(s 1 ) and define X 0 = {x =(x i ) i<ω σ x 0 [0, 1) and x i = 0 for each i>0}, X n = {x =(x i ) i<ω σ x i = 1 for each i n 1,x n [0, 1) and x i = 0 for each i>n} for each n>0 and X = n<ω X n. Here [0, 1) = {t t R, 0 t<1}. Then (1) X is a closed subspace of σ. (2) R + is homeomorphic to X. P roofof(1). Let x =(x i ) i<ω σ X and n be the first element of ω such that x n 1. Then, since x/ X, there is an m>nsuch that x m 0. Then U = p 1 n ([0, 1)) p 1 m ((0, 1]) is a neighborhood of x such that U X =. P roofof(2). For each x R +,n x ω and a x [0, 1) are uniquely defined such that x = n x + a x. Let us define f : R + X by f(x) =(x n ) n<ω as follows: 1. if n x = 0, then { a x if n = n x, x n = 0 if n>n x. 2. if n x > 0, then 1 if 0 n<n x, x n = a x if n = n x, 0 if n>n x. Let us define g : X R + by g(x) =n + x n if x =(x i ) i<ω X n. Then it is obvious that fg = identity and gf = identity. And f and g are continuous. Therefore f is a homeomorphism. We only give a proof of the following. (3) f is a continuous function from R + to σ. P roofof(3). Let x = n ω such that n>0. Then f(x) =(x i ) i<ω,x i =1ifi n 1 and x i =0ifi n. Let U be an arbitrary neighborhood of f(x) inσ. Then there are an m>nand an ɛ>0 such that ɛ<1 and V = i=0 n 1 p 1 i ((1 ɛ, 1]) m i=n p 1 i ([0,ɛ)) U. 1 It is known that the realline R can be embedded as a closed subspace of the σ-product of countably many copies of the closed unit interval. But the author can not show any literature that this fact and the proof of it are written in it explicitely. Therefore she gives a simple proof of this lemma here.

4 288 KEIKO CHIBA Then (n ɛ, n + ɛ) is a neighborhood of x in R + such that f((n ɛ, n + ɛ)) V. Therefore f is continuous at x. In other cases, the proof is similar. (ii) follows from Example 2 below by this lemma. Example 2. R + T (ω ω ) is not strongly paracompact. Proof. Let U n = (( 2/3) + n, (2/3) + n) be an open interval in R + with end points ( 2/3) + n, (2/3) + n for each n>0 and U 0 =[0, 2/3). Then {U n n ω} is a countable open cover of R +.Forn ω, let us put G n = U n [ω n,ω ω ]. For each n ω and α<ω n, let us put G n,α = U n {α}. Put G = {G n n ω} {G n,α n ω and α<ω n }. Then G is an open cover of R + T (ω ω ). ( ) Let H be an open cover of R + T (ω ω ) such that H G. Then H is not star countable. Proof of ( ). Let us choose H 0 Hsuch that 0,ω ω H 0. Then H 0 G 0. Let us put Λ={α α <ω ω, (R + {α}) H 0 }. Then Λ [ω m,ω ω ] for some m. Let us put K α = {x R + x, α H 0 } for each α Λ. Then K α is open in R +,K α and K α R + for each α ω m. Since R + is connected, clk α K α. Let us choose a point c α clk α K α and put y α = c α,α for each α such that ω m α<ω m+1. Let us choose a set H α Hsuch that y α H α for each α. Then (1) H α H 0. Let us denote d =2/3+m and put z α = d, α. Then it is easy to see (2) z α / {G n n ω} for each α such that ω m α<ω m+1. Choose H α Hsuch that z α H α. Then (3) H α U m+1 {α} for each α such that ω m α<ω m+1. Therefore we have (4) H α H β if α β and ω m α, β < ω m+1. Now y α,z α [c α,d] {α} or y α,z α [d, c α ] {α}. Here [c α,d]={x R c α x d}. Since [c α,d] {α} is compact and connectd, there is a finite subfamily {H α,i i =1, 2,...,n α } of H such that H α H α,1,h α,1 H α,2,...,and H α,nα H α. Then there is an n such that Γ n = {α ω m α < ω m+1,n α = n} is uncountable. If there is an uncountable subset of pairwise distinct elements of {H α α Γ n }, then H is not star countable by (1). In case any pairwise distinct elements of {H α α Γ n } is countable, let {H αi i =1, 2,...} be pairwise distinct subfamily of {H α α Γ n }. If there is an uncountable subset of pairwise distinct elements of {H α,1 α Γ n }, then there is an i such that {α Γ n H αi H α,1 } is uncountable. Therefore H is not star countable. In case any pairwise distinct elements of {H α,1 α Γ n } is countable, let {H αi,1 i =1, 2,...} be pairwise distinct subfamily of {H α,1 α Γ n }. If there is an uncountable subset of pairwise distinct elements of {H α,2 α Γ n }, then there is a j such that {β Γ n H αj,1 H β,2 } is uncountable,..., proceeding this process, finally, if any pairwise distinct elements of {H α,n α Γ n } is countable, there is an α Γ n such that {β β Γ n,h α,n H β } is uncountable by (4). Hence H is not star countable. Since strong paracompactness is hereditary with respect to closed subspaces, by Example 2 we have Example 3. R T (ω ω ) is not strongly paracompact.

