Souslin Problem

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

405 Keith J. Devlin H,~vard Johnsbraten

The Souslin Problem

Springer-Verlag Berlin-Heidelberg • New York 1974

Dr. Keith Devlin Seminar fer Logik und Grundlagenforschung der Philosophischen Fakult~t 53 Bonn BeringstraBe 6 BRD Dr. Havard Johnsbr&ten Matematisk Institutt Universitetet I Oslo, Blindern Oslo 3/Norge

Library of Congress Cataloging in Publication Data

Devlin, Keith J The Souslin problem. (Lecture notes in mathematics, 405) Bibliography: p. 1. Set theory. I. Johnsbraten, Havard, 1945joint author. II. Title. III. Series: Lecture notes in mathematics (Berlin) 405. QA3.L28 no. 405 [QA248] 510'.8s [5!i'.3] 74-17386

AMS Subject Classifications (1970): 02-02, 02 K05, 02 K25, 04-02, 04A30 ISBN 3-540-06860-0 Springer-Verlag Berlin • Heidelberg • New York ISBN 0-38?-06860-0 Springer-Verlag New York • Heidelberg • Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1974. Printed in Germany. Offsetdruck. Julius Beltz, Hemsbach/Bergstr.

Acknowledgements

The last five chapters of this book are based almost exclusively on a set of notes written by Devlin during the S11mmer of 1972.

During this time, he was living with,

and almost en-

tirely supported by his wife's parents, Mr and Mrs H. Carey of Sidcup, Kent, to whom thanks

should now be put on record.

The manuscript was typed by Mrs. R. Msller of the University of Oslo, for whose care and patience we owe a considerable debt.

CONTENTS

Chapter I. II.

Preliminaries

I

Souslin's Hypothesis

8

The Combinatorial Property

23

Homogeneous Souslin Trees and Lines

32

Rigid Souslin Trees and Lines

46

Martin's Axiom and the Consistency of SH

5~

VII.

Towards

68

V~I.

Iterated Forcing Jensen Style

74

How J e n s e n K i l l e d

97

III. IV. V. VI.

IX. X.

ConZFC

Con(ZFC + C H + SH) : A False Start

~

a Souslin Tree

Con(ZFC+CH+SH)

113

APPENDIX. Iterated Souslin Forcing and ~ ( ~ )

122

REFERENCES

426

GLOSSARY OF NOTATION

128

INDEX

CORRECTION

A n Israel£ mathematician

called Uri Avraham has pointed

t~at~ in Chapter

IX, the claim straddling

false as stated.

To be precise,

define,

for C a certain

order-embedding

~a+2

(although

on page

should have

In fact,

given an Aronszajn

closed unbounded

subset

tree ~, we

of ~ ,

a ('generic)

to all of ~° As we have set things up this

a slight modification

to the definition

105 would in fact make our claim said is not that our particular

all of @, but

105 and 106 is

of TIC into Q, and claim that this order-embed-

ding can be extended is false

pages

out

that it easily

this is perhaps

in terms of our original

true).

of

What we

embedding

extends

to

gives rise to such an embedding.

more easily definition

seen

(via Theorem II.5)

of special

Aronszajn

(Page

15)= By means of our embedding Um< Am, where

of TIC into Q, we may write TIC =

each A m is an (uncountable)

~:aml~<~1 > be a or~e-one enumeration ~<~I > be the monotone e a c h ~ x e T c , let s x = able,

let

lysTlc

1- , , ( W z ~c~< & )~'-u { U - -, m - , m

Keith J. Devlin 1 4.9 • 74

of A m , each m, and let <m~l

+ I I x<~yJ.

is clearly

of T. Let

of C. For each m< ~I' and for

<sn(x)In< ~> enumerate

~n,~m = [Sn(a~m) Ira< ~I

Bonn,

enumeration

antichain

Since

each S x is count-

it. For each ~ m <

an antichain

)' so ~ i s special.

Q,ED.

~, the set

of T. But ~ =

Introduction

In 1920, the awakening of interest in Set Theory was heralded by the app~arence

of a new journal in mathematics

-

Fundamenta Mathe-

maticae.

It is fitting that in the very first volume appeared the

statement

of a problem which was to play a prominent

opment of that subject.

Indeed,

role in the devel-

as with the continuum problem,

re-

search into the simple question there raised by M. Souslin was to have far-reaching

ramifications

in constructibility

in several branches

theory and in the theory of forcing,

subjects whose existence However,

notably

the very two

stemmed from the work on the continuum problem.

As with the continuum problem, bers.

of set theory,

Souslin's problem concerned the real num-

instead of asking

"how many" reals there are, Souslin

asked whether a certain set of conditions uniquely characterized real numbers.

More precisely,

Souslin's problem was this.

are given a dense linearly ordered set

X , complete

say, Dedekind cuts) and with no end-points, erty that there is no uncountable intervals

of

(Probl~me

3, M. Souslin [Su I]. )

undecidable

X .

Must then

X

collection

Suppose we

(in the sense of, X

has the prop-

of pairwise disjoint open

be (isomorphic

to) the real line,]R ?

It turned out that this problem was

on the basis of the usual

of set theory.

and that

the

(i.e. Zermelo - Fraenkel)

axioms

It is the purpose of this short book to provide the

proofs of this undecidability

and of the undecidability

even when the continuum hypothesis

is assumed.

these results are due solely to one person,

of the problem

The vast majority of

Ronald Jensen.

(In fact,

Chapter VI is essentially the only place where results appear which are not his.)

For the most part, these results appear here for the first

time in print. The exposition falls naturally into two parts. ters, we discuss the non-provability

of Souslin's hypothesis

assumption that Souslin's problem has a positive This part which also contains problem~ sonable

In the five first chapanswer)

in

(Chapter II) a basic discussion

should be fairly easy to read.

The prerequisites

(i.e. the ZFC ~ CH. of the

are a rea-

(e.g. Ist year graduate level) aquaintance with the method of

forcing and a knowledge

of constructibility

sketch briefly the material we require.

theory.

In chapter I we

It is natural to include in

this part some discussion of the notion of a Souslin line (i.e. a "continuum" which does possess the properties

stated in Souslin's problem,

VIII

but which is not isomorphic such a discussion.

to

~R ), and Chapters

IV and V provide just

The second part - Chapters VI - X

tirely with the consistency

of Souslin's hypothesis.

-

is concerned en-

Strictly speaking,

the forcing theory outlined in Chapter I is all that is required for an understanding

of the proofs, but some considerable

kind of material

is really necessary.

part tend to be extremely long and technical. VIII

-

aquaintance with this

In particular,

the proofs in this

We expect that chapters

X will only be read in detail by the highly motivated

that motivation may be ~ ).

stuff of "first-year graduate

courses"~

The book comes from two main sources.

For the first part of the book,

a set of lecture notes written b y Jensen in Kiel in 1969. cond part,

on a (much larger)

by Jensen some years earlier. Jensen manuscript,

Con ( Z F C + S H +CH) however,

These

set of notes written

For readers who have previously

seen the

let us say now that the proof of

which we give here is essentially the same as the

proof described there. ferent,

For the se-

a set of notes written by Devlin in Sidcup in 1972.

latter were based exclusively original

(whatever

These three chapters are certainly not the

The organisation

of the proof is somewhat dif-

and the length is considerably

reduced.

Also

(hope-

fully) we have found all the errors which the original manuscript

con-

tained. We have tried to adopt the terminology rent set-theoretical and use either

QED

usage. or else

this is not really a textbook

We use |

and notation most common in cur-

'iff' to denote

'if and only if'

to denote the end of a proof.

Since

(rather an account of Jensen's work) we

have not included any exercises.

The reader who completes the entire

book will not need any further exercise We thank Thomas Jech for reading the last five chapters some improvements.

Keith J. Devlin & H~vard Johnsbr~ten Department

of Mathematics

University

of Oslo

June,

1974

and suggesting

Chapter

I. PRELIMINARIES

I. Set Theor~ We shall be c o n c e r n e d set theory ~De1~,

including

~e1~,

its p r e d e c e s s o r s , numerous

class

with

~

is a ~imit dinal

~

If

is a set,

X

theory,

is said

LST

written

denotes

note f o r m u l a s

E in

.

if

order

(¥~E a)(~

(set m e m b e r s h i p ) . LST

.

If

F ~ A×B)

B

then

denote

denotes say

a

a = Us X

EX)(~

is a

of a limit

or-

~) .

The language

use

Of set

with ~, ~

X c B , then

In p a r t i c u l a r ,

the

and we say

(with equality)

and

in set

the set of all

we

We g e n e r a l l y

F c A×B E F~.

employed

A subset

language

For

On

Otherwise .

material.

a, ~, ¥, ...

its cardinality.

~x E A I (~Y E X ) E ( x , y ~

by d e f i n i t i o n

for some

lim(a)

to be cofinal

, is the first

nary p r e d i c a t e

the set

succ(m)

to

to be an ordinal not equi-

for cardinals.

written

the r e a d e r

with

In general,

a = ~+I

IX1

identified

ordinal.

If

ordinal,

backgrou~

is d e f i n e d

being r e s e r v e d

ordinal,

We r e f e r

will be that c o m m o n l y

and a c a r d i n a l

of all ordinals.

successor

of choice.

is, accordingly,

w i t h any smaller

ordinals,

the theory Z F C - Z e r m e l o - F r a ~ n k e l

for any r e q u i s i t e

our n o t a t i o n

An ordinal

with

the axiom

or ~Je1~

the most part, theory.

entirely

if

the bito de-

F"X

denotes

F :B ~ A

(so

F"X = ~F(x) I x E X )

2. C o n s t r u c t i b i l i t y We shall assume ning

the c o n s t r u c i b l e

ticular, If

the r e a d e r

X

we

hierarchy,

shall assume

is a set, Def(X)

are f i r s t - o r d e r Z EDef(X)

iffZ~X

is f a m i l i a r w i t h

definable

as d e s c r i b e d

an a c q u a i n t a n c e denotes over

and for some

the basic

with

in,

results

say,

the f o l l o w i n g

the set of all subsets X

from p a r a m e t e r s

formula

EDeT~.

in

~ ( V o , . . . , v n)

of X . of

concerIn par-

material. X

which

Thus, LST

and

-

some

Lo

= ~

Lk

of all sets,

, if

universe

the universe,

that all sets are

is the class is d e n o t e d

< "L L a

and

fines


with

L = U E0nL a

by

LST

is

in the form

We refer

L -definable a

there

to

The c o l l e c t i o n the f o r m u l a Vx~(x

V = L

L-definable

that for any limit

V = L

E L ) asserts

as the axiom

of

well-ordering,

ordinal

a ,

L

=

by the same f o r m u l a w h i c h de-

For further details,

he will

V ; hence

is a c a n o n i c a l

the p r o p e r t y

L .

In particular, Theorem

There


in

constructible.

constructibility. L

by the recursion:

lim(k)

(which can c l e a r l y be w r i t t e n


I a E On) , is d e f i n e d

-

) ;

= U ~xL~

The c o n s t r u c t i b l e

(L

I ..... Xn)]

;

L + I = Def(L

we refer

find a proof

the r e a d e r

of the f o l l o w i n g

to [Del]. results:

1

ZFC

is consistent,

we prove u s i n g ZFC

-

xl, .... x n E X , Z = {z E X I <X,E> ~ ( z , x

The c_onstructible hierarchy,

If

2

so is

the a s s u m p t i o n

ZFC + V

= L .

(Hence,

any result w h i c h

V = L

will be c o n s i s t e n t

relative

to

.)

Theorem Let

2

(Condensation

lim(m) .

Theorem Let

If

X~

Lemma)

Lm

then

X ~ L~

for a unique

6 ~ m .

3

a I = w~

If

X~La2

If

A

, a2 = w~ .

, then

is a g i v e n

sets a n a l o g o u s l y allowing

A

XNLal

If

X~Lal

= LB

for some

set we can define to the d e f i n i t i o n

to figure

of the d e f i n i t i o n s .

, then

La+I[A]

for some

(LaEA] I a E 0 n )

of the c o n s t r u c t i b l e

will

6 ~ at°

B ~ ~I

a hierarchy

in the c o n s t r u c t i o n

(Thus

X = L~

as a u n a r y consist

of

hierarchy,

predicate

of all subsets

by

in all of

-3-

L [A]

which

structure

are f i r s t - o r d e r

(LaCA], E , A N L a [ A ] )

are g i v e n in [DeI], rems 2 and 3

where

.

from m e m b e r s

Full d e t a i l s

proofs

of

L [A]

of this

of the f o l l o w i n g

in the

construction

analogues

of theo-

may be found.

Theorem

4

Suppose

A ~ w L [ A ] . If

nals

definable

X-~L

[A]

, then

X T L ~ [ A 0 y]

ordi-

for unique

¥ < 8 _< ~ •

Theorem Let

5

ml = "I

for some

' ~2 = w

8 ~ ml

If

, A ~ ~I X
2[A]

"

If

then

X~L

XNL

I[A]

then

I[A] = Ld[A]

X = for some

3. F o r c i n g For the most part, the m e t h o d

it will be a s s u m e d

of forcing,

fix the notation,

as d e s c r i b e d

however,

that the r e a d e r

in,

say,

chapter

is f a m i l i a r with 17 of [Jel].

we shall give here

a very brief

(partially

set).

To

survey

of

this method. Let

~

say

= <~,~>

p

is no

be a poset

is an e x t e n s i o n p E~

in

p

q

and

patible,

of

We say

tension

D .

iff

D ~ p,q

Plq

~ .

if

E P

We r e f e r

in

~

to the

in

~ P

if there

~

is an ini-

.

K ~

p E •

q

are

otherwise

p

and

we say

incompatible

chain c o n d i t i o n

is ~

, we

~ q E K) ; s i m i l a r l y

is a cardinal,

mI

p,q E ~

is a t o m l e s s

if every p

If

and

has no p a i r w i s e

extensions X

P

, we say

If

(c.c.c.).

be a poset,

in

extension;

• P

P

extension

is dense

If

p ~ q .

(Vp E K)(Fq E ~ ) ( q ~ p

written

incompatible P

iff

a common

chain c o n d i t i o n

Let

~

have

chain c o n d i t i o n nality

q

w h i c h has no p r o p e r

tial s e c t i o n section.

of

ordered

final

has an ex-

compatible q ~

iff

are incomhas

subset

the

of cardi-

as the countable

if every

p E •

has

two

.

any set.

A

set

G ~

is said to be

X-~ene-

-4-

ric for

~

if

G

that w h e n e v e r . set

is a p a i r w i s e

D E X

is a dense

It is p r o v e d

in [Jel]

G

X-generic

which

is

for

transitive

ting poset

, if

E M

the c o n s t r u c t i b l e a ~eneric The m e t h o d within uses

Jech

ting poset,

there

a n a t u r a l way.

for

p E ~

.

all r e g u l a r X ~ map

w :P

one-one subset

open subsets

the o r d e r i n g

of

is an element complete M (~) ran(u)

therefore,

map here, P

of a

boolean

•

the

.

M 0B) is a d e f i n a b l e recursively

by

class

outline

by

of

(called

of

M[G]

this m e t h o d If

~

of

below,

with

[p] = [ q E P boolean

topology.

= interior BA~P)

, whose

algebra

, then ~ .

~

of

, and the

of

The is a

is a dense

in the usual wa~.

subset

ordering

u E ~

in

that

([p]))

is i s o m o r p h e d

is a dense

ZFC

P

I q~P]

(Recall

range

as a poset

BA(~)

ZFC

is a split-

associated

so that

of

BAOP)

BA(~)

.

= [BA~P)] ~

and

If is an

M-

The b o o l e a n u n i v e r s e

such that

by a simple r e c u r s i o n

~ = [(~,~) I x E a ]

,

taking as an open basis

this

lies in

M

ZFC

is a m o d e l

(complete)

BAOP)

P

of

(M (~) is defined

of

M[G]

(closure

that

algebra which

, then

~(p) into

M

~

of its closure.)

is the set of all f u n c t i o n s ~ ~

M[G]

algebra

is just the b o o l e a n c.t.m.

D0G

is the i n t e r i o r

whence

such

and a split-

the p r o p e r t i e s

of the form

under

X

, then

ZFC

c.t.m,

that

P

P

(when we r e g a r d

assume,

is the i d e n t i t y

•

d e f i n e d by

BAOP)

We u s u a l l y

of

of

on

BA(~)

open iff

order-embedding of

by

~

On n ~ = O n ~ M[G]

boolean

a topology

We denote

of

all of the details.

of

~

set for

, is a

We b r i e f l y

of all subsets

~ BA(P)

M-generic

the proof

to supply

of

of

! • , then there is a

M

allows us to discuss

is a complete

is r e g u l a r

(c.t.m.)

M ) such that

Define

the c o l l e c t i o n

section

section

.

M tJ (G]

of forcing.)

[Jel]

~

is an of

final

Ix o~(~)I

model

(In particular,

the m e t h o d

leaving

of

of f o r c i n g

M .

G

closure

extension

initial

that if

Given a countable P

compatible,

dom(u)

on rank w i t h i n M.)

(M-definable)

is a one-one

c M 0B) and

map

embedding

of

~ defined M

into

-5-

M (~)

.

For each sentence

meters

from

value

II~11E •

II~]I = ~

•

tion on that

~(~)

~

of the l a n g u a g e

we can define

in a n a t u r a l

way.

The r e l a t i o n s h i p

M ~B)

implies

M ) a unique

If

~

Ilx=yll = ~

If we wish,

]Ix=y]] = ~

(in

boolean

is a t h e o r e m defines

we can f a c t o r

x = y °

of set theory w i t h para(truth- 1

of

ZFC

, then

an e q u i v a l e n c e

M (~)

rela-

by this r e l a t i o n

In this case we say

so

M ~B) is n o r -

malised° Suppose n o w that tion of

~

G

is

generated

by

(and conversely). the

c.t.m.

N~G]

will become NEG]

true

M (~)

.

a name

name for

a .

KEG]

to some

is just

consider

names

actual

choice

Gof

G),

for

and if

a E M

is called

formula M~G]

~

~

of names

is a dense

subset

of the l a n g u a g e

~

iff

each p a r a m e t e r

for

We define

~

.

and sentences P ~ II~II the precise

of

G

~

In m o s t nature

M ~B)

~

Each member of m e m b e r s

of

of

language

arguments"

and

closure

~

of

of

M

of the

of

M ~B) is ea-

determined

by this par-

language

for

~

by

we do not have

p II-- ~

of iff

to w o r r y

of v a r i o u s

by

language

If- , b e t w e e n m e m b e r s

the b o o l e a n v a l u e s

.

from M[G],

from

in the f o r c i n g

IP

P

seen that for any

is obtained

for

M U

seen to be

with parameters

~

relation,

an a r b i t r a r y

is i n d e p e n d e n t

, it is easily

by its name

of the forcing

•

into

formula

is easily

the f o r c i n g

, where

the f o r c i n ~

of

of

of set theory,

in

"forcing

on

M ~B)

and the e l e m e n t s

The l a n g u a g e

~

(~pEG)(p~!I~I1)

replacing

class

, the element

ticular ~

~

it w i l l be observed,

a .

Since

.

the c o n s t r u c t i b l e

sily seen to be a name for collection

of

(an atomic

II~II E G - )

use

G = [(b,~) I b E ~ ]

(which,

ultrafilter

sec-

to denote

~[G]

for

a factorisation

, the final

class w h i c h d e t e r m i n e s

Since

a name

G-

of the e q u i v a l e n c e We u s u a l l y

The e l e m e n t

Then

G--equivalence

a °

only.

.

M-complete

just in case

for

, it s u f f i c e s t o

~

in the obvious way

We call any m e m b e r

a E M[G]

[G]

determines

Z~C

thus c o r r e s p o n d s

for

G , will be an

G-

of in

M-generic

about

sentences;

-6-

rather

a knowledge

of

That we can k n o w the f o r c i n g P II- ~ that

I~

language

iff

M[G]

and of the f o r c i n g

without follows

I= ~

considering from

for all

relation

is adequate.

the b o o l e a n

the easily

valuation

established

M-generic

sets

G

II- ~

~

]I-- ~

of

result

(for P )

that

such

p E G .

As a m a t t e r (Clearly, lation

of notation,

11-- ~

II-

is

generic sion

M[G]

of

M

nals

a

M

, M

There

are several

The f o l l o w i n g Lemma

we shall

are

If

M

c.t.m,

give

(Vp E~)~II--~. the re-

used fact.

of

to be cardinal

criteria which

iff

by its definition,

only be i n t e r e s t e d

is a cardinal"

in cardinal ZFC

iff M~G]

exten-

if, for all ordi-

~"a

cardinal

absolute

, a generic

absolute

the ones we shall use.

is a (splitting)

is a r e g u l a r -c.c.", iff

poset

uncountable

then for all M[G]

I----"~

is a cardinal".

absolute

For proofs,

dinal

extension

~

closed there

~P

is

p E]~

of ]P

M

extensions.

see

[Jel].

tions

is dense.

dense.

]P

For h i s t o r i c a l

"c~1-dense"

as

"s-closed"

and

M

in

G M

and

~ E M

~'~P s a t i s f i e s

a _~ ~ , M

whenever

~"a

is

is a cardi-

~-generic

, then

poset.

•

is a d e c r e a s i n g ~

P < P~

of fewer

Clearly, reasons

ZFC

M[G]

for IP.

is a car-

.

~ < B

of any c o l l e c t i o n

of

, and if

c.c.c,

(p~ ] ~
such that

M

a (splitting)

intersection of

in

M

, such that

satisfies

be a cardinal, if, w h e n e v e r

cardinal

c.t.m.

is a cardinal",

if

absolute

in a

a E M

In particular,

Let

that,

a frequently

is a

is said ~"q

Note

or

6

~

nal"

~ •)

M-definable,

extensions.

in

we write

II11 =

iff

For the most part,

If

~

if

~P

than is

~

refer

to be

sequence

from

is

~-dense

dense

~-closed

we s o m e t i m e s "G-dense",

IP

is said

initial

then

]P

IP,

iff the secis

to "~1-elosed"

respectively.

~__~-

~and

Lemma If

7

~

is a (splitting)

is a regular

uncountable

then for all for

~

.

we shall

forth

supress

the explicit

(harmless,

Note

that in the context

are often referred

Lemma Let poset

we shall see

8

maximum

of

, and if

ZFC M

, whenever

if

with

mention

and usually

M

and

~ E M

~ "P is ~-dense", G

is

M-generic

a ~ ~ , M :="a is a cardinal"

only be concerned

a unique

proof,

M

and m o r e o v e r

sers have

Finally,

in

for all

is a cardinal",

also

c.t.m.

a < ~ , [Ma] ~ = [Ma] M[G]

Since

the

in a

cardinal

In particular,

r~[G] ~ " a

make

poset

a < ~ , then

splitting

posets

member,

~M[G](a)

=

we shall hence-

of this adjective.

true anyway)

iff

We shall

assumption

also

that our po-

~ .

of forcing,

the elements

of a poset

to as conditions. require

the f o l l o w i n g

result

of Solovay.

For a

[Shl].

(Product

~I' ~ 2

be posets

PI × ~ 2

(q1'q2)

Lemma)

~I ~ ~ 2

eric

for

~2

Furthermore, (p,~) EG } M[G1]-generic

c.t.m.

on the cartesian

~-*

for

in the

P] ~I iff

ql

GI

&

"

M-generic

is

G

is

M-generic

M-generic for

~2

for

' and

PI

for

of

Then for

' and in this case we have if

of

product

P2 ~2 q2

is

M

ZF

In

~I

M

and

~2

G] × G 2

is

~I

and

G2

, define by

(pl,P2)

M-generic

is

M[G1]-gen-

M[G I × G 2] = M[GI][G 2] PI

x ~2

' G2 = {q E ~ 2

G = G I ×G 2 .

' then

a

.

G I = {p E ~ I

I (~,q) EG}

is

I

Chapter II SOUSLIN'S HYPOTHESIS I. S o u s l i n lines Let

S = (S,~)

be a l i n e a r l y o r d e r e d set.

d e n s e l y o r d e r e d set if w h e n e v e r such that

x ~ z ~ y .

Let

x,y E S

S

We say and

S

is dense,

x ~ y

there is

S

is complete if e v e r y subset

S

S

has a least u p p e r b o u n d in

w h i c h has an upper b o u n d in lub(A) ) .

S

is complete,

in

S

has a g r e a t e s t lower b o u n d in

then every subset

Let us also r e m a r k that

A

of

val if

S

I i ~

and

(denoted b y

S

is complete iff

Vx,y E I

Vz E S I

(x~ z~y

glb(A))

(de-

We say

S

S

is connec-

is an o r d e r e d

I ~ S

~ zE I).

, and

is an interA n interval I

is closed if every point in

is c o n t a i n e d in an open interval d i s j o i n t f r o m

D

of

S

(This is just the usual n o t i o n of dense subset u n d e r the t o p o l o g y on

S ,

We say

S

S

A subset

D .

set.

if every open interval of

I .

S- I

c o n t a i n s an element of

of course.)

S

S

of

h a v i n g a lower b o u n d

is complete and has no end-points.

is o p e n if it has no end-points.

is dense in

S

S

ted in the t o p o l o g y induced b y the ordering. c o n t i n u u m if

A

Note that there is an element of d u a l i t y here.

If

conversely.

z E S

be an a r b i t r a r y d e n s e l y o r d e r e d set for

the rest of these definitions.

noted by

or a

is s e p a r a b l e if it has a c o u n t a b l e dense sub-

The f o l l o w i n g e l e m e n t a r y fact is well known.

Theorem Every separable ordered continuum

S

is i s o m o r p h i c to

~R, the real

line w i t h the usual ordering.

One p r o v e s this t h e o r e m b y s h o w i n g that the c o u n t a b l e dense subset of

S

must b e i s o m o r p h i c to the set of rationals,

D e d e k i n d c o m p l e t i o n of D

D

is

S

itself.

segment of

D .)

Q ~ and that the

(The D e d e k i n d c o m p l e t i o n of

is o b t a i n e d b y adding a lub to e v e r y cut in

cut is a p r o p e r initial

D

D

w i t h o u t a lub.

A

-9In 1920 M. Souslin

[Sul]

raised the question as to whether one could

obtain the same result with the separability the following weaker assumption,

requirement

replaced by

which we will call the Souslin proper-

t_Z: Every family of pairwise

disjoint

open intervals

in

S

is countable. "Souslin's hypothesis" tive answer,

is the assertion that this question has a posi-

i.e.

Souslin's hFloothesis

(SH) :

Every ordered continuum with the Souslin property is isomorphic

to

The major aim of this book is to show that Souslin's hypothesis

is nei-

ther provable nor refutable

in

ZFC , even in the presence

of

GCH

]R.

or

its negation. Theorem 2 The following

(i)

are equivalent:

Souslin's hypothesis,

(ii)

every densely ordered set with the Souslin property (order-)

(iii)

embedded into

can be

]R ,

every densely ordered set with the Souslin property has a countable dense subset.

Proof:

(i) ~ (ii) and (iii) - (i) are trivial. S

be a densely ordered set, and let

ing given b y (ii). For each pair < qj

Let



i,j E w , choose

if possible,

D = [dij I i,j Ew]

otherwise

dij

is a countable

For (ii) - (iii),

f :S -

]R

be an embedd-

be an enumeration

dij E S

such that

of

Q .

qi
is chosen arbitrarily. dense subset of

let

S .

Then |

-10-

Let us call an ordered continuum a Souslin line if it has the Souslin p r o p e r t y but is not separable. only if Souslin's hypothesis Let

S

be a Souslin line.

w h i c h tell us that

SH

Thus there

is false. We state some simple properties

is, to some extent,

Theorem 4 b e l o w we will show that nality

wI

S , then

~ < wI

For each point <xi I i E ~ >

(2)

disjoint

S

x E S

from

D

has cardinality

versely,

x 6 S

2w .

there

x

from

S

defined by

fn(X)

= xn .

(Since

S

is complete,

subsets of

For

in

sequence

ISI ~ 2 w . ConS

into

n E w , let

Then the functions

by a theorem of J. Ball

~w

(D) ~

fn

2w.)

sequence fn : S ~ S be

are not all con-

[Ba I] .

question w h i c h h i s t o r i c a l l y

Mary Ellen Rudin

[Rul]

has shown that

exists a Souslin line, then there exists a normal Hausdorff

space w h i c h is not countably paracompact. C.H. Dowker

[Do I]

pact and normal

iff

we obtain that in (4)

sequence

D , a set of cardinality

In conclusion we m e n t i o n one topological

if there

of cardi-

as the least u p p e r bound.

with lub x .

SH .

In

Consequently:

exists a strictly i n c r e a s i n g

<x i l i e w>

has been connected with

D

there exists a strictly increasing with

S ,

is an uncontable

by (I), there exists an injection from

For each

tinuous,

[<x~,xv+1> I v < w l ]

of

conjecture.

has a dense subset

open intervals).

the set of all countable

(3)

S

a natural

is a strictly increasing

(for otherwise

family of pairwise

(7)

(x v Iv < ~>

If now

exists a Souslin line if and

There

that a topological X x I ZFC + xSH

is normal

space (I

space

X

is countably p a r a c o m -

is the real unit interval),

the f o l l o w i n g

exists a normal Hausdorff

not normal.

Together with the result of

is provable: X

such that

X x I

is

-11

Recently, ~SH

M.E. R u d i n [Ru 3 ]

, i.e. as a t h e o r e m in

-

has f o u n d a proof of (4) without a s s u m i n g ZFC

alone.

2. Souslin trees A tree is a poset [yET

Iy~x]

~ = (T,~)

is well o r d e r e d b y

the height of

x

we m a y define

ht(x)

For

a E On

We write For

such that for every

, ht(x)

, the

Tla

x E T

~

.

.

be a tree. ~

b y i n d u c t i o n as follows: ~

Tx = [yET1 ht(T)

x ~y]

T .

under

ht(x)

is the set

for the r e s t r i c t i o n of

sup[ht(x) I x E X ]

T

, is the o r d e r type of

~'th level of

we set

Let

x E T , the set

T

x E T ,

~

(Hence

= [xET

lht(x) =~].

Tla = U ~ a T ~ -

X ~ T

is called the length of

For

= s u p e r ( y ) lynx].)

to the set When

x =

we let ~ .

ht(X)

=

(From now on,

w h e n there is no danger of confusion, we denote a tree simply b y

T =

~T,~) • )

Two elements

x,y

of a tree

t h e y are incomparable, subset of x E b

T .

and

T

written

A branch

b

y ~ x , then

are c o m p a r a b l e if x ~ y xly

.

.

If

y~x

; if not,

A chain is a l i n e a r l y o r d e r e d

is a chain w h i c h is

y E b

or

ht(b)

~-closed,

= a , then

b

i.e. if is an

a-

branch.

An a n t i c h a i n is a set of p a i r w i s e i n c o m p a r a b l e elements of

A branch

(resp. an antichain)

in a l a r g e r b r a n c h

A tree T

T

ITI = w I

Ta

(resp. antichain).

Ta

ITI = ~I

is countable . T

T

is A r o n s z a j n or Souslin,

is an antichain~

and every chain in

is called a S o u s l i n tree if

and every chain and a n t i c h a i n in

N o t e that if each

is called maximal if it is not i n c l u d e d

is called an A r o n s z a ~ n tree if

and every level

T .

T

then

is countable.

ht(T)

= wI .

every S o u s l i n tree is Aronszajn.

Since Aronszajn

trees can (in the p r e s e n c e of the axiom of choice) be p r o v e d to exist (c.f.

[ J e l ], [ J e ~

, or [Ru2]). It is often convenient that the trees

satisfy c e r t a i n "normality" properties.

So let us call a tree

T

a

-12-

normal tree of length (i)

(ii) (iii)

ht(T)

T

= ~

has a unique least point,

each non-maximal

and

(v)

If

T

point

if:

(i.e. points

called the root of

T .

has at least two immediate

y

such that

ht(y)

= ht(x) + I

x ~ y ) ,

each

at each greater level

8 -branch has at most one immediate

< m ,

successor w h e n

is a limit ordinal,

each level

is normal,

Tv

is countable.

then

will be uncountable).

ht(T) ~ ml

chains.

without u n c o u n t a b l e

antichains.

Souslin tree.

(otherwise,

by (iii) and (iv), T I

A normal A r o n s z a ~ n tree is a normal mq-tree

without u n c o u n t a b l e

antichain,

m-tree)

x

each point has successors

8 (vi)

(or a normal

,

successors

(iv)

~

A normal Souslin tree is a normal

w I -tree

Then a normal Souslin tree is indeed a

(For an u n c o u n t a b l e

chain gives rise to an u n c o u n t a b l e

again by (iii).)

