Renling Jin and Saharon Shelah- Essential Kurepa Trees Versus Essential Jech-Kunen Trees

8/3/2019 Renling Jin and Saharon Shelah- Essential Kurepa Trees Versus Essential Jech-Kunen Trees

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Essential Kurepa Trees Versus

Essential JechKunen Trees1

Renling Jin2 & Saharon Shelah3

AbstractBy an 1tree we mean a tree of size 1 and height 1. An 1tree iscalled a Kurepa tree if all its levels are countable and it has more than 1branches. An 1tree is called a JechKunen tree if it has branches forsome strictly between 1 and 2

1. A Kurepa tree is called an essentialKurepa tree if it contains no JechKunen subtrees. A JechKunen tree iscalled an essential JechKunen tree if it contains no Kurepa subtrees. Inthis paper we prove that (1) it is consistent with CH and 21 > 2 thatthere exist essential Kurepa trees and there are no essential JechKunentrees, (2) it is consistent with CH and 21 > 2 plus the existence of aKurepa tree with 21 branches that there exist essential JechKunen trees

and there are no essential Kurepa trees. In the second result we require theexistence of a Kurepa tree with 21 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 for


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

and T has more than 1 branches. An 1tree T is called a JechKunen tree ifT has

branches for some strictly between 1 and 21. We call a Kurepa tree thick if it

has 21 branches. Obviously, a Kurepa non-JechKunen tree must be thick, and a

JechKunen tree with every level countable is a Kurepa tree.

While Kurepa trees are better studied, JechKunen trees are relatively less popular.

It is K. Kunen [K1][Ju], who brought JechKunen trees to peoples attention by

proving that: under CH and 21 > 2, the existence of a compact Hausdorff space

with weight 1 and size strictly between 1 and 21 is equivalent to the existence

of a JechKunen tree. It is also easy to observe that: under CH and 21 > 2, the

existence of a (Dedekind) complete dense linear order with density 1 and size strictly

between 1 and 21 is also equivalent to the existence of a JechKunen tree. Above

results are interesting because those compact Hausdorff spaces and complete denselinear orders cannot exist if we replace 1 by , while the existence of a JechKunen

tree is undecidable. In this paper we would like to consider JechKunen trees only

under CH and 21 > 2 .

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

structed a generic Kurepa tree with less than 21 branches in a model of CH and

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

the consistency of nonexistence of JechKunen trees with CH and 21 > 2 (see [Ju,

Theorem 4.8]). In Kunens 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 differences between Kurepa trees and JechKunen 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

21 > 2 that there exist Kurepa trees but there are no JechKunen trees [SJ1], (2) it

is consistent with CH and 21 > 2 that there exist JechKunen trees but there are

no Kurepa trees [SJ2].

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

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

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

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

JechKunen tree or kill all JechKunen subtrees of a Kurepa tree without using large

cardinals. Lets call a Kurepa tree T essential if T has no JechKunen subtrees, and


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call a JechKunen tree T essential if T has no Kurepa subtrees. In [Ji1], the first

author proved that it is consistent with CH and 21 > 2 , together with Generalized

Martins Axiom and the existence of a thick Kurepa tree, that no essential Kurepa

trees and no essential JechKunen trees. We required the presence of thick Kurepa

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

is consistent with CH and 21 > 2 that there exist both essential Kurepa trees and

essential JechKunen 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 1, we prove that it is consistent with CH and 21 > 2 that there exist essential

Kurepa trees but there are no essential JechKunen trees. In 2, we prove that it

is consistent with CH and 21 > 2 plus the existence of a thick Kurepa tree that

there exist essential JechKunen trees but there are no essential Kurepa trees. In 3,we simplify the proofs of two old results by using the forcing notion for producing a

generic essential JechKunen tree defined in 2.

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 JechKunen Trees.

