Math. Log. Quart. 48 (2002) 4, 555 – 558

Mathematical LogicQuarterly

c© WILEY-VCH Verlag Berlin GmbH 2002

Attainment of Tightness in Boolean Spaces

Juan Carlos Martınez1)

Facultat de Matematiques, Universitat de Barcelona,Gran Via 585, 08007 Barcelona, Spain2)

Abstract. We consider some questions of Donald Monk related to attainment of thetightness function in Boolean spaces. Then we prove in ZFC that attainment of tightness inthe sense of the definition does not imply attainment of tightness in the free sequence sense.Our results give rise to full answers to [2, Problems 41, 42].

Mathematics Subject Classification: 06E99, 03E20, 54G12.

Keywords: Boolean space; tightness; free sequence; π-character.

1 Introduction

Our terminology is standard and compatible with that of [2].

If X is a topological space and x ∈ X, we define the tightness of x in X by

t(x, X) = the least cardinal κ such that if Y ⊆ X and x ∈ Y , thenthere is a subset Z ⊆ Y with |Z| ≤ κ and x ∈ Z .

We define the tightness of the space X by t(X) = sup {t(x, X) : x ∈ X}. Then we saythat t(X) is attained, if there is an x ∈ X such that t(x, X) = t(X). A free sequencein a topological space X is a sequence 〈xξ : ξ < α〉 of elements of X, where α is anordinal such that for every ξ < α, {xη : η < ξ} ∩ {xη : ξ ≤ η < α} = ∅, and then wesay that α is the lenght of the sequence 〈xξ : ξ < α〉.

It was proved by Arhangelskiı that for every compact Hausdorff space X,

t(X) = sup {|α| : there is a free sequence in X of length α}(see [2, Theorem 4.20]). It follows from the proof of Arhangelskiı’s theorem thatfor any compact Hausdorff space X such that t(X) is a limit regular cardinal, wehave that if X has a free sequence of length t(X), then t(X) is attained. However,this fact can not be generalized to singular cardinals, since it is known that for everysingular cardinal κ there is a Boolean space X such that t(X) = κ is not attainedbut there is a free sequence of length κ in X (see [2, Corollary 12.4]). Conversely, itwas proved in [3] that for any specific singular cardinal κ of uncountable cofinality,

1)The preparation of this paper was supported by DGICYT Grant PB98-1231.2)e-mail: martinez@mat.ub.es

c© WILEY-VCH Verlag Berlin GmbH, 13086 Berlin, 2002 0942-5616/02/0411-0555 $17.50+.50/0

556 Juan Carlos Martınez

it is consistent with the axioms of Set Theory that there is a Boolean space X witht(X) = κ attained and such that there is no free sequence of length κ in X. Then,our aim here is to show that for any limit cardinal of uncountable cofinality such aBoolean space can be constructed in ZFC.

It was shown in [2, Theorem 12.2] that if κ is a singular cardinal of countablecofinality, then every Boolean space of tightness κ has a free sequence of length κ.So, in order to construct a Boolean space X as desired, we have to assume thatcf(t(X)) > ω. However, it is erroneously stated in [3, p. 30] that we should alsodemand that t(X) is singular.

The construction we present in this paper is a modification of the one given in[3, Section 7]. However, it is not an improvement, since the Boolean spaces con-structed by forcing in [3, Section 7] satisfy other additional properties.

A topological space X is scattered, if every closed subspace of X has an isolatedpoint. The Boolean spaces we define in this paper are scattered. It is a well-knownfact that the notion of a scattered Boolean space is the dual notion of a superatomicBoolean algebra (see [1]).

2 Results

Suppose that κ is a limit cardinal. Let λ = cf(κ). Consider a strictly increasingsequence 〈κα : α < λ〉 of successor cardinals such that κ = sup {κα : α < λ}.For every α < λ we define the space Zα as follows. The underlying set of Zα is{(α, ξ) : 0 ≤ ξ < κα}. Then, we consider the topology induced by the order of κα: ifξ = 0 or ξ is a successor ordinal of κα, we consider (α, ξ) as an isolated point of Zα;and if ξ is a limit ordinal of κα, a basic neighbourhood of (α, ξ) in Zα is a set ofthe form {(α, ζ) : µ < ζ ≤ ξ}, where µ < ξ. Then, we define the space Xκ as theone-point compactification of the topological sum of the spaces Zα for α < λ. It iseasy to check that Xκ is a scattered Boolean space.

Our purpose is to show the following general result, which answers [2, Problem 41]in ZFC.

T h e o r em 1. Let κ be a limit cardinal such that cf(κ) > ω. Then there is aBoolean space X with t(X) = κ attained such that there is no free sequence of lengthκ in X.

P r o o f . For κ a limit cardinal of uncountable cofinality consider the space Xκ.Let λ = cf(κ). We denote by Yκ the topological sum of the spaces Zα for α < λ.Note that a compact subset of Yκ is a finite union U1 ∪ · · · ∪ Un such that for someα1, . . . , αn < λ, Ui is a compact subset of Zαi for i ∈ {1, . . . , n}. Therefore, thefollowing two conditions hold:

(1) For each α < λ, {(α, ξ) : 0 ≤ ξ < κα} is not contained in a compact subset of Yκ.(2) If 〈αn : n < ω〉 is a strictly increasing sequence of ordinals in λ and, for n < ω,

xn ∈ {αn} × καn , then {xn : n < ω} is not contained in a compact subset of Yκ.