Theorem 3 Every S o u s l i n tree

T

Souslin tree

Similarly for A r o n s z a j n trees.

Proof:

T*

.

Remove f r o m

can be normalized,

T

the points

x

i.e.

for which

T

contains a normal

Tx

is countable,

then remove those points with only one immediate The resulting tree, T' , is still of length By induction,

pick out one point

in

To

sulting subtree,

struction applies also for A r o n s z a j n trees.

due to E.W. M i l l e r

ml ' hence Souslin.

in

T" 9 is a normal Souslin tree.

The long list of definitions [Ni I] :

successor.

and one point

every b r a n c h of limit length with extensions

and

T'

extending The r e -

The same con-

|

above is justified by the f o l l o w i n g fact,

-13-

Theorem 4 Souslin's h y p o t h e s i s tree,

to the n o n - e x i s t e n c e

of a Souslin

i.e. there exists a Souslin line if and only if there exists a

Souslin tree. cardinality

Proof:

is equivalent

Furthermore,

of

wI

Suppose S

every Souslin line has a dense subset

S

is a Souslin line.

is countable,

it is not dense in

union of a n o n - e m p t y open intervals.

collection

D

in

By induction, I(U~< a D ~ ) in

IS

assume

and let

Indeed,

T

I(D)

D

o

is the

of pairwise

disjoint

is countable,

since

is defined for consist

We claim that

S

~ < ~ .

Set

S.

I~ =

of one point from each interval and p a r t i a l l y order wI

T

b x re-

is a Souslin tree. with

"normality proper-

hence it is enough to show that every anti-

is countable.

f a m i l y of pairwise

S - D

of

consist of one point from

is a tree of cardinality

T

whence

D

above denotes the topological

T = U <Wl Im

(iii) satisfied,

chain in

Let

D~ Da

Now let

verse inclusion.

ty"

S. )

(~

S,

I(D)

And of course,

has the Souslin property. closure of

Observe that if a subset

But an antichain in

disjoint

open intervals

in

T

is just a

S , hence

count-

able. N o w we look at that

D

D = U~<wl D~ .

is dense in

< w1

S .

If not, let

there is an interval

Ja o J~+1 chain in

for each

a , so

T , impossible:

Clearly,

Jm E I m

tree.

.

containing

For each p .

But

is an u n c o u n t a b l e

Hence we have shown that a Souslin

suppose that



In fact, we may also assume that

lowing strengthening

p E S-D

[J~a<~qJ

line has a dense subset of c a r d i n a l i t y On the other hand,

IDI = w I , and we claim

~I

"

is a normal Souslin T

satisfies the fol-

of (iii): E v e r y point in

T

has

~

many

-

i m m e d i a t e sucCessors. mal S o u s l i n tree

T'

14-

<Such a tree m a y be o b t a i n e d f r o m a n o r b y t a k i n g the r e s t r i c t i o n of

p o i n t s at limit levels in linear o r d e r i n g of such that if and

the point in follows:

b

When

iff

ordinal such that <S, < >

b

be a

b~ <

dm,

b~ = d8

S

be the set of maximal

is a b r a n c h of ~ . when

for all

T , we let

We order

S

~ = m(b,d) 8 < ~

is a d e n s e l y o r d e r e d set.

fices to show that

~

E T~ , x < x' , y < y'

S

w h i c h has height

b < d

a > 0 , let

x',y'

Let

x' <8 y'

T .

For

to the

of order type that of the rationals,

~ < ~ , x,y E T ~ ,

x < m y , then

b r a n c h e s of

T=

T' . )

T'

and

b~ by

be <

as

is the least ba ~ dm.

Then

In v i e w of T h e o r e m 2, it suf-

has the S o u s l i n p r o p e r t y and no countable

dense subset.

For

x E T,

let

there is an

x = x IE T

= [c £ S I b < c < d ] , such that then

I x = [b E S I x 6 b j

and

xj

c I . Ix _

such that

let

~ = ~(b,d).

b m <m x <m d m . )

xI

To every open i n t e r v a l I

When

I

are i n c o m p a r a b l e .

Finally, Let then

suppose that

D

~ = sup[ht(b) [b E D ] Ix

diction:

I =

Then we may take and

J

x E T~

are disjoint,

Hence an u n c o u n t a b l e fa-

m i l y of p a i r w i s e d i s j o i n t open i n t e r v a l s in to an u n c o u n t a b l e a n t i c h a i n in


S

w o u l d give rise

T :

is a countable dense subset of

S .

.

xE~,

Then

~ < wI ,

so if we take

contains an open i n t e r v a l disjoint f r o m

D .

Contra-

I

By this theorem, we may concentrate our interest on the S o u s l i n trees. We shall, however,

r e t u r n to the S o u s l i n line in C h a p t e r s IV and V,

w h e r e we t r a n s f o r m certain h o m o g e n i t y p r o p e r t i e s of S o u s l i n trees to c o r r e s p o n d i n g p r o p e r t i e s of S o u s l i n lines.

-15-

For later use, we note the following definitions.

A normal tree

length

T

wI

is called a special A r o n s z a ~ n tree if

a countable n u m b e r of its antichains. szajn.)

When

say that

T

T = (T,~ T) is

f(x) ~X f(Y)

"

X = (X,_<x )

if there is

We then say

f

is the u n i o n of

(Any such tree is clearly Aron-

is a tree and

X-embeddable

T of

f :T ~ X

embeds

T

in

is a poset,

such that

we

x
X .

Theorem Let

T (i)

(ii)

be a normal T

is

If

~l-tree.

Q-embeddable

T

is

Proof:

T

IR-embeddable,

uncountable of

iff

subset of

is a special A r o n s z a j n tree.

then

T

is Aronszajn,

contains

and every

an u n c o u n t a b l e

antichain

T .

(i)

Let

f

embed

T

ix E T } f ( x ) = q} . T = UqEQA q . are pairwise

in

Q .

Then every

Conversely,

let

f :T ~ Q

let e.g. f(x)

For

-1 ) if of

x

Ao

For each Aq

q E Q,

let

Aq =

is an antichain in

T = UnEwAn

T

and

where the antichains

disjoint.

Define an embedding

x

T

= 0 .

as follows:

x E A 1 , let

does not lie b e n e a t h Go on like this,

is a point of

A2

For each

f(x)

(resp.

= 1

A1

O

~

(resp. f(x) =

lies beneath)

setting for instance

lying above

x E A

a point

f(x)

but b e n e a t h

= -3

if

A o . You'll

never fail! (ii)

If

T

is

be countable, able. levels

Then

~R-embeddable,

hence U

U~

T

is Aronszajn.

inherits

We may assume that

many n o n - e m p t y f

such that

Um's

embeds g(x)

E Q

T

Let

U c T

a tree structure from

otherwise we are already done,

If now

then every b r a n c h of

U~ ~ ~

with new m < w I , for

since then one of the countably

must be uncountable. (and hence also and

must

be uncount-

T,

for all

T

U ) in

f(y) < g(x) < f(x)

Let

U* = Um<wlUa+ 1.

~R , let whenever

g

be x

is

-16-

in some

Um+ I

embeds

U~

antichains,

and

in

y

Q .

is its p r e d e c e s s o r in Hence

U"

Can there be A r o n s z a j n trees

ZFC

is special A r o n s z a j n

3- S o u s l i n trees in

ZFC

Q -embeddable?

are all special A r o n -

But J. B a u m g a r t n e r has p r o v e d ~ u I] that in

that it is consistent w i t h

Hence we m a y

~R-embeddable but not

e m b e d d a b l e n o n - s p e c i a l A r o n s z a j n tree.

g

|

~R-embeddable.

In fact, the A r o n s z a j n trees c o n s t r u c t e d in szajn.

Then

is the u n i o n of c o u n t a b l y m a n y

one of w h i c h must be uncountable.

B y this theorem, no Souslin tree can be ask:

Um .

L

there is an

~R-

L a t e r in this b o o k we will show

to assume that every A r o n s z a j n tree

(thereby o b t a i n i n g

SH ).

L .

Hitherto, we h a v e not said a n y t h i n g d e f i n i t e c o n c e r n i n g the t r u t h or f a l s i t y of S o u s l i n ' s hypothesis. proofs. Later,

We n o w turn towards c o n s i s t e n c y

It was T e n n e n b a u m who first p r o v e d the c o n s i s t e n c y of J e c h i n d e p e n d e n t l y gave a n o t h e r proof.

c.t.m, of

ZFC

CH

in the g r o u n d model. consistent r e l a t i v e to

.

T h e i r result is that any

may be e x t e n d e d b y f o r c i n g to a model of

T e n n e n b a u m ' s proof,

7SH

ISH

.

In

h o l d s in the e x t e n s i o n if and only if it h o l d s Hence ZF .

ZF + GCH + w S H

and

ZFC + ~ C H + I S H

For d e t a i l s of these results,

we

are

refer

the r e a d e r to Jech [Je 2]. We shall h e r e o b t a i n these c o n s i s t e n c y results in a n o t h e r way. s e c t i o n we p r o v e the w e l l - k n o w n result of R.B. J e n s e n that del of

ISH

s i s t e n c y of

.

L

In this is a mo-

In the next section we use this result to o b t a i n the con-

I C H + I SH

r e l a t i v e to

ZFC

.

Theorem 6 Assume

V = L .

Then there exists a S o u s l i n tree.

Hence

SH

fails in

L . Proof:

We are going to define a normal tree

T

of length

~I

such

-17-

that

Tm c_ 2 a

clusion m

for

m < wI

(i.e. the initial

The ordering will be o r d i n a r y insequence ordering).

By induction on

we define the levels as follows:

To

consists of the empty sequence

Tm+ I

consists

of the sequences

tion of sequences) Now,

assume

least

6

~

for

s E Ta

~

alone.

s~(i) and

( "

means concatena-

i ~ 2 .

is a limit ordinal

<w I .

Let

6

be the

such that TI~ E L 6 L5 ~

ZF- , and

L6 ~

m

is countable.

(ZF-

means

L6m

contains only countably m a n y maximal

of

Tim =

m-branch

ZF

minus the power set axiom.)

strictly increasing

sequence

by i n d u c t i o n a sequence n

,

and

Sn+ I ~T Sn

Let

b

ing of

L .

Hence T

sn

T

(s n I n E w>

Then let

Let

such that

Tm



such that An'

b = it [~n ( t ~

that

ht(sn) > ~ n ' Sn)]. )

consist of the sequences

Ub s

for

s £TI~.

is defined.

antichain

of

T .

be a countable, A E M .

cally satisfied meters.

Define

So -->T s ~ and

T

Now

wq , so it remains

is countable. T,A c L

, so

Let

elementary

We also want

since b o t h are definable

Now we use Theorem 1.2.

be a w2

submodel of

T,w I E M ,

A

to

T,A E L

wI M

be a

such b r a n c h in the canonical w e l l - o r d e r -

show that every antichain of

Let

A n , n E w,

SUPnEw ~n = a "

is obviously a normal tree of length

maximal

antichains

Let

lies above a point in

be the least

s

6~ < w~ , so

U T8. Hence, for every s E Tim there exists an 8<m b going through s and m e e t i n g every A n (Such

a b r a n c h m a y be obtained as follows.

for each

Then

(L 2 , E )

such

but this is automatiin

L

w2 It follows that

w i t h o u t parawI O M

is

-q8 -

transitive,

so let

e = wI A M < w I

, the collapsing isomorphism,

(by

x E M .

So

Hence if

induction).

Lwq

ON, wI w(wq) = m .

Thus

= Ty

y

for

~(A)

c

Now,

y < wI .

,

But

e

= ~"(AON)

= AAN

of

Tie

and in

= ANT}a

.

But since

is maximal also in

Note:

By modifying

the

some

A ~ w I , then

~SH

If

A final remark.

e = ~(Wl)

ZF

L6e.

antichain of

T e , every point in

Corollary:

M,

is countable in

is a maximal

~(x) = x

is definable Ty E N

in

for y < ~.

,

N o w we are ready to conclude the proof. Lw2

so

Ty'S

.

Since

AcT,

so

~(A)

able in

T¥ Thus

is defined as the u n i o n of the

TI~

x c M,

= w " ( x N M)

= ~(Uy~.4~l Ty)

= Uy<~(~I)~(T ¥) = TI~ T

then

w(x)

Hence we obtain

~(T)

since

be such that

is defined b y

x E L

in the p a r a m e t e r ~(Ty)

n,~

.

~:M~

for

Let

Since

is u n c o u n t a b l e

Hence

B < 6e .

T , A N Tle

is u n c o u n t in

Also,

is a maximal

L8 . since

A

antichain

Hence

(c.f.

of

lies above a point of A NTl~.Thus A N T I m A = A D TI~,

so

proof we may prove that if

A

is countable. |

V = L[A]

for

[ J e 2 ] ).

is consistent,

then so is

There are two other

existence of Souslin trees.

Martin's

are no Souslin trees, hence

SH .

The axiom of determinateness,

wI

A N Tla E L8 ~ L6m ,by the construction Ta

T.

.

ZF + GCH + ~ S H

.

"axioms" which also decide the axiom

( + 7CH) implies that there

(This will be treated in ChapterVI.)

AD , implies that

thus that there are no Souslin trees,

wq

But we cannot

is measurable, and say that

AD

im-

-

plies

SH ,

19-

since the proof of T h e o r e m 4 uses h e a v i l y the a x i o m of

choice.

4. F o r c i n ~ and S o u s l i n trees Let

M

in

N

be a c.t.m, of .

ZFC,

and assume

We are i n t e r e s t e d in w h a t will h a p p e n to

tensions.

T

is a S o u s l i n t r e e u n d e r forcing ex-

First of all, let us see the result w h e n we force w i t h

self w i t h r e v e r s e d ordering, (T~)

T = ( T , ~ T)

.

i.e. if we force w i t h the poset

In this case the n o t i o n s

T

it-

(~,~)

"compatible" and "comparable"

=

coin-

cides. Thus a set of p a i r w i s e i n c o m p a t i b l e chain in

T .

elements of

Since all these are countable,

~P

~P

is just an anti-

satisfies the c.c.c..

Hence generic extensions are cardinal absolute. N o w let

G

be

M-generic

patible,

G

is a chain in

tial in

T,

so

G

for

~

T .

. G

Since the elements of is a final section of

is a branch.

F o r each

clearly a dense initial section of for an

~ w~ , ~-branch

so b

G

is an

M-~eneric

~P

w~-branch for

T

m ~ w~,

l y i n g in of

if

T . b

is

G

are com-

~ , hence ini-

Dm = U ~ m T ~

M .

Hence

is

GAD a i

F o r convenience, we call N-generic

for

~

.

(In

C h a p t e r VI it will be convenient to change this n o t a t i o n slightly.) In the real world, T h a s countable length, n a m e l y N w I -branches.

w~,

We now p r o v e that they all are

and so

N-generic

T

2~

has

for

T .

Theorem 7 Let

N

Let

b ~ T .

is

be a c.t.m, of

M-generic

Proof:

Then

b

for

T .

ZFC , and let

T

be a S o u s l i n tree in

will be a cofinal b r a n c h of

One half of the t h e o r e m is a l r e a d y proved. is a cofinal b r a n c h of final section of

~ ,

T .

Then

b

T

N .

just in case

b

So suppose that

b

is a p a i r w i s e c o m p a t i b l e

so it c l e a r l y suffices to show that if

is a dense initial section of

~ , then

D ~ Da

for some

D

-

20

-

M

< ~

.

(Notations as above.) A = [xED

A E M in

and

M .

have

A

Setting

I~2yED

y
x],

is a m a x l m a l a n t i c h a i n in

So l e t t i n g

D ~ D~ .

~ = ht(A)

T , hence c o u n t a b l e

= sup[ht(x) I x < A ]

, we must

I

Thus we have f o u n d a m e t h o d of "killing" a S o u s l i n tree, e x t e n s i o n the tree contains a n

w I -branch,

(but is still a normal tree of length

since in the

and is not even A r o n s z a j n This k i l l i n g p r o c e d u r e

w I ).

will be e x p l o r e d f u r t h e r in C h a p t e r VI. Tm

In fact, the c o n s t r u c t i o n of limit levels 6

is a "forcing" construction.

is an L6~

~-branch.

A*n = Is E T I a l

s,

for

bs

as the


An

<TI~, ~ ) lying in

s_ot] , h e n c e an

m-

Hence we could have

L6 -generic b r a n c h going

s E TI~ .

In certain f o r c i n g constructions, (See for instance L e m m a VII.4.) in the usual way,



~t E A n

L6 -generic iff it meets every

d e f i n e d the b r a n c h through

L 6 -generic subset of

The dense initial sections of

are p r e c i s e l y the sets

b r a n c h is

An

in the proof of T h e o r e m

all S o u s l i n trees are preserved. We n o w prove that if we d e s t r o y

CH

all S o u s l i n trees are preserved.

Theorem 8 Let

M

in

M .

to make

be a c.t.m, of Let

~ > (2w) M

2w = n

G

Proof:

be

be regular.

M-generic

~ x w for

We first note that

~

to .

~

T =

Let

in the e x t e n s i o n (i.e.

maps f r o m a subset of Let

ZFC , and assume

]P

]P

is a S o u s l i n tree

be the Cohen c o n d i t i o n s

is the set of all finite

2 , o r d e r e d b y reverse inclusion). Then

T

is S o u s l i n also in

satisfies c.c.c, in

are a b s o l u t e in the extension.

M,

By a standard

M[GJ

.

hence cardinals "pigeon-hole"

proof it is also easily shown that every u n c o u n t a b l e subset of

-21 -

]P

contains

an u n c o u n t a b l e

set consisting

of pairwise

compa-

tible elements. Without

loss of g e n e r a l i t y we may assume that

Souslin tree in malized"

M .

Let

(For if the theorem is true for the "nor-

A E M[G]

ces to prove that name for

A, v chain of T "

B E M

countable

A

be an antichain in

is countable

and let

p* £ G

M[G]

such that

and

in

A ~ B,

P*

It suffi-

Let

A

be a

II-- ''~ is an anti-

so it suffices to show that

B

M .

Pt II--~ E A .

For each

t E B,

pick

Note that if

t ~ t'

and

then tlt' v PlI--"~ ~ ~' , ~,t' E A, and

(For if A

Pt ~ p* Pt

P ~ Pt ' P t ' '

is an antichain" ,

and

Suppose that

B

is countable,

then for some

contains

then so

is uncountable. Pt

antichain of

an u n c o u n t a b l e

Let

C = [Pt It E B]

the set

D

ditions.

But then the set

chain of

T , and a c o n t r a d i c t i o n

of pairwise

[t I p t E D]

If

Proof:

is consistent,

WG

then so is

is a function from

So

compatible

is an u n c o u n t a b l e

is obtained.

I

> mqM[G]

different

ZFC + ~ C H + ~SH .

ZFC + ~CH + wSH .

~ x m

subsets of I

to

2, m .

C

It' IP t II--~' E A ]

T , w h i c h is impossible:

subset

.

Corollary ZF

Pt'

.>

an u n c o u n t a b l e

If

is

M.

are compatible,

pll-

in

T .

Let

F r o m now on we work in such that

is a normal

tree, then it is easily seen to be true for the origi-

nal tree.)

Then

T

and gives rise to Hence

M[G]

is C conanti-

-

In fact,

22

-

an alternative proof of the above theorem may be b a s e d on the

f o l l o w i n g general

result

(implicit in Lemma VI.3

, and not so difficult

to prove directly): Let N ]P, Q

be a c.t.m,

(i)

Zl=

~

(ii)

N~

~

G

is

× Q

in every model,

Hence

MI=PXT

Then

sat. c.c.c.,

satisfies M-generic

Since

c.c.c,

M .

sat. c.o.c., and ~ L I - ~

Theorem 8

as in the theorem.

in

are equivalent:

(i.e. Q

Souslin.

ZFC , and let

be sets of conditions

the f o l l o w i n g

From this fact,

of

c.c.c, for

M I=T

sat. c.c.c.,

is Souslin in sat. so

M[G]

whenever

]P) .

is an immediate T

in

sat. c.c.c.

consequence: M,

and

~

Let

]P, T

be

satisfies

c.c.c., and TII- % sat. c . c . c . . v v ~II--T sat. c.c.c., i.e. ]PII--T is

Chapter III

THE C O M B I N A T O R I A L P R O P E R T Y

I. Statement

A

of ~

By analyzing the proof of T h e o r e m II. 6 it is possible

(that Souslin trees e x i s t - i n L )

to isolate as a p r i n c i p l e

makes the whole thing work.

the "combinatorial

The d e f i n i t i o n

core" that

of this principle,

~

, is

due to Jensen. First

some n e c e s s a r y definitions.

C c k , we say that C

is u n b o u n d e d

C

is closed

or cofinal

is stationarTf (or Mahlo) subset of

in

k k

k

(in

be a limit ordinal.

k ) if

if

Va < k

U ( ~ O C) E C ~

if it intersects

E C

If

for all ~ < k .

(a<~).

A c k

every closed u n b o u n d e d

k .

Assume now that

~ > w

section of less that closed unbounded. unbounded

in

Let

is regular. ~

closed u n b o u n d e d

As a first

subset of

~

obtain the f o l l o w i n g

It is easy to show that the intersubsets of

consequence

~

is itself

it is clear that every closed

must be stationary.

As a second consequence we

simple lemma:

Lemma I Let

An c wI

for

n £ w .

If

UnE w A n

is stationary in

is stationary in

wI

for some

Proof:

An

is not stationary,

If every in

wI

such that

By

so

let

for

UnE w A n

Cn

be closed u n b o u n d e d

n E w .

Then

is not stationary in

wI . |

(J> we mean the f o l l o w i n g principle: There is a sequence

An

n E w .

A n N Cn = ~

~JnEwAn)N ( n n E ~ C n ) = ~ '

w I , then

<S O I a < w 1 >

such that

Sac

~(~)

and

-

IS~I ~ • [a<~1 Later,

for

by the

If

X ~ wl,

is stationary

in

then the set

~I

see how (at the cost of some minor combinatorial

tails) we can split the proof Theorem &

-

a < ~I ~ and:

~XAaESm]

we shall

24

and 5

below.

of Theorem

II. 6

The role of the

L8

into two parts~ 's

de-

namely

will be taken over

S~'s

Theorem 2 is equivalent

to the following

There is a sequence

Proof:

property:

~S a I a < w I)

and if

X ~ ~I ' then the set

ary in

~1 "

such that

Sac

a

{m<~11

X A a = S~]

This proof is due to K. Kunen.

Assume

~

Claim I:

(S~ I a ~ w I ^ n ~ w )

each

There is a sequence

S n _c ~ x w

~a l~n

XA (mxw)

Proof:

Let

and

wI x w

= sn3

Sn I n e w )

.

enumerate

let

X_c WlX~

D = ~ w+v I v < ~I}

in

.

and

m E BOCRD

f~(X)

satisfy the clauses

= f"(X)

Furthermore~ for

f(a)

.

D

in

between

wI

f~ : ~ ( w I) ~

Let

IS~I < • ~

S~ =df f~J'Saso let

m < ~q "

We prove that

a}

= (~0)

, so .

f,(XO~)

By ~

Let

AN C ~ d .

is closed unbounded~

B = ~a I X O ~ E S a ] Then

Let

A = {al X O ( ~ x ~ ) E S

~I "

Then

.

such that

be the order isomorphism

Set

m E D ~ then

= f~1(X) so let

(Sml m < w ~ )

S~

.

be closed unbounded

that if

.

(ordered lexicographically).

~(axw)

is station-

is stationary.

be defined by

Sa c

~<w I

X c w I × w , then

f : ~I ~ wl x w

~(w1×w)

Now,

If

Let the sequence

~.

Then

and:

for

f,(~) , B

C Let

and we note

= ~x~.

Define

is stationary~

= f.(X) A f , ( ~ )

=XN(axw).

-

Since

x n a ~ sa,

Hence

a E AN C,

Claim 2: exists

25

-

f,(Sn a) ~ s a , i.e. and Claim fl

There is an X ~ w1×~

is proved.

no E w

such that

x o (a × ~) E S a .

y c w,,I there

such that for all Y = X"[n o]

and

{a I X N ( a x w )

=

n o

Sa ]

is stationary.

Proof:

Suppose

the claim is false,

be a counterexample

to the claim.

[(y,n) I Y E Y n A n E w ] Hence for all ary,

.

n E ~,

Let

X"[n]

A

set

Sa = S

= Yn

for all

A n = [~ I X ~ ( a x e )

=S~]

for

n E ~ .

is not station-

XA (axe)

is not stationary,

~0 ,,[no]

n E ~ , Yn~

X =

so by Lemma I , U n E w A n = [a l ~ n E ~

[a I X A (a × ~) E Sm] Now,

Then

and let, for

=S~]

contradicting

a < ~I "

=

Claim I

Then the sequence

A

(saia<~1} Y ~ wI

satisfies

and let

C ~ wI

the clauses

in the theorem.

be closed unbounded.

Indeed,

let

By Claim 2, let n o

X ~ wlxw

be such that

is stationary.

Y Na Hereafter, rem.

by

Pick

X"[no] n a

=

~

=

~ E C

and

such that

(XA (a X w ) ) " { n o]

we mean the equivalent

=

tions [a<~1

principle:

There is a sequence

such that for every function lh[a =ha]

is stationary.

now state two stronger versions

:

There is a sequence ISal ~ •

for

[a I X O a E Sa]

(S

There is a sequence

(axw) n o

S ~ ° " [ n o]

~

= Sa . = ~

•

= Sa } Then

I

stated in this theois also equivalent to

(ha l a < ~ 1 )

of func-

h :~I ~ Wl ' the set For completeness

of ~

and for later use we

.

I a<~q }

a < ~I ' and: contains

[mlXO

XO (a×~)

property

We leave it to the reader to prove that

the following

<~*

Y = X"[n o]

If

such that

_~-

IP ('~)

ana

X ~ w I , then the set

a closed unbounded

{S a [ ~ < ® 1 >

S

such that

set. S a c ~(a)

and

-

ISaI ~ w

for

26-

a < w I , and:

closed unbounded

C ~ wq

For all

X ~ wI

such that if

there

exists

a E C , then

a

XO ~ ,

C N a E S~ .

Clearly

--<~+ - ~ *

- ~<~>. Jensen has proved that the arrows

cannot

be reversed.

2. Properties Theorem <~

3

implies

Proof:

of

CH.

Let

~S~I a < w 1)

(~)

there

satisfy the clauses

exists

be the least such into

That

~

~1

"

an m .

~ < ~I Then

of

such that f

~.

To every

a = Sa

is an injection

a ~ w

Let from

f(a)

~(~)

|

is actually

will be (implicitly)

stronger than

CH

(which it strongly

seems to be)

shown in the last part of this book.

Theorem 4 Assume

V = L .

Proof:

By induction

on

as follows.

Set

each

Then

~

~ < wI

E Ca

Now,

and

and

lim(~) , let

Sa,C ~ ~ ~,

that

can find a pair

we define

SO = CO = ~

S a A ¥ I S¥

suppose

unbounded,

a < wI

If

pair such that V¥

is valid.

;

Ca

V¥ E C

S~+ 1 = C~+ 1 = ~ + 1 QS~,C~)

be the

does not satisfy

X A y ~ S¥

X,C ~ w I , C Let

(X,C)

for


is closed unbounded

such that

I ~<w 1)

< <Sm,C~

if no such pair exists,

<SaI a < W l ) (X,C>

a sequence

set O-

in

a

and

S~=Cm

=m.

Then we

is closed and be the


- 27-

least Let

such pair. M ~ L

p h i s m of

be countable. Let ~ w2 M onto some L~ . Since

are definable wI N M .

in

~((X,C))

~(<<S~,Cv)l v < % ) )

= ( X D ~, C n ~)

such that

in

(by absoluteness)

But the sentence

for some

~+

"CA a

A ~ wI

XN~

to show that

in the construction

in

m

of

and

in the real world. In particular,

So

a

a E C

inm "

is a limit By the defini-

~ S~ , a contradiction ~

so as to yield:

A somewhat L .

and

is closed unbounded

is closed,

can be strengthened

also holds in

We now proceed

C

C , therefore,

V = LEAS

we obtain

( X N a, C O ~) = (Sm, C a)

C , and since

the proof

~ =

"

point

In fact,

Let

pair of subsets

L~ ~ hence in the real world.

tion of

Proof:


holds in of

M .

II.6

is closed unbounded

L~ , and hence

X A ~ = Sm .

of

= <<Sv,Cv> l ~ < ~ > ,

X A aN ¥ # S¥

Thus, by definition,

Assume

(X,C~

and

The sentence

is the

C O ~

VY E O N ~

Theorem

(S~ I ~ < w I)

L~2 , they are elements

"(XO m, C 0 a)

that

isomor-

Just as in the proof of Theorem

~(%) = ~,

holds

be the collapsing

more ingenious

~

|

holds

if

proof gives us

(See tDeqS.)

~

may replace

of a Souslin

the condensation

arguments

tree.

5 ~.

Let

Then there exists

(Sml a < ~ 1 )

By induction tree points

a Souslin

be as in

~

(in the form given by Theorem 2).

we shall construct

T = (T,~ T)

of length

are the countable

tree.

the levels

wI

ordinals).

such that

T~

of a normal

T = wI

(i.e. its

-

(i)

T

(ii)

Let

-

= [0}

o

Ta

ordinals, along

28

be given.

The elements

of

T~

are countable

and hence have a canonical well-ordering.

T~

under this ordering,

appoint to each

x £ T~

next two ordinals not already used as immediate x .

This defines

(iii) Assume choose an

Tla

is given for

m-branch

x ~ y

and:

then

b x N S~ ~ ¢ .

successors

bx If

lim(~)

.

going through

Sa

is a maximal

To every

x

~ ¢

tichain of induces

of

Tla. )

A £ Sa

happens

The w e l l - o r d e r i n g

a well-ordering

b x ~ by

antichain of

(This is clearly possible.

whenever

x £ Tla

such that

TI~ ,

We note here

that we could have used the original version of b x0A

the

Ta+ I

when

that

Proceeding

t~,

requiring

to be a maximal

of the points in

of the b r a n c h e s

bx

an-

T~

So we may induc-

tively let the least ordinal not already used be the point

ex-

tending

T

bx .

This defines

Ta , and the c o n s t r u c t i o n

of

is complete. By induction it is clear that wI .

We now prove that

a n t i c h a i n in A = {~<~

I lim(~)^XN

A

Proof:

Suppose

then there But

is Souslin.

~

lim(a)

~ £ A .

A0 m .

If

exists an

x ~ T I~b

parable with

Let

is a maximal

is closed u n b o u n d e d

We show that

is a normal tree of length X

be a maximal

T , and let

Claim:

ation of

T

T

~I "

and that

AO a

Let XD ~

(~v I ~ < ¥)

T Is].

is u n b o u n d e d in be the monotone

is not a maximal

x E T I~

for some XN ~

in

antichain of

enumer-

antichain of

w h i c h is incomparable

v < Y , so, ~ fortiori,

; but this is impossible

a .

x

since

with

Tim, X N ~.

is incom8v £ A .

-

Hence

A

the set

A

is unbounded.

[~ I T I ~ = ~]

~ < ~I

we define

[Yx ~ x E T ~ ]

Set

and

TI~*

~*

For

Now let

~n+1

m E A .

some

Now

= ~n

for

x .

Let

x E Tla .

Then

such that

of

X0 m .

Ya

x E TI~ n

Hence

De-

m = S u P n E w ~n "

Yx E X 0 Tl~n+ I ~ X N TI~ ~ X 0 ~,

comparable with an element

Yx

Y~ =

be arbitrary. Let

~I"

let

Let

~ ~

~o < wl n E w .

subset of

x E Tla,

comparable with

~* = the least ordinal

We claim that n E w .

as follows.

X

= ~*

fine i n d u c t i v e l y

Note that, by construction,

is a (closed and) u n b o u n d e d

be the least element in

TI~*

-

is closed.

Next we show that

For

29

for

so

~ > ~o

x

and

is ~EA.

The proof of the claim is complete. Now we use m E A in

Since

such that

Tim .

point of

3- ~

~.

[~ I X 0 a = S ~ ]

X 0 a = Sa

Then

By construction, Sa

Hence

Sa

Sa

is a maximal

every point in

Tm

is maximal also in

cannot

contain any more elements than

hence

X

is countable.

is stationary,

S

, so

pick antichain

lies above a T .