In this section we are going to construct a model of CH and 21 > 2 in which

there exist essential Kurepa trees and there are no essential JechKunen trees. Our

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

21 > 2 plus GMA (Generalized Martins Axiom) as our ground model, so that in

the ground model there are no essential JechKunen trees, then, we add a generic

Kurepa tree which has no JechKunen subtrees. The hard part is to prove that the

forcing adds no essential JechKunen trees.

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

compatible in P. A poset P is called 1linked ifP is the union of 1 linked subsets

ofP. A subset S ofP is called centered if every finite subset of S has a lower bound

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

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


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Suppose P is an 1linked and countably compact poset. For any

< 21, if D = {D : < } is a collection of dense subsets ofP,

then there exists a filter G ofP such that G D = for all < .

We choose the form ofGMA from [B], where a model ofCH and 21 > 2 plus GMA

can be found.

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

iff p = (Ap, lp) where Ap is a countable subtree of ( 1, then TG will be a Kurepa tree with |I|

branches in M[G]. KI 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 KI as an iterated forcing notion

KI F n(I I, TGI

, 1),

for any I I, where GI is a KIgeneric filter over the ground model and F n(I

I, TGI , 1), in M[GI], is the set of all functions from some countable subset of I I

to TGI with the order defined by letting p q iff dom(q) dom(p) and for any

i dom(q), p(i) q(i). The poset F n(J, TG, 1) is in fact the countable supportproduct of |J|copies of TG. We call 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 KI and KI F n(I I, TGI

, 1) are forcing equivalent because the map

F : KI KI F n(I I, TGI , 1)


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such that for every p KI,

F(p) = ((Ap, lp I), lp I I

)

is a dense embedding.

Lemma 1 (K. Kunen). LetM be a model of CH. Suppose that > 2 is a cardinal

in M andK M. Suppose G is aKgeneric filter over M and TG =pG

Ap.

Then inM[G] the tree TG is a Kurepa tree with branches andTG has no subtrees

with branches for some 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 in M[G]. Since |T| = 1, then there exists

a subset I in M with cardinality 1 such that T M[GI], where

GI = {p G : dom(lp) I}.

Notice that TG = TGI (in fact TG = TG). Since in M[GI] the tree TGI has only |I|

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 |B| = 1, there exists a subset J of I

with cardinality 1 such that B M[GI][HJ], where HJ is a F n(J, TGI , 1)generic

filter over M[GI]. Now I can be partitioned into many subsets of cardinality 1

and for every subset J (I J) of cardinality 1 the poset PJ = F n(J, TGI , 1)

is isomorphic to the poset PJ = F n(J, TGI , 1) through an obvious isomorphism

induced by a bijection between J and J. Let B be a PJname for B. Then (B)

is a PJname for a new branch of T. Forcing with PJ PJ will create two different

branches BHJ and ((B))HJ . Hence forcing with F n( 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 1closed poset of size 1 (hence CH must hold). Then

the tree ( 2 plus GMA and letP = (


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Proof: Let T be a JechKunen tree in M[G] with branches for 1 < < = 21.

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

1 < and for every t T there are at least branches ofT passing through t in

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

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

of generality let us assume that

1P (T is a JechKunen tree and f : B(T)

is a one to one function such that (t T)( )( )(t f())).

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

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

Let r be a condition ofR iff r = (Ir,Pr, Ar, Sr) where Ir is a countable subtree of

(


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Claim 3.2 The poset R is countably compact.

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

that for any finite R0 R and for any t

{Ir : r R

0} all p

rt s are same and all

Art are same for r R0 because R

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

to construct a condition r R such that r is a common lower bound ofR. Let

(1) Ir =

rR Ir,

(2) Pr = prt : t Ir where prt = p

rt for some r R

such that t Ir,

(3) Ar = Art : t Ir where A

rt = A

rt for some r R

such that t Ir,

(4) Sr = Srt : t Ir where Srt =

st Ss and Ss =

{Srs : (r R)(s Ir)}.

