Kurepa trees and topological non-reﬂection

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Topology and its Applications 151 (2005) 77–98

www.elsevier.com/locate/topo

Kurepa trees and topological non-reflection

Piotr Koszmider

Departamento de Matemática, Universidade de São Paulo,Caixa Postal: 66281, São Paulo, SP, CEP: 05315-970, Brasil

Received 2 March 2003; received in revised form 15 August 2003; accepted 15 August 2003

Abstract

A propertyP of a structureS does not reflect if no substructure ofS of smaller cardinality thanShas the property. If for a given propertyP there is such anS of cardinalityκ, we say thatP does notreflect atκ. We undertake a fine analysis of Kurepa trees which results in defining canonicallogical and combinatorial structures associated with the tree which possess a remarkably widof nonreflecting properties providing new constructions and solutions of open problems in topThe most interesting results show that many known properties may not reflect at anyfixed singu-lar cardinal of uncountable cofinality. The topological properties we consider vary from normcollectionwise Hausdorff property to metrizablity and many others. The combinatorial propertirelated to stationary reflection. 2004 Elsevier B.V. All rights reserved.

1. Introduction

A Kurepa tree [23] is a tree with countable levels and with more thanω1 uncountablebranches (for all unexplained terminology see [19] and [4,29,32] in [20]). Kurepa’spothesis asserts that a Kurepa tree exists and will be denoted by KH. S. Todorcevicin his Handbook of Set-Theoretic topology article:

KH is a statement about an incompactess at the first uncountable cardinal of cnatural properties of trees, order types and families of sets.

E-mail address:[email protected]: http://www.ime.usp.br/~piotr/stronamat.html.

0166-8641/$ – see front matter 2004 Elsevier B.V. All rights reserved.doi:10.1016/j.topol.2003.08.033

https://core.ac.uk/display/81998298?utm_source=pdf&utm_medium=banner&utm_campaign=pdf-decoration-v1

78 P. Koszmider / Topology and its Applications 151 (2005) 77–98

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In this paper we would like to present some ‘incompactness of certain natural topical properties’ that follows from KH. We also encounter interesting cases of ‘set theincompactness’. Since in the topological context, the notion of compactness has a dmeaning we opted for the wordreflectionor nonreflectionin the sense explained in thabstract. Of course the above concept of incompactness is the logical one, relatedcompactness theorem from first order logic. The incompactness (i.e., our non-reflectsecond order properties is highly nontrivial and has been extensively studied in comtorics (e.g. [28]), group theory (for a references see [26]) or topology itself [6,7,10,18

The topic of non-reflection of second-order properties at singular1 cardinals is of itsown iterest. This is because at singulars the non-reflection is much harder to obtain some general situations impossible to obtain (see results of [26,28] known asShelah’scompactness at singulars). Our results will be most interesting at singular cardinals.

Let us establish some terminology in order to explain our main results. All ourare of heightω1. We say that a treeT has many branchesiff for every t in T , there is anuncountable branch ofT whose range containst . Our constructions will be obtained foany treeT with many branches although for the sake of applications we will be mainterested in a Kurepa tree. Let KHλ(T ) denotes the statement:There is anω1-tree (T)with exactlyλ-many branches andλω = λ. It is easy to see that if KHλ holds then KHλ(T )

holds for someT that is normal and has many branches, thus by a Kurepa tree wmean a normal tree with many branches (it was proved in [32] that KHλ is consistent withany cardinal arithmetic consistent with 2ω1 � λ for λ of uncountable cofinality). IfT is atree, then Br(T ) denotes a collection of all uncountable branches ofT .

The main object investigated in this paper isF(T ) – some well-founded and cofinal sufamily of [Br(T )]ω, the family of all countable collections of branches ofT , for any normaltreeT with many branches. The familyF(T ) is a weak version of an(ω1, λ)-semimorassfor λ = |Br(T )| (see [22]) which is a weak version of an(ω1,1)-morass (see [33]), buit can also originate from the construction of a Kurepa tree from countable setsbranches as is done in [32]. Already Todorcevic realized (private communication) thaT

is a Kurepa tree “obtained” from(ω1,1)-morass, thenF(T ) is a simplified morass.In Section 2, after definingF(T ), we proceed to a fine analysis of its combinato

properties, which gives the main combinatorial results of this paper. The main resthis analysis is a version of the Pressing Down Lemma for subfamilies ofF(T ) and re-lated concepts. These include, defined fromT , a countably complete ideal L (of light seton [λ]ω. This ideal may differ from the ideal of nonstationary sets (this difference is etial in various topological applications) but the behaviour of regressive functions deon elements of L+, calledforking of preimages, is nontrivial. Forking of preimages isweakening of being constant on a large set, thus

Our version of the Pressing Down Lemma(Lemma19)asserts that any regressive funtion defined on a heavy set i.e., from L+ has forking preimages.

1 A cardinalκ is called singular if sets of cardinalityκ can be presented as⋃

α<λ Aα where the cardinality oeachAα is less thanκ as well asλ < κ . Thus, when a propertyP does not reflect at a singular cardinal, it meain particular that there is a structureS with propertyP which is a union of less than|S| substructures withoupropertyP .

P. Koszmider / Topology and its Applications 151 (2005) 77–98 79

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In the following part of Section 2, we provide various applications of our Pressing DLemma in investigating the familyF(T ) like the fact thatF(T ) is a nonreflecting heavset if KHλ(T ) holds.

A more topologically inclined reader may skip Section 2 at the first reading andirectly to Section 3, referring to definitions in Section 2 when needed.

In Section 3, we define a topology onF(T ) (compare with [22]) that is locally metrizable, regular, zerodimensional, non-σ -paraLindelof. Such a space is necessarily not meable. Furthermore, whenever KHλ(T ) holds thenF(T ) is of sizeλ and every subspacG ⊆ F(T ) of size smaller thanλ is metrizable.

So far, topologically speaking, we obtainKHλ implies the existence of a first countable nonreflecting, nonmetrizable spa

sizeλ, i.e., a nonmetrizable space whose all subspaces of sizes smaller thanλ are metriz-able.

As such spaces of singular sizes were already obtained in [22], our weakeninghypothesis needed in this type of result is probably most interesting in case ofλ = ω2 as itis possible to construct (see Fact 36) a model of CH+ KHω2 +¬NR(ω2)

2 and so it is con-sistent with CH that a nonreflecting nonmetrizable spaces exists but the classical eof Hajnal and Juhasz from the nonreflecting stationary set does not exists. It even ffrom the Engelking–Lutzer and Balogh–Rudin characterizations of paracompactngeneralized ordered spaces and monotonically normal spaces (Fact 36), respective

It is consistent with CH that there is locally metrizable, first countable, nonreflecnonmetrizable space of sizeω2 but there is no such space which is generalized ordespace or monotonically normal one.3

But the most interesting topological results of this paper do not consist of a weakof the hypothesis needed for a nonreflecting nonmetrizable space. The main toporesults of the paper are the consequence of the sensitivity of the topology onF(T ) tochanges of the combinatorial properties of the treeT . We consider two such assumptiowhich onT are mutually exclusive,S(T ) andA(T ). Let S(T ) denote the statement thF(T ) is stationary in[Br(T )]ω andA(T ) denote the statement asserting thatT containsan Aronszajn subtree.4

2 By NR(λ) we mean the existence of a stationary subsetA of {α < λ: cf(α) = ω} such thatA ∩ α is nonsta-tionary for eachα < λ.

3 The examples, due to Aull, Miscenko, Gruenhage; or VanDouwen, of spaces having some of these prconsistent with¬CH + KHλ + ¬NR(λ) for λ regular and obtained from versions of MA are presented in [Let us note that the question whether there is a first countable, normal (regular), nonreflecting, nonmespace of size greater thanω1, known as the P. Hamburger question, is open in ZFC. The strongest negativeobtained under the assumption of the existence of supercompact cardinal says that it is consistent thatno first countable nonreflecting, nonmetrizable spaces of size� 2ω (see [10]).

4 It is implicitly proved in [31] thatS(T ) implies the negation ofA(T ) and that KHλ(T ) + S(T ) is consistentwith any cardinal arithmetic consistent with 2ω1 � λ, for any cardinalλ of uncountable cofinality. On the othehand it can be easily shown (Proposition 43) that if KHλ(T ) holds, then there isT ′ such that KHλ(T ′) + A(T ′)holds. Thus topological examples obtainable fromA(T ) exist whenever KHλ holds and this is not the casfor S(T ). Note that assuming the existence of two inaccessible cardinals, it is consistent that KHω2 holds togetherwith A(T ) for everyT such that KHω2(T ) (see [21]).


