Compatible algebraic structures on scattered compacta

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Topology and its Applications 153 (2005) 710–723

www.elsevier.com/locate/topo

Compatible algebraic structureson scattered compacta

Taras Banakha,b,∗, Oleg V. Gutikb,c, M. Rajagopaland

a Instytut Matematyki, AkademiaSwietokrzyska, Kielce, Polandb Department of Mathematics, Lviv National University, Universytetska 1, 79000, Ukraine

c Department of Algebra, Ya. Pidstrygach Institute for Applied Problems of Mechanics and Mathematof National Academy of Science, Naukova 3b, Lviv, 79060, Ukraine

d Department of Mathematics, Tennessee State University,3500 John Herritt Boulevard, Nashville, TN, 37209, USA

Abstract

It is proved that each hereditarily collectionwise Hausdorff compact scattered space withscattered height admits a continuous semilattice operation turning it into a topological semwith open principal filters. On the other hand a compactificationγ N of a countable discrete spaN whose remainder is homeomorphic to[0,ω1] admits no (separately) continuous binary operaturning γ N into an inverse semigroup (semilattice). Also we construct a compactificationψN ofN admitting no separately continuous semilattice operation and such that the remainderψN \ N ishomeomorphic to the one-point compactification of an uncountable discrete space. To show tψNadmits no continuous semilattice operation we prove that the set of isolated points of a cscattered topological semilatticeX of scattered height 2 is sequentially dense inX. Also we provethat each separable scattered compactum with scattered height 2 is a subspace of a separablscattered topological semilattice with open principal filters and scattered height 2. This allowconstruct an example of a separable compact scattered topological semilattice with open pfilters and scattered height 2, which fails to be Fréchet–Urysohn. Also we construct an exama Fréchet–Urysohn separable non-metrizable compact scattered topological semilattice wiprincipal filters and scattered height 2. 2005 Elsevier B.V. All rights reserved.

* Corresponding author.E-mail addresses:[email protected], [email protected] (T. Banakh), [email protected]

(O.V. Gutik), [email protected] (M. Rajagopalan).

0166-8641/$ – see front matter 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.topol.2005.01.004

T. Banakh et al. / Topology and its Applications 153 (2005) 710–723 711

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MSC:primary 22A26, 54D30; secondary 03E17, 22A15, 54G12, 54G20, 54H10, 54H12

Keywords:Topological semilattice; Topological inverse semigroup; Pospace; Scattered space; Scattered

In this note we pursue the investigations started in [6,1] and consider a classical prof topological algebra [17]: which compatible algebraic structures can live on a givenlogical space, or more precisely, on a given scattered compact space? Let us rema topological spaceX is scatteredif every non-empty subspace ofX contains an isolatepoint. In the sequel all topological spaces considered in the paper are Hausdorff. Othat non-discrete scattered spaces fail to be topologically homogeneous and thus acompatible group structure. On the other hand, there are compatible algebraic strumore general than the structure of a topological group, which can live on non-discattered spaces. We have in mind the structure of a regular topological semigroup.

Let us recall that atopological semigroupis, by definition, a topological spaceS en-dowed with a continuous associative operation∗ :S × S → S. A semigroupS is regular ifeach elementx ∈ S has aninverse, that is an elementx−1 ∈ S such thatx ∗ x−1 ∗ x = x

and x−1 ∗ x ∗ x−1 = x−1. If for each x ∈ S such an inverse elementx−1 is unique,then S is called aninverse semigroup. By a topological inverse semigroupwe under-stand an inverse semigroupX endowed with a topology such that the semigroup opera∗ :X × X → X and the operation(·)−1 :X → X of taking the inverse are continuous. Tclass of (topological) inverse semigroups is quite wide: it contains all (topological) grand all (topological) semilattices. By asemilatticewe understand a commutative semgroup(S,∧) so that every elementx ∈ S is anidempotent, that is,x ∧ x = x. We say thata topological semilattice(S,∧) hasopen principal filtersif for any x ∈ S the upper se↑x = y ∈ S: x ∧ y = x is open inS, see [6].

Observe that each topological spaceX admits a continuous binary operation∗ :X ×X → X turning X into a regular semigroup: just takex ∗ y = x for everyx, y ∈ X. Thesituation with the structure of an inverse topological semigroup or a topological semilis much more subtle.

Considering scattered compacta, we see that in the metrizable case, each mescattered compactum, being homeomorphic to a subset of the real line, admits a consemilattice operation max (or min). The simplest examples of non-metrizable scacompacta such as the one-point compactificationαD of an uncountable discrete spaceD

or closed segments of ordinals also admit a continuous semilattice operation.This observation led us to the following natural question (see [6]):does every scattere

compact space admit a continuous semilattice operation? Surprisingly, the answer to thquestion is negative. We shall give two different counterexamples but first prove a poresult: we shall show that each hereditarily collectionwise Hausdorff scattered cospace with finite scattered height admits a continuous semilattice operation turninga topological semilattice with open principal filters.

