Essential Kurepa Trees Versus

Essential Jech{Kunen Trees1

Renling Jin2 & Saharon Shelah3

Abstract

By an !1{tree we mean a tree of cardinality !1 and height !1. An !1{tree

is called a Kurepa tree if all its levels are countable and it has more than

!1 branches. An !1{tree is called a Jech{Kunen tree if it has � branches

for some � strictly between !1 and 2!1. A Kurepa tree is called an essential

Kurepa tree if it contains no Jech{Kunen subtrees. A Jech{Kunen tree is

called an essential Jech{Kunen tree if it contains no Kurepa subtrees. In

this paper we prove that (1) it is consistent with CH and 2!1 > !2 that

there exist essential Kurepa trees and there are no essential Jech{Kunen

trees, (2) it is consistent with CH and 2!1 > !2 plus the existence of

a Kurepa tree with 2!1 branches that there exist essential Jech{Kunen

trees and there are no essential Kurepa trees. In the second result we

require the existence of a Kurepa tree with 2!1 branches in order to avoid

triviality.

0 Introduction

Our trees are always growing downward. We use T� for the �th level of T and use

T �� forS

�<� T�. For every t 2 T let ht(t) = � i� t 2 T�. Let ht(T ), the height of T ,

be the least ordinal � such that T� = ;. By a branch of T we mean a totally ordered

subset of T which intersects every nonempty level of T . For any tree T let m(T ) be

the set of all maximal nodes of T , i.e. m(T ) = ft 2 T : (8s 2 T )(s 6 t ! s = t)g.

All trees considered in this paper have cardinalities less than or equal to !1 so that,

without loss of generality, we can assume all those trees are subtrees of (!<!11 ;�),

1Mathematics Subject Classi�cation Primary 03E35.2The �rst author would like to thank Mathematics Department of Rutgers University for its

hospitality during his one week visit there in October 1992, when the main part of the paper was

developed.3Publ. no. 498. The second author was partially supported by the United States{Israel Bina-

tional Science Foundation.

1

where !<!11 is the set of all functions from some countable ordinals to !1. Hence every

tree here has a unique root ; and if ftn : n 2 !g � T is a decreasing sequence of T ,

then t =S

n2! tn is the only possible greatest lower bound of ftn : n 2 !g. We are

also free to use either 6T or � for the order of a tree T , i.e. s 6T t if and only if

s � t.

By an !1{tree we mean a tree of height !1 and cardinality !1. Notice that our

de�nition of !1{tree is slightly di�erent from the usual de�nition by not requiring

every level to be countable. An !1{tree T is called a Kurepa tree if every level of T

is countable and T has more than !1 branches. An !1{tree T is called a Jech{Kunen

tree if T has � branches for some � strictly between !1 and 2!1. We call a Kurepa

tree thick if it has 2!1 branches. Obviously, a Kurepa non-Jech{Kunen tree must be

thick, and a Jech{Kunen tree with every level countable is a Kurepa tree.

While Kurepa trees are better studied, Jech{Kunen trees are relatively less pop-

ular. It is K. Kunen [K1][Ju], who brought Jech{Kunen trees to people's attention

by proving that: under CH and 2!1> !2, the existence of a compact Hausdor� space

with weight !1 and cardinality strictly between !1 and 2!1 is equivalent to the exis-

tence of a Jech{Kunen tree. It is also easy to observe that: under CH and 2!1> !2,

the existence of a (Dedekind) complete dense linear order with density !1 and cardi-

nality strictly between !1 and 2!1 is also equivalent to the existence of a Jech{Kunen

tree. Above results are interesting because those compact Hausdor� spaces and com-

plete dense linear orders cannot exist if we replace !1 by !, while the existence of a

Jech{Kunen tree is undecidable. In this paper we would like to consider Jech{Kunen

trees only under CH and 2!1>!2 .

The consistency of a Jech{Kunen tree was given in [Je1], in which T. Jech con-

structed a generic Kurepa tree with less than 2!1 branches in a model of CH and

2!1> !2. By assuming the consistency of an inaccessible cardinal, K. Kunen proved

the consistency of non{existence of Jech{Kunen trees with CH and 2!1>!2 (see [Ju,

Theorem 4.8]). In Kunen's model there are also no Kurepa trees. Kunen proved

(see [Ju, Theorem 4.10]) also that the assumption of an inaccessible cardinal above

is necessary. The di�erences between Kurepa trees and Jech{Kunen trees in terms

of the existence have been studied in [Ji1] [Ji2] [Ji3] [SJ1] [SJ2]. It was proved that

the consistency of an inaccessible cardinal implies (1) it is consistent with CH and

2!1>!2 that there exist Kurepa trees but there are no Jech{Kunen trees [SJ1], (2) it

is consistent with CH and 2!1 >!2 that there exist Jech{Kunen trees but there are

2

no Kurepa trees [SJ2].

What could we say without the presence of large cardinals? Instead of killing

all Kurepa trees, which needs an inaccessible cardinal, while keeping some Jech{

Kunen trees alive, or killing all Jech{Kunen trees, which needs again an inaccessible

cardinal, while keeping some Kurepa trees alive, we can kill all Kurepa subtrees of a

Jech{Kunen tree or kill all Jech{Kunen subtrees of a Kurepa tree without using large

cardinals. Let's call a Kurepa tree T essential if T has no Jech{Kunen subtrees, and

call a Jech{Kunen tree T essential if T has no Kurepa subtrees. In [Ji1], the �rst

author proved that it is consistent with CH and 2!1>!2 , together with Generalized

Martin's Axiom and the existence of a thick Kurepa tree, that no essential Kurepa

trees and no essential Jech{Kunen trees. We required the presence of thick Kurepa

trees in the model in order to avoid triviality. In [Ji3], the �rst author proved that it

is consistent with CH and 2!1 >!2 that there exist both essential Kurepa trees and

essential Jech{Kunen trees. A weak version of this result was proved in [Ji1] with

help of an inaccessible cardinal. This paper could be considered as a continuation of

the research done in [Ji1] [Ji2] [Ji3] [SJ1] [SJ2].

In x1, we prove that it is consistent with CH and 2!1>!2 that there exist essential

Kurepa trees but there are no essential Jech{Kunen trees. In x2, we prove that it

is consistent with CH and 2!1 > !2 plus the existence of a thick Kurepa tree that

there exist essential Jech{Kunen trees but there are no essential Kurepa trees. In x3,

we simplify the proofs of two old results by using the forcing notion for producing a

generic essential Jech{Kunen tree de�ned in x2.

We write _a in the ground model for a name of an element a in the forcing extension.

If a is in the ground model, we usually write a itself as a canonical name of a. The

rest of the notation will be consistent with [K2] or [Je2].

1 Yes Essential Kurepa Trees, No Essential Jech{

Kunen Trees.

In this section we are going to construct a model of CH and 2!1 > !2 in which

there exist essential Kurepa trees and there are no essential Jech{Kunen trees. Our

strategy to do this can be described as follows: �rst, we take a model of CH and

2!1 > !2 plus GMA (Generalized Martin's Axiom) as our ground model, so that in

the ground model there are no essential Jech{Kunen trees, then, we add a generic

3

Kurepa tree which has no Jech{Kunen subtrees. The hard part is to prove that the

forcing adds no essential Jech{Kunen trees.

Let P is a poset. A subset S of P is called linked if any two elements in S are

compatible in P. A poset P is called !1{linked if P is the union of !1 linked subsets

of P. A subset S of P is called centered if every �nite subset of S has a lower bound

in P. A poset P is called countably compact if every countable centered subset of P

has a lower bound in P. Now GMA is the following statement:

Suppose P is an !1{linked and countably compact poset. For any � < 2!1,

if D = fD� : � < �g is a collection of � dense subsets of P, then there

exists a �lter G of P such that G \D� 6= ; for all � < �.

We choose the form of GMA from [B], where a model of CH and 2!1 > !2 plus

GMA can be found.