But then

X = S~a

X

~ and

|

obtained b y forcin~

In this section we show that every countable model of tended to a model of ~J*

~j

(and hence of

~SH ) .

ZFC

can be ex-

The similar result for

is also valid and will be shown in Chapter VIII, but we prove the

result for ~

s e p a r a t e l y since the extension is so simple:

quence can be defined from any standard generic

subset of

the ~ - s e wI .

Theorem 6 Let

M

be a c.t.m,

from a countable

of

ZFC.

ordinal into

In

M,

let

JP

be the poset of all maps

2 , ordered by inclusion

the standard poset for adding a generic

subset of

wI ) .

(i.e.

~P

is

Let

G

be

-

M-generic

for

~

.

~M[G](~> : ~M(~)

(ii)

if

~

-

Then:

(i)

M

30

~[G] ~

~,

M

M[G]

and

2 w = w I , then

and

have the same

cardinals,

(i±i) M[o] I= <>. Proof:

~

is clearly

Furthermore,

wfl-closed,

~

trivially

2w = ~I ' then

w2

(Sal ~ < w I)

is not a limit ordinal, ~,

from

a

dompa=

let

jm

onto a+w

We show that

Let

(S

I a<~

1)

Sa = ~ .

If

For every limit ordi-

Pa 6 G

such that

M[G]

Suppose

=1}

and that

C _c mq

~

in

is any closed u n b o u n d e d set.

M[G]).

is closedunbo~ded.

(iN ~ = ~ )

Vp' <_ p

S q _< p'

p' < p .

Set

q !I- ~ a o = domp'

define sequences

and sequences

G. )

It is enough to show

.

W o r k in

M

now.



n E w

for

£ c(in ~ = s¢)


and for each

(and hence that this

(Here "S " is an a b b r e v i a t i o n

the defining formula in terms of

Po < p'

as follows.

Z o ( M ) - definable b i j e c t i o n

satisfies

P II--~a £ C

setence holds in

duction,

G

be such that

We claim that

So let

are also absolute.

a < wq , lim(a)

p ll-2_~v I A ¢ _ ~ V I

that

from

exists a unique

We set for

is given~

p 6 G

There

= {.','<~ll::~+j~(','))

S

X c__ ml

set

be a canonical,

w . .

(2w) + - c.c. , so if

(iii).

We define the sequence

nal

satisfies

and all larger cardinals

So it remains to prove

a

so (i) is clear.

By in-

of ordinals

of conditions

such that

-

3q

-

(a)

Vv < a n

(Pn II-- v El(

or

Pn II--v~]~)

,

(b)

Vv < a n

(Pn II--r E 6

or

Pn II--v g ~ )

,

(c)

8n > an

(d)

an+ I = d o m q n ,

(e)

Pn+1 --< qn

^

qn < Pn

^

qn If--~n E C

(a) and (b) can be satisfied since, by

^ d o m q n >- 8n '

wfl-closedness, we can

easily arrange for countably many statements to be decided by one condition. of

p

Let xa

(c) is also all right, since every extension

forces

C

to be unbounded in

a = SUPnEwa n =

and let

iv
Pw = U n E w P n ° Ca

V

Pw II-- " i N v = Xa A C O v = Ca condition (c) guarantees that

iff

q < Pw

by:

j~1(n) E X a .

Then, setting V

and

V

Define

wq

domq

,T

; ~v
Since Pw II--VEC

-

r If--~ E G

forcing is absolute with respect to Hence we are done, since

q < p'

and

iff

closed",

By (d), d o m p w =a-

= a + w , q~a = p , and

Then we readily get that

(using only the facts that

clearly

q(a+n)

= fl

q [I--iN v = Sv a r < q,

and that

Zo-sentences

from

q ll--~aE6

(iNa

M ). = ~a).

C h a p t e r IV

H O M O G E N E O U S S O U S L I N TREES AND LINES

q. A h o m o g e n e o u s S o u s l i n tree An a u t o m o r p h i s m of a poset

X = <X,j>

such that f o r all

x,y E X,

phism from

X,

in

X ) .

X

to

x ~ y

is a b i j e c t i v e map

iff

~(x) ~ a(y)

or a map such that b o t h

a

o :X ~ X

(i.e. an isomor-

and

G -I

embed

X

In this and the next chapter we will c o n c e n t r a t e our interest

on a u t o m o r p h i s m p r o p e r t i e s of certain S o u s l i n t r e e s , c o n s t r u c t e d f r o m O . Clearly, all

if

T

is a tree and

x E T , h t ( o ( x ) ) = ht(x)

if f o r all

x,y E T

with

which interchanges

F o r a moment,

x

o .

is an a u t o m o r p h i s m on So let us call a tree

ht(x) and

= ht(y)

y

(i.e.

let the r e q u i r e m e n t

T , t h e n for T

homogeneous

there exists an a u t o m o r p h i s m o(x)

= y

and

a(y)

= x) .

(iii) in the d e f i n i t i o n of a normal

tree be s t r e n g t h e n e d so that the n u m b e r of i m m e d i a t e s u c c e s s o r s of any n o n - m a x i m a l point be f i x e d (for i n s t a n c e

w , or

be p r o v e d that any two countable normal trees are isomorphic. that

(Let

UnE w b n = T , and

< d n l n E w>

r e s p o n d e n c e b e t w e e n the b r a n c h e s

s i m i l a r l y for bn

and

T'

dm

It can easily

T, T' of the same length

be a sequence of

a structure p r e s e r v i n g and m i n i m a l way. T ~

2 ) .

T'

a - b r a n c h e s such D e f i n e a cor-

b y i n d u c t i o n on

T

in

This will give an i s o m o r p h i s m

(A d e t a i l e d proof m a y be f o u n d in [Ku I] p . I 0 2 . ) )

fact, it is easy to see that if

n

By this

is a normal tree of l e n g t h

ml

'

then

i) ii)

Tla if

is h o m o g e n e o u s for all

a < wI ,

a < ~ < mq , t h e n any a u t o m o r p h i s m on

can be extended to an a u t o m o r p h i s m on

TI~

Tlm+1 .

H e n c e we m a y be t e m p t e d to claim that every normal tree of l e n g t h is h o m o g e n e o u s

(or, equivalently,

wI

that any two normal trees of length

-33 -

w1

are isomorphic).

will show.

This is not so, however,

So, in order to construct

as the next chapter

a homogeneous

Souslin tree, we

need some strategy for what to do at limit ordinals.

Theorem I Assume

<>.

Proof:

Then there is a h o m o g e n e o u s

(S a i a < w1 )

Let Since

~

be the sequence given by < > .

implies

CH , H I

(the set of all h e r e d i t a r i l y

table sets) has cardinality A ~ wI . on

Hence

Souslin tree.

LwI[A]

a < w I , define

w I , and can be

= H I , and

6a

coun-

"coded" b y some

w~ LAj = w I .

By induction

to be the least ordinal

6 > m

such

that

Set

(i)

LsLAJ ~ Hwl ,

(ii)

S~,(6

M a = Lsm[A]

a < wI .

I ~<~)

E L6LAJ

.

= L 6 a [ A n 6 a]

We now prove a useful

Then

M~

general

is countable

for

lemma.

Lemma 2 Let

T

be a normal tree of length

and let

C ~ 91

for all

~ £ C : If

generic for Then

Proof:

T

~q

be a closed u n b o u n d e d TI~ E M~

and

such that

x E T

~

set of limit ordinals

x E T~,

then

{Y I Y < T X ]

x E Hwl , such that is

N~-

Tla .

is a Souslin tree.

Let

X

be a maximal

is countable.

Let

antichain in B ~ wI

N* =
T .

We will show that

be such that

Define models

C,X,T,A

N v ~ N*

for

follows: N o = the smallest

N ~ N*

such that

B E N

E L[B]

X . Set

v < wI

as

Nm+g

Let

= the smallest U<~N

N W N*

Nm

=

for

~

be the least ordinal

be the collapsing

map

lim(a)

L~ [BN ~v]

.

The set

By elementary a

By

there

~

~

£ C

~L~

for

an ordinal

Bn ~ = S a

Since

S~ £ M a ,

point we use

~>

~

Claim:

6 Ma

m

such that

B N m E Mm

for some ¥o

able in Secondly,

Y < w. 1

w LEBNm3 ~

it follows

= wI .

B O m . T 6 Mm

that

L8 [BN ~] , we obtain assume

w L[BNY]

< wI

~

= ~g , put

able in

for sufficiently

definable

.

in

~v '

and

Thus

w [BAli

in the parameter

and

Let in

assume

be the least

N*

t < wI

be the least •

But since

< T , so

a

8a E Ma

y < wI .

large

~ ~ so

is countable, So

B O ~ £

-

If for some is count-

L[BAD]

and hence

w L[BAa]

Hwl-

is uncount-

~ = w L[BAY] , whence ~

B O ~ .

is

From the fact t h a t .

such

Nm ~< N* , ¥o E Na.

Lt[B N ~]

for all

Y < wg , w L[BnY]

impossible"

in

is countable

in the parameter

L[BO~]

~

ma = m

First,

Yo

Let

is definable

so

such that

M a ~< H~I

in

.

.

Since

definable

,

than earlier.)

= wg

ordinal

[BN ~]

Note that this applica-

w LLBAY]

Yo < a'

=

(This is the only

We split the proof into two cases.

Hence

Nv = L ~ v [ ~ ( B ) ]

n (Y) = Y N L~ [BN ~ ]

Proof:

ordinal.

Then

is closed unbounded

in the whole proof.

is more indirect

~

v

.

exists

tion of

v < ~I

closed u n b o u n d e d

CO ~

v < ~

n v :Nv~N

[BA ~ ] = id ~ L ~

is obviously

equivalence,

, and let

such that

Y ~ Lwg[B] , then

[my I v < ~g}

hence

~

We note that and

N

U Nv ~ Nv ) for

v < ~I ~ there is

Y £ Nv

Na U [N~] ~ N ,

.

not in

(with

for every

so if

such that

is

£ M~ .

= wq , Hug By

-

elementary

equivalence,

onto each

e < Ps

~[B0s]

, so

35

-

s

in

(the image of

Lps[B 0 s] .

Ps E M s

~1 ) is mappable

This means that

also in this case,

Ps ~

and the claim is

proved. Then,

recalling

that

B 0 s E M s = LSs[A] , we have that

Lp [BN ~] ~ L 6 [BN s] ~ M s . w~(X)

= X 0 TIs E Na ~ Ms

is a maximal For

x E Ts,

antichain

.

in

{Y I Y < T X }

X O Tls , so

T .

T

is defined by induction

T

shall be countable

[s~(i) least

(in

[ U d Id

is an

A iE2J

normal

the set Proof:

.

s-branch

on

T

By induction

so

in

X

s,s'

Souslin

Ts

in

is

s

such that

TIs E M s

s = ~+ I

branch

lim(s) Tls .

Tls,and Then

T

such that

of =

inclusion. s < ~1"

P , Ts = Let

b

Define and

T = Us<®1T a

We show that

E T

Ud

.

T.

ht(s)

for every

for some

of

tree

The points

will be ordinary

through

wI .

For each pair [v I s v j s ~ j

TIs ;

antichain

X = X0 Tls,

Now assume

segment]

tree of length

Claim 1:

is a maximal

on the levels

Ms-generic

only on an initial

for

lies above a point

Thus

it will follow that

LLA])

X NTls

Ms-generic

of the homogeneous

T o = [~J , and if

I sETp

x

0,1 - sequences

dom(s) , and the ordering

As usual,

and

(Lemma 2)

We turn to the construction

By induction,

~ Ms'

equivalence,

is

X 0 Tls

TIa+I , and hence also in Q.E.D.

= TI~ENs

TIs .

This means that

countable.

~s(T)

By elementary

the branch

hence it intersects X ~ TIs

"

So

Ns =

Ub

be the Ts = differ

is clearly a

is homogeneous.

ht(s)

= ht(s'),

is finite. on

s = ht(s)

= ht(s')

Trivial

for

0

- 56

and successor and

a .

For

s' are equal except

-

lim(a) , s,s' for an initial

induction hypothesis holds. Now, for finite

ht(aa(S))

(So

ca

and

°a(S)v

v ~ s(q-Sv)

=

changes

s

Claim 2:

c a : 2 <W1 - 2 <wl

if

~ ~ a

if

v E a .

at f i n i t e l y many places.

<

w

T

is closed under the maps

Note that

2 <wq

immediately

imply that

T

we note that for

tended in b = Ca"(d) = ~,

Ca"(D)

a ~ ~q '

).

Finally,

DAd

by:

(for

ca

This follows d i r e c t l y from the definition 2

on which the

So Claim q is clear.

define

= ht(s)

segment,

s

Ua<wl 2 a .)

means

lal

a ~ wq

£ T~ , we know that

T

with then

d

is dense.)

Ma-generic

arbitrary.

oa"(D)Nb

T .

Claim I and

is homogeneous.

lim(a) , each

is, in fact,

of

If

= ~,

D

a-branch w h i c h is ex-

for

Tla

.

(For let

is dense in

w h i c h is impossible

Hence Lemma 2

gives that

T

Tla

and

since

is Souslin.

|

Theorem Assume

~.

Then there is a Souslin tree with exactly

wI

automorph-

isms.

Proof:

We show that the h o m o g e n e o u s has exactly for

wI

a,~ < w I ,

Souslin tree

automorphisms. a # ~ .

So

T

Of course, has at least

The converse direction is an immediate the f o l l o w i n g Claim:

If

c

T

constructed c[a]~T wI

consquence

above

f o[~]~T

automorphisms. of

CH

and

claim. is a u t o m o r p h i s m

of

T,

then there is an

a < mq

-

such that for all then

x E T

-

and

v

<

~

if

,

m ~ ~ < ht(x)

,

c(x)~ = x v .

Proof:

The claim may e q u i v a l e n t l y be stated as :-

If (*)

37

~

is an a u t o m o r p h i s m

D : [y I V z > y is dense in

Clearly,

T

w h i c h is dense in

I~yED

(i.e.

VxET

Souslin,

- z~=G(z)v)}

~y~x

y ED)

(*) , since then

.

D~

[y I ht(y) > a],

T .

assume

y<x}

T , then

V~<~(ht(y)~
the claim implies

Conversely,

of

D

is dense in

is a maximal

X ~ TI~

T .

Then

antichain in

for some

a < ~

.

X =df ~x E D I

T .

This

Since

a

T

is

satisfies the

claim. Suppose

(*) is false.

xo E T

such that

z v ~ a(z)v).

Dy

Vy ~ x o

Define for

Dy = ~z I ( z > y Then

Then there exist an a u t o m o r p h i s m

and

B ~ wI

T

for

for

submodel lim(k)

.

N v , and

There exists

m = am

such that

~

N~ ~ Ma

.

~(~)

so

=DynTl~

By elementary all

y ~ xo

for

zv~(z)v~

v (zly)] .

= L L[B ] LBJ, 2

N*

with

of

N*

NO

is the

B E N o , Nv+ I containing

As before, ~ v : N~ "-~

let

a~

=

~m(T)

is

N v U ~Nv}, be the

LBD ~ ~

S~ = B ~ ~ E N a,

F u r t h e r we get

where we

and as before

= TI~

and

Ut¢y) < ~ .

equivalence with

N*

of

least ordinal not in

E Ma,

%

T,G,A E LEBJ.

submodel

elementary

N k = Uv< k N v

< w I (ht(y)~
y ~ xo .

letting

and

elementary

the smallest

~

y E T,

We now p r o c e e d as above,

smallest

> y

and

^ ~v(ht(y)~
is dense in

now require

~z

a

ht(y)

there is an < a,

D y N Tla

x ° E T[a is

such that for

-

dense in is

Tla

.

Mm-generic

for each

38

N o w let

for

y < x

-

x > x ° , ht(x)

Tla , and thus i n t e r s e c t s

with

a r b i t r a r i l y large

y ~ x°

v < a

For a normal tree

T

of length

r e s t r i c t i o n of

to the set

is c l o s e d u n b o u n d e d and TIC

T

[YlY <xj

D y 0 TI~ E M~

such that

~I ' and UaC B T a .

is a normal

x v ~ o(x) v , w h i c h contrav's

is finite ~

B ~ ~1 ' let

TIB

|

be the

It is obvious that if (Souslin) tree,

C ~ wI

then so is

.

Now, let

T

be as before,

and let

C ~ wI

easy m o d i f i c a t i o n of the proof of T h e o r e m 3 ~I

Then

This m e a n s that there exist

dicts the fact that the set of such

T

= ~ .

be c l o s e d unbounded. yields:

TIC

has exactly

automorphisms.

Finally, If

M

a remark about forcing w i t h Souslin trees.

is a c.t.m,

then of course

M

of ~

ZFC, "T

and

T

is a normal Souslin tree in

satisfies c.c.c."

But

T2

(i.e.

the usual product ordering) n e v e r satisfies c.c.c, in

M .

of this is an easy exercise.)

T2

lapse cardinals.

So, since T,

T

Hence, b y f o r c i n g w i t h

T2

c~lapsed

(The proof we may colBy the

M[b]

with

b

in one f u r t h e r extension.)

be the above defined h o m o g e n e o u s Souslin tree, M

of

ZFC .

T* = is E 2 Then

M[b]

ht(o(s))

~

Let <w 1

T % T*

= ht(s)

and

.

b

T.

N - g e n e r i c for

We now give an example where we a c t u a l l y must collapse cardinals.

c.t.m.

with

is the same as f o r c i n g twice w i t h

does not satisfy c.c.c, in

cardinals ma 2 be

M,

T×T

(This fact can also be said in another way:

product lemma, f o r c i n g w i t h

T

An

be

N - g e n e r i c for

Let

c o n s t r u c t e d inside a T .

Define in

N[b] :

I {v I s v = I] is finite] The i s o m o r p h i s m

o ~ M[b]

o(s) v = Isv- (Ub)vl

for

is g i v e n by: v < ht(s)

If now

s E Ta ~

then there

Choose

s'

x c s

there

o'~I x .

E Ta

Hence

a E B

s'

such that

exists

a

E Ta

we

y ~ s'

and

can f i n d s E Ta ~

s u c h that

Is, N ~ " I s ~ ~ such

I s , = Nycs, ~

~n~alogously if

is an

that

Then for

y E Yx'

Oxcs c'' I x

a closed

.

I s , ~ a"I s .

Iy _c

i.e.

= ~" ( N x c s I x ) = c " I s.

unbounded

then there

every

is an

B ~ wI s'

E Tm

.

Then

s u c h that s u c h that

I s , ~ ~-1"I s .

Let

C = AOB

.

s',s"

E Ta

if

E Is, ~

b

implies Is,

Let

s E Ta

such that then

b o s,

Is'

~(b)

where

c ~"I s --

E Is~ ~

so w e m u s t

a E C and

so

have

Is, , -c a-l"Is,

b

s"

there

E Is

= s .

Thus

Then

"

Hence

are

b ~ s"

Is,, = I s , so

= c"I s

(ii)

in the

Q.E.D.

theorem

Claim:

If

~

a ~ wI

s u c h that

now follows

from

is an a u t o m o r p h i s m if

b

E S

6).

the

on

and

(Lemma

S ,

then

~ ~ v ~ domb

there

is an

, then

~(b)v

=

b

Proof:

Let

: TIC ~

TIC

B y the p r o o f all claim

With yield

Lemma Let

a

be be

an

sntomorphism

as in L e m m a

of T h e o r e m

s E TIC,

if

a ~

3

on

6, w h e r e

there

is an

~ ~ ht(s) ,

then

S , C

and let is c l o s e d

a ~ ~1

unbounded.

such

that

~(s) v = s v .

The

is i m m e d i a t e .

trivial

changes,

the p r o o f s

of t h e two p r e c e e d i n g

lemmas

the f o l l o w i n g

7 C _c ~1

for

be

closed

unbounded.

If

a

is an a n t i l e x i c o g r a p h i c a l

-44-

a u t o m o r p h i s m on : ~S,~>

~

TIC ~ t h e n there is a u n i q u e i s o m o r p h i s m

<S,~>

such that

~ " I s = Io(s)

for all

s E TIC

.

Lemma 8 Let

~ :<S,~)

~

<S,~>

such that if

s E Ta

Is , .

~

set

Hence

~(s)

s'

assume

: T~C ~ on

and

s' E Tm

a u t o m o r p h i s m on

and let

a : <S,~)

there exists a closed u n b o u n d e d TIC , and

TIC

.

~"I s =

TIC

if we

~

~

<S,~)

C ~ wI

.

such that

reverses the l e x i c o g r a p h i c a l o r d e r i n g

On the o t h e r hand, b y the p r o o f of T h e o r e m 3,

a < ml

Thus let

s,s'

such that if

E TIA

a ~ v < ht(s) , then

such that

~(s)
is complete.

with

C~w I

o " I s = Is, .

is reversible,

there is

Then

~ E C , there is an

such that

S

By Lemma 8


T h e n there exists a closed u n b o u n d e d

is an a n t i l e x i c o g r a p h i c a l

= that

Now,

.

s
and

~ ( s ) v = s v-

s~a = s'~a .

c o n t r a d i c t i o n : The p r o o f of T h e o r e m 4

I

Theorem 9 Assume ~

Proof:

.

Then there exists a S o u s l i n line

(i)

S

is h o m o g e n e o u s ,

(ii)

S

has exactly

~ii)

S

is reversible.

wI

S

such that

automorphisms,

We only have to make some minor changes of the p r o o f of T h e o r e m I, 3 and 4. and

~

For

= 1-s v

s E 2 <~1 , let for

v < doms

M s - g e n e r i c for

Tla

sist of all b r a n c h e s U-b

only on an initial

(with

b'

i E2 } . MS

such that

segment.

be d e f i n e d b y

. Define

{~] , Ta+ I = I s " < i > I s E T S and be

~

T

as follows: For

as before). U b'

doms

To =

lim(~) , let Let

differs f r o m

It follows that

= doms

Ta

b

conU b

or

- ~5

(I)

If

s,s'

£ T

and

(q) it f o l l o w s

s,s'

£ T,

est

~ < ht(s),

ht(s)

ht(s)

= ht(s) ,

s v* = s v

From

Define

with

ht(s*)

where

-

for that

o : T "-~ T

s* = s' V

then

for

V

s* 6 T ,

v < ht(s') ~

h t ( s ' ) _< v < h t ( s ) T

= ht(s') ht(x)

> ht(s'),

is h o m o g e n e o u s .

Indeed,

For

let

~x

be the l a r g -

= s~8

or

x~8

x 6 T

s u c h t h at

x~

let

= s'~B

.

by

{ i s'

if ~ v < x~x if v <

o(x) v =

and v and

x~x = x~8 x = s' ~Sx'

otherwise. Then

o(s)

By Lemma 2 (2)

= s'

and

,

is S o u s l i n .

If

T s,s'

o(s')

6 T , ht(s)

t h e n t h e r e is S

= S

I

= s .

for

Furthermore, = ht(s')

v < ~ % > ~

we h a v e

= a

and

lim~

,

such that Or

s

= I -

S t

for

% > ~ .

H e n c e we o b t a i n (3)

If

a ~ wq

is c l o s e d u n b o u n d e d

automorphism s u c h that or Thus

T

If we let

mq

is c l o s e d u n d e r

ble.

|

T~A , then there i) h t (s)

> v ~ a ~ s

o

is an

is an

a < mq

> v ~ m - s v = o(s v) = 1-o(s

)

automorphisms.

S = S T , it f o l l o w s

Souslin line with S

either

ii) h t ( s )

has

on

and

exactly

wj

the m a p p i n g

that

S

is a h o m o g e n e o u s

automorphisms. s'-~s

.

Hence

But in this case, S

is r e v e r s i -

Chapter V

RIGID SOUSLIN TREES AND LINES

1. A rigid Souslin tree A poset ism on

X = <X, ~ > X .

is called rigid if

We now prove that in

idIX

is the only automorph-

L , rigid Souslin trees exist.

Theorem 1 Assume ~

Proof:

.

Then there exists a rigid Souslin tree.

Let

A ~ wI

IV.I.

and

Ma

(m < w I)

We define a standard tree

T~+ 1 = [ s ~ ( i ) with

Is E T

lim(~)

To define

T

^ i E 2]

as follows.

for

a < wq

As an induction hypothesis,

T~

we force over

= <~,~p>

E M a,

M

A p:a-

G c]P

be the least

rify.)

(in L[A])

is countable in

thesis that

TI8 E N B

Define for

Now let

assume

N -generic

Ma+ I , G E Na+ 1 .

for all

lim(B)

set for

ii) iiD iv)

bn

b n ~ bm each

bn

for is

a-branch

of

Tla,

n ~ m, N -generic for

if

nq,...,n m

are distinct,

bn I

x bn 2 × ... Xbnm

is

~

.

will be trivial to ve-

n E

is an

.

(Hence the hypo-

Claim: Each

TIa E M a .

(pn~qn)

bn = [Pn I p E G ]

i)

a < w

TI~)],

~-~ dom(p) ~ dom(q) ^ Vn E dom(q)

Na

= {d} and

O

with the set of conditions

p ~q

Since

.

T

defined as follows:

: [plSa c ®(lal<~

Let

be as in the proof of Theorem

Tla, then

M~-generic

for

-

(T[a) m

v)

iv).

-

(with the product

T~a = UnE

Proof:

47

ordering)

bn

i), ii) and iii) clearly follow from iv). Let

n1~...~n m

be distinct.

and closed u n d e r extensions.

Let

Then

D*

is a dense initial

Then


s E Tla

D'

.

XbnmnD

Then for some

~P,

p E G ND*.

so let

-

Define (Pn~S)}

is clearly a dense initial

p E G O D'

be dense

.

segment of

D' = ~p E~P [ S n E dom(p) Then

D _c (TIa) m

Let

D* = {p E~P l
To prove v)

We prove

segment of

~P .

Let

n E dom(p) ~ s c Pn E b n ,

so

s E b n ~ and the claim is proved. Set

Ta+ I = ~Ub n I n E w J

still normal.

Hence

.

By v) in the claim,

T = Ua<91 T a

Ti(a+l)

is

is a normal tree of length

wI • By iii) of the claim and Lemma IV.2 ~ T

is Souslin

(and stan-

dard). It remains to prove that trivial a u t o m o r p h i s m E LIB]

Set

follows: that

No

B E No

including

~

is rigid.

T .

; N +I

av

Let

LLBJ [B] .

92 is the smallest

Suppose

B c_ 91

is a non-

be such that

elementary

submodel

elementary

N a = U
for

of

T,o,A

Nv

as

N*

such

submodel

lim(a)

be the least ordinal not in

N~ = L ~ v [ B N a ]

~

Define i n d u c t i v e l y

is the smallest

N v U [Nv} ~ and

v < 91 ~ let nv :N

N* = L

on

T

Nv

of N*

For

and

the collapsing map.

As in the proof of Lemma IV.2

there is an

a = am

such that

-

Nm --c M m .

We know that

By elementary

be such that

x o x°

Then

m j n

-

~a(T)

equivalence,

x ° E TI~

is

4 8

a~(Tlm)

x = Ub m

Now,

Nm-generic

By the product Tim,

hence

is non-trivial,

and

= Nm[bmJ , so

is

Pick

~(x)

for

(Tlm) 2

.

is

such that

of

= Ub n

"

so let

x ETa

b n E NaLbm]

lemma (Lemma 1.8), b n

b n ~ Ma[bmJ

wa (~) = ~ ( T I m )

By the construction

Uv~ m a ( x ~ v ) E Zm[xJ bm × b n

and

~(x o) I x o .

~(x) # x .

such that

= Tim

Tm

there

Thus

a(x)

=

.

by iv) in the claim. Mm[bmJ-generic

Contradiction ~

for

I

In fact, the rigid Souslin tree just described was the tree w h i c h Jensen defined in his first construction

of a Souslin tree in

L .

We

also note that the last part of the above proof may easily be m o d i f i e d to give the following result: Souslin tree, gid.

and

C ~ w~

If

T

is the above constructed rigid

is closed unbounded,

then

TIC

is also ri-

We will at once use this result to obtain a rigid Souslin line.

2. Rigid Souslin lines Theorem 2 Assume ~

Proof:

.

Then there exists a Souslin line

(i)

S

is rigid,

(ii)

S

is not reversible.

Let

T

be the above rigid Souslin tree.

fined in the b e g i n n i n g line. let

Now,

let

~(b) ~ b

C ~ wI

S

.

c

of section IV.2.

be a non-trivial

By Lemma IV.6

and an a u t o m o r p h i s m

~

such that

Let Then

S = ST S

automorphism

as de-

is a Souslin on

S , and

there is a closed u n b o u n d e d on

TIC

such that

~"I s =

-

I~(s)

whenever

identity, Let

so

a E C

s E TIc

~"I s = Is

49

-

.

Since

whenever

be such that

easily finds a a(d) g Is,

d E S

where

TIc

s E TIC

such that

s = d~mETIC

e(d) ~ d , .

But,

is the

.

~ ~ max[dom(b),dom(a(b))j

e(d) E o " I s = I s , a contradiction. S

is rigid, ~

.

Then one

dom(d) > a ,

since

and

d E Is,

Hence we have p r o v e d that

is rigid.

If now

S

was reversible,

then, b y Lemma IV.8 , there w o u l d

have b e e n an antilexicographical

automorphism

some closed u n b o u n d e d

But then

is impossible,

since

C ~ wI id

~

on

TIC

for

~ = ida(TIC) , which

is not at all antilexicographical

I

One might be t e m p t e d to conjecture not be reversible.

that a rigid ordered continuum can-

This is not so, however.

We now construct f r o m ~

a rigid Souslin line w h i c h is reversible. Theorem 3 Assume

~

.

tomorphisms,

Proof:

Then there exists a Souslin tree n a m e l y the identity on

T

T

= dom(s)

and

with exactly two au-

and the m a p p i n g

s E 2 <wl , ~

We recall at once that for dom(s)

T

s~ = q - s v

for

s~'~s

.

is defined by:

v < dom(s)

The tree

is d e f i n e d as the rigid tree, but with the f o l l o w i n g modifi-

cation:

If

we now define every branch

lim(~) , then T

to be

{s I s E bn}

Souslin b y Lemma IV.2. that

T

~ , G , b n (n E ~) [Ub n I n E w ] is

are as before.

U [~n

Ma-generic

for

InEw] T

But

" Of course,

, so

T

is

It is also clear from the c o n s t r u c t i o n

is closed under the a u t o m o r p h i s m

sr~s

.

As in the final part of the proof of Theorem I, the assumption that

~

is different b o t h from the identity and the m a p p i n g

-

s~s

-

readily leads to a contradiction.

as before. ther

~(x)

Then for every = x

x o I o(x o) ~ Xo that

o(x) ~ x ,

or "

By m o d i f y i n g the proof, r e p l a c e d by

TIC

o(x)

For let

x E T m , a(x)

= ~ .

Then,

hence

contradiction :

T

50

Let

taking o(x)

be

E Mm[x] , hence ei-

x ° E Tim

x E T m,

= x .

m = mm

be such that

x • x o , we have

But then

°(Xo)

= Xo ' a

|

one finds that this theorem is also true with

whenever

C ~ wI

is closed unbounded.

Theorem 4 Assume ~ .

Proof:

Then there exists a Souslin line (i)

S

is rigid,

(ii)

S

is reversible.

Let

T

be the tree of Theorem 5.

note above and Lemma IV.6

Set

we get that

as in the proof of Theorem 2.

to

such that

S = ST . S

<S, ~ > .

From the

is rigid,

Furthermore,

obviously gives an i s o m o r p h i s m from which is isomorphic

S

(S, ~ ) ( S

exactly

the m a p p i n g

br~b

onto a continuum

is obtained by remo!

ving the right-hand

sides of all gaps from

Chapter IV, section 2.)

ST

as defined in

|

3. S o m e final remarks We know from deed,

if

T

~

that there are two n o n - i s o m o r p h i c

Souslin trees.

is the rigid Souslin tree from section 1, then

In fact, by an easy argument, that there are

2 wl

Jech

non-isomorphic

LJe 3]

has p r o v e d that

Souslin trees.

to the '~classical" result of Gaifman and Specker

~

In-

T (0) ~ T ~1 implies

This is analogous

LGa Sp]

that

(in ZFC)

-

there are

2 wq

different

51

-

i s o m o r p h i s m types of A r o n s z a j n trees.

this context we also m e n t i o n that Devlin [De 2] of B a u m g a r t n e r

and proved that,

assuming

morphic A r o n s z a j n

trees w h i c h are

For normal trees

T

automorphisms

T .

on

of length Hence,

~(T)

= ~1

IV.