We remind that a topological spaceX is collectionwise Hausdorffif for each closeddiscrete subspaceD of X there is a discrete collectionUxx∈D of open subsets ofX suchthatx ∈ Ux for eachx ∈ D (a collectionF of subsets of a topological spaceX is discreteif each pointx ∈ X has a neighborhood meeting at most one member ofF ). A topological

712 T. Banakh et al. / Topology and its Applications 153 (2005) 710–723

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spaceX is hereditarily collectionwise Hausdorffif each subspace ofX is collectionwiseHausdorff. Observe that the hereditary cellularity of a hereditarily collectionwise Hdorff spaceX equals its cellularity.

Finally, we recall the definition of the scattered height of a scattered compactum,is defined by transfinite induction. For a topological spaceX let X(0) = X andX(1) be theset of all non-isolated points ofX. By transfinite induction for every ordinalα > 1 definetheαth derived setX(α) of X lettingX(α) = ⋂

β<α(X(β))(1). By thescattered heightκ(X)

of a scattered topological spaceX we understand the smallest ordinalα such thatX(α)

is finite. Thus the one-point compactification of an infinite discrete space has scaheight 1. Conversely, each scattered compact spaceX with κ(X) = 1 is a finite disjointunion of one-point compactifications of infinite discrete spaces and thus is heredcollectionwise Hausdorff. As we shall see later this is not true for scattered compascattered height 2.

Theorem 1. Each hereditarily collectionwise Hausdorff compact scattered spaceX withfinite scattered height admits a continuous semilattice operation∧ turning X into a topo-logical semilattice with open principal filters and also making any given elementx0 intothe zero ofX (that isx0 ∧ x = x0 for all x ∈ X).

Proof. The theorem will be proved by induction on the scattered height ofX. If κ(X) = 0(which means thatX is finite), then the theorem is trivial.

Suppose thatκ(X) = n for somen ∈ N and the theorem has been proved for all scattecompacta with scattered height< n. First we consider the case when the setX(n) consistsof a unique pointa∞. Let x0 be a fixed point ofX.

First we consider the subcasea∞ = x0. Note thatD = X(n−1) \a∞ is a closed discretsubspace ofX \ a∞. Taking into account thatX is zero-dimensional and hereditarcollectionwise Hausdorff, we can assign to eacha ∈ D a closed-and-open neighborhoUa ⊂ X such thatUaa∈D is a discrete collection inX \ a∞. Let Ua∞ = X \ ⋃

a∈D Ua

and observe that maxκ(Ua): a ∈ X(n−1) < n. By the induction assumption, for eaa ∈ X(n−1) the spaceUa admits a continuous semilattice operations∧a turning it into atopological semilattice with open principal filters. Moreover, we can assume thata∞ isa zero ofUa∞ . Now define a semilattice operation∧ onX letting

x ∧ y =

x ∧a y if x, y ∈ Ua for somea ∈ X(n−1),

a∞ otherwise.

Let us show thatX endowed with the operation∧ is a topological semilattice with opeprincipal filters and zeroa∞.

We shall verify the continuity of∧ at the point(a∞, a∞). Take any neighborhooW ⊂ X of a∞ = a∞ ∧ a∞. Since the collectionUaa∈D is discrete inX \ a∞ andX \ W is compact, the setE = a ∈ D: Ua ∩ (X \ W) = ∅ is finite and hence the sU = X \ ⋃

a∈E Ua is an open neighborhood ofa∞ with U ∧ U ⊂ W . This means that∧ iscontinuous at(a∞, a∞). The continuous of∧ at other points ofX ×X easily follows fromthe definition of∧. This finishes the proof in the particular case|X(n)| = 1 andx0 = a∞.

Now suppose thatx0 = a∞. Find a closed-and-open neighborhoodU0 of x0 such thata∞ /∈ U0. The preceding argument supplies us with a continuous semilattice operati∧1


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onX\U0 turning the scattered spaceX\U0 into a topological semilattice with open princpal filters anda∞ for a zero. Sinceκ(U0) < n we can apply the induction hypothesis to fia continuous semilattice operation∧0 turning the spaceU0 into a topological semilatticewith open filters andx0 as a zero. Now define a semilattice operation∧ onX letting

x ∧ y =

x ∧0 y if x, y ∈ U0,

x ∧1 y if x, y ∈ X \ U0,

x0 otherwise.

It is easy to see thatX endowed with the operation∧ is a topological semilattice with opeprincipal filters and zerox0. This finishes the proof in the particular case|X(n)| = 1.

Now let us pass to the general case. WriteX(n) ∪ x0 = x0, . . . , xm. For everyi m

find a closed-and-open neighborhoodUi of xi in X so thatX = ⋃mi=0 Ui andUi ∩ Uj = ∅

for distinct i, j m. Since thenth derived set of everyUi consists of at most one poinwe can apply the previous argument to construct a continuous semilattice operation∧i onUi turningUi into a topological semilattice with open principal filters and zero atxi . Nowdefine a continuous semilattice operation∧ onX letting

x ∧ y =

x ∧i y if x, y ∈ Ui for somei m,

x0 otherwise.

It is easy to verify that∧ turnsX into a topological semilattice with open principal filteandx0 for zero.

Now we construct examples of scattered compacta admitting no continuous semoperation. In fact, one of these examples has so wild topological structure that it admpartial order turning it into a pospace, locally compatible at each point.

