Set Theory and Model Theory

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

872 Set Theory and Model Theory Proceedings of an Informal Symposium Held at Bonn, June 1-3, 1979

Edited by R. B. Jensen and A. Prestel

Springer-Verlag Berlin Heidelberg New York 1981

Editors

Ronald Bj6rn Jensen All Souls College Oxford OXl 4AL, England Alexander Prestel Fakult~,t fL~r Mathematik, Universit~t Konstanz Postfach 5560, 7750 Konst&nz, Federal Republic of Germany

AMS Subject Classifications (1980): 03Cxx, 03 Exx

ISBN 3-540-10849-1 Springer-Verlag Berlin Heidelberg New York ISBN 0-387-10849-1 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 "Verwertungsgesellschaft Wort", Munich. © by Springer-Verlag Berlin Heidelberg 1981 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2141/3140-543210

FOREWORD O n the o c c a s i o n informal from

symposium

I.-3.

of J ~ n e

all d e d i c a t e d some

Each traces

Scholz.

concept book

The

directly

ontological too polite

that

o n them. prove

faith.

study

of

on them.

with more of P a r i s We

hope

set

fear,

Heinrich

have

a

today,

'real

rather

set

faith

content' than

its to

to

stronger

non-standard Some

of u s

immediate

the

some

models

will

and

but

provided

among

of

he not

us of

'taking and

be

to s h e d

light

of

able of

logic will

the g r e a t

perhaps For

this

encouraged

however,

a first

were

numbers

even

some

unsolved point

reason those

all e v e n t u a l l y

Recently,

volume

natural

tools

theory).

tried,

rewards.

and Harrington

the

he h a s m a d e around

turned

to

the b e a u t i f u l

vindication

of h i s

follow.

are grateful

to P r o f e s s o r his

association

the

should

of

to this

importance

Hasenjaeger,

(and in t h e p r o c e s s

in s o m e

on the

theorists of

that

undecidability

theory

contributions

that

direction

he h a s

that more

the

- his humor, our

deep

is an e a g e r

- a conviction

insistence

the primacy

We

made

contri-

from Hasenjaeger

in l o g i c

however,

qualities which

bears

platonists

content,

learned

the

of P r o f e s s o r

teacher,

must

its

theoretical

accuse

mathematical

of n u m b e r

to work

theorem

We

so, w o u l d

to their s o l u t i o n

fields

of

the e l e m e n t a r y

a lifelong him

as c e n t r a l

He recognizes

a major

of us

to H a s e n j a e g e r ' s

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

questions way

to d o

of

are

place

assistant.

call ourselves

by

took

witness.

questions.

road'.

All

himself

by his

an

at B o n n

proceedings

to m a t h e m a t i c s ,

judged All

held

co-workers his

Hasenjaeger

advocated

be

of m o d e l

bears

or

that mathematics

value.

preponderance

is r e l a t e d

view

should

or esthetic

the

approach

birthday

the meeting

extended.

or a n o t h e r

influence.

sense

a problem

this

easy

own

and

was

in t h e s e

Since

students

time

a l l of u s w o u l d

a profound

that

regard

day

in h i s

not

revised

former

60-th

theory

published

Hasenjaeger.

at o n e

of t h e p l a t o n i c

difficulty

feels

of us,

Though

we retain

which

are

Each was

Hasenjaeger's and model

The papers

been

of H a s e n j a e g e r ' s

proponent

- and

have

to t h i s v o l u m e

Hasenjaeger.

the

1979.

theory

to P r o f e s s o r

of t h e p a p e r s

butors

of G i s b e r t

on set

Hasenjaeger

tolerance, with

him

his

for

patient

the many

human

attentiveness

so p l e a s a n t . R. B. A.

Jensen

Prestel

-

TABLE

KEITH

CONTENTS

J. D E V L I N Morass-like

HANS-DIETER

DONDER, Some

SABINE

constructions

of

~2-trees

in

L

DONDER

Coarse

H.D.

OF

morasses

R.B.

in

JENSEN,

applications

37

L

B.

KOPPELBERG

of the c o r e

55

model

KOPPELBERG A lattice boolean

ALEXANDER

o n the

isomorphism

types

of

complete 98

PRESTEL

Pseudo

TASSILO

structure

algebras

real

closed

127

fields

VON DER TWER

Some

remarks

arithmetic

on the m a t h e m a t i c a l

found

by Paris

incompleteness

and Harrington

of P e a n o ' s 157

Morass-Like

Constructions

of ~ 2 - T r e e s

in L

by

Keith

J. D e y l i n

(Lancaster,

U.K.)

Abstract

Using the

a simplified

construction

Kurepa

~2-trees

techniques lin

~2-tree

principal, quent not

were

version

of a m o r a s s ,

expository

we

fine

originally

developed

uniform

that

the

filter

R. L a v e r

result

in nature.

reduced

on ~,

Rather

theory

to c o n s t r u c t

systems

power

tree

a much the

required Souslin

of c o u n t a b l e

by us in o r d e r

is a K u r e p a

obtained

here.

structure

show how

of d i r e c t e d

T such

this

the

as limits

to o u r work,

include

of

T°/D,

~2-tree. simpler

present

where

account

These

a Sous-

D is a non-

However, proof,

and

trees.

to o b t a i n

for

subse-

so we

do

is l a r g e l y

§

We w o r k ventions. nals IXI

are

initials

Most

the

refer

require

Our of

the

here

von Neumann

cardinality

of the

set X.

results

I] is n o t

require

importance

[De

and

concardi-

ordinals;

It t u r n s

Section out

(in p a r t i c u l a r ,

2 provides

that we morass

to k n o w

V : L,

do n o t

axiom

something

(M7)

of the

(in L). trees

o f the v a r i o u s of w h a t

is a p o s e t

ordinals

and

of constructibility,

b u t we d o n e e d

concerning

resum6

need.

notation

to d e n o t e

I] for d e t a i l s .

o f the m o r a s s

of the m o r a s s

etc.

the a x i o m

theory we

required),

terminology

A tree

are

the u s u a l

e,B,y,

full power

a quick

adopt

We use

to o u r m o n o g r a p h

construction

and

ordinals

o f the m o r a s s

the

[De

theory

ordinals.

of o u r

an o u t l i n e

from

in ZFC s e t

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

denotes

and we

I. P r e l i m i n a r i e s

is

fairly

definitions

standard,

to this

but

in v i e w

paper we

present

w e need.

~ =
~T > s u c h

that

the s e t ~ = {y 6 T I y


A

is w e l l - o r d e r e d is c a l l e d is

by


the heigh_tt o f x in ~, d e n o t e d

the s e t T e = {x 6 T l h t ( x )

that

T 1 = $, a n d

dinal ~e

< such

denotes

is d e n o t e d

that

(Ve)

subset

T be

An

A tree (i)

between

a tree.

~ is n o r m a l

if e < ~ < ht(T)

a tree

whose

antichain

For e < ht(~),

and

(ii) one

The

(under

e'th

level

we

order-type

set

set Tie , though

its


I such

least

Tie = ~ < e

carTS.

in g e n e r a l

we

domain.

is e

o f T is a p a i r w i s e

least

o f ~ is the

of T is a m a x i m a l , (under

linearly

ordered


incomparable

subset

an

of T.

iff:

every

element

of T

•

sions

The w i d t h

of T to the

A branch

o f ~. A b r a n c h

e-branch.

(ITel < ~).

by ht(x).

of x

The _height o f ~T is the

by ht(~).

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

do n o t d i s t i n g u i s h Let

= a}.

order-type

has

infinitely

many exten-

e

on TB; if a < ht(~)

extension

is a l i m i t

on T

. e

ordinal,

any a-branch

of Tie h a s

at m o s t

An ~ 2 - t r e e

is a n o r m a l

An 8 2 - t r e e

is A r o n s z a j n

is e a s y shown

to c o n s t r u c t

that

the

Assuming Laver seen

V = L,

([La]) that

An ~ 2 - t r e e Assuming Silver

V = L, ([Si])

The a tree

to c o n s t r u c t

it is e a s y

has

shown

said

initial

it has

iff

a Souslin

sufficient

it has

CH

definitions

CH,

Silver

(See

has

[Mi].)

of c a r d i n a l i t y ~2-tree.

here.

(See

part

to be t h i n

~2" [De

I].)

It is e a s i l y

at l e a s t

is not

are

~2 m a n y

Kurepa

~2-branches.

~2-trees.

sufficient

for

non-standard.

(See

[De

] ( 3 a 6A) (x _
of T d e t e r m i n e d

If A is a s u b s e t

r

by A. A s u b s e t

U of

if (V x 6U) (3 y 6 T) (x <

y T

& y~U)

.

I].)

this.

set {x6T

it

is A r o n s z a j n .

to c o n s t r u c t

that

here.

~2"

Assuming

[Je].).

no a n t i c h a i n

CH is not

H2-tree

is K u r e p 9

A = the

that

Souslin

(See

of CH is n e c e s s a r y

it is e a s y shown

~2 and w i d t h

no ~ 2 - b r a n c h .

H2-tree.

iff

following

T, we

it has

is S o u s l i n

has

every

iff

~ of h e i g h t

an A r o n s z a j n

assumption

An ~ 2 - t r e e

tree

a tree

T is

of

§ 2.

The other

concept

results,

jecture

of

he

of model

Some

a morass

morasses

theory

under

[De

I] w e

tunately,

for

our

present

the

definition

of

not

adequate

native tive but

us.

construction

we

give

of

committing

of

here

We

be

fact

to p r i n t

assume

the

as

there.

is

self-contained.

The

in

[De

13).

Let

us

(i) ed

I]

either

for

an

in

this

is n o t

full volume This

part

of

[De

v < ~2

suitable

admissible

or else

Jv I= " t h e r e

is e x a c t l y

one

the We

material,

V = L

way the

"new"

I].

ignore

assume

some

the

the from

in

of

fine

alter-

book

[De

purpose

construction.

the

same

the

following

theory

notation account

of

the

morass

on.

iff: {T 6 vlT

is

admissible

is

unbound-

and uncountable

cardinal." J

Let

S be

the

set

of

all

suitable

A

=

{~

s~

:

{vcsI~

ordinals.

Let

and

The

for

~ 6 A,

following

Ires},

set

lemma i s e a s i l y

v

=

2],

completion,

use

now

is

alterna-

structure

construction

I]

an

additional

morass

13 of

to h a v e

This

off

Con-

Unfor-

[De

forthcoming

serves

with

7 of

can

is

in L.

fragment

of a m o r a s s . in the

Cardinals

simply

given

a proper

Amongst

In C h a p t e r

enough

(standard)

reader

V : L.

it

familiar

6 and

Two

a morass

construction

is

Gap-2

Jensen.

construct

require

least

We

the

Ronald

construction

require.

this

ordinal

the

v is

in v;

(ii)

in C h a p t e r s

(Chapter

call

we

reader

described

Except

we

by

assumption to

given

at

as

the

and

that

theory

to p r o v e

purposes,

Rather

will

Theor~

introduced

how

superior)

the

the

show

a morass,

(and v a s t l y

in v i e w

was

used

our monograph

for

Morass

~].

proved:

For

~ 6 S,

set

~v

: el

Lemma

2. I

(i)

w I 6 A;

(ii)

A N w I is u n b o u n d e d

(iii)

Swl

(iv)

Sd

(v)

if ~, e 6 A a n d [ < ~, t h e n m a x

(vi)

for any ~ 6 S

is c l o s e d is c l o s e d

in el;

and unbounded in s u p ( S

'

S~

in ~2;

); (S[) < ~;

R 4 is u n i f o r m l y

El-definable

from

in J4. Q

4

4

L e t 4 be any J8

iff

there

a bounded J8

En-formula r_~ular

of s e t

an

n(4)

be

u. We

o v e r J6'

over

integer

J6"

n 2 1 such

o v e r J6' w h i c h

over

J8 b y m e a n s

f r o m J6).

maps

over

of a

Otherwise

~ is

respectively.

Hence

Let

over

say 4 is E - s i n g u l a r n

is d e f i n a b l e

~ is s i n g u l a r .

t h a t m is s i n g u l a r least

into

s a y ~ is s i n g u l a r

as a c l a s s

(with p a r a m e t e r s

J~ o r E n - r e g u l a r

such

6 ~ 4, w e

f, d e f i n a b l e

f which

theory

L e t 4 e S. T h e n

the

For

of 4 cofinally

is s u c h

over

ordinal.

is a f u n c t i o n

subset

if t h e r e

limit

there

B(4) that

be

is

the

(by V = L) least

such

u is E n - s i n g u l a r

a B ~ 4 6. L e t

over

J6(~)"

Define

n(~)-i

p(4)

Since

: P8(4)

4 is E n ( ~ ) _ l - r e g u l a r

A(4)

'

over

JB(~)'

fact,

p(4) < 4 ,

then

since

Using

the

that

~4 = ~i 4 in c a s e

cases, since

fact

Jp(4) ~

B(T) < ~ ,

~

is the which

"~

4

The

definition

cases Let

is r e g u l a r " .

largest implies

We now define,

p(4) < B(4), j

that

we have

we have

J

p ( 4 ) > 4,

Notice

cardinal

n(4)-i = AB(4)

also

in J4'

p(4)

p(4)->4. b

T is n o t

if T 6 S

that

~4

a cardinal

A 4,

~ 6 S,

o f 10(4) d e p e n d s

S - P.

then

in J4'

upon

a certain

parameter

the n a t u r e

o f 4,

p(4) 6 J

and

there

to c o n s i d e r .

R=

in all

so

p(T)<~.

for e a c h

P = {4 £ Sin(4)

in

"4 is r e g u l a r " .

it f o l l o w s

that

If,

= 1 & B(~)

is a s u c c e s s o r

ordinal},

pC4)" are

two

In case

Lemma For

v 6 P,

be that

y such

that

y + 1 = B(~).

2.2

n(~) any ~ 6 S, PB (~) < evL e t n = n(v),

Proof: that

f[~]

we m u s t f = fn

have

f { J

in ~.

. But

e many

of A~

onto

k is E n ( J 8) proves

the

I.

that

sets

Either

Hence

P~

g such

~ : °~v . Let

g[~]

then by

p~_<~,

= JB"

for each

and k[a]

: ~.

and

Now

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

f(~)

~ is r e g u l a r i n

such

JB'

so

it f o l l o w s

that

and

let

there

is a

a partition

f< be

the

<j-least

Set k = U{f~I~ 6 dom(f)}.

is E n ( J B) and

gok[~]

of

= JB"

Then

This

lemma. D

, B = ~(~)), n = n(v),

Lemma

2.3

There

is a q 6 J

such

p = p(v),

that every

element

Y

parameters By

in J lemma

El-definable there

map

~ 6 P.

Set ~ = ~

Proof:

a E n (JB)

amenability,

e in J~,

gok

be

let < A ~ I ~ < ~> be

~ edom(f).

Hence

f c~xd

B = v or else

if p ~ > m,

(~ × m ) 6 J n ~ J s .

into

Case

~ = ~(~)'

is c o f i n a l

Z n ( J 8) m a p

map

let 7(m)

A = A(~),

of J 7

7 = 7(~)-

is J - d e f i n a b l e 7

from

U {q}. 2.2

in JB

there

is a p 6 J

from parameters

is a r u d i m e n t a r y

function

such

that

every

element

in J~ U {p}.

Since

J~ = rud(Jy)

f and

an e l e m e n t

q of J

such

of JB is

that

Y p : f(J

,q). We s h o w Y Let x 6 Jy. Then

unique

x in J~ s u c h

JB = m~j <~

sm(jy), %

we

that

for some that can

find

Jy,

p = f(Jy,q),

with

parameters

this

above.

E ° formula

(3y6 J~)

( 3 y 6 sm(jy)) Since

q is as

~ and some

~ 6 Ja,

[JB ~ ~ ( y , ~ , p , x ) ] .

Since

an m < w such

[sm(6y)

that

b ~ (y,~,p,x) ].

can be w r i t t e n

x,z,q.

x is the

Hence x i s

as a f i r s t - o r d e r

J]'-definable

statement

from ~,q.D

in

Let q = q(v) J

is

from

parameters

q 6 JT such

that

every

element

in e U {q}.

Y

Set p(m)

By

= .

2.

Case

That

completes

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

in Case

I.

u 6 R.

lemma

element <J

<j-least

-definable

J

Y

be the

2.2,

let q(m)

of Jp(m)

be the

<j-least

is E l - d e f i n a b l e

q 6 Jp(~)

from parameters

such

that

every

in a m U {q}

in

p(~) ,A(m)>. Set ]
>,

[

,

This The

completes following

Lemma

the d e f i n i t i o n two

lemmas

are

if Y < p(~),

if ~ : p(V).

in Case

2.

clear:

2.4 J

<<8(~) ,P(q) ,A(~) ,p(n)>

I q 6 S

n v> is u n i f o r m l Y ~ l

~ 6 S.[3

Lemma

2.5

L e t ~,~ 6 S,

and

suppose

o: <Jp(v) ,A(v)> is such Then

that

o(p(~))

o is u n i q u e l y

(i)

~ 6 P ++ T 6 P;

(ii)

~(~

(iii)

V < p(m)

(iv)

~ < 0(~)

(v)

~6P

(vi)

o(q(v))

) = ~ ;

÷

++ T
o(y(~))

p(~) ,A(T)>

= p(T).

determined

÷ 0(~)


= T; :

= q(T). [3

y(T);

by o ~ e

• Moreover:

~({~

}) for

of

Lemma Let

2.6

m 6 S, ~ _< p(v) , A _ c J ~ ,

o: where

p(~) 6 r a n ( o ) .

that

~ = p(~);

Proof:

Notice

and

<J[,A>

Then

A(~)

= #, so A = ~,

each

~ < ~,

~I

there

A : h(~). first

suppose

<Jp(m),A(~)>,

is

a

(necessarily

Moreover,

o(p(5))

that

<J~,A>

amenable.

and

if m 6 R,

is

<Jp(v),A(v)>

then

~

3x

unique)

5 6 S such

= p(m). For

llm(p(v)),

if

~ 6 P,

so

lim

then

(~),

and

for

(x = A(~) N J o ( q ) ) ,

SO

<J~,A> Set

~ = e

Case Set J

b

Bx

(x = A N J

).

, ~ = 8(~) , n = n(~) , p = p(v) , A = A(m) , p = p(~) , q = q(u) .

I. ~ 6 P y = y(v).

. Since

Thus

B = P = Y + i, A = A = ¢,

p 6 ran(O),

we

have

q,y,~,~

6 ran(o).

and

~ is r e g u l a r

Let

over

q = -l(q),

Y = d-l(y), and

q 6 ran

Claim

A:

Proof: ~(~)

=

~.

B:

Proof:

Claim

5 6 S and

from

~ = o I J ~.

Thus

~:

J~

Jy

= ~. immediately

5 is r e g u l a r ~:

J~

q is the

Let

q,~.

Let

from

~:

J~

41 Jp

and

o(~)

= ~,

[]

is J ~ - d e f i n a b l e

Proof:

~

follows

Since

C:

~ = -i(~).

(~).

This

Claim

J~

~ = o-i(~),

x 6 J~. Set

over

~ Jy

and

<j-least from Then

s = <~>.

J~. ~ = ~ + v = y and

element

parameters ~(x) 6 Jy, Let

~ be

of J~

such

v ~ y ÷ ~(5)

that

every

= ~.~

element

of

in ~ U {2}so for

a formula

some of

~ 6 e, ~(x)

is J - d e f i n a b l e

set

such

theory

that

(i)

J

~ Vz By Vy'

[y'

: y ÷÷

%(y',z,q)];

Y (ii)

J

~ Vz Vy

[~(y,z,q)

÷

(B~)

(z = <~>)];

Y (iii)

(Vy6 J

) [y = ~(x)

+÷ J

Y Let

t be

the

<j-least

~ %(y,s,q)]. Y

elemen{

of J

such

that

J

~ ~(~(x),t,q). Then Y t is J - d e f i n a b l e f r o m ~(x) ,q. But ~(x) ,q 6 r a n ( ~ ) < J . H e n c e Y Y t6ran(~). By c h o i c e of t,t _<js , so t 6 J . Thus t = <~> for some Y

~6~. By

(i) above, (Vy 6 J ) 7 --i q

Applying

and

[ = o

7

u

$ 6 [,

this m e a n s

~(x)

++ J

~ %(y,t,q)]. Y

----1

setting

(Vy6 J:) Since

[y =

(t) = <~>,

we get

[y : x ÷÷ J- b ~ ( y , t , q ) ] . Y

that x

is J - - d e f i n a b l e Y

from

parameters

in

{~}. Hence

U {q}. and

We

let q'

every show

element that

<j p h a v e

of J- is J - - d e f i n a b l e Y Y q is < j - m i n i m a l w i t h this

property.

this

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

property

also.

are $ C ~ such that { is J--definable Y

_

and s e t t i n g from

~,q'.

Hence

Then,

from ~,~'.

w

q' : o(q

) ' ~ = $(~),

every

we h a v e

q'

Applying

in

Suppose

not,

there

~: J-< J

Y Y <j q and q is J y - d e f i n a b l e

is J - d e f i n a b l e f r o m p a r a m e t e r s Y Y in ~ U {q'}, s i n c e in any J - d e f i n i t i o n of an e l e m e n t f r o m p a r a m e t e r s Y in [ U {q} we can r e p l a c e q by its d e f i n i t i o n from ~,q . This c o n t r a dicts

the d e f i n i t i o n

Claim

D:

Proof:

~ is not

For Xm =

of J

of q - D

El-regular

e a c h m6~, {x 6 J~

element

from parameters

over

J~l"

set

I x is E m + l - d e f i n a b l e

from parameters

in ~ U {q} in

J-}. Y Then

X m "~m J~

is the

largest

and

there

c~rdinal

is a J--definabley in J~

and

map

~ c X m "(i J~'

of ~ onto

Xm.

Since

Xm D ~ is t r a n s i t i v e ,

so

10

set

~

= X

m

map of

m

n ~.

e onto

Since

~m'

we

~ is r e g u l a r have

~ m < ~"

s u pm_ ~< " ~

= ~.

Clearly,

Claim

8(~)

: ~ + i, n(~)

E:

Proof:

By

Claim

F:

Proof:

claims

q([)

By

: q and

completes

Case

2. ~ £ R.

By t h e

unique

B A @ such

°:

J~ ~n

Claim

J~

G:

Proof: o(])

such

so

H:

Proof:

find

now

two

into using

So by

<J m<~

X

= J-,

so we

m

are

: ~ + i, y(~)

claims

= p(v).

lemma

Set

] = o-i(~)

theory

~

~-

= [.

not.

cases

to

so

done.

[]

: ~.

E,

p(])

=

D

in C a s e

([De

A and

].

if ~ < p 1],

A = Ag n-l,

page

and

and 100)

set

5 = ~ if

there

is

a

an e m b e d d i n g

~.

~ = ~ and J]<J

O:

J~


. Moreover,

o(~)

And = ~.

if ~ < p, The

claim

then follows

Then,

o:

J~ is

map

over since

f such

J~. ~ is that

the f[[]

largest is

cardinal

cofinal

in ~.

in J~, There

we are

consider.

first

v, w h i c h G~del's

p(~)

the

that

a En_l(J~)

applying

P,

: q.

o(p(~))

of

~ is Z n _ l - r e g u l a r

Suppose by

C,

a J--definable y

m

Suppose

can

q(~)

p = p n-l,

(oIJ~):

immediately,

E,

that

If v = p , t h e n

claim

is

: p(v).

structure

~ 6 S and

= ~,

Claim

fine

by

there

[]

~ = o-i(~).

~ = p.

J- and y

is E l ( a n,+ ly)

: i, ~ 6

o(p([))

proof

But

I m < w>

implies

the

q = o-l(q),

D.

C and

which

That

Set

B and

claims

= ,

<[m

over

that

~ < p. T h u s


pairing

see

that

since

function

~ < @ and

we

~(f)

~ is can

o(~)

maps

a subset

regular code

= ~.

f as

in J~.

If f £ J~,

then

of e c o f i n a l l y Thus

f { J~.

By

a E n _ l ( J ~)

subset

of

11

[, now,

to c o n c l u d e

that

~)(~) D E n _ l ( J ~) { J [ .

Thus

p n-l<~,

contrary

n-i to p~ = p>~. Suppose partition

now that u = p. Thus ~ = ~. In J~,

of [ into [ m a n y

let k ~ 6 J~ be the < j - l e a s t Clearly,

k is a En_l(J{)

= ~ = p~-i Claim

sets map

of A~ onto

function

and c < ~, so this

I: q is the < j - l e a s t

J~ is E l - d e f i n a b l e

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

such

from p a r a m e t e r s

~ 6 dom(f),

Set k = U { k ~ l ~ 6

that k[~]

of J~ such

I ~ <~> be a

~. For each

f(~).

is impossible.

element

let < A ~

dom(f)}.

= [. But

[] that e v e r y

element

of

in [ U {q} in <J~,A>.

Proof:

Let x 6 J-. Then o ( x ) 6 J , so for some ~ 6 e, o(x) is E l - d e f i n a b l e P P from q,~ in <J ,A>. Set s = <~>. Let % be a E o - f O r m u l a such that: P (i)

<J

(ii)

<J

(iii)

( V y 6 J ) [y = o ( x ) P

P P

,A> b Vz 3y Vy'

[y' = y ÷÷ 3 u % ( u , y ' , z , q ) ] ;

,A> b Vz V y [ 3 u # ( u , y , z , q ) ÷÷

<J

P

+

(~)

,A> ~

(z = <~>)];

Hu%(u,y,s,q)].

Let <~ be the l e x i c o g r a p h i c o r d e r i n g on L × L i n d u c e d by <j. C l e a r l y J L <~ is A 1 and u n i f o r m l y £ 1 p for all limit p> O. Let be the <~-least

pair

such

that

<Jp,A> Then < t , U o >

b ~(Uo,~(x),t,q).

is E l - d e f i n a b l e

a(x),q 6 ran(d)

~ i < J p ' A > ' t ' U o 6 ran(o).

t = <~> for some ~ 6 a. By (Vy 6 Jp) Applying

-i,

The m i n i m a l i t y

So as

t ~ j s, t 6 Je.

(i) above,

~ = o-l(t),

[y = x ÷+ <J~,A>

x is E l - d e f i n a b l e

in <Jp,A>. Since

[y = d(x\) ÷÷ <Jp,A>

and s e t t i n g

(Vy~J~) Hence

from o ( x ) , q

we have = 3u¢(u,y,[,q)].

from p a r a m e t e r s

of q is p r o v e d

b 3u¢(u,y,t,q)].

in [ U {q} in <J~,A>.

just as in C l a i m C.

[]

Thus

12

Claim

]

J:

Proof:

By

is n o t Claim

J-

In-regular

"(w × (J- x { q } ) ) .

P,A

In particular,

there

Since

~ = p~-l,

Claim

K:

is

{ = B(]), claims

n = n(~),

By

lim(B)

+

lim(~).)

[]

L:

q = q(~)

and

Proof:

By

a II(<J~,A>)

~ : A ~n - i , t h i s

Proof:

Claim

H and

claims

J.

so o ( p ( ~ ) )

o(~)

:~.

then

That

=

completes

Lemma

Let

p(~).

5 6 S[,

such

Let

~'

that

Case

the

[ onto

~.

~ : p(~), notice

A : A(]). that

~:

JB

~i JB'

so

: p(~). If

: p(v)

~ : ~ and

proof

of

v < p,

by

p(~)

lemma

p(C),

and

suppose

<Jp(~) ,A([)> ~i

~[[].

then

claims

v < p and we

G and

= ,

the so

fact

again

have that we

get

2.6.

] is

a limit

point

of S-.e L e t

<Jp(~),A(v)>

: p(u).

Then

v'e

<Jp(~) ,A(~)>

c~J-cd' v -

Set

a < a,

o(p(~))

= sup

that

Proof: =

of

[]

~ 6 S,

o':

such

a subset

2.7

o:

be

map

q : q(~).

= ,

from

is I n (S~).

~ 6 R,

o(p(~))

I a n d K,

If u = p,

map

(For ~ 6 R,

p(5)

o(p(~))

J~.

I,

: h-

P

over

S

~i

o'(p(~))

and

and

there

is

an e m b e d d i n g

<Jp(~') ,A(~')>

: p(m').

B : B(~) , n = n(v) , p = p(v) , A = A(m) , q = q(~) , p : p(~) , ~

= A(~),

I. ~ 6 P.

Set

~

:

q(]), ~ : p(]).

y = y(v) .

13

For each m £ ~, set X m = {x 6 J

Y

I x is E m + l - d e f i n a b l e

from parameters

in ~ U {q} in

J } Y Thus X m ~ m

Jy.

S i n c e @ is the l a r g e s t

X m @ ~ is t r a n s i t i v e , a Jy-definable

is E l - d e f i n a b l e

t_] X = J¥, m<~ m

f r o m p in Jp,

so as v is r e g u l a r so S U P m < ~ m

Jy'

over Jy, m m < ~ .

= ~. But for each m,

m

so ~m 6 ran(o).

m. Thus 9' = ~ and the l e m m a is t r i v i a l l y

2. ~ 6 R. JS~ n ~ is E 1 ~({~})

in Jm and ~c_ X m < i

so let Um = X m N ~. T h e r e is, by its d e f i n i t i o n ,

m a p of e o n t o Xm,

B u t by c h o i c e of q,

cardinal

Hence

oil]

m

is c o f i n a l

in

valid.

Case

as

J and S~ N ~ is ~I ~({~})

(°IJ~) : J~ ~l J~ and o(~)

= ~, ~' = sup(S

c l o s e d in sup(So) , we h a v e ~' 6 S Let ~ = sup(o[~]),

~ = An J

by the same d e f i n i t i o n . A ~').

So,

S i n c e S~ is

I

. S i n c e ran(u) c- J

~

, p, ~ £ J q . By

E 0-

absoluteness,

o: < J ~ ' A > ~ O So,

as o is c o f i n a l

<J

,A>.

in q,

Set

X = hn, ~ "(~ x <Ja × {q}}).

Let

~: <Jy,B> ~ <X,~A X>.

Thus

~: < J y , B > 4 1 <Jn,A>.

%

C l a i m A: ran(u) c X . Proof:

L e t x 6 ran(6).

T h e n x is E l - d e f i n a b l e

from parameters

U {q} in <J

in

,A>. Let x = o-l(x). A n a r g u m e n t as in c l a i m I of P l e m m a 2.6 shows t h a t x is E l - d e f i n a b l e f r o m p a r a m e t e r s in ~ U {q} in <J~,A>.

So for some i 6 w, z 6 J~, x = h ~ , ~ ( i , < z , q > ) .

Applying

14

e:

<J~,A>

x

= h


<J~'A>

9(i,). A-

Claim

B:

Proof:

xn

Let

: h

and

setting

Hence

x 6 X.

z = o(z) , w e

get

[]

v : ~' ~ 6 X ng.

Then

,~(i,).

= hT,A~n J

for

Since

(i,)

some

lim(~)

. Since

z 6 J

there

and is

somc

i 6 w,

a T < u with

n = sup(e[~])

we

can

pick

here

T

so

that

T

= e(~)

T

for

some

~ < ~.

@ = sup

=

Set

[in h

T

~ ,ANJ

[~ N h-

sup

"(w × (J

× {q}))],

,,(~× <J[× {~}>)3.

-

T,ANJ-

~u

Now,

AN

= AN

J

J

N J

= AN

J

T

T

, so

h

~ T,ANJ

6 J

@

by

amenability.

So

as

T

< ~ and

~

is

regular

in J

, @ < m.

Similarly

@ < ~.

But

clearly,

P e(~)

= @.

Hence

< @ = ~a([)

Thus Now

XA

< sup(e[[])

: ~'

~ c v' .

let

~ 6 ~'.

For

some

[ < ~,

~ < 6 = e([).

~ 6 j5 ' f:

~ o~to

~.

Since

(gIJ~) : J~ ~ O

f:

6.

by

claim

A,

Thus

a o~to ~' c X N Now,

ran(o)

But ~.

suffices By

~1 <J

ran(o)

B,

<J~,A>

-i~,

now

to

Eo-absoluteness ~:

as

f = e(f)6

~cX,

J

is

, and

~ = f[e] c X .

,A>, X M1 <J ,A>, and r a n ( o ) ~ X .

<1 < x , ~ n x>. s o ,

claim

So

~ < ~ there

Hence

D

e':

By

f 6 X.

J~'

Since

41

setting

that y=

,

<Jy,B>

= ~ 1o e , we h a v e

<Jy,B>.

= idly' , so prove

e'

Hence

4 0 <Jp,A>.

e' IJ~ p(~'),

= idIJs, B = A(~'),

giving and

eIJ~ce' -l(p)

It = p(v,).

~ 6 X.

15

So by

the

fine

structure

theory

n-i Y = PR'~ , B = An-16' ' a n d q,

= -l(q)

,

p,

there

a mapping

z:

-i

-i

=
= e,

Suppose

= .

By

the

-l(q)

Claim

C:

Just

Claim

D:

Then

Js, < n

is n o t

claim

So,

~:

<J

v'

~'.

Claim

E:

Proof:

Claim

By

F:

Proof:

But

8'

2.6

lemma

~

such

such that

that w c_ ~.

Set

~'

q'

over

Js'"

,A> a n d × {q'})).

= 6(~'),

n : n(v'),

C and

D.

Hence

as

.

Hence

= o(~')

it

suffices

to

show

J6'"

over

41 < J

: v'.

= q(v').

In-regular

: v' n h q , ~

p =

-l(p)

2.6.

~'

(v)

= 7-

that

that

-i

Hence

v' < y.

[]

"(~ × ( J

~' c X Thus

x {q})).

= ran(~), there

n-I 871 y = PS' ' B = A , so this

we

is

map

have

a II(<Jy,B>) is E n

~' 6 R, y : p ( ~ ' ) ,

(J6

,).

map

of

[]

B = A(~').

[]

= q(~').

"(~ x ( J

parameters

lemma

B,

lemma

"(~ × ( J

claims

q'

B,

B,

to s h o w

i.e.

claim

H in

By definition,

= h¥, B

from

The

,B>

= ~' n hy, B

onto

Jy

y

J6

6'

p =.

p = ,

claim

in Claim

By

by

e>,

Moreover,

is I n _ l - r e g u l a r

Proof: as

using

in o r d e r

as

~'

But

[ : ~ and

= q(Y'),

~'

)>.

= .

And,

above,

that

, -i (~

p'

~ = p.

p'

Proof:

(v)

a unique

~ -i ( p ) .

=

Suppose v < @. T h e n ~ < @ a n d p = < q , ~ p'

is

X = h

× {q'})).

in a U [q'}

completes

is p r o v e d .

the

[]

,~ "(~ × (J~ × { q } ) ) . Hence

every

in < J y , B > .

proof,

u

member

And

an

So, of J y

argument

applying

~

-i

,

is E l - d e f i n a b l e as

in C l a i m

C in

18

Suppose iff for

now

that

v is a limit

all T o - f O r m u l a s

¢(Vo,Vl)

ordinal

and X c J

of set theory,

. We w r i t e

with

X 4Q J

parameters

from

X,

X ~

(¥~)

Clearly,

(B8 > ~)

if X
unbounded

in v,

We w r i t e Define

ole

, then

then

~: J~ ~ Q J~

Jv ~

(V~)

Conversely,

iff ran(T)

~: < J p ( m ) , A ( ~ ) > ~ = p(T)

~Q

most

one ~ y

bedding J

P(~

~:

= h

ran(o) mined

It s u f f i c e s

iff e

1 <J

such

,A(T)> (o~J):

J

~

p(~),A(T)

< ~

and

T

there

is

that

4Q J T

partial

to show

that

= ~. Let

<Jp(v) ,A(v)>

0(Y) ,A(~)

= h by

with

and X N v is

J~.

~ on S by ~ T

and

~ is a w e l l - f o u n d e d

is a tree.

$(8,J8).

if X
'

Clearly,

( B S > e)

.

relation

= idI~ ~, o(p(v))

iff

X ~i J~"

X ~Q J

a binary

an e m b e d d i n g

$.(8,J 8)

ordering if ~ 6 S

~ ,

~

~i <J0(T) 'A(T)>

"(~ × ( J - × {p(~)})),

p(~)

2.5,

the m a p

show

and [ < e,

that

there

= [. T h e n

there

as above.

Now,

is at

is a E l - e m -

so

"(w × (J- × {p(~)})).

T a n d [. H e n c e

on S. We

Thus

is c o m p l e t e l y

ran(0)

is e n t i r e l y

determined

by

deter-

~ and ~. H e n c e

so is ~.

By

lemma

if v ~ T ,

so we m a y

denote

{0 T I ~ T }

and

Lemma

if by ~ T

{~ T I v ~ T }

o which

. Set are

~ ~r :

testifies

this

fact

(a ~T Iv) U {}.

is u n i q u e ,

The

systems

commutative.

2.8

Let ~ T .

Then

maps

S

~ (v+l)

into

S

N(T+I)

in an o r d e r - p r e s e r v -

T ing f a s h i o n (i)

such

that:

if y = m i n ( S

) , then

z T(y)

= min(Se

); T

(ii)

if y i m m e d i a t e l y

succeeds

6 in S

n (~+i) , then V

immediately

succeeds

~

(6)

in S

N (T+l); T

~

~T

(y)

17 (iii)

if y is a l i m i t

point

of S

point

of S

A (m+l),

then

z

(~) is a limit

~ (~+1). T

Proof:

This

following We wish

follows

case.

trivially

Suppose

to s h o w

that

from

lemma

~ is a limit

T : z

(~)

2.1

point

is a limit

(vi),

except

of S

and

point

of S

in the

that

~ = p(~).

. This

follows

T

easily

from t h e

fact

that

where

the Q - e m b e d d i n g

Lemma

2.9

T ~T,

Let

~ 6 S~_~

(g

~J ) : J

condition

4Q J ~ .

This is

is r e q u i r e d .

T, ~ : ~--T~(~). T h e n

the only point

[]

~ ~,

~-

I~ = ~-~T I~' and

g-

(A(~))

T

By

Proof:

lemma

2.4,

-

(p(]))

= p(y)

UTT

~

(p(~))

Lemma

= p(~).

The

~

lemma

follows

= A(Y)

and

TT

'

immediately.

2.10

If T 6 S is a limit

point

of 4, then

T =

k_]

~[~]

and p

J

=

p(T)

Proof: and

t_]

o

v~T We

<X

Iv ~ T >

~:

lemma

o(p(~)) since

X

p(m)

commence

<Jp(T),A(T)>.

2.6

I~ ~ T }

•~

that

Set X = U{X

<J~,A>

there

in < J p ( T ) , A ( T ) > . T)

Iv ~ < } .

~ = sup

for

all

Then Pick

~ such {~

V~T

let X

Suppose = ran(o

Z 1 submodels

not,

T) •

of

X K1 <Jp(T) 'A(T)>"

v-~T,

some

we

we h a v e

Let

with

~i <Jp(z) 'A(T)>'

~6 e

clearly

A = A(~), have

X A JT ~ Q JT"

contrary ~ 6 ~T,

~ = p(~),

that

I~)-~T],

5 in ~ ,

for

~ ~T

of

Thus

[~\~I~ ~ T ) .

X>.

is a u n i q u e

T succeeds

let x 6 Jp(T).

chain

= <X,A(T)N

Since

a T : sup

. For e a c h

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

N JT
~,p(T) 6 r a n ( o

].

by p r o v i n g

{e

= p(~).

• = ~- . Thus VT NOW

[J

let [ = sup

Thus

By

vT

to the

(by the

we have

T).

And

~-~T

assumption

above).

x 6ran(o

~£S~.

Hence

x is E l - d e f i n a b l e

and

and

on T.

from

p(T),

Then,

since

This

shows

18

that

J

once.

T is

not Let

~ ~T

Now

]. T h e ~T [J p ( ~ )

maximal

in

~ = sT,

such

corresponding

result

for

that

S

and e

, then

pick

~ 8.

T is

I 6 S

We

a limit

, I > T.

place

point

Given

ourselves

in

of

~ follows

at

~.

8 < e we

show

that

Jl"

X be

the

Let

e l e m e n t a r y submodel o f < J p ( T ) , A ( T ) > c o n t a i n i n g

smallest such

~

2.1 1

Proof: is

~~

o

Lemma If

=

p(T)

that

X n a is

return

to

the

transitive. real

Since

world. Let

a = ~i'

q:

<J

,A>

X n ~ 6 ~.

p(T)

Let

= <X,A(T)

there

and 8,

~ = X N ~.

n X>.

By

lemma

2.6

P there q

is

= o.

a unique

~ 6 S-

Since

= ~ > 8, w e

~

such

that are

p =

p(~),

done.

A

= A(~).

of

S

. Let

~

[~]

Then

~ ~T

and

[]

~T

Lemma Let

2.1 2

~ ,

Then

and

]~'

Proof:

and

Clearly,

Hence

(q~IJ]):

now.

[]

suppose a~,

~ is

limit

point

~'

= sup(~[-~]).

~J~ = o ] ~ I J ~ .

(o~IJ~): J~

a


J

J~

4° J

,. T h e

,. B u t

lemma

follows

is

easily

cofinal from

in lemma

~' 2.7

~9

§ 3.

We assume

tree

use

A New

the

V = L

Construction

above

theory

of

a Souslin

to

construct

on

T,

%-Tree

a Souslin

~2-tree

in

L.

We

throughout•

We

define,

of

height

by

recursion

trees

T T , T 6 S . T T will

be

a normal

N T), and we shall use the members of S N T T T to index the levels of T ~ (so T ~ = U{T ~ ] u 6 S N T)) • If T < ml' TT v ~ T w i l l h a v e w i d t h ~ l , a n d if T 6 S i, T T w i l l h a v e w i d t h ~2" O u r f i n a l

~2-tree,

T,

For if

otp(S

will

~ 6 S,

~ ~ max(S We

carry

be we

a subtree

denote

of

by

s(~)

), w i t h

s(~)

out

construction

the

the

tree

the

immediate

undefined

if so

U{T <

I T E S 1}. successor

v = max(S

as

to

of

~

in

S

,

).

preserve

the

following

conditions: (PI)

If ~ 6 S

N T,

T T will

be

an

end-extension

of

T~

(i.e.

we

shall

T have

T ~ = U{TTy

I Y E S~

N ~}); T

(P2)

T~

:

S(7)

- Y;

Y (P3

If

~ ~,

~ = ~T

and

The position

~T~ (~)

definition of

T in

embeds

~.

T ~ into

then

•

of

~T[T

T T falls

We

T T so

that

whenever

~ 6 Sa_N T

T-~ ] c T T ~" into

consider

three

these

cases,

cases

in

depending order

of

upon

the

increasing

complexity.

Case In

I.

T is

a limit

this

case

we

normal

tree

of

set height

point

TT =

in

~

~)~

otp(S

T~)].

N ~)

on

By

lemma

T such

2.10,

that

this

~ TI~

defines

embeds

a

T ~ into

T

T T

-5.

(to

satisfy

(P3))for

each

~ ~T.

We

must

check

that

if

n 6 S

N T, T

then

point

T T is

of

3,

an e n d - e x t e n s i o n

and

by

lemma

of

2.]0•

T n.

if

But

by

~)--.l~c i s

lemma

2•11,

sufficiently

~ is

a limit

large,

20

D

: zmT ({)

and

the

Case

for

some

result

~6

follows.

2. T is m i n i m a l

There

are

three

Case

2.1

T is

Then

we must

Case

2.2

S~

D ~,

There

whence

by

lemma

is n o t h i n g

2.9,

further

o~T

~-

to

check•

to

check.

= omT

T~ '

in ~ .

subcases

initial

to

consider.

in S T

set

T is

T ° = @.

a limit

There

point

is n o t h i n g

of

for

us

S T

Set

T T = U{TVI~ 6 S

D T~ There

is n o t h i n g

to

check.

T

Case

2.3

There

Case

T = s(~).

are

three

2.3.1

~ is

subcases

to

consider.

initial

in S Y

TT

Set

= T -

Case

2.3.2

Using

the

member

of

~.

There

is

nothing

to

check.

~ = s(~). ordinals T~

of

T -~,

to f o r m

T TV,

appoint and

set

infinitely

many

T T = T ~ U T T~.

extensions

There

of e a c h

is n o t h i n g

to

check.

Case

2.3.3

~ is

a limit

point

of

S T

First

that

note

by

lemma

2.11,

T is

maximal

in

S

,

so

by

2.1

(iii),

T

• ¢ S 1

Hence

~ is

a

a countable

limit

through

containing

point The

T~,

extensions

actual

following of

the

choice case.

countable

ordinal, x.

of e a c h of

the

Suppose

countability

of

and

limit for

Using branch

each

the

Hence

x 6 T ~ we

ordinals

in

b x to obtain

branches

b

T v 6 Jp(~). the

ordinal.

ordinals

X

is

not

(Notice in

otp(S

can

pick

T - ~,

we

T T~,

Whence

important that

TN U)

S D ~ ± in o r d e r

a branch appoint

to

b

x

one-

T T = T~ U TT.

except

we make

is

in

the

extensive construct

use the

21

trees

T T,

J p(m)

in

J

However,

•

case,

T £ S n ~i'

let

p(v)"

to

certainly

the

Q

countable

is

set

disjoint

completes

of

will

all so

'

from

the

only

each

is

in

as w e

that

show

T~

later.)

segments

of

easy

to

pick

branches

of

Q.

this

case.

the

lies In

initial

member

definition

"accidental"

occur,

thin it

be

this

T ~ which

There

is

nothing

check

is

that

b

in

lie to be

x

for

us

check.

Case As

it w i l l

Q be

eventually That

it

so

3.

in

Case

T immediately

Case

3.1

2,

T is

there

succeeds are

initial

{ in

three

in

~.

subcases.

S T

T T = @. T h e r e

Set

Case

3.2

T is

is

nothing

a limit

to

point

check.

of

S T

TT :

Set

embeds

U{T~I~ 6 S

T~ into

=

The only

thing

that

whenever

~ 6 S~T n T and

T T s o~

~T[TT~]c_ T T. ~T[~

n T}.

Now, b y

U{~TI5

lo~una 2 . 8 ,

] ~ 6 Sa

n ~}.

The

to

~ will

be

result

follows

u

~T~

= ~- (~), TT

a limit

point

then

of

S T,

so

immediately

using

subcases

consider.

% Lemma

2.9.

Case

3.3

T =

Then

~ =

s(5),

Case

3.3.1

s(~). where

~ is

~T(~)

initial

= ~.

in

There

are

three

to

S T

Let

TT

= T - v.