In fact, Jech [Je 3]

length

o(T) w = o(T)

.

2w

, there are

2 wl

if

T

e(T)

denote the number of

is rigid,

Souslin tree

is either finite,

then

T

has proved that if

c(T)

= 1 , and

constructed T

or

T

with

and

2 ml

~(T)

in Chapter

is a normal

tree of

2 ~ ~ o(T) ~ 2 wl

of arbitrary p r e s c r i b e d

provided

non-iso-

Q-embeddable.

He has also proved that it is consistent

a Souslin tree between

o(T)

has extended a theorem

]R-embeddable but not

wq , let

for the homogeneous

w I , then

<>

In

and

that there is

cardinality

~w = ~ .

Jensen has proved that if we assume

, then there is a homogeneous

Souslin tree

(It is doubtful w h e t h e r this

is p r o v a b l e >+

T

such that

from

Define

<>.)

We scetch the proof.

6o , M o

chy of models

N'

~(T) ~ w 2 .

Let

(So l a < ~ 1 )

as in the proof of Theorem IV.1.

is defined as follows:

N'

O

A new hierar-

= L 6, [AJ

O

satisfy

where

6'

is

' O

the least

8

countable in case

such that N'

O

A tree

"

lim(a) : An

generic for

L6[AJ T

o-branch

TIo

and

~

ZF- + "8 a is countable".

is defined as before, b

of

b E N' . o

Tlo

on

Tla,

ded to an a u t o m o r p h i s m lim(S),

if

~ E Ms

on

then

a

is extended iff

TI(a+I)

Furthermore,

for

is an extendable

automorphism

on

(This is so b e c a u s e

M s ~ Hwl . )

of automorphisms

T .

Now,

)

and

A .

Let

is

T

on

let

TIS

a

c E Na

N* = L

Let

LIB] [B], ~2

B ~ wI

o < S < w I with

extending

then

M s I=

~"

be a sequence

code up

N v ~ N*,

is

can be exten-

T~m,

(av I ~ < Wl)

M o-

is a normal

We claim that there is an a u t o m o r p h i s m

(~ < ~1 ) , thereby p r o v i n g the theorem. (av I V < ~ l

i.e.

.

"there are u n c o u n t a b l y m a n y automorphisms

on

b

It also follows that if is extendable,

M o is

except for the

It easily follows that

Souslin tree which is homogeneous. an a u t o m o r p h i s m

Thus

o~,

p #o v

T , Nv

be as

-

in the proof of Theorem EC ~ B~ ~,CN~ the closed

~S~. Let

set

induction

on

phism on

TI¥ o

M¥o

be the monotone

a sequence

.

~P~ I v < w1>

such that

enumeration

NyvC My~

: Po

.

is an automor-

a ~(TI¥o)

for

TIyv+ 1

such that

~v+1 o pv

"

PK = ~ < ~

Now let

P~

for

p = U <~1 p~ .

of

By

from all

on

~ < Y~+I

Pk E Nyk

C ~ wI

As before we obtain

different

for

can be proved that for

.

is an automorphism

Pv+q ~ ~ ~(TIY~+I)

there is a

<¥v I v < w q >

we define in

-

By ~ +

[¥ E C I ¥ =ay] v

Pv+1 E MyV+l

IV.I.

52

~ < ¥o and

lim(k) Then

It

p ~ u~

v < ~I ' and we are done.

Since the above constructed we get the following ous, reversible The notion

result:

Souslin

for all

x E T .

that

~

implies

with

~I

T

such that ~,¥ < ~, ht(s 1)+ht(s

that

T~

~ < a,

Tx = [yET

Ta

and

consist

is defined

let

lie in

S o ~ S 2 = s*

2)

s*

if

T

I x~yJ

T~

s1~s 2 = s . of all

s1~s 2

If

a homogene-

automorphisms. to the following.

is isomorphic . )

as follows

iff there are

to

Tx

Jensen has proved Souslin tree for

T

lim(a) : If Mm-generic

So,S I E Tla a = ~+ y

with

s~'~s ,

and

s2

for some

sq,s 2 E TI~

and

= ~ .

A complete Boolean there

can be proved

~2

be the union of a

The last concepts we shall mention

- [~,M]

exists

that there is a totally homogeneous

s E 2~

let

, there

tree may be strengthened

totally homogeneous

(Recall

for all Let

~+

line which has at least

automorphisms.

branch.

Assuming

of a homogeneous

Let us call a tree

v+ a = ~

tree is closed under the mapping

exists

(if

is isomorphic ~

~

is called homogeneous

an automorphism

I~I > 4) that to

element

in

Souslin

algebra iff

der the reverse

algebra

with

are concerned with Boolean

~

G

such that

is homogeneous

algebras.

if to any o(b)

= b'

iff for all

~Ib

(the algebra obtained by intersecting

b ).

A complete

~

has a dense

ordering.

Boolean

algebra

~

b,b'

E

It b E~, every

is called a

subset which is a Souslin tree un-

(More on this concept

in Chapter VIII.)

-

5}

-

Now it is easy to prove that the canonical complete Boolean algebra constructed from the totally homogeneous Souslin tree of the last paragraph is a homogeneous Souslin algebra with at most Using the trees of Theorem I

and Theorem 3

wI

automorphisms.

we obtain in the same way

a rigid Souslin algebra and a Souslin algebra with exactly 2 phisms. w2

Finally, the definition of the homogeneous Souslin tree with

automorphisms

homogeneous. least

automor-

w2

can be modified so that the tree will be totally

Hence we obtain a homogeneous Souslin algebra with at

automorphisms.

C h a p t e r VI

MARTIN'S

We have

seen e a r l i e r

A X I O M A N D THE C O N S I S T E N C Y OF

that,

given a

are c a r d i n a l a b s o l u t e g e n e r i c

c.t.m.

extensions

M

SH

of

NI

ZFC + GCH

and

N2

of

, there

M

such

that (i)

NI ~

2$ = w 2

&

(ii) ~N2(w) =~M(~)

~ SH

and

N2 F

In this c h a p t e r and s u b s e q u e n t f

e ~Si

GOH

ones we shall show h o w it is p o s s i b l e

I

to o b t a i n m o d e l s

NI

and

N2

w h i c h give

ency r e s u l t s

SH

in place

the c o r r e s p o n d i n g

consistI

for

of

~ SH

.

The c o n s t r u c t i o n

w e l l known,

and there are at l e a s t

two good e x p o s i t i o n s

literature,

so we shall only s k e t c h the proof,

and use

of

N I is

already

in the

this s k e t c h

to

I

motivate

the c o n s t r u c t i o n

in the r e m a i n i n g

of

chapters.

N2 Most

w h i c h we shall give

of the r e s u l t s we shall m e n t i o n here

are due to S o l o v a y and T e n n e n b a u m . first established Suppose

the r e l a t i v e

then we are g i v e n a

a cardinal absolute in

~

.

ticular,

Thus,

in

generic N

any S o u s l i n

w h e n we pass

to

same cardinals, N

N

consistency

c.t.m.

M

linity"

by g a i n i n g an u n c o u n t a b l e

branch)

in

turns

N

tree

SH

ZFC of

M

it was

they who

.

, and we w i s h to find such that

SH

holds

to be no S o u s l i n trees.

~

in

M

antiehain

M

and in

M

N

to have

will

the

still be an

can only lose its "Sous(or, a f o r t i o r i ,

an

~1-

to c o m m e n c e by seeing h o w

the S o u s l i n i t y of single S o u s l i n However,

In par-

to lose its " S o u s l i n i t y "

~1-tree

So it is not u n r e a s o n a b l e

out to be r a t h e r easy.

l a t e r chapters,

of

w i l l have

it is clear that an

, so a S o u s l i n

we can d e s t r o y

of

Since we w i l l r e q u i r e

in

.

M

extension

~1-tree

N

In p a r t i c u l a r ,

there w i l l h a v e tree in

in full d e t a i l

tree in

to s i m p l i f y

M .

This

the n o t a t i o n

this s i m p l i f i c a t i o n w i l l be e x t r e m e l y h e l p f u l . ) ,

(In let

-

us first of all amend Aronszajn

we simply r e p l a c e (chapter II).

~T

than "depth",

and

terms of

rather

that n o w a S o u s l i n Thus,

if

T

force w i t h

tree)

by

To o b t a i n

re-

the a m e n d e d d e f J n i t ~ o n ,

in the o r i g i n a l d e f i n i t i o n s

still

talk about

"height" h o w e v e r ,

this w i l l m e a n than

~T

just what it did before,

")

The a d v a n t a g e

tree is a s p l i t t i n g poset

is a S o u s l i n tree in a

N

of

T , we do not d e s t r o y any cardinals. question,

but in

of this a m e n d m e n t

satisfying

c.t.m.

rather

is

c.c.c.

ZFC

, then if we

It turns out that

for we h a v e

already proved

in

( T h e o r e m II.7)

I N

be a

(Thus, limit b

of an

everywhere

C h a p t e r II the f o l l o w i n g l e m m a

Let

(and h e n c e

~T

we also a n s w e r our o r i g i n a l

Lemma

~1-tree

so that they "grow d o w n w a r d s "

as p r e v i o u s l y .

(We shall

~T

-

our d e f i n i t i o n of an

tree and a S o u s l i n

ther than " u p w a r d s "

55

c.t.m,

of

ZFC

in the real world, ordinal,

and so

T

~

, and let is an

has

if we take a S o u s l i n

be a S o u s l i n

a-tree

for

a-branches.)

Let

T

just in case

b

tree

T

in a

tree in

~ = w~

2~

w i l l be a c o f i n a l b r a n c h of

Hence,

T

e.t.m.

a countable

b c T . is

M

N.

Then

M-generic

of

ZFC

for

and

N

form a g e n e r i c e x t e n s i o n for in

T , then

T

N

N

M[G]

killin~ ly,

~ .

T .

by t a k i n g a set

tree in

We r e f e r

N

extension either:

it is dead and e v e r a f t e r r e m a i n s

that a h o p e f u l m e t h o d

iterate,

in some manner,

one at a time.

tial that the e n t i r e

G

w i l l be,

to this p r o c e s s

N

of

N[G]

once a S o u s l i n dead~

as clear-

, then tree is

This last o b s e r v a t i o n of

SH

w o u l d be

to

1, k i l l i n g all the S o u s l i n trees in

in o r d e r for this to work,

iteration

N-generic

, since

for o b t a i n i n g a m o d e l

lemma

Now,

M[G]

G

This term is j u s t i f i e d because,

any f u r t h e r g e n e r i c

will not be a S o u s l i n

sight,

M

w i l l not be S o u s l i n in

the S o u s l i n tree

killed,

of

, a c o f i n a l b r a n c h of

if we m a k e

means

M[G]

can be "handled",

it is e s s e n -

to some extent,

with-

-

in the original set theory,

Well,

any finite

number

We describe

be much easier their

M :

-

we want

our final model

and the only way we k n o w how

of forcing.

ged.

model

56

in the case of such)

associated

(complete)

is just a disguised this is clearly

not

below.

boolean

with

(and hence

is easily it turns

~ , concerned,

algebras.

form of forcing

irrelevant

this

However,

the posets,

(Since

BA(~)

as far as killing

arran-

out to

but rather

forcing

with

- see chapter

T

of

is by the method

extensions

successively,

this a r r a n g e m e n t

to consider

to do this

of two generic

formed

to be a model

I -

is concerned.)

This

N

will not be apparent next lemma, vantage

nificant

Let

but already

will be quite

to denote

Lemma

in the case

"complete

of two extensions

for doing

~

significant.

boolean

strengthening

"successive

in the

extensions"

the ad-

From now on, we shall use

algebra".

of lemma

described

The f o l l o w i n g

lemma

"BA"

is a sig-

1.8.

2 (Solovay) M

M~Bo )

be a

c.t.m,

be such that

M , a ~ique such that

of (in

m2

~A

M (~°)(]B1)

ZFC M)

.

Let

IB°

be a

I~ I is a BAI~O

BA

= ~

in

.

~ M (]B2) , and a canonical

Let

Then there

domOB 2) -- ~:x~M(]Bo)l

(with

M .

IIx~:~l]Po

complete

]BIE

is, in

= "n:})

embedding

e :]B ° - ]B2 . Furthermore, (i)

if

Go

]BI/G ° is

whenever is

M-generic

~i ) , then there

if that ~o that

G2

for

(the BA in M[Go]

such that (ii)

]Bo, ]BI, ]B2

is

]Bo whose

for

~2

e = idI~ ° in the above) and there MEGo][GI]

is an

GI

name

in the

°

which

is

we have:

M[Go]-generic ]Bo-forcing

is

M-generic

for

language for

~2

. then

(assuming

G o = G 2 0~ o

MEGo]-generic

= MEG2]

as above,

and

G2

is a set

M[Go]KG I] = M[G 2] M-generic

are related

set

GI

for simplicity

is for

M-generic ~i/Go

for

such

-

Proof:

This

of

fix E ~ II~ ° = ~] sense

of

~2

a unique and

such that The rest

2

trees

in one go.

, and,

the m a x i m u m

to define

principle

z E M (~°)

for

llb2 = ~I~ ° = b °

enables

~i)I~ ° = ~

is routine,

us to handle

any finite

That we do not destroy to r e m a i n

nitely

Souslin

trees,

M (~o)

to find

z E dom(~ 2)

element

b 2 E ~2

o

trivial.

number

any cardinals

so when we come

we require

in the

~

but not

I

.

11b2

and

the

x v y

such that

be that unique

e(bo)

describe

= {xEM(~o)

' let

this

|

of Souslin in the process

to deal with

the f o l l o w i n g

infi-

lemma:

3 (Solovay) M, ~ o ' ~I' ~ 2

(in

be related

M)II'~BI has

The obvious

Suppose

now we use lemma

of Souslin (~nl n ~ ) complete

trees

in

of

BA's

idea works.

M .

morph

embeddings

Then

~

~

Thus,

~

to form

We call

~

w

are in [Jell,

system for

(in

I='~ o has

~ '~2 has

with

we construct

is best

an

an

m~

is to regard the direct

union

86.1

~)

of is de-

with

this

as follows,

Iso-

function

~n"

Let •

Lemma

, enm

M) associated

of course.

the c.c.c."

w-sequence

n+1

described

c.c.cJ'

w-sequence

(enm I n ~ m ~

is the identity

algebra ~

M

M

successively

BA

enm

If

, then

(where,

This

is a boolean

(One way

= BA(~) ) .

in

A natural

so that each

= Un~w~ n

completion.

to deal

union.

2.

The details

enm: ~ n - , ~ m

is its direct the system

2

: ~

and a commutative

fined by composition). system

as in lemma

the c.c.c."l~o

Proof:

set

dom~B2)

for example,

of the proof

clearly

Let

simply

of

but for

and

e .

We shall

(in the sense

is obvious,

Let

and

(by normality)

b o E ~o

Lemma

~2

' use

llz = x v y

Given

Lemma

-

is just lemma 85 of [Jel].

construction

many

57

~w

on

be its

as a poset of the chain

and

-

(~n] n < w ) identity, plete

we call

the

embeddings

en

(~Bn)n~ here

In the case where BA

, will

.

the embeddings with

, the direct

Suppose

algebra

-

, together

:~n-~

~ (enm)n<m~)

as the limit

~

58

limit

then we take

(murderous)

were not

the

the c o r r e s p o n d i n g

in our iteration.

all of our previous

enm

of the system

(in If

com-

M) G

•

is

as defined M-generic

work be incorporated

for in

w

M[G]

?

lemma, Lemma Let

The positive

answer

in c o n j u n c t i o n

with

]Bo, ]B2

be

in

BA's

~ .

]~1 i s

in

part

[SoTe]

of lemma

by our next

2.

generic

~

Since

systems

of any

ultrafilters

(limit)

length,

chain bra of

BA's

~8

of the Proof:

such

union

~ ¥ 's .

that

Then

of

cf(~)

Theorem

the canonical ~o

~I

name

is obtained (in

M) for

M-

|

we state

and let

(a)

(~8

~ a

By a combinatorial when

speaking,

is g u a r a n t e e d

construction

ordinal,

' (b) y < a ~ ~ y

is the direct

(in

will work

by the next

equally

it in a fairly

well

general

for

way.

- Tennenbaum)

be any limit of

such that

em-

~ }~(]B2)

any cardinals

limit

be a complete

]B I c M (IBo)

Roughly

on

the above

5 (Solovay ~

M(]Bo)(]B1)

through

does not destroy

lemma.

Lemma

and

~2

e :]Bo-~]B 2

is an element

for details.

by f a c t o r i n g

That

M , and let

Then there

a BAI~O = ~

See

Proof:

Let

the second

is supplied

4 (Solovay)

bedding

M )

to this question

49.

¥ <5
has

c.c.c.,

] 8 < y)

has

(~y] ¥ < a)

be an increasing

is a complete subalgeY and (c) y < ~ & lim(y) ~ T

Let

~

be the direct

union

c.c.c.

argument.

= w I , of course.)

(The only n o n - t r i v i a l The details

case

are given

is

in [Jell,

-

Using

the above lemmas,

c.t.m, N

of

of M

ZFC + GCH

M

-

it is clear h o w one could,

(at m o s t

them all, k i l l i n g

M

, we d o n ' t

stage, we simply a r r a n g e be i n c l u d e d

ing"

we m a y w e l l

it is p o s s i b l e

to arrange)

any n e w S o u s l i n

the title

.)

in M

carry out the m u r d e r at e a c h

this w i l l not imply

introduce new Souslin (and e s s e n t i a l l y change

with

w .)

trees

ling S o u s l i n

(ostensibly)

trees,

a certain

(cf.

that

This is d e s c r i b e d f u l l y in [Jel]

to

our r e f e r e n c e SH

to [Jell will h a v e

, but r a t h e r

to a (stronger)

In order to m o t i v a t e

we prove

the f o l l o w i n g

doing

the f o r m u l a theorem, which

to o b t a i n

s o m e t h i n g more

SH

in

than just k i l -

trees at all,

but

of posets.

Theorem 6 Let

(ii)

M

be a

There

c.t.m,

of

ZFC

is a (splitting)

such that:

to

so as to include

the "usual proof"

and in face n e e d not c o n s i d e r

class

tree

IX".

A x i o m somewhat,

we are

N

just a m a t t e r of " b o o k k e e p -

tells us that in any i t e r a t i o n we p e r f o r m in o r d e r some model,

that

in the process.

the s t r a t e g y

trees w h i c h may appear.

"Model

(in

to go t h r o u g h

a l t h o u g h we a r r a n g e for some S o u s l i n

k n o w n as M a r t i n ' s Axiom.

tion of M a r t i n ' s

rather

S o u s l i n trees

i n s t i g a t e d by a d d i n g a g e n e r i c

Of course,

that he does not r e f e r

principle

extension

(Note that since we are do-

actually

r e a d e r s who h a v e b e e n f o l l o w i n g

observed

is a

it so that the S o u s l i n tree c o n c e r n e d w i l l

to c o n t i n u a l l y

is e q u i n u m e r o u s

under Now,

BA

For at e a c h stage,

w × w

many)

, and p r o c e e d s

in some f i n a l m a s s - m u r d e r ,

s u b s e t of the f i n a l

However,

M

then off s u c e s s i v e l y .

ing e v e r y t h i n g in

be killed,

w2

M

To do this one s i m p l y c o m m e n c e s w i t h an e n u m e r a t i o n

) of all of the S o u s l i n trees in

SH

assuming

, obtain a cardinal absolute generic

in w h i c h all of the

are killed.

59

.

The f o l l o w i n g are e q u i v a l e n t :

poset

~

in

N

of c a r d i n a l i t y

wI

-

(a)

M

~'~

satisfies

(b)

if

G

(c)

for

some

is

generic

Proof:

(i) - (ii).

P

Let

~

, then

@M[G](w)

of c a r d i n a l i t y

, then

~

;

for

X E M

for

-

c.c.c."

M-generic set

60

G ~ M

= T

=

@M(w)

w I , if

G

;

is

X-

, and

set

.

be a S o u s l i n

tree

in

M

x = { u B ~ ~ ] ~ < w 1} (ii)

- (i).

shall,

in e f f e c t ,

proof. wI be

We g i v e

Let If

~

a forcing

give

is

F G E M[G]

T = [FG~a

I G

order

by

T

clearly

gl ~ T g2

Claim

I

In

, define

M

some

T' =

Hence such

Conversely

Then for

, so we f E M

that

.

p

Then

M[G] ~

,

follows

Claim

M

Let

X E M

domain

FG: wMI " 2

FG(~ ) = I ~-aEG.

&

~<w~] "

, and

Thus

set

partially

~ = ( T , ~ T)

of h e i g h t

~

is

.

, X -c T'

G

f E M[G]

Hence

.

on

But

since

f = FG~a

Hence

f E T'

look,

as a s u b s e t

. Pick

, so

and

MFG~ of

w

I= in

, there

is

11

= F~a

G

~

But

f

be

p E ~

(since

antichain

' IXl M = ~IM

and

M-generic

f E T .

immediately

"Every

M-generic

can code

Let

"f = F G ~ ~"

~

has

In the r e a l w o r l d ,

P

some

f E T'

~ = F

the c l a i m 2

pl~"~

suppose

that

~

, with

gl ~ g2

v

p E G

algebraic

If I ( ~ P E ~ ) ( 3 a < ~ I ) [ P I ~ = F ~ ]

, f = FG~a

"fE 2a ^ ~ <W1" MKG]

we

.

f E T .

a < w~

VIII

, we let

G

G .)

for

iff

assume P

for

(in the r e a l w o r l d )

T E M

Suppose

for

such

M-generic

boolean

We m a y

function

for any

is

a tree

(ii).

M-generic

the c h a r a c t e r i s t i c

(Thus

In c h a p t e r

an a l t e r n a t i v e ,

be as in

G _c ~

proof.

of

Hence

. a ( w~

so

~

with

p E G.

T = T' E M

, and

on

~T = ~ ) " ~

In

is c o u n t a b l e " . M

, for each

g E X

,

-

choose

p(g)

Since that

~

E•

has

and

M

and there

is

^ g2 = F

p E G .

Then,

or

if b

of

of

does not But

whose

split,

the fact

if we are

which satisfies

c.c.c.,

set

G

whenever

Now,

if you l o o k b a c k

•

=

with gl

of T"

"T embeds a Soussuffices Then,

can only be one function extends P

P

of

is a p o s e t I~ " @(~)

SH

=

observe

, then in M

@(~)"

it ~I

, and w h e n -

(i.e.

in M) a

.

that n o w h e r e

of a m o d e l

of

did we make use of the fact

tree adds no n e w s u b s e t s

in v i e w of the above,

subset

|

of c a r d i n a l i ~ y

w I , then there is ~

cofinal branch

g , w h i c h clear-

is splitting.

M

to s h o w

clearly,

M-generic

to our s k e t c h of the c o n s t r u c t i o n

that f o r c i n g w i t h a S o u s l i n leads us,

for

M ~

lies on a u n i q u e

and for w h i c h

X-generic

ZFC + SH , y o u w i l l

for

S u p p o s e not.

g

that

is a set of c a r d i n a l i t y w h i c h is

p I~"gI

' so e i t h e r

it c l e a r l y

to o b t a i n a m o d e l

that:

X

poset.

then there

will be the case

ever

such

M I----"X is not an a n t i c h a i n

characterictic

ly c o n t r a d i c t s

By the above,

Thus

M-generic

^ g2 = F G ~ a ( g 2 )

Hence

is a s p l i t t i n g

T .

]P

gl = F G ~ a ( g l )

G

In v i e w of the above

g E T

g1' g2 E X

p _< p(gl) , p(g2).

the p r o o f by s h o w i n g that

lin tree" T

we can find

V ,!

p(g) I~--"~=FG~a.

is proved.

We c o m p l e t e

that

o

such t h a t

Pick

g2 c_ gl

and c l a i m 2

< mI

in

F

c_ g2

-

a(g)

c.c.c,

gl ~ g2

61

to f o r m u l a t e

of

~ .

This

the f o l l o w i n g proposition

for c o n s i d e r a t i o n : :

Whenever c.c.c.,

•

is a p o s e t of c a r d i n a l i t y

and w h e n e v e r

there is a set Clearly

e - SH

of an i t e r a t e d

G

X

which

is a set of c a r d i n a l i t y

X-generic

for

P

And it should be c l e a r n o w forcing argument

~I

satisfies ~I

, then

. that we can,

of the type o u t l i n e d

above,

by m e a n s obtain a

-

model

in w h i c h

which

makes

not

add

tainly

reals

will

add

seen

that

this

is

lead

to the

just

wI

thus

replace

It is tion

reals

ZFC

Cohen

® ®

final we

, only

stages. (In

prove

the

Hence, the

case

there

is n o t h i n g

that

it be

smaller

cannot

of

that

the

ceravoid

~ , it is e a -

® ~ 2w ~ ~2

:

along

lines

special

than

we m a y

we m o s t

we

consideration

that

iteration

although

stages,

outright

Further

of that

sucessor

model.

can

these

about

the

car-

continuum.

We

by: ~

is a p o s e r

than

2w

rJel]

that

CH

sets"

for

"X-generic

¢ +2!~ = ~2

will

the

at l i m i t

less

the

show

of

c.c.c.,

of

true

~

satisfies

in

analysis

forcing~)

~enever

of

or

-

A further

some/all

in our in

in

shown

guments of

at

realisation

dinal

:

true.

SH

new

2~ = w 2

sily

is

either

new

having

®

62

type

and

, then

outlined

' ~ + 2w = w3

of e a r d i n a l i t y whenever

there

(This

countable

' etc.

G

is

can

(And

2 w , which

of c a r d i n a l i t y

X-generic

just

X .)

one

than

is a set

is a set

~ ~ .

above,

X

less

By

the u s u a l

iterated

establish

clearly,

for

P

construc-

forcing

the

.

ar-

consistency

~ + 2w = w 2

is

just

®.) Finally,

we

quirement

obtain

that

because,

by

theorem,

one

lemma ~A

Martin's

P

have

a simple sees

an e x t r e m e l y

concerned,

~ ~ MA

which ZFC

powerful

.

, by

less

of

the

(For

omitting

than

2w

downward

from

.

¢

We m a y

the do

re-

this

Lowenheim-Skolem

a "direct"

of i m p o r t a n c e

.

In

to

tained.

In p a r t i c u l a r , < 2 w Lebesgue

the u n i o n

of

the u n i o n

[MaSo], it is

measure

< 2 w Lebesgue of

assumption

is

relative

and

cardinality

application

that

, MA

proof,

see

[Jell,

89.)

is

on of

Axiom

< 2w

in v i e w

several shown zero

sets

sets

far of

as

the

the

if has

sets

MA

continuum

consistency

consequences

that

measurable

meager

as

of

holds,

Lebesgue

(All

of are

then

measure

is L e b e s g u e

is m e a g e r .

MA

of

is NLA ob-

the u n i zero,

measurable, these

are

-

trivial highly of

if

2 ~ = w I , of course,

significant

NA

we shall

when

to establish,

SH+2 w = w 2

ZFC

trees Let

some

simple

to Kunen,

Baumgartner,

discussions,

the c o n s i s t e n c y

(We shall prove

of

the c o n s i s t e n c y

of

chapters.)

combinatorial

facts

be an infinite

set,

n

integer.

is w e l l - d i s t r i b u t e d

iff,

E Y

[~] = ~ .

such that

U N

to the e x i s t e n c e

integer

m

iff there are = ~ .

buted for all T

for every finite

members

of

Clearly,

, we shall

y c Xn

set

Y'

(~i) ..... (~m)

Clearly,

a positive

of an infinite

are d i s t i n c t

sitive

If

that every A r o n s z a j n

concerning

Aronszajn

in general. X

(~)

strong manner,

.

SH + 2 w = w I , in the r e m a i n i n g First we require

implies

of an a p p l i c a t i o n

in v i e w of our above

in a fairly to

, but they become

As an example

is due v a r i o u s l y

It will,

relative

by d e f i n i t i o n

M A + 2~ ~ w I

This result

and R e i n h a r d t - M a l i t z .

-

2w ~ w I . )

show that

tree is special.

serve

63

set

We say

u ~ X

there

this c o n d i t i o n

y, c y

, then

say that

E Y

such that

is

(~)

is e q u i v a l e n t

such that w h e n e v e r

~] D ~]

thus

Y ~ Xn

= ~ .

Given

(~), a po-

Y ~ Xn

is

m-distributed

I ~i~

j ~m

~ [~i ] q [~j]

will be w e l l - d i s t r i b u t e d

iff it is

m-distri-

m e w •

is a tree

and

z E S c T ,

S (z)

will denote

the tree

m

~y Es l y~_~z Let

or

T I, ~T2

domain

be

z_~y~ ~1-trees.

[(x,y) I ( ~ w l ) ~

~-~ x --~I x' & y --~2 y' wl-tree. case

)

~T1 ® ~T2

xETI~ & yET2~ (It is clear

Similarly

TI®...

T I = ... = T n = T~ here,

The f o l l o w i n g

simple

will denote

®T n

]

that

the

wl-tree

and o r d e r i n g T I ®T 2

will

for any p o s i t i v e

we write

simply

Tn~

l e m m a is the crux of our proof

(x,y)~(x',y') in fact b_~e an

integer

for

with

n .

~TI®''" ® Tn~

:

Lemma 7 Let

T

be an A r o n s z a j n

tree,

n

a positive

integer.

Let

S

be a

In

-64-

final

section

extensions

in

(V8 ~ a ) ( S ! z)p

Proof:

We

of

Tn

S .

such that For every

every

z E S

z E S

there

has uncountably is an

purposes Claim

regard

n-tuples

of this

proof.

I

Let

as m a p s

z E S , i < n

.

with

same

Suppose Pick

level

not,

of

ht(w)

such

2

= ht(w')

Let

that

= ~

that

z ° = z no

I, p i c k

.. = h t ( w n )

, such

v E S , v ~ z~ = ~

i

be

that

, hi(v)

the l e a s t

3

Let

such

that

This

follows

z E Sla m

be

For

and

claim

<j ~n

, and

are

2

z k+1

.

z ko,z~

(z~"k)O

of (z~"k)

--< zko ' h t ( w o ) = "" / wj(k)

.

i ~ n

.

Pick

, (v"k) O

By c a r d i n a l i t y

that

= w

wi(k)

, z~ +I

= v

1

m < ~

there is

a mZ

on

.

. <

W

m-distributed. induction

con-

~ v"n

0

by a simple

S

z ko' z~ ' are d e f i n e d .

all

such

is

= ~

as r e q u i r e d .

~ wi(k) For

each

S(~ am )

,

is a b s u r d .

, elements

is a tree

Set

E S

Hence

, which

and

i ~ n

such.

z E S .

from

~

= ht(z~)

~

on

property.

= w'(i)

(Zo"n) D (z1"n)

= ht(wo) as

are

Zo,Z I E S , Zo,Z I ~ z ,

k < n

O~i

are

~ z I , w,w'

Wo,...,w n E S ' Wo'''''Wn

there must

Claim

of

z I = z nI

Suppose

of c o u r s e ,

siderations, Let

and

w,w'

k _< n

zko'z~ _< z , h t ( z ~ )

z° o = Zlo = z

(wi"k)

on

there

the a b o v e

w(i)

are

and

by i n d u c t i o n

, whence

By claim

There

h t ( z o) = h t ( Z l )

that

Set

T h e n if

~1-branch

the

Zl,...,z p

with

, we m u s t h a v e

is an

for

~ zj(i) @ z k ( i ) .

be m a x i m a l

z 6 S .

construct,

such

p

n

p < ~

that

I ~ j
as s t a t e d .

[w(i) I w E S (zl)] Claim

and

and l e t

Zl,...,z p

and

S

domain

F o r any

Zl,...,z p E S , Zl,...,z p ~ z , such

We

such

is w e l l - d i s t r i b u t e d ) .

shall

the

~ < ~I

many

m

.

I

-

Suppose now that if

B > az

then

--

hence Corollary Let

z ES

-

is given.

S (z)

'

65

is

Set

z = SUPm<~am~ .

m-distributed

for all

Clearly,

m

, and

B

well-distributed.

|

8

~, n,

S

be as a b o v e .

our f u n c t i o n

• : wI - wI

(~ > ~)~s(z) ~(~)

such

Immediate.

Theorem

9

Assume

MA+

Proof:

Let

that whenever

increasing,

z E S~(v)

continu-

, then

I

2w > w I .

~ T

iff if

u

E u

the

point

of

.