By a pospacewe understand a topological spaceX endowed with a partial ordersuch that for every pointx ∈ X its lower set↓x = y ∈ X: y x and its upper set↑x =y ∈ X: y x are closed inX. Two elementsx, y of a partially ordered set(X,) arecomparableif x y or x y. We define a pospace(X,) to be locally compatibleata pointx0 ∈ X if for every neighborhoodU of x0 there is a neighborhoodV of x0 suchthat for everyx ∈ V there isz ∈ U such thatz is comparable withx0 andx. We say thata pospace(X,) is locally compatible at a setW ⊂ X if it is locally compatible at eachpoint ofW .

Each semilattice operation∗ :X × X → X on a setX induces a partial order definedby x y iff x ∗ y = x. This order is locally compatible at a pointx0 ∈ X if the mapsx0 :X → X, sx0 :x → x ∗ x0, is continuous atx0.

Our first counterexample is based on

Theorem 2. Suppose a topological spaceX contains a topological copyW of [0,ω1)

whose complementN = X \ W is countable and dense inX. Then

(a) X admits no partial order turningX into a pospace locally compatible atW ;(b) X admits no separately continuous semilattice operation;(c) X is homeomorphic to no topological inverse semigroup.

Proof. We shall identify[0,ω1) with W = X \ N .


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(a) Suppose is a partial order onX such that(X,) is a pospace locally compatibatW . Let

N1 = x ∈ N : [0,ω1) ∩ ↑x is uncountable

, (1)

N2 = x ∈ N : [0,ω1) ∩ ↓x is uncountable

. (2)

We can find a countable ordinalα such that

[α,ω1) ∩ ↑x = ∅ for everyx ∈ N \ N1, (3)

[α,ω1) ∩ ↓x = ∅ for everyx ∈ N \ N2. (4)

Since the sets[0,ω1)∩↑x, x ∈ N1 and[0,ω1)∩↓x, x ∈ N2, are closed and unboundein [0,ω1), the intersection

C = (α,ω1) ∩( ⋂

x∈N1

[0,ω1) ∩ ↑x

)∩

( ⋂x∈N2

[0,ω1) ∩ ↓x

)(5)

is unbounded with respect to the well order of[0,ω1) (here we assume that the intersectof the empty collection of subsets of a setS is equal toS).

We claim that there are three pointsa, b, c ∈ C such thatb /∈ (↓a) ∪ (↑c). Indeed, ifCcontains two incomparable elementsx, y put a = x, b = y, c = x. If any two elements oC are comparable, take any three distinct pointsa, b, c ∈ C ordered so thata b c.

Note thatU = X \ ((↓a) ∪ (↑c) ∪ [0, α]) is a neighborhood ofb. Using the density oN and the local compatibility of(X,) at the pointb find pointsx ∈ N \ (↓a ∪ ↑c) andz ∈ U such thatz is comparable withx andb.

Let us show thatz ∈ W . Assuming the converse, we would getz ∈ N1 ∪N2 according to(3) and (4) and the fact thatz is compatible withb ∈ (α,ω1). On the other hand the equali(5) implies thatz /∈ N1 ∪ N2 sinceC a /∈ ↑z andC c /∈ ↓z. Thereforez ∈ W ∩ U andthusz ∈ (α,ω1). Sincez is comparable withx, by (3) or (4) we would getx ∈ N1 ∪ N2.Again (5) implies thatx /∈ N1 ∪N2 sincea /∈ ↑x andc /∈ ↓x. This contradiction completethe proof of the statement in (a).

(b) Suppose∗ :X × X → X is a separately continuous semilattice operation. Dea partial order on X letting x y iff x ∗ y = x. It can be easily seen thatX endowedwith this order is a pospace locally compatible at each point. Now the preceding stacompletes the proof.

(c) Suppose∗ :X × X → X is an operation turningX into a topological inverse semgroup. Denote byE = x ∈ X: x ∗ x = x the set of idempotents ofX. Clearly,E is aclosed subset ofX. Moreover, according to [2, Theorem 1.1.7],E is a semilattice withrespect to the operation∗. Let us show thatE is uncountable. Assuming the converseshall prove thatX is countable. SinceX is a topological inverse semigroup, the operat(·)−1 :X → X of taking the inverse is continuous. Then the functionsr1, r2 :X → E de-fined byr1(x) = x ∗ x−1 andr2(x) = x−1 ∗ x for x ∈ X are continuous retractions ofX

ontoE.It is clear thatX = ⋃

e,f ∈E Ge,f , whereGe,f = r−11 (e) ∩ r−1

2 (f ). To show thatX iscountable, it suffices to verify that for everye, f ∈ E the setGe,f is at most countableA routine verification shows that for every idempotente ∈ E the setGe,e is a closed subgroup of the inverse semigroupX.


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We claim that this subgroupGe,e is countable. Observe that the setW \W consists of atmost one point and the spaceX is locally countable at each pointx /∈ W \ W . This impliesthat the topological groupGe,e, being topologically homogeneous, is locally countabThe subspaceGe,e ∩ W of Ge,e, being countably compact, is totally bounded inGe,e (thelatter means that for any neighborhoodU of the unit of the groupGe,e there is a finitesubsetF ⊂ Ge,e with Ge,e ∩ W ⊂ F · U ). Now the local countability ofGe,e and the totalboundedness ofGe,e ∩ W in Ge,e imply that the setGe,e ∩ W is countable. Since the sX \W is countable, we conclude that the groupGe,e ⊂ (Ge,e ∩W)∪ (X \W) is countabletoo.

