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e ~2)]. It is obvious t h a t ~) t P is dense. T h e r e f o r e t h e r e is a a p 1 e (~) t P) ~ Gp. Let ~ 5 / G p = ~q[Gp] : (2p e Go) ~ ~ A Gp * H, w h i c h e s t a b l i s h e s t h e g e n e r i c i t y of Gp * g . ) a n d F ( z ^ < l > ) a r e i n c o m p a r a b l e in T.
We
shall see t h a t ~ 5 / G p is a d e n s e s u b s e t of Q[Gp] in V[Gp]. Let qo e Q[Gp], t h e n f o r s o m e c a n o n i c a l n a m e q0 of a m e m b e r of Q
q0[GP] = q0 , and for some
po 9 cp,po IF,, .% 9 Q".
DqO
Then
=[p C P : p > p o ^ ( H q t ) p I I - p " q t - > q o
' ' and
9 P *9
Let
(.p,qt) e D ) l e v .
We shall
see that -%~, is dense in P above Po- Let P l ~ P o , then
.%
is
dense
in
P *r
there
is
a
9
such
that
43
a n d so ~)-q0 is d e n s e
in P
a b o v e Po.
S i n c e Po E Gp we c a n
pt pl assume
that
P o E ~)go' h e n c e f o r s o m e q t p o I~-P ..qt ~ qo a n d < p a,qt> E ~ ) " . S i n c e p o c Gp we h a v e q t [ G p ] -~ q[Gp] = q a n d qt[Gp] E ~.~/Gp.
S i n c e H is a g e n e r i c s u b s e t
of ~ G e ] over V[Gp] we have ~ 5 / G p n g ~ r let q[Ce] c D / G p q[Gp]c~)/Gp
there
is
a p c Gp
such
that
n g.
c~).
Since
Therefore
1.5 L e m m a : F o r e v e r y P * Q - n a m e T t h e r e is a P - n a m e Y s u c h t h a t f o r e v e r y g e n e r i c s u b s e t Gp of P , V[Gp] is a Q [ G p ] - n a m e in V[Gp] a n d f o r e v e r y s u b s e t Gq of Q w h i c h is g e n e r i c o v e r V[Gp] , ~[Gp][GQ] = "r[Gp * GQ]. We u s e t h e n o t a t i o n Y[ Gp] = T / G p . Proof: T = ~<
By
induction
on
the
rank
of
the
name
r.
Suppose
:i < io] w h e r e
By t h e i n d u c t i o n h y p o t h e s i s e a c h T i h a s a t r a n s l a t i o n T~ s u c h t h a t T~[Gp] is a Q [ G p ] - n a m e a n d T~[Gp][GQ] = T i [ G P * GQ]. Let vi b e t h e n a m e
of < q i , v ~ >
a n d let ~ = I < P i , ~ i > ;i < i0]; t h i s is c l e a r l y s u i t a b l e . (See I. 1.8). 1.6 D e f i n i t i o n : If P ~ Q (see D e f i n i t i o n I 5.3), G C P g e n e r i c , t h e n let
Q~ G c V[G] b e t h e following p o s e r : its s e t of e l e m e n t s is lq c Q:q is c o m p a t i b l e in Q w i t h e v e r y p
c GI ,
its o r d e r is i n h e r i t e d f r o m Q. S o m e t i m e s we w r i t e Q~ P i n s t e a d Q~ G , (so it is a P - n a m e ) . ~p
1.7 L e m m a : (1) P ~ P * Q ( w h e n P a p o s e t , I ~ - F ' Q a p o s e r ")
(2) (P * Q ) / P , Q a r e e q u i v a l e n t (i.e., t h i s is f o r c e d f o r P ) . N
44 Moreover for any generic
G cP,
(.p,q) c P * Q / G , e q u i v a l e n t l y p N
f, f((p,q)/~)
the function
c G) is a n i s o m o r p h i s m
=q[G]/~
(for
from (P * Q)[G]/~
,
N
onto Q[G]/~.
( w h e r e ~ d e n o t e s t h e r e l a t i o n d e f i n e d in I. 5.5).
N
(3)
If p
posers
in
If)
then
Q is
equivalent
to
p * (Q/P)
1.8 T h e A s s o c i a t i v e Law Leiimla: If P is a p o s e t , ]~-p.Q"Raposet",then(P
w
* Q ) *R,~ a n d P
]~-p"Q a p o s e t
"
*(Q~ *R)~ a r e e q u i v a l e n t .
Iterated forcing. 2.1 D i s c u s s i o n : We s a w a l r e a d y t h a t t w o s u c c e s s i v e e x t e n s i o n s
forcing are equivalent to an appropriate the same question
about
single extension.
an infinite number
of V b y
We w a n t t o a s k n o w
of e x t e n s i o n s
by forcing. The
need for this arose in the following ease.
By a c l a s s i c a l t h o e r e m ,
if a l i n e a r l y o r d e r e d
has no first or last member, order-isomorphic
and has a dense
to the real numbers.
one can replace the last requirement, by the requirement countable.
The
that
s e t is d e n s e a n d c o m p l e t e , countable
t h a t t h e r e is a d e n s e c o u n t a b l e
e v e r y s e t of p a i r w i s e d i s j o i n t i n t e r v a l s
statement
that
terexample
then
these
two
additional
subset,
is a t m o s t
requirements
are
s e t w h i c h is a c o u n -
t o S o u s l i n ' s h y p o t h e s i s is c a l l e d a Sousttn's c o n t i n u u m . proved the independence
i t is
Souslin raised the question whether
e q u i v a l e n t is c a l l e d S o u s l i n ' s hypothesis, a n d a n o r d e r e d Tennenbaum
subset
of S o u s l i n ' s h y p o t h e s i s
Jech and
of t h e a x i o m s
of ZFC b y u s i n g f o r c i n g t o o b t a i n a u n i v e r s e i n w h i c h t h e r e is a S o u s l i n c o n tinuum.
Later Jensen
Souslin continuum. and Tennenbaum
continuum
in t h e c o n s t r u c t i b l e
To p r o v e t h e c o n s i s t e n c y
proceeded
it one can construct Souslin
proved that
there
of S o u s l i n ' s h y p o t h e s i s
a generic
is a
Solovay
as follows: Given V and a Souslin continuum
a generic extension by
universe
C in
V[G] of V i n w h i c h C is n o m o r e a
introduction
of a n
uncountable
set
of
45 p a i r w i s e d i s j o i n t i n t e r v a l s in C. In t h i s c a s e we s a y t h a t we h a v e " d e s t r o y e d " C. U s i n g t h e s a m e m e t h o d continua.
we c a n g o o n d e s t r o y i n g
more and more Souslin
It can be shown that a Souslin's continuum
h e n c e t h e r e a r e a t m o s t 22t~~ S o u s l i n c o n t i n u a when we destroy one Souslin continuum b e s u r e we h a v e d e s t r o y e d
h a s e x a c t l y L# ~ p o i n t s ,
( up to isomorphism).
However,
n e w o n e s will b e c r e a t e d a n d we m u s t
t h e m all. S i n c e we h a v e i n f i n i t e t i m e a t o u r d i s p o -
s a l t h i s m a y b e p o s s i b l e . R e a l l y we c a n s h o w t h a t i t is e n o u g h t o d e a l w i t h s u b s e t s of S o u s l i n c o n t i n u u m shall iterate
of p o w e r t~1, s o t h e r e a r e o n l y 2 t% s u c h o r d e r s .
We
h t i m e s , s o t h a t i n t h e n e w u n i v e r s e , 2 s~ = 2 I~t = ~, so we h a v e a
chance to catch our tail. 2.2. D e f i n i t i o n : We s h a l l c a l l Q = < Q . : j
(or Q =
:j ),
f o r s o m e o r d i n a l a, a system of FS ~finite support) iterated forcing if e a c h Q ,j < a is a n a m e , f o r t h e f o r c i n g n o t i o n P j , of a q u a s i - o r d e r , i.e., []-D ' ' Q is ~j ~i a q u a s i - o r d e r " , w h e r e P j f o r j ~ a is d e f i n e d b y r e c u r s i o n a s t h e s e t of a l l finite functions f
o n j s u c h t h a t f o r a l l i c D o r a ( f ) f ( i ) is a c a n o n i c a l n a m e
f o r t h e f o r c i n g n o t i o n P~ of a p o t e n t i a l
member
of Q. a n d we c a l l P a t h e
d i r e c t l i m i t of Q.
T h i s is c a l l e d i t e r a t e d f o r c i n g ~ t t h finite support s i n c e t h e f u n c t i o n s f we u s e i n t h e Pj's a r e f i n i t e f u n c t i o n s .
T h e Pj's a r e s e t s s i n c e we r e s t r i c t
the
c h o i c e of t h e f ( i ) ' s t o c a n o n i c a l n a m e s . We c a n r e p l a c e " f i n i t e s u p p o r t " b y CS ( " countable
The
support") or "
partial
order
on
Pj
for
j -< a
is
defined
as
follows
f -< g <=> D o r a ( f ) c D o r a ( g ) ^ ( Y i e D o m ( f ) ) [g r i I]-p~ " f ( i ) ---@ g ( i ) "].
2.3 D e f i n i t i o n : F o r fl ~ 7 ~ a we d e f i n e P~,7 b y r e c u r s i o n P~,? is t h e s e t of a l l P # - n a m e s
of f i n i t e f u n c t i o n s f
on Y as follows.
on 7-i$ such that for all
c D o r a ( f ) f ( i ) is a c a n o n i c a l n a m e f o r t h e f o r c i n g n o t i o n P i of a p o t e n t i a l memberof
Qr a n d f
ti
I~-P~ " f ( i )
c Q.".
46 T h e n e x t t h e o r e m is given h e r e w i t h o u t p r o o f . 2.4 T h e G e n e r a l h s s o c i a t i v i t y T h e o r e n ~ (a) Pa ,,m PIi * P
~I~.a
f o r fl ~ a, w h e r e P
is a n a m e of t h e f o r c i n g n o t i o n
~I~,a
P p f o r t h e p o s e r w h i c h is P•,a in V[G] , w h e r e G is a g e n e r i c s u b s e t of P~, a n d w h e r e t h e Qj's, fl -< j < a a r e t r a n s l a t e d t o n a m e s f o r t h e f o r c i n g n o t i o n s P p j , m e a n s t h a t o n e of t h e two p o s e r s is i s o m o r p h i c t o a d e n s e s u b s e t of t h e o t h e r , so t h a t t h e y r e p r e s e n t e s s e n t i a l y t h e s a m e f o r c i n g n o t i o n . (b) Pp,IJ+I ~ Qp o v e r V[G] w h e r e G is a g e n e r i c s u b s e t of P (c) If ( • i :
i - T > is a n a s c e n d i n g a n d c o n t i n u o u s
fl0 = 0 a n d fit = a t h e n ( P ~ . ~ + I the ~ sense) to (Q"
i < a~.
sequence such that
: i < ~,) is a n i t e r a t e d f o r c i n g e q u i v a l e n t ( in This is, in s o m e s e n s e , a g e n e r a l a s s o c i a t i v e
rule. 2.5 T h e d e f i n i t i o n b y i n d u c t i o n
theorei~
(one can construct
Q's by a
g i v e n r e c u r s i v e r e c i p e ). If F is a f u n c t i o n a n d a is a n o r d i n a l t h e n t h e r e is a unique
iterated
O :F(
:i <j))
forcing
(Q.
N3
9j < a 0 ~
a n d e i t h e r ex0 : a
such
that
o r else F ( ( Q .
for
all
:i < a0~)
j < a0
is n o t s u i t -
a b l e f o r Q , i.e., it is n o t a n a m e of a p o s e t in t h e f o r c i n g n o t i o n Pao" N ~
o
This t h e o r e m
is a n o b v i o u s c o n s e q u e n c e
of t h e s t a n d a r d
definition-by-
recursion theorem. 2.6 T h e o r e m : If P a n d Q a r e as in t h e d e f i n i t i o n of P * Q a n d P a n d Q N
N
e~
s a t i s f y t h e e.c.e., w h e r e b y "Q s a t i s f y i n g t h e c.e.c., " we m e a n [~-e " Q s a t i s f i e s t h e c.c.e. " t h e n P * Q s a t i s f i e s t h e c.c.c.