A partial

map

Let

be

Claim

observe

p

Dx =

[p E P

Let

X = [D x I x E T] X-generic map

in

has

from

say

is s p e c i a l .

u

of

is a n e a t

T

, and

such in

~

into

p.e.'s

of

subtree

that:

T , then

conversely;

an e x t e n s i o n

subtree

on e v e r y

they

and level

Q

is an o r d e r - p r e -

T

into

ordered

Q .

by i n c l u s i o n .

c.c.c.

of this

claim

follows

I x E dom(p)}

set

u

of

of a l l

the t h e o r e m

let

tree

same h e i g h t

from a neat

the p r o o f

preserving

u

satisfies

how

the

(p.e.)

the p o s e t

~

Ignoring

of

embedding

serving ~

have

We

subtree

same h e i g h t

every u

tree.

is a f i n i t e

x,y

have (ii)

Then every Aronszajn

be a n y A r o n s z a j n

N

(i)

an

is a s t r i c t l y

is w e l l - d i s t r i b u t e d ) .

Proof:

of

There

.

Since G

for ~

~

into

for

the

from

it.

Clearly

time b e i n g , For Dx

IXI = w I < 2 w .

Clearly Q , so

~

each

let us x E T ,

is d e n s e , by

f = UG

MA

in there

~. is

is an o r d e r -

is s p e c i a l .

-

66

-

We turn n o w

to the p r o o f of the claim.

countable.

We show that

ible elements. mains

X

Firstly,

must

of

~) .

that if

p,q E X

of

Secondly, and

.

of

~

, y E dom(q) of

(Of course,

for example,

in some a r b i t r a r y ,

dom(p)

p,q

be the n u m b e r

Pick

k ~ n

largest

linearly

u°

If

k = n

is a n e a t

member

of

assume,

of

~

Let

m

k

U

p E X

of

X

U

orderings Since

determines, of the levels

e a c h level

every m e m b e r

of

of S

~]

E U

.

Let

such that

S~ ~

and taking

u

X k

of the do-

T-level unique

for that

(k+1)'st l e v e l

of

. of all dom(p)

section

~ ), a final

is countable,

of

we can assume

(5)

of

S

in

S . such

such that

By l e m m a 7, we can find

is w e l l - d i s t r i b u t e d .

Tm

that

w h i c h do not p r o v i d e

P i c k an a r b i t r a r y m e m b e r

for

our l i n e a r

S

has u n c o u n t a b l y m a n y e x t e n s i o n s

~ = hts((~) )

of

We may

o

of

X

X.

such that e v e r y

( k + 1 ) ' s t level

(k+1)'st l e v e l s

of

of

and that

(eg. u s i n g

Tm

Let

subtree w i t h

in a n a t u r a l way

(Throw away those m e m b e r s nice p o i n t s ~ )

of

be the c a r d i n a l i t y of the X

,

each level

k ~ n

levels

lies on a

be the set of all .

k

, that the

the d o m a i n of e a c h m e m b e r of Let

so a s s u m e

with

to

such

of any m e m b e r

a single n e a t

we are done,

by choice of

member.

ordering

has d o m a i n an e n d - e x t e n s i o n

m a i n of each m e m b e r

dom(g)

so that u n c o u n t a b l y m a n y m e m b e r s

subtree

X

and

these o r d e r i n g s . )

in the d o m a i n

have a d o m a i n w h i c h e n d - e x t e n d s levels.

correspond

but c o h e r e n t manner,

of l e v e l s

assume

, but we can e l i m i n a t e

just those i s o m o r p h i s m s w h i c h p r e s e r v e n

(as n e a t

there m a y be several

for each such pair

this a m b i g u i t y by,

are i s o m o r p h i c

x E dom(p)

then

isomorphisms

that the do-

we can ( e q u a l l y clearly)

the i s o m o r p h i s m

= q(y)

be un-

c o n t a i n a pair of c o m p a t -

X

one a n o t h e r u n d e r p(x)

X c ~

we can c l e a r l y assume

of all of the m e m b e r s

subtrees

Let

Pick extensions

B~a (51) ,

-

<~2 >

of

<~)

in

SB

Let

<~2 )

pl,p 2

in

= u ° , and if

simple.

I

complicated of w h a t

x E d o m ( p I)

and we are done.

Since

~ .

of l e m m a 7

is g o i n g

the r e a d e r

on here before

that the proofs gave

we d e v e l o p e d w i t h an eye

above.

X

.

and <~2 >

such that Clearly,

[~i] , [32] are dom(pl) N dom(P2) then x a n d y

Hence

to an

Pl U P2

extends

r E~,

s e e m e d quite l o n g be-

In essence,

is a d v i s e d

of t h e o r e m 9

b o t h were e x t r e m e l y

were

w~

W MI

to

any further.

s o m e w h a t more d i r e c t

to the i m m e d i a t e

concerning

N

that

N

M

the , of

of

~ ~

M

adopt a d i f f e r e n t

next

chapter.

approach.

Baumgartner,

than ours,

which

joint c o n s i s t e n c y SH+ 2~ = w 2 such that

as N

~

then by an a r g u m e n t , so that

if we are to o b t a i n a m o d e l

must

(We m i g h t

future~ )

in the u s u a l way,

we see at once

Hence,

to g a i n a clear p i c t u r e

w h i c h Kunen,

S u p p o s e we o b t a i n a model,

in t h e o r e m III.6, N

S

[~1 },

, y E dom(P2), x , y ~ Uo,

If we f o r m a b o o l e a n e x t e n s i o n

by c o l l a p s i n g

fail in

of

proceeding

~,~le end this c h a p t e r w i t h a r e m a r k SH + CH .

in

Let

the ideas i n v o l v e d w i l l be u s e d a g a i n in a m u c h more

and R e i n h a r d t - ~ a l i t z

of

(~1 >

and t h e o r e m 9

them in some detail.

situation,

also r e m a r k

extends

( k + 1 ) ' s t levels.

in

The p r o o f s

i~ 1] n [5 2] = ~ .

S .

are i n c o m p a r a b l e

c use we gave

<~1 >

be those m e m b e r s

their r e s p e c t i v e

Remark

-

so t h a t

[~2 } E U , be such t h a t extends

67

of

We shall c o m m e n c e

SH

SH + CH

CH as

will , we

to do this in the

Chapter VII

TOWARDS

We begin

CON(ZFC + CH + SH)

this c h a p t e r

is possible

by d e s c r i b i n g

to carry

: A FALSE

a particular

out an i t e r a t e d

forcing

START

situation

argument

where

without

it

intro-

ducing new reals. Let

~

be a poset,

if w h e n e v e r p E •

p = A
any

z-closed

same is true Also,

of m a n y

of course,

z-closed

if

will be BA

E On

, if

Lemma

I

Let

~

IB

is

[bTlv

in

BA(P)

is n e a t l y of chapter

to be n e a t l y

sequence

~

, then

. z-closed,

and

the

of the posets used

in forcing,

z-closed,

is c e r t a i n l y

sharpening

then

of L e m m a

BA(~)

(z,oo)-distributive

< 8,T < y} ~

from

z-closed

~

I.

just in case

be an u n c o u n t a b l e

is said

tree will be n e a t l y

(but not all)

a slight

z-dense

~

is a d e c r e a s i n g

normal

~

in the sense

We shall require

that a

a cardinal.

( P a l ~ < B < ~>

, where

Clearly,

z

, then

regular

1.7.

is a

Note

z-dense

iff for all

B < ~

Av<~VT
cardinal

~

a

that a poset poser.

Recall

and all

¥

: Vf6y8 A v <sbv,f(v).

BA

.

The f o l l o w i n g

are equivalent: (i)

B

is

x-dense;

(ii)

~

is

(a,oo)-distributive;

(iii)

Proof:

for all

~ < ~ , II~%

(Sketch)

(i) - (iii).

D~ : [b' 61B I b'A b : © is a dense Then

d_<

initial 11u:~I!

=

vVI~~

Let

b _< flu 6 ~ I I

v = x or b' _< flu({)

section for

:

some

of

]B

Let

x

~ V

, x:a-V

For each

)71

~ < a ,

for some x({) 6 V]

d ~ b , d 6 N{< D{ .

.

-

(iii) - (ii).

Let

69

-

B < ~ , ¥ E On , [bv~l~ < 8 , v < ¥ ] E ~ .

is easily seen that VfEyB A ~ < B b~,f(~)< A <~v
the opposite

llg : ~ " ~II = b

inequality,

and

define

g E V (~)

llg(~) = ~II = b y e - v~<~bv~

Then by (iii), b ~ I I ( 3 f E ~ ) ( V ~ E ~)(f(v) VfEySA~<sbv,f(v) (ii) - (i). of

~

.

tions)

be

projection

e = id~o

Let

~o' ~I

(i)

f

-

BA's "]B O

M

M .

Let

be posets.

h

(ii)

sections

b ~

I~ <¥]

= ~

, so

then

b ^A

<sb ,f(~)

that

c < b , c E

h(bl)

embedding.

The

e-basic

= A(bo E]BoI bl --
the basic pro,~ection.)

A map (ii)

f :~I

~o

is a neat cover iff:

P -
;

(iii)

p ~of(q)

•

the following

be a

c.t.m,

]Pc

be, in

of

ZFC

stronger version

]P2

of lemma VI.2.

M)

an uncountable

M , a neatly in

regular

M , II'~PI is a neatly

complete

cardinal

x-closed poset, ]Bo = BA~Po)

Then there is, in

and a canonical

e"~o --cP2 if

, ~

be such that,

such that (in (i)

A <sV[b

f E y8 , which implies

IB I = BA(]PI)"Ie ° = ~

poset

Hence if

initial

(Solovay)

]PI' ]BI E ~ o and

By density,

is defined by

(3q, ~I q ) ( P = f ( q ' ) )

Let

v,~.

(possibly with repeti-

, e :IBo ~]B I a complete

is surjective;

Lemma 2

for each

= g(~))ll

be dense

enumerate

~ .

, we call

We shall require

so that

|

h: ~I

(If



for some

nv<~D ~ • ]Bo, IB I

, ~ < 8 < ~

<sb ,f(~) = ~

= c >¢

Let

D

D v , each

VEysA

say. To

•

Let

Let

= b

It

embedding

in

. Let

~-closed poset

M , a neatly e :IB° - ~ 2

x-closed

= BA(IP2)

: and

h :]B2 ~]B o

e(~o)

= 72 ;

is the e-basic projection,

then (h~]P2):]P2"~o

-

70

-

is a neat cover ; (iii)

if

[P~I ~ < Y <~} ~

A~
~2

is decreasing , then

(~o' Bo; ~2' ~2;

system]

is

for

PI

that

Proof:

Go

(E M[Go]),

M-generic

there is

e, h )

for

so related a

Pc

and

G 2 , which is

GI

is

~-happy M[Go]-generic

M-generic for

P2

' such

M[Go][G I] = [G2] , and conversely.

We assume

~Bo

is normalized.

liP E]P II~ ° > 0]

is a set.

[I P E P o

&

Thus, in

Work in

q E Q & &

]P2 is a neatly

~-closed poset.

thus

(in

f :P2 " ]Pc

cover.

Define

]Pc in

sily seen that h

is the

(in

e

e-basic projection,

= p .

f

e

Clearly,

Clearly, for

to

]Bo

then

h~2

]PI"' so q E Q

of course. f

is a neat

p E]P o , e(p) = uniquely by the it is ea-

And clearly,

= f .

if

The rest of

I

and simplicity)

(hmnln<m<w>

and

=

is a neat cover,

and that, using lemma 2, we constructed <enm I n < m < w } ,

sequence from

is a complete embedding.

the lemma follows easily now.

0Bnl n < w > ,

]Bo of course),

by setting,

Since

Suppose now that (for definiteness

p =

This also proves ~

(and extending ]Bo).

if <<pa,qa)I

IP2 , then

]BI)" .

f(
e :]B° -]B 2

[q E]P2 I f(q)
by

(P''q') -<2

In particular,

~2 = BA(]P2))"

Define

have domain

~ Bo there is a unique

p l~o"q = Aa
I

It is easily seen that

is a decreasing for

]P2 Set

sequence from

by the maximum principle

Aa
Let

(where this inf is taken in

p l~o"
such that

M .

P' II'-o"q' --<2q" "

is a decreasing

Aa
M, Q = [p E ~ o

p l~o"qEIP1" }


< k <~}

above,

=

;

[We call any system And if

h(A~<¥p~)

~ = wI

w-chains

in the (]PnI n < ~ },

such that for each

n ,

-

(~n' ~n; Pn+1' ~n+1; clearly, so the

~n'S,

-

en,n+1' hn+1,n)

for each pair

Wl-happy.)

71

is an

n < m < w , (~n' ~n; ~m' ~m;

~ ?

we add no new subsets of

v-chain of systems is

~(Pnl n < w ) l ( V n < w ) ( p •

by

P~w

is a neatly

w .

q ~

(Vn<~)(Pn_n< qn )

~n

Set

fn :Pw ~ n

fn

E ~w

for any given

by

is a neat cover.

~w

tion is satisfied;

hn+1, n

p E ~n

and then extending

~w

each

dings

~w

~n

the process.

~w

To see this, first

(i) of the neat cover definiof

is a simple induc-

n < w , the basic projec-

~n+1' ~ n " )

It follows that if we

enw(p) = ~ q E ~ w I fn (q) -n ~ p ] for each to

~n

if

by density,

h n

end' hun) We call

is ~

then

denotes the

' then we clearly have

~;

~w

It is easily checked

a suitable limit of the

n < w , (~n' ~n; ~ '

at once continue

en~

Furthermore, onto

Hence, not only is

•

(ii) is immediate by the definition

by setting

complete embedding.

that

p = (hno(Pn) , hn1(Pn) .... ,pn,Pn ....>

using the fact that for each

enw: B n ~ w

projection of

into

fn (p) = Pn "

is a neat cover for

=

In order that

(iii), all that is required

define

Pw

and partially order

= BA(~ )

' condition

condition

; and for condition

tion argument,

We first of all set

iterated forcing, we must ensure

(Since

Pn E ~ n

the fact that the

It is easily verified

is completely embeddable

that

tion

which also adds no new subsets

n E ~ n & Pn =hn+1,n(pn+1))}

Wl-Closed poset.

of all define

--W

Can we define a limit oper-

Wl-happy.

be a suitable limit operation for that each

(Then

enm , hmn ) is al-

In fact we can do so very simply, utilising

entire

system.

Now, by Lemma I, we know that if we force with any of

ation , suitable for iterated forcing, of

~1-happy

en~

is a

en~ -basic

h~n~w

~n'S

= fn "

, but also for

~1-happy,

so that we can

, together with the embed-

en~ , n < ~ , the inverse limit of the system

((~n)n<~,

(IPn)n<~,(enm)n<m<~ >" We can then iterate forcing with neatly At successor stages we use lemma 2.

wl-closed posets as follows.

At limit stages of cofinality

w,

-

we take a cofinal verse limit. direct limit.

(~Pa' ~ ) I

-

~-sequence of our system so far, and form the in-

At limit stages of cofinality greater than This is easily seen to be neatly

serve happiness. if

72

(In particular,

a
if we have

), ((eys,hsy)l Y < 6 < B >

structed so far, then if

(~BB,(ev~)v

is easily seen to be a neatly

Very well,

take the

Wl-Closed and to pre-

lim(8),

cf(B) > w , and

denotes the system con-

is the direct limit,

[Aa<~ea~(P a) I ( V a < B ) ( P ~ E ~ a) & (Vy < 8 < 3) (py = h S Y ( p S ) ) > O]

~

~

=

& Aa<~eaB~ a)

Wl-Closed dense subset of

~.)

then, we now know of a situation where we can do iterated

forcing without introducing new reals. of destroying Souslin trees ?

Can we apply it to the problem

Well, a $ouslin tree is certainly not

Wl-Closed ; the best that we have along these lines is the following lemma. Lemma 3 If

T

Proof:

is a $ouslin tree, then

T

is

ml-dense.

We observed during the proof of Theorem II.7 that if dense initial section of

~

then

UB>aT ~ ~ D

D

is a

for some

a < m I.

B

The lemma follows immediately from this fact. Perhaps we can modify the inverse limit to preserve

|

just

~1-density?

In fact this trail is doomed from the start, for Jensen proved long ago that there is no boolean limit operation, ated forcing construction over by being given (generic) new subsets of

w .

L

suitable for an iter-

in which Souslin trees are killed

cofinal branches, which does not introduce

(This is an immediate consequence

of the proof

of the main result in [JnJo~. A sketch of the proof is given as an Appendix.) A next attempt might be inspired by the proof of theorem VI.9. pose we decide to kill Souslin trees now by (generically) embedding them into

Q , the rationals

Sup-

order-

(say by means of some poset of

- 73

countable closed

partial

poset w h i c h does

haps we shall hope

embeddings).

succeed

is dashed

-

Assuming

this w i t h o u t

in p e r f o r m i n g

we can find ~ n e a t l y

destroying

cardinals,

a successful

at once by the f o l l o w i n g

~1then per-

iteration?

This

lemma:

Lemma 4 Let

M

be a

Suppose Then, MiQ]

~

if

c.t.m,

of

is a poset G

ZFC in

, and let

M

is

M-generic

in

M .

T

such that

for

~

be a S o u s l i n M

~'~

is

tree in

M .

Wl-Closed"

, T is still a S o u s l i n

.

tree in

.

Proof:

Work

Suppose

p I~"X

is a m a x i m a l

antichain

of

~". N

Let

(x

I ~<w

I)

enumerate

(P~I ~ < w I )

from

there

_c T]~

is

X

P

T .

Pick a d e c r e a s i n g

such that such that

Po ~ p

sequence

and for each

p~ II--"XN T]~ = V,,X

and

~ < wI for

v

some

y

which

is

This is clearly Let we

since

~

x

is

, p II-"Y E X"

Wl-Closed.

X Clearly X is an a n t i c h a i n of T , so ~<~1 ~ " ' ~ must have X = X for some ~ < ~I But by our c o n s t r u c X

is also maximal,

T , w h i c h means Pa ! ~ " X

chapter.

possible,

with

X = U

tion,

Thoroughly

~ - comparable

that

is countable"

disheartened

now,

so

X

is a m a x i m a l

pa11--"X = ~ " , we are done.

we decide

Since

antichain

pa < p

of

and

|

it is best

to start another

Chapter

VIII

ITERATED

FORCING

In this chapter we shall describe an iterated without

forcing

adding n e w reals

the key.

We shall

including

limit

stages

branches,

seem i m p o s s i b l e

to p r e v e n t

causes

Q , say.

to avoid

cal s t r a t e g y

the g i v e n

Of course,

its S o u s l i n i t y

between

sufficient

put all our efforts

to carry

a suitable

initial generic cardinals)

in w h i c h

sight

tree

How-

Jense~s

a ~iven

or even to force

to it.

There

is nothing

(of n e c e s s i t y

by g a i n i n g

not the

order-embed-

to d e s t r o y

framework.

a generic

(in r e l a t i o n

We thus have

out the iteration,

extension

at first

to be g e n e r i c a l l y

properties

into c o n s t r u c t i n g

combinatorial

of adding

to add new reals.

Souslin

to the itera-

trees,

trees.

we have

chapter~)

it is n e c e s s a r y

to first

set up

This is done by p e r f o r m i n g

principles

to

(Just h o w great

(which does not add new reals

the c o m b i n a t o r i a l

branch,

the r a t h e r p a r a d o x i -

all Souslin

these efforts will be, will be seen in the next In order

(namely

the two approaches.

branch

in

the tree w h i c h we do force with will

example.

in order

Now,

chapter

the process

that tree,

tree

provides

tree.

we adopt in d e s t r o y i n g

in the usual m a n n e r

the J e n s e n

that,

in the last

our desire not

difference

~1-closed)

LemmaV~L3

this might

us from forcing with a S o u s l i n

but it will prosses tion)

we forego

w h i c h will add a cofinal

one) w h i c h

ded into lose

adding reals),

tree must not be to force w i t h

with a n y t h i n g

given

without

through

so that at each stage -

to hold by i t e r a t i n g

tells us that the p r o c e d u r e

Souslin

for carrying

- we are f o r c i n g with a S o u s l i n

there is a subtle

result

method

cardinals.

out the i t e r a t i o n

SH

unless

STYLE

(using posets w h i c h are not

of J e n s e n we m e n t i o n e d

that one cannot get generic

Jensen's

or c o l l a p s i n g

carry

v i e w of the r e s u l t

ever,

argument

JENSEN

~ ~

an

or collapse and

[]

hold.

-'75 -

For further

details

to [De

We

need,

I].

shall here

a n d no m o r e .

There mit

We

ordinals

commence

principles,

ourselves

with

~

we r e f e r

with

proving

the r e a d e r just w h a t we

.

< A K I k<m 2 & l i m ( k ) >

such

that

for all li-

k < w2 ,

(i)

Ak

is a c l o s e d u n b o u n d e d

(ii)

if

cf(k)

(iii)

if

a E AI

Lemma

ric

these

content

is a s e q u e n c e

(Hence

Let

concerning

< ~I

' then

and

cf(k)

subset

otp(Ak)

k ;

< wI ;

a = s u p ( A k N a)

= w I ~ otp(Ak)

of

A m = AxNa

, then

•

= ml " )

I N

be a c.t.m,

extension

N

(i)

M

and

N

(ii)

f o r all

of

of

ZFC + (2W=Wl) M

have

such

the

cardinals

of

(iii) @N(~) = ~ M(W ) & (iv)

Proof:

N

~-

[]

Work

in

~ :

Let

P

dinals

below

and

p ~ q

~2

be M - g e n e r i c it is c l e a r

If that

M

,

and

cofinality

(2X) N = (2k) M

p E P

that

closed,

on

~

that

function;

;

initial

sequences

segment

( i ) - (iii) ~

and

set

there

are

be

whose

of the l i m i t

of

~

the p o s e t

.

For



N = M[G].). only

p

Then,

or-

p,q E P, .

(Let

IPI = ~2

two n o n - s t a n d a r d

parts

. D

lim(k)

, there

is

q ~ p

such

.

I v < w I>

p E •

q E O

Let

k < ~2'

k E dom(q)

that

set of all

are:

and


the

' satisfying

These

(2) S u p p o s e and

be

--~ p ~ q .

the proo f . (I)

cardinalities

.

is a p r o p e r ,

G

is a g e n e -

@N(~I) = @M(~I) ;

domain

say

There

that

same i

+ (2mI=~2)

Then

are

dense

there

is

initial q E ~

sections , q ~ p

of

, such

' of

-

Given

(I) and

To prove

-

the lemma follows

(I), what we do is

ordinals all

(2),

76

v < w2

p E P

that:

with

prove

easily.

by i n d u c t i o n

"for all limit

max(dom(p))

on the limit

ordinals

= ~ , there

is

¢ < ~

q < p

and

such that

m

max(dom(q)) vial,

= v ".

Successor

stages

since we can just add a d i s j o i n t

stages,

we pick a cofinal

To prove members qo = p '

(2), we i n d u c t i v e l y of

~

and a n o r m a l

7(0)

< w I , let

and

(q~l~
Our next result

ensures

that

~ w2 .

¢(v) of

are defined, .

(2).

(q~l~w

I) of

Set

are defined, q~

= max(dom(qv+1) )

¢~v

is as required.

qv'

extension

qv = (U <~q~) U [(¢"v,¢(v)>]

q = qwl

obtain

¢(v+I)

¢ : wi+I

If

tri-

and work along

a sequence

function

be a proper

E D v , and set

v ~ w 1, and

define

are

At limit

below in v e r i f y i n g

= max(dom(qo)) qv+1

u-sequence.

cofinality-sequence

it in the same way as d e s c r i b e d

q~+1

in the i n d u c t i o n

such that If

set

lim(~),

~(v) =sup~
It is easily

seen that

|

[] will r e m a i n valid when we force

to

0~

Lemma 2 Let M

M with

O

be a c.t.m,

~

:

in

N

Trivial.

There

countable contains

ZFC + ~ .

the same c a r d i n a l i t i e s

holds

Proof:

of

N

is any generic

and c o f i n a l i t y

function

extension

as

of

M , then

also.

|

is a sequence

subset

If

of

©(~)

a closed u n b o u n d e d

(W

Iv<wl)

, and for any set.

such that each A ~ wI ,

Wv

[aEm11An~

is a EW

}

-

Lemma Let

77

-

3 ~

be a c.t.m,

of

ZFC + (2 w = Wl) + ( 2 ~ I = w2)

ric e x t e n s i o n

N

of

(±)

M

N

have

(±i)

for all cardinals

and

M

.

There

is a gene-

such that: the same c a r d i n a l i t i e s k

of

and c o f i n a l i t y

function;

N , (2X) N = (2X) Z ;

(ii±) (iv)

Proof:

I=0 ~

~ork

in

M

at first:

such

that

all

~ < {(w)

w

such that (w',B')

B in

Let

P

be the set of all pairs

is a map with d o m a i n , w(~)

some

is a countable

is a c o u n t a b l e P , say

subset

<w',B')

{(w)

subset of

~ <w,B)

G

be

= ~2

and

M-generic P

on

satisfies

(w,B)

incompatible

Hence,

as

is clearly

dard part in ~ork

w ~w

in

N :

Set

~

N : ZIG].

w 2 chain condition, with

(w',B')

~1-closed

For &

(w,B), B' ~ B

= (P,~) ) Then

~I

this last be-

implies

also,

w / w'

the only n o n - s t a n -

is the v e r i f i c a t i o n

W = U[w ! (w,~) E G ]

& (v~(w))(Vb

for all

(ii)

.

, and

that

Q~

will hold

N

Por each

(i)

of the proof

@(v)

Let

and set

the

cause

P

P

of

w' ~ w

& (Y~E ~(w')- ~(w))(VbEB)(bOvEw'(~)) (Let

< w I , and for

@(~i) iff

(w,B>

CS)(bn~

~ < w I , let b E W(a)

8(~)

< ~I

, so

G = [<w,B) I

~ = ~[W]

be the least

6

such that

:

if

~

is countable

in

L6[b,W~a]

if

a

in

L[b,W~]

, then

a

is countable

;

is u n c o u n t a b l e

e~1 , then

~(~))]

Clearly,

in

L[b,~[a]

6 ~ ( +)L[b,W[~]

.

but

( +)L[b,W~]

<

-

78

-

m

For

each

a < ~I

, set

subset

of

countable ses

0*

Let

be given.

, such

the l e a s t such

that

@(6) N ( U b E W ( a ) L s ( a ) ~ b , W ~ a ] )

.

We

show

that

and

We

can e a s i l y

find

A E L[X,W]

ordinal

exists,

Define,

@(a)

=

(Wa[a<w

I)

, a reali-

•

A ~ wI

X E M

W

below set

inductively,

wI

and such

0 = ~

X ~ wI ,

Wl = ~ [ X , W ] that

N

Let

~ be

w I = ~ [xO~'WI~]

otherwise.

submodels

a set

Let

if

y = w~ IX'W]

~ Ly[X,W]

.

, ~ < w I , as

follows. No

= the

smallest

N ~Ly[X,W]

such

N + I = the

smallest

N ~ L y ~ rX,W]

such that

Nk

= U <xN ~ , if

Let

~

:N

~ N

8~ = ~ "(N 0¥) is a n o r m a l

Then

sequence

N a~

in

wI

It is c l e a r

Since

, a simple

~o

such

that

be the l e a s t

such.

all

forcing

for all

: ~

~o

~ < w I , which

is

for

Hence,

for all

~ < w I , XOa~

Claim.

For

~ < w I , 8v ~ 8(a~)

Case

I:

w I = W~[XN0'W~6]

able

in

L[X0~,W~]

But

av

is u n c o u n t a b l e

in fact).

Hence

~

in

< 8(~v)

means

, and

that

shows

a < wI , a ~ ~

Then

~

-

= WlON <~

,

I~<w 1)

x q a v = w (X),

[xn%,w~%]

argument

~o E N

all

Let

E w I , a~ = w v ( w 1)

, W~% = ~ (W) , ~

< wI

E N ;

N~ ~ IN ]. _c N ;

is t r a n s i t i v e .

A q % = ~ (A) X E ~

~,X,W,A

lim(k)

, where .

that

.

that

X Na

there

E W(a).

Ly[X,W]-definable,

is Let so

w ~ ( ~ o ) = ~o < a ~

E W(a~)

Thus,

as

L~v[XO~v,W~

av > ~ , a~

v]

(being

is c o u n t -

"~I"

there,

-

Case 2:

it follows that

LCXO~,W~])

.

Hence

is countable in

.

Since

~[xn%,w~%]

~1

.

By a simple condensation

(V~<~1)(w I is inaccessible

(V~<~I)(~XN~'W~]

L~XNav,W~a~]

On the other hand,

suppose

.

av

<~i ) .

in

Suppose

Then, as in Case I, ~ is uncountable in

a~

<8(~).

L[XN~,

a~ = w (w I) , it must be the case that

a~ =

)~[xn%,w~%] ~[xn%,w[%]

Hence,

Thus, by definition, W~a~

-

(V~<~I)(w~[XO~'W~o] <Wl)

argument,

W~]

79

(~

= ~2

8(~)

J=(V~)(J~J ~ )

~ (a~)L[Xna~,W~a~]

8(a~) ~ ~

This proves the claim.

By the claim,

~ <w I - Aqa~

Since

{a~ I ~ <~i ]

done.

|

We shall need to know that

But

EN

~ Ls(a~)[XOa

, so we see that

,wrap]

~ <w I ~ AN~

is closed and unbounded in

0~

L~[XN~,

+ L[XAa~ W (a~) , [a~]>_ ~ '

Hence

(V~ < W l ) ( X n a ~ E W ( ~ ) )

<~I"

proving

•

But

EW~

.

~I , we are

remains true throughout "most" of our

iteration. Lemma4 Let

~

G ~ w~

be a c.t.m,

of

is such that

ZFC + (2 w =~i) M[G]

is a c.t.m,

having the same cardinalities as

Proof:

+ (2 wl = w2) +

M .

of Then

0* .

ZFC + (2~= ~i ) + (2 wl =i~2) 0~

holds in

The argument is virtually the same as the above. however,

instead of

to be a

~

W

Suppose

M[G] also.

This time,

being a generic sequence, we take it

sequence in

M .

The only difference

in the en-

suing argument is that we cannot apply a forcing argument to our set

X .

bounded set if we set

However, C ~ wI

in

since M

X E M , there is a closed unsuch that

C = C n {av J V<Wl ] ' ~ 5

a 6 C - XN~ EW(~),

" xna ~ w

so

, just as before.|

-

Of course,

~

is simply

a much

which we met in the earlier can be regarded require

the principle

we have used By

H

O < w I , we set

version

Accordingly,

of lemmalZI.4.

lemma

We shall

also

form from that

the set of all h e r e d i t a r i l y

countable

sets.

Hw1(a)

= Hw10 V

is a sequence

table

subset

of

, then

w

.

<Solo<w1>

Hw1(~)

For

and,

such that each whenever

SO

A ~ H~I

[~E wl I AnHw1(o)

= S )

is a coun-

and

o<w I

is stationary

wI .

of the next

principle

result,

formulated

above

we shall h e n c e f o r t h as

understand

by

~ the

~'

5.

Proof;

Clearly,

~

there

h: w I ~-~ Hwl

NH

is

I(~)

~

Let

AAHwI(O)

in

= ANHwI(a)]

=

(h"

O

o <~I

~

S ° = (h"S)

We show that

Let

.

Then =

~ E C

Then

subset

~ E C

and

such

that

[a E w I I AAa = SO] is

~ = c n {o 6 w 1 1 h " o

unbounded

.

~ IANHwI(O) I ~ w . Let C ~ I

We seek an

A = h-~"A

, a closed ANo = S

2 w = w I , so

o < w I , set

realises

that

Let wI .

Then

~,

and unbounded.

stationary

~

For each

(Sol o < w 1 >

be such

= SO

that

Now assume

realises

A ~ Hwl

be closed

such

~

, where

(So I o < w I>

Of ANH

_c Hwl (o) & h"(AO~) wI wI

Pick

o 6

(O) = h"(~NO) =

.

I

We shall served,

3

different

IANHwI(O) I ~

Lemma

of this book.

of the principle

~ , but in a slightly

There

In view

version

previously.

we mean

in

-

stronger

part

as a stronger

80

also require the next result, a consequence

which

is not,

of lemma 4, even if we assume

it will be obM

I= ~

-

Lemma Let N

M

be a c.t.m, is

N[G]

Proof:

of

ZFC +

~1-closed".

