Remarks on the finite derived set property

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 6, No. 1, 2005 pp. 101-106 Remarks on the finite derived set property Angelo Bella Abstract. The finite derived set property asserts that any infinite subset of a space has an infinite subset with only finitely many accumulation points. Among other things, we study this property in the case of a function space with the topology of pointwise convergence. 2000 AMS Classification: 54A25, 54A35, 54D55. Keywords: Accumulation points, Urysohn spaces, product, function spaces. Following [8], we say that a topological space Xhas the finite derived set (briefly FDS) property if every infinite subset of Xcontains an infinite subset with only finitely many accumulation points. The class of spaces with the FDS property obviously contains the class of sequentially compact Hausdorff spaces. As sequential compactness, the validity of the FDS property involves in some cases the cardinal characteristic of the continuum. An extensive investigation on the FDS property was initiated in [1]. In this note we will collect a few more results. In particular, we will establish some sufficient conditions for a function space to have the FDS property. All undefined notions can be found in [6]. A space Xis said to be a SC space if every convergent sequence together with the limit point is a closed subset of X. In a SC space a convergent sequence must have a unique accumulation point. Obviously, every SC space satisfies the separation axiom T1. As usual the formula A⊆∗Bmeans that A\Bis finite. A set Ais a pseudointersection of a collection Sprovided that A⊆∗Sfor every S∈ S. The tower number tis the smallest cardinality of a collection of infinite subsets of ωwhich is well-ordered by ⊇∗and has no infinite pseudointersection. Two sets Aand Bare almost disjoint if A∩Bis finite. A maximal family of infinite pairwise almost disjoint subsets of ωis briefly called a MAD family. A 102 A. Bella collection Θ of MAD families is said to be splitting if for any infinite subset A of ωthere exists some B ∈ Θ and B1, B2∈ B such that |A∩B1|=|A∩B2|=ω. hdenotes the smallest cardinality of a splitting collection Θ of MAD families. sdenotes the smallest cardinality of a splitting collection Θ of MAD families each of which consists of two sets. We have ω1≤t≤h≤s≤c. Theorem 1. Every SC space of weight less than shas the FDS property. Proof. Let Xbe a SC space of weight κ < sand let Abe a countable infinite subset of X. If Acontains an infinite subset without accumulation points in Xthen we are done. On the contrary, the definition of sand an argument analogous to the proof of Theorem 6.1 in [5] sufficie to construct a convergent sequence S⊆A. Since Xis a SC space, the set Shas only one accumulation point in Xand again we are done.  As every Hausdorff space of net-weight κhas a weaker Hausdorff topology of weight κand the FDS property is maintained by passing to a stronger topology, we have: Theorem 2. Every Hausdorff space of net-weight less than shas the FDS property. Corollary 1. Every Hausdorff space of cardinality less than shas the FDS property. Lemma 1. Let Xbe a space with the FDS property and Aan infinite subset of X. If Acontains no infinite subset without accumulation points then every infinite subset of Acontains a convergent subsequence. Proof. Let Abe an infinite subset of Xcontaining no infinite subset without accumulation points and let A′be any infinite subset of A. Since Xhas the FDS property, there exists a countable infinite set B⊆A′whose set of accumulation points is {x0,...,xn}. If Bis a sequence converging to x0, then we are done. If not, there exists an open neighbourhood U0of x0such that the set B1=B\U0 is infinite. If B1is a sequence converging to x1, then we are done. If not we may choose an open neighbourhood U1of x1such that the set B2=B1\U1is infinite. Continuing this process, it is clear that for some k≤nthe corresponding set Bkmust converge to xk. The previous Lemma immediately implies that any countably compact space with the FDS property is sequentially compact. Observe however that the same conclusion is no longer true for pseudocompact Tychonoff spaces, even in the class of spaces of countable tightness. A possible counterexample is the space constructed assuming [CH] in [4], Corollary 2. It is easy to realize that sis the smallest cardinal κsuch that any product of κnon-trivial T1spaces does not have the FDS property. More accurate estimate is in the following: Remarks on the finite