Proof:
L e t [ ~ p i , q ~ " i
c l a i m first t h a t t h e r e i s a p
aPsuchthatp
of c o n d i t i o n s in P * 9" We
[]-p"][i'pic
%][
=#1"
where
47
Gp is t h e
generic
][-p"]li:p,
e G p I ] =~o'"
subset
of P . Let
Suppose
this
is n o t
the
eP) r ]F"supli:p,
B=[r
case,
then
9 G ]=r
S i n c e e a c h i in t h e d o m a i n is c o u n t a b l e a n d li : p , 9 Gel is c o u n t a b l e in
V[Gp] also e a c h ~ E B is c o u n t a b l e in V[Gp] a n d h e n c e in V (since P s a t i s f i e s t h e c.c.c. Bfiap] = B ~ .
Since P s a t i s f i e s t h e c.c.c B is c o u n t a b l e
= Max B,~ < ~1.
P~+I I]- "~+1 c l i
Obviously
P~+l ] ~ - " s u p ~ i :p~ c Gp~ > ~"
on
the
:p~ c Gp] "
other
hand
, so l e t hence
we
have
]~p " s u p [ i : p~ 9 Gp] ~-- ~'" which is a c o n t r a d i c t i o n .
Nowweknowthat
forsomep
EPp
]~p " ] li " p i c GpI ] = ~l " let Gp b e a
g e n e r i c s u b s e t of P s u c h t h a t p c Up. Let A = [i : p , c Gel, t h e n t h e s e t lq~[Gp] : i c A | is a n u n c o u n t a b l e s u b s e t of Q[Gp]. S i n c e Q[Gp] s a t i s f i e s t h e c.c.c,
t h e r e a r e i , j c A s u c h t h a t q~[Gp] a n d q.[Gp] a r e c o m p a t i b l e , h e n c e ~a
t h e r e is a q[Gp] e A G e ] s u c h t h a t q[Gp] :* q.[Gp],q.[Gp].
Since p , , p j e Gp
a n d Gp c o n t a i n s c o n d i t i o n s w h i c h f o r c e q -~ q. a n d q ~- q. we have, b e c a u s e Gp is d i r e c t e d t h e r e e x i s t s a p E Gp s u c h t h a t p ~ p~,p -->p j , p P ]~-"q ~ q i " ~p,,g,~
thus (p,q~
~pj,qj~
c P*
Q and ~p,q~
] ~- "q m q~ " a n d
~ (P,,q~,_
tPj,qj~
a n d so
are compatible.
Within t h e p r o o f of this t h e o r e m we h a v e p r o v e d t h e following l e m m a . 2.7 L e m m a : L e t [pi : i
be
a
P
be
sequence
a poser of
which
members
satisfies
of P .
Then
the for
c.c.c,
and
let
some p E P
P ]~-P " t | i < ~1 :P~ ~ NGp]] = i l l " w h e r e NGp d e n o t e s t h e g e n e r i c s u b s e t of P. 2.8 T h e o r e m : If ~ Q. : i < a t N?,
e a c h i < a Qi s a t i s f i e s t h e c.c.c.
is a s y s t e m of FS i t e r a t e d f o r c i n g a n d f o r ( i.e., I~-P~ "O s a t i s f i e s t h e c.c.c.") t h e n P a
s a t i s f i e s t h e c.c.c. P r o o f : We p r o c e e d b y i n d u c t i o n on a. F o r a = 0 P0 c o n s i s t s of t h e null
48 sequence
only, hence
If a is a s u c c e s s o r
let
we h a v e b y II 2.4 Pa = P~+I ~ P p * P~,~+I ~' P~ * Q 9 By t h e ~p
a = fl + 1 t h e n induction
it satisfies, trivially, the c.c.c.
hypothesis
P~ satisfies the
c.c.c, the last theorem
establishes
c.c.c, and since also Q satisfies the ~~
that
a l s o P~ * Q s a t i s f i e s t h e c . c . c . ~~
The
relation ~ obviously preserves the c.c.c, hence P~ too satisfies the c.c.c.
So assume
a is a l i m i t o r d i n a l .
an uncountable assume
~Pa.
IDompi:i
f a m i l y of f i n i t e s e t s , s o w.l.o.g, b y L e m m a
I 3.10 we m a y
i t f o r m s a A s y s t e m w h o s e k e r n e l is s o m e f i n i t e s e t A i n c l u d e d i n a
and therefore by induction there
L e t l P i : i
i n s o m e ~ < a ( s i n c e a is l i m i t ) , b u t [ p t r ~ : i < wl] ~ P~+I h a s hypothesis
there
are two compatible
is q c P~, p ->p~r~, p j t ~ , b u t t h e n
numbers
Pi~$, p j t ~ , i.e.,
clearly Pi,Pj are compatible
in Pa.
e.g. q kJ (Pi t [ ~ , a ) kJ (P~" t [ ~ , a ) ) is a c o m m o n b o u n d .
w
Martin Axiom and few applications.
What
is t h e
Rowbottom
meaning
of MA ( M a r t i n
independently).
axiom
discovered
by Martin
and
I t s a y s t h a t we c a n f i n d q u i t e g e n e r i c s e t s i n s i d e
o u r u n i v e r s e . As we h a v e n o t e d b e f o r e ( s e e l 1.4 ), if P h a s n o t r i v i a l g e n e r i c s u b s e t (i.e., a b o v e a n y p e P t h e r e a r e t w o i n c o m p a t i b l e
member
of P ) t h e n
w e c a n n o t f i n d a g e n e r i c s u b s e t of P o v e r V. B u t we m a y well f i n d s u c h G ~ P g e n e r i c o v e r s o m e F t ~ V. So i t is p l a u s i b l e t h a t f o r a n y f a m i l y of < ~ d e n s e s u b s e t of P t h e r e is a d i r e c t e d
G c P n o t d i s j o i n t t o a n y of t h e m .
build a model V satisfying such a requirement?
How c a n we
We e x t e n d a n d r e - e x t e n d
the
u n i v e r s e , i n s t a g e a we e x t e n d t h e u n i v e r s e we h a v e g o t Va, t o Va+ 1 = V a [ G a ] b y f o r c i n g b y s o m e f o r c i n g n o t i o n Qa. T h e h o p e is t h a t in t h e e n d Vx = kJ V~ a
some
j,
imj
<X,
Qj = P .
r e q u i r e d ; so t h i s c o n s t r u c t i o n compact
So t h e
< i 0 > b e l o n g s t o s o m e V~, a n d will b e
as
is s i m i l a r in s o m e s e n s e , t o t h e c o n s t u r c t i o n
of
m o d e l s in m o d e l t h e o r y .
generic
object
Gj C P j
49 T h i s is c a r r i e d o u t b y i t e r a t e d f o r c i n g .
3.1 D e f i n i t i o n : ~ I a r t i n s ' s A x i o m f o r ~: MA~. If P is a p o s e r s a t i s f y i n g t h e c.c.c, a n d ~ ) i C P is d e n s e i n P f o r i < ~ t h e n t h e r e is a s u b s e t G of P w h i c h is d i r e c t e d s u c h t h a t f o r e v e r y i < ~ G CI D i # r
3.2 Remark: definition of
In order to have
MA~ it suffices to require that the
MA~ hold only for P such that IP] <- ~. To prove this let P be a
poser satisfying the c.c.c, and let D i c P be dense in P. For i < ~ let f, be a function on P such that forp c P tion on P x P
such
fi(P) ->P and fi(p) c Di. Let g be a func-
that for p,q c P
if p
and
q
are compatible then
g(p,q)>--p,q. L e t P 0 be any m e m b e r of P and let Q be the closure of IP0{ under the functionf,,i < ~ a n d g .
W e have ]Q] <-~. Let us see that D i N Q
is dense in Q. L e t p c Q t h e n f , ( p ) - p
,f,(p) cDi, a n d f i ( p ) c Q since Qis
closed under fi- N o w let us prove that Q satisfies the c.c.c . Let A be an antichain in Q; w e claim that A is also an antichain in P and hence IA ] -
g~,q)~-p,q, but Q is closed under g hence also g(p,q) c Q
and p,q are compatible also in Q, contradicting the assumption that A is an antichain in Q. By the version of
MA,~ for ]P I -< ~: there is a directed subset G
of Q such that G N D, C~ Q # r
<x.
This G is as required.
3.3 D e f i n i t i o n : M a r t i n ' s Axiom KA: (V~ < 2 8~ MAyo is t r u e s i n c e if f o r i < ~
MA r.
D~ is a d e n s e s u b s e t of P t h e n l e t u s p i c k
by induction a sequence
c Dn a n d p ~ + l -->p~, t h e n
: n < a~.
T h e r e f o r e we h a v e t h a t t h e CH i m p l i e s MA, H o w e v e r , u s u a l l y o n e m e a n s b y MAMA w i t h t h e n e g a t i o n of t h e CH.
3.4 T h e o r e m .
If t~0 < h = ~,<~ t h e n t h e r e is a P
t h e c.c.c, a n d s u c h t h a t II-p "2 ~~ = ~, ^ MA "
]P ] = A w h i c h s a t i s f i e s
so
P r o o f : F i r s t we s h a l l s h o w t h a t if k = ~,<x, [P{ -- k a n d P s a t i s f i e s t h e c.c.c,
then
]~-p " k = k <~''. S i n c e P s a t i s f i e s t h e c . c . c , a n d X > l~0, b y I 3.6(i)
is a n u n c o u n t a b l e
c a r d i n a l a l s o in V[ G].
To p r o v e t h e t h e o r e m from/~
we s h a l l g i v e a c a n o n i c a l
t o ~,, w h e r e /~ i s a c a r d i n a l
name
< ~,. T h e c a n o n i c a l
to every function
names
(for this con-
t e x t ) will b e a s follows: such a name has the form
w h e r e j~,n < k,
Pi.n e P.
For each P-name
T, I~-p "T a f u n c t i o n f r o m / ~ t o ~ " N
choose for each i
N
IP~,n " n < col c P ( p o s s i b l y w i t h
s u c h t h a t p~,n I~-/, " T ( i ) = J~,n " f o r s o m e Ji,n < h ( t h i s is p o s s i b l e
a s D~ = [p : p I~-P " T ( i ) = j " f o r s o m e j < A is a d e n s e o p e n s u b s e t o f P , a s P
s a t i s f i e s t h e c . c . c . ) ( if t h e m a x i m a l a n t i c h a i n
is f i n i t e we c a n c h a n g e n o t a t i o n
o r a r t i f i c i a l l y i n c r e a s e it). L e t
t h e n c l e a r l y I]-p "Tt = T". S o we c a n c o n s i d e r o n l y c a n o n i c a l T. W h a t is t h e i r number? from
f o r e a c h i < / z we c h o o s e
~,, s o we h a v e
<~I%]PII%;
(X#0[p[l%)~,~h~<~<x=~. return
and
one from P and one
so t h e
numbers
Now c l e a r l y in VP, ~ = k
of s u c h
names
for /~<~.
is
Now we
to the proof.
Let
(Saa ,:a
s e t s e a c h of c a r d i n a l i t y induction. defined.
a cardinal)
be a partition
k s u c h t h a t i c S , , ~ - * i -> a .
of ~ t o
disjoint
We s h a l l d e f i n e Q , b y
At s t a g e a a l l t h e Qp's , f o r fl < a , a r e d e f i n e d , a n d h e n c e Let (<~ : $ c
P a of q u a s i - o r d e r s
such canonical
names
of a c a r d i n a l
Sa,~)
P a is
b e a l i s t of t h e c a n o n i c a l n a m e s f o r t h e f o r c i n g
o f / z . We s h a l l u s e <~ in t h e ~ - t h s t a g e of t h e c o n s t r u c t i o n
but since for ~ c Sa, ~
order
two r
~
a this does not spoil the induction.
are there?
E a c h Qp, f o r fl < a , is a n a m e
/~ < ~, i n t h e f o r c i n g P t .
By t h e i n d u c t i o n
i P p [ ~ ~, h e n c e , a s we saw, t h e r e a r e a t m o s t k s u c h n a m e s . Qp t h e s e t of t h e s e n a m e s we g e t t h a t
Pa
How m a n y of a q u a s i hypothesis
If we d e n o t e w i t h
is a s e t of f i n i t e s u b s e t s of t h e s e t
51
U|lfl]
x Qp
" fl < a} which is of c a r d i n a l i t y < A a n d h e n c e
IP=] ~
},. A quasi-
o r d e r o n /$ is a f u n c t i o n on /z x # i n t o ~0,1}, a n d as we saw t h e n u m b e r of c a n o n i c a l n a m e s in t h e f o r c i n g Pa i s ~ A, s i n c e IRa [ ~ A. [<~,<~> where
i c Sa,~, aTzd We define now Qi = I I~-e~ "(]z,) s a t i s f i e s {< 1,~b> otherudse
the
C.C.C.