Case

3.3.2

Then

~ = s(~),

x 6 T T-D % -

where

Y £ TTS'

TY

D

2). has

are

no

non-trivial

checks

to

be

made.

v = s(~) .

(~ U ~ { T [ T ~ 5 ] )

section of

n

There

Use

and is the

infinitely

~(~) x

=

< T y'

let

infinite ordinals many

~.

For n~T(y)

(this in

each

pair

extend

is

clear

this

set

extensions

in

x,y 6 T ~ such ~T(x)

from

the

to e n s u r e TY

in

. There

that

T T ~.

Now

t

construction

that

every

is

nothing

of

element of

a

22

non-trivial

Case

nature

3.3.3

which

~ is a limit

needs

point

to be

checked.

of S T

There

Case

are two

3.3.3.1

subcases

~-

[~]

to c o n s i d e r .

is c o f i n a l

in ~.

YT

For e a c h

through

x 6 T{~,

let 7-TT(X)

T~ d e t e r m i n e d set

infinite

~ -

by extending Firstly,

by { ~ T ( y )

l y
(~ U ~ r [ T ~ ] )

two further

if y ~

be an e x t e n s i o n

using

the

ordinals

we now c o m p l e t e

the

definition

collections

and y # m a x ( S

in T T~ of the b r a n c h

of branches

) and ~y~[Y]

through

in

the o f T~

T ~.

is c o f i n a l

in ~,

and

if

Y x 6 T s(Y)Y , we e n s u r e onto

that

for

and d i s t i n c t

each from

each

rules

laid

There

are no n o n - t r i v i a l

out

3.3.3.2

lemma

lemma

2.9),

n I ~N

in c a s e

Tu e x t e n d s

checks

X

through

extended

of the b

x

T ~,

above,

is m a d e

containing

x,

and e x t e n d

according

to the

to be made.

and

Let

o~iIJ5

~ = s(1).

successor

Io,I 1 is a limit

o ~Ioli o

b

< ~.

~ ~I,

= ~TI~.

of

a branch

2.3.3.

an i m m e d i a t e Each

choice

~[~]

I 6 Sa~,

~II~

be

= ~IiI

T T . The

I = sup

2.12,

~i = S(ll)" ~I

{ ~ y ~ ( y ) [y
of the b r a n c h e s

branch

Case

onto

x 6 T ~ we pick

that

Let

branch

TT

Secondly,

By

the

~io,

= a-~vIJ~. Choose

~o ~ ~ w i t h

of qo in ~ . point

Let

of its

so in p a r t i c u l a r

Moreover,

~o

~o = S(lo)'

level.

~ioli[Io]

(using

Moreover,

is c o f i n a l

in

n1 1 I. H e n c e Let x E there

Case

T{_.

3.3.3.1

By the

is a p o i n t

applied

construction

y(x)

in the d e f i n i t i o n in C a s e

in T ~i ii w h i c h

3.3.3.1

extends

all

of T for T

. D1

, we k n o w

the p o i n t s

that

~ i I (z)

23

for

z
points

z~

(z)

y' (x)

= ~ 1

(= ~ l ( z )

for

each

(y(x)).

Then

y' (x) 6 T ~ I

= ~ii ~ o ~l(z))

T ~ , containing

dx through

Also,

Let

y' (x),

x 6 TV , pick

and

a branch

for

b

all

Pick

a branch

z
let ~

extends

extend

through

the

d x o n T T ~"

T ~ , containing

x,

and

X

distinct

from

the

branches

d

T -

(~ U ~ [ T T ~ ] )

to extend

The

choice

branches

of t h e

Case

2.3.3.

There

are

That

no non-trivial

completes

the

just

x

these dx,

extended, branches

x 6 T~,

checks

and onto

use TT

a n d b x,

the

ordinals

in

.

x 6 T ~,

is m a d e

as

in

to be m a d e .

construction.

Let

T = U{TTIT 6 S

}. T is

clearly

an ~ 2 - t r e e .

Lemma

3. I

T has

at m o s t

Proof:

Suppose

e2-branches b y N T~.

Let

M be

tive.

Let z(U) that

countably

Let

7: M

otherwise,

of T. Let

the

U =

Let

v = Mn

= Jl"

e 2.

Then,

let

~ < e 2 be

{b

least

. Clearly,

M<Je3

such

Clearly,

zI~

I~ < e l} be such U is

that

81 m a n y

that

~ < T < eI ÷ b

a thin



distinct

initial

N T6

segment

6 M a n d M n e 2 is

of T.

transi-

v 6 e 2.

= id~,

It is e a s i l y

T D ~ = T ~,

e2-branches.

and

~-/ b ~<e I

smallest

= U N ~.

many

seen

z ( e 2) that

= ~,

~ is

U N ~ = U N T ~. M o r e o v e r ,

~(T)

= T n ~,

a limit U D ~ is

point a thin

and

of S

, and

e1 initial

segment

of T ~ .

Claim.

8(~)

Proof:

Since

Jl"

show~ t h a t

We

=

I + 1 and z

-i

: Jl~ v is

n(~)

Je3

and

= i. M o r e o v e r , -i

El-singular

(~)

= e2,

over

I +

cf(~) v is i,

= e.

clearly

and

that

regular cf(~)

over

= e.

24

Let X ° be the s m a l l e s t

X ~l Jl such that 6 X, and set

Vo = s u p ( X o N ~). N o t i c e regular

that since X ° is d e f i n a b l e

in Jl'

and v is

over Jl' we m u s t h a v e Vo < ~"

Now suppose

X n ~n+l

Jl and Vn < ~ are d e f i n e d .

Let Xn+ 1 be the s m a l l e s t

X ~n+2 Jl such that 6 X and ~ n -c X, and set Vn+l = s u p ( X n + i N

~)" S i n c e Xn+ 1 is J l - d e f i n a b l e

and ~ is r e g u l a r

over

Jl and Vn < ~' we h a v e Vn+l < m" Let X = n<~k-] Xn. xn ~ = SUPn<~

Clearly,

Vn

X < Jl and 6 X, and

(which is t r a n s i t i v e ,

Jl is the s m a l l e s t

elementary

and has Thus,

in p a r t i c u l a r

are E l - d e f i n a b l e

By the claim,

submodel

a transitive

we have,

since

through

v>6,

there

U. We o b t a i n

our c o n s t r u c t i o n U N v extend By lemmas Jp(v) T~

U N V 6

contains

clearly,

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

ensures

= 6(v)

[]

= I + i. H e n c e

as

p(v)" in T

contradiction

t h a t only

< X n [ n < w> and < V n l n < ~>

the claim.

p(v)

are ~I m a n y p o i n t s our d e s i r e d

w i t h v. Thus X = Jl"

which

extend

a branch

by d e m o n s t r a t i n g

countably many branches

that

through

onto T

2.1

= vy~

of Jl w h i c h

This p r o v e s

T N v, U N v 6 Ji' we h a v e T V, U D v 6 J

Now,

B u t s i n c e Jl ~ M,

intersection

S U P n < v n = v. But

over Jl+l"

of course).

(iii),

2.11,

and 2.10,

d ~ o ~ [ J p ( ~ o ) ]. So we

d~OV[Jp

can pick Vo h e r e

(~o) ]

can p i ~ k V o ~ V

Moreover,

large e n o u g h

~ is a limit p o i n t

by our above

for ~

in ~ and

large e n o u g h claim,

cf(~)

v [ V o ] to be c o f i n a l

to h a v e =

w, so we

in v. N o t i c e

O

-I(TV)

that if Vo - ~ I ~ ~' then

= T vl and ~

VlV

-I(uN~) Vl v

v1 initial

segment

of T

~Vlv[T

] = TVN ran(~

(because iv)) .

~VlV:

J p ( v l ) ~i Jp(v)

and

is a t h i n

25

Let

C =

{~JVoI~

branches

# max(S

through

_)},

T [ which

that

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

only

branches

and

extend

We p r o v e

through

o

for ~ 6 C let B~ be on T$(5)

that

-I(uNv)

ticular

The

for

proof

~i' ~ Vl"

Case

I.

Then

T

~VoVl[b]^ for

v I = v, this

T 1 = S(ml).

YI is a l i m i t

TI

= T2~T I<-]

are e x t e n d e d

2.

complete

There

T2]

are

Since

is true

of the

the r e s u l t cases

), the

on T s(vl) ~i

this

our p r o o f

three

Notice

in par-

lemma.

holds

for

all

to c o n s i d e r •

in 3 . , and

the r e s u l t

follows

easily

from

the

hypothesis.

T I is m i n i m a l

In this

case,

Case

T I succeeds

3.

b 6 B.

on v I. S u p p o s e

point

~ T 2 T I IT

some

will

is by i n d u c t i o n Let

induction

Case

form

all

~ ~ and v I # m a x ( S

VlV are of the

set of

Let B = ~6C BS"

if Vo ~ I which

the

in ~ .

__90 b r a n c h

through

OVl v

-i

(UNv)

is e x t e n d e d

TI Tvl.

onto

T 2 in 4 . T1

Notice

that

T 1 falls

under

Case

3.3.3.1

in the d e f i n i t i o n

T1 Suppose

that

z 6 T

~i e x t e n d s

construction

of T

branches

since

which

bx,

is a t h i n

a branch

, the b r a n c h these

initial

maining

possibilities.

Suppose

first

that

are

all

segment

z = ~

by

eventually of T

-i ~ o V2Vl

a

-i v2 v

of the

- l ( u n v ) , we

from

in J p ( ~ l ) . T h e r e

(x) t w h e r e

x 6 T

By our

z is n o t o n e

disjoint

T2

Since

of the o 1~

are

~ 2~

-i

(Un~),

two re-

-i

(UNv)

~2

T2T I = o

~ 1 ~ -I(uD~).

through

extended

of T

see

that

x must

extend

a branch

through

Vl v

(UN~). form

So,

by i n d u c t i o n

~ ov2[b]^for

some

hypothesis,

the b r a n c h

extended

by x is

b 6 B. H e n c e

the b r a n c h

extended

by

z is

26

of

the

form

~

[b] ^ = ~ Vom I

Now

suppose

that

. ~ ~2~i

for

some

[b] ^

(for this

that

y ~ max

b).

~o~i

y ~ml

such

(S)

and

~7m [Y] 1

7 is

cofinal

Y J~o'

in ml'

then

y 6 C,

~O ~ Y' in w h i c h and by the

z extends

where

x 6 T s(Y)Y

[YlY


case

{y]y


hypothesis

at y,

~ o~i [b]^

z extends

are done.

If

so

induction

b 6 B, w h e n c e

{~y~l (y) Iy
{YlY

(= YY~I


Otherwise,

: ~ oy[b]^

~ z ~o~i [b]^),

and

for

some

again

we

are done.

By

lemma

3.1 we

can find

contains

x. We

to avoid

carrying

assume

that

generality

Lemma

show

that

a point

x of T such

{y 6 TIx

6.1, y}

x along

T already

no ~ 2 - b r a n c h

is a s o u s l i n

as a p a r a m e t e r ,

is A r o n s z a j n .

that

This

however, clearly

~2-tree. we

shall

causes

no

of T

In o r d e r simply loss

of

in our proof.

3.2

T is S o u s l i n . Proof:

Suppose

not.

L e t X be

an a n t i c h a i n

of T of

cardinality

~2"

Let U = {x 6 Xlx It is e a s i l y U

"splits"

seen

within

that x
IU N T6I

~ > 6 such contrary higher

U

~2 m a n y

IUI

= ~2"

(i.e.

z, and ht(y)

U N (TI~)

to the

fact

of T.

that

at m o s t

x 6 U there

as in

lemma

countably

each member

in 2}.

T is A r o n s z a j n ,

= ht(z)) . Thus

exactly

has

successors

Since

for e a c h

= '~I" A r g u i n g

that

levels

that

has

of U has

each member

are d i s t i n c t there 3.1 many

of

y,z 6 U such

is a ~ < ~2 such now,

we

can

extensions

extensions

find a in T

,

in U on all

27

§ 4. A N e w

We need if it h a s

some

Lemma

preliminaries.

the p r o p e r t y

o n all h i g h e r distinct

Construction

levels,

points

that

on a limit

a Kurepa

A tree will

each

but does

of

point

not

level

has

be

called

infinitely

necessarily

have

~2-Tree

distinct

almost many

normal

extensions

have

the p r o p e r t y

that

sets

of p r e d e c e s s o r s .

4.1

Suppose

T is

~3 many

~2-branches.

tO e<~ 2 Te+l

Proof:

=

an a l m o s t

Then

t r e e T ° of h e i g h t every

extensions

tree

there

of h e i g h t

is

a Kurepa

~2

and width ~2'

~2-tree

T'

such

having

that

tU T' e<~ 2 e+l"

By discarding

e < ~2'

normal

all

~2

limit

levels

and w i d t h

~2

of T ° ~

which

e-branch

on T ° . N o t e

of T, w e

such

obtain

an a l m o s t

normal

that

for e v e r y

limit

ordinal

extends

on T °~ h a s

infinitely

many

that

e

T° =

For which

each

limit

is e x t e n d e d

the p o i n t s defines

4.1

quired

and

T'

above,

Kurepa

below

such

that

a tree

trees

3. T T w i l l

We

T° =

element

T'

4.1

which

of T ° w h i c h Since

is c l e a r l y

of a K u r e p a

lemma

e-branch,

<J T' e<~ 2 e+l"

T satisfying

assume

T T, T 6 S, be

a new

= e<~2qJ T +i,

and r e l y u p o n

tree.

and e a c h

all points

construction

construct

construct

section

lies

~ < ~2 n o w

introduce

e<w2<-] T ' ~ + I

s o to t h e

f a c t d o is lemma

o n T e, °

an ~ 2 - t r e e

And

ordinal

of b and

cO2-branches

We

t/ e < ~ 2 Ta+l"

the

tree

b,

extends extend

This

H3 m a n y

Kurepa.

(in L).

all

b.

T has

[]

What we

requirements

itself

of T ° I ~

to e x t r a c t

in

specified the re-

V = L f r o m n o w on. in a m a n n e r

an a l m o s t

normal

similar

tree

to that employed

of h e i g h t

otp(S e

n T) T

in

in

28

whose

levels we s h a l l

index by the o r d i n a l s

in S

~ T. T ~ w i l l have T

width be

N1 f o r

the

~ G S ~ ml

ordinals

in

and

s(y)-y

width

' ~2

(for

y 6 S

for

~ 6 S 1 n ~).

"

The

elemen~of

We s h a l l

carry

T~7 w i l l

out

the

T

construction (i)

so as to p r e s e r v e

If v 6 S

the f o l l o w i n g

n T, T T is an e n d - e x t e n s i o n

conditions: of T ~.

T

(ii)

If ~ T ,

z~I~

embeds

T ~ into T T in such a w a y

J6 S~_N T - and ~-TT (~) = ~'

then ~

T[T ~

]]

that w h e n e v e r

c_ T T~.

T

(iii)

Suppose

T = s(~), w h e r e

~ is a limit p o i n t

of S

, and t h a t T ~ T

is E l - d e f i n a b l e Tm which (iv)

is

over < J D ( ~ ) , A ( ~ ) > .

~l-definable

Suppose

T,~ are as in

Suppose

further

(iii),

through

<JD(v),A(~)>.

Since every

of ~

~,~ of ~(A)

u6B

Suppose

U {p(~)}

idI~

and ~

c l e a r l y have,

u

Define B~JD(~)

u6B It is e a s i l y

~ = s(]),

is E l - d e f i n a b l e

element

of JP(m)

in <Jp(v) ,A(~)>, x÷ , y÷ 6 ~

is

= ~. Let

there

from

are E l - f o r m u l a s

such that for all u , v 6 Jp(~),

~-m~: <JP(~) , A ( ] ) >

iff

~T(~)

~l-definable

<J p(~) ,A(v)> b ~ ( u , y 'p(~)) .

= p(m)

o n t o T T.

over

iff

(P(~))

through

over < J p ( m ) , A ( ~ ) > .

<Jp(~) ,A(v)> ~ % ( u , v , x , p ( v ) )

for u , v 6 J p ( ] )


extends

iff

x , y 6 e-.~ S i n c e =

,A(~) >

and T ~ T ,

T ~ which

and e l e m e n t s

u
p(v)

t h a t T m is E l - d e f i n a b l e

B be a b r a n c h

elements

over < J

Then every branch

and ~

~i <Jp(m) ,A(~)> and IT ~)] = T m

N ran(~5~),

we

'

<Jp(~) ,A(m)> ~ ~ ( u , v , x , p ( m ) ) .

by

iff

<JD(~),A(~)>

s e e n t h a t B is a b r a n c h

b %(u,y,p(5)) . through

T 5 and that ~L~[B] _cB.

29

Now,

the b r a n c h

there

is

shall

ensure

B just

defined

(by c o n d i t i o n that

~-

(iii)

is E l - d e f i n a b l e

above)

(x) e x t e n d s

a point

B on T T.

over

<Jp(5),A([)>,

x 6 T ?- e x t e n d i n g (This

is w h e r e

so

B. We

we

lose

the

TT

full

normality

with

El-definitions

tions

as if they

not be

As

the

Case

~

T~T

rise

to

not w i t h

and we m u s t

"different"

branches handle

so m u c h

different

branches,

though

as defini-

this m a y

of course.)

construction

a limit

T T = _kJ

We d e a l

of b r a n c h e s ,

3, the

I. T is

trees.

gave

case,

in s e c t i o n

Set

of our

point

[TT].

prpceeds

by

cases.

in ~ .

Conditions

(i)

and

(ii)

are v e r i f i e d

as in

TT

section

3. W e

Suppose

then

cheek that

(iii)

and

Y = s(m),

(iv)

where

simultaneously. ~ is a limit

point

of S

, and that T

T ~) i s

71-definable

which

is E l - d e f i n a b l e

qb,t) o f

over

<Jp(x)),A(~))>.

over ÷

o~(A) a n d e l e m e n t s

Let

<Jp(~),A(~)>. ÷

x , y 6 cry s u c h

B be a branch Then

that

for

there

all

are

through

El-formulas !

u,VEJp(,~)

u
iff

<Jp(m),A(m)>

~ %(u,v,x,p(~)),

u 6 B

iff

<Jp(~),A(~)>

b ~(,m,y,p(~)).

T ~)

÷

Pick

~o ~ T

with

x÷ , y÷ 6 a~

. Let

~o = S(~o) " S u p p o s e

To ~

~T,

and

let

O

= S(5).

Define

Then,

u
iff

B 5~Jp(~)

by

ueB~ Thus

iff

-

~J T

o

~T~T

for u , v £ J

<Jp(~),A(~o)>

<Jp(5),A(~)>

B 5 is a b r a n c h

z--V~. [Bs] B

clearly,

through

T~

•

= B~, n r a n ( z s ~ , ) . Also, ~~

[B ] ~

By "

condition

p (5) '

b %(u,v,X,p(5)).

~ ~(u,~,p(~)).

Moreover, ~sv[Bs] (iii)

for

~o

~T~TI~T,

= B n ran(~L~) , and

below

T, let x- e x t e n d M

B- on

30

T .f x

By

condition

= 7-YY (x ~

(iii).

Case There

T,

~TT'-- (X~)

any

T 'To 4T

~T.

Clearly

the

method

of

T Is m i n i m a l

are

Case

below

for

And

2.

(iv)

2.1

three

T is

in

proof

~

for

To ~ T

x extends

establishes

~T'

B on

~T.

Let

T T.This

proves

(iv).

.

subcases

initial

= X~,

to

consider.

in

S

no

checks

T

T T = ~.

Set

Case

2.2

There

T is

are

a limit

point

to

of

be

made.

S T

TT = U{T~I~ 6 S

Set

N T}.

There

is

nothing

to

check.

T Case

2.3

Y = s(~).

There

are

three

Case

2.3.1

subcases

~ is

to

initial

consider.

in S T

Set

T Y = T -m

Case Use of

2.3.2 the

~ =

member

2.3.3

~

finish.

s(D).

ordinals

each

Case

and

of of

is

T -mto

T ~. n

provide

There

a limit

is

point

infinitely

nothing

of

to

many

extensions

on

TT

check.

S T

For use

x 6 T~,

each the

ordinals

branches This

let

In

Zl-definable

over

extensions branches

be

and

this

possibily only

case

distinct

definitions

may, have

more

than

through

to p r o v i d e

arises

we

other

when

extend

branches,

one

than

of

thereby

actual

We s h a l l

of

these

many)

through

distinct

and

x.

Zl-definable

branch

extension.)

each

(countably

T ~ is

each

treating

rather

T~ containing

extensions

some

<Jp(~),A(~)>,

produced with

a branch

T - v

possibility

<Jp(~),A(~)>.

they

x

from

b x on T T

latter

though

b

over

T ~ which

Zl-definitions

is as

associating

branches.

There

branches.

are

(Hence, no

checks

some to

31

be

made.

Case As

3.

in

Case

T immediately

Case

3.1

2 there

T is

succeeds

~ in

three

subcases.

are

initial

in

~.

S T

Set

TT = $

Case

3.2

and

T is

be

done.

a limit

point

of

S T

Set

TT =

U{T~I~ 6 S

N T}.

The

only

non-trivial

check

to

be

made

T

concerns

condition

(ii).

Case

3.3

T = s(v).

Then

~ = s(~),

where

This

z-

([)

is

handled

= v.

There

as

are

in

section

three

3.

subcases

to

consider.

TT

Case

3.3.1

~ is

initial

in

S T

Let

T T = T -m.

Case

3.3.2

Then

~ = s(~)

There

are

q

, y 6 T_, ~

ordinals has

where

checks

to

be

made.

~-(~)

in

3.3.3

= n.

For

each

pair

x,y 6 T ~ such

that

T%

and

T -

infinitely

Case

non-trivial

~ : s(q).

'

x 6 T

no

x


_

let

(~ U ~?7[T~])~ many

~ is

~- (y) TT to

extensions

a limit

point

extend

ensure on

of

now

~- (x) TT that

on

every

M

. Use

member

the of

Tv q

TT.

S T

There

Case For

are

two

3.3.3.1 each

subcases

to

consider.

~-

[5]

is

cofinal

x 6 T~,

let

7-

(x)

be

in

v.

an e x t e n s i o n

on

T T of

the

ordinals

of

branch

through

TT

T ~ determined T

-

by

(v U ~ T [ T ~ ] ) ,

through Firstly,

{~T(y) we

ly
extend

up

to

Using

the

three

further

collections

of

branches

Tv . if

y ~

and

y ~ max(S

) and Y

z7

[y]

is

cofinal

in

v,

then

for

32

each

x 6 T s(Y)Y we e n s u r e

extends

on

Secondly, every

that

branch

through

T ~ which

on T ~, T regarding branches,

as in C a s e

Finally,

we ensure

that

The

only

T,T.

But

problem

T ~ which

Case

3.3.3.2

Then

I 6 S

case we

point

a few m o m e n t s here,

each

is e x t e n d e d

significant

so we

2.3.3

point

we e n s u r e

over

that

<Jp(v),A(v)>

El-definitions

as if they

has

an

defined

above.

of T ~ lies

on at least

one b r a n c h

on T T.

to c h e c k

not

<Jp(~),A(m)>,

is E l - d e f i n a b l e

reflection

shall

i = sup

over

distinct

distinct

through

Tv

TT. if T ~ is E l - d e f i n a b l e

extension

{~y~(Y) IY
the b r a n c h

is c o n d i t i o n

suffice

give

the

(iv)

to s h o w

simple

for

that

the p a i r

there

is no

details.

~5~[~] < ~.

, ~ X, and ~ 5 ~ J ~ T can e a s i l y s h o w that

= d ~ I J ~. And, for e a c h

as in s e c t i o n

x 6 T_ t h e r e

3 in this

is a p o i n t

y' (x) in

T1 which e x t e n d s t h e b r a n c h {x~T(z) Iz
first

x 6 e~ such

that

that

there

iff

for u , v 6 J p ( ~ ) ,

u
Suppose

further

B is a b r a n c h

through

B is a b r a n c h uCB

<Jp(~) ,A(~)> b

and o r d i n a l s

iff

<Jp(~) ,A(~)> ~

~ is a E l - f o r m u l a T V, w h e r e

.

B c- J

of

%(u,v,~,p(~)).

~A)

and y 6 ~

are

p(m) is d e f i n e d by

<Jp(m) ,A(~)> ~- ~ ( u , y , p ( ~ ) ) .

through iff

%(u,v,x,p(m))

we h a v e

iff

that

u 6B

Then

% of ~(A)

for u , v 6 Jp(v) ,

U
Thus,

is a E l - f o r m u l a

T ~, w h e r e

<Sp(~),A(])>

BC_Jp(~)

is d e f i n e d

b ~(u,y,p(~)).

by

such

that

33

We k n o w with

the a b o v e

being

We

t h a t B has

the

not

definition.

one n o w

complete

an e x t e n s i o n ,

to be

above,

let z- (x) e x t e n d TT

in case

ensure

usual,

that branch

T m is E l - d e f i n a b l e

over

we regard

the

above

over

distinct

<J p(~) ,A(~)>

definitions

(this

extension

definition

of B). T x 6 T_

T ~ containing

y' (x)

each

the r e m a i n i n g

an e x t e n s i o n

<Jp(~),A(~)>,

associated

point

For

using

of T ~ has

on T_,

B on T T

through

on T T. Now,

point

is E l - d e f i n a b l e

with

a branch

each

T ~ which

it w,

of T T as f o l l o w s .

choose

that

call

~- (w) e x t e n d TT

associated

the d e f i n i t i o n

considered

in T - V,

Let

let us

every

extends

as if they

ordinals

o n T T and

branch onto

defined

and

that

through

T ~' T where,

as

distinct

branches.

There

The

are no n o n - t r i v i a l

construction

checks

is c o m p l e t e .

to be made.

Clearly,

T =

[J T6S

T T is an a l m o s t

nor-

e1 mal

tree

of h e i g h t

~2

Suppose

not,

branches. branches

case

~2-branch.

one,

this

Define

of the Since

will

exhaust

submodels

NI~

show

transitive.

Set

L e t N I + I < J~3 transitive.

'

argument

sequence

T has ~3 m a n y

as f o l l o w s , such

all

the

~2~2-

that

T has

not need

at least to be o n e -

that

for

I < m2"

el U {T,B}c_ No.

Clearly

N o A ~2

is

mo = No A ~2"

be s m a l l e s t

Set ~ l + l

let N I =

~I = s u p n < l

shows

< b ~ l ~ < ~2 > does

such

t h a t N 1 U {N I} ~ N I + I. T h e n N X + 1 n e2 is

= N I + I A ~2"

n<-7
N

. Thus

NI~ J

and N 1 N ~2 = s u p ~ < l ~3

Set

that

all p o s s i b i l i t i e s . )

Jm3

be s m a l l e s t

lim(1)

We

let B = < b ~ l ~ < ~2 > e n u m e r a t e

following

the

L e t N o ~ J~3

If

and

~2"

of T.

(A s p e c i a l one

and w i d t h

~"

~"

34

Let

Pl:

easily and

We

NI

-- J < ( l )

seen

that

PI(B)

shall

of b~,

Then

"

vl

point

pl(~2)

of S 1

= ~l

It is

Moreover,

Pl (T)

T

I~ < ~i >-

attempt

~ < w2"

= idI~ 1 and

is a limit

=
now

plp~l

to d e f i n e

If we

succeed,

least

ordinal

an w 2 - b r a n c h

we

shall

have

of T, d i s t i n c t arrived

from

each

at our d e s i r e d

contradiction•

L e t x ° be

the

let xl+ 1 be

the

least

in T o not

ordinal

in b o.

in T

such

If xl 6 T i is d e f i n e d ,

that

xl


and

~I+i xl+ 1 { b I ordinal

. If lim(l) in T i w h i c h

otherwise each

T distinct

Lemma For

q

-I

over

the

breaks

sequence

from each

each

of xq,

down.

< x l ] A < w2>

of b6,

I~ < X> is d e f i n e d ,

~ < e2"

then

q
let x I be the

if s u c h

Providing

clearly

So, we m u s t

x I is also

an x I exists;

x I is d e f i n e d

defines show

least

for

an e 2 - b r a n c h that

if

of

lim(1)

defined.

4.2

each

Proof: Pl

extends

the d e f i n i t i o n

i < ~2'

and <x

and <Xq[q < I> is d e f i n e d ,

I < w 2, T I is E l - d e f i n a b l e

Let

I < ~2'

(~) = ~2" JB(1)

and

Hence

set m = ml"

m is r e g u l a r

(by d e f i n i t i o n

= pI(T) 6 J<(l)"

Thus

(with p a r a m e t e r s )

of

Js(~)"

Now, over

But

A(~),

T ~ is E l - d e f i n a b l e

By the

above

lemma

and

it s u f f i c e s

definable above,

over

now

to s h o w

follow

at o n c e

: J<(1)<

J<(1) • But

that And

and

m is not

•

<J

for

can

regular T~ =

T ~ is ~ n ( ~ ) - d e f i n a b l e Thus,

by

the p r o p e r t i e s

p(~) ,A(~)>, as r e q u i r e d •

(in p a r t i c u l a r ,

by the

if we

J~3

<(I) < B(~) • But

over

construction

<Jp(~l),A(~l)>.

this w i l l

-i

T ~ c J p(~)

codes

(iii)),

Q{

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

of the

our

<Jp(~l),A(~l)>.

~(~)) . H e n c e

T ~ £ JB(~)"

over

over

lim(1) , <x

condition

I~ < I> is E l-

same

considerations

show

that

as

o

35

<x

n

lq < I> 6 J8

from now on) We know

(v~)

. It clearly

suffices

to show that

T v, ,

already

that T v 6 JB(v) " Also,

(setting

v = vl

6 as(v)"

= < b [ N TVI~ < v>II

and = pI(B) 6 J~(1) ~ Js(v)"

So it r e m a i n s

to show that

6 Oh(v). Define

Nq',

n < I, from JK(I) ,~l,TV, just

were defined
n

from J ~ 3 ! ~ i ,T,B.

' In < I> 6 J B ( v ) '

(Recall

that

so <m' n

~(I) < 8(~)-)

as Nq

Set vq' = Nq' N v for each

,q < w 2 ,

q < I. Then,

lq < l> 6 J s ( v ) " But we may r e p l a c e

definition

of Nq for q < I. So,

shows

that

(pl IN n) : Nq ~ N q ', and hence

Hence

<mqlv < I> 6 JB(v)'

J~3 by N 1 in the o r i g i n a l

as Pl: N1 = J~(1)'

and we are done.

that

a simple_ i n d u c t i o n

v q = v'q

, for all q < I.

38

References

[Del].

K.J. Notes

Devlin, Aspects of C o n s t r u c t i b i l i t y . 354

Springer Lecture

(1973).

[De2].

........... , C o n s t r u c t i b i l i t y .

[Je].

T.J.Jech.

Trees.

Springer,

to appear.

Journal of S y m b o l i c Logic 36,

(1971),

1-14.

[La].

R. Laver & S. Shelah.

[Mi].

W.J. Mitchell, A r o n s z a j n Trees and the Independence of the Transfer Property.

[Si].

J.H.

Silver,

The

~2-Souslin Hypothesis.

Annals of Math. Logic 5,

(1972), 21-46.

The Independence of Kurepa's C o n j e c t u r e

Two-Cardinal Conjectures

in Model Theory.

Pure Maths. XIII, Part I, 383-390.

AMS Proc.

and Symp.

i~

COARSE MORASSES I N Hans-Dieter

Donder

Mathematisches Universitat

Introduction several retic

Higher-gap

years

transfer

definition morasses Jensen

ago

(see

in

of h i g h e r - g a p

exist in

noticed

L

morasses

that w e a k e r

his notes,

J e n s e n gave

Our a p p r o a c h

son for

treatment

the simple

the

can be f o u n d uses

questions

morass

structure

can be o b t a i n e d

L

This

treatment We

restriction

In addition,

w h i c h do not

the n a t u r a l

rial

questions

some

direct

in

in § 2 that u s i n g properties. gap-1 torial using

L

which

application

the morass

result w h i c h an a p p r o p r i a t e

can b e more

O-principle. one

We also give some

coarse m o r a s s e s

other

to be new.

easily v i s u a l i z e d .

coarse

can get K u r e p a

~-principle

In

just define

the

One reaout

the

is not n e c e s s a r y some proper-

from the axioms

methods

b u t not

direction,

we show

trees with a d d i t i o n a l

prove

result

but we think

that

combinato-

In § 1 we only deal

things This

the r e a d e r

to a n s w e r

In this

applications.

and among

seems

tool

can be p r o v e d b y

of a

L.

of these struc-

seem to f o l l o w

is to try to c o n v i n c e

are

But

for fine m o r a s s e s .

coarse m o r a s s e s

L

that

easily

in

we use

The m a i n aim of this p a p e r in

but a

L.

rather

arguments

different.

of

that J e n s e n has only w o r k e d

gaps.

false

model-theo-

and use its p r o p e r t i e s .

we deal with.

are a c t u a l l y

strong

in [6]. The p r o o f

the fine

axiomatic

in

is the fact

for small

ties of the natural a n d some w h i c h

to prove

is still u n p u b l i s h e d

in § 2 is s l i g h t l y morass

by Jensen

them

and c o n d e n s a t i o n

a thorough

coarse

this a p p r o a c h

axiomatic for

global

introduced

work

structures

coarse d e f i n a b i l i t y

"natural"

This

essentially

BRD

have b e e n

He u s e d L.

just u s i n g

tures.

Institut

Bonn,

morasses

[5]).

theorems

L

a simple

with

combina-

could also be p r o v e d

that

the m o r a s s

proof

38

I.

Coarse Gap-1

for

gap-1

morasses

the m o m e n t

define

coarse

axioms V=L,

have b e e n d e s c r i b e d b y D e v l i n

that the r e a d e r has gap-1

(M0) -

(see

a copy of that b o o k

as a s t r u c t u r e

[I], p.149).

such a structure

in his book.

which

We first

available

satisfies

show

Assuming we can

the m o r a s s

that a s s u m i n g

can be o b t a i n e d very

easily.

There-

we give some a p p l i c a t i o n s .

Assume

V=L.

Let

L

is a model of Y In addition, set S+ =

For

morasses

(M5)

of course,

after,

morasses

ZF-.

Iv~ S I L v ~ v~ S +

~

be

the

Let

S

"there

class be

of all o r d i n a l s

T

such

the class of l i m i t points

is a l a r g e s t

uncountable

that

of

S.

cardinal"~

set

av =

the l a r g e s t

Lv-cardinal

a n d let

S ~v:=C~ ~

=

Now let

~

S+

such

v* = the least

that

v~ S

and for some

~v

in

aVUI,~}
To see Since onto

v ~.

mvUlf}.

set

~

is not a cardinal

Let

M

be

p if

I:

Let

v
and

qv = P v

let

~>v

be

exists, there

is some

the e l e m e n t a r y

M ~ I~v.

Then

~

The same k i n d of a r g u m e n t Remark

x~ L ~

from parameters

such

such a

We define

v
every

qv = <~v "v}

that

Let

that

pC L~

L~-definable

In a d d i t i o n

is not a cardinal.

such

is

= the

v

vg S ~ ~.

of

is as required. yields

Then

v*
a regular

f~ L

submodel

otherwise. cardinal.

w h i c h maps L

~v

generated by

39 The following basic definability Lemma I : Let

h : Lp -----~Zm Lv.

h(~)=v,

if

v~v*,

or

~,~

a relation

iff there exists such that

~v

V,v.

and

f

otherwise.

~( on

S#

rng(h).

Then

~

8 + , ~=~*

and

as follows~

f~L~.

Obviously,

f • L~. - - - ~

f~

= id~

be as above.

So we may set

denote

~=-p

q~

S+~ ~ $ v

~v

Let

such that

o

We now define Let

verified by standard

arguments.

Let

h (q~)=q~

lemma is immediately

We also set ~

f

qv ~ rng(f)

is uniquely

Sometimes

~v(~)

is a tree,

and

Then

HVv = f ~ .

Lv.

determined

we also use

H~v

by

to

= v.



is a commutative

system and

we have

Now let us fix a regular points

in

cardinal

~

and restrict

our attention

to

S + n ~ +. @

Set ~ : I ~ , ~ l ~

s~

,

~:~,

L W

"~

is regular"!

S ~ = I~rll3o~¢ ~ , V ) ~ Note

that

The following (MO)

(a)

S

properties

are immediately

is closed for

verified.

~<~; +

B~ Ca++1 ; (b)

Sa

is bounded

~ = max S ~ = sup(S°~K); S ~C

is club in

in

a ,

if

~

is not a cardinal

40

(MI)

If

vgx,

then

nv~(~ v) = a

Hvx~v=

id~v,

and ~

~v~ ' < v + l " < ' S ~ v

(M2) ¥ ~

and ~ s~¥, ~

fh~V~v (M3)

~v[v<~l

(M~)

~

and

is closed

~vlV~r} [To prove

this,

Choose

~*
in

where

Ly

and

q~

in

in S in

p~ S

such that m

n~(;)

:

unbounded take

<~+I , < , S

= HT~

HgvlV

not maximal

z~

such that L¥~

is regular

ZF-. some

q:
Given any X~L .

-~*
Then ~
by Remark

we find working

such that

Bg X ~ a ~

~

X.

But then collapsing v-4~

such that

(MS)

I~ v IV ~ }

X

and applying

Lemma I

we get

~v = Xna~] unbounded

in

~

~t

=

U v~c

H v'c

v

We now turn to some applications. restrict

ourselves

trary successor

to the case

cardinals

will be discussed

work with

(MO) - (M5). In particular, morass

to arbi-

cardinals

gap-1

morass

defined

which are not given by the axioms

we essentially

use the fact

that our

is universal.

We first

introduce

A V = l~[~vl not a cardinal get

Inaccessible

we

chapter.

the natural

i.e. we use some properties

simplicity

The generalizations

will be obvious.

at the end of this

We shall actually above

~=~.

For notational

some notations.

and B v = Iv[v
Lb

and

For

Note that

v~ 8 +

set

Bv,Av~

6.. is closed enough.

1.

L~

when

So especially

v

is we

41 Lemma 2~

v~ S +, ~¢~, ~

The f o l l o w i n g

SeuS~-~Av,Bv~

strengthening

of

L

~ +

was introduced by D e v l i n i~

[5]. 0 @'

there is a sequence

(i) N

is ~ countable,

(ii) if

X~m~

[~<m,)

whenever

is a

H~-sentence

]m '~ •

#

&<~

is true in

p.r.

C Om~

H~-reflecting

then there is an

s.t.

transitive

there is a club

(iii)

N


s.t.

for

n<~

Let

e<~,.

~4,

and set

Set

C~X~I,

Cn~

N

Which means: <m,,~,(A~)i(

},

s.t. <~,~(AL~)i<

~e now show that a natural s.

~

is true in a structure

[Actually, D e v l i n only requires

moras

closed set containing

)" H~-reflection]

~ #- sequence

~(~) = max(S u[~l)

5(a) = min(S~-(a+1 ))

if

b(~)

is contained

in the

is not a limit point in

otherwise.

By Lemma 2 we have

Lemma 3~ Proof:



satisfies

We have to v e r i f y (ii) Let

some ~Kv

X~ rng H~v

(iii) Let

~ = <~01,Z,(A~)i<

which is true in v~

So~ I

and

~

s.t.

~

Lv

~/. ~

"~

~

"~

Choose

• Then >

Since

is a successor LV

But

(i) - (iii) in the d e f i n i t i o n _ I. XC~

(i) is obvious. s.t.

0 #

be given and let

S

is club in ~/".

-< . Since

is true in

~/la~ "

is a successor in -4 . Hence

H~v

~

~2

Choose

~ ~.

Xg L v . There is

A V -~_v

(*) shows that

is true in in

s.t.

v~ S

of

is as required.

be a

H~-sentencen

there is some ~
: L~ ~

'Z Lv to

s.t.

N

~-v

=

Lv

rng [I~v

we clearly have

" ~ = max S v

So

~

qued

42

D e v l i n showed in [3] that

~ ~

tree w i t h o u t A r o n s z a j n subtrees. the morass gives us a "natural"

implies

the e x i s t e n c e of a K u r e p a

~ e v e r t h e l e s s we shall show now that tree of that kind. We shall use this

fact in § 2, where we shall show that the c o r r e s p o n d i n g = k+>~

has a d d i t i o n a l nice p r o p e r t i e s .

not good enough,

for there are p o i n t s

i m m e d i a t e successors. K u r e p a family. For of

B

(i.e.

L e m m a 4s Proof:

let

gv I m v

v,~

w h i c h have

$

,v~.

~ , 2

F~a

is

Row let T

6~ ~ ) .

Set

Let

~=a~. T

I:~Ig~

T1

shows that

b r a n c h of

~

=

~ = <~,~>

Ta~L ~ So

is an

T U

is an T

~-tree.

Choose

is a s u c c e s s o r in g ~-tree

for

and and

(since g~ L ~

.

~

a.

We h a v e

v~ 8 1 s.t. HVv(~ HVv

So there is some

is u n b o u n d e d in

v = m a x Sa).

IBvlv~ S 1

countable.

Since

ly have

Bv~B~

F =

we h a v e

has an u n c o u n t a b l e branch.

v ~ v s.t.

is a

tree w h i c h c o n t a i n s no A r o n s z a j n subtree.

a<~

T C _ T s.t.

let

~

many

be the c h a r a c t e r i s t i c f u n c t i o n

It is easy to see that

But for

~

I~ Iv~ S

For the first p a r t it suffices to show that

Hence

that

iff

is a K u r e p a

is a K u r e p a family.

tree itself is

take the tree a s s o c i a t e d w i t h that family.

g~(6) = 0

T

v~ S ~

But it is easy to see that

We shall

v~ S ~

The morass

tree for

) = ~

~

and T

.

is e l e m e n t a r y we clear-

gg T

s.t.

is a s u c c e s s o r in ~ , Hvv(g )

Lv

for some

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

But then

to show

T

above

hence

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

~. qued.

As we said before, s u c c e s s o r cardinals.

Lemma 3 and 4 have obvious g e n e r a l i z a t i o n s

For i n a c c e s s i b l e

~

to

the s i t u a t i o n is different.

43 The reason is that we have show that for many the

Sa

(~<~)

[Sa[=a+

inaccessible

to approximate

~

if

a

is a cardinal. But we shall

we only need small segments of

Sg:.

This argument is due to Jensen.

~e first recall a familiar definition. Definition:

~>~

there is some Now let the

is ineffable

A_C~ s.t. ~

iff for every

I~<~IA~=An~I

s.t.

is stationary.

be regular but not ineffable. Let

~ =

= Iv~ S~ I ~

L v, L

~

"~a

Aa_Ca

sho~,~ tb~t c

~_~

be

set

is not ineffable"l

and ~ : ~ , ~ : t ~ ~ t A ~ L I ( ~ = ) . + Clearly, I~[~_~ For

is closed for

S for

v, v~ ~

Let

v
and

hence

Proof:

We clearly have

L~.

v--~ S~, Now set

is club in

~

and

AIav~ rng H~v

v--~v, v~ ~

v~ ~g,

~

~

set

iff

Lemma 5:

and

~. U ~

~

~<~

s.t.

X~

ring H~v,

and set

~=~v"

Then

~v.

So assume

Hpv(~I~) = A~a.

~

L w.

there is a club

We shall derive a contradiction. Since

CC~, C~ L

C = H~v(~), A = Hpv(AK).

elementary embedding

H~v

So we only have to show that

s.t. Then

AB~, ~

C

also gives us that

8

for all

8g ~.

and

~=-/~a"

But the

AnS@A8

for all

BE C.

Contradiction. qued. Lemma 5 shows that the previous arguments go through for

.< . So

we get Lemma 6~ holds i.e.

(V=L).

Let

K

be regular but not ineffable. Then

there is a sequence

CN I~<~)

s.t.

~ ~

Ic

44 (i)

N

is a transitive

(ii)

if

X_~

(iii)


p.r.

closed set containing

there is a club is

C_~

H~-reflecting

It is well known that

~ ~

s.t.

~

for all

~

C~>X~m,

and

IN~I<_~

Cna~ N

n~

is already false for ineffable

~.

IC

So the lemma above gives various dinals

in

L.

Proposition

if

7:

(V=L)

A _ ~ +,

and Proof:

Let

h.

S ~ ~

be regular b u t not ineffable. 2

Then

(vc~ +) s.t.

there is a club

C _ C ~ s.t.

Vm~ C 3 v , ~

A(fv(~):0

and

A~Lv

for some

a<~

set

v~ Sa~ we can define

s.t. 3v,~

v~ ~

be defined as above. For

So by an easy d i a g o n a l i z a t i o n

...., 2

VA~ G For

~

: ~ ~

IAi=Iml

IG~I¢~.

s ~

fv

IA[=~,

= ~A~I_

Then

car-

f (~)=I) Let

G

of ineffable

We now give another one.

there are partitions (*)

characterizations

A(h

(v)=0

now define

f

and

h (~)=I)

s r ---~ 2

V

by

fv(~) 0 Note that given Choose

~

S

~

s.t.

C ~ ~a~l~'~4~. satisfies

if there is no such there is at most one such

~.

A~ L~ .

Let

~

Then

is club in

C

the requirements

for

~'~ ~

s.t. K.

Let

A ~

s.t.

rug H~,~

and set

It is immediate

that

A. qued.

The p r o o f shows that if if

AC~ +,

IAI>_k,

then

K=k ÷

we can s t r e n g t h e n

(*) to~

... +

Actually,

Jensen has shown a long time ago that for

strengthen V~

~-5

morass..

(*) to:

if

~v,~g A ( f v ( ~ ) = 0

A C ~ + , IAl=k, and

there is some

fv(~)=1).

~z=-k 8<~

IAI=~.

one can s.t.

But the p r o o f uses a fine

C

45

For inaccessible Wolsdorf Let some

K

the situation

and Choodnovsky ~

be weakly

A _ ~ +,

IAI=~,

~v~ A

have shown

compact.

Let

~

B

Proposition there

is some

i~2

let

of

stationary"

IAI=~,

V~

A

and

if

intersection

NOW set

W =

~_

and

~ DP and

D

fA (V)'l I~ [ O l ~

let

immediately

s.t. find

s.t. for some

G~ ~(~)

s.t.