Let I be any index set. We write K I for a poset such that p is a condition in K I

i� p = (Ap; lp) where Ap is a countable subtree of (!<!11 ;�) of height �p + 1 and lp

is a function from a countable subset of I into (Ap)�p , the top level of Ap. For any

p; q 2 K I , de�ne p 6 q i�

(1) Ap ��q + 1 = Aq,

(2) dom(lp) � dom(lq)

(3) (8� 2 dom(lq))(lq(�) � lp(�)).

It is easy to see that K I is countably closed (or !1{closed). If CH holds, then K I

is !1{linked. Let M be a model of CH and K I 2M . Suppose that G is a K I {generic

�lter over M and let TG =S

p2GAp. Then in M [G], the tree TG is an !1{tree with

every level countable and TG has exactly jIj branches. Furthermore, if for every i 2 I

let

B(i) =[flp(i) : p 2 G and i 2 dom(lp)g;

then B(i) 6= B(i0) for any i; i0 2 I and i 6= i0, and fB(i) : i 2 Ig is the set of all

branches of TG in M [G]. Hence if jIj > !1, then TG will be a Kurepa tree with jIj

branches in M [G]. K I is the poset used in [Je1] for creating a generic Kurepa tree.

All those facts above can also be found in [Je1] or [T].

For convenience we sometimes view K I as an iterated forcing notion

K I0 � Fn(I r I 0; T _GI0; !1);

4

for any I 0 � I, where GI0 is a K I0{generic �lter over the ground model and Fn(I r

I 0; TGI0; !1), in M [GI0 ], is the set of all functions from some countable subset of Ir I 0

to TGI0with the order de�ned by letting p 6 q i� dom(q) � dom(p) and for any

i 2 dom(q), p(i) 6 q(i). The poset Fn(J; TG; !1) is in fact the countable support

product of jJ j{copies of TG. We say two posets P and Q are forcing equivalent if

there is a poset R such that R can be densely embedded into both P and Q . The

posets K I and K I0 � Fn(I r I 0; T _GI0; !1) are forcing equivalent because the map

F : K I 7! K I0 � Fn(I r I 0; T _GI0; !1)

such that for every p 2 K I ,

F (p) = ((Ap; lp �I0); lp �I r I 0)

is a dense embedding.

Lemma 1 (K. Kunen) Let M be a model of CH. Suppose that � > !2 is a cardinal

in M and K � 2 M . Suppose G� is a K �{generic �lter over M and TG�=S

p2G�Ap.

Then in M [G�] the tree TG�is a Kurepa tree with � branches and TG�

has no subtrees

with � branches for any � strictly between !1 and �.

Proof: Assume that T is a subtree of TG�with more than !1 branches in M [G�].

We want to show that T has � branches inM [G�]. Since jT j = !1 and K � has !2-c.c.,

then there exists a subset I � � in M with cardinality 6 !1 such that T 2 M [GI ],

where

GI = fp 2 G : dom(lp) � Ig:

Notice that TG�= TGI

(in fact TG = TG;). Since in M [GI ] the tree TGI

has only jIj

branches, then the tree T can have at most !1 branches in M [GI ]. Let B be a branch

of T in M [G�] which is not in M [GI ]. Since jBj = !1, there exists a subset J of �r I

with cardinality6 !1 such that B 2M [GI ][HJ ], where HJ is a Fn(J; TGI; !1){generic

�lter over M [GI ]. Now �rI can be partitioned into �{many subsets of cardinality !1

and for every subset J 0 � �r (I [ J) of cardinality !1 the poset PJ = Fn(J; TGI; !1)

is isomorphic to the poset PJ 0 = Fn(J 0; TGI; !1) through an obvious isomorphism �

induced by a bijection between J and J 0. Let _B be a PJ{name for B. Then ��( _B)

is a PJ 0{name for a new branch of T . Forcing with PJ � PJ 0 will create two di�erent

5

branches _BHJand (��( _B))HJ0

. Hence forcing with Fn(�r I; TGI; !1) will produce at

least � new branches of T . �

Next lemma is a simple fact which will be used later.

Lemma 2 Suppose P is an !1{closed poset of cardinality !1 (hence CH must hold).

Then the tree (!<!11 ;�) can be densely embedded into P.

Proof: Folklore. �

Lemma 3 Let M be a model of CH and 2!1>!2 plus GMA and let P = (!<!11 ;�) 2

M . Suppose G is a P{generic �lter over M . Then in M [G] every Jech{Kunen tree

has a Kurepa subtree.

Proof: Let T be a Jech{Kunen tree in M [G] with � branches for !1 < � < � = 2!1.

Without loss of generality we can assume that there is a regular cardinal � such that

!1 < � 6 � and for every t 2 T there are at least � branches of T passing through t in

M [G]. Again in M [G] let f : � 7! B(T ) be a one to one function such that for every

t 2 T and for every � < � there exists an � 2 �r� such that t 2 f(�). Without loss

of generality let us assume that

1P ( _T is a Jech{Kunen tree and _f : � 7! B( _T )

is a one to one function such that (8t 2 _T )(8� 2 �)(9� 2 �r �)(t 2 _f(�))):

We want now to construct a poset R in M such that a �lter H of R obtained by

applying GMA in M will give us a P{name for a Kurepa subtree of T in M [G].

Let r be a condition of R i� r = (Ir;Pr;Ar;Sr) where Ir is a countable subtree of

(!<!11 ;�), Pr = hprt : t 2 Iri, Ar = hAr

t : t 2 Iri and Sr = hSrt : t 2 Iri such that

(1) Pr � P and for every t 2 Ir the element Art is a nonempty countable subtree

of (!<!11 ;�) of height �rt + 1 (we will use some Ar

t 's to generate a Kurepa subtree of

T ) and Srt is a nonempty countable subset of �,

(2) (8s; t 2 Ir)(s � t $ prt 6 prs), (This implies that s and t are incompatible i�

prs and prt are incompatible for all s; t 2 Ir because P is a tree,)

(3) (8s; t 2 Ir)(s � t! Art �ht(A

rs) = Ar

s),

(4) (8s; t 2 Ir)(s � t! Srs � Sr

t ),

(5) (8t 2 Ir)(prt Ar

t � _T ),

6

(6) (8t 2 Ir)(8� 2 Srt )(9a 2 (Ar

t )�rt )(prt a 2 _f(�)).

The order of R: for any r; r0 2 R, let r 6 r0 i� Ir0 � Ir and for every t 2 Ir0

pr0

t = prt ; Ar0

t = Art and Sr0

t � Srt :

Claim 3.1 The poset R is !1{linked.

Proof of Claim 3.1: Let r; r0 2 R such that Ir = Ir0, Pr = Pr0 and Ar = Ar0.

Then the condition r00 2 R such that

Ir00 = Ir; Pr00 = Pr; Ar00 = Ar and Sr00 = hSrt [ Sr0

t : t 2 Ir00i

is a common lower bound of both r and r0. Since there are only !1 di�erent hIr;Pr;Ari's

and for each �xed hIr0;Pr0 ;Ar0i the set

fr 2 R : hIr;Pr;Ari = hIr0;Pr0 ;Ar0ig

is linked, then R is the union of !1 linked subsets of R. � (Claim 3.1)

Claim 3.2 The poset R is countably compact.

Proof of Claim 3.2: Suppose that R0 is a countable centered subset of R. Notice

that for any �nite R00 � R0 and for any t 2

TfIr : r 2 R0

0g all prt 's are same and all

Art are same for r 2 R0

0 because R00 has a common lower bound in R. We now want

to construct a condition �r 2 R such that �r is a common lower bound of R0 . Let

(1) I�r =S

r2R0 Ir,

(2) P�r = hp�rt : t 2 I�ri where p�rt = prt for some r 2 R0 such that t 2 Ir,

(3) A�r = hA�rt : t 2 I�ri where A

�rt = Ar

t for some r 2 R0 such that t 2 Ir,

(4) S�r = hS�rt : t 2 I�ri where S

�rt =S

s�t Ss and Ss =SfSr

s : (9r 2 R0)(s 2 Ir)g.