"
S i n c e f o r a <- i Pa C Pc e v e r y n a m e f o r t h e f o r c i n g Pa is also a n a m e f o r t h e f o r c i n g P~. It is n o w o b v i o u s t h a t f o r e a c h i < k ][-~ "Qi s a t i s f i e s t h e c.c.c, " s i n c e if t h i s d o e s n o t h o l d f o r < / z , < i >
t h e n we h a v e c h o s e n f o r Q~ < 1,~b>
w h i c h o b v i o u s l y s a t i s f i e s t h e c.c.c. T h e r e f o r e P = Px also s a t i s f i e s t h e c.c.c. Our a r g u m e n t a b o v e t h a t I P a l < - k w o r k s also f o r a = } , h e n c e
]PI--A.
It is
t r u e also t h a t 2 ~~ ~ }, since, as we shall see in t h e n e x t l e m m a MA# -------->2#~ > #.
Let I Z < k , a n d l e t A be a q u a s i - o r d e r in V[G] a n d l e t ~),, f o r i
<#, be
d e n s e s u b s e t s of A in V[G]. As we saw we c a n a s s u m e w i t h o u t loss of g e n e r a l ity t h a t
]AI = ~
member
of
D=~
(if ]AI < ]z we c a n i n t r o d u c e m a n y " c o p i e s " of a single
setting
each
:ieDj}c~x~.
of
them For
each
a
than pair
all
others
there
on
lZ).
Let
is a m a x i m a l
a n t i c h a i n _9,j in P w h i c h d e t e r m i n e s t h e m e m b e r s h i p of < i , j >
in A a n d a
m a x i m a l a n t i c h a i n ~ , j in P which d e t e r m i n e s t h e m e m b e r s h i p of < i , j > in ~). uIDomp hence
:p c h
U /~,j U Jc,j] is a s u b s e t of A in V of c a r d i n a l i t y ~. is
regular
(since
sup U I D o m p ]p C U ~ . j U~.,jl = 7 < k ,
h d x > A)
and
and J~,j,~,j cP,
k = A <x
therefore
f o r i , j Cl.~, a n d
t h e n a m e s A a n d ~ ) of A a n d ~) a r e n a m e s f o r t h e f o r c i n g PT" F o r t h e g e n e r i c s u b s e t G of P l e t G~ = [ P r 7 " p e G}. GTiS a g e n e r i c s u b s e t of P7" This c a n be s h o w n in a n y o n e of t h e following two ways. One way is to u s e t h e f a c t t h a t PimP7
*PT,x a n d t h e n G7 is t h e first c o m p o n e n t of G c P T * P T , x
a n d we
h a v e a l r e a d y p r o v e d t h a t it is g e n e r i c in PT" A n o t h e r way is to p r o v e d i r e c t l y t h a t G7 is a g e n e r i c s u b s e t of PT" S i n c e in e o m ) u t i n g A[G] = A a n d ~ ) [ G ] = ~) o n l y G , is u s e d we h a v e A[GT] = A, D I G s ] = D a n d
52
V[G~] ~ "A is a q u a s i - o r d e r on / l ^ ~ )
for i
A
satisfies t h e c.c.c. "' a n d t h e r e is a p 9 G7 s u c h t h a t p I~p7 "A is a quasio r d e r o n / ~ ^ A satisfies t h e c.e.c. "
=/ Let <
A if p e G7 / ~ t ~ otherwise
t h e n I~-p7 "A is a q u a s i - o r d e r in A ^ A satisfies the c.c.c. " T h e r e f o r e t h e r e is a $eS7, ~ such that <~=<.
Since $ e S 7 , ~ ~_>9, h e n c e
p e G7 Q [ G I ] = < l [ G ~ ] = A .
Q h e n c e G = GI+~/G~ is a g e n e r i c
PI+~ = P ~ *
s u b s e t of ~~[GI]Q = A o v e r V[GI].
G~D G7 a n d s i n c e
F o r i e I.z D i c V[GT] c ~ G I ] a n d ~5~ is
d e n s e in Q [ G I ] = A h e n c e G FI ~5i ~ r a n d G is a d i r e c t e d s u b s e t of A.
3.4. L e m m a " MA~ ~ !~ < 2~~
Proof: A s s u m e / z -> 2I%. Let P be t h e p o s e t of all finite f u n c t i o n s f r o m into 10,1], with p r o p e r i n c l u s i o n as t h e p a r t i a l o r d e r . F o r e a c h a e ~!0,1] let ~Sa = l P
eP:p
~a],
and
for
each
n <~
let D n = IP e P : n
eDompl.
Obviously e a c h ~)a a n d D n is d e n s e (and o p e n ) a n d t h e r e a r e L~~< / z s u c h set. By MA~ t h e r e is a d i r e c t e d s u b s e t G of P s u c h t h a t G f~ D a # ~b f o r e a c h a co~0,1] and G Clan
# 0 for e a c h n
<~.
Since G is d i r e c t e d g
= [3 G is a
f u n c t i o n s i n c e G f~ D m # 0 n e D o i n g for e a c h n < a, h e n c e g e ~ 0 , 1 ] . G (3Dg = r
since for e v e r y p
e G p c L/ G = g
hencep
~Dg,
Now
b u t this c o n -
t r a d i c t s G F%D g # ~ which we g e t by MA~.
Some applications of MA + 2 ~~ > ~1. 3.5
Theorem
Assume
MA a n d
let
k
be
a
cardinal
N0 - < k < 2 I%
< A t : i < A> a family of infinite pairwise a l m o s t disjoint s u b s e t s of r (i.e., if
i ~j,i,j
< A t h e n At CIAj is finite), a n d let S be a s u b s e t of h. T h e r e is a
function f
o n ~ into 10,1] s u c h t h a t for all i < k
f t a t =a~ l& i f f i e S ,
w h e r e l& is t h e f u n c t i o n on At with t h e fixed v a l u e 1 a n d =ae d e n o t e s t h a t two f u n c t i o n s h a v e t h e s a m e v a l u e s f o r all n u m b e r s in t h e i r d o m a i n e x c e p t
53
finitely many. Proof: Let P=
~f : f
some i 1 .....
is a f u n c t i o n
such that
Dom(f)
=Aq U "'"
/~ e S a n d a finite W c co a n d f o r 1 ~ g -- n
t J A ~ (3 W f o r
f t Ar ==e 1A~].
T h e p a r t i a l o r d e r o n P is p r o p e r i n c l u s i o n .
Since, for each f
6 P , f - l [ l O ] ] is finite we c a n t a k e f o r e a c h f 6 P a finite
F/f D f - l [ 1 0 1 ] t o p l a y t h e r o l e of W in t h e d e f i n i t i o n of P . To s e e members
that
of P .
P
satisfies
the
c.c.c, l e t < f i
"i <~l>
be a s e q u e n c e
of
S i n c e all t h e f t Wf b e l o n g t o t h e c o u n t a b l e s e t of all finite
f u n c t i o n s f r o m co i n t o | 0 , 1 | we h a v e i ~ j , i , j < Sl s u c h t h a t f i t ;ff, = f j t
Wfj.
O b v i o u s l y f ~ ( j f j is a f u n c t i o n a n d a m e m b e r of P, h e n c e f i a n d f j a r e c o m patible.
We s h a l l s p e c i f y k d e n s e s u b s e t s of P : such
that
~5 N G # ~b f o r e a c h
one
By MA t h e r e is a d i r e c t e d G c P
of t h e
specified
dense
sets D.
Let
g = [J G . S i n c e G is d i r e c t e d e v e r y two m e m b e r s of G a r e c o m p a t i b l e a n d g is a f u n c t i o n f r o m a s u b s e t of c0 i n t o 10,11. We e s t a b l i s h n o w t h e following p r o p e r t i e s of G. 1. D o r a ( g ) = w . is
dense
since f o r
G • ~)n ~ #
For n e~
f EP
there
is
if f
an
let ~ n
~n
= If c P:n
then
f
6Dom(f)l.
kJ l
f e G
such
that
then
g r A~ =ae 1Ac
rL 9 D o m (f),
~,~
Since hence
D o m (g) D Dora ( f ) 9 n . 2.
If
i e S
~)* = I f e P : D o r n ( f ) Dora(f) If
=Aq U""
i e Ill .....
Ar (3 D o m ( f ) c
DA~{. To s e e t h a t D i uA% U Wwherei 1 ..... in]
then
f
eDi.
For
i e S
is d e n s e in P let f /~ e S a n d If
let
e P then
Wisfinite.
i ~ li~ . . . . .
in~
then
LJ Ar (3 A~ tJ W a n d e a c h s e t p a r t i c i p a t i n g in t h i s u n i o n is
54 finite.
It f o l l o w s i m m e d i a t e l y n o w t h a t f
member
of P i s
f tAt =aet&
obviously inDt.
and sinceg 3.
members
Df
g r At o b t a i n s
L e t ~)~,~ = I f
t h a t ~)~,n is d e n s e l e t f
i ~S,
i ~ Ill . . . . .
Since
At
is
We h a v e n o w G ( S D t
9
there
is
(m ~n
f(m)
there
is a n f
9 G (5~)~,
^f(m)
=01.
a s we s a w a b o v e , D o m ( f ) an
m
9 G such
To s e e
b e a s i n t h e d e f i n i t i o n of P .
9
that
(sAt
such
f U I(m,O>{ c P hence f U [(m.O>{ 9 D n . ~ (5 G # r
Let f
the value 0 for infinitely many
(:~rn 9
9 P, and let f
i n], t h e r e f o r e ,
infinite
#r
g ~tl/ =ae l&.
If i ~ S t h e n
of At.
U (At - D o m ( f ) ) x ~i I 9 P a n d t h i s
that
Since
is f i n i t e . m ~n.
and Dn,, is dense. Since there
is a n m
9
m->n
and
= 0, h e n c e g ( m ) = 0. T h u s g ( m ) = 0 f o r a r b i t r a r i l y l a r g e m 9 A t .
I t is n a t u r a l t o a s k Q u e s t i o n 3.6: I n T h e o r e m that f rAt =~ this later.
1A~ f o r i 9
3.6 c o n d i t i o n s ,
is t h e r e
a n d f t A t =ae 0& f o r i
a n f :~ -* ~0,11 s u c h
r
We s h a l l r e t u r n
to
Note however
3.7 T h e o r e m = I n T h e o r e m
3.6 w e c a n o m i t t h e r e q u i r e m e n t
"At 9 w ' "
r e q u i r i n g o n l y ~ I = I%. P r o o f : We l e t P
=
~|0,1~
I f : f is a f u n c t i o n w h o s e d o m a i n is At, U " " " U A~ a n d w h o s e r a n g e is for some n < ~
i 1 9 S .....
i n 9 S, f-l[[0~]
finite I
ordered
by
inclusion If p i 9 P ( i <1~1) ~l
"~S0 f o r i
are
Sl-conditions,
< ~t, ]A'l =sl, and letg
let
At = u D o m p j ,
be a one-to-one
A'=
U At, s o
function fromA*
onto
~1. D e f i n e a f u n c t i o n h o n ~1
h ( a ) = I g ( i ) : i 9 D o m p a , p a ( i ) = 0] Now C = li:g m a p s At o n t o i |
is a c l o s e d u n b o u n d e d
subset
of c~
and for
a 9 C, h ( a ) is a f i n i t e s u b s e t of a . H e n c e b y F o d o r ' s l e m m a f o r s o m e a # fl i n
55
C, h ( a ) = h(fl), h e n c e c l e a r l y Pa U P ~ e P, a n d trivially Pa,Pp ~ P a UPa" So P s a t i s f i e s t h e c.c.c. Now
if
i < A,
i g S,
At N ( U Aj) is infinite, 2'gs ~[~,n = IP e P:A~ N P - l ( | 0 ] ) h a s c a r d i n a l i t y m n ] is a d e n s e s u b s e t of P.
then
So t h e r e is a d i r e c t e d G c P, G n D~,n ~ r f o r i < A, i g s , n < co a n d let f~ be
uf,
andf+be
a n e x t e n s i o n of f * b y f + ( ~ )
=0 for~e
UAt~
f~G
UAt~ES
It is easy to c h e c k f + is as required. A p a r t i a l a n s w e r to the Q u e s t i o n 3.6 is
3.8 Definition: A s e q u e n c e
< A t : i < A>
of infinite
pairwise
almost
d i s j i o n t s u b s e t of r is called a tree if f o r a n y a < A [A a N Ap:~ < AI is l i n e a r l y ordered
by
inclusion,
i.e.,
for
any
Aa N A p ~ A a n A7
~,7
or
Aa n A7 c A a N Ap. An e x a m p l e of a t r e e of 2 t% s u b s e t s of t h e s e t of all finite s e q u e n c e s of O's a n d l ' s is t h e s e t i T$ :j" e ~ 2 ] w h e r e
rf =If
rn :n e~l.