IGI~_~

of some enumeration "A G

is

to find some

s.t.

for some

= n IAh( and

A G h

D_CA t

for some

Now observe

is stationary (I),

l

Ai

for some

i,vl

and

~.

w.l.o.g,

satisfying

v~ B,

ACB,

that

g=gD s.t. l)(i. A1v-g(v)

[Ag(v) Iv~ ID-P ~ . D~ W

For

It suffices

D_CAh(~)

IWI_~-~, we may assume

Now given

E~

Then

is stationary

I D = ~v<~+II~-A~ Since

in ZFC.

on the choice of the enumeration.

h : B ---~ 2

IDI<~

s to A

is proved

fv : ~ ---* 2 (v<~+).

some notation.

For then it is easy to construct

IAI=

Let

result

7 characteri-

It is easy to see that the statement

IAh(V)l

(2)

is

i<2

fv(a)=i

Atv = [~<~Ifv(~)=il"

IBI=~,

(I)

GuI~l.

Then there

s.t. for some

and some stationary

does not depend

Now set

and

A_C~+,

_~

The following

be ineffable.

be the diagonal



: ~ --~ 2 (v<~+).

we now show that Lemma

L.

We first introduce

& G

_~+,

in

~

Vv~ E

Proofs

fv

f~(a)=i

cardinals

8: Let

different. Namely,

(see [8]):

and some unbounded

For the sake of completeness zes ineffable

is slightly

IIDI=~ +

for

~

ID

for all and set

that if there is some for every

D~ W.

p~+

GC_~DP , IGI(~,

(2). So we may assume

we

that this

46 is not the case. Hence we find for every ~

÷

(6<~) s.t.

(where

I

g=gD).

I-~

and

No~ set

h : B ---* 2

(see

then

(2),

pairwise

disjoint

A IA vg(v) ~ v~ GD6 ~ is not stationary

B = U ~ID~

there is some

DE W

s.t.

W, 6 < ~ .

Since

A IA~ (v) I ~

B~

K

is ineffable,

is stationary

for every

h

satisfies

too. qued.

2.

The ~lobal Assume

Let

V=L

~a(a<~) ~

T~=

~

~

=

function

of


T~

more

remark

closely.

~v

I:

gB : ~---~ 2

as in § I. Set For

and set

~

T~=

k-tree

By = I ~ l ~ v l

let

~

gv

is a tree

for

be the charac-

Igv~alv~ ~ a < ~

For this we introduce

Let

~>~

be regular

be the characteristic

T = T F = IgB~aiB~ F, ~<ml

and

is called a rich ~-tree

1

and

is to investigate

a definition.

First

T

k

of height

the trees let us

s.t.

ITal
IFI =

(ii)

for all

where let

XC~,

function

_~(K). of

D.

For

BE F

let

Set

T =
iff the following

conditions

are satisfied.

~IXI<:~

: IF~XI ~

IXI

F~X = I~,~,X I BE FI T

be a subtree

is regular. Clearly,

and let

+

(i)

for

~.

be regular but not ineffable.

a~k.

Definition

(iii)

~>~

be defined Iv~

L

Our main aim im this chapter

that for us a

for all

in

again and let

and

and let

teristic

coarse morass

Then

~

of

~

s.t.

has a branch

T

is a

k-tree

of length

this is not a very elegant definition,

where

k~

k.

but it is good enough

our purposes. We shall prove:

Theorem

I:

Assume

V=L.

Let

~>~

be regular hut ~ot ineffable.

Then

47

there is a rich

~2-tree.

In fact, we shall show that for rich.

Of course,

m~T

we now have

m

as above

we have already

shown this for

to investigate

small subtrees

the tree ~=-~ of

ning.

So we introduce

the "global introduced

set

Since

~

this

in the begin-

now the general f r a m e w o r k which might be called

coarse morass in

L".

~e use the notations

in the context of his higher-gap

we do not give an axiomatic Let

is

in § I. But for

T ~.

will be done in a u n i f o r m way, we do not have to fix

T~

S, S +, S , v~' Pv'

~(v) = m a x ~ S I L

~

morasses

Jensen has

(see

[5]), but

treatment. qv

"v

be defined as in § I. For

is a cardinal"~.

v~ S+-Card

We note that

~(v)

can

also be found as follows. Define

<~(v,i)li
~(v,o)

=

by

v

~(v,E+1)

~t(v,k) = SUD t~(v,~)

Then

~ ( v , k v) = ~(v)

if

v

Then and

q~

since the d e f i n i t i o n

~(v)~v*,

in

V , v ~ S+-Card

) v

f : v

~undefined otherwise

lim(k)

is not a cardinal Let

Defini tion,

S~(v,~)

•

We obviously have tees that

= ~f m a x S ~ ( v ' E ) i f

iff

f

Lv.+2

and

of

v*

guaran-

.

f : ~(V) ----* ~(v).

has an extension

f*

s.t. f* =L~. ---*2 Lv*

rng f*.

An easy argument d e t e r m i n e d by f : v ===~ v

f~

shows and

f*(q~) = qv"

that

H~v

(E I) Let

f = ~==>

f(~(~,i))

= ~

v, v.

if Then

= ~(~f(i))

f*

is u n i q u e l y

So, by slight abuse of notation,

V,v.

we do not d i s t i n g u i s h

Note that

So

~v.

f

and

We also set

We clearly have

f(k~) = k v for

f*.

i(_k~v .

and

given

f(~*)=v*.

48 We now introduce another notation. Let

f : ~ ~

~

and

f(~)s ~(~) --~ ~(~)

~

~+ s.t. ~(~) ~ ~(~).

by

f(~

= f~(~).

Set

~ = f(~).

Define

We have

(E 2) f(¥) , ¥ = = ~ Proof:

Since

~'¢v*

we get

~(~) & ~(~), f(<~*,q~})

the claim is immediate.

it is easy to see that

= <~,q~)

So assume

so we only have to show that Then

p~

is the

parameters I~-

from

since

~L-least a u~p~,

~ = v ~.

f

Now if

is elementary.

Then

We immediately get

~Wg rng(f).

~ = v ~.

Assume first that

pC Lv, s.t. Pv

is definable

and the corresponding

Hence an easy argument shows that

~"~v ~.

in

av~_a~. Lv.

with

statement is true for

f(p~)=~.

If

a~
proof is similar. qued. (E 3)

Let

Then

f : ~

~

~

and

~g S ~ ( V + I ),

where

[=a~.

Let

~ .

f(~)~f(~).

Proof:

By (E 2) we may assume w.l.o.g,

we can consider

f~)

.

Now set

that

~=-v, since otherwise

~ = a~, G = f(~)

and

D = f(~)"

Set X = the Skolem hull of

8~[q~]

in

Lv~

= the Skolem hull of

~ulq~l

in

L

By § I, Lemma I we only have to show that But be know that by applying

f,

X~a v =6 since

and

Lv~

Xnav=5

otp (~[n~)-~.

has definable

and

otp

(Xnv)--~.

So the conclusion follows

Skolem functions. qued.

We now introduce Definition:

Let

some special maps.

v~ S+-Card, a~v

X = the Skolem hull of Let

~ : Lp

~(~)=v.

Set

~--*

X.

and

aulx,qvl

in

x~ L (v). Lv~

•

By ~ I, Lemma I we know that

f = ~I~(~).

Then

f : ~ ~>

Let

v.

L =L~. , where

We set

f(a,x,v)

:= f"

49

Obviously, (E 4)

the characteristic

Let

f(a,x~)

: ~ ~

Then there is some we

property ~

and

of

f(a,x,v)

g : x ~

is

~ s.t. a u l X l ~ rng(g).

h : ~ ~-~ ~ s.t. f(a,x,v) = g-h.

In addition,

IVI~I~I+~.

have

We now come to the main p r o p e r t y ion. Let

4~(Ir)JT(p>

iff there is some (E 5)

Let

where

lim(p)

By

Let

We say that

for all

<~(Y)IY(P>



conver~es

l'(P. be a convergent

R(T) = f ( ~ ( T ) ) ( Y < P ) .

Then

-~-chain

(~(T)IY

is

-(-chain.

(E 2)

<
Choose

v.

and let

(E 3) we know that

easy to see that Now let

-~-chain.

~ s.t. ~(I") -~ D

f : ~ ~

a convergent Proof:

be a

for which we need another definit-

(~(T~)IY(P>

is a

H (T)~R(6>) = f(H~(7~),~(~ ) ) f o r , H T)

be the direct

~ s.t. D(%-)-~ q

tary embedding

for all

limit of

y
h :
4-chain.

ycS~p

It is

.


We can then define an elemens "t.

the f o l l o w i n g

diagrams

commute

L q (y),

Ln(T)

But then

,

where

:

....

(U,E~

(U,E~ = LG,~> 8=~--*,

~

y

L11*

- ;

(U,E~

is well-founded, for some

h(:)=~.

8.

hence we may assume w.l.o.g,

We clearly have

Clearly,

:(7)~<~

__q~ rng(h).

for all

that

But then

1~(Pqued.

To illustrate +

Set

We recall

our methods

we consider

~k(~) = ~a_Ocl lal(kl. the following

definition

For

the two-cardinal

a~ ~k(r)

(see [I])

set

v e r s i o n of

® (a) = ua

.

.

50 ~,k

i There is a sequence whenever

XC~

a~ ~k(~),

if

x~,

a

~ x

@ = @(a)

~

sequence

we used to define

and set

x = x(a) = ~ a , @ , ~ . Note that

N a = rng(f a)

Assume

Bna,

then

Let

finable

in

Claim I~

If

We first

a

and

~

~

and let

f(~(6)) Hence

~

~ ~(6)

(E 2),

(E 3)


counterexample.

L + ,

since

~Na~

Then is de-

We shall show

is a limit point of

B na~

S®

X.

® = @(a). s.t. ~

-~

that

Since ~ .

v.

for

®

is a limit point of

We only have to show

f(~@) : @. and

X~®

~(6)18~p~ 4~(8)18~p~

f(~)~ ~@ .

f(~) Hence

Na

is definable ~

is c l o u d in

~

rng(fa).

• Set

be the increasing C rng(f).

(ES). So let

4~(8)~

and let

is a successor f(~) = ~

enume-

So let

is a convergent

-4-chaln by

of the sequence

But then

and

(~(~) I!~P~

is a convergent

-<-successor

= ~

to show that

Let

8<:p. Now

B~®

that

By definition,

(~(8~)~
Obviously,

lal+~-

0

of

We also know

f : ~ ~

~a.

the minimal

where

Na

---~ B o

ration of

rng(f a) ~

® = @(a)

We now prove Claim 2. We have f=fa

@ = @(a)

fa = f(lal,x,w)

s.t. L - < L K + o

To see this, first observe under

and let

be the

Na

be given and set

Claim 2:

~

the definition

there is some

set

and

he the

~

X~®, ~

This contradicts So let

XC~

Choose

a~ ~ ( ~ )

then

~ =
a~ ~ ( ~ )

(without parameters)

L + •

Let

•

not. Let

is definable

B~a

is a limit point of

satisfies

Proof:

and

for any

.

set

lal+m, s.t.

a ~ rng(f a)

emma

X

<

INal ~ B~

be regular but not ineffable.

: min S o

s.t.

there is an unbounded

Now let

We set


of

-~-chaln. ~

be

@ = ~.

(~18~ by qued

51 The same method if

k
can be used to prove

We now turn to Theorem

~ +,k

for any regular

#c,

I which was stated at the beginning

of this chapter. Proof of Theorem Let

~

I:

be given.

We shall

show that

TK

is a rich

K-tree,

where

T ~ =

is the tree given by the family F K = IBvlv~ ~ i ÷ IF~I = ~ • Now observe that in the proof of Lemma 2 we

Clearly, implicitly Let B~na,

showed:

~

~.

then

If

~@~

have to show

a~ ~ ( ~ ) N a.

(iii).

So

For

and F

® = ®(a)

satisfies

k=~

we proved

It is easy to see that it suffices for the tree

(T',

~CT'

-~ )

s.t. <~,

branch

of length

-<

)

is a

k.

Choose Since

k~

hence s.t. ~

S

f

= f

Let

f(T) = ~. ~d TS"

D(8)~ ~8. convergent of

<~(8)}.

Hence B ~

IPl

f

-~-chain by

~--(~, since is a "cofinal"

to show that = kl.

~

has a

Clearly,

let

there is some

~
s.t. k~ rng(f

"

~.

Then

is elementary

~(8)

(E 5). Let ~

L~W

So let

~
f(~(~))=

Clearly,

.

v

I ~Pl

cf(p)=k.

8~ "~

branch

where

for ~

and

T

T

8<~.

and

is a

-<-chain Then

f(~) = p

and

So let

(of cardinality

"

) = m. fla = id~m.

k-tree,

we get

<~(8)l~<~a)

s.t.

<~(8)I~<~>

is a

be the minimal

is singular". of

f ~ = f (a,T,~)

f(~) = k

But there is a convergent

So let

B

For

f : ~ ~

Since

U

v~ ~

statement

~o

L -~ -~ L ~+

and let

k~.

is singular.

is regular,

Set

f"~ =

p

I. So we only

this in § I o So let

We have

p = min

that

(ii) in Def.

to show the corresponding

T' =

k-tree.

Let

We may assume w.l.o.g, Case I:

where

is a limit point of

~

-<-successor "~

~ = f(~). k).

is regular". Then

52 Case 2:

k=p

In this case,

the argument

is essentially

the same as in the proof of

Lemma 4, § I, so we only indicate

the proof.

T, A [ k ~ L v,

is the sequence

~

.

where

Let

~v

"cofinal" Hence

be minimal

in

B ~

A =
~

for some

s.t. ~

is "cofinal"

~,

~k~

SEn~,

in

T

Choose

v~ 8 k s.t.

we used

rng H~v.

where

where

~=~

Then and

to define ~

is

H~v(~ ) = T.

~ = nvv(~). qued.

We now give an application Proposition

3s

H :

Assume

, ~, Let

X_~ +

and

h(~)})

Proofs

Let

~_C~(~).

v = max where

where

g : k

)~

T

has a branch

whenever

~tree.

g t k

Then there is some

Let

Then

) X s.t.

~-tree

given by the

)T by

FCF

having

S(~B,B~)

family

the analogous

= gB~v

be given and let It suffices

where IF[ =

to shows

k enumeration

so that for every

E~) = g(v).

is regular.

v<~
be the increasing

,F

k

h : k

~ = H"[~] 2 .

of length

h : k

and

and

)T

Now let

, ~

~k<~

H ~ [F] ~

H : IF] ~

is regular.

s.t. H(lh(v),

Clearly,

vck

~(~h(v),

of such a

there h(:)~)

is some = g(v)

v<~
We now prove ~
~--tree hence

to find

We define

Define

every

be a rich

k<_~

branch.

all

for

~ =
For then let

E~ ~

= g(v)

Ip<~[Bnp = ~ I .

Claim:

is a rich

iXl=k,

It suffices

property.

~-trees.

which satisfies:

there are injective H(Ih(v),

there

of rich

the claim.

We show by induction

Then the claim follows

~.

So let

IDl
(ii) for

T.

g,g~ T a,

g~g.

~.

Set

by induction But simple Hence

[~

from condition

D = {dom f[f: T0

hypothesis.

tree arguments [
Hence

that

for

(iii) for the rich for some

[F~DI:k

show that

[~a[
8<m~,

by condition

g~D $ :~D, ~edo

if

53

Corollar~: Hn

: [ +]n+~

and

~

Assume

IX I

,

there is a rich [~]n(1~<~)

A-tree.

s.t.:

is r e g u l a r there is some

Then there are

for every YC_~ s.t.

X C ~ + s.t. ~ I X [ < _ K

IXI =

}YI

and

IX] n+' 2 [YI n

Proof:

Let

H

be d e f i n e d as above and set

qued. This result gives a rather general s t e p p i n g - u p lemma in v a r i o u s kinds of n e g a t i v e p a r t i t i o n relations.

For example,

L

for

recall

the d e f i n i t i o n of the s q u a r e - b r a c k e t relation. DefinitionL XC ~

and

, [kv] ~n

~

v<~ s.t.

iff for all

]XI = k v

and

f[~]n ___+ ~

there is some

v~f"[X] n .

The f o l l o w i n g result gives a p a r t i a l a n s w e r to P r o b l e m 17 in [4] u n d e r the a s s u m p t i o n

V=L.

the s t a t e m e n t

(see [7~).

P r o p o s i t i o n 4:

Assume

regular,

V=L.

~_~v~

kv

Proof:

We may assume that

left side is false. corollary. Then

Let

f ~ Hn

T o d o r e v i c has p r o v e d the c o n s i s t e n c y of

Let

n~1.

80 let

~-~

be r e g u l a r and

~(_~,

Then:

~

is not ineffable,

Hn

s [~+]n+~

f : [~]n ____. ~

since o t h e r w i s e

~ [~]n

the

be as in the

give the r e l a t i o n on the left side.

gives the r e l a t i o n on the right side. qued.

Note that we could also treat finite (.

be 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

k~s ~2.

above. Finally, For

g=~

let

the f o l l o w i n g

result was p r o v e d b y D e v l i n in [2]. P r o p o s i t i o n 5: ir rich. Let

Let M =

F~(~)

[gB]Bg F~.

be a f a m i l y s.t. the ~ree given b y T h e n given any

X(IM S.to

[X[~.~, ]X[

F

regular, there is some by

(.

or

Y__CX s.t.

~.

Proofs

or

and

Y

is wellordered

~, •

(Recall that there is no by

IYI = IXI

X_C~2 s.t.

IXI = K +

and

X

ls wellordered

~,).

This follows immediately from the proof of Proposition 3. qued.

References [I]

KoJ. Devlin,

Aspects of constructibllity,

Springer Lecture Notes in Mathematics 354 (1973) [2]

KoJ. Devlln, 0rder-t~pes,

trees, and a problem of Erdos and

Hajnal,

Periodlca Math. Hungarica 5 (1974), pp.15J-160 [3]

K.J. Devlln, The combinatorial principle

~ @ ,

to appear [4]

P. Erd~s and Ao Hajnal, Unsolved problems in set theory, ln: Axiomatic Set Theory, Proc.Symp.Pure Math.Vol.23,

Part I(1971)

pp. I 7-48 [5]

R.B. Jensen~ The

[~]

L.J. Stanley,

(~,G)-morass

(unpublished

manuscript)

"L-11ke" models of set theory~ forcing,

comblnatorlal

principles and morasses, Thesis, Berkeley (1977) [7]

St.B. Todorevlc, Some results In set theory If, Notices of the AMS (1979), A 440

[8]

K. Wolfsdorf, Der Beweis elnes Satzes yon G. Choodnovsky~ Arch.Math.Loglk 20 (1980), pp. 161-171

[9]

F.G° Abramson, L.A. Harrington, E.M. Kleinberg, W.S. Zwlcker, Flipping properties: a unifying thread in the theory of large cardinals, Ann~of Math.Loglc 12 (1977), pp. 25-58

SOME APPLICATIONS

D.

Donder,

Bonn,

In § I, § 2 a n d questions

about

In § 3 w e p r o v e

R.B.

Jensen,

§ 4 of this

partition

OF THE CORE MODEL

Oxford,

paper we

properties

an a n a l o g u e

a n d B.J.

apply

the

of c a r d i n a l s

of Schoenfield's

Koppelberg,

core model and

Berlin.

K to

ultrafilters.

absoluteness

theorem

for

K.

In that,

[10] M i t c h e l

if < is R a m s e y ,

it h a s

the p r o p e r t y .

Theorem Erd~s

Ramsey

there

In § I w e

K be

cardinals

is a s m a l l e s t

improve

e-Erd~s,

are

inner model

Mitchell's

where

absolute

cf(e)

result

>~.

in K a n d

in w h i c h

to:

Then

< is ~-

in K.

Corollary

Mitchell,

2: L e t

inner model

then

conjecture. lative

that

then

I (Jensen) : L e t

Following

We

proved

< be

in w h i c h

we

e-Erd~s, it h a s

use o w n m e t h o d s Silver

then

showed

to the ~ x i s t e n c e

get:

where

cf(e) > w .

There

is a s m a l l e s t

the p r o p e r t y .

to d e t e r m i n e that

Chang's

o f an ~ 1 - E r d ~ s

the e x a c t conjecture cardinal.

strength

of Chang's

is c o n s i s t e n t We

show

the

re-

converse:

56

Theorem Then

3

< is

In

~-Erd~s

§ 2 we

weakly no

(Donder) : Assume

Theorem every

4

Ketonen is

Theorem Then

on

no

Jensen

and

(Since

results,

proving

filters

are

The

main

Theorem

6

a e,

M ~L[a].

B.J.

existence the

of

non

statement

regular

that

and

there

is

K be

an

infinite

cardinal.

Then

(<,~-regular.

result

under

~L H.

Let

on

the

in

K be

assumption

the

on

~O ~.

Our

written,

assumption

proved

Ketonen's Donder of

such

that

2~

= <.

K.

originally

the wake of was

regular

ultrafilter

Koppelberg

paper

theorem

of

( J e n s e n ) : If then

7

e : ~I"

work has

V = K

this on

(K,<+)-regu

vastly

that

result

extended

all

uniform

these

ultra-

regular).

K

a corollary

Theorem

K is

normal

~O ~,

this

K = ~2'

method.

weakly

assumption

larity

As

this

his

Let

+

[ 8 ] proved

(Jensen) : Assume is

Let

cardinal.

IL ~. on

on

the

~L ~ b e

a measurable

ultrafilter

based

there

the

real

in

5

R.B.

with

to

Let

(Jensen) : Assume

uniform

proof

ourselves

ultrafilters.

model

conjecture.

K.

address

normal

inner

in

Chang's

is

3 reads

0~

does

not

exist

but

a#

exists

for

every

I

E3-absolute.

of

the

(Sensen) : Let Then

§

Ba 6 L [ M ]

proof

we

obtain:

1

A be~ (A(a)) .

2.

Let

M be

a mouse.

Assume

A(a),

where

57

§

4 contains

In

[12]

two

I) ( J e n s e n , Assume

Koppelberg's

V

uniform

results K.

Prikry

= L. over

Let <

(Proposition

Silver)

there

is

s.th.

~ < I < < .

These

results

an

ultrafilter all

Let

~ ~ <

regular

be

Theorem filter

on

some

methods

filters been to

on

written,

theorem

to

not

weakly for

strongly is

compact.

all

i ~ ~

Let

U be

.

inaccessible

l-indecomposable

following

Assume

regular

~ L~

cardinal

cardinal

<

for

l

all

and

let

LCH

~,<

~ ~ cfl < i are

Assume

for

~ ~ K .

all to

these

cardinals

Donder

<

two

theorems:

and

let

. Then

U

U be

is

a uniform

~-decomposable

6 ~ < .

cardinal

D.

4.3.

the

regular

singular

some

< which

Koppelberg)

leading

and

0 @ exists.

i < < s.th.

(B.

for

over

cardinals

6-decomposable The

that

cardinals

4.7.

presented:

20).

Koppelberg)

some

are

l-decomposable

theorem

Then

decomposability.

Silver)

regular

is

generalize

on

cardinals

U

ultrafilter

(B.

J.

be

Suppose

4.3.

for

~ ~w

I and

on

decomposability

and

. Then

2) (J.

Theorem

on

work

has

<

results

be the statement

strong

7L ~ Assume

left

impregnable. obtained

limit

and

U be

LCH

But

a result

limit

cardinals.

let

the

that all

a uniform . Then

case

of

since for

U

uniform the

this

is

ultra-

paper

case

ultra-

has

analogous

58

5 1.

Partition

In

Cardinals

in K

[ I ] Baumgartner

introduces

the

following

useful

defini-

tion.

Let

~ ~ e ~ < .

subset

of

such

(Concerning

the

IA[

our

the

the

< ~

then

subtle

It

and

In core

Theorem

M

mention

type

of A.

(i.e.f(~)

that, On

unbounded

<min($))

for

if A

the

Being

a pleasant

f.

is a s e t

other

there

hand,

of o r d i -

~ denotes

~-Erd6s

<~

K.

We

that

countable

there

is

M = L[f].

Since

~

for

is o b v i o u s l y

some

facts

a strengthening

(~)<w ordinal So

and

~ is

the

especially

property

showed

improve

< be

< be

paper

of

least

~ such

< is R a m s a y

a-Erd~s

cardinals

that

iff

< is

that

they

are

inaccessible.

Mitchell

Let

Let

Baumgartner's

< is ~ - E r d ~ s .

following

such

a closed

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

us

to

a limit

therefore

model

necessary.

order

reader

d is

[ 10]

The

model

is

l.l:

let

relation

if

<-Erd~s.

~ e and

cardinals.

partition

(~)<~,

C is

o f X).

Conversely,

the

the

iff w h e n e v e r

• < is r e g r e s s i v e

notation

refer

e-Erd~s

IDI

that

cardinality

about

~-Erd~s

f:[C] <~

denotes

We

of

is

< and

is a D ~ C

nals,

<

this

~-Erd~s

example ~l-Erd~s

< is

that

an L - g e n e r i c is c o u n t a b l e

cf(~)

shows

~-Erd~s

cardinals

are

Ramsay

Then

~-Erd~s

in

to:

and

and

Ramsay

that

set

> ~.

some

~ = w~.

in M b u t

not

collapsing

~ is

assumption We

construct

in K M. map

in M a w e l l - k n o w n

about

f:~

an

Since~L(e) , ~.

argument

in K.

~ is inner is

Let due

to

59

Silver

yields

that

ric extension

o f L.

It is w e l l - k n o w n reformulated that

this

Therefore

that

can

in M.

of

for

I ~ < is a g o o d

8 <<

relation

~-Erd~s

L e t A I , . . . , A n c < and

KM = L b e c a u s e

be a-Erd~s

indiscernibles

a l s o be d o n e

= C~I LB [ ~ ] for

But

~ cannot

the p a r t i t i o n

in t e r m s

Definition. ~B

< is a - E r d ~ s

set

for

<

M is a g e n e -

in K M.

, (a)<~ c a n b e

structures.

We

show now

cardinals.

Ol=
6 ,~ > ,

.

set o f i n d i s c e r n i b l e s

for O~ (or g o o d

for (}/) iff

for all ¥ 6 I: (G i) ~ y

< C~

(G 2)

I-y

is set o f

Lemma

1.2:

Let ~ ~ e be a limit

model

~=

type


Let

< be a - E r d ~ s .

T h e n C is c l o s e d of t h e

formulas

parameters

ordinal.

a good

< is a - E r d ~ s

set of

indiscernibles

that

f([$,~})

numbers

= O,

if

and otherwise

. Then

in <. L e t

for

a counterexample.

II = a

G i v e n Ot as a b o v e

unbounded

from X have

gives

6 ,~ > h a s

for

iff e v e r y of o r d e r

~.

Proof:

such

indiscernibles

Let

I is g o o d

The opposite

I qC

. Define

the

is t h e

same

be homogeneous

f:[C] <~ --~ <

type

number

with

over

of a f o r m u l a

for

f such

which

that

for Cry.

direction

It is n o w c l e a r is a s t r e n g t h e n i n g

f({~,~})

< ~ }"

for X £ C formulas

than ~

realize

{X < < [ ~ X

I B < <> b e an e n u m e r a t i o n

such that

less

$,~

< ~

set C =

that

is o b v i o u s .

the

of Theorem

following

i.i.

indiscernibility

lemma

60

Lemma Let

1.3:

I be g o o d

that

I'

Proof:

is good

in 0%.

6 E OZ

that

~y

{(~

Using

for

we

This

, 6 , D N <,A>.

there

is I' 6 K such

y 6 I is i n a c c e s -

is a m o d e l

of ZFC.

y E I and n < w

to the

y < Y1 < .... < Y n

so ~ny

'r in the

as

follows.

set of x E % +

which

are

in y U { y , y l , . . . , y n ]

' Yi 6 I.

= C~6 (y,n)

replace

for a 6(y,n)

" CO%-definable"

definition

above

where


by

Yn+l 6 I , Yn+l > Y n "

yields (i) ~ n

6

O~ n+l Y

Y Set


each

that OZ

from parameters

see t h a t we can


. Then

arguments

of OZ

OZy--n is t r a n s i t i v e ,

(GI)

> w

. Set ( ~ =

(6+)Oz.

(~-definable

Clearly,

cf(IIl)

it f o l l o w s

restriction

where

[A] c K

I cI'

O~y--n ~ y +

= the

L

indiscernibility

~+ =

models

~ny

that

for O% and

Since

set

We d e f i n e

such

for OL such

By s t a n d a r d

sible

For

Let A ~ <

C~y =

Czy--n . It is o b v i o u s

U

that

n<w

Let h ~

(i 6 w) be a c o m p l e t e

set of d e f i n a b l e

Skolem

functions

for

O~

1

Since

n YY'

I

satisfies

--n : CZy

(G2)

~ ~ n y~|

(h

~

as

parameters

for from

define

for y ~ y'

(7,y,yl,.

where Arguing

,

we can

..

elementary

by:

,yn )) =

h.m~ 1

$ < y ~ y' < y l ..... Yn

(i) we I. N o w

see

that

set ~yy,

embeddings

n

(7 ' y

,

'Yl

' " "

"'Yn )

' yj 6 I

6 (~Z and is O Z - d e f i n a b l e y7' = nE~ ~_] ~ YY' n U s i n g (2) we get

in n+3

.

81

(3} ~yy, :~.y It is c l e a r

that

J

for

on y in < ~ y

, Uy >

is o b v i o u s .

and ~yy, (y) = y'. N o w d e f i n e

~ y

y 6 ~yT, (X)

=

is n o r m a l

and < ~ y Normality

~ y = id

yy'

X 6 U7 (4) Uy

f~ ~,,

, Uy >

is a m e n a b l e

To p r o v e

y < y'

.

amenability

set U n = U

•

T

It s u f f i c e s

to s h o w t h a t U n 6 ~ But this T Y" O Z - d e f i n a b l e from ~yy, and U nY 6 0 ~ y ÷ .

Since

that z yy'

~

7,

: < ~

N O W set <%,

, Uy, >

< < o Z y , Uy >

Y

is i m m e d i a t e

since U n is Y

n = U ny, of U n is u n i f o r m we get ~yy, (Uy) Y is c o f i n a l we have:

the d e f i n i t i o n

So o b s e r v i n g

(5)

N ~n Y

, Uy>

,

T* = sup I ,

, < ~yy.

, ,

U

, >

I* = I U {y*}. C l e a r l y , O~y, ~ O ~

[ y £ I >

I Y 6 I > , < Zyy,

since ef(IIl)

<~

be the d i r e c t , y,y

_ y' ]y <

> e, and m a y t h e r e f o r e

(3) - (5) go t h r o u g h

. Let

l i m i t of

6 I > . ~y.

is w e l l

be t a k e n as t r a n s i t i v e .

founded, It is

clear

that

for y,T' 6 I* .

<~T'

U U U Let M Y = J T = ~{ j Y [ j T 6 ~y} for y 6 I* ~X ~+i Uy Uy > is r u d c l o s e d we see t h a t J ~ 6 ~y implies

Since Uy J ~ + l ~ O~y

So MY

is a p r e m o u s e

We n o w show:

and we k n o w M T c

~7

' M7

~ ~Y

(6) M Y is i t e r a b l e .

Noting

t h a t ~yy,

iterable.

This

r M Y : MY ----~o M Y*

in t u r n w i l l

follow

it s u f f i c e s f r o m the

to s h o w t h a t M Y* is

fact t h a t U

is Y*

e-complete.

But this

is c l e a r

since

cf(y*)

> ~ and:

62

X 6Uy~

Our

next

.

aim

~ X 6 ~y

is to s h o w

(7) ? (y) D

and

that

M Y { ~n

3 y< y*

I -yc_X

(gZy 6 M Y

for

. We n e e d

first:

y 6 I ,n <

Y Suppose model

not

of ZFC

is a l e a s t But My

and

then

set

satisfying

T < <

such

T 6 ~y

= J U8

6

,

It s u f f i c e s

N ~y

by

~

(y) fl J T + l

~y

< OZy+

c_ ~

this

, since

T _>y

such

Let N 8 be the

that

be the m o u s e iteration

get a 6 M.

T h e n ?(y) (7).

a 6N.

Using

= ~(y)

proves

(8) let

Clearly,

to

So a s s u m e

we m a y

assume

Hence ~y

~y

N M 0 c_~(y)

we be

are the

ready least

order

since

~(Y)

formulas

and

n

such

that

n N O c N c_ Ozy--n , w h i c h

~ < By

Then,

Let

at a

N 6 ~y

8 > <

satis-

is a b s o l u t e

N 6 (~7+

assume

IN to a r e g u l a r

Since

t h e n (~

of m o u s e

we m a y

to f i n i s h

, hence

+l

iterable.

But

is a m o u s e

M@ c_ N 8 a n d c h o o s e

= O~,

Y of a s u b s e t

there

D ~y

. let M 0

8 . T h e n M@ c N 8 or N@ c M 8. If N 8 c M 8 ,

~yy, ( ~ y )

Ny = j U y

the n o t i o n

~ C~y+ of

(y)

a ~L.

B < T < y+

N 6 (~6y--n contradicts

(8).

Eventually,

map

So let a 6 ~

there

and

y 6 I

iteration

of M Y

N MY

This

Since

B 6 ~y

D MY

for y 6 I.

~l -< Y~ < < "

" Clearly,

observation.

for

Since O g is a

in L [< . H e n c e

an e a r l i e r

(y)

iterability,

normal

f~yn

~

• Hence

ZFC + V = K + V • L . M o r e o v e r

in 05

we

that

since

to s h o w

D L~M Y

fies

V = K, U 6 0 L is not

, contradicting

(8) ~ ( y )

~(y)

B = By , U = U nY . T h e n M Y = J ~ .

of y o n t o

< ~i I i < ~ > for (~.

NY .

the p r o o f

such

~yy, (~y)

= ~y, .

fl j U y ~ J~Y ~+i Hence

Set

there

methods

is a

Z l(NY)

since

it is

'

N Y is a m o u s e ,

be a r e c u r s i v e

By s t a n d a r d

that

of L e m m a 1.3. U O~y 6 J~+l ¥ "

enumeration (cf.

the

of the proof

of

first

63

Lemma and C~y

1.2),

if ~

By

one

v 6X i Y

, ~

~i(~'~)

the

for

Now

I'

set

f o r C Tt,

~

=

' (]rAy b

< ~>

X i 6 N Y such Y

that

Xi £ U Y Y

, < ~ >

6 [ X i - ~ in Y

, then

~ i ( ~' T%) "

choice

i i~< ~ X Y*

is

O~y~

we

"

have

Then

~

OZ.

a mouse.

So

Before is

and

choose

Zyy,(X$)

=

X y, i

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

y < y'

since

N Y~

canonically

< v

canonical

y 6 x y, i

since

may

But

I'

and was

I' 6 K is

proceeding

a natural

I ~I'

further

prewellordering

is

defined

the

let

of

I'

the

set

us

good

for

OZy~

from

N Y*

and

we

were

remind

mice

is

hence

N Y*

£ K,

looking

for.

reader

that

the

which

,

defined

there

as

follows

M

It

is

~ N

shown

iff

in

M 8 6 N8

[4

] that

where

~

@ >M,N

restricted

is

to

regular.

the

core

mice

is

a well-

ordering. We

often

use

Let ~

(¥)

M

D

Our = K.

(cf.

the

HM

= KM

<

K

for M

where

aim

a mouse

proof and M.

of HM

simple

M,N

every

=~(y)

next

Call

following

~ N,

D M c N

Then~(y)

V

the

are

y ~ ~,T

fact.

mice

at

To

prove

.

D

M@

c

is

to

describe

M

at

~(y)

4.9

~

Moreover

ZF.

in

K

Lemma

inner

be

an

inner

let

8 > ~,T

is

1

Then be

regular.

of

if an

which

It

M

is

is

satisfy

easily

seen

critical,

iterative

of

M

then at

M,

. Hence

model

models

H M< 6 M .

that,

if M. '

W

this

iff

M.

HM < H i <
Let

all

[ 4])

<. , t h e n H i £ M. and ±
1.4.

respectively.

n N e c N.

< critical

Lemma

<,T

ZF.

~J H I i 6 0 n ~i

is

an

inner

Suppose

KW

% K.

Let

N

84

be

the

Then

4-1east

N

is

core

critical

mouse and

such

Kw

=

that

~J i60n

N{

H 1

W. where

'

N

are

the

~ L and

let

A6~(K

that

M.

i

iterates

of

N

a t K.. l

Proof: (I)

We

first

show:

~ (Ki) N K W c iN _

This

is

there

clear,

is

if K W

a mouse

M6

= L. KW

So

at

assume

a T > <. --

KW such

A6

Since

i) D K W . T h e n M~N,

we

get

1

A6N.. 1

H Ni c K W Ki-

(2)

Let

H Ni

a6

and

let

JU . A s s u m e

Ni =

w.l.o.g,

acy

Ki

a~ and

L.

Then

a6

Z

for

some

y < K.

-

a6 (M)

3U.~+I - sU~ f o r (see

[4

],

some

K.I < ~ < ~"

Corollary

But

and

i

then

5.2.1).

But

M4N,

critical.

Let

Ac<

M

= 3y

hence

is M6

a mouse K W which

W

implies

It the

A6

a 6 KW . This

remains

bounded

to

proves

show

subsets

N.

But

by

As

a corollary

that

of

(2) w e

N

is

K which

know

of

(2).

that

Lemma

are A6

1.4,

in

N.

K W . So

we

It A6

get

is N

the

recursively

enough

follows

to

show

from

following

code that

(I).

result

of

Mit-

chell.

Lemma

1.5:

Assume ~)

L e t ~ (~)

~(~).

such

Then

be

we

a

can

formula define

such an

that

inner

ZFI- V ~

model

W

(~ (~) - - - ~ < ~t~)). = W~

(uniformly

in

that

(a) ~ ( ~ )

holds

in W

(b)

If Q

is

inner

(Hence

W

the

Proof:

If

is

there

an

smallest

is

model inner

a critical

and ~(~) model

core

holds in w h i c h

mouse

in Q, ~(~)

N such

then

W

= WQ

holds).

that

~(~)

holds

in

65

N.

~]

i60n

H Ni

set

<.'

ih6i0 n

W =

H K 'i

set W

= K.

Corollary inner

The

now

Chang

conclusion

1.6:

model

We

Let

turn

~=

= ~I"

to

an

the

Silver

N is

the

~-least

such.

Otherwise

with

that

Chang's

this

and

an

~-Erd@s

be is

the

Then

there

is

a smallest

~-Erd~s.

of

the

indiscernibility

strong

two

cardinals

~A',B',...~

of

of C h a n g ' s

cardinal. implies

On

the

assumption

the

length ~'

with

= ~I

conjecture

other

lemma.

conjecture.

countable

such t h a t

~ ~

consistency

the

cf(~) > w.

a structure

w1-Erd~s

that

and

application

conjecture

show

immediate.

following

there

proved

is

< is



Then

a model

r be

in w h i c h

proposed

Let

where

1

1

hand

and

~ = w2, ~'

= ~.

starting Kunen

from

showed

existence

of O #.

We

strengthen

in S i l v e r ' s

result

cannot

be

weakened.

Theorem

1.7:

Assume

~-Erd~s

in K.

Proof:

Let

A6~(~)

a good

set

of

fice,

Chang's

N K

and

conjecture.

set

indiscernibles

since

there

will

then

<+

(<+)K

and

choose

0[=
<

Let

,£,DNK,A>.

for O t h a v i n g be

such

< = ~2'

It

order

a set

e = w1"

is o u r

type

in K b y

6.

the

Then

aim

This

< is

to

find

will

suf-

indiscernibility

lemma.

Let choose

a set

countable jecture

=

Ec p such -

cardinals there

is ~ ' ~

that

in L [ E ] . P ~such

p < ~+ such K

cL

p-

Set

p

[E]

t h a t 016 KQ

and

~ =
6,<

are

[E],E,K P

t h a t ~'

= ~I'

and the

,Dp>.

K < K ~+" P first By

two

un-

Chang's

P e n~'

Then

is c o u n t a b l e

and

con-

6B

6 ~'.

L e t b':

~-~

~

where ~

a n d b' (~) = ~ .

Our

choice

b' (e) = <. S e t

~ = ~ n ~'.

s e t b = b' ~K. C l e a r l y , For

later

We

Case

K

=

two

transitive.

of E p r o v i d e s so ~ is the

b:---~

use, n o t e t h a t ~ = < K A K

consider

I:

is

Let ~ = P

that

first

e n ~

is t r a n s i t i v e

point

moved

b y b'.

and

Now

. ,SN~,A>

where

Ac_ ~ a n d b: ~ - - ~

~.

cases.

K_.

P S e t ~ = s u p b"s theorem the

of

first

[4 ] s h o w s point

by another

on e. B u t

can be

iterated

~6 K~L[U<] ~.

Case

2: K # K-. P

Then

there

would

first

in MS,

K, w e

( +)L[U] U

normal

that

N 6 K,

In a d d i t i o n ,

hence

since

8 for

also

get ~7£6.

b(~)

for ~.

for ~ of o r d e r

type

But ~,

that

since

@ > 5. B u t also know

@ large

points

( + ) K < <. H e n c e

U

K. B u t

since

then

of o r d e r

iterate

is a m o u s e

b: K---~m K +. <

union

o f the m i c e w h i c h

iteration

that

e is the

points

of

largest

M6 K

enough.

is a g o o d

b: ~ - - g Z w O .

S o we

AM, 6

B u t t h e n ~ 6 M~.

of M up to ~. T h e n

then b"C

Hence U

otherwise

the

L[U]

But

with

on

= <. S o w e h a v e

M e for

iteration

=

the ~ - t h

K is t h e

in K. W e

in K,

C of C is g o o d

cernibles

mouse

KcM

~ is r e g u l a r .

set of i n d i s c e r n i b l e s

M@ K such

for e v e r y

let C = C M ~ be the

segment

a good

with

So a

K ---}El K.

an i n n e r m o d e l

show that M6 K s • Recall

of K. H e n c e

cardinal

Since ~6 Now

has

is t h e n

U

to a ~:

to e, w h e r e

to see t h a t

to <, y i e l d i n g

g i v e M 6 K.

regular

regular

[ 3 ], t h e r e

it is e a s y

that N ~M

are e l e m e n t s M are

of

is a c o r e m o u s e

at a y < 5- W e

which

b y b is t a k e n

certainly

type

----~i K ~ c o f i n a l l y .

K

t h a t b ° c a n be e x t e n d e d

moved

theorem

normal

observe

a n d b ° = b ~K . T h e n bo:

some

final

s e t of i n d i s -

67

§ 2

Regularity

We Let

first

U be

quence

such is

for

6 U

for

have

U

Let Then

f"YcT

W6

definitions.

I~ £ X > s u c h U

is

u

all

some

cardinal

< y

for

closed

U and

X

6 U

this,

define

for

m 6 X.

unbounded

~ < <.

U

W

for

It

is

subsets

=

f:

of for

{~ 6 W I ~

there

normal

< ~

clear

In

weakly

= sup

U

bounded

a weakly

addition, normal

we

U:

{~ < ~ I d 6 X ~ } } .

W 6 U.

To ~ U,

see f is

regressive

{~If(d)

< y} 6 U.

f:

W ~

mod But

U.

then

< by

f(e)

= sup

Hence

there

is

Y D X

= ~ which

{~ < a l e 6 X

y < < such

}.

We

now

state

statement

Theorem weakly

weakly

Assume

normal

a uniform

contradicts

X

non

is

theorem no

model

on

chapter.

with

a regular

ultrafilter

on

< is

(y,K)-regular

Ketonen

(<,

showed

<+)'-regular

Assume

I L ~.

(see

<

is

[ 6 ])

ultrafilter

with

Let

the

same

< > ~ be

(<,<+)-regular.

Let

cardinal.

that on

nL ~ abbreviate

a measurable

< be

+ ultrafilter

this

Let

ultrafilter

2.2:

of

inner

6 U. T

~L ~.

and

normal

Corollary

main

"There

2.1:

Kanamori

the

Assuming

that

T

the

is

iff

< is

that

<.

se-

~ 6 X and

iff

is w e a k l y

principle

Set

a regularity

c_~

u

function

T < u).

diagonalisation

By

(y,<)-regular

(mod U)

Y 6 U,

u.

X c_~ ,

that called

regressive

for

following

Ultrafilters

a regular

q < <.

every

contains

the

on


all

of

familiar

satisfying

and

(i.e.

normal

Y =

mean

a sequence

U

some

ultrafilter U we

uniform

mod

Normality

recall

an

{ v l q 6 u v}

and

for

if

a cardinal.

Then some

every

~ < <.

< > ~ and

<+ , then

property.

cardinal".

there

< + carries

Hence

Then

we

is a

get:

every

uniform

68

In

[8]

Ketonen

O ~ does filter into

not can

two

exist. be

itself.

actually

get

need

used We

new not

Theorem

He

showed

to

find

make

such

entirely

lemma

proved

an

under

that

an

the

stronger

irregular

a non-trivial

elementary

use

of

his

ideas

embedding

for

K.

But

we

arising

true

for

from

K and

the

we

assumption

weakly

extensive

cases

be

2.1

facts

cannot

ultra-

embedding

to

also

normal

show have

that

exclude

of

that to

that

we

L can

consider

the

condensation

the

possibility

(K+) K = K +.

Proof

Let

of

Theorem

K be

which

a regular

is

not

shall

show

We

first

remark

~(<)

tions tary

N K

f 6 K)

that

with

6.12

such

now

sential

carrying

it

for is

of

the

of

[ 4].

So

to

find

some

of

this

a theorem

with

of

several

K by

V

us

[ 3]

sets Set

each


a system

the

whole

K < m < K

IK < m < < + > {T <

ultrafilter

(restricted

M.

there

But is

of

structures

and

to

an

M

Q~

the

in

K

model

is

es-

code

the

proof.

pick and

IK
which

and

DN
f

Y

mapping

<

onto

m.

_Ac

Let

+ <

~+ T

[A],ANT>

~

~ < < set

_0

}.

Then

E

is

+ unbounded

= K by

inner

. For

smallest

QW

T 6 E, QT

such

that

e~Q.

=
>

and

V

func-

elemen-

then

embeddings

+ For

cardinal.

a non-trivial

class

U

cases,

a measurable

a non-trivial

gives

transitive

ultrafilter

Considering

model

ultrapower For

by

y < <.

normal

cardinal.

build

for

into

a weakly

inner

suffices

that

K

any

an

is w e l l f o u n d e d .

a measurable

We

there

that

embedding

Lemma

cardinal

(y,K)-regular

we

over

2.1

closed

6g Then

C T := {~ <
= < N Q~

x 6 Qy { e 6 C Ix6 Q~}

is a final

~Y:

QY ~ ~T

Finally

where

set T

We o f t e n

of C~.