To show that for anye, f ∈ E the setGe,f is countable, fix any elementa ∈ Ge,f andconsider the maph :Ge,f → X defined byh(x) = x ∗ a−1 for x ∈ Ge,f . Sincex ∗ a−1 ∗a = x ∗ f = x ∗ (x−1 ∗ x) = x, the maph is injective. Given any pointx ∈ Ge,f , we get

r1(h(x)

) = x ∗ a−1 ∗ (x ∗ a−1)−1 = x ∗ a−1 ∗ a ∗ x−1 = xx−1 = r1(x) = e

and

r2(h(x)

) = (x ∗ a−1)−1 ∗ (

x ∗ a−1) = a ∗ x−1 ∗ x ∗ a−1

= a ∗ r2(x) ∗ a−1 = a ∗ f ∗ a−1 = a ∗ r2(a) ∗ a−1

= a ∗ (a−1 ∗ a

) ∗ a−1 = a ∗ a−1 = r1(a) = e,

which means thath(x) ∈ Ge,e.It follows from the injectivity ofh that |Ge,f | |Ge,e|. Consequently, the setsGe,f ,

e, f ∈ E, are countable and so is the semigroupX, a contradiction which shows that thsetE is uncountable.

Since the setN is dense inX, we get that its imager1(N) is dense inE = r1(X). Letα = supr1(N) ∩ [0,ω1) (the supremum taken with respect to the natural well orde[0,ω1)). Sincer1(N) is countable, we getα < ω1. Next, since the setE is uncountableand closed inX, Ω = E ∩ [α + 1,ω1) is a closed unbounded subset of[0,ω1) which ishomeomorphic to[0,ω1) while M = E \ Ω ⊃ r1(N) is a countable dense subset ofE.Thus E is a topological semilattice containing a topological copyΩ of [0,ω1) whosecomplementE \ Ω is countable and dense inE, which is impossible according to thprevious statement.Remark 1. In case of compactX the second statement of Theorem 2 can be derivedthe third one with help of a classical result of Lawson [11, Theorem II.1.5] accorto which any separately continuous semilattice operation on a zero-dimensional cospaceK is continuous (moreover,K has a base of the topology consisting of subsemtices). Unfortunately, this Lawson result is not valid in the class of commutative invsemigroups. To construct a suitable counterexample, consider the one-point comcationZ = Z ∪ ∞ of the space of integers endowed with the discontinuous separcontinuous binary operation∗ :Z ×Z → Z defined by

x ∗ y =

x + y, if x, y ∈ Z;∞, otherwise.


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It follows from Parovichenko Theorem [15] (see also [7, 3.5.H] or [8]) that the ordspace[0,ω1] is the image of the remainderβN \ N of the Stone–Cech compactificationof N under a continuous mapf :βN \ N → [0,ω1]. The mapf induces a closed equivalence relation∼ on βN: x ∼ y iff either x, y ∈ βN \ N andf (x) = f (y) or x, y ∈ N andx = y. The quotient spaceγ N = βN/∼ of βN with respect to the equivalence relation∼is a compactification ofN whose remainderγ N \ N is homeomorphic to[0,ω1]. For thiscompactification Theorem 2 implies

Corollary 1. Letγ N be a compactification ofN whose remainderγ N\N is homeomorphicto the ordinal segment[0,ω1]. Then the compact scattered spaceγ N does not admit aseparately continuous semilattice operation and is homeomorphic to no topological insemigroup.

The compactificationγ N has a rather complex structure. In particular,γ N has uncountable scattered height. Next, we construct a scattered compact spaceψN with scatteredheight 2 admitting no continuous semilattice operation.

The construction of the compactificationψN is based on the notion of MAD family. Wremind that a familyF of infinite subsets ofN is calledalmost disjointif A ∩ B is finitefor any distinctA,B ∈ F . Under amaximal almost disjoint(briefly MAD) family weunderstand any maximal element in the set of infinite almost disjoint families of insubsets ofN.

Given a MAD familyF endow the setF ∪ N with the topology generated by the subasen, A ∪ A \ K: n ∈ N, A ∈F , K is a finite subset ofN. It is easy to see that thtopology onF ∪ N is Hausdorff and locally compact. Finally, letψF (N) = ∞ ∪F ∪ Nbe the one-point compactification of the spaceF ∪ N.

Theorem 3. For any infinite MAD familyF onN the spaceψF (N) is a scattered compacspace with scattered height2 admitting no separately continuous semilattice operation

Proof. It is clear thatψF (N) is compact, scattered and has scattered height 2. Obalso that no sequence(xn)n∈ω ⊂ N converges to the point at infinity∞ of ψF (N). Indeed,for any infinite setX = xn: n ∈ ω ⊂ N there is a setA ∈ F such thatA ∩ X is infinite.Then the subsequencexn: xn ∈ A of (xn) converges to the pointA = ∞ and thus(xn)

cannot converge to∞. Now we see that Theorem 3 is a particular case of the followmore general theorem.

We call a subsetA of a topological spaceX sequentially densein X if each pointx ∈ X

is the limit of a sequence(an) ⊂ A. In particular,N is not sequentially dense inψF (N).