3.9 Theorem= A s s u m e MA, let X be a c a r d i n a l s u c h t h a t ~I0 < k < 2 I% a n d l e t < A t : / < k > be a f a m i l y of infinite p a i r w i s e a l m o s t d i s j o i n t s u b s e t s of which isatree.
Let S C k t h e n t h e r e is a n f : w - *
f tAt =ae l & a n d f o r a l l i
ek-Sf
10,1] s u c h t h a t f o r a l l i e S
r A t =ae 0&.
P r o o f : Let P = I f " f is a f u n c t i o n a n d f o r s o m e i 1 . . . . . of ~ D o m ( f )
=At, U""
uA~
/n < A a n d a finite s u b s e t W
U W, f tAt~=ae l & ~ w h e n i ~ e S
t h e n f r At~ =ae 0A~. The p a r t i a l o r d e r on P is p r o p e r inclusion.
and ifi~igS
56 Let
us
prove
sequence
now
of
that
that
the
e.c.e.
of
Let ~f~ :i
P.
For
upper
a number
bound
of
Wi a n d f o r e a c h
m s u c h t h a t fr
= fi(m)
be
f~
U Aa(i,n(i)) U Wi, a ( i , 1 ) , a(i,2) - - . <~.
k~ is a s t r i c t
A=(i,e) N kr c o n t a i n s
c Aa(i,s
satisfies
members
D o m ( f ~ ) : Aa(i,1) U " " such
P
a let
L e t k~ E co b e
g, 1 - < g ~ n ( i ) , for every number
T h e r e is s u c h a k i s i n c e e a c h fr t Aa(i,g) h a s t h e
w h i c h is > m .
value 0 almost everywhere or the value 1 almost everywhere. We
demand
in
addition
that
for
1 --< g,
m --< n ( i ) s ~ m ,
Aa(i,s (-1 let ~- Aa(i,m) ('~ ki a n d Aa(i,~) ('~ Aa(t,rn ) ~ ]c~ w i t h o u t l o s s of g e n e r a l i t y we c a n a s s u m e t h a t f o r all i
iisfixe
f o r i
a n d Wt is f i x e d . We s h a l l n o w p r o v e t h a t f o r i # j i , j
are compatible.
obviously belongs there W~ :
is %. t i c
a
Consider the union ft to P. Suppose
fl ~ W~ : ~.,
CA i : i < t~l) therefore
is a t r e e
m = Max
and fr r k : fjt
such
hence
f ~ is f i x e d f r o m t h a t
U fj.
for
that
some
<--Slfr a n d
If t h i s u n i o n is a f u n c t i o n i t
it is not a function.
~ -> k (6 < ~ )
fl c Aa(j,~.t). L e t m = M a x that
C~k is f i x e d f o r f i x e d g a n d
S i n c e fr t k = f j t
f ~ (~) # f j (fl).
1-
]c
Since
flcAa(~,~)
and
Aa(i,~) N k, s i n c e / c f~ Aa(i,s c o n t a i n s a n u m b e r s o number
upwards
a n d fl < Aa(i,~. ) ~
we h a v e f i ( f l ) =
Aao.~), fl ~ k,
: fj(m)
Since
we h a v e g : s 1 a n d
Aa(j,~t) ~ k a n d , a s f o r i, fj(fl) = f j ( m ) .
]c we h a v e f i ( m )
fr
Since m < k
h e n c e fi(fl) = f j ( f l ) , c o n t r a d i c t i n g
o u r c h o i c e of ft.
L o o k i n g a t t h e p r o o f of T h e o r e m
3.5 i t is c l e a r h o w t o c h o o s e d e n s e s u b -
s e t s ~ ) of P s u c h t h a t if G is a d i r e c t e d s u b s e t of P w h i c h i n t e r s e c t s of icS
them
then
g = U G
has
g rAr =ae 1 A ~ f o r i c ~ - - S g
the
following
properties:
Dom(g)
each one =co
r A / =ae 0 ~ .
3 . 1 0 T h e o r e m : In 3.9 "A~ c ~ " i s n o t r e q u i r e d , p r o v i d e d ]A~ I - S0. T h e p r o o f is a n a l o g o u s t o t h e p r o o f of t h e o r e m
3.7.
for
57
w The uniformization property.
L e t u s p r e s e n t n o w t h e s e t t i n g with w h i c h we s h a l l d e a l with t h e p r o b l e m 3.6. 4.1 D e f i n i t i o n : L e t A = ~ A~: i < a t is c o u n t a b l e a n d infinite. if
for
every
family
of
b e a s e q u e n c e of sets, e a c h of w h i c h
We s a y t h a t (A,h) h a s t h e u n i f o r m i z a t i o n p r o p e r t y functions
(fi
:i
such
that
fr
On
f ~ ( a ) < h ( a ) , t h e r e is a f u n c t i o n f : k / A / -, On s u c h t h a t f o r e a c h i < a f ~ is i<=
a l m o s t i n c l u d e d in f denote
(i.e., i n c l u d e d u p t o a finite n u m b e r of v a l u e s ) .
We s h a l l
kJ At with D(A) , so D o m h = D(A). If h is c o n s t a n t l y / ~ we w r i t e (A,/~)
if / ~ - - 2 w e
writeA.
We shall s a y t h a t A i s
a t r e e if t h e r e is a p a r t i a l o r d e r <
on 0(4) such that: a) ~ D ( A ) , < >
is a t r e e s u c h t h a t f o r all x e . D ( A )
lY : Y < x ] i s a f i n i t e
set linearly ordered by < b) F o r e a c h i A~ is a b r a n c h i n / ) ( A )
, i.e., if x 9 Ai a n d y < x t h e n a l s o
y 9 Ai. S i n c e e a c h A~ is infinite a n d s i n c e b y a) ( D ( A ) , < ~
is of h e i g h t -----co,
e a c h A~ is a m a x i m a l b r a n c h , i.e., it g o e s all t h e w a y up.
We s h a l l d e n o t e ~ D ( A ) , < ~
Tn = I x " I|Y : Y < x l
=nl.
with T . Tn will d e n o t e t h e n - t h
For a branch
level of T,
A we d e n o t e with (A)[~] t h e o n l y
m e m b e r of A in t h e level n , i.e., t h e o n l y m e m b e r of A (3 Tn. 4.2 T h e o r e m = If A = ( A ~ : i < a ) , a ~- co1, is a t r e e in t h e a b o v e s e n s e a n d if D(A) c ~>2 t h e n A d o e s n o t h a v e t h e u n i f o r m i z a t i o n p r o p e r t y . P r o o f : L e t u s d e f i n e t h e f u n c t i o n f i o n A~ as follows. f/((A~)[n]) = ( A / ) [ n + l ] ( n ) . We n o t i c e t h a t (Ai)[n ] is a s e q u e n c e of l e n g t h n . v a l u e of f i
for this member
of D(A) tells u s w h i c h w a y t h e b r a n c h
(A/)[n ], r i g h t o r left, s i n c e (Ai)[n.1] = ( A ~ ) [ n ] . " ~ ( A r
= (A~)[n]^~fi((A~)[n])).
The
goes at
58
A s s u m e n o w t h a t t h e r e is a [ u n c t i o n f for
some
uncountable
y~(',') = f ( x )
set
I ~ a.
u n i f o r m i z i n g all f j ' s , o r e v e n all f ~ ' s
For
each
f o r e v e r y x c A~ of level > n ( i ) .
i c I let n(i)
be
such
S i n c e I is u n c o u n t a b l e
that
t h e r e is
a n ~z* < ~ s u c h t h a t li c I : n ( i ) = n ~ is u n c o u n t a b l e , we d e n o t e t h i s s e t w i t h 11. S i n c e D(A) h a s a t m o s t 2 n" m e m b e r s
of level n ~ t h e r e is a z c T,~ s u c h
t h a t ~i c I x : (A~)[n. ] = z ] is u n c o u n t a b l e ;
we d e n o t e t h i s s e t with I 2. F o r all
i , j ~ I 2 we h a v e (As)in, ] = z = (Aj)[n, ] a n d s i n c e A~ a n d Aj a r e b r a n c h e s h a v e (As)I~ ] = ( A j ) [ ~ ] f o r a l l k - < n ~
Fori
#j
i,j cA eA i andAj
we
are different
b r a n c h e s h e n c e t h e r e is at l e a s t s ~ ~ n ~ s u c h t h a t (A~)[e+i ] # (Aj)[e§ ] s i n c e s < n ~ = r~(i) = n ( j )
we
have
f~((A~)[e]) = f ((A~)[~])
and
fj((A~.)[e] ) = f ((Aj)[e]). By t h e d e f i n i t i o n of f~ a n d f j we h a v e l =
(I
=
((&))[~]~ = (A~)lel^(f((A~)[e])~
( s i n c e (&)[el = (A;)[e] b y t h e m i n i m a l i t y
of s + 1 ) = (Aj)[~] ^ ( f ~ ( ( A i ) [ e ] ) ~ = (Ai)l~+l ], c o n t r a d i c t i n g (A~)[e+l ] # (A~)[~+l]. 4.3 T h e o r e m = A s s u m e MA. S u p p o s e A = ~ A~:i < ~
is a f a m i l y of p a i r -
wise a l m o s t d i s j o i n t c o u n t a b l e s e t s a n d ~, < 3 n~ T h e n (A,lh0) h a s t h e u n i f o r m i zation property provided that: (*) f o r e v e r y c o u n t a b l e A, li < k:A F~ Ai is infinite I is c o u n t a b l e . P r o o f : L e t f~:A~ -* ~ f o r i < A, a n d we shall find f ,
almost extending
e a c h 3'~. L e t P=
~f:
there
are
/~ . . . . .
/~ < A
such
that
R a n g e ( f ) c ~, jr a l m o s t e x t e n d f ~ f o r s = 1 . . . . .
Domf
=A h U"'uA~,
n|
ordered by inclusion. It s u f f i c e s t o p r o v e t h a t P s a t i s f i e s t h e c.c.c., a n d p r o c e e d a s in e.g. 3.5.; b u t t h i s is n o t h a r d ( p r o o f s i m i l a r t o 3.9). 4.4 D i m c u ~ i o n . T h e o r e m 4.2 a s s e r t t h a t if D(A) is a t r e e w h i c h a t e a c h node branches
i n t o a t m o s t two b r a n c h e s
then A does not have the uniformi-
z a t i o n p r o p e r t y ( t h e f a c t t h a t in t h e t h e o r e m D(A) was a c t u a l l y a s u b s e t of Q2 is o b v i o u s l y i r r e l e v e n t ) .
The a s s u m p t i o n in t h e t h e o r e m
t h a t D(A) is a t r e e
59 c a n be r e p l a c e d b y t h e following w e a k e r a s s u m p t i o n . F o r e a c h i < a let A/ be given as a s e q u e n c e of t y p e ~0 which we write as ( a i , 0 , a t A , . - . )
= At . The
w e a k e r h y p o t h e s i s is t h a t A is u n c o u n t a b l e a n d f o r e a c h s e q u e n c e ( b 0 .....
bn_1)
:
of
members
. . . . .
has
. . . . .
f o r all At given ar
.....
D(.4)
of
the two
at most
set embers.
~ . n - 1 t h e r e a r e a t m o s t two possibilities for ar n.
(This a s s u m p t i o n obviously h o l d s if D(A) is a t r e e with b r a n c h i n g as a b o v e a n d e a c h s e q u e n c e (o~,o,acA , 9 9 9 )
is s u c h t h a t ar n = (A,)[~] f o r all n < co).
We shall see t h a t also u n d e r this w e a k e r a s s u m p t i o n A does n o t h a v e t h e uniformization property.
Let 3 be as above. For e a c h i < a B = (B~ :i < a). most
two
let Be = | ( a , , 0 . . . . .
ar
: n
B is obviously a t r e e w h i c h a t e a c h n o d e b r a n c h e s to at
branches,
and
for
i # j B~ ~ t7t
since
At ~ A1.
Let
: A:B~ ~ ~0,11> be a farmly of function which c a n n o t be uniforr~zed, by
the
theorem.