Notice that from the argument above all p�rt 's, A�rt 's and S�r

t 's are well{de�ned. We

need to show �r 2 R. It is obvious that �r is a common lower bound of all elements in

R0 if �r 2 R.

It is easy to see that �r satis�es (1), (2), (3), (4) and (5) in the de�nition of a

condition in R. Let's check (6).

Suppose t 2 I�r and � 2 S�rt . We want to show that there exists an a 2 (A�r

t )��rt such

that p�rt a 2 _f(�). Let r 2 R0 be such that t 2 Ir, let r0 2 R0 and s 2 Ir0 be such

that s � t and � 2 Sr0

s . Since r and r0 are compatible, then there exists an r00 2 R

such that r00 6 r and r00 6 r0. By the facts that

p�rt = prt = pr00

t ; A�rt = Ar

t = Ar00

t ; Sr0

s � Sr00

s � Sr00

t

7

and r00 2 R we have now that there exists an a 2 (A�rt )��rt such that p�rt a 2 _f(�). �

(Claim 3.2)

Next we are going to apply GMA in M to the poset R to construct a P-name for

a Kurepa subtree in M [G].

For each t 2 !<!11 de�ne

Dt = fr 2 R : t 2 Irg:

For each p 2 P de�ne

Ep = fr 2 R : (9t 2 Ir)(prt 6 p)g:

For each � < !1 de�ne

F� = fr 2 R : (8s 2 Ir)(9t 2 Ir)(s � t and ht(Art ) > �)g:

For each � < � de�ne

O� = fr 2 R : (8s 2 Ir)(9t 2 Ir)(s � t and [�; �) \ Srt 6= ;)g:

Claim 3.3 All those Dt, Ep, F� and O�'s are dense in R.

Proof of Claim 3.3: Let r0 be an arbitrary element in R.

We show �rst that for every t 2 !<!11 the set Dt is dense in R, i.e. there is an

r 2 Dt such that r 6 r0. It's done if t 2 Ir0 . Let's assume that t 62 Ir0. Let

t0 =[fs 2 Ir0 : s � tg:

Case 1: t0 2 Ir0.

Find a sequence fps : t0 � s � tg in P such that pt0 = pr0t0 and

(8s; s0)(t0 � s � s0 � t$ ps0 6 ps):

The sequence fps : t0 � s � tg exists because P is !1{closed. Let

Ir = Ir0 [ fs : t0 ( s � tg:

For any s 2 Ir, if s 2 Ir0 , then let

prs = pr0s ; Ars = Ar0

s and Srs = Sr0

s :

8

Otherwise let

prs = ps; Ars = Ar0

t0and Sr

s = Sr0t0:

It is easy to see that r 2 Dt and r 6 r0.

Case 2: t0 62 Ir0, i.e. Ir0 has no least element which is above t.

Let

Ir = Ir0 [ fs : t0 � s � tg:

Again by !1{closedness we can �nd

fps : t0 � s � tg � P

such that pt0 is a lower bound of

fpr0s : s � t0 and s 2 Ir0g

and

(8s; s0)(t0 � s � s0 � t$ ps0 6 ps):

Let

A0t0=[fAr0

s : s 2 Ir0 and s � t0g

and let

St0 =[fSr0

s : s 2 Ir0 and s � t0g:

If the height of A0t0is a successor ordinal, then let At0 = A0

t0. If the height of A0

t0is a

limit ordinal, then we have to add one more level to A0t0. For any � 2 St0 let s

0 � t0

and s0 2 Ir0 be such that � 2 Sr0s0 . Then for any s 2 Ir0 such that s0 � s � t0 there

exists an as;� 2 (Ar0s )�r0s such that pr0s as;� 2 _f(�). Now let

a� =[fas;� : s

0 � s � t0 and s 2 Ir0g

and let

At0 = A0t0[ fa� : � 2 St0g:

It is easy to see that

(1) the height of At0 is a successor ordinal,

(2) for every s ( t0 the tree At0 is an end{extension of Ar0s , i.e.

At0 �ht(Ar0s ) = Ar0

s ;

9

(3) for every � 2 St0 there exists an a� in the top level of At0 such that

pt0 a� 2 _f(�).

Now for every s 2 Ir, if r 2 Ir0 , then let

prs = pr0s ; Ars = Ar0

s and Srs = Sr0

s :

Otherwise let

prs = ps; Ars = At0 and Sr

s = St0 :

It is easy to see that r 2 Dt and r 6 r0.

We show now that for every p 2 P the set Ep is dense in R. We want to �nd an

r 2 Ep such that r 6 r0. If there exists an t 2 Ir0 such that pr0t 6 p, then r0 2 Ep.

Let's assume that for every t 2 Ir0 pr0t 66 p. Let

t0 =[ft 2 Ir0 : p 6 pr0t g:

Case 1: t0 2 Ir0.

Let t0 = t0 h0i, i.e. t0 is a successor of t0. It is clear that t

0 62 Ir0. Let Ir = Ir0[ft0g.

For every t 2 Ir, if t = t0, then let

prt = p; Art = Ar0

t0and Sr

t = Sr0t0:

Otherwise let

prt = pr0t ; Art = Ar0

t and Srt = Sr0

t :

Then we have r 2 Ep and r 6 r0.

Case 2: t0 62 Ir0.

Let Ir = Ir0 [ ft0g. We construct St0 , A0t0and then At0 exactly same as we did

in the proof of Case 2 about the denseness of the set Dt. For every t 2 Ir, if t = t0,

then let

prt = p; Art = At0 and Sr

t = St0 :

Otherwise let

prt = pr0t ; Art = Ar0

t and Srt = Sr0

t :

Now r 2 Ep and r 6 r0. Notice also that Ep is open, i.e.

(8p0; p00 2 P)(p0 6 p00 ^ p00 2 Ep ! p0 2 Ep):

10

We show next that for every � 2 !1 the set F� is dense in R. We need to �nd an

r 2 F� such that r 6 r0.

Let Ir � Ir0 be such that Ir is a countable subtree of !<!11 , Irr Ir0 is an antichain

and for every s 2 Ir0 there is a t 2 Ir r Ir0 such that s � t. For every t 2 Ir r Ir0 let

pt 2 P be such that pt 6 pr0s for every s 2 Ir0 and s � t, let

Srt =[fSr0

s : s 2 Ir0 and s � tg

and let

A0t =[fAr0

s : s 2 Ir0 and s � tg:

If ht(A0t) is a successor ordinal, then let At = A0

t. Otherwise let

At = A0t [ fa� : � 2 Sr

t g

where

a� =[fa 2 A0

t : pt a 2 _f(�)g:

Since Srt is countable and P is !1{closed, then there exists a prt 6 pt such that for

every � 2 Srt there exists an a 2 !�

1 such that prt a 2 _f(�). Let

Art = At [ fa 2 !

6�1 : (9� 2 Sr

t )(prt a 2 _f(�))g:

Then ht(Art ) > � is a successor ordinal and for every � 2 Sr

t there exists an a in the

top level of Art such that prt a 2 _f(�). For every t 2 IrrIr0 we have already de�ned

prt , Art and Sr

t . If t 2 Ir0, then let

prt = pr0t ; Art = Ar0

t and Srt = Sr0

t :

Hence r 2 F� and r 6 r0.

We show next that O� for every � < � is dense in R, i.e. �nding an r 2 O� such

that r 6 r0.

By imitating the proof of the denseness of F� we can �nd an r0 6 r0 such that

Ir0 r Ir0 is an antichain and for every s 2 Ir0 there exists an t 2 Ir0 r Ir0 such that

s � t. For every t 2 Ir0 r Ir0 �x a �t which is an successor of t (for example �t = t h0i).