5 THE STRONG PARACOMPACTNESS OF σ-products 289 It has been already proved that the strong paracompactness is not preserved for product spaces of two spaces (even in the class of metrizable spaces). In fact, the following examples are known. (1) Let S be the Sorgenfrey line. Then S is Lindelöf and thus S is strongly paracompact. But S S is not strongly paracompact(cf. [8]). (2) Let (0, 1) = {x x R, 0 <x<1} be the open unit interval and B(ℵ 1 ) the Baire space. Then (0, 1) B(ℵ 1 ) is not strongly paracompact(cf. Nagata [11], or [8, p.328, 5.3.F(b)]). Example 3 shows that there exists a metrizable space Y with only one non-isolated point such that R Y is not strongly paracompact. As for the strong paracompactness of X T (λ), we have the following. Theorem 1. Let X be a strongly paracompact space. If L(X) <cfλ, then X T (λ) is strongly paracompact. Here L(X) is the Lindelöf number of X and cfλ is the cofinality of λ. Proof. Let G be an arbitrary open cover of X T (λ). For each x X, there are an open set U x in X and α x <λsuch that x U x and U x (α x,λ] G for some G G. Then U = {U x x X} is an open cover of X. Since X is strongly paracompact, there is a star finite open cover H of X such that H U. Put L(X) =τ. Then there is a subcover H of H such that H τ. For each H H, let us choose an x such that H U x and denote α x by α H. Put α = sup{α H H H }. Then α <λ. For each α α, let us choose a star finite open cover H α of X such that for each H H α,h {α} G for some G G. Let us put L = {H (α,λ] H H } {H {α} H H α,α α }. Then L is a star finite open cover of X T (λ) which is a refinement of G. Remark 1. In the above theorem, we cannot weaken the condition, L(X) <cfλ to L(X) cfλ (cf. Example 1, 2 or 3). Remark 2. By improving the proof of Lemma 1, we can show that R can be embedded as a closed subspace of the σ-product of {[0, 1] n n Z}. Here Z = {0, ±n n =1, 2,...} and [0, 1] n is a copy of [0, 1]. 4. Orthocompactness. Proposition 1. Let X be a space and τ an uncountable cardinal. Suppose there is an open cover U = {U α α<τ} of X sucu that every open refinement V of U is not point finite. Then σ = σ(x, τ) is not orthocompact. Here U α X and x α U 0 for each α<τ where x =(x α ) α<τ is the base point of σ. Proof. Let U = {U α α < τ} be the open cover of X satisfying the conditions in this proposition. Let G 0 = p 1 0 (U 0) and G α = p 1 0 (U α) p 1 α (U 0) for each α 0. Let us put G = {G α α<τ}. Then G is an open family in σ such that G σ 1. To show this, let x =(x α ) α<τ σ 1. If x / G 0, then x 0 / U 0. Since U is a cover of X, x 0 U α for some α 0. Thusx p 1 0 (U α). Since x 0 / U 0,x 0 x 0. Thus x α = x α because x σ 1. Hence x p 1 α (U 0 ) and so x G α. Let us put G = G {σ σ 1 }. Then G is an open cover of σ. Claim. Let H be an open cover of σ such that H is an one to one refinement of G. Then H is not interior preserving.