If

O. G

Since

~

= w~ [G] ~ in

therefore, ~I & C

is

that,

in

M

q ~ p

"~ E C

.

&

AR~ = S~"

_c v

Then

and

Pa+1

P

such that

M

holds

, then

Now, ~

&

0~

to hold.

affecting

two initial Our p r e v i o u s

~ ~

tion, w h i l s t

that

~

and

0~

of

there

during

the iteration.

late e v e r y t h i n g

in b o o l e a n

we are thus led n a t u r a l l y

are sets Let

and

~

A =

~ 0 ~ = S a-

|

.

of

Av,

is closed

such that

extension

Po ~P'

Our eventual of

M

M

, having As a

to get

enable us to do this w i t h o u t Accordingly, in

M .

a large part

As usual,

terms.

' q!~

such that

ZFC + GCH

extensions

lemma 6 will enable us to use

We n o w describe

~

are no Sou slin trees.

generic lemmas

~ < ~I

in

so we can induc-

~ E ~

generic

p !~ " A

Working

, as required.

a l r e a d y hold

will r e m a i n valid

~I"

Z,~ E Wl

any of our stated r e q u i r e m e n t s .

assume

of

CN~ = ~v"

is

~ EC"

absolute

M , in w h i c h

such that

from

&

Iv < w I) Suppose,

v < w I , there

Then

M

(S

M[G]

is ~1-closed

w I , so there

aim is to obtain a cardinal

step, we make

subset

P~ I!-"AO~ = ~v

!~ "A0~ = ~

as

p E~

(pvl v < ~ I )

C = Uv<~IC v . in

0 in

such that for some

then we are g i v e n a c.t.m.

the same reals

is

, and for each

such that

and u n b o u n d e d

trees,

on

in

to show that if

is a closed u n b o u n d e d

Uv<~IA v

us that

M-generic

, there

M , we must find a

C

ways

be a poset

M , it also r e a l i s e s

v <~ <w I ~ p~p~

first

P

it suffices

tively pick a sequence

Suppose

Let

.

realises

and

-

6

I:"P

in

81

0

we shall al-

L e m m a 4 will

tell

of our construc-

even w h e n

0 N is lost.

it is c o n v e n i e n t

to formu-

Since we are forcing with Souslin to the concept

of a S o u s l i n

algebra.

-82 -

A Souslin - [©}

If

~

which

al~ebra which

is a

is,

BA

under

is a S o u s i i n

~ ~

that

there

, a Souslin

algebra

is a S o u s l i n

such

and

T

tree u n d e r

The n e x t

lemma

characterises

at l e a s t

under

the a s s u p p t i o n

, we

Souslin

set

subset

~-

T

tree.

is a d e n s e

~

is a d e n s e

say that

algebras

T

of

Souslinises

in p u r e l y

2w = w I , which

we

{¢} •

boolean

shall have

.

terms,

through-

out.

Lemma

Assume

7 2w = ~I

Let

(i) - (iv) h o l d ,

B

]~I

= W1 ;

(ii)

B

is

(iii)

~

satisfies

(iv)

~

is a t o m l e s s .

~

Since

T

is a S o u s l i n is d e n s e

in v i e w

of the f a c t

191 ~

T

in

of

is a S o u s l i n

means

algebra.

~

,

that

ITI w = l~ = w I

Conversely, subset

suppose

T

of

on the l e v e l s T o = [~]

.

b E To

where

y by

•

such

TO

algebra

iff

is l e a s t

b A by such

(b

defined,

let

we

obtain bA-by

consist

the

, which,

see

that that

Construct

is a tree,

T +I

~.

Set

as f o l l o w s . b

on

of these m e e t s If

a

by i n d u c t i o n

enumerate

extend

so c o n s i d e r e d . Ta

.

(~1,oo)-distributive.

I ~ < ~i )

that n e i t h e r

has not yet been

< ~ , are

and

>

BA means

(i) - (iv).

(T,~

~

Again,

as a

by l e m m a V ~ 3 ,

satisfies

Let

c.c.c.

c.c.c,

by lemmaV]I. 1,

that

Souslinise

~

satisfies

is d e f i n e d ,

Let

T

generates

Finally,

~

Let

satisfies

T ~

as f o l l o w s . If

~

that

W l - d e n s e , and h e n c e ,

yet

•

c.c.e.

density

Let

.

(w1,~-distributive;

Suppose

is

BA

where:

(i)

P r o o f:

be a

is

lim(~)

of all n o n - z e r o

T +I , ©

and

and

T~,

meets

-83 -

of the f o r m conditions a Souslin to see

AB
particular,

Remark.

Without

cardinality nality

wI

dense

every

ded for e a c h

' where

- (iv),

tree,

that

PB

in

assumption

satisfies

ther p r o v a b l e

nor refutable

The f o l l o w i n g

lemma

.

of

Hence

seen that

(We n e e d on an

CH

ZFC

clause

s-branch

greater

than

, lemma

(ii)

" By

T

is in f a c t

(ii)

in o r d e r

T

Wl. )

is e x t e n -

will,

in

|

7 holds

f o r any

of a

- (iv)

PY~PB

which

(iii),

the e x i s t e n c e

conditions in

" PB E T 8 &

A n d by c l a u s e

height

w I , of c o u r s e . which

~ lies

a •

not have

the

it is e a s i l y

point

limit

~
BA

BA of

of c a r d i -

of l e m m a

7 is n e i -

.

is f a i r l y

trivial,

algebra,

and let

but w e l l w o r t h

a special

T

~

men-

tion.

Lemma Let

8 ~

be a S o u s l i n

$ouslinise

.

Then:

N

(i)

~

is the r o o t

of

T ; N

(ii)

If

x,y

(iii)

If

[Xnln~w)

(i), (ii)

Proof:

(iii)

The n e x t

lemma

Souslinisations

Lemma Let

are

~-incomparable ~ T

are

uses

T , then

x A y = © ;

An~ w xn E T .

trivial•

that if •

of

A n ~ ~ x n ~ © , then

the u n i q u e n e s s

shows of

and

elements

are

•

of l i m i t

points

is a S o u s l i n

essentially

the

in

algebra,

T,I

then any

two

same.

9 ~B

there

Proof:

be a S o u s l i n is a c l o s e d

As

T'

algebra,

unbounded

is d e n s e

in

and let

set

~

T,T'

$ouslinise

A c ~I

s u c h that

, for e a c h

x E T

a EA

we may

IB .

Then

~ T a = T'

pick

a pair-

-

wise

disjoint

Souslin,

Using to

~I

Y

set

84

Y c T'

.

find

that

B < wI

our a b o v e

such

is c o u n t a b l e ,

x = V [ Y E T'IB I Y ~ x } a

-

such

observations,

that

so we

x = V Y

can f i n d

Similarly,

a

define

Since

T'

B < w I such

for e a c h

y = ~[x ET[S

we m a y

.

is that

y E T'

we

can

~,~'

on

wI

I x~y]

functions

as f o l l o w s : ~(a)

= the l e a s t

B > a

such

t hat

x E Ta ~ x = V[yET'I8

I

y~x~ ~'(~)

= the l e a s t

xpy~

.

Define

a normal

B(O)

0

=

,

B > ~

function

8(2v+1)

=

~(8) = supv<6S(~) Let

A =

quired.

Let

k <~i

, let

, let

(*)

(*),

x E TB(X)

finition

~ ~I

,

=

of

= {XETs(x)

the p r o o f

of

Y

~'

, and

set

for e a c h

I y~x]

and (**)

I

thus:

° B(2v+I)

,

that

A

ordinal.

, and for

•

is as re-

T h e n we

For e a c h each

claim

y E

that:

(**) y E T~(X) ~ y = V X ( y ) being

, let

Yv = for

show limit

I x~y]

v < B(X)

~ , x v = VY v

enumerates

=

We

~ET'~(X)

- x = ~Y(x)

x _< x'

~ y = V [ x E TI~

by i n d u c t i o n ,

~(2v+2)

be an a r b i t r a r y

For each

that

yET'

lim(8)

Y(x)

X(y)

x E TB(k)

We v e r i f y

such

B : ~I

0 ° B(2v)

if

that

[B(k) I X < ~I & l i m ( k ) ]

x E TB(X) T~(X)

such

entirely xv

be

similar.

that

[y E T'~(v) I Y _< x v]

each

v < ~(k)

v < B(X) , then we

.

Let

x'

E Tv

.

By de-

If < y v , n i n < m ) see

that

x =

V

Av
for

E T~(k) Hence,

any And

by

8

f E ~(k) by and

, Av<~(x)yv,f(v)

® , Av<@(k)yv,f(v) the d e f i n i t i o n

of

>¢

~ x Y(x)

~

e

- Av<e(k)Yv,f(v) in this

case.

, x = ~Y(x),

proving

(*). Finally,

let

x E Ts(k)

.

Pick

y E Y(x)

, x'

E X(y)

. (By

-85 -

(*), Y(x) J ~ problem.) have

, and by

Then

x,x'

x' = y = x .

E TB(k)

Hence

T~(X) . Similarly,

We are n o w almost ready

(**), X(y)

inverse

force with S o u s l i n

trary S o u s l i n i s a t i o n the actual

choice

limit

"thin down"

lean algebra. struction

embed

be S o u s l i n

is a complete of

~ , ~'

of

(respectively)

a closed u n b o u n d e d The n o t i o n

subalgebra

set, where

of a nice

embedding

• ~'

and then enough

in its boolimit

expect

con-

to have

which holds

between

The n o t i o n w h i c h we Use

embedding.

is a nice

subalgebra

of

~'

iff

and there are S o u s l i n i s a t i o n s

such that h :~'

algebras

we should

sequence.

or "nice"

algebras.

irrelevant

stage we first

of the inverse

to "Wl-happiness",

subalgebra,

along an arbi-

At a limit

Souslin

chapter,

in the i t e r a t i o n

We

tree w h i c h is still large

of our d e s c r i p t i o n

corresponding

is that of a "nice"

~ , ~'

and carry

our S o u s l i n i s a t i o n s ,

all of the p r e v i o u s

In v i e w

any two algebras

T,

to a Souslin

|

stages we shall

(Lemma 9 will make

using

TS(X)

The idea is this.

at each stage,

limit,

that

our main i t e r a t i o n

At limit

construction.

given in the previous

some p r o p e r t y

Let

algebras.

, proving

, so we are done.

of this S o u s l i n i s a t i o n . )

this limit

to c o m p l e t e l y

x' ~ y ~ x , so we must

and prove

for each stage.

of all form the inverse

, so this causes no

x = y E T~(X)

to formulate

algebras

and

T~(k) ~ TS(X)

lemma for forcing with S o u s l i n use a m o d i f i e d

j ~

~

~E~I

IT

= ~'T~]

is the basic

is d e f i n e d

analogously

T,

contains

projection. in the obvious

way. The next l e m m a r e m o v e s pendence Lemma Let

from the above

u p o n the choice

definition

the ostensible

de-

of S o u s l i n i s a t i o n s .

I0 ]B, ]B'

be S o u s l i n

algebras,

and suppose

]B

is a nice

subalgebra

-

of

~'

If

tively), where

T, T'

then

h :~'

Proof:

-

are a r b i t r a r y

{ a E Wl I T -~

86

= h"T~]

is the basic

By l e m m a 9.

Souslinisations contains

of

9, 9'

(respec-

a closed u n b o u n d e d

set,

projection.

|

The next l e m m a shows

that the n o t i o n

tive,

clearly need

w h i c h we shall

of a nice

subalgebra

is transi-

if this is to play a similar role

to happiness.

Lemma Let

11 9o' 91' ~ 2

bra of nice

~I

and

subalgebra

Proof:

be S o u s l i n ~I

algebras.

is a nice

of

Suppose

subalgebra

of

~o ~2

Let

hij : ~ i - ~ j

Let

Ti

A12

be closed u n b o u n d e d

(Iteration

hloOh21

(2 w = w l ) + (2~I= w 2 ) +

for all

~

o~B,x) nice (i)

~o

(ii)

B

(iii) (iv)

if

is a S o u s l i n

subalgebra.

~+I if

:

~

is a

9

of

subsets

~i of

projections.

' 0 ~ i ~ 2 . wI

such that Let

Let

Aol ,

~ EAol

-

A o2 : Aol DA I 2

= h2o

, so we are done.

~ +

.

~

is a S o u s l i n

algebra

of which,

Then there

I

Let

o

algebra,

be a f u n c t i o n or

if

in the case

is a sequence

~B

~

is a S o u s l i n

algebra,

,(]B [~ < ~ } )

0 < • < ~ ,

~T

for all

for all

is a nice

I ~ ~ w 2)

~ > 0 ; ~ ;

subalgebra

~

of

such that

= ~ , then

~ ~ , ~

;

= o~B

~o

Lemma)

Assume

, x ,

Then

, 0 ~ j < i ~ 2 , be the basic

be any S o u s l i n i s a t i o n

But clearly,

12

subalge-

9 2 .

T a° = h I o "T aI and ~ E A12 - T ~1 = h21 "T a2 "

Lemma

"

is a nice

•

such

is a that:

-

Proof:

8?

-

VJe c o n s t r u c t the sequence

(]B I v < w 2)

proceed,

we c o n s t r u c t

the f o l l o w i n g

(a)

h

:~

(b)

T~ ~ v

' an a r b i t r a r y

(c)

[~ :~

~ ~ , where

-~

also

(~

are o t h e r w i s e

T)

, the basic

by induction. sequences:

projections;

Souslinisation

~

As we

of

has d o m a i n

•

~I

but

V'

"J

arbitrary;

(~) (e)

C T

(v~)

, an a r b i t r a r y

such that for all Tv : h 'Je let

closed u n b o u n d e d

a 6 C

, ~vI~ = YvOa

set in

, ~[~=

wI

~

,

"T ~

<

<-

denote

the b o o l e a n

ordering

of

respectively. Let

(Ax I X < ~ 2

realise

~

Case

Stage

Set

I.

]B

& lim(X)>

realise

(in the form stated

= ~

~

and let

(S

I a < w I>

earlier).

O.

.

O

Case

2.

Set

~v+l

Stage

v + I .

= o(~v'(]Bm IT < v>)

Condition

(iv) holds

by lemmal

and induction.

Case

3.

Let

~ = otp(Ax)

meration fine

~X

Stage

of

k , k < w 2 , lim(k)

, and let

AI .

from

Notice

, cf(k)

<X(~) I ~ <8} that

<~k(v) I ~ < 8 )

~ ~ k

and

= w .

be the n o r m a l 0 < wI

, or, more precisely,

enu-

We defrom

-

Set

8 8

-

C = [Nv<~<e(Ok(v),k(¢)-~)]

the

normal

Por all

enumeration

a < ~I

of

' ~k(v) e~

U [0] C .

.

Let

Notice

= h~~ ( T ) , k ( v )

that

,,~k(~), ca

Iv < ~ >


c o = 0 and

for

be c 1 ~e.

v < ~ < ~ ,

so we may define

T

= [x=<xvl~<e)I(~¥<®l)(v
<

~

<

e

It is easily

.~.

x v=hx(~),~(v)(x~))]

~

Partially-order

~Yx(v) Cy

by

T

seen that ~X(v))

x _<* y ~'~ ( Y v < ~ ) ( X

(T*,<*)

_
is a tree and that

x E T*

(Vv < e)(x v E T c Crucial

Fact

(C.F.)

(I)

T

is a Souslin

(2)

if

There

is a subtree

of

T* such

that:

tree;

x E T , y E ~k(v),

such that

T

Y ~k(v)

xv

, there

is

x'

E

x' <* x , x' = y . V

We leave plete

the v e r i f i c a t i o n

the d e f i n i t i o n

tween condition Let

~

= BA(T)

C.F.

Bk

.

Define

C.~.

until

(But note

(2) and the n o t i o n

~v(b) = Ix ~ I x Using

of

of the

maps

later,

the similarity

of a "neat

~v:~k(v)

and com-

~ ~

be-

cover".) ' v < ¢ , by

v ~X(~) b]

(2),

it is easily

checked

that

~v

is a complete

N

embedding

of

~k(v)

into

~ . (In particular,

to make

the following

the mapping

is one-one.) Pick

]Bk, ~

diagrams

commute

for

Ov

-89

~" ~(~) /

-

N

"~

~

c

gX(T)

x(,~)

~x(~) Define

~: T -* B X

<~"T,<x) T k = ~"T

Let

~(x) = Av<8~X(]l(xv)

by ~"T

Souslinises

"~k(~)

~Bk .

Then

~:
We may as well set

' ~ < ~ ' be the basic projections.

-I h (~(x)) = ( k ( ~ ) ( x )

v < 8 , a<~ I & ~= c

is a nice

subalgebra

fore, ~ v

is a nice

completes

the construction

the

~x

for definiteness.

h : ~k

Clearly, each

, and

_c

of

for each

T k(~) = h v "T aX " a

-

~k

x E T.

' each

subalgebra of

v < 8 •

Hence for Thus

~X(v)

By lemma 11, there-

of

~k

for all

v < k . This

~k

' and it remains

to verify

C. F.

Claim I.

Suppose

(v,y>,

for some

c E C , y ~ ~X(~)c

Then there is

x' ~ ~k(~)c

~k(T),k(v)(x')

= y

By assumption

~Iv)(y)

(~,x>

and

~k(v)

k(T) (x) >X(T) © e -I

seen that

x

v ~ ~ < 8 and

x ~ yX(~) and y ~)ff~(~),X(~).

such that

ql )/y

lies above

are such that

x' ~k(T)x

and

X(T)(X)) hk(T) , k(~) (~-I Since

.

Hence

c E C , it is easily

~k(~)c (in terms of


Hence,

-

by density, ~%(T)

it follows

%(T) (x)

90

that we can find

' such that

Tck(V) = h%(T),k(v)"T%(~)c

x(v)

-

' hx(~),~(~)(x")

x" ^ ~ and

x"

~ T k(~)

v) (y) >%(T)

T %(v)c

= ~ [ I ~ ) (y)

© "

is pairwise

"

x"

C

~hus

Since

disjoint

x'

:

a

so that

in

(X(~)(x")

is as claimed. We c o n s t r u c t

T = Ua<~IT a

by i n d u c t i o n

is a n o r m a l

a-tree

To

of the m a x i m a l

consists

tain

Ta+1

Claim 2.

from

which

satisfies

TI(a+1)

, proceed

For each triple

(v,y,x)

y E ~k(v) , and y -<~(v) xv c+ I such that

Let

I

of

T*

(2) of the C.F.

, say

~ .

To ob-

as follows. such that

' there

be cofinal

we can i n d u c t i v e l y

'

condition

TI~

is

v

s E T* ~+1

<

e

, x E ~G,

s <* x -'

'

sv = y .

(v(n) In < w)

claim

element

on

such that

in

pick,

Yo = y ' Yn ~ % ( ~ ( n ) )

~X(v(n+1)),k(v(n)) hx(v(n)),k(T)

(Yn+1)

(yn)

By construction,

e , with for each

Xv(n)

Define

where

n

v(O)

C+

by

s

enough for

s , and


let

1 '

Yn =

s: e ~ V

this w e l l - d e f i n e s

By

n < w 'Yn E ~ k ~ ( ~ )

, and

is large

= v .

s

=

v(n) ~ T .

is clearly

as required. For each triple s

as claim 2

as above,

guarantees.

s(v,y,x)

for all possible

s(v,y,x)

is i r r e l e v a n t

(*)

For all

v < e '

such that

Y ~(~) In

this

case,

Let

Ta+ I


.

except when

y E ~%(v) cI

consist

we s e l e c t

of all

The actual ~ = 0

the

choice

, there is a pair
• s(v,y,x)

in

such

of

and:

T ~ v , • < e , z E ~%(~)Ici

hx(~),~(~)(z)

be one such

s(v,y,x)

a way

that

for

E Scl

-

some

<~,z)

,

E Sc (*)

by

itself

Finally,

suppose

obtained

as follows.

there is

s E T*a

s ~* x']

is an

Let

<~(n) I n

and the fact pick,

for

Yn,Xn>

and

that

such that G-branch

s <* x

of

TIa

be cofinal

be eofinal that

Tla

in

a

"

Define

s

by

s

Tia

To obtain

in

T* a

~(0)

x E TT(o)

C.F.

The actual

the f o l l o w i n g

ca = a

there <~

of

= ~ . "

Yo = y

E U

Let

By c l a i m l

' Xo = x , (~(n), ' Xn+1 ~* Xn

'

where

the b r a n c h

(Xn I

we pick one such

guarantees,

and put this into

s(~,y,x)

is i r r e l e v a n t

S a _c Tia

, and for every

except

in

case:

Suppose E U

choice

[x'E Tia I

(2) we can inductively

s , capping

(~,y,x>

as claim 3

is

(~,y , x) E U

= ~k(~(n)),k(~)(yn)

Then

Ta ' for each

Ta

such that

= y , and

with

such that

, is as required.

of

X t

(~,y,x>

For each

, s

9

satisfies

n < ~)

(**)

.

with

• < e .

.

is defined.

•

in

~(n) ~ T , all

a

2.

Tia

x

This is

z .

~(T)

, h k ( v ( n + 1 ) ) , k ( ~ ( n ) ) ( y n + I) = Yn

x n E T~(n)

s(~,y,x>

s

1 and

y ~ (~~ )

n < w , Yn,Xn

E U

claims

, and

< w)

<~(n) I n < w)

and

be the set of all triples

, Y E ~k(~) ca

x E ~l~

and

lim(a)

U

-

z E ~'TX(~)Ic I

possible

Claim 3. Let

91

X

and

is an

x'

E S

such that

(~,y,x')

(v,y,x) E U

and

.

m

In this event, for some itself, Let

x'

we select

E S

.

s(~,y,x)

so that

This is possible

s(~,y,x)

by the c o n d i t i o n

~* x' (**)

of course.

T = U <wit a .

Clearly,

T

is a n o r m a l

~1-tree

saris-

-

lying Let

C.F. X

a < ml

~*x)]

' set

,for all

<* x , we have

x' = y'

Let

be the set of all

~ TI8

unbounded

in

~I

Claim 4.

If

m E K

Ix E TIa IY ~ ( ~ ) that

Let

D 0 Z

x E TIa

y ~(~)

y' ~ ( v )

Y(y',x) there

, y ~(v) x

.

such that

is

x'

E Da

[By

"

Since

Y ~k(~) such that

x 6 T

is dense

Clearly,

"D

subset

that

.

with

incomparable x'

E D

and

T

, sa-

of

xv

K

then

is dense

and

' then

is closed and

D for

is dense for Z", we m e a n

Z .]

Pick

y'

E ~X(v)Ia

y' = V Y ( y ' , x ) y'' "

ca = a > 0

y ~(v)

y 6 ~k(~)

xv

Notice

such that

, and

8 < a •

is a dense

x)]

For each

.

, x E TIa

x ] .

.

pairwise

D

a < ~I

and

T .

such that for some

y : VY(y,x)

for some

of

and each

Since

C.F.

Y(y,x)

section

D = [y E T I

, X a : X N (Tla)

y

tisfies

y E ~k(v)l~

Let

be a m a x i m a l

y' ~ ( ~ )

whenever

T .

y E ~k(v)

set of

K

of

I (3x E X a ) ( y ~ *

Y(y,x)

(2),

is Souslin.

initial

Tia

For each

x v , let

(in ~k(v)) x'

, a dense

D a : D N (TIa)

v < 8

T

antichain

a , D a : [y E

y ~(v)

-

We show that

be a m a x i m a l

(3xEX)(y

Let

(2).

92

, there is

By d e f i n i t i o n

x' _<* x

such that

and

of

y" E

Y(y',x)

y" = x'~ .

,

That

is what we required.

Let E K

K = n

K

a closed ~ b o u n d e d

be such that,

~1(v)la = ~1(v) we see that

n a

by

~

for all

Tia = T N V

subset

, S m = D NV

.

of

~I

Since

v < 8 , so, r e c a l l i n g Thus

S

= D N (Tia)

= D

'

Let ca = a ,

that .

8~ ~ , By

-

claim 4, therefore, .

Hence,

means Xa

x 6 T

case

extends

x E T

is a m a x i m a l

-

the special

each

that each

93

extends

antichain

of

(**) a p p l i e d

an element

an element

T .

Hence

k , k < m2' lim(k),

cf(i)

in d e f i n i n g

of

of

D

.

This

X a , whence

X = X

, w h i c h is

countable.

Case 4.

Set

Stage

~k

= Uv
We must prove

bra and that for every ~k

"

Notice

~k

satisfy

Let

However,

providing

c.c.c.,

which

itself

because

implies

enumerate

Ak

80 = 0

v < y

(ii)

~k

is

~X

will of

~k"

closed u n b o u n d e d

Bo = ~I

; Bv+1

= By N[

; By = Nv
(Vv < ~ I ) ( 8 v 6 B v)

Then,

Define

~ x ( v ) l S 7 = ~x(V) n ~ y

-

T~(v) 8y

v 5 ~ < V "

v -< v

(iii)

=

of

<Svlv <w1>

for all

: (i)

To

and

that

the c o m p l e t e n e s s

B ° ~ B I ~ ..., we can pick a n o r m a l

Y <~I

subalgebra

then

Define

QT
such that

alge-

we can find a S o u s l i n i s a -

B v ~ m I , v < w I , as follows.

Since

is a S o u s l i n

is a nice

v

' this will not matter,



sets

~k

that we do not k n o w a u t o m a t i c a l l y

even complete. tion of

v < k , •

that

= wI

=

; Tv+1

.&. v ~ T < a

~

T = Uv<wIT v .

By

m'th level

By+ 1 = h k ( v + l ) , k ( v

T

to be a n o r m a l

by i n d u c t i o n

T X(v+I) By+ I

; T

on

(iii)

-[A]~k(~)x ~ -~ v
above,

as just defined. ~1-tree

, dense

)

v

, if

lim(a)



~k

as follows,

> © I v <~

In fact, in

;

"T 8y+ ~ ( v +1l )

v < wI

x v = hk(T),k(v)(x~)] (ii),

,,TX(r) By

= hk(r),k(v)

T x(v)

T v _c ~ k

sets

{~}

-

;

(as

" x

T~(v)

v

96

Set

is a tree with T

is easily ~k

seen

= Uv<w~k(v)')

-

We

show

that

Let

X

set

X a = XO

D

(Tia)

antichain

antichain .

of

[y E T I a of

T

section

of

TIa

Let

K o = [aEm I :lim(a)

TIa}

, a closed

unbounded

subset

Claim

Por

Let 8¥+i

•

of

.&.

It is e a s i l y

Pick

and let

y

y'

Thus

z E T x(8+I) 88+i

~k(v)

y

Claim

6.

, z E D .

y"

There

~(v)

We m a y

Let E E

y"

initial

initial

~ E K 2 , all

maps

= [the

least

seen

KI

that

[

= y]]

of

Ak

is a closed

(3~ < W l ) ( 3 z

is d e n s e

E T x(v)

in

For

6Tk(T))[ ~k(v)

"

y E Tk(V)l

some

z' E

1

Now,

z'

assume

E T

so we

can

z E T6+ I , w h e r e

= hk(8+1),k(~)(z

~, $

unbounded

set

~ < ~ , and all

.~. z ~ D

~ < w I , let

Define

point

)

Then

8 >y. y"

.

is a c l o s e d

3zE~X(T)I~)[T ~

each

is a d e n s e

is a l i m i t

By+

= hk(y+1),k(~)(z')

z ~k

and

of w I.

a < w I , set

y < w S , y > ~ , so that

Y' i X ( v )

find

all

z'

subset

is a d e n s e

= {y 6 T ~(~)

,

, y'

is a

wI .

,

E

k(a)

.&. h k ( ~ ) , k ( v ) ( z )

y E T k(~)

T8y+i t(¥+1)

D

a E Ko , Da

v < wI , E

.&. z E D

& X

unbounded

Then

a < m1'

.

.&. a = o t p ( A x ( a ) ) }

~

every

, and for e v e r y .

and for any

K I = { a 6 w I I 8a = a

5.

T , and for

Let

I (BxEXa)(y~kx)}

section

For

of

D = [ y 6 T I (Bx 6 X ) ( y ~ k x ) } =

for

-

is S o u s l i n .

be a m a x i m a l

maximal Let

T

94

Y

y E T~ (~)

~ < wI

such

that

, (3T < a ) (

.~. hx(T),~(~)(z)~y] .

be a m a x i m a l on

K2 ~ wI

wI such

disjointed

subset

by: that

Y

_c T k ( V ) I ~ ]

and

of

-

9(v)

: [the

least

$ <~I

(~z E T X ( T ) I ~ ) [ ~ h v K2 = K2

95

.&.

such

[a E ~I I (Vv < a ) ( ~ ( v ) is as r e q u i r e d . ~Yv

= ~

some

~ < a , ~ ~ v , and

Let

, so we

K=KoNKINK

.&. ~ [ ~ T ) ( z )

some

Let

D*

.&. z E Y X ( T ) I a

.&. ~ [ ~ T ) ( z )

D* N V

0

By

S

= D*~ .

of

Bk(a)

otp

(Ax(~))

tree

= a .

(T,!*)

(xv I v < a ) and

=

first

ED

] .

whose such

that

that

for

a = Ba E B

[0]

.

C , a = c]

.

of

K2 ~ K

see

that

"

Thus,

we

in d e f i n i n g there

is

~k(a) 9"T lar,

TI

(~,z> by

~(x)

E D*

x E TI

"

and case

that

of

x

v)

> © , and

, we

that

that

a Souslin

~

Thus, S a = D[

x =

x v E ~k(V)Ty

C . Since by d e f i n i t i o n , so by choice

(*) m u s t for

as b e f o r e .

that

such

C = [0 <~
z

~k(a)

D*~ =

and

enumerate

, a E C .

= Av
~(x)

, where

T
construction

k(~)

¥ E C , ~ < a

special

"

of s e q u e n c e s

by c o n s t r u c t i o n ,

such

is a S o u s l i n i s a t i o n

= AX0

the

, we d e f i n e d

a E K

the

D*a = [ < ~ , z ) I

We r e c a l l

(cvlv<~1)

But look,

. For

zE~ X(~)

I .e.

~ E K

consisted

= Nv
of

set

Ak(a)

some

Let

y'

, z E D a , we have

we can f i n d

v ! ~ < ~ ~ x v = h k ( T ) , k ( v ) ( x ~)

CX(v),X(T)-~U

that By

, y ~X(~)

a E K

~k(a)

elements

E Y

Clearly

a .

To o b t a i n

show

y E T k(v)

[<~,z>IT<w

a < ~I'

such an

Notice

y'

Let

is as c l a i m e d .

, therefore,

Consider

v < a

' (~7 < 9)

= y] We

z E Tk(~)Ia

K2

and for

y E Yv

~(v) < a ) ]

can find

Thus

2 .

ED],

.&.

a E K2

5,

.

for all

.&. h k ( ~ ) , k ( v ) ( z )


Let

= hx(T),X(v)(z)

that

z E D9

claim

y'

-

see

~k(a)~"~

have

every

applied

x E }I

Define Recalling that, I =

that

in p a r t i c u ~ .

So,

as A

a = B a = c I , the d e f i n i t i o n Let there

of

y E T a , ie. y E O"T I . are

v ~ ~ < a

and

Ta

implies

By our e a r l i e r

z E D a R (TX(~)I~)

hx(a),k(v)(y ) !k(v ) hk(7),X(v)(z )

that

T a = ~"T I .

observations, such

that

Thus, y !k(a ) hk(j,~v)(z),

-

which

implies

Hence

Y ~k

Da

.

ing

X

that

.

X = X

~k

°

h

words,

"

But

every

11, of

~ ~k(~+1) clearly

T

y

is

E Ta

~ ~

suffices B X °

be

extends

extends of

and

Tla

y,z

an e l e m e n t

an e l e m e n t and

E T .

of

is a n i c e

to p r o v e

Let

~ ~ ~I

the b a s i c

T , prov-

have

T k(~+1)a

subalgebra

v ~ k ~ ~k(v+1) , therefore,

projection. = h~"Ta

of

of X a,

countable.

~ ~ k

it

is a tree

y E Ta

antichain

that

subalgebra

, we

-

, every

, which

By lemma

:~k

a = Ba ~ ~

Hence

to p r o v e

is a n i c e let

z

©

is a m a x i m a l

It r e m a i n s of

y ^ z ~k

In o t h e r

whence

96

For QED.

and all I

Chapter

IX

HOW JENSEN K I L L E D A S O U S L I N

In this chapter we shall describe ner w h i c h ter.

can be fitted

As we have

Aronszajn

tree into

~ .

this is by no m e a n s if we a p p e a r start,

~Ve commence

appear

subset

C = (C,~)

by setting

subset

~i ]

of

Vie call

Let

M

on

BA(~)

@M .

bounded

Set subset

we call any such

of

Let

C

that

is

to a model

of

~

is

a poset

is a closed u n b o u n d e d iff

a-closed

Define

v'~v

&

and s a t i s f i e s

A'~

A

&

w 2 - c.c.

set algebra.

and c o n s i d e r (v,A)EG]

in

M[G]

Since

.