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In Sections 4 and 5, we study how the statementsS(T ) andA(T ) influence propertiesof the spaceF(T ). It turns out that ifS(T ) holds thenF(T ) is collectionwise normalcountably paracompact, not metaLindelöf and closed sets behave like club sets in[λ]ω.

If, on the other hand,A(T ) holds, thenF(T ) is not normal, notλ-collectionwise Hausdorff (but always (< λ)-collectionwise Hausdorff because small subspaces are metrizCollins space with point-countable base and hence is metaLindelöf.

The assumptionA(T ) provides new consistency results, while the consequencS(T ) are similar to the consequences of the existence of semimorasses alreadygated in [22]. Thus we obtain, underA(T ):

For any singular cardinalλ of uncountable cofinality it is consistent that there isfirst countable, regular< λ-collectionwise Hausdorff space which is notλ-collectionwiseHausdorff.5

Previously known examples of such a non-reflection of the failure of the collectionHausdorff property at singular cardinals obtained by Fleissner and Shelah (see [15])were at strong limit cardinals of cofinality that was the successor of a singular strongWe also obtain the non-reflection of normality:

For every cardinalλ of uncountable cofinality it is consistent that there is a regular nnormal first countable space all of whose subspaces of smaller cardinalities are nor

It was known that such a non-reflection of non-normality occurs forλ regular whenNR(λ) holds (see [18, Theorem 1]). Thus our result provides new example for sincardinals.6

We also prove that if a forcing adds a branch in an Aronszajn subtreeA of T witnessingA(T ), thenF(T ), a peculiar nonreflecting nonmetrizable space, becomes a metrspace. Hence we obtain

Forcing with a Suslin tree makes certain non-normal, locally metrizable, noncollecwise Hausdorff space a metrizable one.

This consistently answered a question of S. Watson (Q. 146 [34] see [24] andHowever a ZFC example of non-normal space which becomes metrizable after addominating real was independently found by Fleissner [13]. Note that in our casedone without adding a real.

The notation is fairly standard, for all unexplained symbols or notions see [19] o30,32] in [20]. Lim(ω1) denotes the set of all countable, limit ordinals. IfT is a tree, thenLevα T denotes theα-th level of T. If t ∈ Levα T andβ < α, thent (β) � t , t (β) ∈ Levβ T

andt � β = {t (γ ): γ < β}. The wordtreemeans, throughout this paper, a normal7 ω1-tree.By branch throughT , we understand anω1 branch, i.e., an increasing functionb from

5 As in the case of P. Hamburger’s question, the question whether there is a regular, first countable(< λ)-collectionwise Hausdorff, non-collectionwise Hausdorff space forλ greater thanω1 is open in ZFC (see [11,15])The strongest negative result says that it is consistent that there are no such spaces that are locallyω2-c.c. andwas obtained under the assumption of the existence of a huge cardinal with a supercompact cardinal(see [30]).

6 Again whether there is a ZFC first countable, regular, nonormal space whose all subspaces of sizeω1 arenormal, seems open.

7 A tree of heightω1 with countable levels is normal if and only if (a) wheneverα < Lim(ω1), t1, t2 ∈ Levα T

and t1 � α = t2 � α, then t1 = t2, (b) for everyα < β < ω1 there aret1, t2 ∈ Levβ T such thatt1 �= t2 andt1(α) = t2(α).


t

the

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implified

ω1 into T , such thatb(α) ∈ Levα T . The collection of all branches of a treeT is denotedby Br(T ). If F ⊆ Br(T ), θ ∈ ω1, thenF(θ) = {t ∈ Levθ T : ∃b ∈ F b(θ) = t}. If a ⊆ Levθ ′ ,θ ′ � θ , thena(θ) = {t (θ): t ∈ a}. Wheneverf denotes a function,f ′′(X) denotes theimage ofX underf andf1, f2 denote compositions off with the projections on the firsor the second coordinate respectively, if the range off is a product of two sets.

2. F(T ) and the ideal of light sets

Definition 1. Let T be anω1-tree with many branches. LetF ∈ [Br(T )]<ω andθ � θ ′ <

ω1. Then

(a) Fθ (T ) = {X ∈ Br(T )]ω: ∀t ∈ Levθ T ∃!b ∈ X t ∈ ran(b)}.(b) F(T ) = ⋃

θ<ω1Fθ (T ).

(c) (θ, θ ′] = ⋃θ<θ1�θ ′ Fθ1(T ), [θ, θ ′] = ⋃

θ�θ1�θ ′ Fθ1(T ).(d) U(θ,F ) = {X ∈ [0, θ ]: F ⊆ X}.

The familyF(T ) may or may not be stationary; we will deal with this problem innext section. The morass-like properties8 of F(T ) are summarized in the following

Fact 2. F(T ) is a cofinal, well-founded family in[Br(T )]ω such thatr(X) = θ iff X ∈Fθ (T ). Moreover there is a collection of1–1 functionsfX :X → ω for X ∈ F(T ), suchthat wheneverrank(X) = rank(Y ), b ∈ X ∩ Y , thenfX(b) = fY (b).

Proof. First let us prove the cofinality ofF(T ) in [Br(T )]ω. Let X ∈ [Br(T )]ω, find θ <

ω1 such that for everyb1, b2 ∈ X if b1 �= b2 thenb1(θ) �= b2(θ). Now it is easy to findY ∈Fθ (T ) such thatX ⊆ Y .

To prove thatF(T ) is well-founded and rank(X) = θ iff X ∈ Fθ it is enough to seethatFθ (T ) is exactly the collection of all minimal elements ofF(T ) − [0, θ). To see thisone needs too observe thatFθ (T ) is formed by incomparable elements and any elemenF(T ) − [0, θ) includes one of them.

Now definefX ’s for X ∈ F(T ). Let fθ : Levθ → ω be a 1–1 function,then putFX(b) =fθ (b(θ)) if rank(X) = θ . �Fact 3. Suppose thatG is a subset ofF(T ) such that there is a branchb ∈ Br(T ) such thatb /∈ ⋃

G. Then the functionfG :G → (Br(T ) − {b}) defined by

fG(X) = c iff X ∈Fθ , c ∈ X, b(θ) = c(θ),

is a regressive function such that for everyc ∈ Br(T ) there isθ < ω1 such that

(i) f −1G ({c}) ⊆ [0, θ ],

(ii) fG(X) = c, rank(Y ) � rank(X), c ∈ Y ⇒ fG(Y ) = c.

8 To see the morass-like character of these properties the reader is referred to the definitions of a smorasses in [33] and semimorasses in [22].


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Proof. Note thatfG is a well-defined regressive function. Now consider the preimaf −1G ({c}) for c’s in the image offG . Takeθ such thatc(θ) �= b(θ), then ifX /∈ [0, θ ], then

fG(X) �= c. SofG satisfies (i).If f (X) = c, rank(Y ) � rank(X), c ∈ Y , then c(rank(Y )) = b(rank(Y )) because

c(rank(X)) = b(rank(X)), sof (Y ) = c and this completes the proof.�Now we are going to define an ideal on[Br(T )]ω or [λ]ω and prove thatF(T ) does not

belong to this ideal while every its subset of smaller size may belong to this ideal. Wneed several definitions and lemmas.

Definition 4. A function r : [λ]ω → ω1 is called a continuous ranking iff it satisfies tfollowing conditions:

(i) a ⊆ b ⇒ r(a) � r(b);(ii) ∀a ∈ [λ]ω ∀α > r(a) ∃b ⊇ a r(b) � α;

(iii) If (an: n ∈ ω) is increasing, thanr(⋃

an) = supn∈ω r(an).

Fact 5. LetT be a tree with many branches. Thenr : [Br(T )]ω → ω1 defined by

r(a) = min{θ : ∃b ∈ Fθ (T ), a ⊆ b

}= min

{θ : ∀b1, b2 ∈ a b1 �= b2 iff b1(θ) �= b2(θ)

}is a continuous ranking. We will refer to this r as to T-ranking.