Theorem 4. If a separable compact scattered spaceX of scattered height2 admits a separately continuous semilattice operation, then the setIso(X) of isolated points ofX issequentially dense inX.

To prove Theorem 4 we need the following fact, see [1, Lemma 1].


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Lemma 1. If K is a compact subset with a unique non-isolated pointe in a scatteredtopological semilatticeX, then the setK \ ↑e is at most countable.

Proof of Theorem 4. Let ∧ :X ×X → X be a separately continuous semilattice operaon a scattered compact spaceX with κ(X) 2. Applying [11, Theorem II.1.5] we concludthat the operation∧ is continuous. Fix any pointe ∈ X. We have to find a sequence(xn)n∈ω

of isolated points ofX, tending toe. If e ∈ X(1) \ X(2), thene, being an isolated point oX(1), possesses a clopen neighborhoodU ⊂ X such thate is a unique non-isolated poinof U . Take any infinite subsetxn: n ∈ ω ⊂ U \ e and observe that the sequence(xn)

converges toe.So it rests to consider the casee ∈ X(2). If e has a countable closed neighborhoodW

in X, then by the compactness ofX, W is metrizable and by the density of the set Iso(X)

in X there is a sequence(xn) ⊂ W ∩ Iso(X) convergent toe. So we assume that no neigborhood ofe is countable and no sequence of isolated points ofX tends toe.

The scattered topological semilattice(X,∧), being zero-dimensional, is Lawson [1Theorem II.1.5] which means thatX has a base consisting of subsemilattices. SinceX(2)

is finite, we can find a clopen neighborhoodY ⊂ X of e such thatY ∩ X(2) = e andY isa subsemilattice ofX. Without loss of generality we can assume thatY = X, i.e., the setX(2) consists of a unique pointe. ThenX(1) is an uncountable compact set with a uniqnon-isolated pointe. Applying Lemma 1 we get that the setX(1) \ ↑e is at most countableThen the setX \↑e also is at most countable and thus can be written asX \↑e = ⋃

n∈ω Fn

whereFn ⊂ Fn+1 are finite subsets. By the compactness of the set↑e, for everyn ∈ ω

find a closed neighborhoodWn ⊂ X of ↑e such that for eachz ∈ Wn bothz andz ∧ e arenot in Fn. Without loss of generality we can assume thatWn+1 ⊂ Wn for all n ∈ ω. Thus⋂

n∈ωWn = ↑e.

Let 〈Iso(X)〉 be the smallest subsemilattice ofX containing the set Iso(X). SinceX isseparable, the set Iso(X) is countable as well as the semilattice〈Iso(X)〉. We claim that thepartial order induced by the semilattice operation is well-founded on the setC = (X(1) ∩↑e) \ (〈Iso(X)〉 ∪ e). The latter means that each subset ofC has a minimal elemenor equivalently, there is no infinite strictly decreasing sequences(en)n∈ω in C. To provethis fact, assume that(en)n∈ω ⊂ C is such a decreasing sequence (i.e.,en ∧ en+1 = en+1

anden = en+1 for all n). For everyn ∈ ω fix a neighborhoodUn ⊂ Wn of en such thatUn ∩ X(1) = en. We can take the neighborhoodsUn to be pairwise disjoint. Next, foeveryn ∈ ω, fix a sequence(xn,m)m∈ω ⊂ Un ∩ Iso(X) convergent toen (such a sequencexists sinceen ∈ X(1) \ X(2)).

Using the continuity of the semilattice operation, for everyn ∈ ω fix a functionfn :ω → ω such thatyn = x0,fn(0) ∧ x1,fn(1) ∧ · · · ∧ xn,fn(n) ∈ Un. We can assume thafn(i) < fm(i) for all i ∈ ω and n < m. Taking into account thatUn ∩ X(1) = en ⊂〈Iso(X)〉 andyn ∈ 〈Iso(X)〉 ∩ Un, we conclude thatyn ∈ Un \ X(1) ⊂ Iso(X) for all n ∈ ω.The spaceX, being compact of scattered height 2, is sequentially compact [14].sequently, some subsequence of(yn) converges to a pointy∞ ∈ X(1). Without loss ofgenerality, we can assume that the sequence(yn) tends toy∞. The continuity of thesemilattice operation and the equalityyn = xk,fn(k) ∧ yn holding for all k n imply thaty∞ = limn yn = limn(xk,fn(k) ∧ yn) = limn xk,fn(k) ∧ limn yn = ek ∧ y∞. Taking into ac-


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count thate is the limit point of the sequence(ek)k∈ω, we conclude thaty∞ = y∞ ∧ e andthusy∞ e. We claim thaty∞ = e.

Assuming the converse we would findm ∈ ω such thaty∞ ∈ Fm. Now recall that for anyk m we haveyk ∈ Uk ⊂ Wk ⊂ Wm. Since the setWm is closed inX, we gety∞ ∈ Wm.Now the inclusiony∞ = y∞ ∧ e ∈ Fm contradicts the choice of the setWm. Thereforey∞ = e and(yn) ⊂ Iso(X) is a sequence of isolated points ofX tending toe which con-tradicts our assumption. This contradiction shows that the partial order on the seC iswell-founded.