Let

Ir = Ir0 [ f�t : t 2 Ir0 r Ir0g:

For every t 2 Ir0 let

prt = pr0

t ; Art = Ar0

t and Srt = Sr0

t :

11

For every �t with t 2 Ir0 r Ir0 we want to construct pr�t , A

r�t and S

r�t . If there is a � 2 Sr0

t

which is greater than �, then let pr�t be any proper extension of prt , let Ar�t = Ar0

t and let

Sr�t = Sr0

t . Otherwise, �rst, pick an a in the top level of Ar0

t , then choose a � 2 �r �

and a p 6 pr0

t such that p a 2 _f(�). This can be done because

1P (8t 2 _T )(8� 2 �)(9� 2 �r �)(t 2 _f(�))

is true in M . Now let

pr�t = p; Ar�t = Ar0

t and Sr�t = Sr0

t [ f�g:

It is easy to see that r 2 O� and r 6 r0. � (Claim 3.3)

By applyingGMA inM we can �nd an R{�lter H such thatH\Dt 6= ;, H\F� 6= ;

and H \ Ep \ O�0 6= ; for each t 2 !<!11 , each � 2 !1, each p 2 P and each �0 2 �.

Since Dt is dense for every t 2 !<!11 , then

IH =[fIr : r 2 Hg = !<!1

1 :

Let

PH =[fPr : r 2 Hg

and let

AH =[fAr : r 2 Hg:

Notice that for any r; r0 2 H and for any t 2 Ir \ Ir0 we have prt = pr

0

t and Art = Ar0

t

because r and r0 are compatible. So now for every t 2 IH we can de�ne pt = prt for

some r 2 H and de�ne At = Art for some r 2 H. It is clear that the map t 7! pt is

an isomorphism between IH and PH , i.e. for any s; t 2 IH we have s � t i� pt 6 ps.

It is also clear that the map t 7! At is a homomorphism from IH to AH , i.e. for any

s; t 2 IH we have s � t implies At �ht(As) = As.

Claim 3.4 For each t 2 IH the set fpt h i : 2 !1g is a maximal antichain below

pt in P.

Proof of Claim 3.4: Let and 0 be two ordinals in !1. Since IH = !<!11 and H

is a �lter, there exists an r 2 H such that t h i, t h 0i 2 Ir. Hence prt h i and p

rt h 0i are

incompatible. So fpt h i : 2 !1g is an antichain.

Suppose that p 2 P and p 6 pt such that p is incompatible with any of pt h i's.

Let r 2 H \ Ep. Then there is an s 2 Ir such that ps = prs 6 p. Since ps 2 PH , then

12

ps < pt implies t ( s. Hence there exists an 2 !1 such that t h i � s. This means

that ps � pt h i, i.e. p and pt h i are compatible, a contradiction. � (Claim 3.4)

We now work in M [G]. Since G is a P{generic �lter over M , then PH \ G is a

linearly ordered subset of PH . Let TG =SfAt : pt 2 Gg.

Claim 3.5 TG is a Kurepa subtree of T in M [G].

Proof of Claim 3.5: Since for every pt 2 G we have pt At � _T , it is clear that

TG � T in M [G]. For any ps; pt 2 G we have pt 6 ps implies s � t which implies

At �ht(As) = As. Hence TG is an end{extension of At for every pt 2 G. This implies

that every level of TG is a level of some At, hence is countable.

We want to show now that TG has at least � branches. Suppose jB(TG)j < �.

Then there exists an � 2 � such that for every � 2 �r � the function value f(�) is

not a branch of TG. So there is a p 2 PH and there is an � 2 � such that

p (8� 2 �r �)( _f(�) is not a branch of T _G):

On the other hand, since H \Ep \O� 6= ;, then there exists an r 2 H \O� \Ep. In

M let s 2 Ir be such that ps 6 p and there is a � 2 Srs such that � > �. Then for

every t 2 IH , s � t, there is an t0 2 IH , t � t0, such that

pt0 a 2 _f(�)

for some a 2 (At0)ht(At0). This shows that

ps _f(�) is a branch of T _G;

which contradicts ps 6 p and

p (8� 2 �r �)( _f(�) is not a branch of T _G):

Hence TG has at least � branches in M [G]. � (Claim 3.5)

Now we conclude that M [G] j= T has a Kurepa subtree TG. �

Theorem 4 It is consistent with CH and 2!1 >!2 that there exist essential Kurepa

trees and there are no essential Jech{Kunen trees.

13

Proof: Let M be a model of CH and 2!1 = � > !2 plus GMA. Let K � 2 M .

Suppose G� is a K �{generic �lter over M . We are going to show that M [G�] is a

model of CH and 2!1 >!2 in which there exist essential Kurepa trees and there are

no essential Jech{Kunen trees.

It is easy to see that M [G�] satis�es CH and 2!1 > !2 . Lemma 1 implies that

there exist essential Kurepa trees. We need only to show that in M [G�] there are no

essential Jech{Kunen trees.

Assume T is a Jech{Kunen tree in M [G�]. We need to show that T has a Kurepa

subtree in M [G�]. Since jT j = !1, then there is an I � � of cardinality !1 in M such

that T 2M [GI ], where

GI = fp 2 G� : dom(lp) � Ig:

We claim that

B(T ) \M [G�] �M [GI ]:

If the claim is true, then T is a Jech{Kunen tree in M [GI ]. Suppose that B 2

B(T )\ (M [G�]rM [GI ]). Then there is a J � �r I such that B 2M [GI ][HJ ] where

HJ is a Fn(J; TGI; !1){generic �lter over M [GI ]. Let _B be a Fn(J; TGI

; !1){name for

B. For any J 0 � � r (I [ J) such that jJ 0j = jJ j there is an isomorphism � from

Fn(J; TGI; !1) to Fn(J 0; TGI

; !1) induced by a bijection between J and J 0. Since

in M [G�], the branches ( _B)HJand (��( _B))HJ0

are di�erent, then T has at least �

branches. This contradicts that T is a Jech{Kunen tree. Let T have � branches in

M [GI ]. Since K I has cardinality !1 and is !1{closed, then it contains a dense subset

which is isomorphic to P = (!<!11 ;�) in M . Hence there is a P{generic �lter G over

M such that M [G] = M [GI ]. By Lemma 3, the tree T has a Kurepa subtree in

M [G]. Obviously, the Kurepa subtree is still a Kurepa subtree in M [G�], so T is not

an essential Jech{Kunen tree in M [G�]. �

2 Yes Essential Jech{Kunen Trees, No Essential

Kurepa Trees

In this section we will construct a model of CH and 2!1 > !2 plus the existence of

a thick Kurepa tree, in which there are essential Jech{Kunen trees and there are no

essential Kurepa trees. The arguments in this section are a sort of \symmetric" to

the arguments in the last section.

14

We �rst take a model M of CH and 2!1 = � > !2 plus a thick Kurepa tree, where

�<� = � in M , as our ground model. We then extend M to a model M [G] of CH

and 2!1 = � > !2 plus GMA by a �{stage iterated forcing (see [B] for the model and

forcing). It has been proved in [Ji1] that in M [G] there are neither essential Jech{

Kunen trees nor essential Kurepa trees. Instead of taking a model of GMA as our

ground model as we did in x1, we consider this �{stage iterated forcing as a part of

our construction because it will be needed later (see also [Ji1, Theorem 5]). Next we

force with an !1{closed poset JS;� in M [G] to create a generic essential Jech{Kunen

tree, where S is a stationary{costationary subset of !1. Again, the hard part is to

prove that forcing with JS;� over M [G] will not create any essential Kurepa trees.

Recall that for T , a tree, m(T ) denotes the set

ft 2 T : (8s 2 T )(s 6T t! s = t)g:

Let I be any index set and let S be a subset of !1. We de�ne a poset JS;I such that

p is a condition in JS;I i� p = (Ap; lp) where

(1) Ap is a countable subtree of !<!11 ,

(2) lp is a function from some countable subset of I to m(Ap).