Proof. The conditions (i) and (ii) are obviously satisfied, the condition (iii) follows frnormality of the tree. �Definition 6. Supposer : [λ]ω → ω1 is a continuous ranking. We say thatF ⊆ [λ]ω isr-heavy iff for everyX ∈ [λ]ω

w(F ,X) = {r(Y ): X ⊆ Y, Y ∈F

}is stationary inω1. The sets which are notr-heavy sets will be calledr-light sets and thefamily of all of them will be denotedLr . We will omit the subscriptr , if the rankingr isclear from the context.

Fact 7. Lr is a countably complete ideal for any continuous rankingr .

Proof. Obviously Lr is closed under subsets. Now let(Fn)n∈ω be a sequence of lighsets, hence for eachn ∈ ω there isXn such thatw(Fn,Xn) is nonstationary, but thew(

⋃n∈ω Fn,

⋃Xn) ⊆ ⋃

n∈ω w(Fn,Xn) is nonstationary, so⋃

n∈ω Fn is light. �Fact 8. If S ⊆ [λ]ω is stationary, then it isr-heavy for every continuous rankingr i.e., Lr ⊆NS for every continuous rankingr .

Proof. If S is not heavy, then there isX ∈ [λ]ω such thatw(S,X) is nonstationary inω1, but by (ii) and (iii) of the definition of continuous ranking (Definition 4) the set{Y ∈


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[λ]ω: rank(Y ) /∈ w(S,X), X ⊆ Y } contains a club set and this contradicts the stationaof S. �

We will see in the last section thatF(T ) may not be stationary, but obviously it is alwaheavy, hence it is possible thatL+

r �= NS wherer is the ranking considered in Fact 5. Nowe will need a series of definitions and lemmas in order to prove our main Lemma 1can be viewed as a generalization of the Pressing Down Lemma that holds for nonsets (see the remark at the end of this section).

Definition 9. Let A be a family of sets. We say thatA has a root∆ iff for every seta inA we have∆ ⊆ a. We say thatA has a saturated root∆ iff it has a root∆ and there existsan infiniteB ⊆ A such that for everyb1, b2 in B we haveb1 ∩ b2 = ∆. We say thatA isann-family for n ∈ ω iff every element ofA has cardinalityn. We say thatA is a regularfamily iff there is a partition ofA = A0 ∪̇ · · · ∪̇Ak and there are sets∆1, . . . ,∆k for k < ω

such thatA0 is finite and eachAi has a saturated root∆i for 1� i � k.

Lemma 10. Everyn-family is a regular family.

Proof. First note that a family which has an infinite pairwise disjoint subfamily is a regfamily with saturated root∅. The proof of the lemma is by induction onn. The aboveobservation takes care of the casen = 1. Suppose now that the lemma is true for all(n−1)-families andn > 1. Let A be anyn-family. We can w.l.o.g. assume thatA has no infinitepairwise disjoint subfamily, and so any maximal disjoint family is finite, thus therefinite setx = {α1, . . . , αm} which intersects all elements ofA. Now A can be partitionedinto finitely many families

A′i = {a ∈ A: αi ∈ a}

for i = 1, . . . ,m. It is enough to prove that eachA′i is regular. For this it is enough t

consider the(n − 1)-families

Bi = {a − {αi}: a ∈ A′

i

}for i = 1, . . . ,m.

Definition 17. Let T be anω1-tree,S ⊂ ω1. We say thatA = (Aθ : θ ∈ S) is a(T ,n)-family(with supportS, denoted by supp(A) = S) iff for every θ ∈ S, Aθ is a nonemptyn-familyand

⋃Aθ ⊆ Levθ T . If A is a (T ,n)-family with supportS ⊂ ω1, than we say that th

(T ,n)-family B with supportS′ is its closure iff

Bθ ={b ∈ [Levθ T ]n: ∃a ∈

⋃θ ′�θ

Aθ ′ , a(θ) = b

}

for θ ∈ S′ andS′ = {θ ∈ ω1: Bθ �= ∅}.


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Lemma 18. Let A be a(T ,n)-family with stationary supportS ⊆ ω1. Let B be the clo-sure ofA and suppose that for stationary manyθ the setBθ contains no infinite disjoincollection of sets. Then there isΓ ∈ [Br(T )]<ω and a club setC ⊆ ω1 ∩ supp(B) such that

∀α ∈ C ∀a ∈ Bα a(α) ∩ Γ (α) �= ∅.

Proof. SupposeA,B are as in the assumptions of the lemma. First let us notesupp(B) is cocountable. By normality ofT , for everyξ ∈ S ∩ Lim(ω1), there isξ ′ < ξ ,aξ ∈ Aξ such thataξ (ξ

′) ∈ B ′ξ , so we use the Pressing Down Lemma to find unboun

S′ ⊆ S ∩ Lim(ω1) andξ0 ∈ ω1 such thataξ (ξ0) ∈ Bξ0. Now it is not difficult to check thaω1 − ξ0 ⊆ supp(B).

Secondly note that it is enough to findΓ ∈ [Br(T )]<ω and an unbounded setC ⊂ ω1 ∩supp(B) such that

∀α ∈ C ∀a ∈ Bα a(α) ∩ Γ (α) �= ∅.

To see this, takeC′ the closure ofC (which satisfies the statement of the lemma) inω1,and, applying normality ofT , check thatC′ satisfies the statement of the lemma as wBy Theorem 10 everyBθ , for θ ∈ S can be presented as

Bθ = B0θ ∪̇ · · · ∪̇B

kθ

θ ,

whereB0θ is finite andBi

θ has a saturated root∆iθ , for 1� i � kθ . By the assumption,∆i

θ isnonempty for 1� i � kθ andθ in some stationary setS′. By thinning outS′, find stationaryS0 ⊆ S′ ∩ Lim(ω1), k ∈ ω such that for everyθ1, θ2 ∈ S0 we havekθ1 = k, |∆i

θ1| = |∆i

θ2|,

|B0θ1

| = |B0θ2

|, for all 1 � i � k. For everyθ ∈ S0 chooseCθ ∈ [Bθ ]<ω of fixed size, sayl ∈ ω such that

(a) B0θ ⊆ Cθ .

(b) |Biθ ∩ Cθ | > |B0

θ | + k + 1 for all 1� i � k.(c) a ∩ b = ∆i

θ for all a, b ∈ Biθ ∩ Cθ and 1� i � k.

Note that sinceT is normal andS0 ⊆ Lim(ω1) for everyθ ∈ S0 there isθ ′ < θ such that|⋃Cθ(θ

′)| = |⋃Cθ |. Thus we can use the Pressing Down Lemma and the fact thalevels of T are countable to find a stationaryS1 ⊆ S0, θ0 � minS1 such that for everyθ1, θ2 ∈ S1.

(d) |⋃Cθ1(θ0)| = |⋃Cθ1|.(e) (Bi

θ1∩ Cθ1)(θ0) = (Bi

θ2∩ Cθ2)(θ0) for 0� i � k.

(f) (∆iθ1

)(θ0) = (∆iθ2

)(θ0) for 1� i � k.

RedefineCθ0 = Cθ(θ0), B0θ0

= B0θ (θ0), ∆i

θ0= ∆i

θ (θ0), for θ ∈ S1. Let Dθ be the family of

all minimal, with respect to inclusion, elements ofB0θ ∪ {∆i

θ : 1� i � k} for θ ∈ S1 ∪ {θ0}.

Claim 1. Dθ1(θ0) = Dθ2(θ0) for all θ1, θ2 ∈ S1.


e

of

By (d) and (e), the restriction of the projection from Levθm to Levθ0 is an isomorphismfrom

⋃Cθm onto

⋃Cθ0. Now by e), the minimal elements ofDθm must be exactly the

minimal elements ofDθ0.

Claim 2.⋃

Dθ1 = ⋃Dθ2(θ1) for all θ1 ∈ θ2 in S1.

By (a) and (d)–(f)|⋃Dθ1| = |⋃Dθ2(θ1)| for all θ1 ∈ θ2 in S1. So it is enough to showthat

⋃Dθ1 ⊇ ⋃

Dθ2(θ1) for all θ1 ∈ θ2 in S1. So it suffices to show that for everyd ∈ Dθ2

we haved(θ1) ⊆ ⋃Dθ1. Note that

(g) a ∈ Dθ1, a ⊆ d(θ1) ⇒ a = d(θ1).