6 290 KEIKO CHIBA P roofofclaim. We can put H = {H α α<τ} {H } where H α G α with α<τ and H σ σ 1. Let H Y 0 = {H Y 0 H H}and G Y 0 = {G Y 0 G G }. Then H Y 0 is an open cover of Y 0 and an one to one refinement of G. Since G Y 0 = {U {x } U U}where x =(x α ) 1 α<τ, by the assumption of U, H Y 0 is not point finite at y,x for some y X 0. Let {α i i =1, 2,...} be infinite elements of τ such that y,x H αi Y 0 for each i. Here we assume that α i α j if i j and α i 0 for each i. Then i=1 H α i is not open in σ. To prove this, assume the contrary. Since y,x i=1 H α i, there is a basic open set W in σ such that y,x W i=1 H α i. Since W is a basic open set in σ, there are an i 0 and open sets W i in X αi for each i i 0, and there are a finite set Λ τ {α i i =1, 2,...} and open sets W β in X β such that y,x i0 i=1 p 1 α i (W i ) β Λ p 1 β (W β) W. Let γ = α i0+1 and let us choose an element a of X γ such that a / U 0 and define x =(x α ) α<τ by x γ = a and x α = x α if α γ. Then x W and x/ i=1 H α i by x/ G γ. This is a contradiction. Theorem 2. If σ(x, τ) is orthocompact for each τ, then σ(x, τ) is metacompact for each τ. Proof. Assume that σ(x, τ) is not metacompact for some τ. Then, by Fact 2, some finite product X m is not metacompact. Therefore there is an open cover U = {U α α<λ} of X m such that every open refinement V of U is not point finite. By the above proposition, σ(x m,λ) is not orthocompact. Since σ(x m,λ) is homeomorphic to σ(x, λ), σ(x, λ) is not orthocompact. Example 4. Let X = ω 1 with the order topology and σ = σ(x, ω 1 ) with the base point x =(x α) α<ω1 where x α =0for each α<ω 1. Then (i) every finite subproduct of σ is orthocompact, (ii) σ is not orthocompact. Proof of (i). Since ω 1 is locally compact GO-space and ω n 1 is normal, ω n 1 is orthocompact by Fact 6. Proof of (ii). Since ω 1 is not metacompact, (ii) follows from Proposition. 5. Star-Lindelöfness. In this section, we shall prove the following. Theorem 3. Let X be a space and τ an uncountable cardinal such that cf(τ) >ω. Suppose there exists an irreducible 2 open cover U = {U α α<τ} of X. Then σ = σ(x, τ) is not star- Lindelöf. Here x α U 0 for each α<τ where x =(x α) α<τ is the base point of σ. Proof. Let U = {U α α<τ} be the irreducible open cover of X. Let us put G 0 = p 1 0 (U 0) and G α = p 1 0 (U α) p 1 α (U 0) for each α with 0 <α<τ. Put G = {G α α<τ}. Then G is an open family in σ. (1) G σ 1 Y αβ for each α, β 0. P roofof(1). In the proof of Proposition, G σ 1 is shown. Let α, β 0 and x =(x γ ) γ<τ Y αβ. Then x 0 = x 0.Thusx G 0. Let us put A α = Y 0α G for each α with 0 <α<τ. Then (2) A α for each α. P roofof(2). Let s U α β α U β and define y α =(yβ α) β<τ by y0 α = s, yα α = s and yβ α = x β for each β 0,α. Then it is obvious that yα Y 0α. Since y0 α / U 0,y α / G 0. Since 2 A cover U is called irreducible if U r {U α} is not a cover for each α<τ.