Consider the subsetL = e ∪ ⋃↑x: x ∈ 〈Iso(X)〉 ∩ ↑e \ e and observe that it iat most countable. This follows from the finiteness of the setX(1) ∩ ↑x for any x e.Observe also that(X(1) ∩ ↑e) \ L ⊂ C.

By induction for everyn ∈ ω we shall construct an open subsemilatticeVn ⊂ Vn−1∩ Wn

of X containing the elemente, a minimal elementen of (Vn−1 ∩ X(1) ∩ ↑e) \ L, and anelementxn ∈ Vn−1 ∩ Iso(X) such that

(a) xn ∧ z ∈ X(1) for eachz ∈ Vn ∩ Iso(X), and(b) xn ∧ xn−1 = xi−1 ∧ xi for everyi ∈ 1, . . . , n − 1 with xi−1 ∧ xi = e.

Let V−1 = W0 and assume that for somen ∈ ω the neighborhoodsVi , and pointsei ,xi are constructed for alli < n. Since the pointe has no countable neighborhood, theVn−1 ∩X(1) is uncountable. Applying Lemma 1 we conclude that the setVn−1 ∩X(1) ∩↑e

is uncountable and that the set(Vn−1 ∩ X(1) ∩ ↑e) \ L, being non-empty and welfounded, contains a minimal elementen. Since en ∈ X(1) \ X(2), there is a sequenc(yk)k∈ω ⊂ Iso(X) ∩ Vn−1 ∩ Wn tending toen. Find a closed neighborhoodO1 ⊂ Vn−1 of e

such thaten∧z = en for anyz ∈ O1. We claim that there is a numberk1 ∈ ω and a neighborhoodO2 ⊂ O1 of e such thatz ∧ yk ∈ X(1) for all z ∈ O2 ∩ Iso(X) andk k1. Otherwisewe would find an increasing number sequence(ki)i∈ω and a sequence(zi)i∈ω of pairwisedistinct isolated points ofO1 such thatzi ∈ Wi for every i ∈ ω and zi ∧ yki

∈ Iso(X).By the sequential compactness ofX, we can assume that the sequence(zi)i∈ω convergesto some pointz∞ ∈ O1. It follows that z∞ ∈ ⋂

i∈ωWi = ↑e and z∞ ∈ X(1). Note that

the sequence(zi ∧ yki)i∈ω ⊂ Iso(X) converges to the pointz∞ ∧ en = en. Taking into

account that no sequence of isolated points tends toe, we conclude thatz∞ ∧ en = e.Recall thatVn−1 ∩ X(1) ∩ ↑e z∞ ∧ en < en and en is a minimal element of the se(Vn−1 ∩ X(1) ∩ ↑e) \ L. It follows thatz∞ ∧ en ∈ L. By the definition ofL this yields thatz∞ ∧ en ∈ ↑x for somex ∈ 〈Iso(X)〉 ∩ ↑e \ e. Sinceen z∞ ∧ en, we geten ∈ ↑x ⊂ L

which contradicts the choice of the pointen.This contradiction shows that there is a numberk1 ∈ ω and a neighborhoodO2 ⊂ O1 of

e such thatz∧yk ∈ X(1) for all k k1 andz ∈ O2 ∩ Iso(X). Taking into account thatX is azero-dimensional Lawson semilattice, find a clopen subsemilatticeVn ⊂ O2 containinge.We can chooseVn so small thatz ∧ yk1 /∈ xi−1 ∧ xi : i ∈ 1, . . . , n − 1 andxi−1 ∧ xi = efor anyz ∈ Vn. Finally, putxn = yk1. This completes the inductive step.

Now consider the sequence(xn)n∈ω ∈ Iso(X). By the sequential compactness ofX

a subsequence of(xn) tends to some elementx∞ ∈ X. Without loss of generality we caassume that all the sequence(xn) tends tox∞. Sincexk ∈ Vk−1 ⊂ Wm for any k > m,we getx∞ ∈ ⋂ Wm = ↑e. Our hypothesis yields thatx∞ = e. Then the sequenc
m∈ω


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elt

(xn ∧ xn+1)n∈ω tends tox∞ ∧ x∞ = x∞. By the choice of the sequence(xn) we getxn ∧ xn+1 ∈ X(1). Sincee is a unique non-isolated point ofX(1) we conclude that therare numbersn < m such thatxm ∧ xm+1 = xn ∧ xn+1 = x∞, which contradicts the choicof the sequence(xi). This contradiction finishes the proof of theorem.

Unlike the compactificationγ N from Corollary 1 which has sizeℵ1, the size of thecompactificationψF (N) cannot be too small. More precisely,|ψF (N)| = |F | a wherea

is the smallest size of an infinite MAD family onN. It is known thatℵ1 a c. Martin’sAxiom impliesa = c but there are models of ZFC witha < c, see [4,16].

We define a topological spaceX to be aψ -spaceif X is homeomorphic to the compacificationψF (N) for some MAD familyF on N.

The following proposition shows that the cardinality restriction onψ -spaces is essentiaWe remind that a topological spaceX is Fréchet–Urysohnif for every pointa from theclosureA of a subsetA ⊂ X in X there is a sequence(an) ⊂ A convergent toa.

Proposition 1. Suppose thatX is a compact scattered space of scattered height2. If X isnot Fréchet–Urysohn, thenX contains a copy of aψ -space and consequently,|X| a.