Otherwise, we would havea(θ0) ⊂ d(θ0), but by Claim 1 there isd ′ ∈ Dθ2 such thatd ′(θ0) = a(θ0) ⊂ d(θ0), henced ′ ⊂ d . We get thatd ′(θ0) ⊂ d(θ0) and this contradicts thminimality of d . This completed the proof of (g). Now suppose thatd(θ1) �⊆ ⋃

Dθ1, i.e.,there ist ∈ d(θ1) − ⋃

Dθ1. The contradiction that we will obtain will complete the proof Claim 2.

Case 1. d ∈ B0θ2

.

By the decomposition of our families, eitherd(θ1) ∈ B0θ1

or d(θ1) ∈ Biθ1

for 1 � i � k

and then∆iθ1

⊆ d(θ1), so in both cases there isa ∈ Dθ1 such thata ⊆ d(θ1), but t /∈ a, acontradiction with (g).

Case 2. d = ∆iθ2

for some1� i � k.

By (d) and (b) there is 1� j � k such that

∣∣(Cθ2 ∩ Biθ2

)(θ1) ∩ B

jθ1

∣∣ � 2.

By (c) takeb, c ∈ Biθ2

∩ Cθ2 such thatb ∩ c = ∆iθ2

= d , andd(θ1) = b(θ1) ∩ c(θ1). Now

∆jθ1

⊆ b(θ1) ∩ c(θ1), so there isa ∈ Dθ1 such thata ⊆ ∆jθ1

⊆ d(θ1) and t ∈ d(θ1) − a, acontradiction with (g). This finishes the proof of Claim 2.

Now it is easy to see that the sets

b′t = {

s ∈ T : ∃θ ∈ S1 ∃d ∈ Dθ s ∈ d, s(θ0) = t}

are uncountable chains fort ∈ ⋃Dθ0, since otherwise ifθ1 � θ2, s1 � s2, s1 ∈ d1 ∈ Dθ1,

s2 ∈ d2 ∈ Dθ2 (θ1 = θ2 is excluded by (d)), by Claim 2 there iss3 ∈ ⋃Dθ2 such that

s3(θ1) = s1, but thens3(θ0) = s2(θ0) = t and this contradicts (d). Letbt ∈ Br(T ) be suchthatbt

′ ⊆ ran(bt ). We putΓ = {bt : t ∈ ⋃Dθ0}. Now, of courseS1 is unbounded inω1 and

if α ∈ S1, a ∈ Bα then there isd ∈ Dα such thatd ⊆ a, but then there isbt ∈ Γ such thatbt (α) ∈ d , soa(α) ∩ Γ (α) �= ∅, and this completes the proof of the lemma.�


e that

e

re

Lemma 19. SupposeT has many branches andG ⊂ F(T ) is r-heavy for theT -rankingr .Letf :G → [Br(T )]<ω be a regressive function. Then there isF ∈ [Br(T )]<ω such that foreachG ∈ [Br(T )]<ω{

r(X): X ∈ G, r(G ∪ F ∪ f (X)

) = r(G ∪ F)}

is unbounded inω1.

Proof. Since the ideal of light sets is countably complete we may w.l.o.g. assumf :G → [Br(T )]n for somen ∈ ω. We will prove the lemma by induction onn ∈ ω. Sup-pose we are done for allk < n. Consider the(T ,n)-family (see Definition 17)

A = {F(θ): r(X) = θ, F = f (X), X ∈ G

}.

Let B be a closure ofA.

Claim 1. If Bθ contains infinitely many pairwise disjoint elements for club manyθ ’s, thenf satisfies the lemma.

Proof. Let C ⊆ ω1 be a club set such that for everyθ ∈ C the setBθ contains an infinitedisjoint family. LetCk

θ ⊆ Bθ be firstk elements of those families fork ∈ ω. By the PressingDown Lemma and the normality ofT , we may find a stationarySk ⊆ C andθk < ω1 suchthat for eachk ∈ ω andθ ∈ Sk we have thatCk

θ (θk) ⊂ BθkandCk

θ (θk) is a disjoint familyof sizek.

Now, letF ∈ [Br(T )]2 be such thatr(F ) > θ ′ = sup{θk: k ∈ ω}, i.e., |F(θ ′)| = 1. Takeany G ∈ [Br(T )]<ω. Let k = |G| + 3. We claim that for everyα ∈ ω1 there is anX ∈ Gsuch thatr(X) > α and

r(G ∪ F ∪ f (X)

) = r(G ∪ F).

Let θ ∈ Sk be such thatθ > ζ = r(G∪F) � r(F ) > θ ′ andθ > α. As θ ∈ Sk , there must beX1, . . . ,Xk ∈ G such that{f (Xi)(θ): 1 � i � k} = Ck

θ , then by the disjointness, for som1� i � k we have

f (Xi)(ζ ) ∩ (G ∪ F)(ζ ) = ∅hencer(G ∪ F ∪ f (Xi)) = r(G ∪ F). This completes the proof of Claim 1.�

We come back to the proof of the lemma. By Claim 1 and the fact thatG is heavy (hencesupp(A) is stationary) we may w.l.o.g. assume that for stationary manyθ ’s the setBθ doesnot contain an infinite disjoint subfamily, hence we may apply Lemma 18 forA, obtainingC andΓ0 as in Lemma 18. Forγ ∈ Γ0 define

Gγ = {X ∈ G: γ ∈ f (X)

}.

Claim 2. Gγ is heavy for someγ ∈ Γ0.

Proof. SinceL is countably complete, it is enough to show thatE = w(G−⋃γ∈Γ0

Gγ ,Γ0)

is nonstationary. We will prove thatE ∩C = ∅. If θ ∈ E, there isX ∈ G such thatr(X) = θ

andΓ0 ⊆ X but Γ0 ∩ f (X) = ∅. ThenΓ0(θ) ∩ f (X)(θ) = ∅, because otherwise there a


e

.

distinctb ∈ f (X), γ ∈ Γ0, γ (θ) = b(θ), γ, b ∈ X, a contradiction with the definition ofr .We conclude that ifθ ∈ E, then there isa ∈ Aθ ⊆ Bθ (a = f (X)(θ)) such thatΓ0(θ) ∩a = ∅. By the property ofC obtained from Lemma 18, this implies thatθ /∈ C and thiscompletes the proof of Claim 2.�

We come back to the proof of the lemma. By Claim 2 fixγ ∈ Γ0 such thatGγ is heavy.Consider a regressive functionfγ :Gγ → [Br(T )]n−1 defined by

fγ (X) = f (X) − {γ }.By the inductive assumption there isFγ ∈ [Br(T )]<ω such that{

r(X): X ∈ Gγ , r(G ∪ Fγ ∪ fγ (X)

) = r(G ∪ Fγ )}

is unbounded inω1 for everyG ∈ [Br(T )]<ω. Now F = Fγ ∪ {γ }.WheneverG ∈ [Br(T )]<ω andα ∈ ω1, there isX ∈ Gγ ⊆ G such that

r((

G ∪ {γ }) ∪ Fγ ∪ fγ (X)) = r

((G ∪ {γ }) ∪ Fγ

)so

r(G ∪ (

Fγ ∪ {γ }) ∪ (fγ (X) ∪ {γ })) = r

(G ∪ (

Fγ ∪ {γ }))so

r(G ∪ F ∪ f (X)

) = r(G ∪ F)

and this completes the proof of Lemma 19.�Definition 20. Let T be a tree with many branches andr theT -ranking. LetG ⊆ [Br(T )]ω.We say that a regressive functionf :G → ⋃

G has forking preimages if it satisfies thconclusion of Lemma 19.

Fact 21. SupposeKHλ(T ) holds. LetG ⊂ F(T ) be such that|G| < |F(T )| = λ. Then thereis a regressive functionf :G → ⋃

G whose preimages do not fork.

Proof. Since|G| < λ, there isb ∈ Br(T ) such thatb /∈ ⋃G. Let f = fG be as in Fact 3

Fix F ∈ [Br(T )]<ω. Findb1, b2 ∈ Br(T ), such that there areθ1 < θ2 < ω1 such that

b1(θ1) = b2(θ1) = b(θ1), b1(θ2) �= b2(θ2),

b(θ2) = b2(θ2), b2(θ2 + 1) �= b(θ2 + 1)

and that

r(F ∪ {b1, b2}

) = r({b1, b2}

)� θ2.