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

Q~

= O n ~ Q~, K T~ = K ~ and T~ : ~ T ~ K T. H e n c e

use the

= ~IQ~e

To v e r i f y

(hence K ~ =

this

The

so the rest

following

2.3:

{a£WN

6 Q~

Qn"

Since

every

Q < Q

in QT

from ~, hence

is u n i q u e l y

= ~Te n Q . But ~T~ is an end e x t e n s i o n

deterof

is a key e l e m e n t

in our proof.

there

is T 6 E such that

}6 U.

-

Suppose

Proof:

that Qn is d e f i n a b l e

+ Let W 6 U, f 6 6~T W = ~ . Then

T

(~)-I(Q)

is clear.

sublemma

C If(a) < T

YT : { ~ 6 W ~ C T Our e a r l i e r gT~ h a s

regressive,

not. IT

For ~ 6 W let h

< f(~)}

Fact

shows

and define

that

inject

f(e)

gT 6 6T~T~ by gT(~)

the gT are p a i r w i s e

< many p r e d e c e s s o r s

(mod U) gT~

in fact gT~(~) < a, there

into ~. For T 6 E set = h~(T~).

distinct

(~ < < ) "

is a c a r d i n a l

mod U. Hence

Since

gT~ i s

y < < such

that

-

Y = {~IgTm(~) <

y}6U.

Choose

$ 6 E such that

T > T ,T { and set

for

~ 6 Y N C- =: Y: T

u

= {~ < ~i , , T ~ £

Then < u Hence

K<+.

( 7)-~ ( K ) £ K ~, K ~ ~ K ~ and ~a = T ~ K ~ a )"

just note

m i n e d by Q N K, we get ~e ~T Q~ N Q~,

z~:~ K ~ - - ~

following

Q~ n Q~ ~ But then ~T

QU 6 QT"

some

segment

in < and for e v e r y

Let ~ £ C , U 6 E N T, ~ 6 rng ~ T. Then ~ £ C , Q~ =

and ~

Lemma

unbounded

Q~, ~ ~ for ~ 6 E, e 6 C T by:

Define

Fact:

is c l o s e d

[ a 6 Y> i s

U is

rug T ~ and g ~

a regularity

(y,~)-regular.

(~) < gT,(~)}.

sequence

Contradiction!

for

U and ~

< ~

for

~ 6 Y.

70

We model T s

return

to

of

+ V = K and

ZF-

: KT s

Case

the

main

Hence ~

I:

is

the

T 6 E

such

of

our

notion

KT c K S T

~ K +"

There

line

proof. of

We

mouse

note

is

that

absolute

KT

is

a

in

KT t

since

s

that

{s 6 C

IK T

J KT

} 6 U.

T Let

~

Set

W

be

be

tion

L e m m a

s Y i -> T

such

KT S

{aE

7s

and

Let

T 6 E,

. Then

~(~)

let

:+ 7~ < ~

2.4:

> T

Suppose -

M.. 1 that

KT

M

= M ~,

2.5: WN C

for

L,

a

since

are

IT 6 r n g

s

since

mouse

KT N s

7~ < ~

There T

K ~

AM, cy.

at

C

set

since

iterate

WN

and

otherwise

ME K { , ~

, s6

all

N

since

for

for

have

a6

the

Hence

and

Bi

7

S

T £ rng

KT ~

at

a

Let

M s with

if

T > T.

a c-~ ,

itera-

M is

= 7~}

, such ~

T: S

KT ~

6 >- ~ , y .

Let

N =

the

large

Let

~ < T

Since

since

contradicts

S

~T

a N f M i . Hence

and

T

of

.

= y s.

M 6K T

arbitrarily

~T S

and

~ < ~

a 6~(~)

- This

iterate

M exists.

N6

K

i-th

:+ i < s

~ l

not

We

M~ be 1

n K TmM ~ - -

then

Lemma

at

Then

N ~ M

~-th

that

7~-

AM, 6 Nc But

}. T h e n for each ~ £ W there is a core mouse s the ~-th iterate of M s is a mouse at a 7 < T . Let s

point

follows Then

such.

J K~

a mouse

Proof: a6

least

{~IK [ ~

Ms ~ K~ Ms

the

M ~ N.

KT

is

a

unique our

T 6 E

But

choice

such

of

K < +t

core

it

(N).

then

model

core

that

mouse

of

ZF-. with

an

M.

that

E U.

+ Proof: fine

Let T ~ 6 E,

~ < K f

. We

6 sT

show ~+

by

that

there

induction

is on

T > ~ ~ < K

with as

-

T~

=

the

least

{a6 WN CTI

T

such

that

T > T,~,

f~ (s) < T s } 6 U;

SUP ~<~

T

and

the

follows:

property.

De-

71

f

(~)

= y~ where

i = the

Yi > sup

{T

least

I~ < { a n d

i such

~6 C

that

. T

Let

T = sup

T ~ and

= {e6

C I

sup

T h e n C is c l o s e d Z

= {~6

T~ < f f

unbounded

= sum

{~16

(~) h a s

the

On>

NOW

is

Lemma

Set

2.6: that

Let

we

Z 6 U such

(a) T

~ E Z set

the y~

are

X6~(y~) This by

Lemma

We may

Set

Now

;f

let

This

some

shows

T = sum ~<<

Z c C, w e

~ Z

limit

each

ordinal

T is

have

,, t h e n

as

I,

since

remuired.

of T 6 E satisfvin~

the

T ~.

l e t A E ~ ( T ) N K.

Then

there

is y 6 T

T-A.

that for

A6

K

. Argueing

as

in t h e

limit is

cofinal

N ~ = M I~ a n d

Be

= { 7 ~ ] ~ < i} w h e r e

iteration

I,

points

a 6
in p a r t i c u l a r

proof

of

Lemma

2.5,

e 6 Z:

= y~}

is

hence

~ 6 Z }. B u t

that

seauence

Since

~ 6 Z

(~)]{ < e a n d for

Z,Z~ E U,

a 6 Z.

Te ~

N N ~ there

Then

~ < 6' < ~,

T e = Y<

and

{<<.

some

the

applies

for

C and

for

(b){% < ~ ] B i For

that

Z = WN

if

a monotone

I-yc

v6 E such

= y<

= sum

sequence.

3.

or

Proof: get

. But

I = { T ~ I 6 < <}

I-yeA

Set

Z }} 6 U.

Hence

Yi"

l e t < [ 6 L 6 < <> b e of Lemma

such

T

a normal

conclusion

<.

T~}.

f~ (~) < T~}

e 6 Z

Hence

form

= sup ~<~

in

{~ < ~ I ~ 6

< ~ and

,(~) < T%'.


(~T n T) ~

Z[T { 6 r u g ir~ a n d

= {~ 6 Z[~ = sum

set

T,T,A6

zl;

in ~.

of

the

such

to A

rug

that

:=

mouse B~

(~)-I

T

, = y<-

M s.

Hence

for

or

- ~cy<_

-6cX_ (A),

B~

since

we

Remember

that

every

have

A

- X. 6 Ne

2.4.

assume

w.l.o.g,

that

A

contains

a final

segment

of B

for

72

every B

e£

Z.

- Tg(a)

Using

c

the

that

I

We

Using A

for

fact

that

are

A6 is

is

now

can

define

But

a

a6

Z.

Y

=

to

finish

Case

fl K

follows:

ready

V

V

over

iff

~-closed,

then

g

regressive is

{~ 6 ZI~ ~ 6 B

g:

bounded

} 6 U,

it

Z ~

mod

U

<

by

follows

such some

that y < <.

immediately

~(T)

3y < ~ since

I

as

I.

By

Lemma

2.6

we

can

define

an

- y cA.

cf(7)

=

< > ~.

Hence

the

ultrapower

of

K

by

V

wellfounded.

Case

2:

Case

2.1:

Set T

we

- y_cA.

ultrafilter

V

(b)

<' (~')

to U. that

W

=

=

T

(<+)K

= <'

[' <_ ~

Y

fa6 n

K=+.

W

not. K~

T

For .

iff to

To

see

~(e)n

show

T 6 E,

. Then

a'

a

T

~ K

I~

Let < +" = ( +)K, ~ b e l o n g s

( +)K

pick

Y =

{~I~

cardinal

For

~ 6 Y we

in

:+ < ~

@ >6

such

~ }¢u.

K T but

not

in

can

define

a non-trivial

by:

V

yields

for

each

e6

Y

well-known.

~

T

is

some

this,

~'

W

that

that

and

{e6

K

such

show

Then

since

K~ .

( +)K}. K

that

~T(X).

K

T 6 E.

that

~6

not. ~ ~

all

<' < T,

such

T > ~,

rng

WTI~'

over

first

~ 6 W

Pick

T 6

Clearly

. We

=

{~6

f~: n

is

T 6 E.

=

Suppose

gument

e6

for

•

KT ~

suffices

K,

+

: K T} 6 U

T

contradicts

V

fe£ n

IK

T

Fix

for

in

ultrafilter X6

< K

a 6 Y with

K~ w h i c h

Set

C

( +)K.

Suppose

Choose

It

{~6

a wellfounded

there

is

{~ < ~ I f ~ (v) 6 f ~ ( v ) } n+1 n let Then

us

assume

there

is

K

~

L,

a mouse

a

£ V

sequence . But

since N

ultrapower

f~ n

then

we

otherwise

such

that

such

that

may the

{f~In n

of

take ar-

< ~} oN.

K.

73

Let is

N4N

such

that

transitive.

-I

(fn)

is

Then

also

choose

NOW

{ f ~ I n < m} U ~ c N

is

a mouse

a s-descending

=+ < ~

~

such

N

and

of

N

= ~.

Let

cardinality

~,

sequence

that

mod

V

( f ~ I n < ~} c_ K 6

_

6 Z and

>_ ~a} 6 U. set

Xn

:

Then

Z =

{~ 6 Y N

and

,

fn* = ~ ( f

) . Then

~6

Xn

X* = z n

*

=

hence

pick

~

W-TIT 6 rna~ ~

{~If ~ (~) 6 f ~ ( v ) } , n+1 n

N ~ ~

where

N6

K=+.

T ~ T such

--

and

~(Xn )

T~_> @~} 6 U.

Let

and

*

*

{~I fn+1 (~) 6 f n ( V ) } ,

But

.

~

{~[T

that

~:

hence

*

fn+1(~)6

fn(~)

Contradiction!

Case In

2.2:

this

case

we

theorem

of

Un

normal

K

is

+

(K+) K =

K

actually

[ 3 ] it in

show

suffices K.

This

that

to is

U N L[U]

show

that

equivalent

is


normal

in L [ U ] .

By

~,UNK> is a m e n a b l e K the following

a

and

+ Claim:

There

are

arbitrarily

large

T < <

such

that


,UNKT>

is

7

amenable

and

UN

K

T

is

normal

in < K

T

,UnK

T

>.

+ Let For W

T

y < K

the =

moment,

{~ 6 C vT :

We

. We

first

~

shall let

PK T = K } 6 U. ~ T~

{X6~(a)nK

see

that

T 6 E be

T

there

arbitrary.

For

~ 6 W

is

T > y

Recall

satisfying

the

claim.

that

define:

I~(X)}.

show

WT = { ~ 6 WTIVT6 6~K} To

show

this

pick

T 6 E,

U. T ~ T.

Then

Y =

{6 6 W - I T 6 r n g

T}

6 U.

Let

a6

T

Let A

A

=
I~ < c~ 6 K ~

= ~ ~(A)"

Note

that

we

Then

be

an enumeration

of ~(a)

is

now

set

V~ = V ~a N KT.

actually

have

* 6 K. : { A v I ~ 6 A v} a V T 6 K=+ for a 6 ~ • d

It

n KT~ a n d

obvious

that

we

can

T

apply

the

method

of

Lemma

2.5

to

Y.

74 find

T 6 E Y

We

such

that

T > ~,

cf(T)

=

<

and

{~ 6 WTI i s a m e n a b l e } 6 U.

=

shall

show

that

By t h e d e f i n i t i o n

T

satisfies

o f V~, T

we

our

Claim.

obviously

have

for

~ 6 Y:

VT~ i s n o r m a l i n .

For

{~16

that Z6

~ 6 Y choose

U.

For

V~

< T~ } 6 U. ~6

is

Z we

normal

We

need

a

Lemma

2.7:

Let

one

V

satisfying

V

is

=+ < ~

6

Set

have on

final

<

such

Z = VT6 e

that

{~6

YN

K~-~' s o

in < K

T

,V~>

VT 6 K6

Pick

{ 6 E,

W -TI T 6T r n g ~ ~ a n d set

Va

and
T

is

> 6~}.

such Then

Then:

= ~(V~). ,V~>

~

T > T,

amenable.

lemma

+

Let

X6

V

normal

us

V a = V e'

= V

T satisfies

of

amenable.

Claim:

We we

get

N ~ M.

a

There

first

<

in < K

how

every

cf(T)>

T

,V>

this

~,

and

T < K


T

,~

finishes

~ , a ' 6 Z. segment

It

of

. Then

is

our

follows

Z.

But

there

is

at most

amenable.

proof.

We

now

immediately

Z 6 U.

Hence

V

know

that = UN

that

every K

T

. So

Claim.

2.7:

Let

suffices

a mouse

show

that mice

AN, 6 ~ ( < ) N

V be

to

is

distinct Then

and

<+

final

our

It

~ K

show

for

Lemma

T

on

first

contains

Proof

K

on

<

in < K

T

,V>

where


T

,V>

is

show:

N = 3~

the N,M

normal

at

claim at

MeN.

<

< such

implies

such

that

that

the ~(<)N

Contradiction!

~(~) n N

lemma. N

= ~W(<) n K T.

Suppose

= / ~ ( < ) N M.

not. Let

Then e.g.

75

We now prove

the c l a i m .

V V S e t M = ] e = U{]~+II- ]~ 6 K T }"

We

closely

we have

McK

iterable.

~(K) N M ~ K

a least mouse,

We

T < <

that

vide

since

cf(T) > ~.

(8) of the e a r l i e r

proof

we

in the

(using KT <

, we know

but

of T h e o r e m

cannot

argument

is n o w e a k l y

first

in L[V].

Then

Assume

is n o w e a k l y

(a) L e t < u

Hence

the

there

is

is a

claim.

that

normal does

the

ultrafilter.

at l e a s t

normal

assumption

tell

ultrafilter

qL ~

A light

us,

mo-

however,

on r e g u l a r

<

2~ = K.

2.8:

2.9:

M is

get

6 _> ~ a n d N = ~

to u s e

I L ~. L e t n be

normal

ultrafilter

a general

discovered

regular

property

by Ketonen

(see

such

that

Let U be weakly Iv 6 X> b e

normal

a regularity

L e t D ~ ~ be u n b o u n d e d

in <.

25 = <. T h e n

on K.

of w e a k l y [8 ]).

normal

on <, K r e g u l a r . sequence

ultrafilters

For completeness,

a proof.

Lemma

Hence

there,

2.1.

prove,

of a n y w e a k l y

foregoing

IL ~ there

(J~) ~ KT.

1.3 As

a n d t h e n ~ ( < ) ~ M = ~(<~. n K r-

So N s a t i s f i e s

the p r o o f

suspect,

K +) <

t h a t V is n o t n o r m a l

that ~(<) N ~

of the

such

was

. V is o - c l o s e d ,

T

the e x i s t e n c e

assuming

We have

of L e m m a

V is w - c l o s e d .

that

which

+

strongly

Theorem

(7),

completes

dification

there

in

6 such

precludes

proof

for ~ < T

since

This

an a r g u m e n t

but M~ K

T

As

Since

follow

for U.

we

pro-

76

Then (b)

{mlsup(DN

U has

u m)

= m} £ U.

a regularity

lulI 5 c f ( 1 )

for

sequence

all

< u ~ l l < <,

lim(l)>

such

that

I.

Proof: (a)

Suppose

Hence

not.

there

6 6 D - y.

(b) L e t

alc

W

Set

Proof

Theorem

fixed

for

Define now

By

luxl

be

the

~o

=
Then

set ~

2.9

{~6

U has

and

{mIDN

= ~.

lall

mod

u m -c T } .

U.

Let

Contradiction!

: I.

the

Define

qE,q~,f

by

= <.

We

~ < < such For

and

<

that

normal

= I for

as : L

-< Q

<

also
in

is

on

<,

a regularity

where

sequence all

[A]

and

set

~ { C that

~ : < N Q~=

AT : Qa N H<}.

the

_ A ]., A N [

such

~ Co a n d ~oa _ c ~,A,' a .r

AO CTIQa

ul

T [ {o} U E, Q



I.

<

is

regular

< u l I l < <,

Let

this

lim(1)>

sequence

be

proof.

since

2~

Then

a regularity

sup

H

QT

weakly

~T ~,%exactly

smallest

~

ZN W

(k) 6 q E } .

U be

that

~T~ ~

then

Z =

regressive

property.

Let

of

f

however,

[A],AN~>.

fir -~ ~

C'T =

Lemma

rest

of

= the

CT'

2.8:

E,fm,A,QT,

possible,

But

is

I > ~

and

desired

= cf(1)

the

is

a6

the

require,

where

a I = I and

for

{E < lil 6 d o m ( f )

that

U,

(DNu m)

follows.

{~I f- I " ~ 6 U}

with

such

sup

= min

= <.

Z6

sup

q

sequence

2~

that

(a I - ~ )

of

v ~

that

: rain

uI =

and

map

{yI6 6 uy} 6 U.

E < < as

= sup

~

=

i such

on

f6 (I)

the

is y < < s u c h

Then

induction

Then

Now

An

< code

[A] , A N < >

~>~Q

For

u

o.

then

~6

C

2.1.

DN

<.

We

This

and

let

C

we

may

set

o

o

define:

mQ. N

set

that

of T h e o r e m

_ 0 ° =
we

T

proof

for

).

It

is

T 6 {o} U E:

clear

that,

if

oS

[-~

emm~q

]o7

~n~ywsqns

BuiMoiio

7 ~q~

~A~q

~M

(" n 9 9 D

l

"03 D

U~

(~ D

" ~9 D

~ = b~

' ~1

pu~

6ux

~

~X9

U Z9

mbx

x

~oq

(o)

D

= bZ ueq~

b , z,D 9 ~ 7I

(q)

O oM~

~ = ° ~ pu~

:poTii]OA

-M

XITs~

= o M u~q~

ax~

'' D 9

=

D

II

(e)

s~oe 9 5u!MoIIO / eqL

+ (~)7o> o ~ U u0 ~M U u o

= ~

0 = ~ ~

z ~eq~

:]~s

= l~

~ON

£IIeU~ 7 ~M ~ M =

~

z0

pu~

:%~s

'eAT~ISUel]

ST

i0

alaq~ ,D9 ~

"~IOa

IeTOnXo

~q~

£eId

~

s~an~onx~s

~q~

'/ooxd

l~ ~

:1~

1o

'Z ~ {0} 9 i ao 7

~u~s~xd

~q%

uI

o~ 9 x

"g% 3 ]eq~

qons

co> I

sT

~a~N%

pu~

o~9

<'!> u~qm

~ ~sooq0"

ST

qons

~)q

pu~

x ~q~ 96A

1

n3 ql~>

A ~]!uT7 ~)

dns

1

( nuo)dns

e

~eq~

n3

X)

o>

" {~n 3 (~) ~0

•~ 9 { ~ =

I D3

~]

{g ~ [g > @ ( i ~ ) q } l -~no~



e UT

~IqeuTl~p-

= ~ u~q& = X 6"g

(g~)q}

ox~q~ %~S

'~>

oS

M

x u~q~

l~ U

~eq~

"~ uI

"s~Inu~xo7

{(x) T~

70

~ou~ H

= x qons x ~q

II~qs sT

~ ~

~AIS

I<X'T>}

:q

H

~.eq

= g~

:7oo~d

:OL'~

,D

~M

0 uaq&

uoI~ex~mnu~

[V]@q3x

0 ~ eq

H9

~oqs

papunoqun

~q;

~0 p u e O

~Iq~uIT~P-

]eq~

• ~D0

0 s!

" ~1 D ~-X

"X3 ~ ~q

eurmaq £ q

= 0 ~S ~ ~o7

"uo!%o~fIq

l

"F/3

'~umi~q

ZL

78

Lemma

2.11:

{sIT

Let

W 6 U,

f 6 ~6

cf(~) +.

Then

there

is T 6 E s u c h

that

> - f(e) } 6 U .

Proof:

Suppose

set

= {~ 6 W f l C ' I T ~ ~ < f ( ~ ) }

YT

not.

(b) (c)

above,

the

K many

predecessors

For

gT

~ 6W

are

let h and

define

pairwise

(mod U)

g

inject

f(a)

into

gT 6 6 ~ T

distinct

mod

(6 < <) . C h o o s e

cf(~) . F o r

by gT(~) U.

Hence

T 6 E

= h~(T~).

some

T 6 E such

gT*

By

has

that

i

[ >T~,T 6 and v Then

=
set

I~ 6 Y> v

We

now

prove

C

T

<~ f o r

I: T h e r e the

all

but

2.8

Y:

g ~ (~) < g ~ , ( ~ )

sequence This

by

for

U.

contradicts

cases,

} But

v

Lemma

closely

~ gT,(~)

< cf(~) ,

2.9.

imitating

the

proof

of

that

of C a s e

uses

the

{~ 6 C'I K T Te # K T ~ } 6 U.

I

in T h e o r e m

sets

C' a n d T

2.1.

the

The

points

repetition

cf(a)

+ in

is

al-

place

of

.

2: C a s e

Case

2.1 : {~ 6 C W =

by T °

I fails.

o

l~ ° # id rK } ~ U. d

{c~ 6 C o l O r ° # i d ~ K Set

Y =

Suppose not.

{~ 6 W I ~

For

~ 6W

Z = {a 6 (W-Y) n C ' T I K T ~ is r e g u l a r KT = K

m K

For

c~ 6W l e t

a cardinal

- Y there

T:a

be the

in K}

that

We

first

first

point

show

that

that

moved Y 6U

e is n o t

a

{c~}Tc~_> y } 6 U. L e t

Ta >_ ya}.

K~---~K<+

~ is n o t

d

is y e < cf(e) + s u c h

"r s u c h

= KT~ a n d

in KTa' s i n c e . Hence

}.

is

i n Kyc~,. Now p i c k

cardinal

tion.

a 6 Y.

Theorem

argument

Case

Let

6 rng ~ ~ and

is T 6 E s u c h

literal, =+

, ~

=:

2.1.

Repeat most

nC{

is a r e g u l a r i t y

sup

Case

~ 6Y~

{~ < ~ I T * , T ~

hence

Theorem

for

a cardinal

Then and

Z 6U-

Let

z r~(~) = <.

in K T, w h i c h

a 6 Z. T h e n But is a c o n t r a d i c -

79 i

This

proves

define

Y 6U.

a non-trivial

X 6V An

founded

In for

iff

imitation

Case

But

any

of

Case

case y < <.

of

2.1

So

we

ultrafilter

Case

we

a 6Y

clearly

have

(6~) K < ~.

over ~(~

) DK

by:

Hence

we

may

6 T°(X) .

ultrapower

2.2:

this

@

for

2.1,

Theorem

2.1

shows

that

some

V

is

not

yields

a well

K.

fails.

have the

{e < < I s

regular}

conclusion

6U.

follows

Hence

from

U

Theorem

2.1.

(y,<)-regular

80 I § 3. Z 3 - A b s o l u t e n e s s

Shoenfield does

showed

hand,

therefore, the

to Z 3I s t a t e m e n t s ,

not extend

other

case

the

if K is t o o

solution

We

3.1:

since

Within

of S h o e n f i e l d ' s

L e t A be ~ i .

than

Assume

I 1 3.

Clearly

if K = L. N o r

t h a n K - e.g.

in L. H i s

3a a ~ L is i t s e l f

K is z ~ - a b s o l u t e .

- e.g.

condition.

extension

Theorem Then

of a ~

that

small

larger

are a b s o l u t e

I rather Ba a ~ K is Z 4

statement

to conjecture

V is t o o m u c h

modest

I that Z 2 statements

can

On the

It is t e m p t i n g , this

cannot

it b e t h e

if O T e x i s t s ,

since

these

however,

limits,

theorem

I Z3

case

be if

O T is t h e u n i q u e we obtain

a

theorem.

A(a) , w h e r e

a# exists

but

L[a]

~ IL ~.

3a 6 K A(a) .

immediately

Corollary

3.2:

Ba(A(a) A a~

This

get:

Assume

exists)

iO T

•

1

L e t A be ~ 2 "

Then

~ 3a 6 K A(a) .

in t u r n y i e l d s :

Corollary

3.3:

Assume

iO % b u t

that

the r e a l s

are c l o s e d

under ~.

Then

1 K is Z 3 - a b s o l u t e .

As a c o r o l l a r y

Theorem M ~L[a].

3.4: Then

Harrington

of the p r o o f

of T h e o r e m

1 L e t A b e ~ [ 2. L e t ~ b e a m o u s e 3a 6 L [ M ]

and Kechris

3.1,

and

we

shall

assume

obtain:

A(a),

where

A(a). proved

this

for t h e c a s e M = O~

(see

[5 ]).

81

The now

proof

on

wise

of

assume

set

Theorem

A(a),

L[a]

K + = L [ V o]

dinal

which

Lemma

3.5:

3.1

I= nL U,

where

is m e a s u r a b l e

There

(a)

~{6 K +

(b)

Let

M.

is

be

an

the

stretches

over

a # exists.

V°

is

normal

in

an

inner

iterable

iterates

several

K+ = K

K ° and

on

Ko

From

if 7 L H.

is

the

Other-

least

or-

model.

premouse

of

Set

sublemmas.

M

~ with

such

that:

iteration

points

<..

1

Then

l

Mi H

= K L [ a ] 6 M.. Ki i

Ki

Proof: Case

i:

K L[a]

= K.

Since

a # exists,

self,

hence

M

of

there K

Vo 3<+ setting o

=

Case

2:

K L[a]

Then

we

can

that

M~/ L [ a ] .

is

into +
a nontrivial

itself.

Hence

embedding

K + = L [ V o]

of and

L[a] we

can

into

it-

take

(K~)K

~ K.

apply

Lemma

(Note

1.4

that

and

the

H i <.

take

the

~-least

are

the

same

core

whether

mouse

M

such

we r e g a r d

the

1

M.

as

the

iterates

or the

mouse

V M = Ja

be

iterates

of

M).

1

M. l

From now V. = • 1 be

on the

~, i

~... i 3

Set

let

C =

iterates

{K. ! i 6 1

fixed

with

On}.

with

iteration

Finally

set

H

1

=

<, 1

K'

K'

<. 1

(2) ~(K< ) n ~ v

:~(K~

!

•

1

(since <. i

is

1

,'p(K~.) n M i = ~ ( K ~ . ) 1

(3)

) n~

•

1

inaccessible

1

in K'.

normal

points

f!. (1)

V

n~{i+l )

= KL [ a ] .

on K.

i

< in M.

and Then

Let

iteration we h a v e :

maps

82

Not

let I be the c a n o n i c a l

< T i l l < ~> the m o n o t o n e

indiscernibles

enumeration.

L e t bij:

for L[a] L[a]

~i

and L[a]

be de-

fined by: b i j t T i = idtTi, Define

b i j ( r i + a) = Tj+ h.

an u l t r a f i l t e r

X 6 U. l

iff

U i on ~ ( T i) n L[a]

T. 6 b..(X) i i]

U i is o b v i o u s l y

normal

by:

(i < j) .

in < L [ a ] , U i > .

It is k n o w n

that ~

is

i

amenable iterate

(setting

Proof:

(T)L[a])

and that < L [ a ] , U i >

of < L [ a ] , U o > , the bij b e i n g

ultrafilters

Lemma

÷=

~i

3.6:

the i t e r a t i o n maps.

We use the

U i to prove.

(r+) K' =

Set rt1 =



is the i-th

(r+)L[a!

(T~)L[a!

is a m e n a b l e

T'i =

' +,K' ~T i) . Suppose

T lI < T~l" Since

we t h e n have U!1 6 L[a], w h e r e

U!1 = U.1 n K'. A

1

standard

argument

shows,

however,

that < K ' . , U ~ > T

over,

cf(Ti)

covering

= T i in L[a],

lemma holds

in L[a] w i t h r e s p e c t

!

in L[a].

of

[ 3 ] t h a t U!l is n o r m a l

U i is m - c o m p l e t e

since

Hence,

is a m e n a b l e .

to K'

in L[a],

in L[U~].

More-

1

i T i is i n a c c e s s i b l e

in L[a]

(see

and the

[3 ]) . But then

we have by a n o t h e r

This contradicts

theorem

our a s s u m p t i o n

that L[a] ~ ~L Z.

Now

set W = ~ ( K ~ ) n K'.

For X 6 W set X :

y

~oi(X) . C l e a r l y

K' = x ~ .

Lemma

3.7:

There

is i

such that if i O

< i < j, then O

-

-

bij (~. N V ) = ~ N Vb. for X 6 W, v 6 On. lj (~))

Proof:

Set

~=

{~I~ = T

class.

Let D 6~.

and cf(~) > <}. C l e a r l y , r

By a F o d e r

type a r g u m e n t

is a s t a t i o n a r y

for e a c h x £ L[a]

there

is

83

< % such

t h a t b..(x) 13

~D < ~ such The m a p ~' ~ .

that

= x for ~ ~ i m j g D-

~D A i s j < D i m p l i e s

~ ~ ~D is r e g r e s s i v e Let

~G = i ° for D 6 ~'.

i o ~ i ~ j, X 6 N , bij(xnv

~)

Corollary (a) D 6C

on ~,

D £ On.

= bij((~NV

3.8: iff

Let

Pick

We

D) n v

) =

bij(X N V

hence show

D £ ?'

Since

) = XA V

constant

that

such

~ < K < cf(~),

there

for all X 6 W .

on a s t a t i o n a r y

i ° is as r e q u i r e d

that

j,~ < D.

( X N V D) n V b . . ( ~ ) 13

Let

Then:

= ~nVb..(~). 13

i o < i N j. T h e n

b. •(D) E C i]

(b) ~. £ C. 1

Proof:

(a)

To p r o v e hence

is i m m e d i a t e

(b) p i c k

T i 6 C by

Every

by

D = T

the L e m m a

= <

such

and C =

that

D ~

. Then

has

the

f o r m ~(~)

for

some ~ 6 C,

f:K n

<. By L e m m a

3.7 b i j ( ~ ( ~ ) )

= ~ (bij (7)

action

of bij

the o r d i n a l s

is d e t e r m i n e d

(see

upon

a version

[ 5 ]) to s h o w

Lemma

3.9:

" bi

(T i) = N 6 C,

(a).

ordinal

We n o w p r o v e

/~ X6V

Let

that

T•

of Paris' this

action

= < . There

1

"patterns

is

by

of

is v e r y

8 such

for

f 6 W such i ° < i g j. its

that Thus

action

the

on C.

indiscernibles"

simple

lemma

indeed.

that

O

(a) Tio+J

= <~+Bj

(b) b i o + i , i o + J ( <

Proof:

Set C 1•

+Bi+~)

= < +~j+

C A'(Ti,Ti+ I)

=

.

.

It s u f f i c e s

to s h o w

that

(*) b" C ij i+h = Cj+h'But

this

follows

from

(**)

b'! .C,

C,

since

then

13 I

=

3

h" C i+h = b i j b i , j + h " Ci = b i , j + h " Ci = C j+h" ;ij

is

for

i°

_
84 (**)

in t u r n (***)

To

see

follows

from

bi,i+l

"C i

this,

assume

The

case

j =i

For

j = k+l,

= Ci+l"

(***) . We

prove

(**)

by

induction

on

j-i.

is t r i v i a l . k>

i we

b i , k + l " Ci = b k , k + l For

j = X such

Cl,


have

b i k "C i

that

= b k , k + l C k = Ck+l"

lim(1) it

i < I>

is e n o u g h

is the

direct

to n o t e

limit

of

that


i < i>,

< b i h [ C i l i o <- i <_ h < l>.

we

now

prove

of < L [ a ] , U i > . for

an

f:

L[a]

We

(***).

Hence

each

T. ~ L [ a ] , 1 ~

not

constant

on

may

take

But

Set = sup n<w

get

let

is

the

the

form

ultrapower

bi,i+l(f)

(T i)

But

then

~n

Since

bi,i+n:

6 U i.

y ~ rn~bi,i+l)

f is m o n o t o n e Set

T i + 2 , we for

~ ~(~(~))}

y 6 C i + 1 - rng (bi,i+l) • L e t

T i ~ Ti+ I.

applying

6 Xn

has

{~ < T i l L [ a ]

So

T i} 6 U i

on

, f is

an X 6 U. a n d 1

fn = b i , i + n ( f )

for

we

n ~ ~.

have:

X 6 V.

xi+ n} £ U i + n

(n < ~ , X 6 V) .

implies

fn+l(Ti+n)

6X6~V x N

Yn = f n + l ( T i + n )' 6 . Then n

@ = sup

of L[a]

)

6 ~n

{v < T i + n l f n ( ~ )

in t u r n (3)

But

we



everywhere.

6 C n Ti+ 2 = x/~ 6v

{m < T i l f ( ~ )

then

This

X 6 U.. 1

it to be m o n o t o n e

fl(Ti)

(2)

any

f:

that

and

by contradiction. (Ti) , w h e r e

(i)

element

(Ti)) (

y = bi,i+l(f)

Since

recall

f 6 L[a]

~(bi,i+l(~)

argue

we

f"

£

n

Ti+~'

E C;

Ti+n+ IWC

(n < ~).

an = bi+n+l,i+m(y hence

since

6 C C since

f W is m o n o t o n e

n)

= f

(Ti+n).

it is a l i m i t and

nsup <~

Ti+n

Finally point

set of C.

: Ti+ ~ . H e n c e

85

cf(~)

= Ti+ ~ in L[a].

L[a]

since

(T+

'

But

)L[a]

lemma

for K'

Now

fails

define

+ w) K' by L e m m a (Ti+

=

i+w

6 is i n a c c e s s i b l e

in L[a].

in K' 3.6

and

~ > Ti+ w in

Hence

the c o v e r i n g

Contradiction!

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

relation

~ -

on the

set of

in-

M,e,B

creasing

tupels

<
• '''

f r o m C as follows:
> ~ <<.

30,.--,Jm-1

(a)

ik < ~ ~

ik = Jk

(b)

e_< ik ~

( e < Jk and

(c)

e

(where

<

i k < i h and

y : B"

Corollary

I~,/~),

methods

3.10:

Let

(ik-yB)

=

n

= m and

~k-/B)

k-

[Y//B] +

By s t a n d a r d

iff

>

we

(?B~) < B) •

then

<~>,<~>

get:

6 [C] n such

that

<~> ~ <~>.

Let

f. 6 W, 1

f• :


1

~

L[a]

Now

for

K

4~

=

~ ~(~(~))e-~L[a]

ordinals.

> ~ , B , O n R M such iteration

~

Let

some

n. For

= ~o,~+B

(f)

Now define

(x 6 H)

that

only and

an

.

iterable

L~[M]

premouse

is a d m i s s i b l e .

set C = {Kili < ~}.

~ = ~M,a,B

n ~]<~) £ H.

The

~p(~(~))

Set H = HM,~, ~ = L~[M],

of M and

as b e f o r e .

9or

~

let M be an a r b i t r a r y

arbitrary

(~[C

Then

1 ~ , . . . , m .

= ~+~w"

Set W = W M = the f£W

set ~ =

and < ~ ] f

~ is the

least

Let <Mi,~ij, ~ = ~M,e,B

on

e,B

be

the

[C] <w

~,C n ~ £ H; h e n c e

set of

YZoi(f)

H-language

symbols

an a d d i t i o n a l

Then

where

< and

f 6 M such

and ~

= ~

that

f: <

n

~ K

n. T h e n

6 W> 6 H.

infinitary

predicate

Define

at some

~

= ~M,~,8

of ~ are =,6.

constant

~6 has

~. As a x i o m s

we

as

follows.

constant

names

take

together

ZF-

x

86

with (a) v 6 x<-~z6~X ~ = z (b) ~ _c w

(el L - [ ~ ] K" (d)

~ A(~.)

If < ~ > , < ~ > and

f,

6 [C n 7] <w,

6 W such

1

that

f.

1

:

K

n

~ <

(i = 1 ..... m) , then for e v e r y

An

immediate

Lemma

consequence

3.11:

There

this

gives

But

Corollary

3.12:

K and Z M , ~ , 6

Proof:

Let

lemmas

~.

is

that ~ M , e , 8 is c o n s i s t e n t .

is < M , e , B > 6 K such

that M,a,6

are

countable

in

is c o n s i s t e n t .

<M,~,B>

= b-l(e),

there

foregoing

is < M , e , 8 > 6 K + such

There

M , ~ , B 6 X. L e t b:

But M,~,8

of the

formula

be as in the

H~-~X

6 = b-l(6).

are

countable

is n o t h i n g

<M,~,8> 6 H L[Vo]= K0

where

lemma.

Let

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

Then

H = H~,~,~

in K +. We

claim

to be proved.

Otherwise

K

argument

. The

X ~ H,

same

~ = w in K + such

Set M = b - l ( M ) ,

and~,~,~ that

that

is c o n s i s t e n t .

< M , ~ , B > 6 K.

If K = K +,

K + = L [ V o] and shows

that M , e , 8

are

countable

K0

in K.

We

are n o w

and ~ = ~ M , ~ , 8 theorem part L~[a]

ready be as

to f i n i s h

in C o r o l l a r y

let 0~6 K be a m o d e l

of O i i s b A(a).

the p r o o f

transitive. It s u f f i c e s

Let

of ~ . a be

to show:

3.12. We m a y

of T h e o r e m By

the

assume

3.1.

Let M,~,8

Barwise compactness that

the ( ~ - i n t e r p r e t a t i o n

the w e l l

founded

of ~. T h e n

87

Claim:

L[a] ~ A(a)

Let ~' be the Z F - l a n g u a g e w i t h the c o n s t a n t ~ and ordinal c o n s t a n t s (~ £ On). Define a class S of ~ ' - s e n t e n c e s • Let 6 [C] n, fi 6 W, fl: Kn

~

<

(i=l

.....

as follows: m).

Then:

iff

3 < ~ > 6 [C N ~ ] n ~ < ~ > ~ < T > and L~[a] b ~(~(~)))

I n d i s c e r n i b i l i t y a r g u m e n t s show that this is a c o r r e c t d e f i n i t i o n and that (i) S is a consistent, (2)

r~(~)q 6

(3)

rBx ~(x) I 6 S

S

iff

d e d u c t i v e l y closed class of sentences

L~[a] b ~(~) iff

Bt £ T

where T is the class of (4)

rBx 6 On ~(x) I 6 S

Now l e t ~ b e

iff

for ~ <

r~(t)l 6 S L-terms

3~

r~(~)1 6 S

the term model of S. By

the rank of

[t] in 6 ~ is v w h e r e

rrn(t)

equivalence

set of a term t). Hence ~ i s

(4) , 6 ~ is well founded and = _vI 6 S ([t] being the isomorphic

to a t r a n s i t i v e

model Q. But then a is the Q - i n t e r p r e t a t i o n of ~ and L ~ [ a ] < Q by Hence Q = L[a] and L[a] I= A(a).

(2).

This finishes the proof of T h e o r e m

3.1.

In c o n c l u s i o n we prove T h e o r e m 3.4, m e n t i o n e d at the outset. a,M be as in the h y p o t h e s i s of that theorem. Assume w.l.o.g, is the ~-least core mouse. M ~ L[a]. We first show that a Suppose not. Then the c o v e r i n g

lemma holds for L[a].

Let

that M

exists.

But L[a] I= ~L H,

since K L[a] ~

K; hence in L[a]

to K' = K L[a].

Hence the ,covering lemma holds with respect to K'. This

is nonsense,

the c o v e r i n g lemma holds w i t h respect

since the mouse M enables us to c o n s t r u c t a n o n t r i v i a l

88

El-embedding iterable

of K'

premouse

into used

itself.

Thus

in the above

a ~; exists. proof w h i c h

M is then-the shows

3 a 6 L[M]

A(a).

8g

§ 4

At

Decomposability

first we

filters.

repeat

of U l t r a f i l t e r s

some

definitions

L e t U be an u l t r a f i l t e r

and elementary

on some

results

cardinal

K. L e t

for u l t r a -

6 be

a car-

dinal.

U is c a l l e d
Iv < 6 > 6 U 6 s u c h

U is c a l l e d that

~<~6 av 6 U

dingly tions

see

[ 9

An

and

~
for all

uniform

iff

-

with

iff t h e r e A

DA

is a s e q u e n c e

and

~<~6 A

is a p a r t i t i o n

~ <6

~ y6S

a m {U.

6-decomposable.

= ~.


Iv < 6> s u c h

U is

For more

6-descen-

on these

no-

[ 12].

U'

iff

for a l l A , B 6 U

is b e l o w

is a f u n c t i o n

6-decomposable

implies

iff t h e r e

S c6

iff U is cf

] and

ultrafilter

there

that

incomplete

6-decomposable

incomplete

U is c a l l e d

iff

~-descendingly

U in the R u d i n - K e i s l e r - o r d e r i n g

f: K ~ U U '

there

~ = B.

such

is a u n i f o r m

t h a t U'

=

{f"AIA 6 U } .

ultrafilter

U'

on

U'

~RK U

U is

6 below

U in

§ 2 we

call

the R u d i n - K e i s l e r - o r d e r i n g .

Let

! be

a sequence for v 6 X quences

a cardinal.
and

I~ 6 X > {~lq 6 u

for U are

Slightly

extending

a l-regularity } 6U

just

for all

the n o t i o n

sequence

q
Hence

of

for U iff X c < , the

the r e g u l a r i t y

sequences

U is c a l l e d

(y,l)-regular

u

K-regularity

c~ se-

for U in the s e n s e

of § 2.

Let y be a cardinal. set H cU

U is

such

that ~ = I and

(y , l) - r e g u l a r

iff

iff

there

for all H ' c H H ' > y i m p l i e s

there

is a l - r e g u l a r i t y

is a s u b -

DH'

sequence

= ~.


6X>

g0

for U s u c h

t h a t u-- < y

(l,K)-regularity y',l'

be

for all ~ 6 X .

of § 2 c o i n c i d e s

cardinals

such

of U i m p l i e s

the

is

is c a l l e d

just what

that y ~y'<

iff

it is then

y ~cf(6)

~ 6 ~l.

~ R K U be

U is 6 - d e c o m p o s a b l e

of U iff T <<.

The

a uniform

T h e n U is w e a k l y

a weakly

g < f

the

here.

theorey,

of U

see e.g.

for r e g u l a r

~-deocmposable

Let

(~,l)-regularity

(e,l)-regularity

6-decomposable

(mod U)

normal

ultrafilter

following

The

in m o d e l

ultrafilter

iff

If g 6 < < is a f i r s t

normal

of U.

introduced

of

[2

cardinals

for all

6 such

]. I.

that

((y,l)-regular) . Then

(~,l)-regular).

for all g 6 K K

(See § 2).

it is

I = < the n o t i o n

i. T h e n

(l,l)-regular

(y,l)-regular, L e t U'

for

the n o t i o n

l'~

A-regularity

If U is

l e t U be

with

(y',l')-regularity

U is l - d e c o m p o s a b l e

Now

Hence

lemma

on <.

f 6 K K is a f i r s t

implies

idr<

g"y ~T

is a f i r s t

function

of U,

function

for s o m e y 6 U

function

t h e n U' =

and

o f U.

(g"AIA 6U}

is

~ R K U on <.

is a t h e o r e m

of K a n a m o r i

and

is p r o v e d

in

[6]-

Lemma for U.

The

4.1:

T h e n U is

next

f r o m the

Lemma

4.2:

lemma

Let

that O

do~s

cardinals

ultrafilter

is a r e s u l t of

[ 9

for all

of Kunen

] and

6 be a c a r d i n a l

T h e n U is

following

a uniform

(~,l)-regular

theorems

posable.

The

Let U be

on

K without

cardinals

and

function

i
Prikry

the p r e c e e d i n g

first

and

follows

easily

remarks.

a n d U an u l t r a f i l t e r

which

is

6+-decom -

f(6)-decomposable.

result was not exist

(L C H) . In

proved

(,O ~) 1976

in

1974

a n d all

it w a s

under

limit

proved

the

stronger

cardinals

without

are

assuming

assumptions

strong L C H

limit

gl

applying This

results

proof

result

on

now

analogous

gives

with

those

the

of

§ 2 with

stronger

nO ~ i n s t e a d

results

of

§ 2 the

of

~L H

following

decomposability.

Theorem4.3:Assume form

to

7L ~.

ultrafilter

dinals

6 ~K~

Proof:

Evidently

on

U

Let

<.

< be

Then

is n o t

u-decomposable.

Now we

Case

regular

a regular

U is

6-decomposable

a-complete

proceed

cardinal.

by

for

any

for

Let

U be

all

regular

cardinal

induction

on

6 >~

a unicar-

and

hence

K.

+ I: L e t

composable uniform U'

is

all

is

2:

by

Let on

the

U

is

and

6 <<

be

that

y ~d'

show

that

duction

for

But

since

there

is

hypothesis

2.1 Then

the

Then But

by

first

cardinals

Now

U'

(y,<)-regular

is

there

is

regular

a uniform

limit

ultrafilter But

some

hypothesis for

a uniform

Then

by

by

assume

for

then

the

that

some

cardinal

~RK

Lemma

4.2

U has

y
a

U' Now

<< the

on

6',

which

U'

~RK

U

such

now

U 6 ~RK

~RK let

But

U'

ultra-

preceeding

ultrafilter

cardinal. U~,

is

U be

then

normal

a regular

6-decomposable.

let

But

a weakly

there

<+-de-

6-decomposable

function.

is

U is

§ 4 •

and

i <~.

. Then

induction

U is

of

~
all

<

remarks

the

then

cardinal

no

on

preceeding

beginning

limit

for

uniform

there

< is a

is

<.

~ ~<.

U has

all

then

theorem

a cardinal. ~

in

U be the

U on

all

a regular that

let and

l-decomposable

on

by

~RK

for

<. A s s u m e

function.

remarks by

U and

inU

6-decomposable.

Remark: I.