Notice that from the argument above all prt s, Art s and S

rt s are welldefined. We

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

R if r R.

It is easy to see that r satisfies (1), (2), (3), (4) and (5) in the definition of acondition in R. Lets check (6).

Suppose t Ir and Srt . We want to show that there exists an a (Art )rt such

that prt a f(). Let r R be such that t Ir, let r R and s Ir be such

that s t and Sr

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

such that r r and r r. By the facts that

prt = prt = p

r

t , Art = A

rt = A

r

t , Sr

s Sr

s Sr

t

and r R we have now that there exists an a (Art )rt such that prt a 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 )}.

For each < define

O = {r R : (s Ir)(t Ir)(s t and [, ) Srt = )}.


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Claim 3.3 All those Dt, Ep, F and Os are dense in R.

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

We show first that for every t


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and s Ir0 be such that Sr0s . Then for any s Ir0 such that s

s t0 there

exists an as, (Ar0s )r0s such that p

r0s as, f(). Now let

a =

{as, : s s t0 and s Ir0}

and let

At0 = At0

{a : St0}.

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 endextension of Ar0s , i.e.

At0 ht(Ar0s ) = A

r0s ,

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

pt0 a f().

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

prs = pr0s , A

rs = A

r0s and S

rs = S

r0s .

Otherwise let

prs = ps, Ars = At0 and S

rs = St0 .

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

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

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

Lets assume that for every t Ir0 pr0t p. Let

t0 =

{t Ir0 : p pr0t }.

Case 1: t0 Ir0 .

Let t = t0 0, i.e. t is a successor oft0. It is clear that t

Ir0. Let Ir = Ir0 {t}.

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

prt = p, Art = A

r0t0

and Srt = Sr0t0

.

Otherwise let

prt = pr0t , A

rt = A

r0t and S

rt = S

r0t .

Then we have r Ep and r r0.

Case 2: t0 Ir0 .


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Let Ir = Ir0 {t0}. We construct St0, At0

and then At0 exactly same as we did in

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

let

prt = p, Art = At0 and S

rt = St0.

Otherwise let

prt = pr0t , A

rt = A

r0t and S

rt = S

r0t .

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

(p, p P)(p p p Ep p Ep).

We show next that for every 1 the set F is dense in R. We need to find an

r F such that r r0.

Let Ir Ir0 be such that Ir is a countable subtree of


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By imitating the proof of the denseness of F we can find an r r0 such that

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

s t. For every t Ir Ir0 fix a t which is an successor of t (for example t = t 0).

Let

Ir = Ir {t : t Ir Ir0}.

For every t Ir let

prt = pr

t , Art = A

r

t and Srt = S

r

t .

For every t with t Ir Ir0 we want to construct prt , A

rt and S

rt . If there is a S

r

t

which is greater than , then let prt be any proper extension ofprt , let A

rt = A

r

t and let

Srt

= Sr

t . Otherwise, first, pick an a in the top level of Ar

t , then choose a

and a p pr

t such that p a f(). This can be done because

1P (t T)( )( )(t f())

is true in M. Now let

prt = p, Art = A

r

t and Srt = S

r

t {}.

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

By applying GMA in M we can find an Rfilter H such that HDt = , HF =

and H Ep O = for each t


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Claim 3.4 For each t IH the set {pt : 1} is a maximal antichain below

pt in P.

Proof of Claim 3.4: Let and be two ordinals in 1. Since IH = . Then for

every t IH, s t, there is an t IH, t t

, such that

pt a f()

for some a (At)ht(At). This shows that

ps f() is a branch of TG,

which contradicts ps p and

p ( )(f() is not a branch of TG).

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

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


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Theorem 4. It is consistent with CH and 21 > 2 that there exist essential Kurepa

trees and there are no essential JechKunen trees.

Proof: Let M be a model of CH and 21 = > 2 plus GMA. Let K M.