PutG = {b1, b2}. Note thatr(F ∪ G) = r(G) � θ2. Now, suppose thatr(X) > θ2, we willshow thatr(F ∪ G ∪ f (X)) > r(F ∪ G). By the definition off we have thatf (X) = c,wherec(r(X)) = b(r(X)), butc(θ2) = b(θ2) = b2(θ2), c(θ2 + 1) = b(θ2 + 1) so,

r(G ∪ F ∪ f (X)

)� r

(G ∪ f (X)

)> θ2 � r(G) = r(G ∪ F). �


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e

r

Note that Lemma 19 is a weak version of the Pressing Down Lemma that holdheavy sets. Namely, suppose that a regressive function is constant on a stationary sG.PutF = f (X) for anyX ∈ F . Obviously for eachG ∈ [Br(T )]<ω, X ∈ G we haver(G ∪F ∪ f (X)) = r(G ∪ F), and{r(X): X ∈ G} are unbounded inω1, sinceG as a stationaryset it heavy (see Fact 8).

3. Topology on F(T )

In this sectionT also stands for anω1-tree with many branches.

Definition 22. We define a topology onF(T ) by declaring the basis of open sets toB(T ) = {U(θ,F ): F ∈ [Br(T )]<ω, θ < ω1} ∪ {∅}.

Fact 23. B(T ) is a basis for topology onF(T ).

Proof. U(θ1,F1) ∩ U(θ2,F2) = U(min{θ1, θ2},F1 ∪ F2). �We will show that the space(F(T ),B(T )) has many properties of the space(F ,B(T ))

considered in [22], whereF is a stationary(ω1, λ)-semimorass althoughF(T ) may notbe stationary. This possibility having some of the properties of a stationary set whihaving the others will be essential in Section 5. In general the topological propertF(T ) are supposed to resemble the properties of the ordinal topology onω1, with the sets(θ, θ ′] playing the same role as the corresponding subsets ofω1.

Fact 24. For eachθ1 < θ2 < ω1 (θ1, θ2] is clopen.

Proof. First prove that(θ1, θ2] is open. LetX ∈ (θ1, θ2], so we haveθ1 < r(X) �θ2 < ω1, and by the normality of the tree there are distinctt1, t2 ∈ Levr(X) T such thatt1(θ1) = t2(θ1). Find b1, b2 ∈ X such thatti ∈ ran(bi) for i = 1,2. Now it is easy to sethat X ∈ U(r(X), {b1, b2}) ⊆ (θ1, θ2]. For closure note that(θ1, θ2] = F(T ) − ([0, θ1] ∪⋃

θ2<θ<ω1(θ2, θ ]). �

Fact 25. B(T ) is a basis of clopen sets, hence(F(T ),B(T )) is a zero-dimensional, regulaspace.

Proof. First we prove that the setsU(θ,F ) are closed. TakeX ∈ F(T ) − U(θ,F ). Ifr(X) > θ then(θ, r(X)] ∩ U(θ,F ) = ∅ andX ∈ (θ, r(X)] and the set(θ, r(X)] is openby Fact 24. Ifr(X) � θ , then there ist ∈ Levθ T andb ∈ F, b′ ∈ X, b �= b′ such thatt ∈ ran(b) ∩ ran(b′), b /∈ X. ThenU(r(X), {b′}) ∩ U(θ,F ) = ∅.

So we are left with the proof that our space is aT1 space. LetX �= Y ; X,Y ∈ F(T ).Suppose thatr(X) < r(Y ), Then we can separateX andY by Fact 24. So, supposer(X) =r(Y ) = θ . Take t ∈ Levθ T such that there areb1 �= b2, t ∈ ran(b1) ∩ ran(b2), b1 ∈ X,b2 ∈ Y . ThenU(θ, {b1}) andU(θ, {b2}) separateX andY . �


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,

sat

Fact 26. [0, θ ] is a metrizable space as a subspace of(F(T ),B((T )).

Proof. It is enough to see that{U(θ1,F ): θ1 � θ , F ∈ [Br(T )]<ω} is aσ -discrete basis fo[0, θ ]. Clearly it is a basis asU(θ1,F ) ∩ [0, θ ] = U(min(θ1, θ),F ). Now fix θ1 � θ anda ∈ [Levθ1(T )]<ω it is enough to show thatBa = {U(θ1,F ): F(θ1) = a} a discrete familyof sets. It is clearly disjoint. Now letX /∈ ⋃

Ba if r(X) > θ1, then(θ1, θ ] isolatesX from⋃Ba . If r(X) = θ2 � θ1, andX /∈ ⋃

Ba , it means that|a(θ2)| < |a| which implies that[0, θ2]∩U(θ1,F ) = ∅ for anyU(θ1,F ) ∈ Ba which completes the proof of the discreteneof Ba , asX ∈ [0, θ2]. �Fact 27. (F(T ),B(T )) is locally metrizable.

Proof. For eachθ < ω1, F ∈ [Br(T )]<ω we have thatU(θ,F ) ⊂ [0, θ ], now useFact 26. �Fact 28. Suppose thatG is a subspace ofF(T ) such that there is a branchb ∈ Br(T ) suchthatb /∈ ⋃

G. ThenG is metrizable.

Proof. Define a functionf :G → (Br(T ) − {b}) by

f (X) = c iff r(X) = θ, c ∈ X, b(θ) = c(θ).

Note thatf is a well-defined regressive function. Now consider the preimagesf −1({c})for c’s in the image off noting thatf −1({c}) = U(∆(b, c), {c}) where∆ is the first or-dinal whereb andc split. We will show that they constitute a partition ofG into clopen,metrizable sets, which is sufficient condition for the metrizability ofG.

To see thatf −1({c}) is metrizable for everyc in the image off , note thatf −1({c}) ⊆[0, θ ], for someθ < ω1. Hence, using Fact 26, the preimages of single branches are mable. Since{f −1(c): c ∈ Im(f )} is a disjoint family, it is enough to show that this isfamily of open sets, but this follows from Fact 3(ii).�Fact 29. SupposeKHλ(T ) holds. ThenF(T ) is a space of sizeλ such that wheneveG ⊂ F(T ), |G| < λ, thenG is contained in a clopen metrizable subspace and hencitself metrizable.

Proof. If |G| < λ, then⋃

G is a proper subset of Br(T ), sayb /∈ ⋃G so, by Fact 28

the spaceG is metrizable. Now note thatH = {X ∈ F(T ): b /∈ X} is clopen. It is closedbecause the setsU(θ, {b}) are open forθ < ω1. To see that it is open, letX ∈ H, let c ∈ X

be such thatc(r(Y )) = b(r(Y )), thenX ∈ U(r(Y ), {c}) ⊆ H. �Fact 30. If KHλ(T ) holds, cf(λ) > ω1, thenF(T ) is weakly collectionwise Hausdorff.

Proof. LetX be a closed discrete subspace ofF(T ) of size� λ. Since all small subspaceare metrizable, hence they are collectionwise Hausdorff we may w.l.o.g. assume thX isactually of sizeλ. By Fact 26 the set[0, θ ] ∩X can be separated for everyθ < ω1 < λ, soif cf(λ) > ω1 one of those sets is of sizeλ, and this completes the proof.�


ble.

a.

f

s

ts

Fact 31. (F(T ),B(T )) is notσ -paraLindelöf, hence nonparacompact and nonmetriza

Proof. SupposeF(T ) wereσ -paraLindelöf, then there would beV = ⋃n∈ω Vn a refine-

ment ofB(T ) such that eachVn is locally countable. By Fact 7 there isn ∈ ω such that⋃Vn ∩F(T ) is heavy. Define a regressive functionf :

⋃Vn ∩F(T ) → [Br(T )]<ω by

f (X) = a if ∃V ∈ Vn U(r(X), a

) ⊆ V.

Now apply Lemma 19 forf , obtaining anF ∈ [Br(T )]<ω as in the statement of this lemmNow take anyX ∈ ⋃

Vn such thatF ⊆ X (by heaviness of⋃

Vn). Now, sinceVn is locallycountable, there isG ∈ [X]<ω such that∣∣{V ∈ Vn: V ∩ U

(r(X),G

) �= ∅}∣∣ � ω.