For any p; q 2 JS;I de�ne p 6 q i�

(1) Aq � Ap,

(2) for every t 2 Ap r Aq either there is an s 2 m(Aq) such that s � t or that

� < ht(Aq) and � 2 S is a limit ordinal imply

� 6=[fht(s) : s 2 Aq and s � tg:

(3) dom(lq) � dom(lp) and (8� 2 dom(lq))(lq(�) � lp(�)).

Lemma 5 (CH) JS;I is !1{closed and !1{linked.

Proof: We show �rst that JS;I is !1{linked. For any p; q 2 JS;I, if Ap = Aq, then

the condition (Ap; lp [ lq) is a common extension of p and q. Because there are only

!1 di�erent countable subtrees of !<!11 , it is clear that JS;I is the union of !1 linked

sets.

We now show that JS;I is !1{closed. Let fpn : n 2 !g be a decreasing sequence in

JS;I. Let A =S

n2! Apn and let D =S

n2! dom(lpn). For each i 2 D let

l(i) =[flpn(i) : n 2 ! and i 2 dom(lpn)g:

15

De�ne a condition p 2 JS;I such that

Ap = A [ fl(i) : i 2 Dg and lp = l:

We claim that p is a lower bound of the sequence fpn : n 2 !g. It su�ces to show

that for any n and for any t 2 Ap r Apn either there exists an s 2 m(Apn) such that

s � t or that � < ht(Apn) and � 2 S is a limit ordinal imply

� 6=[fht(s) : s 2 Apn and s � tg:

If t 2 A, then there is an k > n such that t 2 Apk . Hence either there is an s 2 m(Apn)

such that s � t or that � < ht(Apn) and � 2 S is a limit ordinal imply

� 6=[fht(s) : s 2 Apn and s � tg

because pk 6 pn. If t = l(i) for some i 2 D, then, by assuming hlpn : n 2 !i is not

eventually constant, there is a k > n and there is a t0 2 Apk r Apn such that t0 � t.

Hence either there is an s 2 m(Apn) such that s � t0 � t or that � < ht(Apn) and

� 2 S is a limit ordinal imply

� 6=[fht(s) : s 2 Apn and s � t0g

because pk 6 pn. �

Remark: Again, we may consider the poset JS;I as a two{step iterated forcing

JS;I0 � Fn(I r I 0; T _GI0; !1), where I

0 is a subset of I, TGI0=SfAp : p 2 GI0g for a

generic �lter GI0 of JS;I0 and Fn(I r I 0; T _GI0; !1) is a countable support product of

jI r I 0j{copies of T _GI0. The map

p = (Ap; lp) 7! ((Ap; lp �I0); lp �I r I 0)

is a dense embedding from JS;I to JS;I0 � Fn(I r I 0; T _GI0; !1).

We now de�ne S{completeness of a tree T . Let � be a limit ordinal and let T be

a tree with ht(T ) = �. Let S be a subset of �. Then T is called S{complete if for

every limit ordinal � 2 S and every B 2 B(T � �) the unionSB 2 T�, i.e. every

strictly decreasing sequence of T has a greatest lower bound b in T if ht(b) 2 S.

16

Lemma 6 Let M be a model of CH and let JS;I 2M where S � !1 and I is an index

set in M . Suppose G is a JS;I{generic �lter over M . Then the tree TG =S

p2GAp is

(!1 r S){complete in M [G].

Proof: Let � 2 !1 r S be a limit ordinal and let B be a branch of TG ��. We need

to show that t =SB 2 TG. The set B is in M because JS;I is !1{closed and B is

countable. Let p0 2 G be such that B � Ap0 . It is clear that

p0 B � T _G:

Let

DB = fp 2 JS;I : p 6 p0 and t =[

B 2 Apg:

Then DB is dense below p0 because for any p 6 p0 the element p0 = (Ap [ fSBg; lp)

is a condition in JS;I and p0 6 p (here we use the fact that � 2 !1rS). Since p0 2 G,

then there is a p 2 G \DB. Hence t =SB 2 TG. �

Lemma 7 Let M be a model of CH. In M let U be a stationary subset of !1, let T

be an !1{tree which is U{complete and let I be any index set. Let K 2 M be any

!1{tree such that every level of K is countable. Suppose P = Fn(I; T; !1) 2 M and

G is a P{generic �lter over M . Then

B(K) \M [G] �M;

i.e. the forcing adds no new branches of K.

Proof: Suppose that B is a branch of K in M [G]rM . Without loss of generality,

let's assume that

1P _B 2 (B(K)rM):

By a standard argument (see [K2, p. 259]) the statements

(8p 2 P)(8� 2 !1)(9t 2 !�1 )(9p

0 6 p)(p0 t 2 _B)

and

(8p 2 P)(8� 2 !1)(8t 2 !�1 )(p t 2 _B �!

(8� 2 !1 r �)(9 2 !1 r �)(9tj 2 ! 1 )(t0 6= t1)(9pj 6 p)(pj tj 2 _B))

for j = 0; 1, are true in M .

17

Let's work in M . Let � be a large enough cardinal and let N be a countable

elementary submodel of (H(�);2) such that K;P; _B 2 N . Let � = N \ !1 2 U (such

N exists because U is stationary). In M we choose an increasing sequence of ordinals

f�n : n 2 !g such thatS

n2! �n = �. Again in M we construct a set

fps : s 2 2<!g � P \N

and a set

fts : s 2 2<!g � K \N

such that

(1) (8s; s0 2 2<!)(s � s0 $ ps0 6 ps $ ts0 6 ts),

(2) (8s 2 2<!)(ps ts 2 _B),

(3) ht(ts) > �jsj,

(4) (8i 2 dom(ps))(ht(ps(i)) > �jsj),

where jsj means the length of the �nite sequence s.

Let p; = 1P and let t; = ;, the root of K. Assume that we have found fps : s 2

26n g and fts : s 2 26n g which satisfy (1), (2), (3) and (4) relative to 26n . Pick any

s 2 2n. Since the sentence

(8p 2 P)(8� 2 !1)(8t 2 !�1 )(p t 2 _B �!

(8� 2 !1 r �)(9 2 !1 r �)(9tj 2 ! 1 )(t0 6= t1)(9pj 6 p)(pj tj 2 _B))

for j = 0; 1, is true in M , then it is true in N . Since ps; ts 2 N , then in N there exist

p0; p1 6 ps and there exist t0; t1 2 ! 1 , t

0 6= t1, for some 2 � r �jsj+1 such that

pj tj 2 _B

for j = 0; 1. Again in N we can extend p0 and p1 to ps h0i and ps h1i respectively so

that

(8i 2 dom(ps hji))(ht(ps hji(i)) > �jsj+1)

for j = 0; 1. Since T is U{complete and for every f 2 2!, for every i 2S

n2! dom(pf�n)

we have [fht(pf�n(i)) : n 2 ! and i 2 dom(pf�n)g = � 2 U;

then the condition pf such that dom(pf) =S

n2! dom(pf�n) and

pf(i) =[fpf�n(i) : n 2 ! and i 2 dom(pf)g

18

for every i 2 dom(pf) is a lower bound of fpf�n : n 2 !g in P. Here we use the fact

that T is U{complete so that pf (i) 2 T for every i 2 dom(pf). Let tf =S

n2! tf�n.

Then ht(tf ) = �. Since

pf tf�n 2 _B

for every n 2 !, then

pf tf 2 _B \K�:

It is easy to see that if f; f 0 2 2! are di�erent, then tf and tf 0 are di�erent. Hence

K� is uncountable, a contradiction. �

Lemma 8 Let M be a model of CH and 2!1 = � > !2 and let JS;� 2 M where � is

a cardinal in M such that !1 < � < � and S is a stationary subset of !1. Suppose

that G is a JS;�{generic �lter over M . Then in M [G] the tree TG =S

p2GAp is an

essential Jech{Kunen tree with � branches.

Proof: It is easy to see that TG is an !1 tree. We will divide the lemma into two

claims.

Claim 8.1 For every � 2 � let

B(�) =[flp(�) : p 2 G and � 2 dom(lp)g:

Then

B(TG) = fB(�) : � 2 �g

and for any two di�erent � and �0 in � the branches B(�) and B(�0) are di�erent.