Suppose G is a Kgeneric filter over M. We are going to show that M[G] is amodel of CH and 21 > 2 in which there exist essential Kurepa trees and there are

no essential JechKunen trees.

It is easy to see that M[G] satisfies CH and 21 > 2 . Lemma 1 implies that

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

essential JechKunen trees.

Assume T is a JechKunen tree in M[G]. We need to show that T has a Kurepa

subtree in M[G]. Since |T| = 1, then there is an I of cardinality 1 in M such

that T M[GI], where


We claim that

B(T) M[G] M[GI].

If the claim is true, then T is a JechKunen tree in M[GI]. Suppose that B

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

HJ is a F n(J, TGI , 1)generic filter over M[GI]. Let B be a F n(J, TGI , 1)name for

B. For any J (I J) such that |J| = |J| there is an isomorphism from

F n(J, TGI , 1) to F n(J

, TGI , 1) induced by a bijection between J and J

. Sincein M[G], the branches (B)HJ and ((B))HJ are different, then T has at least

branches. This contradicts that T is a JechKunen tree. Let T have branches in

M[GI]. Since KI has size 1 and is 1closed, then it contains a dense subset which

is isomorphic to P = ( 2 plus the existence of

a thick Kurepa tree, in which there are essential JechKunen 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.


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We first take a model M of CH and 21 = > 2 plus a thick Kurepa tree, where

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

JechKunen trees nor essential Kurepa trees. In stead of taking a model of GMA as

our ground model like we did in 1, 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 a 1closed poset JS, in M[G] to create a generic essential JechKunen

tree, where S is a stationarycostationary 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) denote the set

{t T : (s T)(s T t s = t)}.

Let I be any index set and let S be a subset of 1. We define a poset JS,I such that

p is a condition in JS,I iff p = (Ap, lp) where

(1) Ap is a countable subtree of


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Define a condition p JS,I such that

Ap = A {l(i) : i D} and lp = l.

We claim that p is a lower bound of the sequence {pn : n }. It suffices to show

that for any n and for any t Ap Apn either there exists an s m(Apn) such thats t or that < ht(Apn) and is a limit ordinal imply

=

{ht(s) : s Apn and s t} S.

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

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

=

{ht(s) : s Apn and s t} S

because pk pn. If t = l(i) for some i D, then because oft Apn , there is a k > nand there is a t Apk Apn such that t

t. Hence either there is an s m(Apn)

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

=

{ht(s) : s Apn and s t} S

because pk pn.

Remark: Again, we may consider the poset JS,I as a twostep iterated forcing

JS,I F n(I I, TGI , 1), where I is a subset of I, TGI =

{Ap : p GI} for a

generic filter GI of JS,I and F n(I I, TGI , 1) is a countable support product of

|I I|copies of TGI . The map

p = (Ap, lp) ((Ap, lp I), lp I I

)

is a dense embedding from JS,I to JS,I F n(I I, TGI

, 1).

We now define Scompleteness 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 Scomplete if for

every limit ordinal S and every B B(T ) the union

B T, i.e. everystrictly decreasing sequence of T has a greatest lower bound b in T if ht(b) S.

Lemma 6. LetM be a model ofCH and letJS,I M where S 1 and I is an index

set in M. Suppose G is aJS,Igeneric filter over M. Then the tree TG =pG Ap is

(1 S)complete in M[G].


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Proof: Let 1 S be a limit ordinal and let B be a branch ofTG . We need

to show that t =

B TG. The set B is in M because JS,I is 1closed and B is

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

p0 B TG

.

Let

DB = {p JS,I : p p0 and t =

B Ap}.

Then DB is dense below p0 because for any p p0 the element p = (Ap {

B}, lp)

is a condition in JS,I and p p (here we uses the fact that 1S). Since p0 G,

then there is a p G DB. Hence t =

B TG.