We
denne
gi(aLn ) = j',(ar 0 . . . . .
a
fa~ly
as
a~,=). We shall see t h a t this family c a n n o t be u n i f o r -
mized. S u p p o s e g : D ( A ) -* ~0,11 u n i f o r m i z e s this family. Define f follows,
f((a~, o .....
a n y value. i
at.,_1>)
We shall see t h a t f
=g(a~,,~_ 0 if n < 0 a n d f ( ( ~ )
by .....
the :
we have, f o r all n > k
definitions
of
o n D(B) as c a n be given
is a n u n i f o r m i z a t i o n of ( f , :
we k n o w t h a t f o r s o m e k
hence,
follows.
i < a>.
gr
Given
=g(a%n )
gr and
f
. . . . .
Let u s r e m a r k now t h a t t h e t h e o r e m is t r u e also f o r a t r e e
D(A) w h i c h h a s
a b o u n d e d b r a n c h i n g a t e a c h n o d e (and h e n c e also in t h e m o r e g e n e r a l c a s e m e n t i o n e d above). We shall show it h e r e f o r t h e c a s e w h e r e
D(A) ~ ~>q, w h i c h
is sufficiently g e n e r a l to exhibit t h e p r o o f of t h e g e n e r a l case. F o r a n a t u r a l n u m b e r k let
d~(k) d e n o t e t h e c o e f f i c i e n t of 2 ~ in t h e b i n a r y e x p a n s i o n of k,
de.(k ) E 10,1~ . We define now on t h e b r a n c h Ai~ b y s e t t i n g fr162
= d~((At)[n+l](n)), for t~ = 0,1. Let f ~ be t h e f u n c t i o n
uniformizing the functions ~f~ thatlca,
II[ >N0
two f u n c t i o n s f o a n d f l
(At)[n- ] = z
"i < a )
for g = 0,1. Let
fori EIandf~(n)
n~
=f~(n)
a n d I be s u c h forn>n
~
60 the proof of the theorem).
W e get n o w a contradiction f r o m the fact that for
i,j e [ i ~ j there is a least level w h e r e .4/ a n d Aj go to their separate ways, while at e a c h n o d e
(from level r~* and above ) f 0 a n d f i determine
com-
pletely the w a y the b r a n c h goes in the next level so A t a n d A 3. m u s t go the s a m e way.
We
deal
here
exclusively
with
trees
A
and
other
systems
where
ID(A)I =M0. In 3.8 w e dealt with a different definition of a tree, n a m e l y w e called a family Ir of subsets of ca a tree of for all x,y 9 ICif n 9 z ~r~ = y
F l Y then
t~z. In this case let us define <* on ~ by /e < ~ 1 6 3if /e < s
every A c 6) if IZ e A
then also k 9
a n d for
It is easily seen that <* is a partial
order; the set I/c : k < * h i is linearly ordered by <* which coincides with < o n this set a n d that each A c ca is a b r a n c h in the tree
T h u s a tree in
the sense of 3.8 is also a tree in the present sense.
Now l e t A b e a t r e e a s d e f i n e d h e r e w i t h ]D(2~) I = t~0 w i t h <* a s t h e p a r t i a l order
relation
numbers
of t h e t r e e .
in s u c h a w a y t h a t if k < * ~
a n d k <~z k c A ~ . r~ < * k
Identify the members
hence
a n d k <* n we g e t a l s o k e Aj. A~ f~ n = Aj N n .
t h e n k <JZ. N o w s u p p o s e n 6 A ~ N A j
SinceA~ isa branch
would imply n
of D ( A ) w i t h t h e n a t u r a l
of t h e t r e e we h a v e k
k <'~z.
<*r~ orn
S i n c e a l s o Aj is a b r a n c h ,
<*k. n cAj
S i n c e t h i s w o r k s i n b o t h d i r e c t i o n s we h a v e
T h u s a t r e e in t h e p r e s e n t
s e n s e is i s o m o r p h i c t o a t r e e in
t h e s e n s e of D e f i n i t i o n 3.8.
4.5
Theorem:
A~ = [a~j:j < ID(A)I :~0.
~]
be
Assuming a
sequence
have
that
la~,n : < a , , o . . . . . members.
+ 2 I% > 8z.
for
As a c o n s e q u e n c e
since A? does not have it.
A : < A ~ : i < t~l>,
infinite
sets
such
that
subset ~t c A which has the pro-
o r d e r i n g of e a c h A~, i < t~1, of o r d e r t y p e
every
%,n-,>
Let
of t~1 c o u n t a b l y
T h e n t h e r e is a n u n c o u n t a b l e
perty that for an appropriate we
MA
=
sequence
.....
bn_,>,
< b 0 ..... bn_1> i <S,]
has
at
A does not have the uniformization
the
set
most
two
property,
61 P r o o f : G i v e n 24 we o r d e r e a c h m e m b e r of A i n o r d e r t y p e ca a n d d e n o t e g
Jai,o,ai, l, 9 9 9 ~.
with
At
Let
P = {W C I~1 9 W
is
finite
and
for
all
bn-l>
#
a n d P is p a r t i a l l y o r d e r e d b y s t r i c t i n c l u s i o n . By d e l e t i n g a t m o s t ~o m e m b e r s of A we c a n m a k e s u r e t h a t f o r all i < Sl
II~ <.~ : <~j.o . . . . .
~ ; . ~ - $ :
~.~-~>! I - - ~
[For every finite sequence < b 0 .....
b ~ _ l > of m e m b e r s of D ( A ) ,
are
if
S0
such
= < b0 . . . . .
sequences,
and there
0 < I IJ < ~1: < aj,o . . . . .
aj,r,-l>
b ~ - l > l l <-s0 we delete all t h e A i ' s w i t h j i n t h i s s e t , so a l t o -
g e t h e r we d e l e t e d c o u n t a b l y m a n y s e t s . ]
L e t u s s e e n o w t h a t P s a t i s f i e s t h e c.c.c. L e t < Wf : t < l~l> b e a s e q u e n c e of m e m b e r s of P . number
k
r = 1 .....
such
n(t)
that for allt
s,t <~i Ws U
that
=n
ar
we
< a#,o,a#, 1 . . . . .
sequences
and for eachr
have
to
hence show
<%0
Ws that
and for
Wt all
exactly
m = 1 .....
n < a~, 0 .....
a~i,/r
these
= 1 .....
n the sequence
(~,.o . . . . .
~,.~_,> = < b o . . . . .
of all At f o r i e Ws a r e
m + 1
hence
a,,m_l> = < b o . . . . .
m e m b e r s s i n c e Ws e P .
are
compatible.
sequences
To
for
prove bin_l>
ai,rn_l,ai,m>
tuples
for
I~
of t h e A t ' s f o r
i e Ws
since
for
ai&,~_,> =
for
..... =~,~-i> = < 0 o ..... Om-~>ll <2.
m < k t h e n t h e i n i t i a l m + 1 - t u p l e s < at, o . . . . . are
ar
W i t h o u t l o s s of g e n e r a l i t y we c a n a s s u m e
k(t) =k
and
li,~,,~ :~ c w~ u w~ and i e Ws U Wt
be the least
is f i x e d f o r all t
Ws U We c P
Wt e P
the
a r e all d i f f e r e n t .
/ ~ ( t ) ] - Let k ( t )
Let u s w r i t e Wt a s litl ,i t . . . . .
iai,m : i E Ws U Wt
and
= la~.rn : i e Ws
and
bin_l> I
b ~ - l > I , and the latter set has at m o s t two
If m ->/c t h e n s i n c e t h e i n i t i a l s e q u e n c e s of l e n g t h /r all d i f f e r e n t , a n d t h e
s a m e h o l d s f o r all i e Ift
b i n _ l > c a n b e a n i n i t i a l s e q u e n c e of At f o r a t m o s t o n e i e Ws a n d
62 at
most
one
i 9 Wt
bm_l>]<2
(bo . . . . .
b
Fori
so
hence
I Ii 9 Ws U Wt : < a~.o . . . . .
that
CQ.m-1) =
[[a,. m : i 9 If, U If,. a n d < a , . o . . . . .
a,.m_,)
=
_,)ll _-2
prove that~
b e t h e s u b s e t of P , ~
is a d e n s e s u b s e t of P .
s e q u e n c e s < a~,0,a~, 1 . . . . .
=[WcP:(aj
Let IfcP
>i)j
~_ W]. L e t u s
a n d let lc b e s u c h t h a t t h e
a~,a_l> a r e all d i f f e r e n t f o r d i f f e r e n t i 9 If. T a k e a
f i x e d i 9 If ( t h e c a s e w h e r e If = 0 is trivial ). By o u r a s s u m p t i o n t h e r e a r e ~1 j's such that
aj.k-1)
= 4ar
.....
a ~ . ~ _ l ) , h e n c e t h e r e is s u c h
a j > i. We s h a l l s e e t h a t If D ~J ] 9 P , t o p r o v e t h a t we h a v e t o s h o w t h a t f o r
b,n_l>
all < b o . . . . .
(*) I~a~,m : i c W U I J l a n d
a~,m-1) =
.....
bm-l>]'~2-
If m < k t h e n s i n c e t h e first k m e m b e r s of Aj a r e t h e f i r s t k m e m b e r s of A~, w h e r e i 9 If, t h e l e f t s i d e of (*) r e m a i n s u n c h a n g e d
if If U ~J~ is r e p l a c e d b y
W a n d s i n c e t h e i n e q u a l i t y h o l d s f o r If, a s If 9 P , it h o l d s also f o r If U IJ ]- If m >- k t h e n , s i n c e t h e s e q u e n c e s
i 9 If
there
is
(a~, 0 . . . . .
a~,n_l) = < b o . . . . .
at
a i . ~ _ l > a r e all d i f f e r e n t f o r
most
b,n_l>,
hence
one
i 9 If
there
are
two
such i
that Ifu|Jl
w h i c h s a t i s f y t h i s e q u a l i t y a n d (*) follows i m m e d i a t e l y .
By MA t h e r e is a d i r e c t e d s u b s e t G of P w h i c h i n t e r s e c t s e a c h D r . fore I= have
(ar
t J G is a c o f i n a l s u b s e t o f R 1. We t a k e n o w A = 4 A ~ : i e I ) . to
.....
prove
that
=i,m-,)
=
( b0 .....
bin_l)
Since I=
uG
Then
we
(bo .....
for
all
( b 0 .....
bm_,)~]
this set has three
and
G is d i r e c t e d
have
(b 0 ..... bin_l) <2.
members
that
for
and
some
aQ,m,a~s,m,a~s.m, i:,i2,i 3 c I .
we [:ave il,i2,i 3 c W f o r s o m e
]la~.m : i c W
We still
]lat.m:iEI
Assume
There-
and
< ai. 0 . . . . .
Wc G cP. ar
=
b i n - l ) ~l ~ 3, c o n t r a d i c t i n g If c P .
4.6 T h e o r e m : It is c o n s i s t e n t
that
there
is a t r e e A = 4 A a : a < ~ I >
s u c h t h a t I D(A) ] = I~0 a n d A h a s t h e u n i f o r m i z a t i o n p r o p e r t y .
63 Proof: Let
T be a tree,
IT] =~0, Tn is f i n i t e f o r n
x ~_ Tn ] lY c Tn+l : y >x]] >_2nz. We s h a l l o b t a i n a g e n e r i c e x t e n s i o n o f which there
is a t r e e A of l e n g t h ~1 w h i c h h a s t h e u n i f o r m i z a t i o n
We s a w a b o v e t h a t if t h e b r a n c h i n g does not have the uniforlnization
of a t r e e a t e a c h n o d e is b o u n d e d property.
~z e a c h n o d e b r a n c h e s ' , t o 2 n~ b r a n c h e s
Vin property. the tree
H e r e we s h a l l s e e t h a t if a t l e v e l
the tree may have the uniformization
p r o p e r t y . T h e r e is a g a p h e r e w h i c h o n e s h o u l d n a r r o w o r e v e n e l i m i n a t e .
First one introduces
t~1 g e n e r i c b r a n c h e s
of T b y t h e f o l l o w i n g s e t Q0 of
conditions Q0 = I f : f is a f i n i t e f u n c t i o n o n r
x r such that:
• D o m ( f ) -* f ( a , n ) e rn ^
t,J G , A a = I F ( a , n ) ' n
andA=
the tree which has the uniformization
property
a,~<~ 1
subset
A,t#A#,
since
the
If G is a g e n e r i c s u b
( i t is o b v i o u s t h a t f o r a # fl,
If CQo:~n[
c D o m ( f ) ^ f (a,n) # f ( f l , n ) ] ] o f Q0 is d e n s e in Q0)-
T h e n e x t s t e p is t o c a r r y o u t a n i t e r a t e d different system g =
i < r
g,:A~ -, 10,1] g e t u n i f o r m i z e d . E v e n t u a l l y
we obtain that A has the uniformization a s i n g l e s t e p of t h i s i t e r a t e d
f o r c i n g so t h a t i n e a c h s t e p a
property.