7 THE STRONG PARACOMPACTNESS OF σ-products 291 y0 α / U β for each β α, 0,y α / p0 1 (U β) and therefore y α / G β for each β α, 0. Since yα α / U 0,y α / p 1 α (U 0) and hence y α / G α.thusy α / G. (3) G ( 0<α<τ A α ) σ 2. (3) is easily proved. (4) {A α 0 <α<τ} is discrete in σ. P roofof(4). Let x =(x α ) α<τ be an arbitary element of σ. If x G, then G is a neighborhood of x such that ( G) ( α A α )=. Ifx/ Gand x/ α A α, then x/ σ 2 by (3). Put V = σ σ 2. Then V is a neighborhood of x such that V ( α A α )=. In case x/ G and x A α for some α. Then x β = x β for each β 0,α. Since x/ G,x / σ 1 by (1). Therefore Q(x) = 2. Hence x 0 x 0 and x α x α. Let W 0 and W 1 be open neighborhoods of x 0 and x α such that x 0 / W 0 and x α / W 1 and put V = p 1 0 (W 0) p 1 α (W 1). Then V is a neighborhood of x such that V ( β α A β )=. Let us put H 0 = σ 0<α<τ A α and H α = σ (( β α A β ) F α ) where F α = {x Q(x) α} for each α with 0 <α<τ and put H = {H α α<τ}. Then H is an open cover of σ and, by (2), H α for each α<τ because H α A α. (5) F σ with F ω st(f, G ) σ. P roofof(5). 3 Let F σ such that F ω. Then x F Q(x) ω. Hence there is an α<τsuch that F F α. Therefore F H β = whenever β α. Thusst(F, H) σ. To construct an example to show that the question for star-lindelöfness is no, we define the space X below. The space X: We consider N c with the product topology where c =2 ω and N = {1, 2,...}. We may assume that N c = {f : R N function}. Let us put X = {f N c f 1 (1) }. For each α<c, let π α : N c N be the projection to the α-coordinate, and put U α = πα 1(1). Then (1) U = {U α α<c} is an open cover of X, (2) U α β α U β for each α<c, (3) X is separable. (1) is obvious. To prove (2), let us define x α =(x α β ) β<c by x α α = 1 and xα β = 2 for each β α. Then x α U α β α U β. P roofof(3). Since X is an open subset of N c, it follows from the fact that N c is separable (cf. [8, p.77]). Example 5. Let σ = σ(x, c) with the base point x =(x α) α<c where x α : R N be the constant function with its value is equal to 1. Then (i) finite subproduct of σ is star-lindelöf, (ii) σ is not star-lindelöf. Proof of (i). Since X is separable and finite product of separable spaces is separable, finite subproduct of σ is separable and therefore star-lindelöf by Fact 7. Proof of (ii). (ii) follows from Theorem Normality. In this section we shall prove the following. Example 6. There exists a family S of normal spaces such that each finite subproduct of S is normal, but σ = σ(s) is not normal. To show this, we use the following lemma. Lemma 2([1, Theorem 1]). If X is a space such that X Y is normal for any perfectly normal space Y, then X is perfectly normal. 3 This proof is improved by Dr.Yan-Kui Song.

8 292 KEIKO CHIBA Proof of Example 6. Let S = {[0, 1] α : α<ω 1 } where [0, 1] α is a copy of [0, 1] and σ = σ(s ) with the base point x =(x α ) α<ω 1 where x α is an arbitrary element of [0, 1] α for each α. Then σ is not perfect 4 because the base point is not a G δ -set. By Lemma 2, there exists a perfectly normal space Y such that σ(s ) Y is not normal. Let us put S = S {Y }. Then S has the desired properties because σ is homeomorphic to σ Y. Here the base point of σ is x,y where y is an arbitrary element of Y. Acknowledgments. The author would like to thank Prof. H. Ohta for his helpful advices and she is also grateful to the referee for his valuable comments. References 1. K. Chiba, Two remarks on the normality of product spaces, Reports of Faculty of Science, Shizuoka University, 11 (1976), K. Chiba, On σ-products, Math. Japonica, 32, No.3 (1987), K. Chiba, Covering properties in products, Math. Japonica, 34, No.5 (1989), K. Chiba, The submetacompactness of σ-products, Math. Japonica, 36, No.4 (1991), K. Chiba, Covering properties in σ-products, Math. Japonica, 39, No.2 (1994), H. H. Corson, Normality in subsets of product spaces, Amer.J.Math., 81 (1959), E. K. van Douwen, G. M. Reed, A. W. Roscoe and I. J. Tree, Star covering properties, Topology and Appl. 39 (1991), R. Engelking, General Topology, Polish Scientific publishers, Warszawa (1988). 9. S. Ikenaga, A class which contain Lindelöf spaces, separable spaces and countably compact spaces, Memories of Numazu College of Techonology 18 (1983), A. P. Kombarov, On the normality of Σ m-products, Soviet Math. Dokl., 14, No.4 (1973), J. Nagata, On imbedding theorem for non-separable metric spaces, J. Inst. Polyt. Osaka City Univ. 8 (1957), B. M. Scott, Orthocompactness in normality in finite products of locally compact LOTS s, in Set- Theoretic Topology(G.M.Reed, ed.), Academic Press, (1977), B. M. Scott, More about orthocompactness, Topology Proc. 5 (1980), H. Teng, On σ-product spaces I, Math. Japonica, 36, No.3 (1991), Department of Mathematics, Faculty of Science, Shizuoka University, Ohya, Shizuoka, Japan address: smktiba@ipc.shizuoka.ac.jp 4 a space X is called perfect if each closed set is a G δ -set.