derived set property 103 Theorem 3. his the smallest cardinal κsuch that there is a family of κSC spaces with the FDS property whose product does not have the FDS property. Proof. Let κbe the cardinal in the statement of the theorem. As there exists a family of hmany compact sequentially compact Hausdorff spaces whose product is not sequentially compact [7] and any compact Hausdorff space with the FDS property is sequentially compact, we immediately see that h≤κ. For the converse inequality, let λ < hand let {Xα:α < λ}be a family of SC spaces with the FDS property. Put X=Q{Xα:α < λ}and denote by πα:X→Xα the usual projection mapping. Fix a countable infinite set A⊆X. If there exists some αsuch that the set πα[A] contains an infinite subset Bwithout accumulation points in Xα, then we are done by considering the set A∩π−1 α(B). In the other case, with the help of Lemma 1, we have that for each αevery infinite subset of πα[A] contains a convergent subsequence. Now, by arguing as in [7], we may construct a convergent sequence S⊆A. Since any product of SC spaces is still a SC space, the set Shas only one accumulation point and we are done.  Remark 1. Corollary 1 provides a consistent negative answer to question 1.9 in [1]. For this, it is enough to consider a model where ω1=p<s. The next assertion improves Theorem 1.6 in [1]. It should also be compared with Corollary 1. Theorem 4. Any Urysohn space of cardinality less than chas the FDS property. Proof. Let Xbe a Urysohn space without the FDS property and fix an infinite set Asuch that every infinite subset of Ahas infinitely many accumulation points. Let A∅=A. For any s∈<ω2, we may select an infinite subset Asof Aaccording to the following rule: if t∈n2 then At⌢0and At⌢1are infinite subsets of Atsuch that At⌢0∩At⌢1=∅. For any f∈ω2 we can fix an infinite set Bf⊆Asuch that |Bf\Af↾n|< ω for each n < ω. For any f∈ω2 let xfbe an accumulation point of the set Bf. It is easy to check that the map f→xf is injective and consequently |X| ≥ c. The previous construction with some minor modifications provides a better result for Lindel¨of spaces. Theorem 5. A Lindel¨of Urysohn space of cardinality less than 2thas the FDS property. Proof. Let Xbe a Lindel¨of Urysohn space and assume that Xdoes not have the FDS property. So, we may fix a countable infinite set A⊆Xsuch that every infinite subset of Ahas infinitely many accumulation points. For any α∈tand any f∈α2 we define an infinite set Af⊆Ain such a way that: (1) if β < α and f∈α2 then Af⊆∗Af↾β; (2) if f, g ∈α2 and f6=gthen Af∩Agis finite. 104 A. Bella Put A∅=Aand assume to have defined everything for each β < α. If αis a limit ordinal and f∈α2, then take as Afany infinite pseudointersection of the family {Af↾β:β < α}. If α=γ+ 1 denote by f′and f′′ the only two elements in α2 such that f′↾γ=f′′ ↾γ=gfor some g∈γ2 and define Af′ and Af′′ by selecting two infinite subsets of Aghaving disjoint closures. By the Lindel¨ofness of X, for any f∈t2 we may pick a point xf∈T{A′ f↾α:α∈t} (here A′is the derived set of A). As the mapping f→xfis injective, we see that |X| ≥ 2t. The next result is an attempt to weaken the separation axiom in the previous theorem. Its proof mimics Theorem 2.5 in [2]. Theorem 6. Any Lindel¨of SC space of cardinality not exceeding thas the FDS property. Proof. Let Xbe a Lindel¨of SC space and assume that X={xα:α∈t}. Let A be a countable infinite subset of X. If Acontains some non-trivial convergent sequence, then thanks to the SC property we obtain a subset of Awith only one accumulation point. In the opposite case, for any α∈twe may select an open set Uαand an infinite set Aα⊆Ain such a way that: (1) xα∈Uα; (2) Aα∩Uα=∅; (3) Aα⊆∗Aβwhenever β≤α. At the first step, since Adoes not converge to x0there exists an open set U0such that A\U0is infinite and we may put A0=A\U0. At step αlet B be an infinite pseudointersection of the family {Aβ:β < α}and select an open set Uαsuch that xα∈Uαand Aα=B\Uαis infinite. Since Xis a Lindel¨of space, the family {Uα:α∈t}has a countable subcover V. Since tis a regular cardinal, we have V ⊆ {Uα:α≤γ}for some γ∈t. To finish, observe that the set Aγhas a finite intersection with each member of Vand therefore it does not have any accumulation point in X. Corollary 2. Every countable SC space has the FDS property. Recall that a space Xis said to be ω-monolithic if any separable subspace of Xhas a countable net-work. Lemma 2. Every Hausdorff ω-monolithic space has the FDS property. Proof. Let Xbe a Hausdorff ω-monolithic space and fix a countably infinite set A⊆X. Assume that Ais not closed and discrete and let xbe an accumulation point of A. Since the subspace Y=Ahas a countable net-work, there exists a decreasing family {Vn:n∈ω}of open neighbourhoods of xin Ysuch that T{Vn:n∈ω}={x}. For any nselect a point an∈Vn∩(A\ {x}) and let B be the set so obtained. All accumulation points of Bmust lye in Vnfor each n and therefore the set Bhas at most only xhas accumulation point.  