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

T be an 1tree which is Ucomplete and let I be any index set. Let K M be any

1tree such that every level of K is countable. Suppose P = F n(I , T , 1) M and

G is aPgeneric filter 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]M. Without loss of generality,

lets assume that

1P

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

(p P)( 1)(t 1 )(p

p)(p t B)

and

(p P)( 1)(t 1 )(p t B

( 1 )( 1 )(tj 1 )(t0 = t1)(pj p)(pj tj B))

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

Lets work in M. Let be a large enough cardinal and let N be a countable

elementary submodel of (H(), ) such that K,P, B N. Let = N 1 U (such

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

{n : n } such that

n n = . Again in M we construct a set

{ps : s 2


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and a set

{ts : s 2


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It is easy to see that if f, f 2 are different, then tf and tf are different. Hence

K is uncountable, a contradiction.

Lemma 8. LetM be a model of CH and 21 = > 2 and letJS, M where is

a cardinal in M such that 1 < < and S is a stationary subset of 1. Supposethat G is a JS,generic filter over M. Then in M[G] the tree TG =

pG Ap is an

essential JechKunen 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 let

B() =

{lp() : p G and dom(lp)}.

Then

B(TG) = {B() : }

and for any two different and in the branches B() and B( ) are different.

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

D, = {p JS, : dom(lp) and ht(lp()) > }

is dense in JS,, then B() is a branch of TG. For any two different , the set

D, = {p JS, : , dom(lp) and lp() = lp(

)}

is also dense in JS,. So the branches B() and B() are different.

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

{B() : }.

Without loss of generality, let us assume that

1JS, B (B(TG) {B() : }).

Work in M. Let be a large enough cardinal and let N be an elementary submodel

of (H(), ) such that ,S, B, B = {B() : }, JS, N and if p N JS,, then

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

countable ordinals {n : n } such that =

n . We now want to find a

decreasing sequence {pn : n } JS, N such that p0 = 1JS, and for each n

(1) ( dom(lpn))(t Apn+1)(pn+1 t B() B),


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(2) ( dom(lpn+1))(t Apn+1 Apn)(ht(t) ht(Apn) and pn+1 t B),

(3) ht(Apn) n.

Assume we have found {p0, p1, . . . , pn}. We now work in N. Let

dom(lp) = {k : k }

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

there is a t Aqk such that

qk t B(k) B.

Assume, in N, that we have found {q0, q1, . . . , qk}. Since the sentence

qk (t TG)(t B(k) B)

is true in N (because it is true in H() and k N), then there is a t (


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and

Ap = (

n

Apn) {a : dom(lp)}.

By the construction of pns we have

{ht(t) : t Ap and p t B} = S.

Pick any t Ap. If t = a for any dom(lp), then we can find a 1 such that

t Ap. Extend t to t delta1 . Define p such that

Ap = Ap {u : t u t}

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

ht(a) = , then b = a). Define p such that

Ap = Ap {u : t u b}

and

lp = (lp (dom(lp) {})) {(, b)}.

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

a =

{t Ap : p t B}.

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

fact S, is a limit ordinal and ht(Ap) > . Hence

p B TG B Ap.

This contradicts that

p B is a branch of TG.

(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

|K| = 1, then there is an I such that |I| 1 and K M[GI], where


Notice that GI is a JS,Igeneric filter over M. Since JS, is forcing equivalent to

JS,I F n( I, TGI , 1) and TGI is (1S)complete in M[GI] (notice that S is still

stationarycostationary), 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 = TGI has only |I| branches in M[GI] and K is a


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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 21 = > 2 with


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For any r, r R, let r r iff Ir Ir, and for every t Ir

pr

t = prt , A

r

t = Art and S

r

t Srt .

Claim 9.1 The poset R is 1linked.

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

Claim 9.2 The poset R is countably compact.


For each t )}.

For each < 2 define

O = {r R : (s Ir)(t Ir)(s t and [, 2) S

rt = )}.