U'

and 4.2

remarks

< be

first

is

Lemma

(~,l)-regular

remarks

~,

by

6-decomposable

filter U

and

ultrafilter

6 ~<+

Case

< be

Clearly

a modification

of

this

argument

also

would

cover

case

92

Case

3:

makes

Set

for

B+

extensive

lowing

use

some

of

singular

the

methods

cardinal of

[8]

8.

and

Here

[12].

We

the

proof

need

the

fol-

iemma:

Lemma

4.4:

filter

Let

on

(i~

8 be

U

is

If

IL ~

cardinal

Then

one

of

and the

U

a weakly

following

normal

ultra-

conditions

holds:

(~I' B + ) - r e g u l a r

U {v~U

Clearly

a singular

S+ . A s s u m e

(ii)

~+.

< =

is

~-decomposabie

this

lemma

U does

not

have

and

so

(~,B)-regular

function,

then

lemma

implies

settles

a

A V regular}

case

3.

first

Let

function,

6-decomposable

there

is

U be

a weakly

all

normal

B-

a uniform

then

for

=

by

ultrafilter

lemma

~ ~ 5+ . If

V ~RK U o n

4.1,

U is

U has

a

Now

(i)

B+ .

on

first

of +

4.4

such

that

that

w1 ~ef(6)

V

and

and

hence

6 = w.

U are

But

(ii)

6-decomposable

and

the

for

induction

all

6 ~ 5

hypothesis +

show

that

The

V

and

proof

U

of

model

results

[12].

For

are

lemma

from

some

6-decomposable

4.4

[13]

is

and

ultrafilter

the

for

crucial

on

results

U on

< let

all

regular

step

and

K.

Prikry

of


6 ~

is

based and

the

on

J.

core

Silver

from

ultrapower

of

<

K mod

U

f mod

U.

Lemma

4.5:


Let

seuqence

and

for

Let

f EK K let

C6 6
Let

6 <<

U be

and

such

the

denote

function

with

a ~-indecomposable

cf(6)

that

[f]u 6 < > / U

=

U q<~

6

~. ~

=

Let

<6

n

equivalence

constant

value

ultrafilter

In < ~ >

6. T h e n

the

in

be

on

q

of

6.

<,

a strictly

K~/'~" u <[C6

class

~ regular increasing

]lq < m >

converges

to

[c6]. This as

lemma

result 6.

The

has

been

next

proved

lemma

is

by

J.

theorem

Silver

and

can

be

5 in

[12]

and

due

found to

K.

in

[12]

Prikry.

93

Lemma

4.6:

Let

< be

a regular

t e r on

K. L e t C ~K
of the

following

cardinal

closed

conditions

and

{61 [C 6] 6 C } 6 U

(ii)

{~IU is c f ( 6 ) - d e c o m p o s a b l e }

of

in K a n d in K.

lemma

4.4.

6+ =

(8+) K.

B+ =

quence,

there

C 1 6 i ~ cf(1)

(iii)

o t ( C I) < 8

(iv)

I:

Then by

[ida<] U

Then

one

covering

lemma

the

that

oB-sequence

a sequence

arguments there

B is s i n g u l a r

is a ~ - s e q u e n c e

in K is j u s t

< C A l l < 8+ A L i m ( 1 ) >

a DB-se-

such

that

1 =

(ot(X)

is

the o r d e r

type

of X)

• 6 C I ~ C T = C 1 n ~.

Now we have

Case

in

below

ultrafil-

6U.

±t is s h o w n

that

exists

C 1 is c l o s e d

(ii)

[73],

(~+)K i m p l i e s

i.e.

(i)

By standard In

unbounded

normal

holds:

(i)

Proof

and U a weakly

to c h e q u e

{l 6 B+Icf(l) lemm~a 2.9

two cases.

= ~} 6 U .

(b) , U is

(~I' B+ ) - r e g u l a r

a n d the

lemma

4.4

is

proved.

Case Now

2:

{I 6 B+Icf(1)

assume (i)

that

the

>w} 6U. following

condition

{61U is c f ( ~ ) - d e c o m p o s a b l e }

If for all

regular

~<5

cf(~)-decomposable,

then

t h a t t h e r e is a r e g u l a r cf(6)-decomposable 2.9.

(b) U is

such

that

lemma

<~.

(~+, B + ) - r e g u l a r But

6U.

is a 6 < 8+ s u c h 4.4.

v < ~ such

cf(6)

~ + , cf(6).

J {~ < 81v r e g u l a r

there

holds.

But

that

v
is e s t a b l i s h e d .

So a s s u m e

that

for a l l

then

{61cf(6) < ~ } 6 U

and hence

a n d U is

6 < ~+ s u c h

t h a t U is

and by

6-decomposable

(i) and

lemma

for a l l

6 _< B

then

and U v-decomposable}

= B and

the

lemma

is p r o v e d .

94

(Evidently

Now

this

assume

case

that

(i)

(ii) {6 I L i m ( @ ) ^ normal. X =

Hence

Now fine

I

Hence

T < B X

for o t ( C I) >~.

C'

are

[id~B+]U . S i n c e

lemma

of

tensive

>w} 6U

regular

B+ . For

in

~+. For

U is w e a k l y

~ < B. The

set

~ < B+ d e f i n e

each

i < B+, Lim(1)

and X T n C ~

of

the

ultraproduct

U is w e a k l y

B+ and

we

implies,

assume 6U.

de-

= ~ for all

normal,

so o n l y

the

with

limit

C~ : I for

it c o n t a i n s

C~

from

4.4

Since

C'

is c l o s e d

of D,

This

X T is a s t a t i o n a r y

v, t h e r e

and

this

shows

is c o m p l e t e . of K e t o n e n

such

from

that

shows

that

The

set

is a s t r i c t l y

<6qln < ~ >

~ 6D ~X % . But

of ideas

that

{61U is c f ( 6 ) - d e c o m p o s a b l e }

of c o f i n a l i t y

C~ D X T = ~.

lemma use

in

since

all

the m o d i f i e d

important.

implies

contradicts proof

subsets

~ of e l e m e n t s

4.5

to be

< ~ . Since

{~I [C~] 6 C ' }

elements

of type

for some

is a D E - s e q u e n c e

= i<8+]~ C ~

unbounded

4.6 D =

6U,

8+ .

{I 6 B+Jcf(1)

only

Lim(1)>

cardinals

oS-sequence

Then

= u and

Cl\(yX+1)

define

closed

ot(Cyl)

for o t ( C I) < v

I <

nonlimit

is s t a t i o n a r y

T

C

cardinals

(i)) .

= ~}.

that

< C ~ l l < B+A

Now

true.

= v} is s t a t i o n a r y

71 6 C 1 such

C~ :

is not

assuming

U is c f ( 6 ) - i n d e c o m p o s a b l e }

{P 6 X L o t ( C p )

for s o m e

occur,

U is ~ - i n d e c o m p o s a b l e

{p < 8 + I c f ( p ) X~ =

cannot

[8].

~U,

by

lemma

increasing

6 = ~

lemma

sequence

a n 6 X T. N o w 6U

is i m p o s s i b l e

of this

below

containing

{I < B+I6 6 C ~ }

(ii)

proof

unbounded

and a n d the

makes

ex-

gS

LeZ

~ ~K

cardinals

Theorem

be

i ~K

4.7:

cardinal

cardinals

a n d L C H ~K b e

such

H ~cf(1)

Assume

<. A s s u m e

Theorem

4.7

that


~L H. L e t U b e L C H w 2 <. T h e n

follows

immediately

the s t a t e m e n t

are

strong

a uniform U is

that

limit

from theorem

limit

numbers.

ultrafilter

6-decomposable

all

on

a regular

for a l l

4.3 a n d the

6 ~<.

following

two

results.

Lemma

4.8:

Assume

U a uniform

Lemma

4.9:

ultrafilter

on

K such

t h a t U is

of cardinals Then

of Lemma

U is

4.8:

show

28 =

8+ . T h i s

theorem

2.8 U c a n n o t

limit

(~,cf(1))-regular.

t h a t U is an

Let

that

implies

B÷ ~ =

a first

and

(~,8)-regular.

< l _
8 is s i n g u l a r ,

cardinal

standard B÷ B =

function

6 -decomposable

covering

B+

and

for a s e q u e n c e

2@ =

lemma

4.1

lemma

B+.

ar-

T h e n by

shows

that U

(~, 8 ) - r e g u l a r .

Proof

of

lemma

ing sequence

4.9:

of elements we may

H =

that

<6

I such Let


I~ < c f ( 1 ) >

that

/q U v = ~ v6S

be

for e a c h

Iv < c f ( 1 ) >

v
let x

increas-

~
be an injective

for all

~ 6K

a strictly

<6

=

is

sequence

infiniteS ccf(1) H

>

{~ < c f ( 1 ) IH 6 U

W.l.o.g. }. T h e n

and

{y 6 ~ ( c f ( 1 )

cardinality

the m a p p i n g

to

f o r U.

of U such

assume

6~(cf(1))

Let w.l.o.g.

converging

a 6 -decomposition

has

have

such

cf(1)

strong

U 6~ = I u b e v
Since

guments

X

T h e n U is

cardinals,

for ~ < c f ( 1 ) .

is

B+"

K,l be

Iv < c f ( 1 ) >

Proof

on

8 be a s i n g u l a r

Let

ultrafilter <~

7L H. L e t

13H 6 K

K. L e t

H ~ x

H

of

(y = XH) }

K onto

H.

be the p a r t i t i o n Now

let

for e a c h

of x 6H

K generated ~

x

be

the

by common

g8

refinement

of the p a r t i t i o n s

taining

the sets

we

that

show

partition UY N U

some

by

the

{y 6 Y I H q

Y _>I. S i n c e gives

Since

rise

K ~Px"

~ be

for x E H

Let ~ x =

such

that

H 6 UY N U

y 6 ~ x for some

definition <~

> for ~ 6 X

_(U Ny)_c

to a l - d e c o m p o s i t i o n

for U. Then


Since

Clearly

for e a c h

m. H e n c e

U~

con

Next is a

~
y _cpq for

is a ~ - d e c o m p o -

cardinality we o b t a i n

for U.

the p a r t i t i o n

, H is an e l e m e n t

x containing

pq~} has

Ux < _ 6 max(x)

UY 6U.

u 6x

of Ux,

and

{YI~ ~ Y c-P x and y 6 ~ x } .

Ux is a l - d e c o m p o s i t i o n

for e a c h

determined

q <6

sition,

U • = x~

of <. L e t Y c U

6U.

uniquely

P x and

< p ~ I q <~

> 6 v" This

shows

i = ~* = ~ and U*

97

References

[I]

J. Baumgartner, and H i n t i k k a

I n e f f a b i l i t y properties

(eds.), Logic, Found.

87-106, D. Reidel [2]

of Math.

[4]

T. Dodd andR.

[5]

L. H a r r i n g t o n and A. K e c h r i s , ~

[8]

of AMS 220

J. Ketonen,

singletons

Logic

and O ~. Fund. Math.

393-399

47-76 and large cardinals,

Trans.

(1976), 61-73

K. Kunen and K. Prikry, Symb. Logic

W. Mitchell,

[11]

J. Paris, (1974),

filters and i r r e g u l a r ultrafilters,

Nonregular ultrafilters

[10]

36

On d e s c e n d i n g l y

(1971),

incomplete ultrafilters,

650-693

Ramsey cardinals and c o n s t r u c t i b i l i t y ,

Patterns of indiscernibles,

to appear

Bull. London Math.

Soc.

6

183-188

K. Prikry,

On d e s c e n d i n g l y complete ultrafilters,

mer School in Math.

[13]

appear in Ann. Math.

Strong c o m p a c t n e s s and other cardinal sins, Ann.

J. Ketonen,

Journ.

[12]

(1976),

Logic 5 (1972),

of AMS 224 [9]

Jensen, The core model, to

A. Kanamori, W e a k l y normal

Math.

unpublished

167-171

Trans. [7]

Theory,

(1973)

T. Dodd and R. Jensen, The core model,

[6]

and Comp.

Model Theory, North Holland,

[3]

(95),

II, in Butts

(1977)

C.C. Chang and H.J. Keisler, Amsterdam

of cardinals

Logic,

Lecture Notes in Math. Vol.

Springer,

New York

P. Welsh,

C o m b i n a t o r i c a l Principles

thesis, O x f o r d

(1973),

(1979).

C a m b r i d g e Sum337,

459-488 in the core model,

D. Phil.

A LATTICE

STRUCTURE

OF

COMPLETE

ON T H E BOOLEAN

Sabine II.

ISOMORPHISM ALGEBRAS

K o p p e l b e r g I)

Mathematisches

Institut

der

K~nigin-Luise-Str. D-IOOO

For

every

type of

of

B.

complete

B.

Let

Boolean

T(A) ~ T(B)

where

the

order

inherited

of

is

B,

relative

most

for

arbitrary

lattice;

To the

33

(cBA)

B,

up

isomorphism,

to

let

T(B)

be

the

isomorphism

a direct

factor

{T(B

6 B}

B [ b = Ix C B I x ~ B]

B and

hence

ordered

which

r b)]b

has

a cBA. set,

as

(T(B),S), proved

a greatest

is

the

in

element

endowed "type

[13],

T(B)

with

1.31

and

the partial

structure" or

[10],

a smallest

22.6 ele-

• (B ~ O ) .

Our

and

by

1.4),

:

algebra

a partially

also

ment

Berlin

Let

T(B)

(see

is,

FU

24/26

Berlin

algebra if A

TYPES

general

B is

both

Theorem

algebras

Theorem

structure

Stone

space

how

(T(B) ,~)

Heyting get

result

of

R(B)

B we of

=

type

B in its

prove

structure

section dual

satisfying

global

R(B),

and

the

of

(T(B) ,S)

lattice

(a--~b)

in T h e o r e m

sections

2:

(T~B),~) is

(T(B) ,~)

v (b--~a)

A that a sheaf

linear

like

a distributive

are

= 1 for

(T(B) ,~) of

looks

Stone

algebras

arbitrary

is

isomorphic

orders

over

where

{x 6 Blf(x)

=x

for

every

automorphism

f of

B]

I) The author gratefully acknowledges partial support by the Forschungsinstitut fur Mathematik, ETH ZUrich

a,b. to the

99

is the c o m p l e t e

Our

starting

Theorem this of

point

paper.

It s a y s

a homogeneous of

first version this

theorem

section

3.7

version

of

about

the

R = R(B)

[5].

paper,

complete

version

of

B =C I where

such

~c)

Solovay, the

~

T(A) ~

Y(B)

and

then

aware

by Solovay after

and

to t h e

paper

stronger

above)

cBA.

families

in

- More

in

results (where

the

technique

[11]

and u s i n g

established

lattice. this

condition

precisely,

(Ri)i6 1 of

that

R ~B

is a c o m p l e t e

chain

fact

a

seen the first

situation

only

countable

the

is q u o t e d

having

as d e v e l o p e d

answering

finishing

of

of

B is a p o w e r

thus

after

result

cBA's

The

for

and for

Solovay's

and

(ai)i6 1 of

~

,

T(B)

~

proved

reduct

something

isomorphism

types

, is a c a r d i n a l to be

is e x p r e s s i b l e s ~ t

I-~ (ai,<) (Ri) iEI of)

the Boolean

power

of an

a c B A C.

is d e f i n e d

above

became

t h a t T(B)

(.the t w o - v a l u e d

has

T of all

rations

defined

[--~Ri i6I

d w.r.t,

in fact,

class

are

Only

of B d e f i n e d

the present the

order,

was

that

denotes

q-structure

shows

structures

interesting

considerably

forcing

C is a r i g i d

there

the most

Solovay,

B satisfying

at t y p e

[3].

author

subalgebra

of V (R) h e

that

of

obtained

of B.

is w e l l - o r d e r e d ) ,

of T(B) . A p p l y i n g

R ~

where

has

Boolean-valued

shows

T(B)

result

Conversely

structure

ordinals

paper

look

is a l i n e a r

the

algebras

algebras

if T(B)

(and h e n c e

this

elements

is p e r h a p s

of

C inside

Boolean

that

a closer

version

this

preliminary

having

3 which

cBA

invariant

a preliminary

of

is t h e

Theorem

of

is an u n p u b l i s h e d

two-stage

method

for

C in s e c t i o n

a question

of

subalgebra

iff

of

more:

it is w e l l - k n o w n

cBA's,

algebra

endowed

in t h e

T ( A x B) ; h e n c e

the

sense partial

by t = s Q

r

for

some

r 6 T.

with of

that

two

[13].

order

opeHere,

~ of T

100

Now,

for

T(E),

a fixed

where

E is

b 6 B satisfying bra

of

cBA

T the

B,

a cBA

let To(B)

be

such

for

O < d ~ e,

structure

that

E } d ~B

the

~ b.

of w h i c h

has

class

each

isomorphism

e 6 E~{O},

T~(B) been

of

is

then

there

types

are

a cardinal

completely

d 6 E, subalge-

described

by

Solovay.

Another

consequence

without

rigid

remarks

in

quoted

In

factors.

4.6

above

contrast

to

forcing.

proofs

of

R(B)

In

be

that

section

(this

is,

who

the

proof

of T h e o r e m C:

cBA

C such

that

not

too

satisfying Theorem

or

in

apply

more

in

[ 9 ] and

the

some

E in

we

The

on

B

results

paper.

of B o o l e a n -

sometimes

literature.

facts

cBA

elementary

4.

technique

automorphism

to

every

a forthcoming

the

the

for

with

readable,

found with

simple R(B)

course,

the

1.5.

Souslin

in

give

This

group

applies,

of

cardinal

B or

algebras

number

In

in

C,

that,

be

completely

another

condition). in

section

section

compared by

3 we

4 two

embedded

with

countable

into The the

first

these

A

and

give

the

B

a homogeneous

that 2 in chain

a homogeneous second

read

A).

applications

into

Proposition

the

embedded

problem

Theorem

Theorems

B cannot

solves

for

the

a cBA B satisfying

E which

decomposition

2 contains

is

chain

every

4 may

and

of

that

3 and

completely

D ; remember

countable

a product background

in g e n e r a l .

be

mainly

sections

there

the

to

intuitive

above,

B can

showing

rise

Section

(T(B,~)

cBA

(Theorem

gives

interested

to

of

remarks

the

C mentioned

every

large

that

of

1.1

structure

of T h e o r e m

such

not

B e ~B

Theorem

Solovay

paper to be

[7]

some

of

after

that

do

connected

is m a i n l y

about

implies

by

that

together

immediately

we

the

is

[13].

1 we m a k e

immediately

gives

in

of

reader

that

published

facts

decomposition

The

this,

facts

product T(B)

Note

To m a k e

to

in

method

paper,

contained

contained

Solovay's

4.7,

elementary

example,

with

and

will

valued

for

of

of

B,

[I],

V = L

condition cBA

application

version

is

of

is

[3]:

101

if B is nite,

BA's on

a cBA

then

are

a cBA and ~

IAut

that

B1 ~ =

Aut(B) , the

iAut

by

are

by +,-,-,0,1,

denoted write

their

a Sb

automorphism

group

of

B,

is

infi-

B I.

abbreviated

. We

instead

such

underlying

instead

the

of

sets.

The

infinite

a + b if

a,b

finite

joins are

operations

and

meets

disjoint

by

and

~a i i6I call

of

[a i if t h e a i a r e p a i r w i s e d i s j o i n t . If ~b i = I, w e i6I i6I (bi)i6 I a p a r t i t i o n of B. B is t h e n i s o m o r p h i c to t h e p r o d u c t a l g e b r a

(B ~bi) , a n d e v e r y p r o d u c t d e c o m p o s i t i o n of B a r i s e s , u p t o i s o m o r i6I p h i s m , in t h i s w a y . E s p e c i a l l y , w e h a v e t h a t t h e d i r e c t f a c t o r s of B are,

up

set

D of

is

some

We

use

to

isomorphism,

B\{O]

said

to be d e n s e

d 6 D such

that

d ~b.

standard and

The

author

ful

comments

pointing

set-theoretical

IXI

is

on

out

the

of

this

I. D e c o m p o s i t i o n s

B and

B,

C are

in

of

correspondence

B.

A sub-

b 6 B\{O} , t h e r e

the

set

of n o n n e g a t i v e

and

M.Rubin

mostly

his

for

help-

to R . S o l o v a y

for

simplification methods

improving

paper.

automorphism

to be

totally

B ~b

different.

of

R(B)

algebra

into

products

different

~ C ~ c.

group if

x,y 6 B are

of B.

there totally

are

If C is no

a cBA,

b 6 B\{O}

different

and

if B ~x,

Let

R(B) = {r 6 B l f ( r )

the

of

X.

about

the

R instead

each

B ~b

a considerable

B be

totally

of B,

2 and

let A u t

B by

algebra

C - but

T(B)

that

w is

to M . R i c h t e r

and

said

algebras for

a set

R(B)

such

We write

of

Theorem

section

for

if,

of

c 6 C\{O} are

in B

notations;

gratitude

proof

moreover

relative

cardinality

her

an e r r o r

C;

results

a cBA

the

expresses

in T h e o r e m

For

the

is

integers

the

exactly

=r

for

if B is of

every

fixed.

invariant

f 6 Aut Clearly

elements

B}, R is a c o m p l e t e

of B.

sub-

102

1.1. and

Let

a 6 B.

assume

that

interchanging f(a) # a,

a 6 R

O <x

x

and

iff

~ a, y

and

different.

f 6 Aut

So

of

B

each

r i is

1.2.

For

f(a)

such

such

On

~ a and,

totally

a 6 B,

different

is

f(a) of

clear

R which

is

a 6 R iff

1.3.

For

x ~ r. g(x)

=x;

f 6 Aut

greater

Let

g 6 Aut(B

Then

If

g(x)

(ri)i61

rj

are

Fy.

since

There

z ~-(x+y)

hand,

symmetry,

a 6 R. B

or

have B;

proves ~ r).

and

equal

R(B

put

let

f(a)

f(a) = a.

totally

-a

-a

-If

and

hence

f 6 Aut

that

totally

is

i # j,

a parthen

{ r i J i 6 I} ~ R .

6Aut

B}.

Moreover,

if

hence

a ~r.

to

i.e.

a,

r 6 R and

So

a is

a ~r,

the

then

smallest

element

r r) = R ~r:

x 6 R.

To

since

a partition

r r) . S ~ n c e

prove

f 6 Aut

B

the

such

x 6 R.

This

of

we

R,

first,

that

x 6 R(B

g 6 Aut(B

converse,

shows

have

suppose

g ~f

[ r) , w e

let and

~r) . So get

x 6 R ~r ~B f(y) = y f o r

x 6 R(B

~r) .

B =]-~ Bi, iCI

where

Bi~B

f(x) = r r. y ~-r.

~ r i.

Let

R i = R ( B i) . S o R ~

(R [ r i ) i61

i.e. gives

each rise

product to

B

totally

(ri)i61 for

a 6 R

shows

be

are

different

[ rj = - r i , j#±

an

which

a and and

is

let

6RIa_
g = f~(B

Choose

= f(x) = x,

is

• - a = O,

different:

a = a.

f 6 Aut this

B ~x ~B

other

[{f(a)If

a ~ a and

for

r 6 R we

Let

totally

f(z) = z f o r

from

a :Jl{r

and

are

let

that

~ f(r) = r

-a and

the

by

r i and

a =

It

that

B . f(a)

that

a and

O < y ~-a

a contradiction.

different

tition

Then

decomposition

a product

= I~R(B i61

rri)

=]~Ri, i61

B ~ ~-~B i g i v e n b y a p a r t i t i o n i61 decomposition R(B) ~ J J R ( B i ) . i6I

of

R ~B

103

1.4.

For

assume, are

each by

of

cBA

defining

minimal

proper

isomorphism B

hence

a cBA.

i6I If

I={1,...,n},

cardinal

and

we

for

are

For

t 6 T,

(e ® t) Define,

~

~

any

the

means

and

that

of

s_
a cBA

of

T:

t

freely the

of

BA's

isomorphic

that

T(B)

is

a set

sets

and

speak

class

about

of

all

(ti)i6I

where

in

T ( B i)

®

of

B. to

even

we

= t i for

t n instead

of

than

classes

types

define

may

B which

rather

isomorphism T,

We

iI] = ~

associative, e e (s ~

iff

t=s

cBA

the

and

T(B)

Q t i 6 • by i6I

G ti. i6I

If

~ is

t) = (~ ®

and

distributive

s) ~9 (~ ® t) ,

(~B)

laws ~t =

By

of

type

< is

for t has

we

such

the

and that

choose

induction,

x 6T.

a direct

give

reflexive

x,y 6T

some

factor

of

that

< is

proof

transitive. t =x

an,

(D s,

bn,

cn,

Now s =y d n in

type

s.

suppose ~

t. B

bn Pan)

=s,

s,t 6T

Let

such

B be that

_> b o _> a I > b I _> . . . .

= Cn + a n + l T(B

Let

rbn)

e =

=t,

T(B

I-~ an = ~ b n " n6~0 n6w

[Cn)

=x,

NOW

For

a partial

an = d n + b n

ao

a

i 6 I.

I = ao

So

of

i 6 I.

t i = t for

commutative

69 x

reader,

Choose

T(B

a

in ~ ,

each

s.

rank,

~ t i where iEi

clearly,

type

set

type

let

of

t_<s.

isomorphism

etc.

s and

convenience

order

the

t i ~_) ...

® , e.g.

(8 e t)

the

family

write

s_
be

T be

obvious

and

for

be

t i = 4(l--~Bi) i£I we

some

(~ ,

to

may

Let

® t =

There

T(B)

T(B)

types.

is

let

set-theoretical

class;

where

B,

T ( B ~ d n) = y .

104

ao

:

e -~ [

Cn "~ ~ d n

nEa) bo

and

hence

1.5.

= e 4

~ Cn n6w

s =

~(B

be) @ ~

~x

~

~ ®y

t =

T(B

be) +<0 ® x

O

~ ~y

Let

B

is

T has

B be

a fixed

fixed,

cBA; =

T(B)

element

T(b) 6 T t o

b 6 B gives

implies

Suppose

x.b =0.

different

do

1.6. we

We

do

b,

not

identify

clear

know (note

s , t 6 T,

(i.e.,

and

we

different)

is

instead

of

T(B

r b).

-b

R wiht

(T,~)

= 0 Tiff that

element

T : B --~T w h i c h we

Put

Choose

are

x 6 B

totally

a subset

of

which

of

following

is

the

a lattice, not

s,t

imply

are

7(a) = s,

the s O

totally T(b) = t,

onto

a ~b

that x

in

O <x is

we

following t 6 T,

~a

x ~ a.

shall

way.

R

totally

contradicts

diagram

in

that

such

commutative

T

clearly

different,

T,

and

0 T = 0 T.

is

prove

T a) ~ T(b) , w h i c h

s,t 6 T does

such

a smallest

S T(b).

the

T = {t 6 T i t N T(B) },

injective:

a by

R inside

inf(s,t)

if, a , b 6 B

T(b)

Clearly,

a map

and

have

whether that

T. and

~(a)

from

characterize

yet

T [R

Sinceb

hence

sequel,

may

for

x ~-b.

therefore

the

fectly (I)

from

may

in

So

by

I T = T(B)

. Now

T(a) ~ Y ( b ) .

b 6 B write

{~(b) ib 6 B}.

and order preserving

- We

[ dn n->l

for

abbreviate

a greatest

Assigning

but

$

s : t.

T(B)

If

nE<J

always

Although are

per-

generally):

different then

a,b

are

totally

105

(2)

if

s,t 6 T

Call

t 6 T

sup(s,t)

and

inf(s,t)

complemented

: s ~

t : I T. R :

which

follows

t 6 T has

{r 6 T i r

1.1.

a pseudocomplement

t*

which

1.7.

We

proceed

t ^ b

is

is

even

to the

Namely,

in

show

Let

that

define,

for

infimum

t A b ~y

f(x-b)

argument,

that

complemented It

is

in T

also

equals

inf(s,t)

s ~

t.

: 0 T and

in T } ,

not

(i.e.

of

b 6 R and

t and

b

where

x 6B,

is w e l l - d e f i n e d , be

f(x) since

such

and

hard

to

a greatest

see

that

element

every

disjoint

in

t 6T,

an

(T,<),

element

but

we

t ^ b 6 T

will

not

(in

use

let

f : B rx--~B

and

s C T

exists

R.

t Ab=m(x-b) To

is

is

from

t)

this).

there

sup(s,t)

Then

easily

from

fact,

if

= 0 T then

using

an

suppose

isomorphism.

= f(x.b)

re(x) = t . y 6 B

totally

different

f-i

instead

of

f,

that

T (y) = t.

Now

Sf(x.-b):y-b

is

such

%y--b,

from

shows

y.-b,

that

f(x.b)

_
f(x.b)=y-b,

The

hence

same

T(x.b)=

m (y-b) .

I .8.

For

t E T,

let T ~t =

So

if

b 6B

and

decomposition the

product

~(b) : t , of

of

R gives

a family

{s 6 T l s _
~b) = T

rise of

to

partial

order).

i 6 I,

~ b i and

Ti = T ( B i )

Bi = B

We

now

a product

partially

coordinate-wise let

Pt.

Let :T

that

decomposition

ordered (bi) i6I r b i.

prove

sets be

Define

is

each

product

of

(where

T

endowed

a partition

with of

R.

the For

106

%° : T --~ T~i i E I

i£II

t ~-~ (t ^ bi)

F

Since

[ I Ti f o r t 6 T. ~ is o n e - o n e : f o r t 6 T, i61, l e t x 6 B s u c h t h a t r(x) = t. B y x = I x . h i , w e s e e t h a t t = T(x) = i6I ~T(x-b i) = ~ (t ^ b i) m a y b e r e c o n s t r u c t e d f r o m ~(t) . ~ is o n t o : i6I i6I l e t t i 6 T i f o r i 6 I, i.e. t i S b i . C h o o s e , f o r i 6 I, x i 6 B s u c h t h a t

Y(xi)

t A b i ~b i for

= t i and

i 6 I,

x i ~ b i 6 R ~ B.

T ( x o b i) = T ( X i) = ti; Choose

~(t)

x,y 6 B

such

so

~(t)

that

6

Put

x

i xi i6I = ( t i ) i 6 I. -~

T(y) : t,

=

and is

x ~ y and

t = T(x).

For

i 6 I,

order-preserving:

T(x) = s.

Then,

t A bi =

let

for

s ~ t.

i 6 I,

S A b i = T (x'b i) < T (y'b i) : t A b i. Finally,

<0(s)_<%0(t)

(i.e.

s ^bi

_
i E I)

implies

s _< t,

since

s = i ~ i ( s ^ b i) < ~ (t ^ b i) = t . iEI This

shows

that

1.9.

Define,

<0 is

for

our

an

isomorphism

cBA

from

T onto

~-~T(B i61

~ b i)

B,

a =

[

{x 6 B l x

is

an

atom

of

B],

h =

[ {x E B I B

~x

is

atomless

and

homogeneous}

r =

[ {x 6 B I B

~x

is

atomless

and

rigid}

d = -(a+h+r) (a B A

C

every

c 6 C\{O}).

totally

is

called

rigid a,h,r

different.

By

B [a

is

cardinality lyl = Izl. I ~ <.

So

the

T(B

set

~ a)

CI : I a n d

hence

d

are

homogeneous

elements

of

if

R and

C ~C

~ c for

pairwise

1.8.

isomorphic of

IAut

and

T =T(B)

Now,

if

is

~T(B

to of

r a) x T ( B

the

power

atoms

of

isomorphic

~h)

set

x T(B

algebra

rr)

x T(B

P(<)

~d).

where

< is

B~

for

y , z E P(<) , ~(y) = T(z)

to

the

well-ordered

set

of

the iff

cardinals

107

Next,

(see

[7]) ,

TTH
B rh ~

i6I where

e a c h H i is a h o m o g e n e o u s

and H o are t o t a l l y

different.

ordered

chain

structure

number

a greatest

of H.

cBA

and

< a cardinal,

element

It is n o t

T ( H K)

(which d e s c r i b e s

is a w e l l -

fully

the

hard

to p r o v e

that,

for a n y

cardi-

~,B < c(H) .

a cardinal,

by

each

Cj

=

is r i g i d ,

are t o t a l l y

different.

C is r i g i d ,

T ( C <)

itself

greater of

and

one

c(H) ~ ~ ~ ~}.

~ 6 K, w e h a v e

Since, that

for e v e r y

(T(H <) , ~ ) ~

(K,~).

[7], B ~r

a complete

than ~

~C< j j6J 3 <j

is a c a r d i n a l

It c a n be p r o v e d

is a r e t r a c t

of

lattice;

< and (Z(c) d e n o t e s , b y C.

If C s a t i s f i e s

T ( C <) ~ (K,~) (c) . H e n c e on T ( B [d) , w h e r e the existence

1.10.

1.8,

and d-d' : O for d,d' 6 D, d # d ' },

x 6 H K , H K r x ~ H ~ for e x a c t l y

power

i # j, H i

~,~ > O ,

L e t K = {0,I} U {el~

hence

for


atomless

H ~ ~ H B iff

where

and,

J J T(H i ) . i6I

= sup{ Jml+Jm ~ H ,

the S o u s l i n

Again

K i is a c a r d i n a l

of T ( B [ h)) : let

c(H)

nals

with

By

--

T ( B ~ h) ~

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

cBA,

of

It is e a s i l y

established

j ~ j', Cj

in an e l e m e n t a r y

complete

lattice

here

K is t h e

set of

for

any

the

countable

the difficulties

algebras

for

the

B ~ d is a c B A w i t h o u t these

and,

was

that

way

that,

if

(K,<) (c) and cardinals

~ not

s t r u c t u r e O6, the B o o l e a n chain

condition,

in d e s c r i b i n g homogeneous

first

the

and Cj,

proved

following

in

T(B)

then

concentrate

or r i g i d

factors

[I].

conditions

on a cBA

-

108

B

are

equivalent:

a)

T =T(B)

b)

T

c)

T = R(B)

d)

T(x) = T(x)

e)

each

is

by

each

for

where

A

is

2.

The

We

assume

bra

B.

But

a cBA

T(B)

notation

Let,

method

of

Stone

P 6 X, We

get

X where

predicate

Theorem global The

of

a quotient

then

over

space

<.

A.

are

clopen

subset

establishes

a cBA

that

hard

not

{~ ®

t-

and

to

see,

t holds

e ® t = t for

the

an

of

R = R(B) the

this

for

each

t : ~(A)

T,~)

a complete

subalge-

element

r 6 R

section,

topology.

of

is

smallest

ultrafilter

usual

(Tp,~)

so

x 6 R is

rest

is

the

I;

and

of

R},

We

shall

define,

a topology

on

logic

one

for

S = U Tp. p£X (as d e f i n e d in [6])

~ is

language

of

first-order

isomorphic

to

(r($) ,<) , t h e

We

is

iff

$ = (S,z,X,~)

the

condition

it

a sheaf

sections

(Tp,<)

is

chain

factors.

x 6 B,

structure

of i - s t r u c t u r e s

with

binary

prove:

(T,<)

essential

R with

a cBA:

yields

section

for

countable

T(B)

X : {p ~ R I p the

the

T(B)

rigid

for

T.

always

that

of

that,

x ~ r.

is

Solovay's without

Recall

in

satisfying

structure

the

satisfying

B

equivalence,

lattice

of

a complement

factors,

t 6 T(B).

algebra

x £ B

a cBA

above

algebra

Boolean

for

t 6 T has

rigid

the

a Boolean

a complete

Moreover, without

is

is

~ -structure

of

all

of ~. point

linear of

in

orders X,

np

this

representation

and, being

for the

s , t 6 T, canonical

of

(T,<)

{p 6 X I ~ p ( S ) map

from

is

that

the

stalks

< ~p(t) } is T to

Tp.

This

a

109

Theorem

B.

and

dual

lattice

the

operation

If

its ~

is

(T,<)

a distributive (T,>) of

algebra

in

2.1.

1.7,

In

T(x) = t.

the

We

R imply

we

of

there

from

B rx.b

from

z.-c ~-c.

2.2.

Let

onto

some

p E X be

s ~

tiff

s ~

~p

a reflexive

Clearly,

is

equivalence s ~pt

iff

s , t E T, zp(t)

is

Heyting

(T,<)

is

algebras. in

a

(T,<)

(T,<) ,

linear

Heyting

t E T,

the

fact

such

such

b E R by that

that

that

t Ab = ~(x.b)

s ^b

~ t Ab

T(x) = s,

z ~y.b and

and

in

T and

T(y) = t.

some

x - c ~ c is

where c ~b

By

isomorphism

totally

t ~

b E p P

such

that

s ^b

~ t A b

s.

and

transitive

relation

on

T,

~p

is

an

and

there

is

some

b E p

such

that

s A b = t ^b.

let = {s' E T I s '

~ p t} '

T p = { z p ( t ) It 6 T} np(s) So

(Tp,~)

and

is

2.3.

s

mapping

ordered and

Tp

set,

has

~p:T

--~Tp

a smallest

is

Z p ( O T)

an

order-preserving

and

a greatest

Wp(IT).

Let

the

iff

a partially

surjective

element

to


s,t ET.

following

In

the

rest

construction

of

Section

and

2,

notations:

we

shall choose

frequently x,y 6 B

f

different

s,t 6 T define

some

t and

relation

Both

~ z.c ~ y.c.

For

there

for

z = z.c $ z.-c

f(x.c)

fixed.

p

and

I.

pseudo-complementation

x,y E T

z E B

s ~p t i f f P

use

Now

Hence,

algebras

O and

[8].

let

B F z.

with

s , t E T - i.e.

sometimes

is

Stone

t A b E T

s A C ~ t ^ C:

~ t Ab

for

defined

shall

s Ab

For

sense

are

lattice

relative

t) V ( t ~ S) = I T h o l d s

(S ~

in

is

such

refer that

110

T(x) = s , maximal T ( y i)

T(y)=t. with

for

Let,

respect

i 6 I;

by

to

for

Zorn's

the

i # j,

XO

=

lemma,

properties:

{ (xi,Yi) li 6 I}

O < x i ~ X,

x i • xj : Yi " Yj = O.

[ Xi i6I

x 1 = X"

F=

-x o

a family

O < Yi ~ Y a n d

T(X

i) :

Put

Yi

Yo

:

Yl

: y " -Yo"

[ i6I

be

So T ( x o) for

some

d 6 T

and

= :(Yo)

x l , y I are,

= d by

maximality

of

F,

totally

in

1.2.

different.

Hence b = xl, disjoint

are

2.4. So

Let

by

p6X

f =-b

6 p

the

be and

definition

fixed.

s ^ f = dA

c ~ p.

proved

So

for

Note

that,

s,t,

the

depend

on

the

d 6 T defined

P

by

is

proved

symmetry

2.3

for

or

c £p.

Assume

that

b {p.

yields = Xo.f ,

b ~ p;

(Tp,_<)

s <_ t or P

choice

that

b £p

=

t ^f

of

by

symmetry,

is a l i n e a r

x,y,F,xo,xl,Yo,y

x,y,F, ....

from

order.

I are

t _< s f o r p

t _
not

every

uniquely

p 6X

Moreover

for

does, each

determined

by

of

not

course,

p 6 X,

the

type

satisfies

~p(d) this

This

given

t.

p 6 X,

that

in

b.c =O,

s _
although

fact

xl,Yl

f = T(yo.f ) S ~(y-f)

that any

of

= Xo.f Sxl.f

s -<

have

By

xl.f =O. x.f

We

c = ~I

the

: min(~p(S) case

c ~ p implies

,Zp(t)) ;

b ~ p by

the

above

d A --C = t A --C.

lines,

and

it

follows

111

2.5.

Let

s,t 6 T be

fixed

u : By

the

want

definition

to

set

show

{p 6 Xl

cl(f) ~ u .

2.1,

as

in

u

f £ p}

of

=

it

is

even for

{a 6 RE

there

2.1

is

that

claim

that

p 6 u,

choose

clear

that

clopen.

f 6 R by

s ^ a ~t

a subset

2.3.

Put

u

an

First,

is

open

subset

of

the

clopen

abbreviating

cl(f) , w e

u = cl(e). f 6 p proves

{p 6 X I ~p(t)

~p(S)

W.l.o.g.,

i.e.

Clearly,

} and

In

{p6X

that

=

see

that

X.

We sub-

s ^ f S t ^ f implies

of

[RM.

M

such

e is

e 6 M,

an

So

in X,

Now

f 6M,

analoguous

p # q

~ =

largest ~u.

I Zp(S)=~p(t)

for

that

the

cl(~)

s ^ f ~ t ^ f.

u ~cl(~).

=

~

that

by

satisfying

suppose S

^ a} ,

{ a i l i 6 I}

s ^ e S t ^ a;

cl(e) , w h i c h

2.6.

X

in

s ~pt}.

~p, is

b , c 6 R as

Let M

By

{p6x{

of

that

and

~ a i. It f o l l o w s i6I element of M. W e

suppose

f ~

way,

} are

we

and

p 6 cl(f)

see

that

clopen

T p N T q = ~.

p 6 u.

subsets

of

By

X.

Put

UTp.

p6x Let

z : S --~X be

defined z(x)

For

t 6 T,

g 6 R

{bt,glt

checked

that

onto

For

by

X.

2.7.

Let

=

6 T , g 6 R} ~ is,

x 6 S there

~ rn

p be

iff

x 6 Tp.

is

the

{ ~ p ( t ) I p 6 cl(g) }. is

a basis

w.r.t,

z homeomorphically

n = bt,iR;

= p

let bt,g

Then

by

an

map

is

this

of

a topology

topology,

an

injective

which

open

n of

subset

continuous

assigns

to

S.

It

a continuous

a neighbourhood

onto

of

of

X:

open

every

x

in

open S which

let, map

p 6 X

is

map

from

is m a p p e d

if x = Zp(t) ,

from

the

easily

n onto

X.

~-structure

S

112

(Tp,<) . T h e x < y in there

underlying

are

sheaf

The

neighbourhoods

of

g(p) 6 T p Since

details

F(~) ~ ~ Tp, p6x

give

2.9.

For

show

F(~)

examples

t 6T,

that

be

(p) . S u p p o s e

show

sense

of y s u c h

that

of

is

~=

[6].

g : X --~S s u c h the

where

s , t 6 t. that

(S,~,

Recall

that

of

all

F(~)

as

a substructure

sections

of

global

an

for

every

X,p) that

and

Then

is a a global

~ o g = idx,

set

(if n o n - e m p t y )

x,y 6 Tp

i.e.

sections

of

~ .

of t h e p r o d u c t

~ -structure.

In 2.9

$ .

be defined

by

= Zp(t) .

g t 6 F($),

bs,g

bt,g

2.7

consider

gt : X - - ~ S

-i

y = Zp(t)

2.5.

the

map

F(~)

and

from

and

X in

of g l o b a l

let

q 6 X satisfying and

2.6

denotes

we may

is T p = z

of x a n d

follows

in

pl61x(Tp,<) ; so

Zp(t) ; l e t

p=q

as

over

~t(p) We

bs,g

proved

p 6 X.

~(p)

x = ~p(s)

@ is a c o n t i n u o u s

for

structure we

< ~q(t)

of ~ - s t r u c t u r e s

section

of

p(p) = (Tp,<) ; e.g.

q 6 cl(g) , Zq(S)

2.8.

set

i.e.

that

g t is

a neighbourhood

g 6q.

By

zp(t)=~p(S).

of x.

Zp(t) = ~ q ( S )

Let,

by

a continuous

map:

let

S o g 6 p a n d x = ~q(S)

and

Tp ~Tq=~

for

p#q

gt(p) for we

= x

some

get

2.5,

v = {p' 6 X I ~ P, (s) : ~p, (t) } = el(B) . So p E v and

g.

of p in X i n t o

~ 6p. the

We

prove

that

neighbourhood

gt m a p s

bs,g

the neighbourhood

of x in S:

and gt(p') since

g E p' .

2.10.

Proof

of T h e o r e m

= ~p, (t)

A.

We

: T -~ F (~)I r t ~--~gt

J

= ~ p, (s) 6 b s , g

prove

that

cl(g.~)

if g • B 6 p' , p ' 6 v

=

113

is

an

isomorphism

of ~ - s t r u c t u r e s .

Since

s ~ t

iff

F(~) :

~(t)

6 F(~)

for

every

for

t 6 T,

F(~) # ~.

Now

if

s ~ t,

in

-i

assume

u = X, are

us ~ ut

Us( p ) = Z p ( S )

Conversely, that

T

~ ~p(t)

that

a = I and,

be

given.

also

Utp(P)

by

s £ M,

that

u ~Up = Utp

~Up.

Since

p E X

By

holds

This

- So

remains

to

choose is

I ~ i ~ n

is

the

of

X,

hence

terminology

that

such

open

p 6 X;

one-one

prove

an

there

bl,...,b n 6 R are

in

~

tp 6 T

compactness

for

means,

s ~ t.

= ~(p) , t h e r e

I R : b I + ... S b n a n d c l ( b i) ~ U p i .

it

For

= np(tp)

= ut(p)

u s ~ ot"

order-preserving;

u E F(X,S)

in

and

~

that

is

exists

and

Let

Up

of

p

such

bl,...,b n 6 R such

Pi E X

pairwise

~

2.5,

~(p) = Z p ( t p ) . S i n c e

neighbourhood

choose

of

both

onto.

u s ~ ot-

that

satisfying

disjoint,

the

isomorphism

type t = is

in

that

T.

We

claim

p 6 Upi

ut(p)

and

: ~p(t)

~_t_!~" P r o o f properties

that

of

resp.

of

Define

B.

largest

u : {p 6 X [ ~ p ( S ) SO u

and

which

v

are

ut;

likewise,

and

with

which

Os

with the

on

v

coincides

docomplement (Tp,<) , b e i n g

with

of

fact,

for

the

other

On

prove B.

that

~(OT)

elements

subsets ~s

on

u

function is

(tpn A bn) p 6 X,

e.g.

hand,

b i 6 p,

we

have

t A b i = tpi A b i and

(F(~) ,<) and

of

~(I T)

F(~) . L e t

is

a lattice

are,

the

and

u s w.r.t,

X.

Clearly,

with

~t

u s v o t 6 F(~)

supremum

UlT

a linear

of

on

u

qt"

order,

and

of

is

v

which

u s and

with

F(~)is

on

the

ot

ut;

is

course, u t be

function the

the is

a distributive

with

the

the

elements

}. u s ^ u t 6 £(~

infimum

coincides

on v ~ u

a distributive

of

o s and

~ ~ p ( t ) }, v : {p 6 X l ~ p ( t ) ~ p ( S )

clopen

coincides

We

in T h e o r e m

the

~

: UtPi(p) .