Remarks on the finite derived set property 105 Now, we look at the FDS property in function spaces in the topology of pointwise convergence (see [3]). For this reason, in the sequel all spaces are assumed to be Tychonoff. To begin, observe that a space of the form Cp(X) may fail to have the FDS property: it suffices to take as Xany discrete space of cardinality at least s. A space Xis said to be ω-stable if all continuous images of Xthat can be injectively mapped into a second countable space have a countable net-work. The class of ω-stable spaces contains all pseudocompact ([3], 2.6.2c), all Lindel¨of Σ ([3], 2.6.21) and all Lindel¨of P spaces ([3], 2.6.28). Thus, we have: Corollary 3. If Xis either a pseudocompact of a Lindel¨of Σor a Lindel¨of P space then Cp(X)has the FDS property. Proof. The result follows from Lemma 2 and the fact that a space Xis ω-stable if and only if Cp(X) is ω-monolithic ([3], 2.6.8).  Corollary 4. Every normed linear space in the weak topology has the FDS property. Proof. Let Xbe a normed linear space and U∗be the unit ball in the dual space X∗. The well known theorem of Alaoglu says that U∗is compact in the weak∗topology on X∗. The result follows by observing that the space Xin the weak topology is just a subspace of Cp(U∗).  Theorem 7. The space Cp(X)has the FDS property in the following cases: (1) Xhas density less than s; (2) Xhas a dense set which is the union of less than hpseudocompact (or Lindel¨of Σ, or Lindel¨of P) subspaces. Proof. Observe that if Yis dense in Xthen there is an injective continuous mapping from Cp(X) into Cp(Y). Therefore, if Cp(Y) has the FDS property then the same happens to Cp(X). Item 1 follows from Theorem 1. If Y=S{Yα:α < κ}then Cp(Y) is homeomorphic to a subspace of the space Q{Cp(Yα) : α < κ}. Taking into account Theorem 3, item 2 follows from Corollary 3.  Corollary 5. If ξis an ordinal then Cp(ξ)has the FDS property. Looking again at Corollary 3, one may wonder whether the second iteration of a function space, namely Cp(Cp(X)), on a compact space Xhas yet the FDS property. This is not the case in general because the FDS property of Cp(Cp(X)) implies that the compact space Xmust be sequentially compact. We do not know if the converse always holds, although it happens for sure in the class of compact ω-monolithic spaces. However, the FDS property of Cp(Cp(X)) is far to imply either ω-monolithicity or ω-stability of the base space X. For instance, if Xis a subspace of the Sorgenfrey line of cardinality ω1then Xis neither ω-monolithic nor ω-stable, but by assuming ω1<sCp(Cp(X)) has the FDS property. 106 A. Bella Of course, Corollary 3 and Theorem 7 leave open the general problem to characterize the spaces Xfor which Cp(X) has the FDS property. Some specific questions are: 1. Is it true that any Hausdorff space of cardinality less than chas the FDS property? 2. Find a compact sequentially compact space Xsuch that Cp(Cp(X)) does not have the FDS property. References [1] O. T. Alas, M. Tkachenko, V. Tkachuk and R. Wilson The FDS-property and the spaces in which compact sets are closed, Sci. Math. Japan, to appear. [2] O. T. Alas and R. G. Wilson, When a compact space is sequentially compact?, preprint. [3] A. V. Arhangel′ski˘ı, “Topological Function Spaces”, Kluwer Academic Publishers, Dordrecht (1992) [4] A. Bella and O. Pavlov, Embeddings into pseudocompact spaces of countable tightness, Topology Appl. 138 (2004), 161–166. [5] E. K. van Douwen, “The Integers and Topology”, Handbook of Set-theoretic Topology (K.Kunen and J.E. Vaughan Editors), Elsevier Science Publishers B.V., Amsterdam (1984), 111 –167. [6] R. Engelking, “General Topology” Heldermann-Verlag, Berlin (1989). [7] P. Simon, Product of sequentially compact spaces, Rend. Ist. Mat. Univ. Trieste 25 (1994), 447–450. [8] D. Shakmatov, M. Tkachenko and R. Wilson, Transversal and T1-independent topologies, Houston J. Math. 30 (2004), 421–433. Received December 2004 Accepted January 2005 A. Bella ([email protected]ict.it) Dipartimento di Matematica, Citt´a Universitaria, Viale A.Doria 6, 95125, Catania, Italy.