Since for eachV ∈ Vn there isU(θ,H) ∈ B(T ) such thatV ⊂ U(θ,H), there isα ∈ ω1such that

if V ∩ U(r(X),G

) �= ∅, thenV ⊆ [0, α].Now, by Lemma 19, takeY ∈ ⋃

Vn such thatr(Y ) = r(Y ) > α, r(G ∪ f (X) ∪ F) and

r((

G ∪ f (X)) ∪ F ∪ f (Y )

) = r(G ∪ f (X) ∪ F

).

Hence, there isZ ∈ F(T ) such thatr(Z) = r(G ∪ f (X) ∪ F) such thatG ∪ f (X) ∪F ∪ f (Y ) ⊆ Z. Note that sincer(X) � r(f (X)), andF ∪ G ⊂ X we have thatr(Z) =r(G ∪ f (X) ∪ F) � r(X), alsor(Z) = r(G ∪ f (X) ∪ F) < r(Y ), andf (Y ) ⊂ Z, thus

Z ∈ U(r(X),f (X)

) ∩ U(r(Y ), f (Y )

).

SoU(r(Y ), f (Y )) ⊆ V ⊆ [0, α], for someV ∈ Vn by the choice ofα but r(Y ) > α by thechoice ofY , a contradiction. �Fact 32. SupposeG ⊂ F(T ) is heavy. Then there is no disjoint cover ofG by elements oB(T ).

Proof. Suppose the opposite, letV be a disjoint cover ofG by elements ofB(T ). Define aregressive functionf :G → [Br(T )]<ω such that

∀X ∈ G ∃V ∈ V U(r(X),f (X)

) ⊂ V,

Now apply Lemma 19 and findF as in this lemma. Now findX ∈ G such thatF ⊂ X. Letθ be such thatV ⊆ [0, θ ] if X ∈ V andV ∈ V . Apply Lemma 19 to findY ∈ G, r(Y ) > θ

such that

r(f (X) ∪ F ∪ f (Y )

) = r(f (X) ∪ F

).

But r(f (X) ∪ F) � r(X) becausef (X),F ⊆ X, so there isZ ∈ F(T ) such thatr(Z) =r(X) � θ < r(Y ) andf (X),f (Y ) ⊆ Z. HenceZ ∈ V1 ∩ V2, whereV1,V2 ∈ V containrespectivelyX and Y . But V1 �= V2 becauser(Y ) > θ , a contradicting the disjointnesof V . �Fact 33. SupposeGi ⊂ F(T ) are heavy fori = 1,2, then there are no open disjoint seU1,U2 such thatGi ⊂ Ui for i = 1,2.


or

st

]),

-a

remis notntablectf non-Balogh

ar,

sace.e.

Proof. Suppose the opposite, that there areU1,U2, disjoint, such thatGi ⊂ Ui for i = 1,2.Let fi :Gi → [Br(T )]<ω be regressive functions such that

U(r(X),fi(X)

) ⊆ Ui for X ∈ Gi .

Apply Lemma 19, we obtainFi ∈ [Br(T )]<ω as in this lemma. Now findX1 ∈ G1 such thatr(X1) > r(F1 ∪ F2) and

r(F1 ∪ F2 ∪ f (X1)

) = r(F1 ∪ F2)

and then findX2 ∈ G2 such thatr(X2) > r(F1 ∪ F2) and

r(F2 ∪ (

F1 ∪ f (X1)) ∪ f (X2)

) = r(F2 ∪ (

F1 ∪ f (X1)))

.

Hencer(F2 ∪ F1 ∪ f (X1) ∪ f (X2)) = r(F1 ∪ F2). So, there isZ ∈F(T ), r(Z) = r(F1 ∪F2) � r(Xi) for i = 1,2 and(F1 ∪ F2 ∪ f (X1) ∪ f (X2)) ⊆ Z so

Z ∈ U(r(X1), f (X1)

) ∩ U(r(X2), f (X2)

)henceU1 ∩ U2 �= ∅, a contradiction. �Fact 34. If KHλ(T ) holds andλ > ω1, thenF(T ) is not a generalized ordered space na monotonically normal space.

Proof. First we will prove thatF(T ) does not contain a closed copyX ⊆ F(T ) of astationary setS ⊆ κ for any uncountable, regular cardinalκ . Suppose the opposite. Firconsider the caseκ = ω1; then |X| = ω1, hence by Fact 29 and becauseλ > ω1, X is ametrizable subspace ofF(T ), but stationary subsets ofω1 are not paracompact (see [17hence nonmetrizable, a contradiction. Ifκ > ω1, then considerXθ = X ∩[0, θ ] for θ < ω1.Since the ideal of nonstationary sets ofκ is κ-complete, someXθ corresponds to a stationary Sθ ⊆ S, but againXθ is metrizable by Fact 29, andSθ is not paracompact asstationary subset of an uncountable cardinalκ , a contradiction.

In order to conclude thatF(T ) is not a generalized ordered space we use a theoof Engelking and Lutzer (see [12]) which says that a generalized ordered spaceparacompact iff it contains a closed copy of a stationary subset of a regular uncoucardinal. In order to conclude thatF(T ) is not monotonically normal (that implies in fathatF(T ) is not linearly orderable (see [16]) we use the analogous characterization oparacompactness of monotonically normal spaces that was recently obtained by Z.and M.E. Rudin (see [1]). �

As a corollary of the above results we obtain the following

Theorem 35. SupposeKHλ holds forλ > ω1. Then there is a zero-dimensional, regul(weakly collectionwise Hausdorff, if cf(λ) > ω1) locally metrizable, non-σ -paraLindelöf,hence nonparacompact, nonmetrizable spaceF(T ) of sizeλ such that every one of itsubspaces of size less thanλ is metrizable and included in a clopen metrizable subspMoreoverF(T ) is not a generalized ordered space nor a monotonically normal spac


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arity

Fact 36. Under a large cardinal assumption, it is consistent thatKHω2 holds, i.e., thereare spaces as in Theorem35but there are no nonreflecting nonmetrizable space of sizω2which is a generalized ordered space or is monotonically normal.

Proof. The characterizations of paracompactness in generalized ordered spaces otonically normal spaces from [12] to [1] imply that the nonmetrizability of a nonreflecnonmetrizable space of sizeω2 yields its subspace which is homeomorphic to a stationsubsetS of a regular cardinal. As all subspaces of cardinalityω1 of such a space must bmetrizable and so first countable, it follows that the stationary setS is in ω2 and consistsof ordinals of countable cofinality. Moreover, the nonreflection of nonmetrizability impthatS ∩α is nonstationary for eachα ∈ ω2. So, it is enough to prove that KHω2 is consistentwith the nonexistence of such setsS which we denote by¬NR(ω2).

Assume thatκ is an inaccessible cardinal andλ > κ is a weakly compact cardinal. Firwe follow [2] to obtain a model whereκ is inaccessible and for every stationary setS ⊆ κ+there is anα < κ+ with cf(α) = κ such thatS ∩ α is stationary inα. This is done by Levycollapsingλ to κ+ (with conditions of cardinalities less thanκ). Let us call the obtainegeneric extension the intermediate model. The cardinalκ remains inaccessible, since tforcing isκ-closed. Similarly as in [2, Corollary 7.9] (where, however, a Levy collapsω2 instead of ourκ+ is described), the stationary reflection is obtained at anα < λ whichis an inaccessible cardinal in the ground model (see [2, Lemma 7.6]). The fact thforcing isκ-closed implies thatcf(α) = κ in the intermediate model.

Now one Levy collapsesκ to ω1 (with finite conditions), i.e.,κ+ becomesω2. It isknown that there is a Kurepa tree in the final model (see [19, VII 8.9]). As the forhas cardinalityκ , and the ideal of non-stationary subsets ofκ+ is κ+-complete, any stationary subsetA of κ+ in the generic extension contains a stationary intermediate msetA′. Using the stationary reflection in the intermediate model, there areα ∈ κ+ suchthat cf(α) = κ andA′ ∩ α is stationary inα. But as the final Levy collapse isκ-c.c., theintermediate model setA′ ∩ α stationary in an ordinalα of cofinalityκ remains stationaryin the generic extension (see [2, Lemma 7.5] or [19, Ex. VII H.2]), i.e., in particularA ∩ α

is stationary inα which completes the proof of¬NR(ω2) in the final model. �

4. The space F(T ) when the set F(T ) is stationary

In this section we begin to study the question of how properties ofT may affect topo-logical properties ofF(T ). We assume in this section thatS(T ) holds i.e., thatF(T ) is astationary set in[Br(T )]ω. In the next section we will assumeA(T ), i.e., thatT containsan Aronszajn subtree. The paper [31] implicitly contains the following two propositio

Proposition 37 [31]. Let λ be an uncountable cardinal. It is consistent(with λω = λ, ifcf(λ) > ω) with any cardinal arithmetic consistent withλ � 2ω1 that there is a Kurepa treewith exactlyλ many branches such thatF(T ) is stationary.