Theorem

claimed

...

u = u t . In

q(p) = O t p i ( p ) .

: ~p(tpi)

smallest F(~).

(tpl A b I ) ~

with

function the

u s and

ot

on u

~s ~ u t

relative

lattice

lattice

of

and

since

6 F(~ pseuevery

u s A ~t,

114

Os V O t are holds

computed

in e v e r y

gebra:

for

and h e n c e

such

in

that

lattice

structure

(Tp,>)

of

algebra

(F(~) ,<)

of g l o b a l

2.12.

Let

s i in T)

and

of is

(a ~ b )

Moreover,

F(~)

t 6 T,

~t

and

in F(~)

of the

same

v (b ~ a ) = I is a S t o n e

subset

(ot)~(p) = O and

sheaf

(ot)~v is,

up

show

that

of X.

Let

if p ~ w . (ot)~=o

I

T"

to i s o m o r p h i s m ,

(X,n,S,~')

arguments

al-

where

U' (P) =

(F(~) ,>)

is a

algebra.

a family

For

is a c l o p e n

(F(~) ,>) , w h i c h

sections

a Heyting

(si)i6 I b e exists.

F(~).

equation

(ot)~(p) = ] if p 6 w

if U(P) = (Tp,<) . So the

Stone

The

w : {p 6 X I o t ( p ) = O T p }

is the p s e u d o - c o m p l e m e n t

The dual the

(Tp,<)

ot 6 F(~),

(ot)~ 6 F(~) (ot)~

coordinate-wise.

in T s u c h

that

V (t A S i) a l s o i61

V s i (the s u p r e m u m i61

exists

of the

and

t^ k/si = V ( t ^ s i ) , i6I since

(F(f) ,V,A,~)

i6I

is a H e y t i n 9 a l g e b r a .

tv

As

i =

/k (tvsi)~

i6I if

AS i6I

i exists.

~ tij j6J

i6I

since

the of are

as w a s

s,t 6 T

infimum

of

chosen then

let A ~ B cBA's

such

and

of T

(R is the

noticed

in 1.6)

of the

form

can be p r o v e d set of

and

these

for T,

complemented laws d o

not

cBA's.

and

choose

x,y,..,

in 2.11. that

Thus,

× E, A' ~ B ' suppose

different. ThenB~D,

since

in 2.3.

follows

D~D',

Then

b y 2.4

e.g.

T(y) = t, x o _ < x

the

x E' w h e r e C and C'

T(B) = T ( A )

AY(A')

d = T(x o) = T(yo)

is

and the d e s c r i p t i o n

b u t x - - x o, Y ' - Y o

yields

× C' ~ D '

B~B',

as

if T(x) = s,

y(x o) = ~(Yo)

T(x o) = s ^ t. T h i s

× C~D

IJl ~ 2

sublattice already

laws

II, ~ ,

s and t in T w h i c h

Os ^ o t g i v e n

rent,

are

of T,

Let

no distributive

~/ ~ t i f ( i ) for f6J I i6I

in a r b i t r a r y

2.13.

i61

course,

R is a c o m p l e t e

elements hold

:

Of

By duality,

are t o t a l l y

following A,A'

and yo < y

fact:

(and h e n c e

resp. =T(D).

diffe-

E and E'

B,B'...) are t o t a l l y

115

2.14.

Let

(V,A,V,*)

be

a Stone

a subalgebra

of V and

of V in

Let X be the

x
[4].

iff x ^ b _
epimorphism sheaf

over

(a) b o t h algebra

to 2.10

Vp.

which

of

Sk.

to see

{x*Ix 6 V } . is c a l l e d

For

p6X

repeat that

hard

the

Sk is

skeleton

and x , y 6 V ,

the w h o l e

~c

is the

to p r o v e

that

put

construc-

=p : V - - * V / ~ p = V p

and V ~ F ( ~ c) w h e r e It is not

Then

is an

"canonical" the

following

equivalent.

stalk

Vp

lattice

are

Stone

(a~b) v (b~a) = I for of ~ c

is a l i n e a r

= ~ ' (p) } is a c l o p e n

algebras

a,b 6 V;

order

subset

and V is a H e y t i n g

of X

and,

for

(i.e.

~,~' 6 F ( ~ c) ,

Sc = ~ Vp p6X

is a H a u s -

space).

course,

not

isomorphism,

every

the

B A and,

if R(B)

Theorem

C and

3. W e a k l y In this cBA's.

Stone

is finite,

homogeneous

Our m a i n

we define goal

(Solovay).

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

in 3.2

every

cBA B:

stalk

can have,

Sk(T(B))

=R(B)

up

to

is a c o m p l e t e

Tp of ~ is w e l l - o r d e r e d

by

in 3.1

Every

some

3.1.

For

a cBA,

Then

the

following

homogeneity,

a property

of c e r t a i n

weakly

several

of t h e s e

homogeneous

cBA

is, up

to i s o m o r p h i s m ,

cBA.

equivalences

although

weak

is

a power

ty and

some

(a)

cBA's

C

give

for

satisfying

1.9.

section

first

algebra

f o r m T(B)

Theorem

We

space

Sk=

b 6 p. We m a y

algebras

stalks

satisfying

{p 6 X I o ( p )

Of

are

some

and

algebra

Stone

2.6

V a n d its d u a l

(b) e v e r y

dorff

in 2.2,

X with

conditions

a Boolean

for

of S t o n e

algebra

equivalent

to b e i n g

equivalences

let R = R ( B

a power may

and T = T ( B )

conditions

formulations of

a homogeneous

be k n o w n

and

homogeneicBA,

to the reader.

be d e f i n e d

are e q u i v a l e n t

of w e a k

as in S e c t i o n

a cBA

satisfying

I. any

116

of

them

will

a)

R = {0,1}

b)

if

B ~x ~B

c)

(T,5)

d)

if

is

weakly

are

x,y 6 B

(i.e.

a,b

are

~y

a linear

a~b:

if

6 R,

O
O < a ~a

c) : l e t family

and

i ~ j implies

Ix i = a i61

or

not

exists

O < a,b 6 B and ~b

w.r.t,

£ R and By

the

r 6 T

a,b

a b : O,

Zorn's

iYi = b . i6I

d)

~

is

being

3.2.

On

B~H

f)

(T,~)

g)

If

other

< where

H

is w e l l

ITI ~ 2,

Proof.

e)

~

f)

g)

~

e) : l e t

so

H

is

=

@(x

of

i)

is

:

some

different,

R has

let

By

0 < r ~ s,t.

b)

at

least

F = { (xi,Yi)

the

four

Li E I}

0 < Yi ~ b '

and

then elements. be

a maxi-

T ( x i) = T(yi) ,

maximality

of

F,

Then

@T(y

they

i) ~ ( b ) .

the

Then

satisfy

following

r

and

-r

infT(r,-r)

conditions

cBA

and

are

totally

different

= O T.

are

equivalent:

K a cardinal

ordered

then

T has

a

was

shown

in

as

g) . L e t

to 6 T

subset

cardinal

< I.

a homogeneous

homogeneous.

a dense

that

i61

O
T,

hand,

D : is

0 < y ~b,

different)

O < x i ~ a,

~x i = a . i6I

r 6 R and

elements

the

e)

~ a,

trivial.

a) : s u p p o s e

and,

O <x

totally

hence

lemma,

conditions:

Assume

such

are

i61

~ d)

that

totally

x i . xj = Yi " Yj = O.

<(a)

c)

such

order

a , b 6 B.

mal

homogeneous:

there

s,t 6 T~{OT} , there

Proof.

~

called

a,b 6 B\{O},

and

b)

be

of

< such

in Now,

least

by

1.9.

Hence

that

f)

~

H be

g)

element

is

a cBA

to.

trivial. such

that

T(H) = to,

g) ,

{X 6 B ~ { O ~ T B.

positive

(X) : t o } there

x~ 6 D

for

is

a partition

~ < K,

and

B~

(x~)~<<

of

U B~xa~H

<.

B

for

117

3.3.

Trivially,

We

begin

in

3.3

with

to

geneous

(*) and

to

get

cases.

be

uncountable

it

turns

numbers

or

In

out and

that

use

C

A

T

B

is

in

a cBA

3.5

C of

as

but

w.l.o.g,

i.e.

is w e a k l y

Theorem

but

not

T +=T~{OT}

proof

is

Since

C.

homogeneous. So

a power

will

rise

a measure the

type

t.

we

assume

of

a homo-

b = I.

Our

be

easy.

to to

cardinal

the

s ~

In

case

B

unit

element, relative

to

a

split

@ of

(3.5),

up

in

B will

where

interval

positive

6 = ~,

of

real

~-additive

homogeneous.

t = t for

t = t,

will

invariant

a strictly

s ~

every pass

proof

closed

and

smallest

for

that

that

By

holds

frequently

algebra

fact

(*)

no

shall

isomorphic

gives

has

we

a certain

we

may

Cn+l)

= S

s,t 6T choose

implies

~

• s ~ t:

a sequence

(Cn)n6 W

satisfying I = c o ~ c I ~c

7(Cn) This

cBA

homogeneous

well,

(3.4),

T : B--*T

We

shall

b >O,

and the

hence

converse,

ordered,

assume

case

measure;

the

a contradiction.

where

B bb

of

a homogeneous

that

want

B ~b,

of

is w e a k l y

linearly

suitable

in

B

is

algebra

choose

proof

that

i.e.

power

(T,~) we

two

the

3.5

BA,

each

shows

2 ~

= t

... T(c n -

that

* s=~( [c~.-%+~)~(c/=t. n6~

Call

t 6T

a-divisible

such

that

e e s = t and

almost

such

that

e

e ~

t = ~ e s

e s ~ t.

implies

a-divisible,

(where

If

e ® t = e e

then

t is

e is

a cardinal)

a-divisible

and

t is

if

if

there

a-divisible,

(e e s) = e ® s = t.

a-divisible:

there

let

C be

If

a cBA

is

is some

then

some

s E T

s 6 T,

e e t = t,

every

t' ~ t is

of

t.

type

By

assumption, D = {x 6 C ~ { O } I T ( x ) is

a dense

subset

{ c i [ i 6 I} ~ D;

of

suppose

C.

Let

%(ci)

= e e s f o r s o m e s 6 T}

(ci)i6 I be = e e si.

a partition

Then

of

C

such

that

s #OT, for almost

118

t =

Finally, T(c)

if

is

let

joint

that

some

is

factor.

above

of

is

We

3.3.

A:

~ _> e,

joint elements T (b) < t.

By

Let

Zorn's

of

3.5.

B

definition

and

B:

s~gt = t,

Hence,

for

then

and

Note

that

the

more

general

@

last

for

is

we

~ < e we

then

are

k 6 e

n-di-

disjoint

such

that

get

2 k pairwise

and

have

the

n < 2 k. dis-

same

type,

Surely

lemma,

amaximal

~ = II

since

pass

s-divisible then

cardinal

6 >e,

otherwise,

have

show

T (b i

smallest

if

B has

to

~<

such

B ~b

6 but

where

no

t 6T +

~ ® T (b) = T (b) : T h i s w a s

e < el -< 6 a n d

: T(b).

we

6,

the

a n d e I _< 6,

-<e ® T ( b )

start (for

that

t=IT,

family

= t.

contradicting

(bi)i61

Put

b = - [ b i. i£I is s m a l l e r t h a n 6.

<- (~+I) ® t

of

(*)

pairwise

dis-

By maximality This

in

,

yields

=t,

out

with

several

s ~ 0 T would

s =n

such

® s'

and

assertion

is

conditions

remarks:

e ~s
if

s,t 6 T

contradicting

it is e a s i l y

checked

6=e) .

that

s = s'

a,c 6 B

b 6B

simple

imply

T(a)
' implies

exists

t =n

Let

c 6C, argument,

there

non-zero

T(B) = t o ;

t6T+;

r < s < t in T;

if

~O.

6 be

T(b)

~Y(b)

s = OT

=t~gs

there

foregoing

k times, are

~ g s i. i6I

n 6~{O}and

rigid,

dl,d2

let

if O < e < 6

of

We

a < b inB,

if

~ ®

< 6.

6 =~.

t~s

=

the

6-divisible.

satisfying

~ + I <max(e,~) Case

C,

b 6B~{O}

IT=~®t since

which

<~®T(b)

e I <6. by

by

is n o t

and

assume

Moreover,

for

Choose,

si

factors,

device

almost

may

each

T(b) 3.4.Case

= T(d2)


~

hence,

C ~c

Theorem

not

for

6-divisible.

proved

Since

~9~ i6I

rigid

and

splitting

proof

T(b) = t o . SO,

=

s _<2 k e s < T ( c ) .

to 6 T

rigid

i)

without

el,...,e2k n e

the

®~(c i6I

T(dl)

above

Hence

for

no

that

the

Now,

a cBA

c 6 C~{O}.

elements

s.

=

n-divisible

_
Applying

say

is

almost

visible: dl,d2

C

~(c)

but

and that

with

a < c,

~(a) = r ,

a < b < c and

~(c) = t ,

n 6 e\{O},

then

s =s'

proved

[13]

as

a much

in

longer

then

Y(b) = s

2.34

proof.

under The

much

unique

s £ T

119

satisfying

n

e s = t will

be

denoted

by

~ ®

t.

Denote

by

Q the

set

of

n

rational

numbers

q

satisfying

0 Nq N 1.

For

q = ~ 6 Q and

t E T,

let

n

q ®t

1

=k

e ([ ® t ) .

Then (q+q')

® t = (q ~ t) ~

q ® t ~q'

®t

in

(q' ~ t)

Tiff

q

~ q'

if

q+q

in

Q

'EQ

etc. 1

(I)

For

0 T < t 6 T,

let

X = { b l , . . . , b n]

T ( b i) = t where

would

So

1 n+l For

for

get

® I

T

T, By

k 6 {0,I,...,n-I] that It

Define

that

(X h a s

Clearly

~

injective. s , t 6 {q ®

=

to

be

and,

0 T <~_m__ ® n+l

subset

finite

by .

that

by

d

there

that

of

B

and

we

IT < t:

satisfying

6 = w) . So

maximality, .

e t ~ I T = (n ® t) ~

is

q 6 Q 1 [ ~

let

maximal

(n ® t) (~ d = I

t % d.

If

could

conclude

~

d Q

is

t <

®

IT1 q 6 Q} and

~(s Q and

Q

I n+l

®

IT ,

t ~d.

U : B ~ [0,1 ] b y

and ~(q t) = ~(s)

dense

~=~

d < - [1 ®

t = s ~

®

IT;

we

k+l t ~ d" = -" n

IT,

prove ~

I T.

a contradiction.

by

by in

e I T) = q + ~(t)

if

for s ~

monotonicity T resp.

[O,1]

q 6 Q. t 6 T

By

(2),

- this

and

the

fact

for

arbitrary

~

holds

ST

[o,i] O < b

additive

in B, measure

O < p(b) on

B,

follows and

by

(I).

So

p is

a positive

for

s,t 6T.

oT:

B

is

that

T

For

d

Choose

k+l s <-~--

I T and

let

IT_
follows

are

® I T < t:

n E ~{0}.

_

d" = ~ i ®

order-preserving

Moreover,

where

s < q

k e I T ~ s. S o n k l e t ~ ~ I T ~ d' = s,

otherwise,

d'

such

IT
w.r.t,

sup{q 6Qlq

I TI q 6 Q}

Define

such

disjoint

<~ : T ~ [0,I] = {x 6 ]RI 0 _<x _< I} <0(t)

{q ®

n 6 ~

a maximal

(I),

I T < t:

follows

some

< t.

d 6 T.

k+l --6-- ®

is

$ bn)) .

.

(n+1)

s < t in

some

be

I ~ i Sn

d = T(-(b I $

we

(2)

for

there

finitely

120

(3)

T(b)

since We

~ was

show

that

T(C)

iff

~(b)

=

~(c)

one-one.

that

disjoint

=

~

is

o-additive:

elements

of

B

and

let

(an)n6 ~ be

a : ian.

~ ( a ) = s u p ( s o + ... + a n ) . n6~

For

Let

a

sequence

a n = ~(an) ; we

every

of

pairwise

have

to

prove

we

show

n,

s o + ... + a n = ~ ( a o + ... + a n ) ~ ~ ( a ) . Let

~ E [0,1 ] s u c h

z(a) By

S o.

If

not,

induction,

that we

may

choose

by

bo S c

such

that

T(bo)

instead

of

c,o,

~o,ao

~i

Now

We

p(a)

have

let

S ~(c)

thus

o-additive says

that

family

of

Since

B

b

=

B

is,

up

hence B

Corollary.

a power

of

the

Proof.

to

such

get

that

satisfying

T ( a o) S T ( c ) .

~ ( b o) = ~ o "

~ T (b n) n6~

B

is

So

Using

that

~ ( c ) : ~.

p ( b n) = ~ n :

there

c -- b o ,

b I ~ c °- b o

:

is

some

~-~o,

such

~i,

that

al

p ( b I) :

T (b) S T (C)

Theorem

carrying

theorem

Theorem

proves

C

to

which

be

all C,

we

atoms know

cBA.

of

strictly

[14] of

weakly

again

Theorem

~x

in

homogeneous

for

positive these

a finite measure

or

countable

algebras.

homogeneous,

contradicts

algebras

B

is

(*)

in

3.3.

atom

of

R

d)

and

even This

C.

r { R = R ( B ) , r > O.

~{x 6 BIB

a

a product

different

a homogeneous

of

cBA

structure

T + = {I T} and

a

isomorphism,

in

Let

supremum By

c 6 B

bn S c

~ o + ~i S ~ w e

totally

a = is

and

~ T (a n ) = n{~

The

pairwise

Case

get

n C ~;

Then

that

measure.

homogeneous,

is

:

proved

assumed

concludes

ibn. n6(0

we

every

Choose

disjoint

= m(ao) and

for

= o.

is

3.6.

o 6 Q.

z ( a o) = ~ o ~ o = ~ ( c ) ,

T (a)

and

assume

pairwise

first,

etc.

s o + ... + a n S o

Then

r

is

an

a)

through

iff

Moreover

is

homogeneous}

R.

that

conditions

e)

B [r

121

through

g) r

is

by

Theorem

4.

Some

4.1

of an

C.

Then

space

and

atom

The

Let

the

X I is

3.2

of

are

R

second

applications

Theorem.

set.

3.1

X be

algebra

equivalent. ~r)J

iff

IR(B

iff

B ~r

is w e a k l y

iff

B ~r

is

assertion

of

Theorem

an

arbitrary

If

~

: 2

(by

homogeneous

a power

follows

1.3)

of

a homogeneous

cBA

immediately.

C

C = R O ( X I)

homogeneous.

Now,

topological of

is

regular

space

open

a cardinal

and

subsets

such

that

I an of

infinite

the

IUI ~

product

for

set

U of

pairwise

disjoint

open

subsets

of

X,

then

[Vl ~ 2 e f o r

set

V

pairwise

disjoint

open

subsets

of

XI

(so,

if

every

every +

then

of

c(RO(XI))

Proof.

We

regular c ~v.

X i =X

non-void

such

,

prove

that

subsets

of

C

X I.

is w e a k l y Choose

homogeneous.

non-void

basic

Let

u,v

be

open subsets

non-void be_u,

E.g.

where

are

~ ~

~ (2~)+).

first

open

c(RO(X))

for

open

b =

~ Xi x ~ ui i£I o i61 o

C :

~ Xj J~Jo

i 6 I,

subsets

regular

open.

that

I o N 1 1 =~

u.,ik = V j k

for

i ~k

I o and of

Let

x

X

~Vj JEJ o Jo

for

d =

finite

i E Io,

subsets

j 6 Jo"

I o = {il,...,in],

: J o N J1,

_<m,

are

We

for

I Sk

_
and

e =

~ j~(JoUJl

Xj × ~ v j )

j6J o

× ~vj, j6J 1

ui,v j are that

Choose

Jl = {J{ ..... j~}.

Now,

~ Xi × ~ u i × ~ u i, i~ (IoUI I) i61 o i61 I

I and

assume

Jo = {Jl,-'-,Jm}"

Ii : {i~ ..... i ~ } ,

vj~ = Uik

may

of

ui,v j Ii,J 1 ~I Let

122

are

regular

ecc~v.

By

open

subsets

of

C,

C

C ~d

-~ R O ( e )

-~ C 5 e .

is

homogeneous

it

is

(cf.

and

homeomorphic

b is

c(RO(XI)) theorem

product

of

D.

Proof. and

4.3.

by

the

c(C) A ~ 2 . an

ccc

4.4. be I<<

this

if B

d c_bcu,

there

show

that

where

have

open The

Stone

of

4.1

of and

B.

v c_X n a n d

in

cardinal.

is

C

is

of

so

the

a free

Then

C

is

a cardinal

2~

isomorphic

that

b

n 6 ~;

of

e is

ICl _< ( I - I B i )

C

fact

If

each

[2].

infinite

C.

is

for

assertion

3.13

into

C

subset

some

completion

Then

the

open

the

_< (2~) + a n d

to

R O ( X I)

completely

gene-

B.

the

example

B

satisfying

a cBA

an

that

~ u) = c(C)

second

C be

show

basic

see

embeddable

space

by

let

i is

c(C)

c(C

a non-void some

To

completely

in

ccc

embedded

the

other

hand,

cBA

satisfying

[I]

showing (the

into

Martin's ccc,

the

that,

if V = L

countable

chain

a homogeneous

axiom

and

algebra

cBA

condition) C,

e I < 2 ~ hold

C = R O ( X ~)

is

and

satisfies

]CI ~ (w. IBI) W = IBi.

< be

K-closed

an

if, is

dense

<-closed

It

proved

is

on

and

completely

, we

is

contains

a cBA

B,

with

is

infinite

Let

is +

I copies

If,

and

of

follows

there

that,

is

that

cBA.

Erd@s-Rado-theorem;

B be

the

theorem

assumed,

B

X be

Compare

such

B

c(B) _<~

Let

the

rated

and

u

to

5u) _ < c ( R O ( X I ) ) .

the

Let

a homogeneous

to v × X I for

I copies

homogeneous that

But

by

of

sufficient

~b) < c ( C

follows

Theorem

such

1.9).


-~ RO(d)

a power

u 6C~{O}

4.2.

X I such

Hence,

Theorem

even

homeomorphic

infinite for

some

any

d £ D

subset, in

[12]

cardinal. desc@nding

A

subset

sequence

that

d ~d~

then

B

(l,~)-distributive

that,

if

B has

a dense

a cBA

(d~)~
such

is

for

D of

every

in

e < i.

<-closed

for

B

is

said

to

D where If

B has

every

subset

a

I < <.

D then

B

123

may

be

sed

subset.

We

completely

give

4.5.

Moreover,

an

analogue

Theorem.

K-closed

embedded

Let

subset

if of

B.

a rigid

< is r e g u l a r

this

B be

of

into

result

a cBA, Then

cBA

and

and

~ : <.iDI

be

has

a dense

then

<-clo-

c(C) ~ ( ~ ) +

4.2:

< a regular

B may

C which

cardinal

completely

and

D a dense

embedded

into

a homoge+

neous

cBA

some

~

Proof.

and

the

a dense

Z = ~.<,

We may

basis that

C with

we

assume

sets

<-closed

have

that,

if

c(B) ~ ~

for

I B 6 D.

B is

{d 6 D l d ~ e}

for

isomorphic

e 6 D.

Let

to

~ be

RO(D) an

where

infinite

D has

as

cardinal

a such

< ~ 4. L e t

be

endowed

is

a <-closed

dense

with

subset

the

of

C.

We

where

turn

B

the

question

a cBA

and

b,c 6 B and

b ~ c,

then

B ~c

x ~ c .- b. If B ~ cBA's

the

Clearly,

a rigid

presentation If B h a s

that

fb

C : RO(E) ; w e by

analoguous

about

with

order

the

a few

to

of

f to B 5 c

function

(Bi)i61 TAut i6I and

of A u t ( C n)

Bi

interchanges

(see

of

remarks. for

pairwise

a

that

4.1.

of A u t

Firstly, B ~b,

f(x) = x

B if

let

for

totally

different

[7]).

[7].

IBI ~ IAut BI:

b = b I $ b 2 and b I and

E is

proof

in

f 6 Aut

So E

f is o n e - o n e .

given then

that

cardinals

JAut(cn) I = ICJ

in

b y D I.

The

given

satisfying

f to

a family

[2].

that

2 ~ n < w,

factor,

satisfying

is

mapping

assume

in

possible trivial

induced

may

3.13

tAut B ~bl ~ bAut B ~cl:

the

cBA

no rigid

b2 6 B

begin

extension

~B i where i6I then Aut B ~

If C is

b I and

be

Let

completely

is

6 Aut

order.

<<}

partial

c(C) ~ (2~) + f o l l o w s

is

to

l{i~Jxi~IB}l

coordinate-wise

partial

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

such

E such

c(C) S (2~) +.

E : {x~D~[

4.6.

subset

for

follows

from

b 6 B,

choose,

T ( b I) = T(b2) . L e t

b 2 and

fb(x) = x

for

the

fb 6 A u t

x ~-b.

re-

by B

3.3,

124

The

function

mapping

b to fb is o n e - o n e

b = -[{x £ Blfb(y) =y An

arbitrary

c B A B has,

since

for e v e r y

y ~x}.

u p to i s o m o r p h i s m ,

the

form

~i ~j B = A x ~T H i x ~ E j x D i6I j6J = AxH×ExD where

A is a t o m i c ,

ly of p a i r w i s e is a f a m i l y ~j are

cardinals

position

such

of

E.

for

different totally

that

homogeneous

Theorem

(i.e.

totally

A ~P(~)

of p a i r w i s e

cBA without

4.7.

i.e.

E is r i g i d ) ;

atomless

0 < ~i,

or r i g i d

e is f i n i t e , hence

cardinal

different

L e t B be a cBA.

4.6,

some

~j for

a;

(Hi)i{ I is a f a m i -

homogeneous atomless i 6 I,

cBA's;

rigid

j 6 J;

(Ej) j6 J

cBA's;

v i and

D is an a t o m l e s s

factors.

Then Aut

B is f i n i t e i f ~

IH × D I = I and

IAut B I : e!.

~j

If A u t

: I

or

in t h e d e c o m -

IEjl : I for

B is i n f i n i t e

j 6 J

, [Aut BI ~ :

[Aut B I • Proof. (~)

We have Aut

B ~Aut

If e is f i n i t e

A × ~Aut(H i6I and B ~ P ( ~ )

morphism,

the group

and hence

A u t A, A u t ( H × D)

So e < ~.

vi) ~j i x ~ A u t ( E j ) × Aut(D) . j£J x E where

E is r i g i d ,

of all p e r m u t a t i o n s

If 2 ~ Zj a n d

and,

I < IEjl

of ~. C o n v e r s e l y ,

for e v e r y

for

some

A u t B is,

j 6 J, A u t ( E ~ j)

up

to iso-

suppose are

Aut B

finite.

j 6 J,

W ~ IEjl : IAut(E~)I ~ [Aut(E~J) I by

4.6 a n d

factors.

since

So,

E

if

3

is i n f i n i t e .

IH x D I > I, w e

~ 1H × D[ ~ A u t ( H

H x D is an a t o m l e s s

cBA without

get x D)

by 4 . 6 . Now

let A u t

B be

infinite.

By

(~)

it is s u f f i c i e n t

to p r o v e

rigid

125

IAut B1 e = IAut B I for for

some

i 6 I, B ~ = B

the which

A u t ( B e) ~ A u t

B.

3.6,

is a t o m l e s s .

R=R(B)

r 6 R~{O} O
gives

For

=

i.e.

and B = E

to an e m b e d d i n g B has

r 6 R\{O},

from R~{O}

× D. of

If B = H i ± (Aut B) e i n t o

no homogeneous

factor.

By

let

to t h e

= ~(r)

if ~(r'

for

class every

of

cardinals.

r' 6 R s u c h

Call

that

So X : {r 6 R ~ { O } I r

is a d e n s e less,

rise

B : Hi

IAut B ~ r I .

function

~-homogeneous

Sr.

B :A,

So let B = E × D,

~(r) is a m o n o t o n e

cases

subset

we may

of R. L e t

assume

that,

is ~ - h o m o g e n e o u s } P ~X

be a partition

if ~(p) = y

for

of R.

s o m e p 6 P,

Since

R is a t o m -

{p 6 P1~(p) = y}

is

infinite. Now Aut B ~

since

B ~p

and B ~p'

(~)

since

4.8.

IAut Bi

every

Let,

~ Aut(B~p) p6P

factor

are =

totally

~(p) p6P

~(p)

=

for p # p ' .

So

tAut Bi ~

Occurring

in t h e d e c o m p o s i t i o n

in

(~)

occurs

of B g i v e n

E~ : ~ { E ~ J I % = I } , It m a y b e p r o v e d ,

different

by considering

infinitely

often.

in 4.6,

E 2 : ~{E~JI~j

2}.

the

B : H × D and B = E t h a t

cases

B :A,

iB1 ~ iAut BI if iBi = IA ×H ×E2 ×DL and iAut BI ~ .

References [I]

B. and

Balcar,

P.

v

consistency

of S y m b o l i c

v

Stepanek: results

Logic

42

Embedding

theorems

an o r d i n a l

(1977),

64-75.

for B o o l e a n

definable

sets.

algebras

The Journal

126

[2]

[3]

W.W.

Comfort,

S. N e g r e p o n t i s :

Heidelberg-New

York

]974.

E. van Douwen,

J.D.

Monk,

algebras, ETH

preprint

G. Gr~tzer:

[5]

S. Grigorieff:

[9]

Model

J.D.

bras.

Colloquia

nite

and Finite

Rasiowa:

Theory.

questions

Berlin-

about B o o l e a n

f~r M a t h e m a t i k ,

completeness

Monk:

81

sheaves

1978.

extensions

in

447-490. of structures.

73-89.

societatis

groups Janos

of B o o l e a n

Bolyai,

10.

algeInfi-

951-988.

An A l g e b r a i c

Approach

1974.

M.

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

automorphism

for

(1975),

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

Amsterdam Rubin:

and g e n e r i c

101

(1973),

mathematica Sets,

Basel-Stuttgart

submodels

of M a t h e m a t i c s

Mathematicae

R. McKenzie,

H.

Lattice

Intermediate

Annals

A. MacIntyre: Fundamenta

[8]

Some

by the F o r s c h u n g s i n s t i t u t

General

set theory.

[7]

Rubin:

of Ultrafilters.

Z~rich.

[4]

[6]

M.

The T h e o r y

to N o n - C l a s s i c a l

of B o o l e a n

Logics.

algebras

groups,

to appear

in: A r c h i v

Boolean

algebras.

Berlin-Heidelberg-New

from

their

f~r m a t h e m a t i s c h e

Logik. [10]

R. Sikorski:

[11]

R. Solovay, Souslin's

[12]

S. Tennenbaum:

Problem.

P. Stepanek:

Annals

Cardinal

Iterated

Cohen

of M a t h e m a t i c s

collapsing

extensions 94

and o r d i n a l

A. Tarski:

[14]

D.A.

Cardinal

Vladimirov:

Algebras.

Boolesche

New York

Algebren.

definability,

]949.

Berlin

and

(1970), 201-245

preprint. [13]

York

1972.

1964.

PSEUDO REAL CLOSED FIELDS

*) Alexander Fakult~t

Prestel

fur M a t h e m a t i k

Universit~t D-7750

In h i s fields

Ax

famous

considered

ly i r r e d u c i b l e These

fields

by Frey The

The

above

version:

It has b e e n in

pseudo ordered

later

this

[Ba]

real

in

the p r o p e r t y

that

out

(prc) (L,<)

pseudo

, then

fields. if

(K,<)

in w h i c h

K

a

of f i n i t e every

[Jr]

absolute-

K-rational

algebraically

to be v e r y

K

has

by Jarden

extension

characterization

closed

over

of p a c - f i e l d s

L

for o r d e r e d

extension

with

studied

for e v e r y

closed

theory

called

turned

definition

elementary

defined

and e x t e n s i v e l y

algebraically

l ized

been

on the

K

variety

of p a c - f i e l d s

equivalent

[Ax]

fields

affine

have

[Fr]

class

paper

Konstanz

Konstanz

point.

closed

(pac)

and W h e e l e r

[Wh].

interesting.

c a n be p u t field

L

into

of

K

is e x i s t e n t i a l l y

the , if

closed

of p a c - f i e l d s

which

Basarab

an o r d e r e d

calls

is e x i s t e n t i a l l y

Basarab

closed

it is a l g e b r a i c a l l y

following K

is

in

L

genera-

field

(K,<)

in e v e r y

closed~)

*) T h e o r i g i n a l i n t e n t i o n of the a u t h o r h a s b e e n to p u b l i s h in this v o l u m e a s t r a i g h t - f o r w a r d p r o o f of t h e e l i m i n a t i o n of q u a n t i f i e r s for the e l e m e n t a r y t h e o r y of a s p e c i a l c l a s s of p r c - f i e l d s . D u r i n g his s t a y in B r a s i l in s u m m e r 1980 the a u t h o r o b t a i n e d v a r i o u s r e s u l t s on p r c - f i e l d s . T h e r e f o r e he d e c i d e d to g i v e h e r e a comp r e h e n s i v e i n t r o d u c t i o n to the t h e o r y of p r c - f i e l d s . +)

W e h a v e b e e n i n f o r m e d b y S. B a s r a b t h a t t h i s n o t i o n of p s e u d o r e a l c l o s e d f i e l d s h a s a l s o b e e n i n t r o d u c e d b y K . M c K e n n a in his t h e s i s , i n d e p e n t l y of [Ba].

128

One algebraic

which

answer (K,<)

in

. Hence

hand

most

no.

can

be

a prc-field

a problem):

_ a

K

ordering

<

over

closed

K

in

K

of

two

in

squares

in

One every if

field.

we

every

orderings

on

K

Basarab's

and is

irreducible plane

as

poly-

curve

(K,<)

algebraically

is

other

stated

field

Then

(Theorem of

L

of

over

this

sentence:

shows

. In

case

determined

ordering

(3.1))

that

prc

real

a

do

is

not

fix

admits

is

a sum

is

clear

<

this

L

of

K

be

(or one

if

K

is

is all closed.

a certain ordering

the

that

a formally

to which

one

is

a prc-field,

algebraically

from

case

paper

again

(prc)

exactly

This in

closed

K

this

Let

field

in which we

is

suitably:

extension

K

of

a prc-field

pseudo

that

definition.

fact

definable

number) we

that by

come the

an

existen-

sentence: 0

case

the

function

existential

of

in

here

The

< ).

the

closed

On

been

irreducible

(K,<) . T h i s

K'

K

point

tial

in

call

crucial

uniquely

to

has

a prc-

_ a = 0

definition

extend,

to

absolutely

results

K

back

absolutely

an

this

a

Unfortunately, (1.1))

2 orderings.

The

the

also

orderings

of

least

ordered

too,

algebraically

positive.

x 2 +y2

closed

The

is

[B]

by

existentially of

at

~

I).

Theorem

(in

denoted

main

the

Then

(compare

that

a prc-field

Lemma

ordered

extends

extension

extends

is

every

.

the

algebraic

one

real

of

K

such

K

(L,<) , h e n c e

is w h e t h e r

([Ax] ,§ 14,

admits

Obviously,

( B xy)

holds

be

ask

(K,<)

arguments

defines

(again

(L,<).

(/-~)

to

a prc-field

pac-fields

uniquely

a 6

X 2 + y2

curve

of

standard

= K is

Let

questions

constructed

nomial . The

for

By

K'

natural

(K',~)

holds

is

field K

the

extension

property the

of

Extension K'

< a

Theorem

turns

out

iff above

to

be

( B xy) also

a = x

holds

a pac-field.

2

for Thus

+ y

2

non-real we

may

fields drop

the

K'

. In

this

condition

129

of

K

being

are

special

formally cases

real.

Now pac-fields

of p r c - f i e l d s

which

(of c h a r a c t e r i s t i c

seems

to b e n a t u r a l

zero)

in t h i s

approach.

The Extension we

investigate

we deal with rings.

We

having

only

several

also

show

every

real

closure

condition'

of

induce

of

K

closures

that

class

2

(Theorem

has

paper we make

and results.

Keisler

or t h e

We

(2.1))

sufficient on

(1.4)

have that

has

In S e c t i o n

exactly

of K - r a t i o n a l

to g u a r a n t e e K

and t h a t

and

(1.6)).

K

point

points.

that

all

if

in This

orderings

is d e n s e

In s e c t i o n

in all

4 we prove

is e l e m e n t a r y . free use the

thesis

of

standard

reader

[v.d.D]

model

to the b o o k of v a n

den

theoretic

[Ch-K]

of C h a n g -

Dries.

O. P r e l i m i n a r i e s

In t h i s algebraic in t h i s

results

paper

For we deal

. Two

a field

L/K

we give

which

will

only with

extension L

section

will

have

K

linearly

disjoint

integral

domain.

frequently

we denote

fields

of

by

is

K

if t h e

By T h e o r e m

used.

K

its

of

zero,

a common

tensor

2, Ch.

collect

All

some

fields

algebraic

(relatively)

LI, L2

and

considered

zero.

characteristic

if . K

fields over

be

characteristic

regular

extension

some definitions

call

algebraically subfield

product

III of

closure.

we may

LI

K

@ K L2

[La] w e h a v e

2

K

if a n d o n l y

a simple

I

n orde-

a field

is a p r c - f i e l d

an i n f i n i t y

refer

excellent

which

curve which

topologies

3 . In S e c t i o n

of p r c - f i e l d s .

of p r c - f i e l d s

plane

of p r c - f i e l d s

in S e c t i o n

properties

(Propositions

notions

be p r o v e d

of o r d e r i n g s

is a l r e a d y

different

its r e a l

In t h i s

number

irreducible

'curve

the

theory

in S e c t i o n

a finite

abso~ly

will

algebraic

the m o d e l

every

K

Theorem

Since

a field closed

are

in

called

is an

130

(O.1)

L/K

is

a regular

field

extension

if

and

only

if

L @K ~

is

a

field.

From

Theorem

(0.2)

Let

LI/K

are

We

4, Ch. III

linearly

L2

a subset

V

of

of

some

over

K

we

absolutely

is

Of

an

By

By

over

~m

I c

called

absolutely in

and

only

we

a

V

(see

of

K

is

of

a

iff

b - a 6 P

uniquely

denote

the

determined

well-known;

for

(0.4)

Let

V

F(V)

its

ordering

up

an

is

its

irreducible.

In

by

70).

defined

real to

closure

LI is

the

get

and

a regular

set

of

function

field

that

is

I

K[X] this

L2

is

still

case

Hence

we

also

over

K

. From

K

is

say

that

the

give

field

p

extends

are

(K,P)-rational

point

over to

positive, a

P

defined

of

K in

with ~

with

and

K

such

{O}.

if

that

Actually,

respect

. The of

to

following

P

P

. It

is

fact

is

over

K ,

it.

variety

, and

F(V) if

irreducible

by:

a proof

K

of

P = K~

irreducible

function

o_ff) f1' .... fs

absolutely

extension.

P U-

isomorphism we

is

a subset

, and

absolutely

o_ff K

V

a regular

an o r d e r i n g

convenience

be

V

in

Then

, P D - P = ~ cone

we

if

it h a p p e n s

[La] , p.

we mean

positive

L2)

we

get

F(V)/K

P

Q u o t ( L I @K

I

variety

K-variety.

if

, P. Pep

for

[Ja]

Then

. By

generated

absolutely

irreducible

[La]

be

ideal

If

in

arbitrary.

K[XI,...,Xm]

the

is

I, p . 2 0 3

a K-variety

i.e.

V

be and

prime,

V

(K,P)

K

= Quot(K[~]/I).

an o r d e r i n g

is t h e

L2/K

F(V)

Let

P + pep

Corollary

and

some

ideal

definition

discussion

(0.3)

prime

and

mean

then

a field V

disjoint

of

zeros

[La]

regular

extension

call

prime,

be

in

defined

fl .... 'fs 6 K[~]

such only

that if

f1(a),...,fs(a)

(the V

. Then

residue

has

cla~es

a simple

positive.

~1

I31

Proof:

First

F(V). Then

Let the

classes

assume

that

the p r i m e following

of

there

ideal

I

of

statement

X 1 ..... X m

exists

I

where

d

F(V)

K

. Note

an e x i s t e n t i a l the m o d e l

completeness

Conversely, f. (~) 3 F(V)

is p o s i t i v e . of

then

over

A/M

A/M.

is

Since

power the

V

series

ordering

ordering

Q

The

this

local

closed

If

(K,P) its

(K,P)

on

Quot(A)

transcendence

residue

and

of o r d e r e d

fields

of M

denotes

such

(K,P) .

V

such

that

function

its m a x i m a l +M

that

methods

which

field ideal,

are p o s i t i v e

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

R

by

in

A

By w e l l - k n o w n on

Hence

by

holds

of

in the

fl + M ' ' ' ' ' f s

to an o r d e r i n g = F(V)

point

of

expressed

fields.

it also

a

completion

(K,P).

degree

can be a c t u a l l y

be a s i m p l e

is r e g u l a r .

over

the

language

A

is r e g u l a r ,

of

the

]

statement

ring

to

R

to

) = m - d

, i.e.

~ £(K,P)

(K,P)

ring

V

of r e a l

'equal'

A

of

in the

let

P

3

that

sentence

of

gl .... 'gr 6 K[XI ....Xm]

(F(V) ,Q) by

~ = O Arank(

is the d i m e n s i o n

over

in

by

Q

:

[i (3 ~)

an e x t e n s i o n

be g e n e r a t e d

is s a t i s f i e d

modulo

r fj (~) £ Q A A gi(2) j:1 i=I

V

such

fl,...,fs

to some

one

then

in

extends

induces

6 Q

an

. q.e.d.

Note over

t h a t we

only

(K,P).

The

next

essentially

van

den

proof

is m o d e l

(0.5)

Let PI'

Consider N

Dries'

and

P

on.

the

K

set

the

fact

will

Lemma we g i v e

L2/K

of o r d e r i n g s

ordering

Proof:

assertion

theoretic,

LI/K P2

used

(2.5),

Ch.

remains

II in one

extensions. resp.,

is a c o m m o n

irreducible

in S e c t i o n

an a l g e b r a i c

LI, L2

, there

V

be u s e f u l

be r e g u l a r on

that

2.

[v.d.D for Then

inducing extension

It is

]. S i n c e

his

convenience. to e v e r y the Q

pair

same to

L I ®K L2"

132

where we

let

we have

Q' + Q ' c Q',

x 6 R

R = L I @K L2

. We have

that

" This

Q' • Q ' c Q',

to p r o v e

only

only

hold

certain

K

we may

PI

and ~

is a s i m p l e (~,~)

point

of

V2

we may

, and

assertion.

, i.e.

x 2 6 Q'

for

It s u f f i c e s

of

P

varieties and P

point

of

with

qi(~)

to

, by

F(V)

and

them

all

to s h o w

L2

V2

there pi(~)

I < i < N

such

implies

o v e r K.

~,~

6(K,P) and Now

V = VI x V2

that

of

functions.

are

. Thus

a

finitely

fields

I < i < N).

variety

= LI ®K L2

are

positive

(for

only

defined

are r e g u l a r

positive

= 0

involves

as f u n c t i o n and

with

all

E(p i ® qi)x~

LI

(0.4),

of the p r o d u c t for

sum

VI

qi

VI

this

that

consider

extend

point

Since

assume

Pi

is p o s i t i v e

Hence

by

(0.4)

all

xi : O

such there

Pi ® qi for

all

.

to an o r d e r i n g extends

PI

We XK

arguments Q

of the

as w e l l

now

as

let

= {P 6 X K becomes

K is c a l l e d

la 6 P}

a totally

[M-H],

is c a l l e d

is s u r j e c t i v e ,

P2

we

can extend

domain

ordering

of K}

where

a 6 K ~ {O}.

or

real

if e a c h

L I ®K L2

Obviously,

Q'

Q

q.e.d.

generated

real

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

"

topology,

(2.8),

totally i.e.

integral

disconnected

Lemma

formally

(cf.[Pr])

X K : {PiP

, the Harrison

(compare

L/K

vanish.

1

that

both

is a s i m p l e

By standard

XK

assume

is a s i m p l e

are p o s i t i v e .

x.

irreducible

P2

is an e x t e n s i o n

H(a)

~

R

= 0

. Hence we may

(Pi ® qi ) (~)

1
all

elements

absolutely

such that

that

of

over

Moreover, Since

in c a s e

number

generated

on

third

=

on

N

finite

=

is a p r e o r d e r i n g

Q' D -Q'

the

X (Pi © q i ) x i 2 i=I can

set

Ch.

b y the

compact III or

just

real

if t h e P 6 XK

There

(clopen)

Endowed space,

with the

[PJ, T h e o r e m if

XK # @

restriction extends

is a t o p o l o g y

map

sets

this

order

topology space

(6.5)).

of K

A field

. A field

extension

from

to

to s o m e

XL

ordering

on

XK L

.