We s h a l l d e s c r i b e f i r s t o n l y
f o r c i n g . T h e c o n d i t i o n s w e u s e n o w a r e t h e fol-
l o w i n g . L e t g b e t h e n a m e of a s y s t e m of f u n c t i o n s a s a b o v e i n t h e f o r c i n g Q0We d e f i n e Q ( g ) = I h : h is a f i n i t e f u n c t i o n f r o m t ~ 1 t o co ^ Va,~ff c d o r a ( h ) t h e f u n c t i o n s g , , r | x c A a l h e i g h t ( x ) -> h ( a ) l a n d g ~ t ~x E A#] h e i g h t ( x ) -~ h ( f l ) ] a r e c o m p a t i b l e The partial order relation
o n Q ( g ) is p r o p e r
inclusion.
].
T h i s f o r c i n g is d o n e
o v e r V[G0], w h e r e G O is a g e n e r i c s u b s e t of Q0. L e t G b e a g e n e r i c s u b s e t of Q(g) and H = (JG.
We. c a n n o w d e f i n e t h e u n i f o r m i z i n g f u n c t i o n g:T --, ~0,1]
64
a s follows. F o r x 9 T if x 9 A a a n d h e i g t h (x) -> H(a) t h e n g ( x ) = ga(x); if x b e l o n g s t o n o A a s u c h t h a t H(a) <- h e i g h t ( x ) we d e f i n e g ( x ) a r b i t r a r i l y .
If
x 9 Aa, h e i g h t (x) -> H(a) a n d a l s o , f o r fi # a, x 9 A~ a n d h e i g h t ( x ) -> H(fi) t h e n s i n c e G is d i r e c t e d a n d H = UG t h e r e is a n h c G s u c h t h a t a,fl 9 d o r a ( h ) , . h(a) = H(a) a n d h(fl) = H(fl). S i n c e h 9 Q(g), h(a),h(fl)
It is o b v i o u s f r o m t h e d e f i n i t i o n of g t h a t i t u n i f o r r n i z e s all ga's f o r
I.
L e t u s c o n s i d e r n o w t h e s e t Q0 * Q(g) a n d p r o v e t h a t it s a t i s f i e s t h e c.c.c. Let W be a n u n c o u n t a b l e
E a c h < p , q > c Qo * Q(g) is
s u b s e t of Q0 * Q ( g ) .
f i r s t e x t e n d e d as we s h a l l tell. If two of t h e e x t e n d e d c o n d i t i o n s will t u r n o u t to be compatible then the corresponding too.
F i r s t we e x t e n d p
original conditions are compatible
t o p l w h i c h d e c i d e s e x a c t l y t h e v a l u e of q (q is t h e
n a m e of a f i n i t e s u b s e t of ml x ~), so we s h a l l n o w r e g a r d q a s a f i n i t e f u n c t i o n f r o m o l to m a n d n o t as a n a m e .
T h e n we e x t e n d p 1 t o P2 so t h a t t h e
D o m t p 2 -- | a : ( : ~ n ) [ < a , n > 9 D o m p 2 ] { ( w h i c h is t h e s e t of b r a n c h e s
p " s p e a k " ) will i n c l u d e t h e d o m a i n of q.
on which
Now we e x t e n d q t o q l so t h a t t h e
d o m a i n of q l will b e e q u a l t o t h e Dora ?p2 i.e., P2 a n d q l will m e n t i o n e x a c t l y the same branches name q names
o u t of A.
of t h e r e q u i r e d m e m b e r s
and does not depend indices
of
sufficiently
the
c e ~
l0 . . . . .
about we
c|
can
which
p
extend
Let Ip d e n o t e contains P2
to
and p3(i,c) #p3(j,c)
d i f f e r e n t i e / p ~ a r e i n d i c e s of b r a n c h e s
by adding to the
of V w h o s e i n t e r p r e t a t i o n
o n t h e g e n e r i c s e t G.)
branches
large
Dom(p3) =/p~•
(This c a n b e a c c o m p l i s h e d
i.e., t h e s e t of
information.
P3
so
that
f o r i,j 9
of A w h i c h b r a n c h
o t h e r a t a l e v e l u p t o l e v e l 2 ), a n d c "~ql(i) f o r i
9
is f i x e d
For
a
c > I/p,],
i #j
(i.e.,
off f r o m e a c h
F r o m n o w o n we s h a l l
a s s u m e t h a t all t h e m e m b e r s of W h a v e t h e p r o p e r t i e s of < P a , q l ) .
Using the standard
technique
we c a n d e l e t e m e m b e r s
o b t a i n a s e t W with t h e following p r o p e r t i e s . JC~I; members
finite
sequences
ja=<j~
.....
such that the sets iJ~ .....
j,~
from
W so a s t o
T h e r e a r e a c < ~, a f i n i t e s u b s e t
Jn")
for
a<~
1 of
n
different
( w h i c h we s h a l l a l s o c a l l j a ) a r e
65
d i s j o i n t e d f r o m e a c h o t h e r f o r d i f f e r e n t a ' s a n d a r e d i s j o i n t f r o m J; a f u n c t i o n p on Jx
i0 . . . . .
cl, a function/v
o n I1 . . . . .
o n J a n d a f u n c t i o n h" o n I1 . . . . . pa-p
n ] x 10 . . . . .
n l s u c h t h a t if
=
them
there
are
at
least
above
For
Notice that
.
.
.
e"
.
2 iz d i f f e r e n t
let p:
~(n,i)
members
of
=p.
f o r d i f f e r e n t a ' s t h e initial p a r t s different branchings
way on J
and
Tt+ 1.
Let
:
of t h e b r a n c h e s
a t level i + 1 .
have otherwise
of
of Ti+ 1 w h i c h a r e
with indices
j na w e r e t h e s a m e , b u t t h i s is n o t t h e c a s e f o r t h e p a ' s
introduced same
.
.....
obtained above). Above each
< t ~ j z : 1 _~lZ _< 3 ~ > be a s e q u e n c e of 2 ~ d i f f e r e n t m e m b e r s
jl t .....
: a
The m e m b e r s / ~ ( 1 , i ) , ~ ( 2 , i )
T~ a r e all d i f f e r e n t ( b y t h e p r o p e r t i e s of ~ p , q > of
If = I(Pa,qa>
:1
q . - h = I<j~a,K(/c)> : 1 --
c I, a f u n c t i o n h
s i n c e we
S i n c e all t h e p ~ ' s b e h a v e t h e
disjoint
domains
also
(j
p~ c Q0.
l ~ a < 2 tz
Extend
this
gj~(m,n))
condition f o r all
to a condition p ~ which determines
1<-a<2 ~, l<-m <-i,
1--]c ~n.
the values
of
If we k e e p a fixed a n d
let /c,m v a r y we h a v e h e r e 2 i8 f u n c t i o n s o n a s e t of n , i - ( l I p l )
§
m e m b e r s i n t o IO,1t, h e n c e t h e r e m u s t be two d i f f e r e n t fl # 7 I -< fl,7 ~ 2~e s u c h t h a t gi/(/~(/c , m ) ) = ga.~(/~(/c , m ) ) f o r all 1 < / c _< n ,
1 -< m ~- i.
What
may
prevent and
hence
< p ",q~ U qT>
(p',q,>
is ( s i n c e
and
being
<Pa,qa>
and
We c l a i m t h a t are
a condition (p~,,q,>
compatible.
when
are
each
of
conditions)?
tcl,m,/r 2, s u c h t h a t m ->/~(kl), rn _> ~ ( k a ) iY(kl,m ) = ~ ( k a , r n ) g~(j~(lc~,rn))# g~t(~(ke,m)). This is t h e e a s e w h e r e t h e
There may be such
that
corresponding ~(k~,m)
functions
=;ff(/%,m)
branches
g~. give
of T a n d
with indices j~
this member
values
to
the
same
is a b o v e t h e p l a c e
a n d j~:~ w h e r e t h e u n i f o r m i z a U o n
o c c u r . By o u r c h o i c e of fl a n d 7
member
in t h e
is s u p p o s e d
two to
g~(~(ka,m)) = g~t, (P~(lce,m))' h e n c e
gi~(l~(k~,m)) # gil(p"(ke,m)) that ~p ~
different
is n o t a c o n d i t i o n in
while
y(k~,m)=iff(kz,m
Qo * Q(7),
This of c o u r s e d o e s n o t y e t p r o v e t h e o r e m
), b u t t h i s s h o w s
w h i c h is a c o n t r a d i c t i o n . 4.6. If we w a n t t o c a r r y o n t h e
i t e r a t i o n , s o m e h o w i m i t a t i n g t h e s c h e m e of M a r t i n ' s a x i o m , we s h o u l d do two
66
things: first isolate some property the "reason"
w h y Q ( g ) is c . c . c .
o f
i n V Q~ a p r o p e r t y
T h e n we s h o u l d e n o n c i a t e
ing notion which would ensure the property
a property
w h i c h is of f o r c -
of
at
l a s t we s h o u l d s h o w b y i n d u c t i o n o n a ~ ~2 t h a t t h e i t e r a t i o n of Q ( g ) o v e r a l l names for candidates
has the desired property.
Where by candidate
we m e a n
If T w a s c h o s e n s u c h t h a t : V m 3 n > m : V z 9 Tn[ : l Y c :In+ 1 : y > x ] [ >- 2 ]~yeT:(y<x)]I'*] W h a t we g e t f r o m t h e g e n e r i c i t y of t h e 8 v b r a n c h e s is: for n
every ,a I .....
(i) a~ = (Ar
k < ca
and
a~ 9 Tn
distinct
and W
A~
i
from
A there
are
Wcca lsuchthat:
for every i 9 W
(iii) [ W I > I] 2 IIyly
= Q(ga) uniformizes a candidate g a
ga = Ig~:~ < ~1| such that l~p~" g =
~a
is a candidate '/ W e prove by induction on a ----~e the following condition. (*)=: if /c < co a n d p ~ n > n 1 "~a 0 . . . . .
9 P=; A~~ c A f o r i < coI, E < / c , t h e n Y n 1 9 ca 3 n 9 r
aqr 1 9 T ( a ~ ~ a m f o r ~ # m ) ~W c co1 s u c h t h a t
(i) a~ = (Ace)[n] f o r g < k , i 9 W. (ii) f o r i ~ j c W
( A / ) [ n + l ] # (Ajg)ln+,] o r Ace = A~.
(iii) [ W[ > q ( a 0 . . . . .
a~)
(iv) 9q 9 P=: pC --< q f o r e a c h i 9 W So let us carry the induction.
(i) a = 0 nothing to prove.
67 (2) a = ~ a .
Then
so (*)a g i v e s t h e c o n c l u -
:~fl < a V i : pc 9 P a
sion. (3) a limit, cf a = r let
a = U a~
~z I [ i : n ( i )
= n|]
then
V i 3 n ( i ) : P t 9 P=-r
= Sl t h e n (*)~r
(4) a l i m i t c f a = ~ l
so
for
some
gives the conclusion: so a =
U at
<~1>
ascending
and
C<~1
continuous
f o r e a c h i we l e t h ( i ) = S u p (I0] U D o m p t
down function, tionary.
so b y F o d o r ' s L e m m a f o r s o m e i 0
W.l.o.g. Pt 9 Pa,+, a n d f o r a s t a t i o n a r y
clearly Pt,Pj are compatible
. If Pc t at(0) p j t
(~ i ) t h i s is a p r e s s i n g S = [i : h ( i ) = io~ is s t a -
s e t of j ' s j > a t, f o r s u c h P c ' s aj(0) a r e , b u t t h i s , b y r e n a m i n g
a n d u s i n g (*)c(0) i m p l i e s (.*)=. (5) a = / ~ + 1. r e m e m b e r f such that :Domf
t h a t s i n c e Q ( g ) is t h e s e t of f u n c t i o n s
is a finite s u b s e t of ~1 a n d f o r ~ 9 D o m f , g(~) < ~ a n d (Y ~,~ 9 D o m g ) ( Y n ) [ g (~-) -< n ^ g (~) ~ n
,, (Ar
= (A~)[,,] -, ge((Ar
So f o r e a c h fl 9 D o m p t W.l.o.g.