Proof. ObtainT by forcing as in [31] over a model of GCH. The stationarity ofF(T ) canbe easily concluded from the construction in [31]. For more explicit proof of the station


e

g thatr

ot

.

reyse

ly

of the

n of

f allspacetedwise

of F see [22]. The forcing is of sizeλω and isω1-Baire andω2-c.c.. Hence we havCH+2ω1 = λω1, 2α = λω1 for ω1 � α < λ, and 2α = α+ for λ � α. Now we may obtaina generic extension of the resulting model where the cardinal arithmetic is anythinis consistent with 2ω1 � λ, by forcing with a product of powerfully c.c.c. forcing (foexample by adding some Cohen reals) and anω1-closed forcing. Those forcings do nadd branches throughω1-trees (see [9]), hence Br(T ) and consequentlyF(T ) is preservedmoreover those forcings as proper forcings preserve stationarity ofF(T ) (see [3]). �Proposition 38 [31]. If F(T ) is stationary, thenT does not contain Aronszajn subtrees

Proof. SupposeA is an Aronszajn subtree ofT . By stationarity ofF(T ) we can find amodelM such thatM ≺ H(|Br(T )|+), A,T ∈ M , M ∩Br(T ) = X ∈ F(T ), M ∩ω1 = θ ∈ω1, |M| = ω. Note that by elementarityr(X) = θ . We will show thatA ⊂ ⋃

θ ′<θ Levθ ′ T ,contradicting uncountability ofA. Suppose that there ist ∈ A such thatt ∈ Levθ T . Takeb ∈ X such thatb(θ) = t . ThenM |= A is unbounded inran(b), so by elementarityA isunbounded inb, a contradicting the fact thatA does not contain an uncountable branch.�

Let us note that there are in ZFCω1-trees with no Aronszajn subtrees (see [8]).

Proposition 39. S(T ) impliesNR([λ]ω).

Proof. The nonreflecting stationary set in[λ]ω witnessing NS([λ]ω) is an isomorphic copy(under an isomorphism between Br(T ) andλ) of F(T ) for T such that KHλ(T ) andS(T )

hold. It is stationary sinceS(T ) holds. Now, ifY ⊆ Br(T ) is a proper subset, then theis a regressive functionf from G = [Y ]ω ∩ F(T ) into Br(T ) as in Fact 3, i.e., for everc ∈ Br(T ), f −1(c) ⊆ [0, θ ], hence preimages of points underf are nonstationary (becauthey are of bounded rs). SoG cannot be a stationary subset of[Y ]ω, because it wouldcontradict the Pressing Down Lemma (see [3]).�Fact 40. SupposeS(T ) holds, then(F(T ),B(T )) is a collectionwise normal, nonperfectnormal, nonmetaLindelöf space.

Proof. The proofs of all these properties are based on the Pressing Down Lemmafollowing form: If F is stationary in[A]ω, for an uncountable setA andf :F → [A]<ω is aregressive function i.e.,f (X) ∈ [X]<ω, then there is a stationaryS ⊂ F and a finiteF ⊂ A

such thatf � S = F . This version can be easily obtained from a classical formulatiothe Pressing Down Lemma for[A]ω (see [B]), first by assuming w.l.o.g. thatf :F → [A]nfor somen ∈ ω and then applying the usual PDF finitely many times. The proofs othe above properties are similar to the proofs of corresponding properties for the(F ,B(T )), whereF is a(ω1, λ) -semimorass (see [22]). Since we will be mainly interesin the question of normality of our space, we present only the proof of collectionnormality, for other proofs the reader is referred to [22]. We will need the following

Lemma 41. SupposeK ⊆ F(T ) is a closed subspace of(F(T ),B(T )), thenK is closedunder countable unions of increasing sequences if the union belongs toF(T ). If moreover


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he

able,ace ofece.

S(T ) holds, either there is a clopen metrizable setU ⊇ K or there is a club setC ⊆[Br(T )]ω such thatK = C ∩F(T ).

Proof. Note that if (Xn: n ∈ ω) is an increasing sequence, such thatX = ⋃n∈ω Xn ∈

F(T ), then for everyU(r(X),F ) � X there isn ∈ ω such thatXn ∈ U(r(X),F ), henceXis in the closure of(Xn: n ∈ ω). Hence a closed set is closed under countable unionincreasing sequences if the union belongs toF(T ).

Now suppose thatF(T ) − K is stationary. We will show that there exists a clopmetrizable setU containingK . Fix a regressive functionf : F(T )−K → [Br(T )]<ω suchthatf (X) = F if X ∈ U(r(X),F ) ⊆ F(T )−K and find a stationaryS ⊆ F(T )−K and afinite F ⊂ Br(T ) such thatf � S = F . Note that ifX ∈ K thenF �⊆ X, because otherwiswe would takeY ∈ S such thatr(Y ) � r(X) and we would obtain a contradiction withe definition off . SoK ⊆ ⋃

b∈F {X ∈ F(T ): b /∈ X}. The sets{X ∈ F(T ): b /∈ X} areclopen metrizable by Facts 25, 28 henceK is included in a clopen metrizable set.

Now suppose thatF(T ) − K is nonstationary, whileF(T ) is stationary, in particulaK is cofinal in[Br(T )]ω. Define a sequence(Cα: α < ω1) by C0 = K , Cλ = ⋃

α<λ Cα forλ ∈ Lim(ω1) andCα+1 = {a: ∃ increasing(an)n∈ω ⊂ Cα

⋃n∈ω an = a}. SinceC = Cω1

containsK , and is closed under countable unions of increasing sequences,C is club in[Br(T )]ω. So we have to show thatC ∩F(T ) ⊆ K . It easy to show by induction onα < ω1that

∀a ∈ Cα ∀F ∈ [a]<ω ∃b ∈ K F ⊆ b ⊆ a.

So ∀a ∈ C ∩ F(T ) ∀F ∈ [a]<ω K ∩ U(r(a),F ) �= ∅, hence sinceK is closed, we havethatC ∩F(T ) ⊆ K and this finishes the proof of the lemma.�

Now we will prove that ifF(T ) is stationary in[Br(T )]ω, then(F(T ),B(T )) is col-lectionwise normal. LetU be a discrete collection of closed subspaces of(F(T ),B(T )).Let f :F(T ) → [Br(T )]<ω be such thatf (X) = F if |{U ∈ U : U ∩ U(r(X),F )}| � 1.Sincef is a regressive function, we can find a stationaryS ⊆ F(T ) and a finiteF suchthat f � S = F . Now we will show that all but at most one element ofU are containedin the metrizable setV = {X ∈ F(T ): F �⊆ X} (it is metrizable by Fact 28). Suppose topposite, i.e., there areU1,U2 ∈ U andYi ∈ Ui ∩ {X ∈ F(T ): F ⊂ X} for i = 1,2. TakeX ∈ S such thatr(X) � r(Y1), r(Y2), then∣∣{U ∈ U : U ∩ U

(F, r(X)

)}∣∣ > 1

contradicting the definition off . Now we can use collectionwise normality ofV to con-struct the separating family of open sets.�

As a consequence of the above results we obtain the following

Theorem 42. Let λ > ω1 be a cardinal of uncountable cofinality. Suppose thatKHλ(T )

andS(T ) hold, then there is a zero-dimensional, collectionwise normal, locally metrizcountably paracompact, nonmetaLindelöf, nonperfectly normal, nonmetrizable spsizeλ, such that every one of its subspaces of size less thenλ is metrizable. Moreover thspace is not a monotonically normal space hence it is not a generalized ordered spa


of

t-

-

5. The space F(T ) when T contains an Aronszajn subtree

From now on we will be interested inF(T ) for T with an Aronszajn subtree.

Proposition 43. Suppose thatT is anω1-tree with many branches. LetA be an Aronszajntree. Then there is a treeT [A] with the same number of branches asT such thatT [A] hasmany branches andT [A] containsA as a downward closed Aronszajn subtree.