Proof. By [14], the spaceX, being scattered compact, is sequentially compact. AssumthatX is not Fréchet–Urysohn, we could find a countable subsetA ⊂ X and a pointa ∈ A

such that no sequence(an) ⊂ A converges toa. It is easy to see thata ∈ X(2). Let W ⊂ X

be a closed-and-open neighborhood ofa such thatW ∩ X(2) = a. It follows that theintersectionX(1) ∩ W ∩ A is finite and thus we can assume thatA ⊂ W \ X(1). Let K bethe closure of the setA in X. It follows that K(2) = a, A = Iso(K) and no sequenc(an) ⊂ A converges toa.

We claim thatK is a ψ -space. SinceD = K(1) \ K(2) is a discrete subspace ofK , toeach pointx ∈ D we can assign a clopen subsetUx of K such thatUx ∩ K(1) = x. Weclaim thatF = Ux ∩ A: x ∈ D is a MAD family onA.

First we verify that for any distinct pointsx, y ∈ D the intersectionUx ∩ Uy is finite.Otherwise by the compactness ofK , we would find a non-trivial sequence(an) ⊂ Ux ∩Uy

having a cluster pointa∞ ∈ Ux ∩Uy ∩K(1) which is not possible sinceUx ∩Uy ∩K(1) = ∅.Consequently,|F | = |D| and hence the familyF is infinite.

Next, we show thatF is amaximalalmost disjoint family. Assuming the converse wwould find an infinite setU ⊂ A such thatU ∩ Ux is finite for allx ∈ D. By the sequentiacompactness ofK we could find a non-trivial sequence(an)n∈ω ⊂ U convergent to a poina∞ ∈ K(1). Taking into account thatUx ∩ an: n ∈ ω is finite for all x ∈ D, we concludethat a∞ = x for any x ∈ D and thusa∞ = a ∈ K(2). Therefore, the sequence(an) ⊂ A

converges toa, which is a contradiction.ThusF is a MAD family onA. Now consider the bijective maph :K → ψF (A) defined

by h(x) = x if x ∈ A = K \ K(1), h(x) = Ux if x ∈ K(1) \ K(2), andh(a) = ∞. Observethat h is continuous and thus is a homeomorphism by the compactness ofK . There-fore X contains the topological copyK of the ψ -spaceψF (A) and hence|X| |K| =|F | a.


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Besides the size and the scattered height, there is another essential differencethe scattered compactaγ N andψN from Theorems 2 and 3: unlike the spaceγ N, the spaceψN embeds into a compact semilatticeS so that the remainderS \ ψN is countable. Thisfollows from the subsequent general

Theorem 5. Each scattered compactumK of scattered heightκ(K) 2 is a subspace oa scattered compact semilatticeS ⊃ K with open principal filters such thatκ(S) = κ(K),S(1) = K(1) and | Iso(S)| = | Iso(K)|.

To prove this theorem we need to make some preliminary work. For an infinite disspaceD let αD = D ∪ ∞ be the one-point compactification ofD. The compactificationαD can be identified with the subspaceχ∅, χx: x ∈ D of the Cantor cube0,1D (herefor a subsetA ⊂ D by χA :D → 0,1 we denote the characteristic function of the setA,i.e.,χA(x) = 1 iff x ∈ A). We can think of0,1D as a compact semilattice endowed wthe usual min-operation. In this caseαD ⊂ 0,1D is an ideal in0,1D in the sense thamin(x, y) ∈ αD for eachx ∈ αD andy ∈ 0,1D .

For a subsetA ⊂ 0,1D let 〈A〉 be the smallest subsemilattice of0,1D containingthe setA.

Lemma 2. If K is a subspace of0,1D such thatK(1) = αD, then 〈K〉 is a compactscattered semilattice with〈K〉(1) = αD and | Iso(〈K〉)| = | Iso(K)|.

Proof. First we verify that the semilattice〈K〉 is compact. Given an open coverU of 〈K〉find a finite subcoverV of the compact subsetαD = K(1) ⊂ 〈K〉. Taking into account thaαD is an ideal of0,1D , find an open idealV of 0,1D such thatαD ⊂ V ⊂ ⋃

V . SinceK(1) ⊂ V , the setF = K \V is finite. Consequently, the set〈K〉 \V = 〈F 〉 is finite as welland hence can be covered by a finite subcoverW of U . Then〈K〉 is covered by the finitesubcoverV ∪W of U which yields the compactness of〈K〉.

The above argument implies that the set〈K〉 \ V is finite for any neighborhoodV ofK(1). This observation implies that〈K〉(1) = K(1) = αD. Then Iso(K) ⊂ Iso(〈K〉) andhence| Iso(K)| | Iso(〈K〉)|.

SinceK(1) = αD is an ideal of0,1D , 〈K〉 = K(1) ∪ 〈Iso(K)〉 and thus Iso(〈K〉) ⊂〈Iso(K)〉 which yields| Iso(K)| = | Iso(〈K〉)|.

Now we are able to prove Theorem 5.

Proof of Theorem 5. Suppose thatK is a compact scattered space of scattered heκ(K) 2. Repeating the argument in the proof of Theorem 1 we can reduce the prthe special case when the second derived setK(2) of K consists of a unique point∞.