.

~M .

Let

Clearly,

CG

In fact,

N-generic

subset

G

be

M-generic

is a closed un-

C G = 0[A I (0,A) EG]

G = ~(v,A)I vNA~C

an

ZFC.

adding a

of

G~A}, ~

, as

M[C G] = ~ [ G ] , o~er

so

@M

I be an

N-generic

closed u n b o u n d e d

C

for g e n e r i c a l l y

(v',A') ~ (v,A)

ZFC

wI

CG

as we shall from the very

~ : {(v,A) I v < w I & A

CG = U[vNAI of

of our goal,

~I

the closed

is easily verfied.

Lemma

of

Clearly,

be a c.t.m,

and the reader must bear with us

a certain m e t h o d

and p u t t i n g

N A' = v 0 A .

the g i v e n

to so do.

by d e s c r i b i n g

closed u n b o u n d e d

of the last chap-

the way in w h i c h we shall go about

straightforward, sight

apparatus

tree in a man-

our m e t h o d will be to embed

However,

to be l o s i n g

in fact,

h o w to kill a S o u s l i n

into the i t e r a t i o n

a l r e a d y hinted,

TREE

C-a

(in

~ A .

M[C]),

a < ~j~1 ~

subset

of

Again,

and if

such that

subset ~iM

if

f E M Ya = ~

of

w~~

over

(in M ).

( ¥ v I v < ~ 1 M) is a n o r m a l and

~M .

Let

Then there

is

is the n o r m a l function

A E M

on

( V ~ _ a ) ( f ( ¥ B) = yB)

a < w~

be a such

enumeration M

of

w I , then there .

-

Proof:

Trivial.

In

define

~,

(where

98

-

|

6 : dom(~1) - B A ( ~ )

the ~ats' are calculated

6(~) = { < ~ , A > l ~ v n A }

by

for

BA(~)

in

M ) .

Clearly,

if

ii

is

M-generic

over

~

, then

It will be convenient Suppose

U

P

be a poset

for

~

iff

meets

G

every

. is definable ments.

subset

in

U

Suppose

extension

Let

which

assume

of

a model

G ~ of

is a

U , since

in that

ZF).

which in

r e l a t i o n for U,

recursive

ZF-model

•

definable

for

that this d e f i n i t i o n

U

of

U-~eneric

section

the forcing

and

of genericity.

is f i r s t - o r d e r

this definition,

N ~ H

IN1 = w •

aN = ~I D N . - ~N(X)

Then, = x

~N() =

Lemma

•

final

6 _c ~I

Zo

state-

of genericity

and we are forming

case the two definitions

coincide.

x E Hwl N N "

call a set

compatible

is used when

I~

the n o t i o n

, and is in fact primitive

w2 ' Set

Also,

(but not n e c e s s a r i l y

We shall

of

of course,

is the one which

clearly

U .

Using

We can,

a generic

in

set

is a pairwise

dense

.

for us to redefine

is a transitive

Let

C = 6 ~[C]

C

'I

Let

a N E ~I

~N: N - ~ N

, where

N

is transitive.

' ~N(Wl ) = aN ' wN(Hw I) = Hw I N N

, x E ~(HwI) N N

- ~N(X)

= xDN

,

,

E CON

• We use ~N ( ¢ ) to denote UN"(C~N ).

2 N ~ H

w2 With N , ~ N ( C ) ~

, INI=w,

p E~NN.LetU

Uo

Then there

is a

U-generic

is any

sentence

beany

is

countable

p' _~ p

transitive

of the form

set

p' =

such that : (i)

C O aN

(ii)

if

~

stants

from

{~ I x EN)

subset of

of

V (BA(¢))

U {6}

, then

aN

over which

WN(~) involves

N [ C N aN ] ~

; only con-

~N(~)

iff

-

p'

Proof:

1F~"H

Let

I=~"

w2

p = <~,A>

wN(p) G ,

E G . B

But look, Thus

Set

G

be

in

wI .

since

Hence,

IF~ ~) P' Lemma Let

, ~hioh

II-~ " ~ 2

of

Now,

w~1(B) E @ .

.

E

Since wN(p) aN) = A .

CO ~ = An v . Notice

We n o w v e r i f y

(ii).

( 3 q E G ) ( q IF~N(¢ ) ~N(~)) • Now,

p'

Similarly

(O,B>

is closed

C ~ ~I(Ao

N[C NaN] ~ WN(~) , then that

if

G , (i) holds.

w~1(q)

iff

with

IFC •

(3q EG)(WNI(q)

, s o by e h o i e e

for

~

, proving

, let

U

be a function

of

(ii).

~ , I

3 M

be a c.t.m,

~N(~ M) E UN.Let

~2

.

E G , we also have

by choice

implies

k~""

of

for each countable

EM

p' =

q E G , p' ~

: N < i 2 , so i f

~N(@)

a N , so

E G , whemce

<~,ANaN>

that for any

in

Hence

E G ,

on

I E G ]

and u n b o u n d e d

By ( i ) , N[C naN] I= ~N(~) wN

U-generic

C = D{w~I(B)

p' ~ p .

also

Let

is closed

: <~,AflaN>

-

•

.

and u n b o u n d e d

99

.

C 0 aN

(ii)

there -1

N ~ Hw2 be an

Then there

and such that, (i)

C

ZFC.

In ,U N

M

is a countable

M-generic

is a countable

in

subset N ~ H

of

m2

in

transitive

~

over M

such

such that set withN,

~M . that

Let

x E

x E N

M ~ C J , we h a v e : -

is a

UN-generic

is a map

w ~ WN

subset such

of

that

aN w(C)

over

WN(C M)

= C Da N

; and

and

: NEON aN] ~ Hm2

Proof:

By lemma over

So much

2

and a simple

argument

for

~M

forcing

M . |

(for now)

for

the closed

order for us to state, rest

density

of this chapter

precisely,

in proving.

set algebra. the result

It would

seem

which we shall

to be in spend

the

-

100

-

CRUCIAL LE~A Let

N

Souslin nisation tree.

be a c.t.m,

of

algebra

M , and let

of

~

Let

.

C

is a S o u s l i n

in

ZFC + (2 ~ = ~ )

Let

be,

M-generic

algebra

~f

M~C]

Note

that we do not find our S o u s l i n to

when we try to fit become

M~C]

clear in the next

has d o m a i n

write

T

TB

in a generic tion for it.

.

"

extension

T

cause

either.)

M

, lit

lemma

and vice-versa.

in

Sousli-

@M . Then there M~C]

~'

in

M

,~

is a

, but must any problems

into our i t e r a t i o n

as above.

~

scheme,

We may assume

to consider

we shall

and

T

will

branch

Bearing

. (This will

= ~

of

V) will be

(in

this in mind,

if we of

a special nota-

different

~1-branchl~ T

that

the e v a l u a t i o n

introduce

are quite

a E C ~ lim(~)

has an

we see that any cofinal

over

be a

, an A r o n s z a j n

this will not cause

T, C

over

(So remember,

in

w~

algebra

, since when we come

that

that,

of

~

It will not cause any a m b i g u i t y

We shall also assume

Note

~-value

~

chapter.

= ~

any headaches,

Let

is specialll ~' =

That

M, ~ , TB,

~I

for

lIT

the crucial

F r o m n o w on we fix lIT

and

~*.

I~ , an a r b i t r a r y

such that,

subalgebra

of all pass

IB,

in

subset

nice

first

in

T E M (~) be, with

be an

of

~

+ (2~I= w2)+

animals!)

clearly not

In fact by

II.7

M-generic

on

we prove:

Lemma 4 Let

b

be a cofinal b r a n c h

bounded

subset

of

bounded

in

~I

"

Proof:

In

M~bl

wI

, let

be such that

of

There

T , and let is

(a(~) I v ~ 1 IIa

B E M

)

is a n o r m a l

A E MEbl

, B ~ A , B

enumerate function

on

be a closed unclosed

A . ~iI ~

Let = ~

and un-

a E M (~) and

101

-

~M[b]

=

a

.

We say that an element = ~" for some X

fixes

that fixes

= } ~, then bounded

set

5

There

is a sequence

(ii)

if

B

Let 0*

]

branch

satisfies

.

,

P

p!l-a(v)

of the closed un-

A . I

and

of T .

A E N[b]

Then,

W E M ~B)

Work

such

such that:

T

branch

p E X

T~(v)(if

, A ~ w I , then

a closed u n b o u n d e d

we can find

0~I~ = ~

M

of

contains

be a cofinal Hence,

in

, let

p E Ts(v)

element

of

and

be least

v , if

every

pll--]Ba(~)

v < ~I

T

X_

I (Zv
Clearly,

iff

For each

~(v)

for each

is a limit point

is a cofinal

b

.

M .

a(~)

~ , let

B = [a<~1

(Wa I a < w 1 >

[a E w I I A N a E W Proof:

Then,

Let

fixes

set such that

~or each

T
Lemma

b

.

T

W o r k in

TIS(v)

a(v)

of

disjoint

a(~)

Xv _

p

~ < wI

be a m a x i m a l p

-

in

set

by

VLTI.4

such that,

~ .

Since

B E ~ .

B

M[b]

in is

M ,

II~

~]-dense, o

we can extend q II-w~

p E T

, for some

for each p

every

a < wI

fixes

W'a "

to a

w E M),

each

we can find Pot each

q E T

w E M

p ETa(a)] easily

such that

, a < w]

.

~ < wl

~ ,

each

fixes

W~ (i~.

Then,

as above,

such that

p E T~(a),

set

p ET B W(a,p)

II

P I~ W ~ = ~ .

Then,

•

~ = B(a) ~ w ]

~i o

= that

which

Set

W a = U[W(~,p)

u s i n g l e m m a 4, ( W a l ~ < w l )

seen to be as required.

I

is

|

Lemma 6 There

is a sequence

(ii)

if

b

(W~I a < w 1)

is a c o f i n a l b r a n c h

in

of

M

T

such that:

and

A 6 M[b]

, A ~ HWl,

and

-

102

(v~< ®I)(IA n~®iCa)l ~m) , a closed u n b o u n d e d

Proof:

Fix

This follows

(Wa i a < w 1 >

We say that a M

such

that

As before,

5

as in lemma 6

p E T

fixes

Ii

v

then

{~

~ w11A n v ~ w }

contains

B E M .

from lemma

by an argument

for the rest

Ti~

iff there

as in lemmaV]XI.5.|

of this proof.

is a normal

a-tree

t

in

I!

p II-TI~ = t .

for each

that every

set

-

a < wI

p E TB(a)

the closed u n b o u n d e d

there

fixes set

TI~

is a least •

Let

{~ E m I Iv < ~

ordinal

B(m)

<~(~) I a < m 1 >

- B(v)<~}

•

< w I such

enumerate

Thus,

for all o

a < ml

' p E T~(a)

-

p

fixes

TI

.

For such

a, p , let

II

that normal

~(a)-tree

I~ TI 8 = #

Let

may assume C

Note

a < ml that

a = O

that if

We w i s h a nice that

b

here. segment,

of

and

C .

(¥~

is because

so will

itself

to know about

branch of

II

in all of this.

enumeration Ym = a

V

P i~ T i ~ C a ) :

Let

Since ~

By lemma

=

I, we

~)(~(yB ) = yB ) . We C - y~

be

~ •

differs

M-generic

from

over

C ~)

T , then

~[b]

= U[Tp I P E b

of

a Souslin ~

and,

algebra in

~

M[C] (~)

in

M[C]

it holds

such that

~

with

~-value

~[C]

with

is special.

Our strategy

~,

.

subalgebra

following

= O

(This

is a cofinal

to construct

T

such that

~(O)

such that

is all that we need

htT(P) ~C]

X

be the normal

only on an initial

which

in

, we may assume

(Ya I ~ < m 1 >

can find an

t

Tp be

is to construct

properties:

a Souslin

tree

~

in

the

is

-

(i)

There is a map

I03

h: ~ - T

such that:

(a)

x <~ y - h(x)
(b)

h:F

(c)

if

ONT~ Tyv z
all and

y <~ x , such that (Thus Set

h

resembles

~ = BA(~)

embeds

-

~

in

defined by

~

; v < wI ;

htT(z)

E C , there is

h(y) = z .

a neat cover very strongly.)

[Note that, by (i), there is

~:~

In fact, the restriction

~

.

~(z) : {xE ~I (3yE ~ ) ( x ! ~ y

of

~

to make

~

the identity.

to

which nicely U~<~ITy~

& h(y) = z)} .

we are done, all we shall need to do is replace around

-~

~

is

Thus, when

by an isomorph

We shall not bother ourselves

with this point again.] (ii)

lIT

is specialI~ =

(iii)

The elements htT(x)

of

~

E C , and

are pairs f

(x,f>

such that

is an order-preserving

x E T ,

map of

TxlC

( =

o

U[(Tx) a l ~ < h t T ( x )

& a E C])

into

Q

(the rationals).

(Remark: Since our trees "grow downwards", the rationals "increase" likewise.) (iv)

<x,f> !~ <x',f,>

(v)

h(<x,f>)

-- x iT x,

&

we shall assume

f muf ,

: x .

In order that our definition does not break down, we ensure that at every stage the following

two conditions hold: o

(*)

If

lim(a)

= sup[f(s)

(**) Let

and Is

(x,f> E ~a+1

> °t}

TX

' then for all

t E (Tx)Ya,

f(t)

.

<x,f> E T a + I , x'
E C .

Let

bs,...,b n

be

104

-

-

o

cofinal branches of T x (Tx,)htT(x)

.

Let

i = 1,...,n . ment.)

Then there is an

ZF-

eous recursion,

a

8(m)

(i)

if

a

bi~C

in

qi > sup(f"~i~])'

will have a maximal ele<x',f'> ~

<x,f>

and f'(ti) =

ZF

minus the power set axiom.

and

(N

I ~ < w I>

by a simultan-

as follows.

and for all is countable

LsEb,C qa~ (ii)

be such that

such

<8(v) I ~ <~I )

be the least

ZF--model,

f'

tl,...,t n

.

denotes

We define sequences

Let

ql,...,qn E Q

(Note that each

qi ' i = 1,...,n

As usual,

which extend to points

6 > a

such that

b E W

Ls[,Wa,C 0 a]

is

:

in

L[b,C 0a]

in

L[b,C 0a]

, then

a

is countable in

;

if ~ is uncountable

but

(~+)L[b,C0a]

< ~I ' then

8 > (a+) LEb'C0a]

Let

N

= L s ( a ) [ < N v I ~ <~>,W ,C O a]

The definition

of

~

is by induction on the levels.

As we proceed,

we set

F

< FlY ~ michel+2>

=

T

Na = N

°

Y~+2

[~l~,ml~+2,F~] .

Set

t o : {<~m,~>~

and

~I = {<x,<0,0>> I X E T y I }

= ~

for convenience).

Suppose define

;

~+I

is defined and

~a+2

by forcing over

(where we have assumed that

~I(a+2) Na+2

"

satisfies

llTo= {0)II~

(*) and (**) .

We

-

Let

S = {<x,<x',f>>ix

For each

~ mya+2 &

105

-

<m

x

s = <x,<x' ,f>> E S , let

mya+1 & <x,,f> ~ ~I(~+2)}

x, e

ps = [P I P

.

is a finite function

&

dQm(p) ~ (T ° & ran(p) ~ Q & (Vt E dom(p))(Vs E (Tx,) (tQf(s))} . Regard ~s as a poset under 2 Note that s E Na+2 '

For each Ta+ 2 Let

s E S

as above and each

p E ~s

, we define an element of

as follows. Xs, p

be an

< x , f U X s , p>

Na+2-generic subset of

into

~+2

It is obvious that

Suppose next that (*), (**).

For each

"

Note that

~I(G+3)

x E Tya , let

with

Put

p E Xs, p

( x , f U X s , p> ~

<x',f> .

still satisfies (*) and (**).

lim(a), a < m I , and

We construct

~s

~a

~la

is defined and satisfies

by forcing over

B x = {b I b

~

.

is a cofinal branch of

T°x such

o

that, for some extends

b}

For each <~ Q &

with

Y
z E (Ty)ya

x E Tya , let

& x
& p

px = {((x',f'>,p> I<x',f'>E ~B+ I for some

is a finite function &

dom(p) ~ B x &

° Ic)(t E b ~ f'(t)
Note that

(<x',f'>,p> ~ (x",f">,p'> ~

<x',f'> ~

Regard

ran(p)~ ~x

as

<x",f">& p2p'.

p x E N~ .

For each

x E T

follows.

Let

Xx, u .

which

.

a poset under

~la

y E Tya+1

Set

and each Xx, u

be an

u E ~x

, we define an element of

Na-generic subset of

fx,u = U{f' I (<x',f'>,~> EXx, u] .

~x

such that

~

as u E

Since (**) holds for

(and to some extent since (*) holds also), it is easily seen that

-

fx,u

is an o r d e r - p r e s e r v i n g

U[p

I (3d)(
B x , gx,u(b) <X,fx,u> ~

.

suppose

, where

~.Ve define

For each

Ta+ I

x E T

lim(a), to make

.

TylC

from

?y,x,u(t)

that,

<X,fx,u>

and for each into

that

(y,}y,x,u > ~

~a

b E

Note that

(**) to consider here.

~l(a+1)

a < w I , and

is

defined

(as above).

(*) hold.

' Y
: (fx'u(t)


by taking

gx,u =

Set

u = (<x',f'>,p>

in the above, x

Ta+ I

Clearly

Q .

gx,u : Bx " Q

Put

,

gx,u (b) ,

Noting

into

, y E T

Y6 o

y,x,u

T xIC °

Clearly,

There are no new cases of (*),

Finally,

-

map of

= sup(fx,u"[b~C]) <x',f'>

106

if

t E

if

t E

u E Bx

, define

T o

x Y~ &

b = [sE

is u n i q u e l y

determined

for each pair



by

x It

s].

y , we obtain

as above.

Note

(X,fx, u) .

(*) holds for

~I(~+2)

, whilst no essenZiall 2 n e w case of

(**) arises.

Set

~ = U <w1~ a , a normal

hold.

Let

V~e verify

b

be

It is clear that (i),(iii)-(v)

~ .

b ° = [xE T I (3y E T ) ( Y ~ T X

(ii).

M[C]-generic

E b)]

wl-tree.

on

By iemma VII.4,

Set

T

is a Souslin

tree in

M[C]

, so

o

is

M[C]-generic

}~{[C][b]

Set

ving map from

on

T .

Let

closed unbounded

subset of

b0 o

T b : U[Tx I X 6 b o ~ C

fb = U[f I (x'f> Eb] TM[C][b]Ic

&

into

Q

~I

in

. in

Then

fb

M[C][b]

M[C][b]

] .

Clearly,

Tb =

is an orderpreserBut look,

, so, using

C

is a

fb ' we can

-

107

-

o

easily Q .

define,

Hence,

, an o r d e r - e m b e d d i n g

~ "T

speci ll

is

It remains

Le~a

NEC]Eb]

M[C]~b]

WII

second

in

to check

is special"

:

of all of

, whence,by

Tb

choice

into

of

b ,

.

that

~

of which may appear

is Souslin, rather

We require

strange

two lemmas,

the

at first glance.

7

Let

D E }~[C]

Set

~b = {(x,f> E ~ I x E b]

Proof:

be dense

Since

T

Thus,

if

and

D0~b

and where

Let

b

~"

MrC]

that

, b

is

in

~b

x'

in

~b

E T~ here

I~]B "<x',f'>

T .

"

T .

xo E b

&

<x,f)

is over

M[C], ]B

Pick

of

on

we can find

set term for here.

branch

M[C]-generic

the forcing

x = x°

Then

is dense

x o l~B"(x,f>

, where

assume

be a cofinal

D 0 ~b

is the generic

_(~ <x,f)

x' --
in

b

(x',f') E T~OD"

has

M[C]

,

forcing. E D , , contrary

"

8 b

be a cofinal

x Eb~] that

. C

Let

is an

defined Let

Then

such

in

We may clearly

Lemma

.

Let

is not dense

E ~b

no e x t e n s i o n

to

~ .

is Souslin

<x,f>

<x',f')

in

in

branch

of

b 1,...,b n

T , and set

be cofinal

M'[bl,...,bn]-generic M[C]

p : [bl,...,bn}

as above,

branches

subset

still.

- Q , and set

}~' = M[b]

of

, T ° = U{T x 1

of

T ° , and suppose

WlM

over

~M

, with

Set

~b = [(x,f> E ~ I x Eb] . n ~P = [(x,f> E ~ b I iA1(~qi < Q P ( b i ) ) (

o

VtETxlC)(tEb Then

Proof:

i ~ f(t) < Q q i ) ]

D 0 T~bb is dense

Let

P(bi)

in

T~

.

Let

D E M'[C]

be dense

in

~b

"

.

= r i , i = 1,...,n

~(x,f)~ i = [z E T° I htTo(z)

.

For each

= htT(x)

<x,f>

E ~b

& (3r.~
' let

E T °IC)

-

108

-

(z<

o z' - f ( z ' ) < r ' ) } , i = I ..... n . Set [ u ~ = [u3 1 x.. T i .. × [u~ n . Note that [- ~ is definable in ~'[C] . We

clearly have

Suppose

the lemma is false.

no extension about

in

D O ~

M'[bl,

Then we can find

.

(Let

bn]rC]

compatible with

u o = <Xo,fo>

ht~(u o) = 8. )

so by choice

of

C

M ' [ b 1 ' ' ' ' ' b n]

This is a statement

we can find

v

R o E CM

o

n

M'[D I , .... bn](h t r!~ --><~>)FRoI~- M

n

~

(<~>) E C O

y~ - iA1(~o(gi) < Q r l ) )

.&.

(¥u6 T ~ ) ( u ~ u o & u 6 D

= Yht(u)-

" <~> ~ [ u ~ ) ] } .

S°

is a subtree

Thus °

be the set of all

of

(T°) n .

with successors

in

S

(~)n

Let

S

at all levels of

( V a < ~ 1 ) ( < b 1 ( Y s) ..... bn(Ys)> 6 S) , so

of

& ht(<~>)

e

z 6 S

By (I),

empty subtree

with

(vu~)(U~Uo ~ u~D - iV1(~in~u~i:~):

S° = {(~>6 ( T ° ) n l f ° r s ° m e

(T°) n .

E ~

C , such that

(I) ~o l~c~ Let

= [uE ~b I i=1(bi ~ N~u~ i/~)}

~

such that every member of

S

S

is a non-

has uncountably o

many successors

in

S .

Also,

in the definition

of

S , the sentence M'

, so we can replace

i~C~M'[bl . . . . .

that

S

M'

(by d e f i n i t i o n

of

being forced is here,

to conclude

we have

(2) Since VI.8

Zo(M')

<~> 6 S

&

S E M'

u 6 uo and

T°

is an element

u 6 D

~

~ ~ M

can apply lemma I

to find

So)(T(y8)

So, we have:

(3)

yao = S o &

that,

in

M'

= Yht(u)

in M'[C]

" <~> ~ [ u ~

•

Then,

¢ : ~I ~ Wl

ao <~I

as

C

is

such that

(Vv_> ~o)(Vz C Syv)(8 > v

-

.

, we can apply Corollary in

M'

S ~(B) (z) is well-distributed)

we may assume that

.

& ht(<~))

is Aronszajn

to obtain a normal function

= y6)

Note

I~C M

S ) ,

&

(V~<®I)(Vz~ST(~))(B>v

of

bn] by

s(z) Y8

M-generic ySo = s °

such that By lemma 4, over and

CM

,

we

(¥8 _>

is well-distributed).

-

By Lemma

6, we can find

that for all that

Y~I

a normal

v < w I , Sl0(v)

= al

and

-

function

= S n V0( v

(VS_>~1)(p(ys)

val = 21

Let

a

Now,

Sa+ I J ~ , so by the construction

B > a •

By SyB

If follows,

by

that

< u , and

S

YS

u'

~

E ~b

of

~emma

for some

ht(u')

' u'

, u'

, which

such

~ < ~I

so

~a'

.

SyB+I

nature

D , contrary

the proofs

to

of lemmas

such

for all

8 > a+1

, if

D

is a

B > ~ •

nEu'~

being

u'

of

~,

6 (~b) ,

is well-distrubuted.

B = ht(u') Y~

NyB+I

of the construction

YS

Set

So, as

E ~B

S (~> n [u~

S

(and the first

u = (x,f) ~ u o

E NyB+I

B ) for all

is to say

~a+1

E S~+ I , (~) E ~u~

SyB,

E D .

a > h t ( u o) =

By

= ~ .

will be well-distrituted!

YS

can lie in

on

= B , then

< u

of

(~>

Hence,

induction

and

S °) we can find

SIyB+ I C WyB+1

$<x) ~ ~u'~

By examining

of

(3) and the generic

~

u

Pick

M

So, we have:

y~ = ~ ~ ao,al

E NyB+I

<~> ~ [U~

seen that

and,

(4),

(by a simple

Let

that

in the d e f i n i t i o n

ZF--model,

u'

such

u E (~b)a+1

Let

EVJ0(v)

in

(vBz~I)(SlYs(wyB)

be least

clause

o : w I ~ ~I

= YB ) "

(4)

that

&

109

dense

(2),

(~>

But we have

Hence in

E

just

no extension

~b "

|

4 to 8, we obtain

9

The conclusions

of lemmas

4 to 8

able

model

ZF- + "there

transitive

At last we can prove:

of

are valid w h e n e v e r

M

is an uncountable

is a countcardinal"

+

110-

-

Lemma

10

M[C~

I= "~

Proof:

is a S o u s l i n

Let

J

be

(J(a,~)

Let

X

there

I a E0n)

is a set

is c h o s e n

Let

To

be

set

TI

be l e a s t

such

lemma

we

the l e a s t

3

, and

setting

a = aN

(i)

~ , TI(~+I)

(ii)

C O a

(iii)

there

is

N(T)

= Tim

dom(w')

and

dom(N')

of

NKC 0 a ]

each

Tim

.

apply b x}

~I~

•

x E T a , set

Now, lemma

Tim 7

, to c o n c l u d e

that

that

the

,

Since

~,X

E M[C]

,

E L[Y,C]

and

(X,~>

.

T,T,W

We can as-

E L[Y]

= J(To,
= J(TI,Y)

.

, and

.

~(C M)

E M[C]

of

;

, w' o_ w , such

HM[C] "'2 = W~a

E U ;

that

w'(C)

= CN

@

, and

, X,~

Ya = a "

E dom(w')

Also,

Xa

.

Since

To,Y,C

Clearly,

is c l e a r l y

~la

tree

Then

~Ia anti-

D E

•

in

N

, a cofinal

, so by l e m m a

9

branch we

, ~I~ , bx,D, ~Tb x =Def [(x' ,f'> ~ I ~ set

E

~'(~)=

a maximal

I (BxEXa)(y~x)]

b x = [y E TIa I X < T Y ]

N[C 0 ~]

~

'

is a S o u s l i n to

set

, N ~H M , INI M = w , such that w2 t r a n s i t i v e set U E M , such that,

D = [yE~la subset

any

N E M

over

a] ~

, w(W)

Set

is a d e n s e

For



~ H M[C] w2 = X a = D e f X 0 (~la)

w'(X)

chain

that

~ = NN

N[CN

that for

.

~,X

w I = w~ [Y]

such

~'

M[C]

that

,

Ciearly,

and

in

, such

a countable

is a m a p

1i"'-1:

and

~

ordinal

U-generic

such

L[~]

of

so that

obtain

and

function

enumerates

Y ~ Wl, Y E M

that

< Y , T o , T I) E N

~I

antichain

sume

Applying

.

the u s u a l

be a m a x i m a l

Y

tree"

D N Tbx

is d e n s e

in

Tbx

of

can

I x'

, each x E T a.

-

Again,

for each

x E T

, let

111

-

Bx

be the set of all b r a n c h e s

b

of

o

Tx

of the form

w E (Ty)a

.

b = [z E T °y l w < T o z]

Let

arbitrary.

bl,...,b n E B x

By choice

of

for some

, and let

U , C qa

is

y E T~(a+1),

y
p : [bl,...,b n} ~ Q

be

N[b;bl,...,bn]-generic

over

o

C~

, so by lemma 9

w(T)),

bl,...,bn,

is dense

Claim.

Now,

in

we can apply l e m m a 8

, each

D E

y~ = a

I u ED]

dense u

in

(and lim(a))

By t h e

px

Thus

Let

that

X = X

0 = ~[Y,C] P ~ ~2

Whichever

Set and

of these

:N'-~ Lo[Y,C] X , ~ E L0[Y,C] D E N'

was

constructed

For such

Dx E ~ a

DOT~bx:

"

generically

x , let

By o u r

over

D x = [(u,p)

arguments

above,

E

Dx i s

u' E D , whence u E ~ a ' u <~ u' for some T X Hence X is a m a x i m a l a n t i c h a i n in , which

is countable.

The~remains

Hence

Y,W E N , so there in

wI &

to prove

H M~J2 ' we see that

YO a E W

w ' ( ( y v I v
.

in .

N' ~ N

that

A E N

only the

.

Then,

0 ' = On O N

.

and

(w'-1~N ')

and

Wo

therefore,

- -

such that

recall

is cofinal

N a]

N' c

Since

a , whence And,

J

In particular,

- Y N v E ~Vv"

is closed and u n b o u n d e d

and

is closed under

X a, ~la E N'

is a set

P = ~2

we must have

Lo[Y,C ]

vEA

, N' = L 0 , [ Y A ~ , C

, or else

is the case, Since

.

0' = ~"(0 N N )

P' = w(O)

, and it suffices

and u n b o u n d e d A Na

that

of the claim.

either

Now,

of

to conclude

(and

as above.

~

, x E T

if

N , b x , T I a , Tx

this later.)

, so

claim,

an e l e m e n t

, proving proof

~x

.

extends

x E To , p

(We prove

for the posets

~x

cN, YIa , D N ~Tbx,

p, C Oa,

T~bbx

to

in

N

w(A)

a E A . that a.)

_c L0,,[Wa,C N a ] , where

~

"A is closed = A~ ~ ,

Thus,

as

N~(

Ya = a

(because

So,

IWaI L[Y]

as

0" = m a x ( o ' , 8 ( a ) )

-

Thus,

if

red.

We prove

Suppose

p' < 6(a)

first

-

N' _c L S ( a ) [ W

, then

that

112

,C N a] E N a _~ N ~ , as requi-

p' < 8(a)

that

a

is countable

in

L[Y n a,C N a]

•

Then

a

is !

countable Hence

in

LS(a)[YN

p' < 6(~)

Now suppose

inequality

Finally,

a 2 w~ [YNa'CNa]

of

here

8 , we have is easily

we show that

(V8 < W l ) ( W ~ [YoS'C08] L [ Y 0 8,C 0 8]) least

, by d e f i n i t i o n

of

8 .

but

there

since

are no further

< m I) , then

w~ [Y08'CoS]

(~+)L[YNa,CNa]

p' ~ w~ [YNa'CNa]

established,

, so the above

such that we have

table

in

absurd.

~ E dom(w')

L[YNa,C |

a = w~

fla]

p~=

.

@

Again

(the first

~"(0 O N ) .

possibilities.

)

Well,

(V8 < w l ) ( w I is inaccessible

cases

suffice.

= wI

Since

, whence

, we have

< Wl

< 8(a)

Otherwise, 8

is thus

•

able,

But

.

that

by d e f i n i t i o n

a , C N a]

8 < a .

Then,

a ~ w~ [YNS'CN~]

if

let

if

in 8

be

HM[C ]_de fin_ ~2

a

is uncoun-

= w I , which

is

Chapter X

CON(ZF) We piece

together

CON(ZFC+GCH+SH)

-

all of our p r e v i o u s

results

to prove

the f o l l o w i n g

theorem: Theorem 0 (Jensen) Let

M

dinal

be a c.t.m, absolute

(i)

@N(~)

generic

(iii)

N

=

extension

@M(~)

k

of chapter V i i

set f o r c i n g

in

M

poset,

I

There

is a Cohen e x t e n s i o n

(ii) (iii) (iv)

M

such that:

, (2k) N = (2k) M ;

and

M*

have

.