Claim 9.3 All those Dt, E

p, F

and O

s are dense in R.


Note that |R| = 2. Also note that M[G][H] = M[H][G] because P is 1

closed. By the construction ofP 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 = R,

i.e. R is the poset used in th step forcing in the stage iteration.

LetH be a Qgeneric filter over

M[G] such that

G

H =

G+1.

Since Dt is dense for every t


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and let

AH =

{Ar : r H}.

Notice that for any r, r H and for any t Ir Ir we have prt = pr

t and Art = A

r

t

because r and r are compatible. So now for every t IH we can define pt = prt for

some r H and define At = Art for some r H. It is clear that the map t pt is

an isomorphism between IH and PH , i.e. for any s, t IH we have s t iffpt ps.

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

any s, t IH we have s t implies (At)ht(As) = As.

Claim 9.4 For each t IH the set {pt : 1} is a maximal antichain below

pt in P.

Proof of Claim 9.4: Same as the proof of Claim 3.4.

The next claim is something different from Lemma 3. Let TH = {At : pt H}

where H is the Pgeneric filter over M[G].

Claim 9.5 TH is a JechKunen subtree of T in M[G][H].

Proof of Claim 9.5: By the proof of Claim 3.5, we know that TH is a subtree ofT

with more than 1 branches. It suffices 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, lets assume

that1P ( 2)(B = f()).

Let B be a Pname for B and let

DB = {r R : (s Ir)(t Ir)(s t and ht(B A

rt ) < ht(A

rt ))}.

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

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

tree T. We assume also that the Pname B is in M[G]. Hence the set DB is in M[G].

Let

EB = {prt PH : r DB H and p

rt ht(B A

rt ) < ht(A

rt )}.

Subclaim 9.5.1 DB is dense in R.

Proof of Claim 9.5.1: Let r0 be any element in R. It suffices to show that there

is an element r in DB such that r r0. Lets first extend r0 to r such that for every


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s Ir0 there is a t m(Ir) such that s t. Let t m(Ir). For every Sr

t let

a (Ar

t )rtsuch that pr

t a f(). Since we have

pr

t (u T)(u f() B)

and P is 1closed, then there is a u a in 2 plus the existence of a thick

Kurepa tree that there exist essential JechKunen trees and there are no essential

Kurepa trees.


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Proof: Let M be a model of CH and 21 = > 2 such that in M, 2 + GMA.

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

costationary subset of 1. Suppose that H is a JS,generic filter over M[G]. Then

by Lemma 8, the tree TH =

{Ap : p H} is an essential JechKunen 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 |K| = 1, then

there exists an I such that |I| = 1 and K M[G][HI], where

HI = H JS,I = {p H : dom(lp) I}.

Since JS, is forcing equivalent to

JS,I F n( I, THI , 1))

and by Lemma 6, the tree THI is (1 S)complete, then by Lemma 7, there are nonew 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 1closed and has size 1. So by Lemma 2,

the poset JS,I is forcing equivalent to ( 2 in which there exist

essential JechKunen trees and there are no essential Kurepa tree without requiring

the existence of a thick Kurepa tree. Let M be a model ofGCH. First, increase 21 to

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

model CH and 21 = 3 hold and there is an essential JechKunen tree. It can be


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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 21 > 2 that there exist JechKunen trees and there are no

Kurepa trees. The model for that is constructed by taking Kunens model for non

existence of JechKunen trees as our ground model and then forcing with a countable

support product of2 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 JechKunen tree in the resulting model. In [Ji3],

this pruning technique was also used to construct a model ofCH and 21 > 2 in which

there exist essential Kurepa trees and there exist essential JechKunen trees. Here

we realize that the JechKunen tree obtained by forcing with that carefully pruned

tree in [SJ1] and [Ji3] can be replaced by a generic JechKunen tree obtained by

forcing with JS,, the poset defined in 2. 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 Levy collapsing order, denote a poset defined