Proof of Claim 8.1: Since in M , for every � 2 � and for every � 2 !1 the set

D�;� = fp 2 JS;� : � 2 dom(lp) and ht(lp(�)) > �g

is dense in JS;�, then B(�) is a branch of TG. For any two di�erent �; �0 2 � the set

D�;�0 = fp 2 JS;� : �; �0 2 dom(lp) and lp(�) 6= lp(�

0)g

is also dense in JS;�. So the branches B(�) and B(�0) are di�erent.

We now want to show that all branches of TG in M [G] are exactly those B(�)'s.

Suppose that in M [G] the tree TG has a branch B which is not in the set

fB(�) : � 2 �g:

19

Without loss of generality, let us assume that

1JS;� _B 2 (B(T _G)r f _B(�) : � 2 �g):

Work inM . Let � be a large enough cardinal and letN be an elementary submodel

of (H(�);2) such that �; S; _B;B = f _B(�) : � 2 �g; JS;� 2 N and if p 2 N \ JS;�, then

dom(lp) � N . Let � = N \ !1 2 S. In M we choose an increasing sequence of

countable ordinals f�n : n 2 !g such that � =S

n2! �. We now want to �nd a

decreasing sequence fpn : n 2 !g � JS;� \N such that p0 = 1JS;� and for each n 2 !

(1) (8� 2 dom(lpn))(9t 2 Apn+1)(pn+1 t 2 _B(�)r _B),

(2) (9t 2 Apn+1 r Apn)(ht(t) > ht(Apn) and pn+1 t 2 _B,

(3) ht(Apn) > �n.

Assume we have found fp0; p1; : : : ; png. We now work in N . Let

dom(lp) = f�k : k 2 !g

which is an enumeration in N . Choose q0 = pn > q1 > � � � such that for every k 2 !1

there is a t 2 Aqk such that

qk t 2 _B(�k)r _B:

Assume, in N , that we have found fq0; q1; : : : ; qkg. Since the sentence

qk (9t 2 T _G)(t 2_B(�k)r _B)

is true in N (because it is true inH(�) and �k 2 N), then there is a t 2 !<!11 \N = �<�

and there is a q0 6 qk such that

q0 (t 2 T _G and t 2 _B(�k)r _B):

Since

q0 Aq0 � T _G;

then there is a qk+1 6 q0 such that t 2 Aqk+1. Since N j= \JS;� is !1{closed" and

fqk : k 2 !g is constructed in N , then there is a q 2 JS;� in N such that q is a lower

bound of fqk : k 2 !1g. Let � = maxfht(Apn); �n+1g. Notice that � 2 � because

pn 2 N . Since in N

q _B is a branch of T _G;

then

q (9t 2 (T _G)�+1)(t 2 _B):

20

Hence there is a �q 6 q and there is a t 2 !�+11 \N such that

�q t 2 _B:

We can also assume that t 2 A�q.

We now go back to M and let pn+1 = �q. This �nishes the construction of

fpn : n 2 !g.

Let p 2 JS;� be such that

dom(lp) =[

n2!

dom(lpn);

for every � 2 dom(lp)

lp(�) = a� =[flpn(�) : n 2 ! and � 2 dom(lpn)g

and

Ap = ([

n2!

Apn) [ fa� : � 2 dom(lp)g:

By the construction of pn's we have

[fht(t) : t 2 Ap and p t 2 _Bg = � 2 S:

Pick any t 2 Ap. If t 6= a� for any � 2 dom(lp), then we can �nd a 2 !1 such that

t h i 62 Ap. Extend t h i to �t 2 !�1. De�ne �p such that

A�p = Ap [ fu : t � u � �tg

and l�p = lp. If t = a� for some � 2 dom(lp), then simply extend t to b� 2 !�1 (if

ht(a�) = �, then b� = a�). De�ne �p such that

A�p = Ap [ fu : t � u � b�g

and

l�p = (lp �(dom(lp)r f�g)) [ f(�; b�)g:

It is easy to see that �p 6 p and ht(A�p) = � + 1. Let

a =[ft 2 A�p : �p t 2 _Bg:

21

It is also easy to see that for any q 6 �p the element a is not in Aq. Here we use the

fact � 2 S, � is a limit ordinal and ht(A�p) > �. Hence

�p _B \ T _G �_B \ A�p:

This contradicts that

�p _B is a branch of T _G:

� (Claim 8.1)

Claim 8.2 TG has no Kurepa subtree in M [G].

Proof of Claim 8.2: Suppose that TG has a Kurepa subtree K in M [G]. Since

jKj = !1, then there is an I � � such that jIj 6 !1 and K 2M [GI ], where

GI = fp 2 G : dom(lp) � Ig:

Notice that GI is a JS;I{generic �lter over M . Since JS;� is forcing equivalent to

JS;I �Fn(�r I; TGI; !1) and TGI

is (!1rS){complete in M [GI ] (notice that S is still

stationary{costationary), then by Lemma 7, the set of all branches of K in M [GI ] is

same as the set of all branches of K in M [G]. Hence K is a Kurepa tree in M [GI ].

But by Claim 8.1, the tree TG = TGIhas only jIj branches in M [GI ] and K is a

subtree of TG. Hence K has at most !1 branches in M [GI ]. This contradicts that K

is a Kurepa tree in M [GI ]. �

Lemma 9 Let M be a model of CH and 2!1 = � > !2 with �<� = �. In M let

((P� : � 6 �); ( _Q� : � < �)) be a �{stage iterated forcing notion used in [B] for

a model of GMA. Suppose that G� is a P�{generic �lter over M . In M [G�] let

P = (!<!11 ;�) and let H be a P{generic �lter over M [G�]. Then in M [G�][H] there

are no essential Kurepa trees.

Proof: For any � < � the poset P� can be factored to P� � P� and G� can also be

written as G� �G� such that G� is a P�{generic �lter over M and G� is a P�{generic

�lter overM [G�]. Suppose T is a Kurepa tree inM [G�][H] with � branches. Without

loss of generality, let's assume that for every t 2 T there are exactly � branches of

T passing through t in M [G�][H]. In M [G�][H] let f : !2 7! B(T ) be a one to one

function such that for every t 2 T and for every � < !2 there exists a � 2 !2 r �

22

such that t 2 f(�). Notice that !2 here can be replaced by any regular cardinal �

satisfying !2 6 � < �. Without loss of generality, let us assume that

1P ( _T is a Kurepa tree and _f : !2 7! B( _T )

is a one to one function such that (8t 2 _T )(8� 2 !2)(9� 2 !2 r �)(t 2 _f(�))):

We want now to construct a poset R0 in M [G�] such that a �lter �G of R0 obtained

by applying a forcing argument similar to GMA in M [G�] will give us a P{name for

a Jech{Kunen subtree of T in M [G�][H].

Let r be a condition in R0 i� r = (Ir;Pr;Ar;Sr) where Ir is a countable subtree

of (!<!11 ;�), Pr = hprt : t 2 Iri, Ar = hAr

t : t 2 Iri and Sr = hSrt : t 2 Iri such that

(1) Pr � P, and for every t 2 Ir the element Art is a nonempty countable subtree of

(!<!11 ;�) of height �rt +1 (we will use some Ar

t 's to generate a Jech{Kunen subtree of

T ) and Srt is a nonempty countable subset of !2, (the requirement \Sr

t � !2" makes

R0 di�erent from R de�ned in Lemma 3,)

(2) (8s; t 2 Ir)(s � t$ prt 6 prs),

(3) (8s; t 2 Ir)(s � t! Art �ht(A

rs) = Ar

s),

(4) (8s; t 2 Ir)(s � t! Srs � Sr

t ),

(5) (8t 2 Ir)(prt Ar

t � _T ),

(6) (8t 2 Ir)(8� 2 Srt )(9a 2 (Ar

t )�rt )(prt a 2 _f(�)).