M

will be as above. that

@ , of the p r e v i o u s

M*

~

M

We first require

the same

for all c a r d i n a l s

@I,:*(®)

then,

we shall also assume

Lemma

M

M

of

;

of the chapter,

iteration lemma holds

(i)

of

N

is a car-

I= SH

For the whole

closed

There

ZFC + ( 2 w = ~i ) + (2 wl = w 2)

For all c a r d i n a l s

(ii)

sults

of

of

of

M

I= ~ &

By the re0*

, so the

a lemma concerning

chapter.

such that:

cardinals

and c o f i n a l i t y

M , (2~) M* = (2~) M

function;

;

@M(w ) ;

:

M* = M~(C l ~ < w ~ > ]

, where, f o r

each

~<~

, if

we s e t M

M[(C

(v) (vi) (vii)

the

]~<~)]

, Cv

is an

My-generic

subset

of

wl~ over

M

= M

~ v ;

M* k t : + O ; for all if

a < W2M

~ < w~

Souslin

tree

and in

, M

]---- 0 +

O*

T

is a S o u s l i n

M*

.

; tree in

M s , then

T

is a

-

Proof:

114

This is a classic example

-

of the kind

c h a p t e r VII.

We iterate

M .

stages of c o f i n a l i t y

At l i m i t

first a p p e a r mit

to be)

(we d e s c r i b e

of c o f i n a l i t y

the closed

we

It w i l l be clear that

by

~

so if

, then

(i) holds, fices

M*

(iii) and

whence

take

~

~

~

s e r v e d at limit

.

stages

also cause no problem, we are g i v e n

w2

, m < ~2

assume

that

for any < ~ , A B)

satisfies

a dense

stages

~I

M).

Well,

by

~ 2 - c.c.

,

~ 2 - c.c. will be pre-

since

w 2 - c.c.

, so we are t h r o u g h . ]

It suf-

to w 2) of the argu-

~I

has

@ .

determined

of course.

of c o f i n a l i t y

of

M

is

in [Je I], T h e o r e m 49,

the

¢

be the

We must check that

in the i t e r a t i o n has

clearly,

•

subset w h i c h

e x t e n s i o n of

(from

Successor

stages

[For s u p p o s e

We can assume,

by d i s c a r d -

that they are of the f o r m ~ < wI

= A B 0~

~ < B < ~12 ' ( ~ ' A a 0 A B)

the d e f i n i t i o n s

sequence

described

a < ~
is p r e s e r v e d at limit

at limit

w 2 - c.c. 0 n

' for some f i x e d

der to f a c i l i t a t e

below);

so c o n s t r u c t e d .

ing some of them if n e c e s s a r y , < v , A m)

of the inverse li-

(iv) are immediate.

members

(what w i l l at

iteration

generalisation

And

times in

Let

is a g e n e r i c

we see that if each stage

w2

in

the d i r e c t limit.

w i l l have

of S o l o v a y - T e n n e n b a u m

then so does

we take

(ii) w i l l be i m m e d i a t e

to show that

a straightforward ment

w

a simple m o d i f i c a t i o n

d i r e c t l i m i t of the entire

~1-closed,

set a l g e b r a

this " m o d i f i c a t i o n "

wI

of f o r c i n g d e s c r i b e d

.

We may l i k e w i s e But

extends b o t h

It r e m a i n s

that we

(~,A)

and

to check that

stages of c o f i n a l i t y

this a r g u m e n t

then we see that

~ .

~2-c.c.

It is in or-

(ostensibly)

modify

of the inverse l i m i t as it was d e s c r i b e d

in

c h a p t e r VII.

Suppose

then

have d e f i n e d

~

is a limit

(~

I~ <~),

ordinal,

(~viv<m>,

cf(~)

= ~ , and that we

<e ~ [ ~ < ~

< ~ > ,
-

(where we use corresponds we take

the same n o t a t i o n

to

~

~
some

countable

~ PvE~

e ~(p~) # p~

, cM~

; (ii)

set

via lemma VII.2,

(We call

limit,

X

at stage

a •

condition

(iii)

Clearly,

a support

for

(in

p

that

stages

-

or

the idea of an

source

is p r e s e r v e d of trouble

cover r e q u i r e m e n t s .

is

But

to v e r i f y

inductive

construc-

a ; and by the very d e f i n i t i o n limit

v <~

~)

Wl-happiness

out a simple

for

in such a situ-

modified

the only p o s s i b l e

all we need do is carry

tion sequence,

(p~)

For ]Pa'

; (iii)

is such that

PT = A < e

check

of the neat

tion of length

• < a

Pv+1

such that:

~ Pv = h~v(PT)

(it appears)

we should

; thus

etc.).

p = (p~[~ < a > <~

X ~ ~ , if

Since we have

inverse

~<~

, then either

~ E X •

ation.)

this,

,~

as in C h a p t e ~ VII

the set of all sequences

(i)

else

115-

of the itera-

in such a c o n s t r u c t i o n

cause no

problems. We shall prove

that the

mit operation.

w 2 - c.c.

Before we do so however,

time off to m e n t i o n

mit of a cofinal More

using

of

on

proach we have adopted,

]P

w-inverse

and the above

whatsoli-

, as in c h a p t e r

<]Pvl~<w2>

convenient,

the

as the inverse

limit

v < w 2 , ]Pv ~]P'~ .

and slightly more

this li-

let us n o w take

~P~iv < a >

if one defines

the "original"

by an easy i n d u c t i o n sufficient

if we defined

w-subsequence

precisely,



under

that it would make no d i f f e r e n c e

ever to the c o n s t r u c t i o n

VII.

is p r e s e r v e d

as above and instead,

then

It is, however,

to stick

equivalence

to the apwill there-

fore not be required. Assume

then that

We show that sider

the case

~a

Pv

satisfies

satisfies a = w

ment may be m o d i f i e d

first,

w 2-c.c.

w 2-c.c.

for each For clarity,

and then indicate h o w

for the more g e n e r a l

case.

~ < a • we conthe argu-

116

-

In

ZFC

, we may define functions

= <~,A) For

, let

p

E

l~V

f(p) = v , h(p) define

,

~o (p) = Po ; If

-

p E n+Iv

ci(p)

,

f, h

= ~ NA

i

<

m

on

~

thus:

if

. by induction,

,

thus:

a1(P) = ; ~i+I (p) = (ai(P)'Pi+1>

, define

p

Co(p),...,an(p)

"

similarly.

Recall now the proof of lemma VII.2. For each

n < ~ Set

'

let

~P# n

=

[P=
IP i ) }

•

Thus

an ']P#n ~ ]Pn ' and clearly

""

• ' Pn > I (Vi -
E

P -n <# q -- an(P)
is a sort of representa-

tion of

~P in a form as if lemma VII.2 had not been involn ved at all (ie to obtain P # from ~ we unravel the dell• n n nition

(using VII.2)

For each

of

n < w , let

]Pn' ~n-1 .... ~PI )"

~P*n = [pE]Pn# ~ (3~ < ~ 1 ) ( ~ X o , . . . , X n C ~ ) _

If(P°) = ~ ~ h(P°) =X° ~ o
"] ] } " #

Claim I

IP* •

is neatly

Wl-Closed

it holds for

quence from that

an(pm)

~ *n+1

n .

"

l~n"ff~m~n+1)

q = Am< ~ an(pm)

on

Let

(in

n .

For

(pmlm<w>

For each

SUPm<~ Vm ' X = nm<~ Xm

ly have

]Pn •

This is proved by induction Assume

in

n

&

Hence

BA(~)

) , then

q E ~#n "

is a decreasing

1

Clearly,

se-

be such Set

~ =

Now, by induction hypothesis,

~ )"

p = E ~ +

Vm, X m

h (Pn+1) m = Xm"

use the m a x i m u m principle to find a unique (in

it is trivial•

be a decreasing

m < w , let

= ~m

q l~n"
"r = Am<wPn+ Im

n = 0

if

Since we clear-

sequence from ~", r

such that

q I~n

q II-n "f(r) = ~ & h(r) =X".

But clearly,

p = Am< w pm

(in BA~P:+I)),

-

117-

so we are done. Claim 2.

]Pn

is dense in

We prove claim 2 nothing

n .

For

n = 0

Assume it holds for

n .

Let

By induction hypothesis

and claim I, we can find

p o _
v1 -> ~o

p I < ~ pO , and

q1

We

q2

here.

v2 -> Vl

such that

h(q2) = ~2"

On(

for some

finish by setting p~ E]P*n "

Since

qience from

p2

v1 = f(pO)

]P* n '

h(Pn+1) .

=

We may

pl E ]P* n

'

(in C) &

Let

v2=f(plo ).

Similarly, pick p2 E ]P* 2 -<#n p I ' n ' P q2 ql ) I~n" < (in C ) & f ( q 2 ) = ~2 &

X 2 _c ~2 "

Proceed inductively now, and

pW = Ai< ~ p i (in BA(]Pn# )) an(pW ) 1 ~ n " ( q i J i < ~ >

Setting

Hence

By claim I ,

is a decreasing qW

such that

se-

On(p w)

~ = suPi< ~ vi ' X = Oi< w Xi,

an(p~ ) N_n,,f(q(U) = ~

f(p~) = ~ .

~o &

X I ~ vI .

@", we can find a unique

we clearly have

pOE

Again, we can find

for some

1~n,,q~ = Ai< w qi (in ~)" •

clearly,

Let

there is # p E IPn+ I .

~n(pl ) l~n ''ql -< Pn+1

such that

& h(ql) = XI"

may assume

and

~n (p°) 1 ~ n " f ( P n + 1 ) =

Vo < Wl ' Xo --~ Vo .

clearly assume

f(ql) = ~I

"

by induction on

to prove.

Xo" for some

]P# n

&

h(q w) = X".

(pW qW) E ]Pn+1

Since

But (pW qW)

<# --n+1 p ' we are done. Set

]P# J ( V n < w ) ( p ~ n + 1 E]P n#] ' and let p w = [p = < p n J n < w ) <# q ~-~ ( V n < ~ ) ( p ~ n + 1 <# q~n+1~ Define ~:]P#-]P by -~ --n " w w ~(p) = (~n(P)I n < w ) Clearly, ~ : ~ # ~ w So, to prove W

that

]Pw

has

~2-c.c.,

it suffices to show that

Set P*w = [P~Pw# l(vn<w)(p~n+1 Claim 3.

~*

is dense in

W

Let

p E ]P#

~ P*n )} "

]P# . W

be given.

W

<# pO P I -~

]P~~ does.

such that

Set

pO = P . Using claim 2 , pick . n+l <# pl ~2 E ]PI ; and in general pick p -~

-

pn

pn+1 ~n+2 E P *n+1

such that

Ai~ w pi~n+1 Define IP* w

(in BA~Pn~))

q E WV

and

N o w let dinality

by

q -~ ~# p A w2

As

q~n+1

assume

that

vaires

Hence,

as

through

IW~(~)I

which

under

lim(~)

3

f(po)

= ~

qn = n

Clearly,

of

that

for all

p E A , let

•

q E

We

p E A , where

(Xi(P) l i ~ w )

A , (Xi(P)li ~ w )

= Xi)

, where

The proof

A

be such

varies

is to p r o c e e d

(xili~)

through

~(~).

that for all is fixed.

is p a i r w i s e

But

compatible,

is complete• the above p r o o f for the case

[Perhaps

a ~ w I , establishing

of car-

A c p*

= 2 w'w = w I , we may assume

, w ~ a ~ wI

]P# w

(Vi ~ ~)(~i(p) If-i "h(Pi+ I ) = ~ " . )

seen h o w to m o d i f y

formal p r o o f

subset

we may assume

these conditions,

is absurd•

It is easily

each

~

n ~ w .

incompatible

and

p E A , (Vi ~ w ) ( X i ( P ) clearly,

n ~ w , let

I 9 qn E~P*n

= qn ' each

For each

h(Po ) = Xo(P) p

For each

, as required.

By claim

is fixed here•

"

By claim

•

be a pairwise

may f u r t h e r

that

118-

the easiest way

by i n d u c t i o n

analogues

to express

on the limit

of claims

a

ordinals

I, 2, and 3

at each

step.~ Suppose n o w that A

be a p a i r w i s e

For each As

p

for

induction a)

s

.

subset

of

s(p)

a •

subset

A , there

w I ~ a ~ w 2 . Let

of ~ a

of c a r d i n a l i t y

that

P

support for

are at most

Hence we may assume countable

~

s(p)

= wI = s

~2-c'c"

p . possi-

for all

(and cofinal,

satisfies

w 2.

by the

for all

But since we need only n o w concern our-

the s u b - p o s e r

otp(s)

= w , and

be a countable

is a fixed

hypothesis

selves with since

through s(p)

p E A , where

, cf(a)

incompatible

p E A , let

varies

bilities

lim(a)

[p E Pal

is a c o u n t a b l e

limit

support ordinal,

(p) = s]

, and

we can obtain

the

119

-

desired

contradiction

tion "along"

VIII.6

the proof

are valid.

the n o t a t i o n

the previous

of lemma

Well

, (vi) by lemmas

l e m m a Vil.4.

Using

by an a r g u m e n t

as above

(eg. by induc-

s ).

That c o m p l e t e s (v)-(vii)

-

I (i).

(v) holds by lemmas VIII.2

VIII.2

and VIII.4

The lemma is proved.

of lemma

chapter

We must check

, and

that and

(vii) by

|

I, we can restate

the "crucial

lemma"

of

as follows:

Lemma 2 Let

~ < ~

be, with algebra lIT

and let

~-value ~'

in

~

in fact

= N

such that .

tree.

~

in

Ma

Let

Then there

is a nice

T E M~B)a

is a S o u s l i n

subalgebra

o£

~'

and

out our m a i n i t e r a t i o n w i t h i n in order).

N*

The next lemma adapts

.

(By l e m m a

lemma 2

to

situation.

Lemma

3

Assume with Bra

algebra

|

carry

I, this will be quite this

be a S o u s ! i n

, an A r o n s z a j n

M +I

is s p e c i a i [ ~

We shall

~

V = M* m-value

~'

Let ~

~

be a S o u s l i n

, an A r o n s z a j n

such that

~

tree.

is a nice

algebra

and let

Then there

subalgebra

of

T E V (~) be,

is a Souslin

~'

and

alge-

!!3 is speeial!~'

=

Proof:

Let

U

U,~

_c Hwl

since Souslin

be an a r b i t r a r y

U

.

Thus for some

is a Souslin algebra

in

M

find a S o u s l i n

algebra

subalgebra

~'

of

Souslinisation

and

~'

~

a < w 2 , U,~,T

tree in and

of

M

We may assume

E M

Then,

, we see that

T E M (~) in

.

Ma+ I

~

By lemma 2 such that

II~ is special1~'

= ~

~ Let

is a we can is a nice U' be an

-

arbitrary U'

Souslinisation

will r e a l l y

a Souslin a nice Ma+1

of

~'

(ie.

subalgebra

of

in

' we clearly have

tree

M*)

~'

in

~+I

(ie.

and

in

~

Since

By lemma M*),

will

so

~'

still be

lIT is speciallp'

lIT is specialll ~'

= ~

I (vii), will be

(in

M*)

= I

in

, so we are done.|

Theorem O.

4

There is a (i)

M*

(ii)

if

Proof:

-

be a S o u s l i n

algebra

The next l e m m a p r o v e s Lemma

120

BA

~*

in

1= "[~*[ G

is

Assume

M*

= w2

such that

&

~*

M*-generic

V = M*

lin a l g e b r a

.

~

satisfies

for

~*

Define

a,8~

, the sequence

5-,-~

function < ~2 •

n o w that

"

v .

f(~(v)

tree in and

~

,(v)1 )

quence w h i c h Uv<~2]B v

•

Let

(-)I

~B

~

I ~ ~ v)

is

with

~

3.

gives

(i) is immediate.

llThere is a n o n - s p e c i a l

enumerates

be a b i j e c t i v e a,B < w 2 , a,8

as follows. Souslin

functions

Since

is a sequence of

•

~

Aronszajn

Let

QB

from

we let

~(~

I v <w 2)

We prove

(ii).

> 0 .

•< ~

an A r o n s z a j n ~ ,

the m a i n be the se-

us in this case.

tree]~*

Suppose

of S o u s l i n

] ~ < ~ ) )'

Otherwise

to

holds

for each

value

a(~ v ,QB

~

algebra.

for w h i c h we can apply

lemma VIII.12.

lemma VIII.12

Then

so that for any Sous-

be the inverse

•

~1-dense";

1= SH .

~-,-~

subalgebra

by lemma

a function

lemma,

f

is

(f(~,~) I v < w 2)

V(~V ) o, then we obtain '

This d e f i n e s

~*

M*[G]

be an a r b i t r a r y

a nice

T

f~B(v)o,(V)1)

iteration

(-)o'

and

&

such that for all

a function

~ < w2 with

w2

Let

o(~,~)

algebras If

on

Define

we can let

, then

a function

{X E V ~B) J t!X ~ ~1×~11~ = ~ ] pairing

c.c.c.

Let

~*

Suppose Now

=

-

II(~X)[X ~ ~I ×~I then

X

V (~*)

is one)Ifp * = ~ we can pick

so

~ < w2 X = f(~

, so by the maximum

X E V (~*)

clearly

course,

X E vOB~ )

for some

~ < ~2

fiX is an Aronszajn

1IX is specialICV+1 which

is speciall~*

such that

is a non-special

so that ,T)

-

and if there is a non-special

and (in particular) Pick

121

= ~ .

'

Let

treel~V

Hence,

is a contradiction. = ~

Then

= ~

Aronszajn

principle

, and we are done.

: ~

t r e e l ~ * > @.

fiX ~ ~I × ~II~ ~ = ~ v = ~,~

Then,

, so by construction,

IIX is speciall~* Thus

for

II~ ~ ~I × ~II~* Aronszajn

tree

= ~ , of

IIEvery Aronszajn tree |

Appendix ITERATED In [Jn Jo], Jensen constructible forcing.

and Johnsbr&ten

ing over

Jensen developed L

~(w)

give a generic

construction

of a non-

w-iteration

of Souslin

of an

is a specialisation

of an earlier

in order to show that iterated

must introduce

and because

cumbersome,

FORCING AND

A I3 set of integers by means

Their construction

tion which

der,

SOUSLIN

new reals.

the construction

Souslin

For the convenience

in [Jn Jo]

which

forc-

of the rea-

is (necessarily)

we outline here a simplification

construc-

fairly

just gives the above

result. In

L,

we define

a sequence

(i)

T°

is Souslin;

(ii)

if

bo

is an

define

(iii)

~I of

branch TI

of

in

Lib o]

if

is an

Lib o] - generic branch

define

a subtree

bI

is Souslin

if

(bo,b~,...)

C

of

T2

~I

is Sous-

~I , then we can cano-

in

L[bo,b I]

such that

L[bo,b q] ;

is any such sequence

T°,T I,...,

then

regardless

of branches

L[
by forcing

over

of what we do at limit L~

iteratively

if the method we use to destroy given a branch, As we proceed

Souslin

then the iteration given

x E Tn

such that: tx

such that

of the respec-

7] = L[a]

for some

W°

This shows that, SH

of

such that:

similar~

tive trees a

in

~2

wq-trees

T ° , then we can canonically

Lib o]

(iv)...< (w)

(i)

of

lin in

~2

obtain

L-generic

a subtree

nically

(w)



is a

ht(x)-tree

must

we shall

stages,

destroying

trees results introduce define

if we try to

Souslin

trees,

in their being

new reals.

a subtree

tx

of

T n+1

-

(ii)

-,

x
(iii)

x,y

t x = ty lht

in

Tn ,

-

;

are incomparable

z
then

in

Tn

Tn

z

t x O ty = t z

will consist of

~-sequences

dering, ~ n ' will be the usual one. then

and

is the largest

(unless

T211, .... then

T°I2,

then

of natural numbers,

We define TI12,

T°I1

then

recursive bijection

Stage O.

For each

n E w , set

Tno:

Sta~e 1.

For each

n E ~ , set

T 1 = [(n,i) I i E w ]

for

= O, when

a+2

For each

•

x E ~+1

, i E w , set

{-,-~

first, then

: w xw

TI11,

etc.

~w

.

[] , t(n ) = ~.

n

n E w , set

and the or-

T212,...,

Fix some canonical,

Stage

ht(z)

t At : []) ; x y T n+1 = U tx . xET n

(iv) Each

123

Tn

, t(n,i > = [] .

= [x~
~+2

tx^ : t x U [ y ~ ( ~ i , j > >

&iEw]

+1

'

and

I y e t x & ht(y) = 6

& jE®]. Stage

a ~lim(6).

We first define

T °6 "

that

T°16 E LO,

L~

lal LO = w.

bx

be the

~

ZF-,

< L - least

L

T6O = [Ub x Ix E T°I a] . x E Tna by

x E T na'

Assume now that

define a poset f

f-

dom(f')~

the


i E w .

don(f)

with

°

Set

= ~x

in

&

each T n+1 ~

T°16

L~

1 =

and let

Define

tx

and

~

.

is a branch of n~

&

for

Tnl~+1,

The elements of

ran(f) ~ t x , such that

The ordering is defined by

subset of

[Ub[IxET

x,

be least such that

as follows.

dom(f) ~ w

x E T°16, let

through

T n6 is defined.

(ViEdom(f))(f'(i)~n+qf~))

b~

be least such

For each

~ = ~n+I,6

= ht(f(j)).

L~-generic

Clearly,

t X _c UiEmb ~

Let

~

i,j E dom(f) ~ ht(f(i))

~ = ~o ,6

- generic branch of

tx = Uy< n x t y "

are finite maps

and

Let

Set t

X

. Let

f'

G = Gx

bSl = If(i) I f EG], , i # j - b ~ I b~

i E w]

i

°

8'

be each and

-

Sta~e Let

a+1~

lim(a).

x E T na'

For each

i E w.

each fixed

T n+1

i E ~,

b kx

tx~
L.

Suppose

course) • T°

b°

i ~ j ~

txqi)N

L,

in

~I .

Assume

the smallest :N ~

M

A N (TIa)

L~1,a

in

T°

T°

(necessarily

~ [b°] = w~,

so

~1

Thus

Then

~ < ~1,m"

±s ~ i ~ a l

UxEbltx

a branch of

bo,bq,..,

in

Let

M~

f(x) = the

L~2[b o] be

L~ .

But

T°la+1

~

E L~I

is

bo~EL~q

L~1,a, whence

Thus

is a branch of

a

is countable in

, so

lies in T.

Now,

A = A n (TI~),

Tq , then

~2 =

~I = UxE b t x , o a ~ w.

as follows: Tn-least

Given

y E Tn

is well-defined,

etc.

n E ~,

such that

and that

a = [~n,i~ I (n,i> E b n n T ~] . we show that

Clearly,

L[
x E T n+q , x E ty.

ht(f(x))

It

= ht(x) + q

L[a] _~ L[] .

_c L[a] .

bo

We show that

x.

Conversely,

T

L[bo,b I] , etc.

for some f

in

Let

L~[bo~].

A N (TI~)

bI

a branch of

f

wq-tree

Work i n L ~ . W r i t e

a = ~I o M "

and hence in

if

Note that as

is an arbitrary sequence of branches,

define a function

is easily seen that

Let

TI~+~,

Similarly,

T° , b I

ht(x) ~ q , let

for all

But then

in

L[] = L[a] L,

TIa

But look,

is a Souslin tree in

Suppose now that

L[b~.

of

w(T) = Tla , ~(A) = A N (TI~) , and

L~[bo~a] , hence also in

and we are done •

In

~(w I) = a,

L ~ [ b o ~ ] _c L q,m

~ n (~I~)

Set

L-generic,

is clearly an

is a maximal antichain. A,T E M .

for

will be a Souslin tree

Tq = UxEbotx .

such that

Hence

c U x x -- jE~b~i,j~

by

is a maximal antichain of

uncountable

t

L[b o]

A _c T

L~[bo~a] .

tx~ =

By the genericity of the

L[bo] . We show that, in fact, Tq is Souslin in for

i Ew].

tx~<j > = t x .

Clearly,

is any branch of

~I ~ T 1

is Souslin in

as follows:

(above), we clearly have

and

Define

T na+1 = {x~< i> I x E T an&

are as above.

That completes the construction. in

-

n E w , set

We define

t x U [Ub~i . ,j~ I J E m] , where construction of

q24

-

Working

in

L[a] , define

(i)

Let

<

(ii)

Let

xq = that pair

(iii)

Let

~+q

Let

-

a sequence

each

rl

x~+1

(iv)

= ,

q 2 5



of sets as follows:

n E w •

= f,LX~n+J ) , each

such that

~n,i~

E a,

n E w , whenever

each

~ < wq

n E w.

and each

is defined. x n~ be the unique

ever

n [x~ I ~ < 6 ]

each

n E w,

sucessor

is a branch

when

of

of

[x~I ~ < ~ ] TnI~

which

on

T n~ when-

extends

a < wq , lim(~) , and each

x~,

on

mn _a,

~ < ~, is

defined.

(v)

Let

L y < wq

breaks

down,

be the first point where the above definition and set

But by a simple induction and

b'n = bn

for each

on

n .

b n!

=

6, Thus . .

[xn I ~ < ¥] ( • n E w)Cx~Ebn).'" ~ "
> EL[a]

Hence

y = wT,

, and we are done.

REFERENCES Ba 1

J. Ball:

A Note on the Separabilit~ of an Ordered Space,

Can. Journ. Math. 7 (1955), 548-551. Bu I

J. Baumgartner:

Decompositions and Embeddin G of Trees, Notices

Amer. Math. Soc. 17 (1970), 967. De I

K.J. Devlin:

Aspects of Constructibility,

Springer Lecture

Notes 354 (1973). De 2

K.J. Devlin: 76(1972),

Do I

Note on a Theorem of J. Baum~artner, Fund. Math.

255-260.

C.H. Dowker:

On Countably Paracompact Spaces, Can. Journ. Math.

3 (1951), 29-224. Fe I

U. Felgner:

Models of

ZF

Set Theory, Springer Lecture Notes

223 (1971). Ga Sp

H. Gaifman and E.P. Specker:

Isomorphism Types of Trees,

Proc. Amer. Math. Soc. 15 (196¢), I-7. Je I

T.J. Jech:

Lectures in Set Theory, Springer Lecture Notes 217

(1971). Je 2

T.J. Jech:

Trees, Journal Symb. Logic 36 (1971), 1-14.

Je 3

T.J. Jech:

Automorphisms of w I -trees, Trans. Amer. Math. Soc.

173 (1972), 57-70. Jn I

R.B. Jensen:

Souslin's Hypothesis is Incompatible with

V =L ,

Notices Amer. Math. Soc. 15 (1968), 955Jn 2

R.B. Jensen:

Automorphism Properties of Souslin continua,

Notices Amer. Math. Soc. 16 (1969), 576. Jn Jo

R.B. Jensen and H. Johnsbr&ten: Constructible

Ku I

G. Kurepa:

A~

A New Construction of a Non-

Subset of w , Fund. Math., to appear.

Emsembles ordonn@s et ramifies , Publ. Math. Univ.

Belgrade 4 (1935), 1-138.

-

Ha So

127

-

D.A. Martin and R.H. Solovay:

Internal Cohen Extensions,

Annals of Math. Logic 2 (1970), 143-178. Hi 1

E.W. Miller:

A Note on Souslin's Problem, Amer. Journ. Math. 65

(1943), 673-678. Ru 1

H.E. Rudin:

Countable Paracompactness

and Souslin's Problem,

Can. Journ. Math. 7 (1955), 543-547. Ru2

H.E.

Rudin: 1113-1119.

Souslin's Conjecture,

Ru3

M.E. Rudin:

A Normal Space

X

Amer. Math. Monthly 76 (1969,

for which

X xI

is not normal,

Fund. Math. 73 (1971), 179-186. Sh 1

J.R. Shoenfield: Unramified forcing, in Axiomatic Set Theory, Proceedings of Symposia in Pure Mathematics, Amer. Math. Soc., 357-381.

So Te

R.H. Solovay and S. Tennenbaum:

Iterated Cohen Extensions and

Souslin's Problem, Annals of Math. 94 (1971), 201-245. Su 1

H. Souslin:

ProblSme 3, Fund. Math. 1 (1920), 223.

GLOSSARY

SYMBOL

OF

NOTATION

APPROXIMATE MEANING

PAGE

V

the universe

Va

the a'th hierarchy

dom(f)

domain of the function

f~A

the restriction

of

IXl

the cardinality

of the set

(p(x)

the power set of

level in the cumulative

f

f to A

X,

i.e.

X [Y I Y_cX]

2~

If I f : a ~ 2] or the cardinality this set~ according to context

2
U!3
On

the set of all ordinals

lim(a)

of

is a limit ordinal the rational

and real numbers

(as ordered sets)

CH

the continuum hypothesis

GOH

the generalised

ZF-

the theory

A

A

is a model of

f : A ~==> B

f

is an isomorphism between

A~B

A

is isomorphic

A-~B

A

is an elementary

LST

the language

F"X

the " F - i m a g e "

Def(X)

the set of all "definable"

L

the constructible

L

the a'th level in the constructible hierarchy

2

V =L

the axiom of constructibility

2


the canonical well-ordering

p-qo

ZF

continuum hypothesis without the power set axiom

to

A

and

of

B

B

B submodel

of set theory of

I

X

I subsets of

universe

X

I 2

of

L

2

-

129

-

T,[A],T,[A],V =S[A]

relative

plq,

incompatibility, posets

=iy

constructibility

concepts

incomparability

in

chain condition

2 3, 11

C.C.C.

countable

c.t.m.

countable transitive model

M[G]

the generic

BA(:~)

the boolean algebra determined by the poser

4

the

4

extension of

3 4

M

by

G

4

[p]

n(m) v

~-valued

extension of

the canonical map of

II~oll ll-

N

M

into

N (~)

4

the boolean value o£

5

a name for

5

a

in the forcing language

the forcing relation

5,6

lub (A) gZb(A)

the least upper bound of A

8

the greatest

8

SH £

the Souslin hypothesis

ht(x), ht(~)

the height of

Tc~

the

11

~1~ (or ml~)

UB U~E C T a , for

38

lower bound of A

9

11

~'th

11

x,

level of

T

C c m1

11 11

~IC (or mlC)

<TIC, _

38

Tx

[y~m I x<_y]

11

O.

23

O* <>+

25 25

BA

complete boolean algebra

N.A

Nartin' s axiom

[] HmI

,

(~)

Hmfl

56 62 75 80

-

15o

-

the "closed set forcing" poset

97 I05

INDEX antichain, qq

-dense

poset,

6

antilexicographic automorphism, 40

direct union, 57

Aronszajn tree, ql

direct limit, 58

atomless poset, 3

m - distributed, 63

automorphism of a poset, 32

(~,oo)-distributive,

(e-) basic projection, 69

X - embeddable tree, 15

(a-) branch,

68

11 final section of a poset, 3

cardinal absolute, 6

forcing language, f. relation, 5

chain, 11 countable chain condition

generic branch of a tree, 19

(c.c.c.), 3 - chain condition,

generic extension, 4

3

generic set, 4, 98

closed set, c. interval, 8 closed set of ordinals, 23

-happy,

closed set algebra, 97

height of x

closed set forcing, 97 -

~ - happiness,

70

in T, 11

homogeneous tree, 32

closed poser, 6

homogeneous line, 39

comparable, 11 compatible, 3

initial section of a poset, 3

complete ordered set, 8

interval, 8

conditions, 7

inverse limit, 7J

condensation lemma, 2 constructible hierarchy, c. universe, axiom of constructibility,

iteration lemma, 86 2

cut, 8

killing a Souslin tree, 55

Dedekind completion, 8

~'th level of T ,

dense subset of a poset, 3

length of

densely ordered set, 8

lexicographic automorphism, 40

ql

T , 11

-

132

Martin's axiom, 62

-

tree, 11, 54/55

maximal branch, m. antichain, 11 well-distributed, neat cover, 69 neatly

~ -closed, 68

nice subalgebra, n. embedding, 85 normal tree, 12 normalised boolean universe, 5

open set, o. interval, 8 ordered continuum, 8

product lemma, 7

reversible line, 39 rigid line, 48 rigid poser, 46 root of a tree, 12

separable, 8 Souslin algebra, 52, 82 Souslin hypothesis, 9 Souslin line, 10 Souslin property, 9 Souslin tree, 11 Souslinisation, 82 special Aronszajn tree, 15 splitting poser, 3 standard tree, 39 stationary set, 23

63