Proof. Let T be as above, and suppose it hasλ many branches. Consider the familyω1 copies ofT indexed by elements ofA. i.e., {Ta: a ∈ A}. Construct a treeT [A] whoseunderlying set isA ∪ ⋃

a∈A Ta and the order is defined bys1 � s2 iff si ∈ Taifor i = 1,2

anda1 = a2, s1 �Ta1s2 or s1 ∈ A ands2 ∈ Ta ands1 � a, i.e., we glueTa abovea, outside

of A. Then obviouslyT [A] is has exactlyλ.ω1 = λ many branches andA is its downwardclosed subtree. Also, ifb is a branch ofT [A] it cannot have range included inA, soA isan Aronszajn subtree, and this completes the proof.�Lemma 44. Let T be a tree with many branches andA ⊆ T a downward closed uncounable subtree ofT . LetC be a club subset ofω1 andS ⊆ C, then

K(S,C,A) = {X ∈ F(T ): r(X) ∈ S, ∀b ∈ X b

(min

(C − (

r(X) + 1)))

/∈ A}

is a closed subspace ofF(T ).

Proof. Let X /∈ K(S,C,A). First suppose thatr(X) = θ ∈ S. Then there isb ∈ X such thatb(min(C − (θ +1))) ∈ A and thenU(θ, {b})∩K(S,C,A) = ∅. If r(X) = θ /∈ S andθ ∈ C,take t ∈ Levθ T ∩ A andb ∈ X such thatt ∈ ran(b), thenU(θ, {b}) ∩ K(S,C,A) = ∅. Ifr(X) = θ /∈ C, then, sinceC is closed, there isθ ′ < θ such that(θ ′, θ ] ∩ K(S,C,A) =∅. �Fact 45. SupposeT is a tree with many branches andA ⊆ T its downward closed Aronszajn subtree. ThenK(S,C,A) as in Lemma44 is heavy, providedS ⊆ C is stationary.

Proof. Let X ∈ [Br(T )]ω, find θ ∈ S such that for allθ ′ � θ we haveb(θ ′) /∈ A (this canbe accomplished, sinceA is an Aronszajn tree). Now for everyθ ′ ∈ S − θ we may findY ∈Fθ ′ ∩ K(S,C,A) (this can be accomplished by the definition ofK(S,C,A)), hence

w(K(S,C,A),X

) ⊇ S − θ

and henceV is heavy.

Fact 46. Let T be a tree with many branches and suppose thatA(T ) holds. ThenF(T ) isnot normal.

Proof. Let A denote an Aronszajn subtree ofT . Let C ⊆ ω1 be such a club set that

∀α ∈ C ∀β < α ∀t ∈ Levβ T ∃b ∈ Br(T ) t ∈ ran(b) & b(α) /∈ A.


f

d

.

ts

C can be found by the standard closure argument. LetSi ⊂ C for i = 1,2, be stationarydisjoint sets. By Lemmas 44, 45K(Si,C,A) are closed and heavy subspaces ofF(T ). Ofcourse the sets are disjoint. Fact 33 implies that they cannot be separated.�Fact 47. SupposeKHλ holds. Then there is a Kurepa treeT with exactlyλ many branchesand with an Aronszajn subtreeA ⊆ T , such thatF(T ) is not collectionwise Hausdorff.

Proof. Let KHλ(T ′) andT = T ′[A]. Let aθ = Levθ T − A, and let tθ be any element oA ∩ Levθ T for everyθ < ω1. Let fθ be a 1–1, onto function such that

fθ :{Y ∈ [

Br(T )]ω: ∃X ∈ Fθ Y ⊆ X, Y(θ) ⊆ aθ

}→ {

b ∈ Br(T ): b(θ) = tθ , b(θ + 1) /∈ A}.

By the construction ofT [A] (see the proof of Proposition 43)fθ as above may be founfor everyθ < ω1. Now for everyθ < ω1, Y ∈ dom(fθ ) fix a setXY ∈ Fθ (T ) such thatY ∪ {fθ (Y )} ⊆ XY andXY (θ + 1) ∩ A = ∅ and finally put

D = {XY : Y ∈ dom(fθ ), θ < ω1

}.

We will prove thatD is a discrete, closed subspace ofF(T ) that cannot be separated.For discreteness and being closed, letX ∈ Fθ (T ). Findb ∈ X such thattθ ∈ b. The set

U(θ, {b}) separatesX from all but possibly one element ofD. Indeed, ifr(XY ) < θ , thensinceXY (r(XY ) + 1) ∩ A = ∅ we have thatb /∈ XY . If r(XY ) = θ , then sincefθ is a 1–1function, then there is at most oneXY such thatfθ (Y ) = b.

To see thatD cannot be separated, it is sufficient, by Fact 32, to prove thatD is heavy.So suppose thatV is a cover ofD. Let Γ ∈ [Br(T )]ω, then there isθ < ω1 such that forall θ ′ > θ if b1, b2 ∈ Γ are distinct, thenb1(θ

′) �= b2(θ′) and Γ (θ ′) ⊆ aθ ′ . HenceΓ ∈

dom(fθ ′), so{r(X): X ∈ D, Γ ⊆ X} = [θ ′,ω1], sow(D,Γ ) is stationary. �Fact 48. SupposeKHλ(T ) holds andA is a downward closed Aronszajn subtree ofT . ThenF(T ) is a Collins space with point countable base, hence it is a metaLindelöf space

Proof. Using the terminology of [27] a space isCollins iff everyone of its pointx has acountable open baseBx with the property that, ifU is a neighbourhood of a pointy, thereis a neighbourhoodV of y such that for allx ∈ V , there isB ∈ Bx with y ∈ B ⊆ U . So forX ∈ F(T ) let us put

BX = {U(θ,F ): X ∈ U(θ,F ), θ � sup

{θ ′: ∃b ∈ X: b(θ ′) ∈ A

}, F ∈ [

Br(T )]<ω}

.

SinceA is an Aronszajn tree,BX is countable for anyX ∈F(T ). AlsoBX contains the se{U(r(X),F ): F ∈ [X]<ω} sinceA ∩ Levr(X) T �= ∅ becauseA is downward closed, thuBX is a basis atX.

Now takeY ∈ U and letV = U(r(Y ),F ) such thatV ⊆ U and that there isb ∈ F suchthatb(r(Y )) ∈ A. Consider anyX ∈ V , we know thatF ⊆ X and in particularb ∈ X, hence

r(Y ) � sup{θ ′: ∃b ∈ X b(θ ′) ∈ A

},

thusU(r(Y ),F ) ∈ BX. So putB = U(r(Y ),F ) and note thatY ∈ B ⊆ U , henceF(T ) is aCollins space. Note that we have proved more, namely, thatY ∈ B ⊆ V sinceB = V . This


and

lly-

lly

al,

e

force

repa-d very

f Set-

ology,

(Eds.),

ixth

condition is sufficient forF(T ) to have a point countable base (see [5], Theorem 15Lemma 7) and this completes the proof.�

As a consequence of the above results we obtain

Theorem 49. Suppose thatKHλ holds, then there is a zerodimensional, regular(weaklycollectionwise Hausdorff if cf(λ) > ω1), nonnormal, noncollectionwise Hausdorff, locametrizable, metaLindelöf but non-σ -ParaLindelöf space of sizeλ such that any of its subspaces of size less thanλ is metrizable and included in a clopen metrizable space(hencethe space is(< λ)-collectionwise Hausdorff). Moreover the space is not monotonicanormal and hence it is not a generalized ordered space.

Fact 50. Suppose there is a Suslin treeS, then there a nonnormal, regular, zerodimensionlocally metrizable spaceF such that

S � F is metrizable.

Proof. Take anyω1-treeT ′ with many branches. Then as in Proposition 43 obtain a treT

that containsS as an Aronszajn subtree. Hence, by Fact 46, the spaceF(T ) is not normaland by Facts 25, 27 it is regular, zerodimensional, locally metrizable space. Nowwith S. It is well known that forcing withS adjoins a branch throughS (see [19]). Call thisbranchb. Now note that

S � b /∈⋃

F̌(T )

hence, by Fact 28,S forces thatF̌(T ) becomes a metrizable subspace ofF(T ). �

Acknowledgement

I would like to thank Professor S. Todorcevic for his helpful comments during the pration of this paper. I am also grateful to the referee for a detailed report which helpemuch in improving the presentation.

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Kurepa trees and topological non-reﬂection

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