133

I. P s e u d o

In

this

vestigate are

section

algebraic

supposed

existentially

K

the

the

holds

is c a l l e d

closed

polynomials

is

L

of with

(1.7))

for

also

~

here

fields.

real

closed

real

means

with

. The

from is

in

P 6 XK

K

parameters

XK

closed

we

in

L

K

extension

L

every

existential

(prc)

if

of

K.

sentence

finite even

# O]

sentence K

. As we this

in t h e

~

is

, i.e. will

pl,...,pr

see

already

a formula are

later

implies

language

in

that

augmented

K

by

predi-

.

observe zero)

that

by

[Wh],

is

also

Theorem

2.2,

a prc-field.

the

real

fields

every

real

every

In t h i s

closed

completeness

of

the

fields:

Let

~

real

of

L

tension

of

sentence

k

infields

regular

from

the m o d e l

fix

all

is

= 0 ^ q(~)

from

us

that

K

that

follows

L

Recall

and

pac-field

case

we

K

have

•

Among

in

of p r c - f i e l d s

zero.

pr(~)

holds

case

(of c h a r a c t e r i s t i c =

such

totally

coefficients in

each

First

X K

existence

pseudo

= O A...A

fields

existentially

cates

of

in e v e r y

closed

[Pl (~)

in

language

(Theorem

the

fields

type

(3xi ..... Xm)

which

closed

characteristic

Existentially of

prove

properties

to h a v e

A field

we

real

the ~

and

be

unique

with hence

some

. Consider

the

field the

ordering

parameters in

K k

closure

. To

with

K

to

from

K

which

together

following

theory

of

obtain with

class

of

field

L

further

ordered

of

is prc.

real

respect

. Then holds

an in

examples

a system

K

(Pi)i£i fields:

closed

to

some

ex-

existential L

of

This

also

holds

prc-fields of

orderings

let of

134

M : { (K,(Pi)i61) It is clear chains. tially then

that

M

Thus ~ere closed,

every

is c l o s e d

are

i.e.

for

hat

K

is prc.

of

Now

(L, (Qi)iCi)

tence

with

We thus

(1.1)

parameters

(linearly

of

M

this

K

which

language

let

M

K

L

(K, (Pi)iC I) augmented holds

[Ch] , Ch.

the

. Hence in

L

by

in

real

ordering

every

Pi

holds

I). reguto

existential

also

,

III,§

be a t o t a l l y

Qi°f

holds

are M - e x i s t e n -

which

(compare

an e x t e n s i Q n to

which

extends

from

ordered)

in

orderings

on

that

K

We w i l l (K, (Pi)i6 I)

If w e

field there

is

prc

see

later

pi }

senK

.

k

and e a c h

is a r e g u l a r

and e v e r y

in g e n e r a l In case

XK

after

need

(Pi)iEl

extension

ordering

(the r e m a r k

system

not

Pi

K

of

k

extends

to

K

Proposition coincide

is finite,

of

(1.6))

such .

that,

with

equality

will

hold

let

but

k = ~ of

does

~

in the

above

, we o b t a i n

not

construction

a prc-field

have

/~-

gives

an a l g e b r a i c

K

as an element.

and

p

be the

which Hence

will K

be u n i q u e -

is not

real

closed.

The n e x t We

in

(1.6)).

ordering

ly o r d e r e d

k

above, X K

extends

(Proposition

fields.

L.

proved

To e a c h

unique

parameters

belongs

from

of

in the

(K,(Pi)i6 I)

. Choose

THEOREM.

{PilPi

in

(even

To p r o v e

lar e x t e n s i o n obviously

K

(K,(Pi)i6 I)

sentence

holds

unions

pi }

(K', (P i' )iCI ) C M

Pi ) w i t h

also

Pi e x t e n d s

under

elements

if some

each

(K',(P i' )i6I ) claim

is r e g u l a r ,

existential

predicates

We

I K/k

could

theorem have

equally

well

taken

characterization

this

as d e f i n i t i o n .

of p r c -

135

(1.2)

THEOREM.

K

is

irreducible

variety

(K,P)-rational

Proof:

Let

K

be

some

gl,...,g r 6

given

by

(B~)

obviously is

absolutely

know

that

the

with

existential

pl,...,pr

holds

in

L

. Fix

F

the

function

prime

ideal

(0.3)

we

fact

know

that

rational

{f

that

point

for

shows

point

= 0

that

of

V

is

(a,b)

^...^

~

P on

pr(~)

also

ideal

regular

real

given

a simply

be

point.

generated

by

sentence

in

pr(~)

I f(x,y)

the

know

6 XK

. Hence, Thus

in

^ bg(~)

. Since (0.4)

in

V

we

K

extension.

Assume

the

. Let

F : K(x,y).

K-variety

that by

V

V F/K By has

we

Obviously

defined is

(0.4)

by

the

regular,

by

and

a simple

assumption,

particular

-

sentence

- I = O]

irreducible.

we

By

holds

also

: O } . Since

real

in

~

K

% 0 ]

. Then

, y 6 L of

(0.3).

regular

: 0 A yq(~)

~ K

L

by

over

by

holds

holds

V

V

Thus

: 0 ^ q(~)

= O

has

a K-rational

of

of

pr(~)

V.

absolutely

existential

F(V)

absolutely

totally each

to

those

6 K[X,Y]

which has

F(V)

is

~

over

every

: O]

a totally

= 0 A...A

field

is

Pl (~)

be

6 K[~]

some

L/K

K-rational

This

I =

L

if

K

XK

the

field

extends

: 0 ^...A

[Pl (2)

prime

F(V)/K

sentence

, q

P 6

Then

function

let

(B~y)

is

gr(~)

P 6 XK

[Pl (~)

the

= O A...A

irreducible,

each

(3x)

let

only

over

each

m]

the

and

defined

for

and

if

K[XI,...,X

in

Conversely, that

V

point

prc

[gl (~)

holds

a prc-field

there

the (K,P)is

a

have

I = 0

K q.e.d.

136

In curves. Z

CK

the

For

: For

following,

convenience

every

zero

f(x,y)

= 0

Z

is of

Theorem

(1.2).

C~

such

ProofS)

:

Consider

f(X,Y)

= abX2y 2 +

inspection

~[Y].

Clearly,

(x,y)

be

for

each

of

that

Let

, i.e.

Easy

the

irreducible

subset

. Note

i__~s c 6 K 6 P

this

criterion

'curve

f(X,Y) P 6

Z

only

to p l a n e

condition'

6 K[X,Y]

, there

which

are

x,y

has

a

6 K

with

.

PROPOSITION.

a,b

apply

introduce

(K,P)

a fixed

instead

(1.3)

in

will

we

absolutely

simple

Here

we

for

plane

f

as

simple

a K-rational

point

K

. For

P 6 XK

defined

zeros of

just

write

satisfies

CK

all

we

by

I = a X 2 ( b y 2 + I)

polynomial

f has

, we

a,b

have

CK by

6 K~

{O}

c 6 P

there

iff

= H(c)

curve

aX 2 + b Y 2 -

of

C KZ

each

N H(b)

Z = XK

prc-field

satisfy

~at

the

. If

every

K

H(a)

XK

in

in

this

X

(K,P) curve,

the

+

polynomial

(bY 2 - I)

proves for

all

i.e.

irreducibility P 6

f(x,y)

over

XK

. Now

let

= 0

. The

ele-

ment c

then is

satisfies

not

~)

assertion

of

the

= ab

- a2b2x2y 2

proposition.

Note

that

possible,

Using that

the

:= a b ( a x 2 + b y 2)

~K

c = 0 q.e.d.

the

language

of

order

implies

K

be

a SAP-field

to

T h i s p r o o f is d u e t o J. thank for many helpful

spaces,

Kr. A r a s o n , discussions

Proposition (see

[Pr],

(1.3)

tells

Proposition

whom the author would on this subject.

like

us

(6.6)).

to

137

(1.4)

PROPOSITION. for

Proof: the It

all

P 6 XK

Given

suffices

the

to

(K,P)

set

also

of

also

g(K)

for

all

x 6 K

fields

there

(ordering) where form

Q 6 XK the

junctive

fact part,

p(~)

holds we

in

find

=

in

such

all

Inspection c 6 Q

may

zero

of

(x,y)

of

f

some

(K,Q).

(O,/~ 2 - c g 2 ( O ) ) . But

for

no

has

[Ka]

K[X] zero

, Theorem such

that

Quantifiers

for

(K,P)

I

for

the

if

g

formula

(K,P)

by


real

.

-6 2

closed

only

positive

the cone

a zero

that

P

g2(x)

in

in d i s j u n c t i v e

implies

(K,P),

. Hence

O

has

in

which has /k in ( K , P ) , t h e n

I)

involving

only

this

qr(~)

in this

proves zero

62

some

(K,Q) , normal

dis-

c 6-

(K,Q).

c 6 Q c

Hence

implies

Proposition

(1.3)

:

6 Q

we

. Now

.

know

that

consider

g

the

has

a zero

polynomial

6 K[X,Y]

f

(K,Q) Q

,

. Iterating

qr(~)

that s 6

H

Q 6 XK

I < P _

6

~ [X]

each

that

that

If

6

6 Q ^...^

+ cg2(X)

K[X]

is

in

g

K

and

in

A...^

such

assume

over

f

if

a zero

ql (~)

= y2

, there

in

induced

g 6

quantifiers

Taking

6 ~

that

Q £ XK

f(X,Y)

dense

contained

monic

g 6 P of

P ' q 1 ' .... q r

iff

We

(see

is

uniformity

a predicate

(K,Q)

has

where

c 6 K

(K,Q).

the

is

without

in

g

0

there

and

A ql (~)

0

(K~P)

for

is

say

c 6 Q

Hence

K

(K,P)

irreducible A in (K,P) . If

arbitrary.

that

that

to

Elimination

g

holds is

respect

Thus

a formula

which

prove

approach

by

of

to

a zero

not.

~

. Then

every

not

. Now

is

coefficients

with

has

does

have

that

does

g2(K)

we

K

show

C. Z

satisfy

K

.

P £ XK

completion

a zero

Let

, we by

is of have C KZ

absolutely g

irreducible.

. Thus

the there

(a,e)

simple is

is

If

a simple

zero

a K-rational

point

138

g

2

(x)

-I • ( 2

=c

_y2)

<

2 P

a contradiction.

(1.5)

q.e.d.

PROPOSITION. is

Proof:

a sum

Let

Hence

of

two

a E ~P P6Z

f(X,Y)

Obviously

Let

= y2

are

squares

in

. Consider

is

6 K

K

. Then

every

element

of ~ P P6Z

.

the

absolutely

irreducible

polynomial

K[X,Y]

a simple

x,y

Z CK

satisfy.

+ X2 _ a 6

(/~a ,0)

there

K

zero

such

of

that

f y2

in

(K,P)

+ x2 :

a

for

each

P 6

Z

. q.e.d.

(1.6.)

PROPOSITION: induce

Let

different

K

Z

..satisfy

topologies

CK

. Then . If

the

on

K

Z

the

same

topology

depends

only

orderings

is

finite

P 6 XK we

get

Z = XK

Proof: (K,P I) By

Assume =

PI,P2

(K,P 2)

Proposition

completion

6 XK

since (1.4)

of

the K

(K,~).

induce

completion

is

dense

Therefore

it

in

(K,PI) . H e n c e

is

real

closed.

on on

K

. Then

the

(K,P I) Let

P

topology. is

also

be

the

the

i

unique

ordering

of

Now

Z

let

Since

P

and

the

logies

on

K

, by

Theorem to a

I

(4.1)), for

is

a sum

all of

(K,P I)

be

. Then

finite

elements the

get

of

Q

. But

square

a 6 K

Z

and

there

Theorem

by

= K N P : P2

induce

which

then

PI

assume

Approximation is

two

and

Q

there 6 Z

we

is

is

pairwise for

be

P 6 XK ~ different

V-Topologies

P-close

Proposition

cannot

some

to

-I

(1.5),

P-close

"

to

Z

.

topo-

(see[Pr-Z], and

a 6 ~-~Q Q6z

O-close . Thus

-1 q.e.d.

Remark: that

If w e

there

is

drop no

finiteness

a 6 K

such

of that

Z

, by H(a)

the

proof

n z : ~

of

. Since

(].6) for

we

obtained

SAP-fields

139

the

sets

tells the

H(a)

us

that

best

we

z = {plp Xk

CKZ

can

a base

get:

Let

satisfies

CK

XK

to

Xk

Xk

implies

with

is

(1.1)

and

Z

topology in

the

is

Choose

the

and

its

is

on

we

this

subset

(Pi)i6i

get

is

where

proper

K

map

the density

of

such

a field

restriction

compact,

, this

k = ~(X)

system

the

XF

. Actually,

a dense

proof

. Since

XK

XK

example

of k}

Z = {Pili6I}

Z % XK

that which

from

of

z

in

.

Let

XK

be

a prc-field

if

and

closed

consider

(9.9)).

is c o n t i n u o u s

THEOREM.

Harrison of

ordering

. By Theorem Z

the

density

us

[Pr] , T h e o r e m

z = {Pili6I}

of

implies

non-archimedean

(compare

(1.7)

form

in e v e r y

finite,

say

only

( K , P I , . . . , P n)

if

extension

X K = {PI,...,Pn}

. Then

K

is e x i s t e n t i a l l y

(L,QI, .... Q n ) s u c h

that

L/K

i_ss

regular.

Note language fixed

of

'existentially

fields

augmented

closed' by

is u n d e r s t o o d

predicates

here

Z1,...,~n

in

for

the

the

n

orderings.

Proof: of

that

Assume

that

(K,PI , . . . , P n)

proof

there

such

is p r c that

that

and

L/K

is an e x t e n s i o n

(L,Q I ..... Q n ) s u c h Consider

K

( L , Q I , . . . , Q n)

is r e g u l a r .

(K',PI' , .... P~)

K'/L

an e x i s t e n t i a l

let

is r e g u l a r

sentence

~

the

an e x t e n s i o n

By Theorem

(1.1)

and

its

of

and

of

be

K'

satisfies

C ~,Ir,~

type

r

(3~)

where

P'qi

holds and

[p(x)

in

: 0 A

6 K[~].

A i=1

qi(~)

6 P

'

Assume

for

all

] ui

~

(K',PI' , .... P n ) . W e

qi(~)

6 ~

holds

choose 1 < i < r

in

(L,Q1,...,Qn).

some

~ E K'

Then

such

. By P r o p o s i t i o n

that

(1.6),

it

also

p(~) P] , . . . .

= O 'Pn

1

induce Theorem

different

topologies

(see p r o o f

of

(1.6))

on

K

for

. Hence each

by

the

I < i < r

Approximation

we

find

-

has

the

same

sign

with

respect

to

P. J

as q i ( 5 )

c

6 K

which

I < j < n

. Thus

1

for

all

140

ciqi(a)

is

a sum o f

existential from

two

sentence

0

squares

in

(in

language

the

K' b y P r o p o s i t i o n of

(1.5).

fields

with

Now t h e

parameters

K) -~

r

(3~) Lp(x) : o ^

-)

A

<3uiv i) ciqi(x) : u 2 +

v 2]

i=1 holds

in

holds

in

K' . S i n c e K

. But

J

K'/K

then

is r e g u l a r

~

clearly

and

holds

K

in

is a p r c - f i e l d ,

9 also

(K,P I ,... ,Pn ) q.e.d.

2. T h e

elementary

In t h i s orderings.

section

Since

characteristic

The

n

theory

we

may

zero)

of p r c - f i e l d s

consider be

0

with

only prc-fields

too,

the

case

n orderings

having

exactly

of p a c - f i e l d s

n

(of

is i n c l u d e d .

structures

under

consideration

now

are of

the

type

(K, P 1 , . . . , p n where

K

called

an

is a f i e l d n-ordered

First expressed

by

of n - o r d e r e d

observe

fields.

slightly

(C) :

every

6 XK

.Such a structure

that

the

curve

set of e l e m e n t a r y

What we will

stronger

actually

'curve

sentences use

there

are

x i , Y i 6 (K,P i) w i t h

(xi,Y i)

# 0

for

are

a,b

be

6

K

with

can b e

in the

in t h i s

language

section

is the

condition'

that

all

{PI ..... mn } CK

condition

irreducible

there

will

field.

absolutely

such ~~y f

P]'''''Pn

a recursive

following

For

and

I < i < n f(a,b)

f(X,Y)

£ K[X,Y]

f ( x i , Y i) = 0

, and e v e r y = 0

, monic

and

h(X) h(a)

Y

and

6 K[X]

% 0

in

.

~

{0}

,

141

The

class

of n - o r d e r e d

mentary

class

express

(C) b y a f o r m u l a

predicates

since

fields

for

HI,...

since

fixed

H '

f ( x i , Y i) = O

The fields

free

next

free

which

theory

class.

of n - o r d e r e d

formula

formula

The

c a n be

and

of

of

f

(C)

and

fields,

is an e l e -

h

we may

augmented

the n o r d e r i n g s .

that

by

This

(2.1)

THEOREM.

Let

set of

shows

of t h i s

is c l e a r

be

(C)

in

K

by

(K,P i) w i t h in

(K,P i) b y

.

that

and

class

We denote

formulas

satisfies

xi,Y i 6

I < i < n

is p r c

(K, P I , . . . , P n )

(K, P I , . . . , P n )

are

expressed

can be expressed

each

K

theory

already

"there

for

prc-fields.

by the recursive

the

class

of n - o r d e r e d

X n = {PI,...,Pn} will

be

called

it b y P R C n

expressing

if and o n l y

if

the

It is a x i o m a -

the

an n - o r d e r e d

is

curve

field. K

condition(C).

Then

is a

prc-field

XK = {P1,...,Pn}

Proof:

First

f

h

and

for

in p a r t i c u l a r

such

tized

and

stand

condition

degree

language

over ~"

theorem

( K , P I , . . . , P n)

an e l e m e n t a r y

in the

~f ~-~ (xi,Y i) # O "

and

some quantifier

(total)

the

n

"f is i r r e d u c i b l e

some quantifier

satisfying

let

K

satisfy

be a p r c - f i e l d

the

assumptions

with

of

(C)

XK = {P1,...,Pn} Then

. Assume

the e x i s t e n t i a l

sentence (Hxy)

holds

in

K,

[f(x,y)

because

Quot(K[X,Y]/(f)) a totally

real

Next, Ci x I ' ' ' ' ' X n } Now

let

certain

L

= O ^ h(x)

# O]

it o b v i o u s l y

of the v a r i e t y regular

assume

(C) h o l d s

for

function

by

over

f

( K , P I , . . . , P n)

from Proposition

be a totally

existential

defined

in the

K

field , and

L = L/K

is

extension.

that

, we know

holds

real

sentence

regular ~ (in the

(1.6)

that

extension language

of

. Since

(C)

implies

XK = {PI,...,Pn} K

. Assume

of fields)

with

that

a

142

parameters

from

subfield L

of

itself

(as

L

is

K~

which and

Xl,...,Xm,X,y

of

L

(i)

Xl,...,Xm,X

(ii)

y

Y

K

over

are

in

in

K

K

to

be

it

is

holds

. Hence

. We

are

extension

we

(K ~,

PI

in

may

going

to

some assume

embed

that L

'''" ,Pn ~ ) of

IKl+-saturated.

possible

such

Then

~

will

find

independent

K ( x I ..... X m , X )

that

to

generators

that

algebraically

, such

already

over

over

K

over

~

.

regular,

algebraic

monic

generated

assume

hence

is

. Then

elementary

we

L/K

L

generated

some

Since

is

in

, finitely

into

( K , P I , . . . , P n) in

holds

finitely

a field)

hold

K

and

f(x,y)

= O

[Sch],

Theorem

over there

und

f

K is

is

f6 K ( X l , . . , x ~ [ X , Y ] ,

irreducible

over

K ( x I ..... x m) (For

a proof

of

now

like

embed

apply

to

the

the

(K~,Pi~) I

we

(C) in

K(~) to

have

let

now

going

Qi' : Qi

to

then

is

to

p(~)

There

is

no

algebraically

^

finally

a simple

be

to

embed

realize

% O

embed

QI,...,Qn

N K(~)

a formula

to

care

~

(K~,PI~

of

non-trivial

every

the

qj(~)

zero

the

the ,

°

.

.

L in

K~

real

of

subset

in

order

closure

too.

into

is

then

irreducible

. However,

each

would

and

PI,...,Pn

diagram

finite

V) . W e

absolutely

of

,Pn ~)

Ch.

into

orderings

extensions

. Consider

there

r /k j=1

of

the

(K(~),QI' , .... Q n , )

(K(~),QI', .... Q n ) . S i n c e have

of

take

l

only

image

on

3D,

= K ( x I .... ,x m)

the

order

assumptions

Hence

where

see

first

f(X,Y)

to verify

are

fact

condition

polynomial

We

this

to

L resp.

( K ~ , P I ~ , . .. ,Pn ~ ) the iKI +

of

structure --

saturated

this

diagram,

we which

type

E vj

equality

independent.

to

Separating

take by

care the

of,

since

index

of

Xl,...,x n Z

and

are

143

replacing

p(~)

@ O

by

m --- ~ ( l ) ^ . . . A

where

each

6 ]1 1

p2(~)

we may

assume

that

m (n)

~(i)

is

of

the

type

(i) (Z) 6 ~ ^ .A p ~ i ) ( Z ) Pl i "" .

6

i

1

Since

( 3 ~ ) ~ (i)

nation

of

Quantifiers

satisfies

(C)

( 3 ~ ) ~ (i)

also

in

(K ~ , P i ~)

(i) for

holds

, by

real

Proposition

. Thus

in for

is n o n - e m p t y .

intersection

(L,Q i)

for

holds

V-Topologies

in

By

holds

closed

fields.

(1.4),

K~

each

I < i < n

[Pr-Z], sets

in Since

~(i) the

(1.6)

Theorem

by

Elimi-

(K~,PI~,...,Pn~) in

(K~,Pi~) . Now

defines

P.~-z o p e n

and

an o p e n set

set

defined

the Approximation

(4.1))we

is n o n - e m p t y

(K~,Pi~)

is d e n s e

(K~,Pi~) , since

Proposition

(see

of t h e s e

it a l s o

too.

may

Theorem

conclude

Hence

~

by

that

the

is r e a l i z a b l e

in

(K~,PI*,...,Pn~)

Now

we

can

( K ~ ' P I ~ .... 'Pn ~) absolutely validity

identify . Finally

irreducible

!

(K(~),QI' , .... Q n )

with

we will

in

apply

polynomial

(C)

f(X,Y).

some

substructure

( K ~ , P I ,. . . ,Pn ~)

This

is p o s s i b l e

of

to t h e

since

the

of

(3xy)

now

carries

for

all

[f(x,y)

over

~f = 0 A ~

from

(x,y)

(L,Q i)

al,...,a N 6 K ~

we

to

find

% O]

(K~,Pi~)

for

a,b

such

£ K~

each

I < i < n

that

f(a,b)

. By

(C),

= O

and

N

h(a)

=

H (a-aj) ~ 0 j=1 therefore assume that

K(x,a,b)

is

isomorphic

. Using a to

again

IKI +-

saturatedness

is t r a n s c e n d e n t a l L

over

K(~)

of

K~

. But

we may

then

. q.e.d.

It s h o u l d the of

above some

proof

be mentioned is

absolutely

the

that

following:

irreducible

the

original

consider

variety

V

L

geometric as

. Find

the

idea

function

a generic

behind field

curve

on

V

144

and

spezialize

spezialized of

K

that

should

Lemma

a decidable

set.

(2.2)

LEMMA. PRC

Let

holds

Proof:

there

is

regular,

and

that

By

(0.5)

tensions

argue

by

K'/K

)

product

we

know

L = Quot(K'

~K

K") is

orderings

of

PI ' '''" 'Pn'

every in

the

sentence

also

i

in

the

closure fact

orderings.

be

of

two

regular

holds

K".

Since

K'

~K

K"

is

PRC

models

field

n

of

which

and

K'

are

K

( K " , P I " , . . . . Pr~')

integral of

ex-

from

in

K'/K

an

extension

K"/K

domain and

K"

common

ex-

resp.

with

) trivially

holds

it

~K

<0

by

parameters

K'

PI '~,... 'P n "

and

(K' 'PI ' ''" " 'Pn

way

if

L

real

theorems

")

are

and only

on

the

substructure

with

aregular

that

each

the

existential

sentence

QI '''" 'Qn

existential

(I .7), <0 t h e n

that

respect

K"/K

is

proof

common

and

if

in

(K",PI,...,Pn

n-ordered

existential

tensor

find

for

that

(0.2)

holds

Theorem

the

we

Now which

an

an

above

to

the

so

point

the

K * had

and

(K',P 1 '''''Pn

in

in

(K',PI',...,Pn')

Then

Consider

are

into

in d o i n g

a simple

that

(K,P I , .... P n ) s u c h tensions.

have

rejected

K(x)

will

difficulty

imply

. Assume

n

is

of

next

The

still

difficulty

embedding

The form

suitably.

curve

. This the

it

parameters

holds

(K",P I

in

'''''Pn

from

K

(L,Q I ,... ,Qn ) . B y ) " Similarly

we

back. q.e.d.

(2.3)

THEOREM.

The

subset

of

PRC

consisting

of

boolean

combinations

n

of

Proof:

existential

Follow

[v.d.D] and

the

essentially

Replace fact

effectively

sentences

that given

OD n it

is

is

the by

proof

PRC n

decidable

algebraic

decidable.

number

of

, and whether

Theorem use

(2.1) , Ch.

Lemma f

6 ~[X]

(2.2), has

II

in

Theorem a zero

(1.1), in

an

field. q.e.d.

145

N o w we which

consider

do not h a v e

an e l e m e n t a r y

among

any

class

the m o d e l s

algebraic

since

( K I , P I , . . . , P n)

extension

we may

of PRC n such

(L,QI,...,Qn)

express

. This

the m a x i m a l i t y

of

ones

again

is

(K,PI,...,P ~

by (A) : E v e r y

monic

each

I < i< n

By the use sees

that

tences,

(A)

The

see

(compare

[v.d.D

are

real

closed

Proof:

OD n

ordered

fields.

companion

Hence

OF

zero

PRC

and

Ch.

closed

= OD

(C) w i l l coincides

n

weaker

fields

one

be d e n o t e d with

sen-

by PRC n-

v a n den

Dries'

of OD n

II). closed

fields

and PRC I-

n model

field n)

companion to s h o w

of OF n, that

(k,Pl,...,pn) in the

(1.1)

is a m o d e l

Then

there

in w h i c h

PRC

the

theory is also

n

of

n-

a model

there

are M - e x i s t e n t i a l l y

class

6 M

we o b t a i n

of PRC

n

is a f i n i t e for

(A) d o e s

(Bx) in

(K,P I ..... Pn ) }

not

extension

irreducible

hold

of

( K , P I , . . . , P n)

algebraic

sentence

it a l s o m u s t

that

. Assume

a certain

the e x i s t e n t i a l

( L , Q I , . . . , Q n)

contradiction.

fo__~r

n

(C) and h e n c e

f 6 K[X]

(K,P i)

axiomatization

algebraically

it s u f f i c e s

of T h e o r e m

( K , P I , . . . , P n)

in

set of e l e m e n t a r y

{(K,P I ..... Pn ) I (k,p] ..... pn ) s u b s t r u c t u r e

(K,PI,...,Pn).

Since

a

fields.

an n - o r d e r e d

in the p r o o f

nomial

(A)

are

extensions(K,P1,...,P

M =

fies

n

by

(1.2),

is the u n i q u e

of

Given

of

], T h e o r e m

PRC

a recursive

a seemingly

PRCo-models

THEOREM.

has

for real

by

theorem

provides

that

(2.4)

As

next

Note models

closed

axiomatized

in the

This

which

of Q u a n t i f i e r s

can be e x p r e s s e d

theory

OD n

f 6 K[X]

is linear.

of E l i m i n a t i o n

As we w i l l theory

irreducible

hold

in

( L , Q I , . . . , Q n)

non-linear

f(x)

satis-

poly-

= 0 holds.

(K,P I , .... Pn ) , a

146

It

remains

(K,P1,...,pn) a regular

c

show

of

sentence

holds

in

that

PRC

(L,QI,...,Qn)

extension

existential also

to

K

with

be

. Now

is m o d e l

n two

models

Theorem

parameters

(K,PI,...,Pn)

. This

complete. of

(1.7)

from

proves

PRC n

. By

implies

K

which

model

Let (A) ,

that holds

L

is

every in

completeness

(L,Q I ..... ~) of

PRC n q.e.d.

In by

[v.d.D

] van

den

Dries

definitions

of

OD n

new

predicates

he

adds

for

the

of

pseudo-finite

showes

theory that

Kiefe's

Quantifiers.we constants

Let

augmented

W

each

for

be

I

there

Elimination

natural

then

fields.

In

do that

not

adding

to

just

sufficient.

elementary

theory

in

c I .... ,c n

, axiomatized

by

(A),

each

I < i<

n

THEOREM.

Proof:

By

PRC

the

K(i)

common

(K(i)

of

PRC

existential R

n

for

holds

in

of

sentence

. If

then

K (2) .

by

(2.16) obtain

n

new

the

and

Kiefe van

., Xm)

[K]

den

Dries of

individual

language

of

(m+1)-ary 6 9.

1

i

admits

~=~(3Y)Xo +xlY

Elimination

criterion

for

situation:

of

n-ordered

predicates ^

/~

c.

6 K.

j$i

±

]

i = K

1,2

(1)

(in t h e ~

holds

in

: O ]

Quantifiers.

elimination

we

have

Let (i) Cn '

. Furthermore

and

+. " "+ x m y m

quantifier

K

(2)

augmented K (I)

t

and

let

language) we

have

(i) (Wm

let

to

(R,...)

be

with show

) )m>1

R : ~

n

The

Elimination

(C) , -c.

_(i) ,p~i) c#i) PI . . . . . . . . . .

'

substructure

from

n

following

=

be m o d e l s

[ Wm(Xo''"

a well-known

consider

OD

, and

(VXo ..... Xm)

(2.5)

extension

used

Example

suffice

an

of Q u a n t i f i e r s .

those

predicates

constants

is

-

for

to

less

show

the

by

m >

m

now

Kiefe's

PRC n

fields,

are

that

admits

predicates

will

makes

which

shows

be

a simple parameters

that

it

also

a

147

Passing field.

Since

f 6 R[X]

K (2)

to

closed c! I) 3

W (1) m

the

algebraic

isomorphic.

and

R sign

Lemma

just

of

that

(By)

To

that

that we

R

is

Hence

the

in

a square

K

K (2)" " . It in

K

(I)

is and

isomorphism

this: we

the

just

of

Then

(I)

a

algebraically

product

negative.

is

polynomials

R

an

R

in

of

such

coincide. be

same

a zero

assure

a square

is

have

can

c is

, the

that

constants

write

those

ac

is

. Thus,

if

in K (2) . H e n c e

a

c. 2

.

c.'s 3

totally a

is

and

~(a)

and We

K(2)/R may

and

atomic

formula

not

an

atomic

assume

regular,

that

unnegated

those

in

are

atomic

formulas involving

~

~

we (Bx)~

formulas.

involving any

are

Wm,

some we

in

the

where

~

If

we

could

predicate

could

is

deduce

W

m

from

R (2)

formula

..... a m ( X ) )

it b y

ao(X)

eliminate

proceed

is

holds

all

+ a1(x)y

the

Wm (ao (x) ..... we

and

a

negated

~

eliminate

replace

(A)

(2.2).

existential

Wm(ao(X)

R

K(1)/R

by

To

to

fact,

R

assume

closures

assume

assume

may

behaviour.

equivalently

(2,2)

. In

, o(ac)

replace

Lemma

K (2)

by

since of

may

whenever

hence

same

can

a £ K (I) . L e t

negative

over

we

belong

some

conjunction

some

If w e

orderings,

c! 2) 3

Now,

the

also

forces

and

situation

on

this

K (I)

the

coincide

that

positive and

have

W (2) m

we

K (I)

in

algebraic

quotients

in

the

are

and

of

a zero

consider

which

field

have

be

respects

the

which

well-known

Now

to

as

+...+

negated

am(X)

follows:

am(X)y m

= 0

formulas

)

First

we

fix

some

x

6 K (1)

which

satisfies

4.

148

Now

-~

= a

Wm(ao(X)

(x)

+...+

a

0

(x)Y m

means

6 K(1) [Y]

that

does

the

not

polynomial

have

any

f(Y)

zero

in

=

K (I)

. We

m

split By

.... , a m ( X ) )

f

into

(A),

each

irreducible of

the

fi's

factors

over

must

without

be

K (1) , s a y

f

zerosin

= fl"'''"

fr

(K(1) ,P (I))

for

"~i some

~.

. By

Elimination

of

Quantifiers

for

real

closed

by some quantifier

free

fields,

this

1

can be

expressed

in

(K (1) , p (x 1) . ) )

formula.

21.

H e n c e we s e e existential

that

we may r e p l a c e

formula

fl,...,fr

such

~

expressing:

that

fi

Wm(ao(X), ....

"there

does

not

are

have

am(X))

(coefficients

a zero

in

by an of)

polynomials

( K ( 1 ) , P (I))~.

and

1

f = fl

"'''"

,i

fr

" q.e.d.

3.

In

this

prc-field real

section

is

again

hilbertian

which

are

prc

extension

(3.1)

but

need

V

a simple method

there real

exist

L

first

be

an

treat

W

each

defined P

Then,

by

the

image

under

6 XK

the

is

point

K

Remark

be

of

for

will which

K

in

an at

, W

L-rational

end

of

Then

a

of

field

k

k

such

section).

every

algebraic

too.

construct

extension, defined

over

. Using

an

absolutely

from

to a

V

[L:K] L

Well's

V

defined

K-rational .

= n

which

. has

descent

irreducible

(K,P)-rational

W

have of

say

6 XL

a simple

point

this

of

a countable

real

the

Q

will

is

arbitrary

variety

has

k

extensions

a finite

a

case

a prc-field.

each

extension

algebraic

a prc-field

a morphism on

that

For

K

algebraic

real

irreducible

over

an

many

case

we

assumption ~

is

§ 1.3)

, and

are

every

show

closed.

K

absolutely

[We],

also

Let

of

extensions

that

(see

THEOREM.

We

(see

prove

. We

(L,Q)~rational

variety for

not

not

extension

Let

prc

field

EXTENSION

Proof:

we

Algebraic

point over

point

L

whose

.

149

Let By

Pi(X)

of

L

the p r i m e = x(i)

over

the p r i m e over and

with

K

into

ideal

L (i)

ideal

of

V

be g e n e r a t e d

I< i < n K•

say

we d e n o t e

Pl = id.

an a b s o l u t e l y

. If we

g~i)

of

take

(i) 6 L ( i ) [x#i)

'

variety

V (I)

base

L

over

. Then

d e t ( s ~ i))

(i)

n z : j=1

(*)

XI

defines

K

a(i) 3

an i n v e r t i b l e

the p r i m e

ideal

an a b s o l u t e l y already

which

determined

over

+'''+

(n) al

+"

where

(X/Y)

that

f

s3

fs

I

each

of

P

assumption• Q's V

(i)

variety i

x...x

V

(i)

X# i) ,...X m ,_(I)

'''"

of the

V (n) . Let

defined

(i)

, say

g~1)

I

generate

gr

'''"

absolutely

Sl,...,an

be

for e a c h

I < 1 <m

. Call

substitution.

J

# O

a

and h e n c e

substitution

results

K

from

this

variety , take

fsj

W

. In o r d e r

as g e n e r a t o r s

which

satisfy

to see

for the

J

thus

that

W

J

defines is

the u n i q u e l y

linear

system

(I) f (I) (X/y) an sn = gs

.+ a

(n) n

the

f

sn

(i) (X/Y) gs

:

substitution

(*)

. Now

that

a simple

it is e a s i l y

seen

. 6 K[Y]

P 6 XK

embeddings Q

"

indicates

It r e m a i n s for

linear

polynomials

a 1) fsl

gl (i) '''" 'gr(i)

then

ideal

embeddings

YIj

irreducible

defined

different

'

the p r i m e

product

gl '''''gr 6 L[XI,...,X m]n

for e a c h

n

irreducible of

Then

. X (i) ] ''"

(n) (n) • "" 'gl '''''gr generate

the

irreducible

new variables

• ''''gr

by

This for

Pi to

some map

. Say has

means each

. Fix which

L V

to s h o w

there

I < i < r

has

P 6 XK L

into

pl,...,pr

a simple

that

W

and w r i t e K

are

K

for

correspond these

(L,Q)-rational is a s i m p l e

. The

(K,P)-rational

remaining

point

embeddings.

for

By

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

(x~ i) '''''x(i))6m

'non-real'

. The

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

'real' point

(K,P)

point

embeddings

in

150

Pr+1,...,pn conjugate simple

map with

(~i)

(Here

x

means point

in

K

Since

The

first

that

the type We

have

to

sequence Hence such

It

is

of

that

W

of of

is

is

preimage

. This

choose

an

and

now

clear

and

has

finishes

the the

are

arbitrary conjugate

to

Pl "

over

that

the

corres-

its

a K-rational under

which

the

is c o n j u g a t e

simple

there

its V

is

in p a i r s

"

coordinates

point

on

W

substitution proof

for

.

(*)

finite

.

for

an

arbitrary

proved

3~

in t h e

of

first

L

order

(4.1)

~ (~,~) that

elements

K(~)

on

a prc-field

point

Theorem

show

V (i)

K).

property

<0 -= V ~

for

over

(4.1),

of

we

conjugation

in t h e

proof

6 K(V-~T)

come

case

P3 '

coordinates

the

axioms

this

if

proof

of T h e o r e m

K

. In

They

V (j)

K

L/K

The

: K

for

an L - r a t i o n a l

show

to

K(/-I)

(y11,...,Ymn)

m

extensions

of

respect

~(i)) '''''m

ponding

form

into

(x I(i) ,. .. 'x m(i))

point

point

L

is ~ (u,v)

next

where these

of

L

by

holds

prc

language

these

is

axioms

. Then

fields. can

be

is

the

above

proof.

in

K(~).

Hence

it

we

also

by

be

Let

use

we will a set

seen

to be

free

a finite

Thus

(4.1)

As w i l l

chosen

<0hold in L.

makes

expressed

a quantifier

K(~)

L/K

In T h e o r e m

can be

of

axioms ~

extension

section.

being

all

prc

algebraic

of

from

the

formula.

~

be

a finite

extension

find

~

holds

in

of

in L

K

K(~) .

q.e.d.

The we

have

Theorem

only

seen

so f a r

(3.1),

( L , Q I , . . . , Q n)

algebraic

van

are den

which

PRC1-extensions

real Dries

are

closures proves

algebraic

of

the

of k

a given In

existence

extensions

of

(real)

[v.d.D] of

field , Ch.

II,

PRC n - models

a countable

k

.

151

hilbertian

field

topologies

on

Modifying theorem

k

THEOREM.

which

Proof:

We

a,b

Dries'

n I

PRC1-extensions

Let

about

6 R

to

construct

orderings

inducing

such

an

L

proof

we

will

of

k

still

which

hilbertian

we

different

is

obtain are

fields

real in

not

the

real

refer

closed. next

closed.

the

reader

field

and

to

K = ~ m ~

Km

that

(see

, we

take

construct

f(a,b)

= 0

Sl .... 'ar km

. We

and let will

(f,h)

= 0

number

of

(R,Q)

elements

a PRC1-subfield

such

(K,P)

of

(R,Q)

occurs

K m c . . .c

{0}

, and

(a,b)

Y

that

there

~

, such

% O

. The

enumeration

often.

We

are

going

of

k

. The

with

R

. Thus field

its

If

either

h

Km+ I D K m % 0 r H j=1

or If

such . To

f

both that

take

Irr(ai,k)

ensure

Km

are

looking

(unique)

constructed

Km+ I = K m

every

we

already

then

in

infinitely

we

g =

h 6 R[X]

chain

[Ro])

h(a)

that

monic ~f ~

and

pair

Km

together

(f,h)

hilbertian

finite

irreducible,

every

K Ic . . . c

that

every exists

pairs

f(a,b)

a countable

Assume

Km

there

all

extensions

pair

k

To

real

Sl,...,ar

with

as w e l l

a countable

k

absolutely

k = K° c

field

be

of

6 R ~

such

finite

k

closure

is

chosen

in

den

enumerate

is

will

van

omits

f £ R[X,Y]

in

n =

information

Sl,...,ar

m-th

for

algebraic

a real

of

admits

.

(3.2)

are

which

. But

slightly

Concerning

[Ro]

k

that

ordering

Km do

has

will

P = K

have

we its

n Q

are

also

of

and no

zero

in

Km

6 Km+ I

the

elements

in

.

the

coefficients

a,b

assume

be

consider

coefficients

there

6 k[Z]

a hilbertian for

. Then

not

have

care

g

is

g Km+ I

has

, we

with

no too.

zero

152

Now and is

the

let

By

section and of

(Km,K m

n Km

S Q)

f 6 Km[X,Y]

Theorem

such

that

Km

is

we

algebraically Let

U

enlarging

proof

and

Function

set

U

its

6 Km[X]

Implicit

a Q-open

(R,Q).

h

if

for

and

that

an

closed

From

there

Km•

F = Km(t) [Y]/(f ~Y))

an

we

may

assume

f(t,Y)

such has

has

Ch.

on

that

there

a zero the

f

in

inter-

II

in [ v . d . D

extension

(Km~ ,P*)

that

a zero

following

know

that

(1.16),

elementary

and

the

we

, f(a,Y)

6 Km* ~ K m

. From

(R,Q)

assumption

6 U

Theorem

is

the

a

t

element in

to

each

necessary,

non-empty.

know

applied

. By

K m (t) in

diagram

]

is

(Km~ , P*) . we

conclude

K*F m

K* m

F

K m (t)

1

K m

(i)

f(t,Y)

is

(ii)

f(t,Y)

has

(iii)

g(Z)

has

(i) and

no

is

follows

from

~m

implied

zero

follows

K ~F m

F/K m

a zero

Km*/Km(t)

(iii)

and

irreducible in

O

K * [Y]/(f(t,Y)). m

the

regular. the

h(t) ~

(Km~ , P~)

in

from

K • , and m

facts (ii)

facts

that

t

follows

is

from

transzendental the

choice

over

of

t

K

m

.

that

--~ K * [ Y ] / ( f ( t , Y ) ) m

is by

over

algebraically the

closed

regularity

of

in

K ~F/F m

Km *F

. This

(cf. (0.2))

last and

assertion

the

is

regularity

of

(cf. ( O . 3 ) ) .

It formula

is

clear

~

in

the

how

to

express

language

of

(i) , (ii) , a n d ordered

fields.

(iii) Hence

by we

an know

elementary that

the

153

sentence means Km

H t ~

that we

, h(a)

z e r o in

can

, and

Km(b)

~

(Km~,P~)

a 6 K

f(a,Y)

and t h u s

such

m

that

has

a zero

Km[Y]/(f(a,Y))

. Now

finally

PRC1-model

in

find

# 0

If w e is a

holds

also

in

f(a,Y) b 6 R

let

This

is i r r e d u c i b l e

such

that

g

over

has

no

K m + I = Km(b)

take

K = ~ K m , it b e c o m e s m6~ that al,...,a r ( K

such

(Km,K m D Q).

clear

that

(K,K D Q)

q.e.d.

Remark:

The

assumption

be henselian

with

prc-extension K

must

respect

.

be real

(4.1)

THEOREM.

Proof:

We will

m > 0

we have

(Hm) : F o r

~~y f

Y

there

fixed

total

and

(1.4)

of p r c - f i e l d s

we prove

K/k

and §8

Let

a real

in

[Pr],

k

algebraic

the

field

is e l e m e n t a r y

just one

of p r c - f i e l d s

that

K

absolutely that

% 0

are

The

class

class

show

theorem.

c a n be a x i o m a t i z e d

irreducible

there

£ K

degree

problem and

are

the

in t h e

f(~,b)

if for

f 6 K[Xl,...,Xm+I,Y]

, and = O

collection

is to e l i m i n a t e

arbitrary

if and o n l y

~ , y 6 (K,P)

P 6 XK

with

to e x p r e s s

only

is a p r c - f i e l d

for e a c h

~,b

It r e m a i n s of f i e l d s .

is e s s e n t i a l :

of f i e l d s .

, such

(~,y)

to s o m e v a l u a t i o n ,

section

The

every

in

to be h i l b e r t i a n

closed.

final

language

k

Then by Proposition

4. T h e

In t h i s

on

with

every and

f(x,y)

, monic = 0

(Hm)

#

O

.

in the

for p o l y n o m i a l s

coefficients

~

and

h 6 K [ X I ,- . . , X m + 1 ] ~ { O } , h(~)

of all

every

the

clause

language f

of

154

"for

each

P 6 XK ~f ~-~

and

By

Elimination

"there this

are

..."

formula

express

in

the

to

qi(~)

field

P6

generate fact

XK

This

is

K every

2 squares

by

Proposition

(1.5).

see

of

that

the

: O

P it

of

this

take

this

to

6 P

"

coefficients. such

by

that

equals as

that

So we all

cannot

a formula,

fact

elementarily

take

suffices

qr(~)

K

3 squares

replace

. If w e

-q] (~) ..... - q r ( ~ )

express

We

on

can

type

integral

ordering

sum

we

involving

we

with

that

formulate

f(x,y)

fields

formula

clause

no

. To

a prc-field

can

with

= 0 v ql (~) £ P v . . . v

in

we

closed

form,

some

that

Now

real

free

means

of

6 (K,P)

"

polynomials there

a preordering

~,y

normal

: p(~)

that

negative.

prc-fields.

O

for

language

are

formalize are

•

a quantifier

p,ql,...,qr

have

(~,y)

conjunctive

each

are

Quantifiers

by

in

"for

where

of

there

we

a sum

the

of

a first

there

use

are

axiom

for

sums

s.

1

N

of

squares,

range

over

not all

all

zero,

possible

such

that

r+1-fold

E i=I

b.s. i 1

products

of

= O

, where

elements

the

b. 1

of

{ I ,-ql (~) ..... - q r (~) } " It prc.

As

remains in

generated, a

that

totally

plane

Assume

that

and

(3x,y)

Q

(2.1)

real

regular

extension

6 XL

here,

the

I

[f(x,y)

f

that ~f

= 0 ^ ~

K

satisfying

have

to

K • of

K

algebraically

absolutely

P~ n K = Q Q K

(x,y)

% O

]

we

do

monic

in

irreducible. . Then

is

into not

easier.

independent

and

(H m)

a finitely

. Since

is m u c h

irreducible

is

embed

all

L = K(Xl,...,Xm+1,y)

situation

are be

(o.3), such

we

extension

Xl,...,Xm+

. By

~

field

Theorem

curves

= 0

every

of

f 6 K[XI,...,Xm+I,Y]

f(x,y)

P~ 6 X K ~

that

elementary

to

let

prove

proof

IKl+-saturated

restrict

and

the

to

the

over Y

K

such

Let validity

of

,

155

carries for

over

real

and

closed

h(~)

* 0

K*

is

h(~)

* 0

since and

from

L

fields.

to

(K~,P-~) b y

By

(Hm) , t h e r e

. In g e n e r a l ,

~,b

IKL+-saturated, we for

algebraically to

(L,Q)

all

the E l i m i n a t i o n are

depend even

over

K ,

6 K*

on the

find

h 6 K[~] ~ {0}

independent

~,b

with

choice

f(a,b)

of

h.

~ , b 6 K* w i t h

. Hence

and

of Q u a n t i f i e r s

thus

However,

f(~,b)

a I ..... am+ I K(a,b)

= 0

is

= O

are

isomorphic

. q.e.d.

It s h o u l d not

necessary

But

it m a k e s

be mentioned

to p r o v e life

easier

interesting

in its

to r e s t r i c t

to p l a n e

being

the

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

axiomatizability

in w r i t i n g

own right. curves,

down

of the

such

to h y p e r s u r f a c e s class

axioms,

We do not know whether as w e d i d

in T h e o r e m

is

of p r c - f i e l d s .

and

it is

it is p o s s i b l e

(2.1)

in c a s e

of

X

K

finite.