IDompt(~)]
Pt(fl) is ( t h e n a m e of ) a finite f u n c t i o n f r o m r 1.
is c o n s t a n t on
depends
Acg(i < m l :
g
m <_ n qc ][-Pp " g ~ ( r n ) The
number
of
l a~o[ > q ( a o . . . . . >q(a 0 .....
ak).
only
so l e t D o m p c ( f l ) = IA~r : ]c < s < / c ( 0 ) J w.l.o.g.
pc(]g)(A~r
we c a n find qt 9
g
l) = gd(A~)[,,l)
we
apply
such that
to
k(0),
= c c ( g , m ) ' ' w h e r e Ar~ = A T of c o u r s e possible
functions
am(0) ) so f o r s o m e c,
pcrflcPp 0andq0.
f o r e a c h Y i 9 W0, k - s
is
and Clearly
and
c c ( g , m ) 9 10,1~.
q (a k .....
ak(0) )
as
W = | i c W0:cc = c~ h a s c a r d i n a l i t y
Now q = q ~ U I ~ P , U P c ( f l ) > ] ieWo
e x e m p l i f y ( *)a.
(*)~
9
and r, ar
W
68 w
M a x i m a l a l m o s t d i s j o i n t f a m i l i e s o f s u b s e t o f co.
B y a m a d ( m a x i m a l l y a l m o s t d i s j o i n t ) s u b s e t of/9(co) we m e a n a n i n f i n i t e subset
F of SP(co) s u c h
f o r a l l x , y 9 F Ix l --~0
that
e v e r y z 9 ]9(co) t h e r e e x i s t x 9 F s u c h t h a t
Ix n y l < ~ a n d f o r
I z (3 x I = S0-
5.1 T h e o r e m : T h e r e is a m a d s u b s e t of SP(a)) of c a r d i n a l i t y 2 s~ P r o o f : R e p l a c e b y t h e s e t of all n o d e s of t h e f u l l b i n a r y t r e e , i.e., w i t h t h e s e t of a l l f i n i t e s e q u e n c e s
of O's a n d l ' s .
e v e r y t w o of t h e m a r e a l m o s t d i s j o i n t .
T h i s t r e e h a s 2 ~~ b r a n c h e s
E x t e n d t h e s e t of all b r a n c h e s
and to a
mad set by Zorn's lemma. s u b s e t of SP(r
5.2 T h e o r e m = No c o u n t a b l e
is m a d .
Let < at: i < co> be a sequence of infinite pairwise almost disjoint subsets of co. For e a c h i < co we have
ai -
gaj
=a
t -
K2aj
j<~ set
aj Aa~
is f i n i t e
and
at
is i n f i n i t e
there
is a n
Xj 9 a j w h i l e ) Q ~ aj h e n c e Xj ~ X / , t h e r e f o r e
j
an infinite set.
Since fori
the intersection
<j
X~ ~ a j
~
ai,
a n d since each
j
X~ 9 t - U a j .
For
t h e s e t b : IX~ : i < w ] is
we h a v e b (3 a j c I X 0 . . . . .
of b w i t h e a c h a j is f i n i t e a n d t h e r e f o r e
Xj]. T h u s
l a i : i < co| is n o t
mad.
If t h e c o n t i n u u m
hypothesis
L9%= ~1 a n d of n o o t h e r
h o l d s t h e n t h e r e a r e m a d s e t s of c a r d i n a l i t y
cardinality.
If t h e c o n t i n u u m
h o l d we a r e f a c e d w i t h t h e q u e s t i o n w h e t h e r t i e s -> 11t b u t
hypothesis
does not
t h e r e a r e m a d s e t s of c a r d i n a l i -
< 2 I%.
5.3 T h e o r e m = M a r t i n ' s a x i o m i m p l i e s t h a t e v e r y m a d s e t is of c a r d i n a l i t y ~0.
Proof: Let F be an infinite set of infinite pairwise almost disjoint sub-
69 s e t s of ~0, I F I < 2 M~ . L e t finite subset b Da
Forp
=~a,t?
and for each u ct
b (3u
order.
of F t .
PF = I < a , t ) : a is a f i n i t e s u b s e t of c0 a n d t is a
The meaning
and q =~b,s> =a
(%~
~
of t h e c o n d i t i o n ~ a , t >
in P f w e
definep
~q
if
is e a s i l y s e e n t o b e a p a r t i a l is t h a t t h e s e t w w h i c h we c o n -
s t r u c t will i n c l u d e a , b u t f o r e v e r y u c t w (3 u = a F5 u , s o n o n e w m e m b e r s of u will b e i n w ( o t h e r t h a n t h o s e a l r e a d y i n a ) .
If C a , t >
Py t h e n , o b v i o u s l y , a l s o < a , t
are in
and (
U s~
c_Pyhence
are compatible Therefore inoompatible members or P ,
must have different first components.
S i n c e t h e r e a r e o n l y ~0 f i n i t e s u b s e t s
of ~ e v e r y s u b s e t of P F of p a i r w i s e i n c o m p a t i b l e
members
must be countable,
and the c.c.c, holds.
[
v US
U |gClu
k--ran,
we
u Es;(b
aFletDu
:
:u cs~
shall
see
U [/c~) f ~ u
is f i n i t e h e n c e that
=b
Flu
sincek
By M a r t i n ' s a x i o m t h e r e
v-U
s is i n f i n i t e .
U|/cl,s)_~b,s~.
To
Ig U s h e n c e k
is a d i r e c t e d
subset
that
s, let
flu.
G of P F w h i c h i n t e r s e c t s
e a c h D u a n d ~ ) n - L e t A b e t h e u n i o n of t h e f i r s t c o m p o n e n t s let
Let k cv-U prove
We s h a l l s e e t h a t
of G. L e t u c F ,
A (3u
=a
Flu
and
e a,s~ c G a CA; a s s u m e t h a t m c A
A N u
rn E A N u - - a
is
finite.
Clu,
See
then m c b for some
wehave-nc:-n the
other
hand
rn e ( A
i n f i n i t e since f o r e v e r y n tains a number
~ n.
f~u)
Nc
~u
G contains
f~c,
,hence
-no
w h i c h is a c o n L r a d i c t i o n .
a member
(a,t>
On A is
of ~ ) n h e n c e A c o n -
70 T h u s we h a v e c o n s t r u c t e d every member
a n i n f i n i t e s e t A w h i c h is a l m o s t d i s j o i n t f r o m
of F, a n d h e n c e F is n o t m a d .
5 . 4 T h e o r e m : If V d o e s n o t s a t i s f y t h e c o n t i n u u m c a r d i n a l s u c h t h a t t~1 <-- A < 2 t% i n V t h e n
hypothesis
V can be generically extended
c . c . c , e x t e n s i o n w h i c h p r e s e r v e s a l s o 2I% t o o n e e x t e n s i o n a m a d s u b s e t of/9(r
a n d A is a by a
V[G] w h e r e t h e r e is
of c a r d i n a l i t y X.
P r o o f : We s t a r t
with a sequence
i n V ( A a : a < ~,~ of a l m o s t d i s j o i n t
s u b s e t of ~ ( t h e r e is s u c h b y T h e o r e m 5.1). Now we p r o c e e d w i t h a s y s t e m of iterated
f o r c i n g of l e n g t h ~1 a s f o l l o w s . A t s t e p a we a s s u m e
constructed s e t s of ~.
a sequence
(At:
i < A + a>
F a=~At : i < A + a ]
We t a k e
t h a t we h a v e
of p a i r w i s e a l m o s t
disjointed
sub-
a n d we u s e a t t h i s s t e p t h e f o r c i n g
n o t i o n PF,, w h e r e P F is a s i n t h e p r o o f of T h e o r e m
5.3, a n d we i n t r o d u c e
by
AA+a of co a l m o s t d i s j o i n t w i t h e a c h m e m b e r of F a. ( t h i s f o l l o w s i m m e d i a t e l y f r o m t h e p r o o f of t h e o r e m 3). S i n c e PF,, is a c . c . c .
means
of i t a n e w s u b s e t
forcing our iterated
(a,t)
f o r c i n g is c . c . c .
For
PF, we c a n u s e , i n s t e a d of p a i r s F a, s u c h p a i r s
w h e r e a is a f i n i t e s u b s e t o f ~ a n d t is a f i n i t e s u b s e t of
w h e r e t is a f i n i t e s u b s e t of A + a a n d e a c h i 9 t s t a n d s f o r t h e c o r r e s p o n d i n g At. T h u s e a c h
PFo will c o n s i s t of s e t s of V ( w h i l e i t s p a r t i a l o r d e r is n o t n e c e s -
s a r i l y i n V), a n d t h e c a r d i n a l i t y iterated hence
of
PF,, is t h e r e f o r e
f o r c i n g we c a n u s e o n l y s t a n d a r d
M a x (a,A) = A. F o r t h e
n a m e s i n t h e s e t P of c o n d i t i o n s ,
I P ] = A. S i n c e P is a c . c . c , f o r c i n g s t a n d a r d
arguments
] p ( ~ ) l ( v [ g ] ) = (h) I%(10 __ (At%vlc] _< (2t%) I%(I0 = 3 It(V), h e n c e
in
show that
V[G] 2 I% is t h e
s a m e a s i n V.
Finally let us prove that
A Cco, A 9 V[G]. F o r e a c h n
IA a : a
is a m a d s u b s e t
9
Qn b e a m a x i m a l a n t i c h a i n
of ] ~ ) .
Let
in P of c o n d i -
t i o n s w h i c h d e c i d e r~ 9 A, w h e r e A i s a n a m e of A. S i n c e P s a t i s f i e s t h e c . c . c . N
N
N
IQnl =t'lo. L e t Q = U Q~, I Q I =t~0- F o r e a c h q 9 Q D o m ( q ) setof
~,,hence
is t h e c o m p o n e n t
t h e r e is a n a
<~1 such that for allq
of G i n t h e i t e r a t e d
is a f i n i t e s u b -
9 QDom(q)
ca.
If Ga
f o r c i n g u p to; a n d n o t i n c l u d i n g , a w e
71 h a v e A 9 V[Ga]. We s h a l l s e e t h a t A f~ Aa is i n f i n i t e f o r s o m e b -> A + a w h i c h establishes
t h a t tA, : i < A + r
for every ~ < A +a then
is m a d .
We s h a l l s h o w t h a t if A C~ A a is f i n i t e
A ~ A a is i n f i n i t e .
This follows from the following
lemma, which ends the proof.
5.5 Lemma Bcc0be
: L e t F b e a s e t of p a i r w i s e d i s j o i n t s u b s e t s
such that for every subsetF
t of F
B-U
of c0, a n d l e t
F t is i n f i n i t e .
Let G be a
g e n e r i c s u b s e t of PF o v e r V a n d l e t A b e t h e u n i o n of t h e f i r s t c o m p o n e n t s the members
o f G, t h e n A C~ B is i n f i n i t e .
Proof: For every n shall see that infinite hence
ulml,t>
hence
let Dn
is d e n s e . it contains
= I
Let < b , t >
an m ->n.
> >
A Da
of
c
m eB
By o u r a s s u m p t i o n
SinceDn
and A contains
ePr:(am->n)
U [ml,t>
e~bn
is d e n s e G c o n t a i n s
an m 9
minn.
Nai.
We
B - U T is
and since m ~ U t some
9
Since this holds for every
n < ~ A f~ B is i n f i n i t e .
We
shall
equivalent.
prove
that
If o n e c a r r i e s
essentially
all
countable
o u t t h e p r o o f of t h e l a s t t h e o r e m
w h e r e a = t~1 , b u t o n e s t a r t s w i t h a s e q u e n c e disjoint subsets equivalent
of ~ t h e n
forcing
each
PF is a c o u n t a b l e
t o t h e a d d i t i o n of a C o h e n R e a l .
notions
are
for the case
of p a i r w i s e a b o u t
forcing
Thus the iterated
and
therefore
a d d i t i o n of t~1
C o h e n R e a l s y i e l d s a n e x t e n s i o n of V w i t h a m a d s e t of c a r d i n a l i t y t~i. B u t t h e iterated
extension
of t~1 C o h e n R e a l s is t h e s a m e a s t h e s i m u l t a n e o u s
addition
of t~i C o h e n R e a l s .