For any r; r0 2 R0 , let r 6 r0 i� Ir0 � Ir, and for every t 2 Ir0

pr0

t = prt ; Ar0

t = Art and Sr0

t � Srt :

Claim 9.1 The poset R0 is !1{linked.

Proof of Claim 9.1: Same as the proof of Claim 3.1. � (Claim 9.1)

Claim 9.2 The poset R0 is countably compact.

Proof of Claim 9.2: Same as the proof of Claim 3.2. � (Claim 9.2)

For each t 2 !<!11 de�ne

Dt = fr 2 R0 : t 2 Irg:

For each p 2 P de�ne

Ep = fr 2 R0 : (9t 2 Ir)(prt 6 p)g:

23

For each � < !1 de�ne

F� = fr 2 R0 : (8s 2 Ir)(9t 2 Ir)(ht(Art ) > �)g:

For each � < !2 de�ne

O� = fr 2 R0 : (8s 2 Ir)(9t 2 Ir)(s � t and [�; !2) \ Srt 6= ;)g:

Claim 9.3 All those Dt, Ep, F� and O�'s are dense in R0 .

Proof of Claim 9.3: Same as the proof of Claim 3.3. � (Claim 9.3)

Note that jR0 j = !2. Note also that M [G�][H] = M [H][G�]. By the construction

of P� there exists an � < � such that those dense sets Dt, Ep, F� and O� are in

M [G�], the tree T is in M [G�][H] or _T is in M [G�] and

1P� _Q� = R0 ;

i.e. R0 is the poset used in �{th step forcing in the �{stage iteration.

Let U� be a Q�{generic �lter over M [G�] such that G� � U� = G�+1.

Since Dt is dense for every t 2 !<!11 , then

IU� =[fIr : r 2 U�g = !<!1

1 :

Let

PU� =[fPr : r 2 U�g

and let

AU� =[fAr : r 2 U�g:

Notice that for any r; r0 2 U� and for any t 2 Ir \ Ir0 we have prt = pr

0

t and Art = Ar0

t

because r and r0 are compatible. So now for every t 2 IU� we can de�ne pt = prt for

some r 2 U� and de�ne At = Art for some r 2 U�. It is clear that the map t 7! pt is

an isomorphism between IU� and PU� , i.e. for any s; t 2 IU� we have s � t i� pt 6 ps.

It is also clear that the map t 7! At is a homomorphism from IU� to AU� , i.e. for any

s; t 2 IU� we have s � t implies At �ht(As) = As.

Claim 9.4 For each t 2 IU� the set fpt h i : 2 !1g is a maximal antichain

below pt in P.

24

Proof of Claim 9.4: Same as the proof of Claim 3.4. � (Claim 9.4)

The next claim is something di�erent from Lemma 3. Let TH =SfAt : pt 2 Hg

where H is the P{generic �lter over M [G�].

Claim 9.5 TH is a Jech{Kunen subtree of T in M [G�][H].

Proof of Claim 9.5: By the proof of Claim 3.5, it is easy to see that TH is a

subtree of T with more than !1 branches. It su�ces to show that TH has exactly !2

branches.

Suppose that TH has more than !2 branches. Then there is a branch B in

M [G�][H] which is not in the range of the function f . Without loss of generality,

let's assume that

1P (8� 2 !2)( _B 6= _f(�))

where _B is a P{name for B and let

D _B = fr 2 R0 : (8s 2 Ir)(9t 2 Ir)(s � t and ht( _B \ Art ) < ht(Ar

t ))g:

Since M [G�][H] = M [G�][H][G�] and P� is !1{closed in M [G�][H], then B is in

M [G�][H] because any !1{closed forcing will not add any new branches to the Kurepa

tree T . We assume also that the P{name _B is inM [G�]. Hence the setD _B is inM [G�].

Let

E _B = fprt 2 PU� : r 2 D _B \ U� and prt ht( _B \ Art ) < ht(Ar

t )g:

Subclaim 9.5.1 D _B is dense in R0 .

Proof of Claim 9.5.1: Let r0 be any element in R0 . It su�ces to show that there

is an element r in D _B such that r 6 r0. Let's �rst extend r0 to r0 such that for every

s 2 Ir0 there is a t 2 m(Ir0) such that s � t. Let t 2 m(Ir0). For every � 2 Sr0

t let

a� 2 (Ar0

t )�r0tsuch that pr

0

t a� 2 _f(�). Since we have

pr0

t (9u 2 _T )(u 2 _f(�)r _B)

and P is !1{closed, then there is a u� � a� in !<!11 for every � 2 Sr0

t and a pt 6 pr0

t

such that for every � 2 Sr0

t

pt u� 2 _f(�)r _B:

Without loss of generality, we can assume that there is a 2 !1 such that ht(u�) =

and

pt _B di�ers from all _f(�) below

25

for every � 2 Sr0

t . Let

Ir = Ir0 [ f�t : �t is a successor of t for t 2 m(Ir0)g:

For every t 2 Ir0 let

prt = pr0

t ; Art = Ar0

t and Srt = Sr0

t :

For every �t 2 Ir r Ir0 let

pr�t = pt; Ar�t = Ar0

t [ fs : s � u� for some � 2 Sr0

t g and Sr�t = Sr0

t :

Now it is easy to see that r 6 r0 and r 2 D _B. � (Subclaim 9.5.1)

Subclaim 9.5.2 E _B is dense in PU� .

Proof of Subclaim 9.5.2: Let p0 2 PU� . We need to show that there is a p 2 PU�such that p 6 p0 and p 2 E _B.

Since p0 2 PU� , then there is an r 2 U� such that p0 = prs. Since D _B is dense and

r 2 U�, then there is an r0 6 r such that r0 2 U� \D _B. Since prs = pr

0

s and r0 2 D _B,

then there is a t 2 Ir0 such that s � t and

pr0

t ht( _B \ Ar0

t ) < ht(Ar0

t ):

Hence we have pr0

t 6 prs = p0 and pr0

t 2 E _B. � (Subclaim 9.5.2)

We prove Claim 9.5 now. Assume B be a branch of T and B is not in the range

of f . We want to show that B is not a branch of TH . Suppose B is a branch of TH .

Then there is a p 2 H such that

p _B 2 B(T _H):

Since E _B is dense in P, then we can �nd a prt 2 E _B such that prt 6 p. Hence we

derived a contradiction because we have

prt _B 2 B(T _H);

prt T _H ��rt + 1 = Ar

t

and

prt ht( _B \ Art ) < ht(Ar

t ):

Hence in M [G�][H] the tree TH has !2 branches because B(TH) � f 00!2 (the range of

f). � (Claim 9.5)

By Claim 9.5 there are no essential Kurepa trees in M [G�][H]. �

26

Theorem 10 It is consistent with CH and 2!1 > !2 plus the existence of a thick

Kurepa tree that there exist essential Jech{Kunen trees and there are no essential

Kurepa trees.

Proof: Let M be a model of CH and 2!1 = � > !2 such that in M , �<� = � and

there is a thick Kurepa tree. Such model exists by Lemma 1. In M let

((P� : � 6 �); ( _Q� : � < �))

be the �{stage iterated forcing notion used in [B] for a model of GMA. Suppose G�

is a P�{generic �lter over M . Then

M [G�] j= CH + 2!1 = � > !2 +GMA:

In M [G�] let � be a cardinal such that !2 6 � < � and let S be a stationary{

costationary subset of !1. Suppose that H is a JS;�{generic �lter over M [G�]. Then

by Lemma 8, the tree TH =SfAp : p 2 Hg is an essential Jech{Kunen tree in

M [G�][H]. It is obvious that the thick Kurepa trees in M are still thick Kurepa

trees in M [G�][H]. We need only to show that there are no essential Kurepa trees in

M [G�][H].