In a f o r t h c o m i n g elementary In f a c t , ordered

subclass

we will fields.

jectured

by van

Last but many

that

inspiring

of

show The

paper we the

that

its

existence

den Dries

not

class

least,

discussions.

(see

show

there

of p r c - f i e l d s

theory of

that

such

[v.d.D]

we would

is a c e r t a i n

which

is the m o d e l a model , p.

like

is m o d e l

companion

companion

complete. of p r e -

has b e e n

con-

Roquette

for

73).

to t h a n k

Peter

REFERENCES

[Ax]

J. Ax: The e l e m e n t a r y theory of finite Ann. of Math. 88(1968), 239-271

[Ba]

S.A. Basarab: D e f i n i t e functions on algebraic v a r i e t i e s ordered fields. To appear in Revue R o u m a i n e Math. pures appliqu6es.

[ Ch]

G. Cherlin: Model t h e o r e t i c Lecture Notes in Math. 521,

[ Ch-K]

C ± Chang-H.

[v.d.D]

L. van den Dries:

[mr]

G. Frey: P s e u d o a l g e b r a i c a l l y closed fields real valuations. J. Alg. 26(1973), 202-207

[Ja]

N. Jacobson: L e c t u r e s in abstract van Nostrand, New Haven, 1964

[Jr]

M. Jarden: E l e m e n t a r y statements over large Trans. Amer. Math. Soc. 164(1972), 67-91

[Ea]

I. Kaplansky: P o l y n o m i a l s in t o p o l o g i c a l Bull. Amer. Math. Soc. 54(1948), 909-916

[Ki]

C. Kiefe: Sets d e f i n a b l e over finite fields: their functions. T r a n s . A m e r . M a t h . Soc. 223(1976), 45-59

[La]

S. Lang: I n t r o d u c t i o n to algebraic Interscience, New York, 1958

[M-H]

J. Milnor-D. Husemoller: S y m m e t r i c b i l i n e a r Springer, B e r l i n - H e i d e l b e r g - N e w York, 1973

[Pr]

A. Prestel: L e c t u r e s Matem&ti~ca 22, IMPA,

[Pr-Z]

A. Prestel-M. Ziegler: of t o p o l o g i c a l fields. 318-341

[Ro]

P . Roquette: N o n s t a n d a r d aspects of H i l b e r t ' s i r r e d u c i b i l i t y theorem, in: Model theory and algebra, 231-275, L e c t u r e Notes in Math. 498, Springer, 1975

[Sch]

W. Schmidt: E q u a t i o n s over finite fields. An e l e m e n t a r y approach. Lecture Notes in Math. 536, Springer, 1976

[we]

A. Weil: studies,

[Wh]

W. Wheeler: M o d e l - c o m p l e t e theories of p s e u d o - a l g e b r a i c a l l y closed fields. Ann. Math. Logic 17(1979) , 205-226

Keisler:

Model

Model

fields.

algebra; selected Springer, 1976 theory.

theory

topics.

North-Holland,

of fields.

algebra

over et

Amsterdam,1973

Thesis, with

Utrecht,

1978

non-archimedean

III.

algebraic

fields.

fields.

zeta-

geometry.

forms.

on formally real fields. Rio de Janeiro, 1975

Monografias

de

Model t h e o r e t i c m e t h o d s in the theory J. reine angew. Math. 299/300 (1978),

Adeles and algebraic Princeton, 1961

groups.

Institute

for advanced

Some

Remarks

on the M a t h e m a t i c a l

Incompleteness

of P e a n o ' s

Found

and H a r r i n g t o n

by P a r i s

T. v o n d e r

I.

paper

shall

is to s u p p l e m e n t

assume

2.12

give

a simplified

[I]

and

Paris

on the

For m o s t

these

plicitly.

occur,

Finally,

theoretic

central

rSle

strong

they

e.g.

[2],

of r e c u r s i v e

PA satisfied

by

of

L "

is a

. The

have

segments

[I] M ~

(c)~

of the m a i n

on n o n - f i n i t e than

~

of by

in PA.

ones

illustrate

exthe

playing

a

axiomatizability of PRA,

-

and

independence

to w e l l - k n o w n

[2].

2.6

ideas

provabilities

shall

and

(numbers

Z -indiscernibles o

other

[I]

of P ~

completions

of

of m o d e l s .

and N o t a t i o n

P A is f i r s t - o r d e r 0,1,+,.,<

models

the

of

famous

avoid

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

[I]

and

the

be r e d u c e d

of

nonexistence

Basic Concerts

will

a couple

initial

exposition

can c o m p l e t e l y

papers

relation

including

significance

in

lemmas

partition

elaborate

results, we

and H a r r i n g t o n ' s

combinatorial

and m o r e

and H a r r i n g t o n ,

Whenever

model

- 2.14)

[2].

Paris'

the p u r e l y

2.9,

Zo

Twer

Introduction

This We

Arithmetic

only

Peano's

Z -formulas o bounded

Zn+1-formulas, Hn-fOrmula.

arithmetic,

n £ ~ Dually,

its

(or b o u n d e d

quantifiers,

formulas,

e.g.

, are of the Hn+1-formulas

language

3 x < y,

shape are

the

L containing or

limited Vx

< y

3 Xl... 3 X m ~ VXl...

VXm~

formulas) . 9 ° is

, where with


158

Models

as w e l l

as t h e i r

is the n a t u r a l are

thought

underlying

model

of PA,

of as c o n t a i n i n g

sets

other

are d e n o t e d

models

~ . Th N

(M)

by M,MI,M'

of the o c c u r r i n g

is the Th(M)

D Hn

etc.

theories

" We

often

n write

M F ~(a,b,c,...)

M p ~(x,y,z,...) For

n 6 ~

notation

n

M

L-variables

does

a,b,c,..,

I ~ M

and

of

is an i n i t i a l

M

V a,b 6 M(a

an e n d - e x t e n s i o n means

that

I ~ M

I ~ ~(~) We have

of

iff I

and

M

if

nal

defining

for

only,

recursive

plus

truth-predicate

Xl,...,x m

of v a r i a b l e s 6 ~o'

substitution-functions, n ~ I

we

need

in for

of

M

beiff

substructure

in

I

Z -substructure n

arithmetic) recursive

induction

of t h e e x t e n d e d

recursive

~(Xl,...,Xm),~

of e l e m e n t s

M

I ~e M

, ~

M

x,y,z,..,

segment

by

s

I

is t h e n

. I~in

M

: of

M

.

.

any primitive

equations

parameters)

~

arith-

in g e n e r a l .

initial

be denoted

is a

for

a substructure.

~ -formulas n I

. Our

~e

(primitive

symbols

is a l s o

will

all

M

s

interval

a 6 I). A n

of

~ ~(a,b,c,...).

their meanings

I is an i n i t i a l

which

i.e.

I

E~ o PRA

for

symbols

of m o d e l s ,

the

instead

L

off by using

marked

denote

M

in

being

< b 6 I ~

M ~ ~(~),

between

n

and

these.

I , which

for

or r e l a t i o n s

will

segment

a,b,c,...6

(M,a,b,c...)

term

for e l e m e n t s

[a,b]

, including

and

or of

not distinguish functions

a,b 6 M

a and b

~

constant

, the d i f f e r e n c e and

For tween

is the n a t u r a l

definable

the m o d e l

formulas

[ x / a , y / b , z / c .... ]

widely

metically

for

tr

o

r~

~n-definabl~

function

axioms

for

language.

for

satisfying our

is a r i t h m e t i c and

shall

I -formulas o

suppressing

denoting

the

G~del

truth-predicates

P A ~ trn( ~I , X l , . . . , X m ) e~ ~ ( x I .... ,Xm)

for

all

their

formulas

(with

the p r i m i t i v e

a fixed

sequence

I .... ,Xm) ~

numeralnumber

tr n

additio-

for

use

with

P R A H tro( ~ , x

notation

axioms

the o p e n

We

with

of

and ~

. For

satisfying

~ ( X l , . . . , x m)

6 Zn

159

The Strong

Partition

(Paris-Harrington

For

a,b,c,d,e

partitions)

Relation

with

£ ~

[a,b]

the same

1.1

The Finite

This

use G~del's codes

with

large,

i.e.

which

1.2

leads

The proof 1.1,1.2

[a,b] ~

3b

sequences

for

P

[a,b]

X ~[a,b]

for

if

is

P(~)

6 ~

can be f o r m a l i z e d

PA U T h H I ( ~

Matiyasevic's

the p a r t i t i o n homogeneous one obtains

in the

¥n'

Bn~hierarchy

lenghts

to

also b o u n d e d

the

as set-

arithmetically

and hence be p r o v i d e d of 1.1

relation sets

X

is k n o w n

[a,b] ~

to be

(c)~

by

be r e l a t i v e l y

a relation

[a,b] ~ (c)~

d e Theorem,

H2-sentence,

leads

regard

principle

Ramsey

) by Paris

theorem

, as follows:

and their

that any b o u n d e d

theorem, V 2 ) is the best

whereas

as sets

[a,b] ~ (c)

as a

PA

arithmetically,

H2-formalization

Infinite

of

then build up the i n v o l v e d

combinatorial

the

language

can thus be encoded

< Card(X),

6 ~

in the

Note

Strenghtening

to the v a l i d

Matiyasevic's

set

of

d e

sets,

The r e s u l t i n g

min(X)

of

d-element-subsets

homogeneous

sequences

M ~ PA

of this uses

independent

(c)

to enCode

that the r e q u i r e d

V a,c,d,e

(called

reads:

as usual.

set in an

of PA.

for all functions

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

is called

increasing

notions

the c o n d i t i o n

there

the r e s p e c t i v e

a cardinal.

a theorem

of

Theorem

~-function

the c a r d i n a ~ o f

definable

[[a,b]] d

can be f o r m a l i z e d

of s t r i c t l y

theoretical

X

H b 6 ~

statement

means:

~ 6 [X] d

Ramsey

6 ~

(c) d e

<e

~ c .

for all

V a,c,d,e

~

P from the set

Card(X)

Principle

Formula).

into the set of numbers P

and the C o m b i n a t o r i a l

cf.

[I].

w h i c h was proved

and Harrington. attainable

H2

to be

(or, by

by their m e t h o d s ,

Vl-independences. quantifiers

Like

count.)

(Recall

that

160

It 1.3

V xV

which We

is

again

is

of

Lemma: is

c > d

, we

of be

ing

there

1.1) [a,b]

M ~ A

VX

Vy

steps

partitions lifting

the

Let

us

the

, g:~

~

of

,

[I])

([x,y]

~

(z+1) z z

2.7

- 2.9,

of

Z -Indiscernibles o

~

2.12

lemmas

2.6

a primitive

, so

that

If

M ~

[a,b]

~

which

are

for (c)~

all

in

-

[I].

2.9,

2.12

set

X

recursive

M ~ PA

, then

encodable

a homogeneous

~

- 2.14

single

, d ~ r

~

a d

in

for M

function.

and

all

(cf.

, encoded

are: one,

(i)

is

a,b,c

6 M

,

partitions

the

in

lifting e

which

M

remark , for

followP

,

ugly

X

explicitly

together

superscripts essentially, ~

relative

misprint

in

A

< y)

distinguishes

of

< Card(X)

completely

collecting

subscripts

an

^ g(min(X))

< y ~ g(x)

2.1

properties

mention

2.8

s

by

by

lemmas

in

Amin(X)

> min(X)

additional

and

by

1.2

that

M =

into

2.7

combinatorial

(X E X A y 6 X ^ X

the

Card(X)

is

> c

For

lemmas

denoted

of

that

Existence

~asy

[0,2)

Remark: main

the

r 6 ~

, such

Card(X)

using

expressed

have:

r ~

Xc

the

d E ~

P:[[a,b]]

(by

remark

formalization

(z+1) z z

by

on

the

, henceforth

(z+1) z zomax(x+1,7))

Let

There

replace

equivalent,

Results

can

in P A

~

obtained

meaning [~]

2.1

~

to

(z+1) z

~

, but

provably

Basic

The

1.3

[O,y]

P A i- ([O,y]

2.

[x,y]

equivalent use

V z B y

which

convenient

z 3 y

is

shall

1.4

is

[I]

to in

a fixed d

to

(ii)

in

finite

number

a common

using

from a fixed lemma

proved

~

the to

of

without

condition

obtain

primitive 2.14

one

[I],

the recursive

. There

it

is

g.

161

important

that

ly r e a l i z e d a,b,c

s' d o e s

to hold

may be kept

The main tion leads

[a,b]

to the

homogeneous

~

The

model

next

sharpened, in

[I],

2.2 w i l l

of 2.1

all

if

io < ii

M ~

Vx

Proof: of

<'''<

for

P(u,~,~)

these

X

M ~ Card(X)

in

iff

is the large

[a,b]

observaenough

which i.e.

segments

lemmas

are

sets

of

of

M

it < Jl

[a,b]

(M ~

: there

~

< s

satisfy

M ~ PA

the

,

is a s e q u e n c e

in the

for all

language

of

PA

:

~ ( x , e j I ,. " ' ' e j t )).

2.1

c a n be

t-tuples

is s m a l l e r

(~(x,~) ~ with set for

ei,

2.15

' then

denote

Vx
, so let

~

d

< e i + I , so t h a t

[O,c)

Jt

and

b e the p r i m i t i v e

for

Lemma

of

g

. Then we have

<'''<

(~,~

. Let

r,g

6

2.11

same.

. Let

r~

though

2.10,

= no (2n°s+1)

jl,...,jt

a homogeneous

~ c

simplified,

g(n)

with

~ ( x , Y l , . . . , y t) be

the

, e i 6 [a,b) , e i2

any c o m p o n e n t

= O

initial

of the

, c > d

the a s s u m P t i o n s ,

Zo-formulas So let

by

< e l•o ( ~ ( x , e i l , . . . , e i t )

that

of

but

is n e e d e d .

parameters,

a technically

remaining

given

io,il,...,it,

2t+1-tu~es

means

give

~ ( x , Y l , . . . , y t)

With

with

6 ~ , t > O , r = 2t+I

(eili 6 M, M ~ i < c)

and

of P R A

c,d,e

subsets

to e s t a b l i s h

(c) dd ' a , b , c 6 M

Zo-formulas

with

is i m m e d i a t e -

all p a r a m e t e r s

and H a r r i n g t o n

enough

of w h a t w e n e e d

s,t

function

M ~ [a,b] ~

, since

proof

again.

Let

conditions

this

little

M ~ PA

large

thus

. Now

only

Zo-formulas

i d e a of p r o o f

2.2 L e m m a : recursive

of

e

of P a r i s

in s o m e

and

version

the

finite;

the w o r k

(sets of)

PA

lemma

M ~ PA

(c)~

Z -indiscernibles, o which

arbitrary

standard

existence for

on

for

idea behind

that

not depend

i <~c

~

applied with

than

~(x,~))

u < ~ < ~

any of

according

, be the

~)

, ~ <

defined

by

for a l l

< s) , o t h e r w i s e P

to the p a r t i t i o n

first

P(u,~,~)

to 2.1. c

= I

We have

elements

of

X .

162

By

the

choice

of

2 e i < ei+ I

g

2t+1-tuples

in

faction

Z -formulas o

of

P(u,v,w)

{eili

= O

for

< c}

P

q0,z

(~)

=

sequences

have

(a _>2t+I)

has

a subset

~,w

6 A,

~

hence

for

# ~

, we

are

these

be

least

of

M

Notice , so

possible

The

~

of

the

the

M ~

results

is

~,~

~

2.1

Card(X) a

by

> a

to

. There the

> to ( 2 a ° s + 1 )

< ~

with

, and

result

such

. Now by

choice

of

prin-

(P%0,z(~) l
for ei

all can

.

familiar

with

the

case

of

I -indiscernibles o

the

=

of

deals

on

we

Dirichlet

this

only

.

g

for

the

sequence

,

[X~{a}] t

induction

homogeneity

[X] t

at m o s t

proof

PA

in

are

(Pcp,z(~) J~,z)

, the

. For

, ~

that

the

satis-

that

, so

by

are

on

prove

z < a

and

available

6 A

= O

M

there

~ = min(X)

, M ~

elements

or

the

. Set

in by

2 ~'s

< w

lenght

steps

, and

2t+I

suffice

, z 6 M

~0(z,~)

(which

. Since

that

for

basic

have

P(a,~,~) X

X

= a- ( 2 ~ ° s + I )

distinct

of

Remark:

of

conclude

in

< s

than

principle

at

it w i l l

r~

of m o r e

2t+1-tuples taken

with

: c > g(a)

A

there

s

(Pq0,z(~) I%0,z)

"pigeon-hole" ciple)

to

2t+1-tuples

truth-value

2 a's

c > d >_ r =

. Thus,

below

all

~ ( x , y I .... ,yt ) 6 ~o set

By

existence

with

bounded M =

~

definable will

subsets

remain

can

now

be

so

that:

established:

2.3

Theorem:

(i)

For

all

finite

If

M ~

PA

(eiJi

sets

, M ~

[a,b]

e. 2 l

<

(2)

for

all

ei+1

max(a,c)

of ~

I -formulas o

(c) n

, c > n

a < e 1• < b

< max(a,c)),

(I)

F

'

i+I

< max(a

¥x

, then

is

n 6 IN

there

is

a sequence

, satisfying

'

~0(x,y I . .• . . y t ) 6 M ~

there

c)

'

? , i°

<

ii

<'''<

it

< Jl

<" "'<

Jt <

< e l•o (%0(x,e i I , .... eit) (-~ ~ ( x , e j I ..... e j t ) ) •

163

(ii)

(Paris)

If

sequence

M b PA, M ~ (eili

6 IN)

[a,b]

c (c+1)c,

~

, e i £[a,b)

2 (I) e i

< el+ I, i 6 IN

(2) for

all

c > IN , then

there

is a

satisfying

~ ( x ' Y 1 ' ' ' ' ' Y t ) 6 TO'

i o < ii < ' ' ' <

it'io<

J 1 < ' ' ' < J t inl~:

MI= V x < eio(<0(x,eil,...,eit)~ ~x,ej1,...,ejt)). (3)

Moreover, we h a v e

Remark:

(ii)

proved

there.

need

any of

for I ~

(eili)

e

is s t a t e d Note

2.1

or,

sometimes,

Proof:

(i)

is i m m e d i a t e

(ii) the

Let

T(Y)

G~del

graph have

be

I b PA

[2], w i t h

up to 4.3

none

by

2.2,

modification,

of the r e s u l t s

models

M ~ Th(IN)

a formula of

of PA,

9)

for

all

but

but

not

to f o l l o w

will

only

IN

for

.

taking

into

describing

all the

it < Jl

component

of

i]

.

a slight

arbitrary

for

I = {d 6 MI B i 6 IN d < e

account

the

the p r i m i t i v e

final

remark

. By

<'''<

of

x

n E IN

Jt

in

(insert

(i) , u s i n g

recursive

set of

Z -formulas o

(~(x, (Z)il . . .(z)it~-~ . . . ~(x . . (Z)jl, . .

<'''<

the y ' t h

in

(I),(2),

2.2

numbers

M x <(Z)io io < ii

of

and

for

itself

in the p r o o f

M

that

- 2.3

as in

M ~ p(n),

~ , (X)y

the

lemma

(

z)jt ))

denoting

Zo-arithmetic

2.8

where

in (tr

[I]

~(x,y+1) ,

description

to lift

is the

o

,

of the

superscripts,

truth-predicate

we for

E -formulas) o p(x)

= Hz Vy

<x((¥(y)

The v a r i a b l e M

. Since

for

some

stence give

, ~

d 6 M ~ ~

the p r o o f

(i w i t h

might

M b PA

formula

be as in

z

~ tro(Y,Z))

p(d) of

(3)

even

^ (Z)y2 <(Z)y+1

be b o u n d e d

is not d e f i n a b l e M b p(d) , ((e)ili in

[I]

by

with

. Choose 6 ~ ) will

, which

some

do.

(z) Y < _b)

nonstandard

parameters

e 6 M

is not

^ _a ~

in

satisfying For

to be

element M

, so

the exi-

convenience, altered.

(3), d < e i , ~ a E o - f o r m u l a in the l a n g u a g e o subscript always denoting a natural number):

of

we

Let

of PA.

ei

, I

We h a v e

I ~e M, by

(I),

164

and I b

3 Xl V x 2 ... B x t

iff

3 i 1 > i 0 V i 2 > i 1

. ..

iff

~ ii > i o V i 2 > ii

"'" ~ i t > it-l:

iff

M ~

3 x]

< ei

~ ( d , x ] ,... ,x t)

+1 V x 2

~ i t > it_

< e i

o iff

I

b

3 x I < ei

for

any

effectively

+1 vx 2 < e i

e i , so t h a t (without The

induction known.

I b

without

instead

condition

(2).

The

a system

basic

axioms

guished main

(cf.

[I]

+t

~0(d,~)

formula

,

9(x)

containing

this

by

(2)

there

parameters

the

Strenghened

,

c a n be among

the

I

m

e

M

. Now

into

Finite

sequences

of

full is w e l l -

"smaller"

the h o m o g e n e i t y -

Ramsey_Theorem

Its P r o o f . P A I~ ~ in o u t l i n e the m o d e l s

plus

~ Con(T)

Second

subsequent

3.3

of the p r o o f

and

satisfy contain

as o c c u r the

and b y

P A ~ Con(T) Theorem

Instead,

(theorem

3.2

in

for

~ a

, and

PA

a distin-

technical

a lot of c a r r y i n g

theorem.

the

in 2.3, (ii) . The

Incompleteness

Thus

G~del's

as f o l l o w s :

of w h i c h

(for this

2 is n e e d e d ) ,

counting

runs

E -induction o

can be concluded.

as a c o r o l l a r y

and

by i n s e r t i n g

parameter,

and GSdel's

not

M

E -indiscernibles o

in p a r a g r a p h

in P A is r e q u i r e d , attack

of

is P A I- s

Con(PA) , PA ~ s

in

of the

set of

2.3, (ii),(3))

directly

~0(d,~)

i m p l i e s the same w i t h p a r a m e t e r s , a s

j u s t one

O,1,+,o,<

[1]

exposed

3 x t < e i

it is in

is i n t r o d u c e d

for

in

+t

o

since

[I] f o r

T

infinite

lemma

ground

of

proof

First

...

%' (x)

avoided

Applications

original

< ei o

arbitrary

parameters

Independence

and S o m e

3 xt

(9(x) ~ ~' (x)) . B y t h i s , i n d u c t i o n for I o r e d u c e s to Z o - i n d u c t i o n for I with parameters.

have

elements,

...~(d,~)

V x 2 < el2

V x < ei

is s a t i s f i e d ,

We might

3. T h e

and

< ell

M b 3 xl < eil v x2 < ei2"''
"'"

+2

a bounded

parameters)

latter

~ 9 x]

o

i° 6 ~

found

I

o

o

Thus,

+2

1 :

back~ Con(PA) saying

out proofs

we

[I]) w h i c h then we

can occurs shall

165

obtain

the

independence

3.3 r e l y i n g fied proof which

in

only for

[2]

application proofs the

PA

of

PA.

~ a

As

for

the

the

and

constants

ci,i

(I

the u s u a l

axioms

recursive

definitions

induction

axioms

for

for

all

could

By m e a n s

of 3.1

by

For

6 ~ : If

safely

~

f(a)

Jt ~

replace

condition

we

all

[a,b]

can

in

formal

, contrary

to

T

:

0,1,+,o,<

T includes

theory

and

the

(with p a r a m e t e r s )

in the

in the ~ , the

language

of P A

,

axioms:

~(x'cj1'''''c'3t))"

here

io < ii

state

finite

~

by

x

<'''<

b

a

, we have

= ~b([a,b]

~

in a h a n d y

S ~ T

(d+1) d

~

( ~ , e o , . . . , e k) b S

Since

avoiding

to an

(cf.

it'

the p r o o f

of

2.3,

io < Jl < ' ' ' <

Jt

by

(4).

3.2 L e m m a :

with

down

of them.

well-ordering

~ ( x , Y l , . . . , y t)

it < Jl < ' ' ' <

and the

in

[2]

axioms:

Eo-formulas

< Cio(~(x'ci1'''''cit)

the o n e

boiled

simpli-

+,',

all

Eo-formulas

io < ii < ' ' ' <

a,b

of

in

of

of PA,

(4

(3))

the

this

given

tr °

both

system

system

first-order

for

< ci+ I , i £ ~ ,

We

to g i v e

the

again

the p r o o f

that

is e a s i l y

entails

above-mentioned

for

notice

, thus

still

while

corollary,

to the one

and which

6 ~ . T contains

(3

Remark:

close

M b Th(~)

language

2 ci

(ii),

very

it m a y b e w o r t h

The

. Then we

second way

Definition:

Vx

M = ~

on 2.3,(ii]

we define

language

as an i m m e d i a t e

comes

3.1

(2

s

for

2.3,(ii)

one,

First

2.2

is b a s e d

within

first

on

of

there

, then

, ei

is are

recursive This

is n e e d e d

d £ ~

interpreting

a total

(a+1) a )

there

form what

such

of

that

2.2:

for

all

k 6 ~ , eo,- ..,e k 6 [a,~] ci

function

function

exceeds

f:~

~ ~

given

any p r o v a b l y

166

total

3.3

recursive

Theorem

recursive

Proof:

function

(Paris,

Harrinqton)

function,

Let

g:IN

then

~

IN

, with

that

is

not

provably

d

for

S as

S finite, By

3.2

there

Vn

are ~

a

the

interpreting the

By

compactness,

T'

= T U {c

Take I'

c M', --e

g(y)

M'

= x

3.2,

new

a

constant

IN

6 IN

is

(f(m) let

a provably

(m _> n ~

thus

f(m)

its

~ g(m)) . We

S c

T

(the

so

graph

is

to

show

of

3.1),

< g(a)

that < ci(

c,

> g(m)) .

want

T

, a 6 IN , b = f(a)

total

~ g(c)

e i the

definition

of

ci, the

~ x)

i < k}

i K k, graph

"g(y)

of

g

, ~ x"

.

system

of

~ PA.

is

Now

~

function,

S U {c ~ c o } U { V x

the

~ T')

Vm

> n

a ~ d

arithmetical

M'

I'

dicting

total.

~ c O } U { Vx

a model

IN

B m

PA) :

g:IN

a recursive

6 IN

in

in

If

e o , . . . , e k 6 [a,b]

(IN , a , e ° .... ,e k)

abbreviating

:

B n £

be

Z1-definable g

(provability

< ci(~

this, If w e

M'

then had

~ g(c)

It-formula

g(c)

~ x) ]i 6 IN}

the

c

I' ~ d

by

= g(c) ^ d

and

M' l ~

< c

I ' ~Z

d

M'

consistent.

2.3,(ii),(3) , then

for

1

is

we

some

As

we

determine

had

(contra-

i 6 IN might

an

, since

have

added

o Th(IN ) t o 3.4

T'

, we

Corollary

obtain

(Paris,

Harrington)

: PA

U Th

(IN)

~

~

,

nI noting the

that

recursive

total

proof

there

is

2.3(ii) I ~

just

of

3.4:

M ~

for

, a 6 I < b

had

(Xl-formulas

I ~e

contradicts

M

Let b

£ M

models . If

~

of

were end

the minimality

of

f

(a+1) a

M ~

b'

(a+1) a a

there

is

with

. Hence

remark

any

~

extensions) b

the

IN . S i n c e

[a,b]

6 I

. By exceeds

, a 6 M ~

arithmetic)

to

I ~ Th 1 ( I N ) .

up

that

true

there

going

Th(IN)

so

of

totality

~b[O,b]

too.

PA

which

a ~

the

function,

a smallest (only

states

function

recursive

Second

we

a

I =

to

1.4

provably

N ~

a

. Thus

an

I c

[a,b']

M ~

[a,b']

I ~

~ a

~

, by

e ~

M

,

a (a+1) a,

(a+1) a

, and

by

'

167

Remark:

In

finable iff

Paris

function

there

graph

[2]

of

Y:MxM ~ M

is a n

I c M ~e

Y

a

countable

has M ~ PA

Y(a,b)

, then

indicator

M ~ PA

the

in PA,

and

for a l l

, which

which

sentence

the We

can

just

countable

be proved

MI,M 2 ~ PA

one has

MI ~

M3 ~e M2

. This

of

PA

system

formula

~

initial

segment

account,

in

in

by

of

a turns and D.

3.5 C o r o l l a r y : segments

{X c ~

o u t to b e Jensen

There

a nice

asked

are models

(up to i s o m o r p h i s m )

Let

determined

by

language

Define

M 2'

M I' {c i li 6 ~ of P A

S S y ( M I) = SSy(M2) , h e n c e

if

this

MI

part

of t h e

Note

that

for

any in

[5]

3 b 6 M I a
in t h e d e f i n i t i o n

of an

at all.

are

countable

where

{n 6 ~ MI

the

of G a i f m a n ' s

MI,M 2

SSy(MI) , the

IM ~ ~(n) }

for

is i s o m o r p h i c ~e M2'

models

MI

some

to an

% ~ '

[3]. T a k i n g

this

into

example

for T h e o r e m

2 in

[3] w h i c h

MI,M 2 ~ PA

} , MI,

MI

that

M 3 = {a 6 M 2 1

I c M

is

"only-if"-part

2.3, (ii),

that

if

of 3.3

M2

such

, M I'

initial

~e M 2 and

which

another

to be t h e

~

the

> z

, cf.

of o n e

. Since

Y(x,y)

for:

T' b e as in the p r o o f

countable.

in a n y

satisfies

in M } , t h e n

S S y ( M I) = S S y ( M 2)

an i n d i c a t o r

is an i n d i c a t o r

JX =

M 2 . Furthermore,

if t h e

for unprovability.

shows

>

that

but

by a r e s u l t

[4] t h a t

M I , is

parameters

, then

Ehrenfeucht

the

of

with

MI,M 2 ~ PA

M I'

also

: Y(a,b)

. He shows

, then with

c a n be r e p l a c e d

showed

a,b 6 M

, Thz1 (M I) ~ Th(M2) , S S y ( M I) = S S y ( M 2 ) ,

standard

Proof:

stated

M ~ PA

M I ~ M2,

H. F r i e d m a n

all

requiring

this

P A to be a d e -

V z H y

(c+I) c} c

If

I c M --e

M ~

defines Vx

latter

exploited

easily

for

for

, a 6 I < b

reading:

indicator

that

= max{cl[a,b]~

indicator-property any

such

, I ~ PA

indicator-property.

function

an i n d i c a t o r

Z1-definition

true but unprovable of the

defines

are mutually that

C--e MI' 2

U T'

substructure of

ThzI(MI)

are m u t u a l l y

M

M I ~ a , M 2 ~ ~ s.

~ Th(~)

the r e d u c t s

initial

of

, M I'

M I' , M 2' = ThzI(M2)

initial

to and

segments,

168

and

by

M I' l= T'

As theorem

a second that

PA

i~

~

3.6

Lemma:

provable

The

PA

is n o t

All

sentences

provable,

Let

us

but

for

now

this for

these.

Cf.

one

be

so

Theorem that

a

even

V x3

y[x,y]

(~+I)[

remark the

to

v xV

Ryll-Nardzewski's First

we

n

~

(n+1) n

being

provided

give

to

, n 6

IN , a r e

one

given

in

Wilkie)

, even

axiomatizable.

Remark:

By

using

3.6

is

[2].

fact,

3.7

:

For

S U Th

all

(~)

restricted

the

"from

i~ a

provable,

do

far

is

Ramsey

outside"

on

.

theorem

by

finite

are

Infinite

will

which

by

Paris

and

than

the

simpler

a corollamyof

S ~

PA

there

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

2.3,(i)

is PA

n 6 is

not

n

truth-predicates,

induction

of

a proof In

El-sentences,

(Z)y

following

~I

finitely

~

induction

here

the

true

.

version

But

proved,

with

1.4

using

topic.

, being

y H z[x,y]

a definable

this

once

S I~ s

n

~

superscripts,

(Paris,

PA with

reprove

axiomatizable.

=

n

n

to:

can

e-incompleteness:

the

needs

for

game-theoretic

3.7

use

fixed [6]

can

an

H y[O,y]

that

2 . 3 , (i) Wilkie

we

.

mention

Theorem

of

sentences

are

s .

finitely

form

PA

~

application,

the

in

Proof:

, M2 b

the to

result

can

E -formulas

easily

for

be

fixed

extended r 6

r

does

not

Proof:

imply

Let

a finite

meters)

S

set

induction

all

be £

axioms in

such

i_~f (M, (ei) i E ~

the

a

n

a finite of

Eo-fOrmulas

contained a way ) b

subset

in

of

PA.

By

the

~(x,Yl,...,y S

(which

may

be

proof

of

t) m a t c h e s

2.3,(ii) with

assumed

without

(where

TF

the para-

that

PA U {c~

< ci+1 Hi 6 ~ }

U TF

is

,(3)

the

.

169

last

group

of

then

the

initial

Set

M b

Th(~

according [a,b]

~

to

% ~

the

. By

Remark:

the

As

Paris

model

models than

~

PA

(the

T

of

Proof: are

for

is

a model

Let

S

(eili

6 ~

of

so

(M,eo,...,ek) ~

the

in

number

in

F )

) models

M

S

for

Theorem:

There

not

even

with

such

are

.

S

, so

that

, hence (~)

nl [a,b']

, but

not

s

, n

~

(n+1) n n

would

means there

that are

whereas

not

3.8

segments

in models

of

initial

will

segments

imply

modelling

that

PA

in

, other

3.9, from

PA

nonexistence

paragraph

M ~ PRA

defined

, so in

model by

of

, as

same

~

ce M

argument

as

recursive

that

arithmetic)

(M' (ei)i encoded

of

an

PRA of

. By

( ~ , e o , . . . , e k)

recursive

a sequence

a set

T

of

2

(primitive

, ei 6 M

subset

S

on

~

has

b T

by

an

initial

seg-

Eo-indiscernibles.

3.2

and

S

. Hence,

. Using for

6 ~

PRA

~

~

b

a

there

for tro(~,x)

2 . 3 , (ii) ,(I) ,(2)

~ ~(x) yields

) as d e s i r e d .

3.9

,

with

, a ~ ei < b

U Th

I b

) contained

that ~

6 ~

TF

3.7

nonstandard

a finite

6 ~

element

S

6 ~

a natural

ei,i

initial

6 ~ )

6 ~

any

(eili

formulas

.

element,

extract

(eili

the

all.

(eili

which

be

6 ~}U

[2],

nonstandard

. Thus

Eo-fOrmulas

b

application

M

eo,...,e k

of

at

of

M ~ PRA,

< Ci+lii

any

we

3.1),

be

are

to

by

smallest

there

It-induction

any

a sequence

n

in g e n e r a l

models,

is

the

,

} satisfies i b' 6 I with

above

there

ment

6 M

~

induction

third

For

determined

in

present

Lemma:

M

restricted

pointSout

, are

the

3.1

< e

of

of

nonstandard

element

d

# ~

For

b

minimality

prefix-restricted which

of

2 P A U {c i

) ~

in

2.3, (i)

existence

contradict

T

, a 6 M ~

2.3, (i),

{d 6 Mi 3 i 6 ~

since

3.8

of

segment

), M

(n+1)~

(M' (ei)i 6 ~ I =

axioms

no

recursive

recursive addition.

nonstandard

models

of

PRA

,

170

Proof:

Let

M b PRA,

(M'((e)i)i I 6 ~

6~

, the

) b

T

proof

of

(*)

H x I Vx2...

Set

A =

of

Using

Hence PRA be

we

an

2.3,(ii)

G~del

take

initial ,(3)

numbers

one

e 6 M

segment

shows

a 6 M

as

the

but

on

n)

of

realizes

containing

induction

get

taken

the

3.8,

H x n ~(Xl,...,x

tr °

p(z) by

. By

so

I

that

that

given

for

any

by

(e) i

open

~

,

in

the

PA

set

formula

# ~

. For

of

language I b

M

iff

9(y)

that

the

A =

formula

{n

6 EO

there

3 x1<(e) I Vx2<(e) with

is

parameter

a primitive

with

M b

e

so

IM b

B y({+..~=

x)

A =

p

(M,e)

available

. Here

, p(n)

.

N ~

.

reeursive

that

formula

n 6 a}

~

}(e)

a primitive

recursive

6 ~

M ~

2...

n

6 x

in can

the

p(n)-times n-th b

prime

6 M

number,

so

that

If

addition

follows:

For

6 x of

in

G~del's of

~

and

+

PA

M

, is

models

M

the

that

in

right of

whether

in

IM ~

recursive

n

on

the

[3].

side

M

its

of

Likewise

, we of

could PA

(*),

then

structure

and

~ 6 a

Theorem

there

is

a

6 b}

language

M ~

Incompleteness

proof

of

for

A c as

~

or

decide

compute

of

M ~

Th(I) the

compute

the

the

as

associated formula

recursiveness

~ 6 b

Th(I) , being

6 ~

uses

an

step

a nonrecursive does

nonexistence

in

which

{n

reduction

3.9

the

P_AA g i v e n

representability

Note

{n 6 ~

help

decide

representable

yielding

the

proposed

. But

by

a complete

extension

undecidable.

The

the

were

on

a nonrecursive

Thus

~ A =

sentence

with

First

Remark:

~

any

Zo-Sentence

a device

not

[3] is

JM p n

is

A hold,

even

6 a}

simpler:

for

which

Th(I) which

recursive

arithmetically

induction from

of

to is

however,

just

definable

some is

it

nonstandard

not

a 6 M

available

((M,e)) is o representable even models

and

. But

Thz

for

starts

of

the

from

thus this

in

PRA

.

decisive, in

M ~

basic

PRA

axioms

.

171

for

0,I,+,.,<,-

cf.

plus

open

induction,

as

Sheperdson's

model

shows,

[7].

4.

Generalizations

a

is

a

of

a

H2-statement

Hn+2-formulas

Applying

independent

independent

of

to of

PA

Higher PA

U Th

Prefixes

U Th

I (~)

, and

(~)

similar

we

to

look

for

to

E n-

a

~n+1 So w e

introduce

substructures

modelling

En-indiscernibles proofs

in

replaced M ~

PA

fixed in

by

tr n.

have

n 6 ~

general

rather

try

PA

instead

paragraph

we

generalizations

an

, instead of

shows,

by

infinite

sequence

to

supply

sequences

initial

substructures

way,

of

a result

an

corresponding

the

sequence

by

just

s

of

of

applying

Z -substructures. o will

Zn-indiscernibles

infinite

, whereas

of

of

Eo-indiscernibles

2 yield (This

an

not

as w e l l ,

that

in

if

any

Knight

([6])

Z -indiscernibles o be

closed

for we

so

under

the

tr 0

Zn-indiscernibles, J.F.

for

is

nonstandard

Z -indiscernibles).

of

do,

Using

do

any not

Thus

we

that

the

should

Z -Skolem-functions. n

4.1

Definition:

formulas ~b(V

with

c < a V d

Let two

n > 0

fixed

~ a

, tr

n

the

variables,

(3 e tr

E -truth-predicate n

gn:

(c,d,e)

~

~

~

g i v e n by (n)

[x,y]

~

(z+1) z

gn(a) be

tr

-

g b

the

with

3 e < b

n

relation

~

t h e n becomes

formula

saying:

gn(a)

(c,d,e)))

Hn-definable. ~

required

gn(min(X)) formula

a

n =

< Card(X)" ~

=

Let plus

Z

homogeneous

Ix,y]

E n

. The

(z+1) z

Z

"the

for

n

[x,y]

sets

X

(instead

of

z

(z+1) z)

can the

be

chosen

special

such

case

gn

that = id

in

the

. Set

(n) VxV

z 3y[x,y]

~

(z+1) z z

i

n

(n) is

then

a

Hn+2-sentence

, [x,y]

have

(z+1) z is z

a

H -formula. n

172

The paragraph

second

proof

3 generalizes

4.2 L e m m a : language

For

each

is an i n i t i a l

then

PA U Th

the

to

an

for

independence . First,

n 6 ~ :

of P A so t h a t

there

for

If

all

~

there

~(x,y,z)

a

given

is n o t h i n g

is a

M b PA:(if

substructure

(~)

of

new

in

En+1-formula

M k ~(a,b,c), I ~ PA,

I ~e M with

in

in the

c > ~ , then

I _ c M, n

a 6 I < b),

V x V z ~ y ~(x,y,z) .

~n+] Remark:

Such

indicator,

a

~

this

has

the m a i n

time for

property

of

~ -substructures n

a definition

which

of an

are m o d e l s

of

PA

.

(n) NOW we

show

that

[x,y]

~

(z+1)zz

indeed

defines

such

a

Z -indicator: n

(n) 4.3 L e m m a :

If

then

is an

there

M ~ PA,

a,b

I ~ PA,

£

M,

c 6 M \ ~

I --eC M,

I ~

, n > O, M ~

M

with

[a,b]

~

a 6 I

[a,b]

< b

~

c

(c+I) e,

.

n

(n) Proof:

[a,b]

sequence

c (c+I) c

~

Xo-indiscernibles

of

in 2.3, (ii).

Since

a homogeneous

set

[a,b] as

M

for

being

any

<eo,...,ei>

infinite

segment

in

of

6 ~ ),

partition g~(ei)

primitive

recursive

by

6 ~)

< b

, as

[[a,b]]

d

function,

condition

ei+1,

the

. By 2.3, (ii)

from

all

than

be

arises

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

for

I

a

described

, the

if w e w i s h .

Let

(eili

" So w e h a v e

sequence

is s m a l l e r

Zo-indiscernibility.

determined

this

on s o m e

< ei+1,

(c+I) c c

a ~ ei

2.3,(ii)

(sequence-number)

by

M

some

yields

dominates

i 6 ~

(eili

b y the p r o o ~

(~)* (c+1)Cc

gn

implies

all

ei

initial only

I ~

M n

remains

to b e p r o v e d .

We

show

I ~

M, m

~ n, b y

induction

on

m

.

~ 6 ~m+1

"

Zm By

I --eC M w e h a v e

already

I ~ Z M.

Let

m

< n,

I ~

O

~ =

H x ~ ( y I ..... Y t , X ) ,

I ~NmM .

So

let

Equivalently replacing

M N M ~

yj

by

<eit+1'

~ 6 Nm

Hx

. Clearly

~(al,...,at,x),

3 x~' ( < a l , . . . , a t > , x ) ,

M, m

I ~ ~(~)

~ M ~ ~(~)

a.3 6 I

, a.3 < e . l j ,

~' b e i n g

obtained

, by

eil<'''< from

ei t

$

in

(y) j. W e h a v e g~(eit+1)

< eit+2'

~'

< eit+1'

~'

£ H m ~ Xn

'

173

thus

M ~

3x

trn(r~'n , < a l , . . . , a t > , x )

hence

M ~ 4' (,b)

M ~ ~(a I , . . . , a t , b ) ,

Finally,

4.4 Lenuna: As

all

to its p r o o f ,

homogeneous difficult

X

4.5 T h e o r e m : of

PA U T h

.

taken

- 4.4 w e

all

(~)

b 6 I , for hypothesis

k a that

f(min(X))

4.2

For

remark

M ~

4.3 u s e f u l ,

~

c a n be

to a c h i e v e

Combining

4.2 a n d

n > 0

we only

sets

some

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

to r e n d e r

For

for

entails

3x the

this

<e i + 2 t r n ( . . . ) , t same

b

holds

in

I .

we need

n in the p r o o f

from

of

an i n f i n i t e

< Card(X)

for

any

~

~ s

set,

the

so it is n o t

fixed

function

f:~.

obtain

n > O

an

Moreover,

if

is a t r u e M ~

PA,

Nn+2-sentence

M e

~

,

there

is

independent an

~n+1 I --eC M,

I

N PA,

so

that

I

~

M

and

I

~

~

~

n

n

Remark: case

i-6

For

the f i r s t

M ~ Th(~

Corollary:

For

of i n i t i a l

complete

extensions

stronger

all

Ehrenfeucht result

completions that

for

with

regard

of

n = O

of

4.5,4.3

is o n l y

needed

in the

special

)

infinity

Remark:

part

that PA 4.6

n 6 ~

any nonstandard

Zn-SUbStructures of

PA

and D.

satisfying

of

Th(~)

pairwise

has

an

different

.

Jensen

proved

a nonstandard satisfied

by

immediately

to the r e s u l t

model

model

in

[3] f o r

of P A h a s

n = 0 2

~o

its i n i t i a l

substructures.

generalizes

to m o d e l s

by Gaifman

quoted

in the

of

remark

the distinct Note PA

U Th

following

(~), 3.4.

174

References

[i]

J. Paris,

L. Harrington:

Ar i t h m e t i c ,

A Mathematical

in: H a n d b o o k

Incompleteness

of M a t h e m a t i c a l

Logic,

ed.

in Peano

Jon Barwise,

1133-1142. [2]

J. Paris:

Some

The Journal

[3]

Proc.

1971

of the

Lecture

H. Gaifman: Proc.

Lecture

J. F. Knight: The Journal

[7]

A Note

J.C.

of Number S6r.

Types

Astr.,

Bull.

vol.

43, No.

1976,

Summer

in Math.,

School vol.

vol.

40,

Model

XII,

of A r i t h m e t i c , London

255,

1975,

1970,

128-144.

Models

of A r i t h m e t i c ,

317-320.

Polonaise

No.

2,

Logic,

539-573.

for a Free V a r i a b l e

l'acad&mie vol.

in:

Logic,

in U n c o u n t a b l e vol.

arithmetics,

in M a t h e m a t i c a l

337,

in M a t h e m a t i c s ,

de

in e l e m e n t a r y

and Submodels

Logic,

725-731.

223-245.

in M a t h e m a t i c a l

Phys.,

4, 1978,

of Set Theories,

A Nonstandard

Theory,

Math.,

Models

Omitted

of S y m b o l i c

Sheperdson:

XCII,

on Models

Notes

for Peano A r i t h m e t i c ,

Some p r o b l e m s

Cambridge Notes,

of the C o n f e r e n c e

Sp r i n g e r [6]

Mathematical, Countable

Results

Logic,

D. Jensen:

H. Friedman:

Sp r i n g e r

[5]

of S y m b o l i c

A. E h r e n f e u c h t , Fundamenta

[4]

Independence

1964,

Fragment

des Sciences, 79-86.