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.
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
<jleast
definable
J
Y
be the
2.2,
let q(m)
of Jp(m)
be the
<jleast
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 = dl(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~
~ = oi(~),
x 6 J~. Set
over
~ Jy
and
<jleast 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
<jleast
~ %(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 Jdefinable 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
Elregular
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 Jdefinabley 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
] = oi(~)
theory
~
~
= [.
not.
cases
to
so
done.
[]
: ~.
E,
p(])
=
D
in C a s e
([De
A and
].
if ~ < p 1],
A = Ag nl,
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 Jdefinable y
m
Suppose
can
q(~)
p = p nl,
(oIJ~):
immediately,
E,
that
If v = p , t h e n
claim
is
: p(v).
structure
~ 6 S and
= ~,
Claim
fine
by
there
[]
~ = oi(~).
~ = p.
J and y
is E l ( a n,+ ly)
: i, ~ 6
o(p([))
proof
But
I m < w>
implies
the
q = ol(q),
D.
C and
which
That
Set
B and
claims
=,,b)
<[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 nl<~,
contrary
ni 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. Letbe 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,
~ = ol(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
Inregular
"(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~Jcd' 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 Jydefinable
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 = ol(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
Eoabsoluteness ~:
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
ni Y = PR'~ , B = An16' ' 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').
Inregular
: 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
nI 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
~2tree
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
~2tree,
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
endextension
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 TD % 
where
Y £ TTS'
TY
D
2). has
are
no
nontrivial
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
nontrivial
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 7TT(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 nontrivial
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
e2branches 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 ~,
e2branches.
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
Elsingular
(~)
= 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 that6 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 that6 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 and6 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
~2tree. 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
~2Tree
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
~2branches.
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'
~2tree
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
ebranch
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
ebranch,
<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
cO2branches
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 ( ~ ) > .
~ldefinable
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(~)
~ldefinable
<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
Eldefinitions
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
71definable
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
Elformulas !
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
zV~. [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
= 7YY (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
Zldefinable
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.
Zldefinable
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
Zldefinitions
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
nontrivial
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
nontrivial
~ : 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)>
Eldefinitions
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
~2branch.
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),
114.
[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
~2Souslin Hypothesis.
Annals of Math. Logic 5,
(1972), 2146.
The Independence of Kurepa's C o n j e c t u r e
TwoCardinal Conjectures
in Model Theory.
Pure Maths. XIII, Part I, 383390.
AMS Proc.
and Symp.
i~
COARSE MORASSES I N HansDieter
Donder
Mathematisches Universitat
Introduction several retic
Highergap
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. gap1 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
Oprinciple. 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
modeltheo
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 Gap1
for
gap1
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 gap1
(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
Lvcardinal
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 v4~
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
gap1
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
[email protected]
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. A1vg(v)
[Ag(v) Iv~ IDP ~ . 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~=
ktree
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
ktree
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
~2tree.
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 highergap
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
~Lleast 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) = gh.
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 :
4chain.
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 wellfounded, 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 twocardinal
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([email protected]) : @. and
X~®
~(6)18~p~ 4~(8)18~p~
f(~)~ [email protected] .
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
4chaln 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
Ktree,
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
[email protected]~
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.
ktree,
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' =
ktree.
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~<~)
Atree.
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, 0rdert~pes,
trees, and a problem of Erdos and
Hajnal,
Periodlca Math. Hungarica 5 (1974), pp.15J160 [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 748 [5]
R.B. Jensen~ The
[~]
L.J. Stanley,
(~,G)morass
(unpublished
manuscript)
"L11ke" 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. 161171
[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. 2558
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
eErd~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
eErd~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
E3absolute.
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
lindecomposable
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
6decomposable The
that
cardinals
4.7.
presented:
20).
Koppelberg)
some
are
ldecomposable
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($))
for
if A
the
Being
a pleasant
f.
is a s e t
other
there
hand,
of o r d i 
~ denotes
~Erd6s
<~
K.
We
that
countable
there
is
M = L[f].
Since
~
for
is o b v i o u s l y
some
facts
a strengthening
(~)<w ordinal So
and
~ is
the
especially
property
showed
improve
< be
< be
paper
of
least
~ such
< is R a m s a y
aErd~s
cardinals
that
iff
< is
that
they
are
inaccessible.
Mitchell
Let
Let
Baumgartner's
< is ~  E r d ~ s .
following
such
a closed
D is h o m o g e n e o u s
us
to
a limit
therefore
model
necessary.
order
reader
d is
[ 10]
The
model
is
l.l:
let
relation
if
<Erd~s.
~ e and
cardinals.
partition
(~)<~,
C is
o f X).