Suppose that K is an essential Kurepa tree in M [G�][H]. Since jKj = !1, then

there exists an I � � such that jIj = !1 and K 2M [G�][HI ], where

HI = H \ JS;I = fp 2 H : dom(lp) � Ig:

Since JS;� is forcing equivalent to

JS;I � Fn(�r I; T _HI; !1))

and by Lemma 6, the tree THIis (!1 r S){complete, then by Lemma 7, there are no

new branches of K in M [G�][H] which are not in M [G�][HI ]. So K is still a Kurepa

tree in M [G�][HI ]. But the poset JS;I is !1{closed and has cardinality !1. So by

Lemma 2, the poset JS;I is forcing equivalent to (!<!11 ;�). Hence by Lemma 9, the

Kurepa tree K has a Jech{Kunen subtree K 0 in M [G�][HI ]. Since every branch of

K 0 is a branch of K and the set of branches of K stays the same in M [G�][HI ] and in

M [G�][H], then K 0 is still a Jech{Kunen subtree of K in M [G�][H]. This contradicts

that K is an essential Kurepa tree in M [G�][H] �

27

Remark: It is quite easy to build a model of CH and 2!1>!2 in which there exist

essential Jech{Kunen trees and there are no essential Kurepa tree without requiring

the existence of a thick Kurepa tree. LetM be a model of GCH. First, increase 2!1 to

!3 by an !1{closed Cohen forcing. Then, force with the poset JS;!2. In the resulting

model CH and 2!1 = !3 hold and there is an essential Jech{Kunen tree. It can be

shown easily that there are no thick Kurepa trees in the resulting model. Hence it is

trivially true that there are no essential Kurepa trees in that model.

3 New Proofs of Two Old Results.

In [SJ1], we proved that, assuming the consistency of an inaccessible cardinal, it is

consistent with CH and 2!1>!2 that there exist Jech{Kunen trees and there are no

Kurepa trees. The model for that is constructed by taking Kunen's model for non{

existence of Jech{Kunen trees as our ground model and then forcing with a countable

support product of !2 copies of a \carefully pruned" tree T . The way that the tree T

is pruned guarantees that (1) the forcing is !{distributive, (2) forcing does not add

any Kurepa trees, (3) T becomes a Jech{Kunen tree in the resulting model. In [Ji3],

this pruning technique was also used to construct a model of CH and 2!1>!2 in which

there exist essential Kurepa trees and there exist essential Jech{Kunen trees. Here

we realize that the Jech{Kunen tree obtained by forcing with that carefully pruned

tree in [SJ1] and [Ji3] can be replaced by a generic Jech{Kunen tree obtained by

forcing with JS;�, the poset de�ned in x2. So now we can reprove those two results in

[SJ1] and [Ji3] without going through a long and tedious construction of a \carefully

pruned" tree.

Let Lv(�; !1), the countable support L�evy collapsing order, denote a poset de�ned

by letting p 2 Lv(�; !1) i� p is a function from some countable subset of �� !1 to �

such that p(�; �) 2 � for every (�; �) 2 dom(p) and ordered by reverse inclusion.

Let Fn(�; 2; !1), the countable support Cohen forcing, denote a poset de�ned by

letting p 2 Fn(�; 2; !1) i� p is a function from some countable subset of � to 2 and

ordered by reverse inclusion.

Theorem 11 Let � and � be two cardinals in a model M such that � is strongly

inaccessible and � > � is regular inM . Let S 2M be a stationary{costationary subset

of !1 and let JS;� 2M be the poset de�ned in x2. Let Lv(�; !1) and Fn(�; 2; !1) be in

28

M . Suppose that G�H � F is a (Lv(�; !1)� Fn(�; 2; !1)� JS;�){generic �lter over

M . Then M [G][H][F ] j= (CH +2!1 > !2 + there exist Jech{Kunen trees + there are

no Kurepa trees ).

Proof: It is easy to see that

M [G][H][F ] j= (CH + 2!1 = � > � = !2):

It is also easy to see that !1 and all cardinals greater than or equal to � in M are

preserved. By Lemma 8, the tree TF =S

p2F Ap is a Jech{Kunen tree. We now need

only to show that there are no Kurepa trees in M [G][H][F ]. Suppose that K is a

Kurepa tree in M [G][H][F ]. Since jKj = !1, then there exists an I � � with jIj = !1

such that K 2M [G][H][FI ] where FI = F \ JS;I (recall that the poset JS;� is forcing

equivalent to JS;I � Fn(� r I; T _FI; !1)). By Lemma 7, the tree K is still a Kurepa

tree in M [G][H][FI ]. Since the poset JS;I is !1{closed and has cardinality !1, then

by Lemma 2, JS;I is forcing equivalent to Fn(!1; 2; !1). By a standard argument we

know that Fn(�; 2; !1)�Fn(!1; 2; !1) is isomorphic to Fn(�; 2; !1). Hence there is a

Fn(�; 2; !1){generic �lter H0 over M [G] such that M [G][H][FI ] = M [G][H 0]. But it

is easy to see that in M [G][H 0] there are neither Kurepa trees nor Jech{Kunen trees.

So we have a contradiction that K is a Kurepa tree in M [G][H 0]. �

Theorem 12 Let M be a model of GCH. Let � and � be two regular cardinals in

M such that � > � > !1 and let S be a stationary subset of !1 in M . In M let K �

and JS;� be two posets de�ned in x1 and x2, respectively. Suppose that G � H is a

K � � JS;�{generic �lter over M . Then

M [G�H] j= (CH + 2!1 = � > � > !1 +

there exist essential Kurepa trees + there exist essential Jech{Kunen trees ):

Proof: It is easy to see that M [G � H] is a model of CH and 2!1 = � > � > !1.

Since K � and JS;� are !1{closed, then K � is absolute with respect toM andM [H], and

JS;� is absolute with respect to M and M [G]. By Lemma 8, the tree TH =S

p2H Ap

is an essential Jech{Kunen tree in M [G][H]. By Lemma 1, the tree TG =S

p2GAp is

an essential Kurepa tree because M [G][H] = M [H][G]. �

29

References

[B] J. Baumgartner, \Iterated forcing", pp. 1|59 in Surveys in Set Theory, ed.

by A. R. D. Mathias, Cambridge University Press, 1983.

[Je1] T. Jech, \Trees", The Journal of Symbolic Logic, 36 (1971), pp. 1|14.

[Je2] , \Set Theory", Academic Press, New York, 1978.

[Ji1] R. Jin, \Some independence results related to the Kurepa tree", Notre Dame

Journal of Formal Logic, 32, No 3 (1991), pp. 448|457.

[Ji2] , \A model in which every Kurepa tree is thick", Notre Dame Journal

of Formal Logic, 33, No 1 (1992), pp. 120|125.

[Ji3] , \The di�erences between Kurepa trees and Jech{Kunen trees",

Archive For Mathematical Logic, to appear.

[Ju] I. Juh�asz, \Cardinal functions II", pp. 63|110 inHandbook of Set Theoretic

Topology, ed. by K. Kunen and J. E. Vaughan, North{Holland, Amsterdam,

1984.

[K1] K. Kunen, \On the cardinality of compact spaces", Notices of the American

Mathematical Society, 22 (1975), 212.

[K2] , \Set Theory, an introduction to independence proofs", North{

Holland, Amsterdam, 1980.

[SJ1] S. Shelah and R. Jin, \A model in which there are Jech{Kunen trees but there

are no Kurepa trees", Israel Journal of Mathematics, to appear.

[SJ2] , \Planting Kurepa trees and killing Jech{Kunen trees in a model by

using one inaccessible cardinal", Fundamenta Mathematicae, to appear.

[T] S. Todor�cevi�c, \Trees and linearly ordered sets", pp. 235|293 in Handbook of

Set Theoretic Topology, ed. by K. Kunen and J. E. Vaughan, North{Holland,

Amsterdam, 1984.

30

Department of Mathematics

University of California

Berkeley, CA 94720

e-mail: jin@math.berkeley.edu

Institute of Mathematics,

The Hebrew University,

Jerusalem, Israel.

Department of Mathematics,

Rutgers University,

New Brunswick, NJ, 08903, USA.

Sorting: The �rst address is the �rst author's and the last two are the second

author's.

31