Let x∗ be the point in infinity of Xκ. Note that if α < λ and ξ is a limit ordinalof κα, we have t((α, ξ), Xκ) = cf(ξ) < κ. On the other hand, we can show thatt(x∗, Xκ) = κ. For this, consider a cardinal θ < κ. Let α < λ such that θ < κα.

Attainment of Tightness in Boolean Spaces 557

We infer from (1) that x∗ ∈ {(α, ξ) : ξ < κα}. However, since κα is regular, for everyZ ⊆ {(α, ξ) : ξ < κα} with |Z| ≤ θ, we have that x∗ /∈ Z . Therefore, t(Xκ) = κ isattained.

Suppose that there is a free sequence 〈xξ : ξ < κ〉 in Xκ. Let 〈αn : n < ω〉 be astrictly increasing sequence of ordinals in λ such that for each n < ω,

{xµ : µ < κ} ∩ ({αn} × καn) = ∅.Since λ = cf(κ) is uncountable, there is an ordinal ξ < κ such that for each n < ω,

{xµ : µ < ξ} ∩ ({αn} × καn) = ∅.It follows from (2) that x∗ ∈ {xµ : µ < ξ}. Let γ = sup {αn : n < ω}. Clearly,|{xµ : ξ ≤ µ < κ} ∩ ⋃{{α} × κα : γ ≤ α < λ}| = κ. So, we can find a strictlyincreasing sequence 〈βn : n < ω〉 of ordinals in λ with γ ≤ β0 in such a way thatfor each n < ω, {xµ : ξ ≤ µ < κ} ∩ ({βn} × κβn) = ∅. Thus, by (2), we infer thatx∗ ∈ {xµ : ξ ≤ µ < κ}. So, we have proved that x∗ ∈ {xµ : µ < ξ} ∩ {xµ : ξ ≤ µ < κ},contradicting that 〈xξ : ξ < κ〉 is free in Xκ. �

Theorem 1 implies also a negative answer to [2, Problem 42]. Nevertheless, aresult more general than Theorem 1 can be obtained in this case. Recall that if X isa topological space and x ∈ X, a π-base at x is a family F of nonempty open sets inX such that for every neighbourhood U of x there is a V ∈ F with V ⊆ U , and thenthe π-character of x ∈ X is defined by

πχ(x, X) = the least cardinal κ such that there is a π-base at x of size κ,

and the π-character of the space X is defined by

πχ(X) = sup {πχ(x, X) : x ∈ X}.It was shown by Shapirovskiı that for every infinite Boolean space X,

t(X) = sup {πχ(Y ) : Y is a closed subspace of X}(see [2, Theorem 11.8]). Then the following result answers [2, Problem 42] even inthe case of singular cardinals of countable cofinality.

T h e o r em 2. Let κ be a limit cardinal. Then there is a Boolean space X witht(X) = κ attained such that for every closed subspace Y of X and every y ∈ Y wehave πχ(y, Y ) < κ.

P r o o f . For κ a limit cardinal, consider the space Xκ defined above. We putλ = cf(κ) and denote by x∗ the point in infinity of Xκ. It follows from the proof ofTheorem 1 that t(Xκ) = κ is attained. Let Y be a closed subspace of Xκ. Considery ∈ Y . Assume that y = x∗. Let α < λ such that y ∈ Zα. Since Zα is homeomorphicto κα with the order topology, we have πχ(y, Zα) < κα < κ, and thus πχ(y, Y ) < κ.Now, assume y = x∗. Let

a = {α < λ : Y ∩ ({α} × κα) = ∅}.First, suppose that a is finite. Let β be the largest ordinal in a. Then we haveπχ(y, Y ) ≤ κβ, and so πχ(y, Y ) < κ. Now, suppose that a is infinite. Let 〈αn : n < ω〉be a strictly increasing sequence of ordinals in a. For n < ω choose an isolated point yn

in Y with yn ∈ {αn}× καn. Then y is an accumulation point of the set {yn : n < ω},and so πχ(y, Y ) = ω. Therefore, there is no point y ∈ Y with πχ(y, Y ) = κ. �

558 Juan Carlos Martınez

Also, we want to remark that for any scattered Boolean space X, the existence ofa closed subspace Y of X and an element y ∈ Y with πχ(y, Y ) = t(X) implies thatt(X) is attained. For this, put κ = t(X). Since πχ(y, Y ) = κ and Y is scattered,there is a set Z of isolated points in Y of size κ such that y is an accumulation pointof Z. Now if t(y, X) < κ, there would exist a subset Z0 of Z with |Z0| < κ and yaccumulation point of Z0, which contradicts the assumption that πχ(y, Y ) = κ. Thus,the statement of [2, Problem 43] is true for the class of scattered Boolean spaces.

References

[1] Koppelberg, S., General theory of Boolean algebras. Handbook of Boolean Algebras(J. D. Monk and R. Bonnet, eds.), Volume 1, North-Holland Publ. Comp., Amster-dam 1989.

[2] Monk, J. D., Cardinal invariants on Boolean algebras. Progress in Mathematicsvol. 142, Birkhauser Verlag, Basel 1996.

[3] Ros�lanowski, A., and S. Shelah, More on cardinal invariants of Boolean algebras.Ann. Pure Appl. Logic 103 (2000), 1 – 37.

(Received: October 19, 2001; Revised: December 30, 2001)