Conversely,
the
the
iff w h e n e v e r
• < is r e g r e s s i v e
notation
refer
eErd~s
IDI
that
cardinality
about
~Erd~s
f:[C] <~
denotes
We
of
is
< and
is a D ~ C
nals,
<
this
~Erd~s
example ~lErd~s
< is
that
an L  g e n e r i c is c o u n t a b l e
cf(~)
shows
~Erd~s
cardinals
are
Ramsay
Then
~Erd~s
in
to:
and
and
Ramsay
that
set
> ~.
some
~ = w~.
in M b u t
not
collapsing
~ is
assumption We
construct
in K M. map
in M a w e l l  k n o w n
about
f:~
an
Since~L(e) , ~.
argument
in K.
~ is inner is
Let due
to
59
Silver
yields
that
ric extension
o f L.
It is w e l l  k n o w n reformulated that
this
Therefore
that
can
in M.
of
for
I ~ < is a g o o d
8 <<
relation
~Erd~s
L e t A I , . . . , A n c < and
KM = L b e c a u s e
be aErd~s
indiscernibles
a l s o be d o n e
= C~I LB [ ~ ] for
But
~ cannot
the p a r t i t i o n
in t e r m s
Definition. ~B
< is a  E r d ~ s
set
for
<
M is a g e n e 
in K M.
, (a)<~ c a n b e
structures.
We
show now
cardinals.
Ol=
6 ,~ > ,
.
set o f i n d i s c e r n i b l e s
for O~ (or g o o d
for (}/) iff
for all ¥ 6 I: (G i) ~ y
< C~
(G 2)
Iy
is set o f
Lemma
1.2:
Let ~ ~ e be a limit
model
~=
type
Let
< be a  E r d ~ s .
T h e n C is c l o s e d of t h e
formulas
parameters
ordinal.
a good
< is a  E r d ~ s
set of
indiscernibles
that
f([$,~})
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
OZyn 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~yn ~ 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 =
Czyn . 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*
ecomplete.
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_ Ozyn , 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
[email protected] 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 [email protected] c N 8 or [email protected] 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 (~6yn 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
[email protected]
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
41east
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
[email protected]
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
w1Erd~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
[email protected] 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 nontrivial
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 nontrivial
class
U
cases,
a measurable
a nontrivial
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
ith
:+ 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
TA.
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
Iyc
v6 E such
= y<
= sum
sequence.
3.
or
Proof: get
. But
I = { T ~ I 6 < <}
IyeA
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 nontrivial
by:
V
yields
for
each
e6
Y
wellknown.
~
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 sdescending
=+ < ~
~
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
~
WTIT 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 WTIi 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 (WY) 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 nontrivial
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 ith
(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
ji.
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,.,Jm1
(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
(ikyB)
=
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)
Hlanguage
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>
= bl(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 = bl(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 • Let6 [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 Lterms
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
Elembedding iterable
of K'
premouse
into used
itself.
Thus
in the above
a ~; exists. proof w h i c h
M is thenthe 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.
6decomposable.
= ~.
Iv < 6> s u c h
U is
For more
6descen
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
6decomposable
implies
iff t h e r e
S c6
iff U is cf
] and
ultrafilter
there
that
incomplete
6decomposable
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 lregularity } 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
Kregularity
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
6decomposable
(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'~
Aregularity
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
udecomposable.
Now we
Case
regular
a regular
U is
6decomposable
acomplete
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
6decomposable.
let
But
a weakly
there
<+de
6decomposable
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
ldecomposable
on
by
~RK
for
<. A s s u m e
function.
remarks by
U and
inU
6decomposable.
Remark: I.
U'
and 4.2
remarks
< be
first
is
Lemma
(~,l)regular
remarks
~,
by
6decomposable
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,
6decomposable
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)
6decomposable
and
the
for
induction
all
6 ~ 5
hypothesis +
show
that
The
V
and
proof
U
of
model
results
[12].
For
are
lemma
from
some
6decomposable
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
oBsequence
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 DBse
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
6decomposable
(i) and
lemma
for a l l
6 _< B
then
and U vdecomposable}
= 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
oSsequence
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
6decomposable
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
[email protected] =
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 cP 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.
87106, 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.
393399
4776 and large cardinals,
Trans.
(1976), 6173
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,
650693
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
183188
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
167171
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,
459488 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~niginLuiseStr. DIOOO
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
qstructure
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
Booleanvalued
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
twostage
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
settheoretical
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
settheoretical
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(xb)
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(xb) 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
fi
instead
of
f,
that
T (y) = t.
Now
Sf(x.b):yb
is
such
%yb,
from
shows
y.b,
that
f(x.b)
_
f(x.b)=yb,
The
hence
same
T(x.b)=
m (yb) .
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
coordinatewise 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(xb 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,
orderpreserving:
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
wellordered
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 dd' : 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
firstorder
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
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+2sentence
M e
~
,
there
is
independent an
~n+1 I eC M,
I
N PA,
so
that
I
~
M
and
I
~
~
~
n
n
Remark: case
i6
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
ZnSUbStructures 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,
11331142. [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,
128144.
Models
of A r i t h m e t i c ,
317320.
Polonaise
No.
2,
Logic,
539573.
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,
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223245.
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Phys.,
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A Nonstandard
Theory,
Math.,
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Sheperdson:
XCII,
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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, 7986.