by letting p Lv(, 1) iff p is a function from some countable subset of 1 to 2

such that p(, ) for every (, ) dom(p) and orderd by reverse inclusion.Let F n(, 2, 1), the countable support Cohen forcing, denote a poset defined by

letting p F n(, 2, 1) iff 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. LetS M be a stationarycostationary subset

of 1 and letJS, M be the poset defined in 2. LetLv(, 1) and F n(, 2, 1) be in

M. Suppose that G H F is a (Lv(, 1) F n(, 2, 1) JS,)generic filter over

M. ThenM[G][H][F] |= (CH+ 21 > 2 + there exist JechKunen trees + there are

no Kurepa trees ).

Proof: It is easy to see that

M[G][H][F] |= (CH + 21 = > = 2).


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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 =pF Ap is a JechKunen 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 |K| = 1, then there exists an I with

|I| = 1 such that K M[G][H][FI] where FI = F JS,I (recall that the poset JS,

is forcing equivalent to JS,I F n( I, TFI , 1)). By Lemma 7, the tree K is still a

Kurepa tree in M[G][H][FI]. Since the poset JS,I is 1closed and has size 1, then

by Lemma 2, JS,I is forcing equivalent to F n(1, 2, 1). By a standard argument we

know that F n(, 2, 1) F n(1, 2, 1) is isomorphic to F n(, 2, 1). Hence there is a

F n(, 2, 1)generic filter H over M[G] such that M[G][H][FI] = M[G][H

]. But it

is easy to see that in M[G][H] there are neither Kurepa trees nor JechKunen trees.

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

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 letK

and JS, be two posets defined in 1 and 2, respectively. Suppose that G H is a

K JS,generic filter over M. Then

M[G H] |= (CH + 21 = > > 1 +

there exist essential Kurepa trees + there exist essential JechKunen trees ).

Proof: It is easy to see that M[G H] is a model of CH and 21

= > > 1.Since K and JS, are 1closed, then K is absolute with respect to M, and M[H] and

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

is an essential JechKunen tree in M[G][H]. By Lemma 1, the tree TG =pG Ap is

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

References

[B] J. Baumgartner, Iterated forcing, pp. 159 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. 114.[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. 448457.[Ji2] , A model in which every Kurepa tree is thick, Notre Dame Journal of Formal Logic,

33, No 1 (1992), pp. 120125.[Ji3] , The differences between Kurepa trees and JechKunen trees, Archive For Mathe-

matical Logic, to appear.


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[Ju] I. Juhasz, Cardinal functions II, pp. 63110 in Handbook of Set Theoretic Topology,ed. by K. Kunen and J. E. Vaughan, NorthHolland, Amsterdam, 1984.

[K1] K. Kunen, On the cardinality of compact spaces, Notices of the American MathematicalSociety, 22 (1975), 212.

[K2] , Set Theory, an introduction to independence proofs, NorthHolland, Amsterdam,1980.

[SJ1] S. Shelah and R. Jin, A model in which there are JechKunen trees but there are no Kurepatrees, Israel Journal of Mathematics, to appear.

[SJ2] , Planting Kurepa trees and killing JechKunen trees in a model by using one inac-cessible cardinal, Fundamenta Mathematicae, to appear.

[T] S. Todorcevic, Trees and linearly ordered sets, pp. 235293 in Handbook of Set TheoreticTopology, ed. by K. Kunen and J. E. Vaughan, NorthHolland, Amsterdam, 1984.

Department of Mathematics

University of California

Berkeley, CA 94720

e-mail: [email protected]

Institute of Mathematics,

The Hebrew University,

Jerusalem, Israel.

Department of Mathematics,

Rutgers University,

New Brunswick, NJ, 08903, USA.

Sorting: The first address is the first authors and the last two are the second

authors.

Documents

Renling Jin and Saharon Shelah- Essential Kurepa Trees Versus Essential Jech-Kunen Trees