For i = 0,1 let Di = K(i) \ K(i+1) whereK(0) = K . Let f0 :K → 0,1D0 be the con-tinuous function defined byf0(x) = χ∅ if x ∈ K(1) andf0(x) = χx if x ∈ D0 = K \K(1).By analogy, define a functionf1 :K(1) → 0,1D1 letting f1(x) = χ∅ if x ∈ K(2) andf1(x) = χx if x ∈ D1 = K(1) \ K(2). SinceK is a zero-dimensional compact space,function f1 extends to a continuous functionf1 :K → 0,1D1. Now consider the func


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tion f = (f0, f1) :K → 0,1D0 × 0,1D1 and observe that it is injective and hencea topological embedding.

We can think of the productP = 0,1D0 × 0,1D1 as a compact topological semlattice endowed with the usual min-operation. LetS be the smallest subsemilattice ofP

containing the compactumf (K). The previous lemma implies thatS is a compact scattered semilattice withS(1) = f (K(1)), κ(S) = κ(K) and| Iso(S)| = | Iso(K)|. It is easy tosee thatS is a compact semilattice with open principal filters. This finishes the proof otheorem.

Theorem 5 implies that the spaceψN from Theorem 3 is a subspace of a separascattered compact semilatticeS with open principal filters such thatκ(S) = 2. Sinceψ(S)

is not Fréchet–Urysohn,S is not Fréchet–Urysohn either. Thus we get the following sprising result nicely complementing Theorem 4.

Corollary 2. There is a separable scattered compact semilatticeS with open principalfilters and scattered heightκ(S) = 2 which fails to be a Fréchet–Urysohn space.

There are also simple examples of Fréchet–Urysohn separable non-metrizable toical semilattices with open principal filters and scattered height 2. Such an exampbe constructed as follows. LetC be the Cantor cube0,1ω andB be the standard basof the topology ofC, i.e., B = pr−1

n (x): n ∈ ω, x ∈ 0,1n where prn :C → 0,1n,prn : (xi)

∞i=0 → (xi)

n−1i=0 , is the natural projection. For a subsetA ⊂ C denote byχA : C →

0,1 the characteristic function of the setA (that is,χA(x) = 1 iff x ∈ A). Such a char-acteristic function is an element of the non-metrizable Cantor cube0,1C . In this cubeconsider the subspace

S = χ∅, χx, χU : x ∈ C, U ∈ B

⊂ 0,1C.

It is easy to see thatS is a compact scattered subspace of0,1C of scattered height 2. Ishould be mentioned that the spaceS is well-known in topology as an example of a compscattered space with is neither supercompact, nor hereditarily normal, see [13].

Observe thatS is a subsemilattice of0,1C with respect to the coordinate-wise mioperation. Moreover, it can be easily shown that the topological semilattice(S,min) hasthe following properties.

Proposition 2. The spaceS endowed with the min-operation is a Fréchet–Urysohn serable non-metrizable scattered compact topological semilattice with open principal fiand scattered height2.

Now let us pass to open problems, first of which is suggested by Theorem 4.

Problem 1. SupposeX is a compact topological semilattice of finite scattered heighthe set Iso(X) of isolated points ofX sequentially dense inX?

Another question is suggested by Theorem 5.


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Problem 2. Is every (separable) scattered compact space [with finite scattered hesubspace of a (separable) scattered compact topological semilattice?

As we saw in Corollary 2 a separable scattered compact semilattice needs noFréchet–Urysohn. It should be mentioned that the class of compact Fréchet–Uspaces includes all Corson compacta (and consequently, Eberlein, Gulko, Talagranpacta) and all Rosenthal and Rosenthal–Banach compacta, see [10,3,9,12]. The coficationsγ N andψF (N) from Theorems 2 and 4 are not Fréchet–Urysohn.

Problem 3. Does every scattered Fréchet–Urysohn (Eberlein) compact spaceX (with scat-tered height 2) admit a continuous semilattice operation?

In light of Theorem 2 and Remark 1 it is reasonable to ask

Problem 4. Does every scattered compact space admit a separately continuous opturning it into an inverse semigroup?

A semilattice operation is a particular case of amean, i.e., a functionm :X × X → X

such thatm(x,y) = m(y,x) andm(x,x) = x for everyx, y ∈ X, see [5].

Problem 5. Does every scattered compact spaceX admit a (separately) continuous mem :X × X → X?

Following [5] we say that a compact spaceK is hyadicif X is a continuous image of thspace of all closed subsets of some compact space in the Vietoris topology (equivX is a continuous image of a zero-dimensional compact topological semilattice). Beach non-discreteGδ-subspace of a hyadic compact space contains a non-trivial convesequence. Consequently, the spaceβN is not hyadic. In fact each Tychonov spaceX withhyadicβX is pseudocompact, see [5, p. 42].

Problem 6. Is every scattered compact space hyadic? In particular, are the spacesγ N andψF (N) from Corollary 1 and Theorem 3 hyadic? Isγ N (respectivelyψF (N)) a continuousimage of a compact topological semilattice?

Acknowledgements

The authors express their sincere thanks to the referee for very careful reading thuscript and valuable remarks improving the presentation.

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Compatible algebraic structures on scattered compacta

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