5 . 6 T h e o r e r I x E v e r y two c o u n t a b l e condition there are two incompatible
Proof: Let P
forcing notions where above every
conditions are equivalent.
be the set of all finite sequences of natural n u m b e r s
ordered by proper inclusion a n d let Q be a countable forcing as in the theorem. W e shall s h o w that P and Q are equivalent by constructing an iso-
72 morphism
F of P o n a d e n s e s u b s e t of (~. We s h a l l d e f i n e F(W), f o r 7 / c P , b y
induction
o n t h e l e n g t h of W- F ( 0 ) = 0, w h e r e 0 is t h e m i n i m a l m e m b r
Let
Q = Iq~:n
< ~].
We s h a l l
take
care
to
satisfy
the
of Q.
following in the
definition. (i) F o r e v e r y 7) c P I F ( w A < i > ) : i < ~ ] is a m a x i m a l s e t of incompatible
members
of Q g r e a t e r t h a n F(~7).
(ii) F o r e v e r y n t h e r e is a n 77 c P of l e n g t h n + 1 s u c h t h a t F(W) -~ qn. We a s s u m e , predense
as
an
induction
hypothesis,
that
IF(q):
length
(W) = n {
i n P . If (i) h o l d s , l e t q c Q, t h e n b y t h e i n d u c t i o n h y p o t h e s i s
is a n W of l e n g t h n
such that
F(W) is c o m p a t i b l e
l e n g t h (~) = n + l ]
is p r e d e n s e
and
F(W)
is
members
members
compatible
of
Thus the set IF(t) :
t o o . To d e f i n e F(W ^ < i > ) , f o r a l l i < r
s e t ~9 of p a i r w i s e i n c o m p a t i b l e
(7) = n
w i t h s o m e F(W ^ < i > ) .
there
L e t q* >- q,F(w).
w i t h q.
S i n c e , b y (i), I F ( w ^ < i > ) : i < ~ ] is a m a x i m a l s e t of i n c o m p a t i b l e Q a b o v e F(~?), q t is c o m p a t i b l e
is
define a
of Q a b o v e F(W) a s f o l l o w s . If l e n g h t h
with
qn
t o - > F(W). L e t t~,s 1 b e two i n c o m p a t i b l e
to~-F(w),qn,
let
members
b y i n d u c t i o n , l e t tn+l,Sn+ 1 b e two i n c o m p a t i b l e
of Q g r e a t e r
members
otherwise
let
than t o and,
of Q g r e a t e r t h a n s n .
T a k e T = Itn : i - n < c01, T is a s e t of p a i r w i s e i n c o m p a t i b l e
members,
F(~?) is c o m p a t i b l e
Let T be a maxi-
withgn then each member
m a l s e t of p a i r w i s e i n c o m p a t i b l e countable, ~F(w^):i
since <~]
Q
is,
members
and
we
a b o v e F(W) w h i c h i n c l u d e s T. T is
define
= T. S i n c e (ii) h o l d s [ F ( w )
Using the method
of T h e o r e m
of T is -~ gn.
a n d if
the
F(W ^ < i > ) ' s
so
that
: w c P ] is a d e n s e s u b s e t of Q.
5.4 we c a n e x t e n d
V so t h a t in t h e e x t e n -
s i o n f o r e v e r y I s u c h t h a t 1~0 < X --< 2 s~ t h e r e is a m a d s u b s e t of c a r d i n a l i t y X. To d o t h i s l e t ( k a " a < / z > For a
b e t h e c a r d i n a l s t~0 < h < 21% in i n c r e a s i n g
a mad
subset
IAta : i
order.
of p(co) a s f o l l o w s .
lAta : i < k . ] i s a f a m i l y of p a i r w i s e a l m o s t d i s j o i n e d s u b s e t s of ~. We e x t e n d V b y i t e r a t i n g /~. ~1 t i m e s a n d a t t h e s t e p of o r d i n a l / z , i A~+ 1 -- r a s a n a l m o s t d i s j o i n t s e t of IA/a : j < k a + i ] .
+ a we a d d t h e s e t
]]I PROPER FORCING @1 I n t r o d u c i n g p r o p e r n e s s 1.1 D i s c u s s i o n : W h e n we i t e r a t e
we a r e
faced
with t h e p r o b l e m
of
o b t a i n i n g f o r t h e i t e r a t i o n t h e g o o d p r o p e r t i e s of t h e single s t e p s of i t e a t i o n . U s u a l l y t h e w o r s t p o s s i b l e v i c e of a f o r c i n g n o t i o n is t h a t it c o l l a p s e s t~t. T h e v i r t u e of n o t c o l l a p s i n g l~1 is n o t i n h e r i t e d b y t h e i t e r a t i o n f r o m its s i n g l e coml~onents.
As we saw, t h e v i r t u e of t h e c.c.c, is i n h e r i t e d b y t h e FS i t e r a -
t i o n f r o m its c o m p o n e n t s .
H o w e v e r t h e c.c.c, is t o o s t r o n g a r e q u i r e m e n t .
shall look for a weaker requirement
w h i c h is m o r e n a t u r a l l y c o n n e c t e d
We
to the
p r o p e r t y of n o t c o l l a p s i n g I~l, a n d w h i c h is i n h e r i t e d b y i t e r a t i o n s .
We s h a l l n o w s t u d y a c e r t a i n g e n e r a l i z a t i o n unbounded
a n d a s t a t i o n a r y s u b s e t of ~l.
of t h e c o n c e p t s of a c l o s e d
They were introduced and investi-
gated by Jech and Kueker. 1.2 D e f i n i t i o n : F o r A u n c o u n t a b l e
If c
S # 0 ( A ) is c a l l e d c l o s e d if it is c l o s e d u n d e r
sequences. s ~
SI~o(A ) = Is :s CA,]s] <--1%1.
let
If c
~t%(A)
is c a l l e d
unbounded
u n i o n s of i n c r e a s i n g
(in
~ I % ( A ) ) if f o r e v e r y
S R o ( A ) t h e r e is a t ~ t~ s u c h t h a t t ~ s.
1.3 I ~ m m a : T h e i n t e r s e c t i o n of t~0 c l o s e d u n b o u n d e d of
c0-
SR0(A) is a c l o s e d u n b o u n d e d
s u b s e t of
SI%(A).
P r o o f : S i n c e e a c h s e t lqt is c l o s e d t h e i n t e r s e c t i o n closed too. Let us prove now that
A W i is u n b o u n d e d
we h a v e t o p r o v e
of a t D s ,
the
existence
s u b s e t s Iri, i < o,
too.
FI ~ is o b v i o u s l y Let s c
SI%(A);
t c F~ W~. We shall d e f i n e a
74
sequence
~sa:a
a = r 9k + s (J s a c
< co~ ) of m e m b e r s sa
is
an
of
arbitrary
~ t % ( A ) a s follows,
s o = s, f o r a > 0,
member
which
~t%(A); it e x i s t s a s W~ is U n b o u n d e d .
of
W~
We t a k e n o w t =
p
includes U sa- Obvi-
a
ously f ~s o =s.
F o r a f i x e d i
a < c o ( k + l ) + i, h e n c e , fore f =
b y t h e d e f i n i t i o n of s~(k+l)+,
+s
s a = s~(~+i)+~.
then There-
U s a = U so.~+~- T h e s e q u e n c e ( s o . ~ + ~ :/c < co) is a n o n d e c r e a s i n g a
~-sequence
/,
of m e m b e r s
of Wi, a n d s i n c e W, is c l o s e d also its u n i o n t is in ~ .
T h u s t c f~ W~, w h i c h is whaL we h a d to p r o v e .
1.4 D e f i n i t i o n : B y t h e l a s t l e m m a we k n o w Lhat t h e c l o s e d u n b o u n d e d subset
of
subsets
of
filter with closed
Se0(A ) generate
filter, n a m e l y t h e f i l t e r of all
S~0(A) which include a closed unbounded DM0(A ). A s u b s e t of
unbounded
subset
We s h a l l n o w p r e s e n t and
an sl -complete
of
S ~ o ( A ) is c a l l e d
set.
We d e n o t e t h i s
stationary if it m e e t s e v e r y
S 0(A), i.e., if it m e e t s
every
t h e l e m m a w h i c h s a y s thaL f o r ~] = Sl
member
S#o(A)
D 0 ( A ) do n o t d i f f e r s i g n i f i c a n t l y f r o m ~1 a n d t h e f i l t e r Do, g e n e r a t e d
the closed unbounded
is a c l o s e d u n b o u n d e d
u s u a l s e n s e of a c l o s e d u n b o u n d e d s u b s e t of
by
s u b s e t s of co1.
1.5 L e m m a : A s u b s e t of r unbounded
of
subset
s u b s e t of ~1 (in t h e
of a n o r d i n a l )
iff it is a c l o s e d
SB0(~I).
Proof: Easy.
We s h a l l n o w i n t r o d u c e a m o r e r e s t r i c t e d
s e t of g e n e r a t o r s
for
D 0 ( A ).
75 1.6 Definition: M will d e n o t e a n a l g e b r a , with u n i v e r s e A, a n d with c o u n t a b l y m a n y f u n c t i o n s . Let
S m ( M ) = [s : s ~ A , I s I ~ o ,
s is c l o s e d u n d e r t h e o p e r a t i o n s of M], i.e.,
S m (M) is t h e s e t of c o u n t a b l e s u b a l g e b r a s of M. Srn (M) is o b v i o u s l y a c l o s e d u n b o u n d e d s u b s e t of
--~-~0(A).
1.7 L e m m a : F o r e v e r y c l o s e d u n b o u n d e d s u b s e t W of
S~0(A ) t h e r e is
a n a l g e b r a M o n A s u c h t h a t S m (M) C W. P r o o f : We shall define, f o r e v e r y finite s e q u e n c e ~ = ( a 0 . . . . . of m e m b e r s
of A, b y i n d u c t i o n
s ( a ) ~ [a o . . . . . sible b e c a u s e
on t h e
length
a~_ 1] a n d s ( a ) ~ s ( ( a 0 . . . . . W is u n b o u n d e d .
Sn = s ( ( a 0 . . . . .
s(~)~
W such
that
a ~ _ l ~ ). This is o b v i o u s l y p o s -
We define n o w n - p l a c e f u n c t i o n s F~, s < r
f o r all ~
LetM=~A,F~tn<~,r
n,
an_it
an_l~ ) =
~. L e t S = | a o , a l , . . - |
an_l~ ) = {F~(a 0 .....
|F~(a 0 .....
a n _ l ) : s
b e a s u b a l g e b r a of M. D e n o t e
a.n_l) : s < (hi. We have:
a) Sn C S, s i n c e s is a s u b a l g e b r a . b) Sn C Sn+ ,, b y d e f i n i t i o n of s ( ( a 0 . . . . .
a n t ).
c) an < S~§ d) S~ r W. By (a) a n d (c) we h a v e S = (J S~; b y (b) a n d (d) w e g e t S E
W. T h u s we h a v e
show Srn (M) C W.
We h a v e n o w s e e n t h a t t h e filter
D 0 ( A ) is g e n e r a t e d b y all s e t s SIn(M)
w h e r e M is a n a l g e b r a on A as a b o v e . We shall now see one u s e of t h i s f a c t . 1.8 Theorem= L e t P • V be a p o s e r w h i c h s a t i s f i e s t h e c.c.c., let ~, b e a n uncountable
c a r d i n a l , a n d l e t G be a g e n e r i c s u b s e t of P o v e r V. E v e r y
76 closed
unbonded
unbounded
subset
s u b s e t of
B
of
~B0(X)(v[G]) in
S 0(k)(v) in V. T h u s
c l o s u r e s of t h e g e n e r a t o r s of
V[G] i n c l u d e s
a
closed
D~o(k)(v[c]) is g e n e r a t e d
by the
&0(k)(v).
P r o o f : By w h a t we h a v e p r o v e d a b o v e we h a v e t h e following in V[G]. T h e r e is a n a l g e b r a M = < N , F ~ > n , ~ < ~ s u c h t h a t S r n ( M ) c B.
In V t h e f u n c -
tion /~
h a s a n a m e F n. B e c a u s e of t h e c.c.c., b y L e m m a I. 3.6 (ii) f o r all ~g
a 0 .....
a n _ 1 < X we k n o w in
I~(a o . . . . .
V the
countable
set
of p o s s i b l e
values
of
an_l). We d e f i n e t h e f u n c t i o n s F~, k so t h a t t h e s e ~0 v a l u e s a r e
IF~,~(a 0 . . . . .
a n _ l ) : k < c0].
a0 . . . . .
< x
a~-i
~(a0
So
.....
we
a~-l)
c
have,, I~.~(a0
for
all
.....
n , g < co
a~-l)
:~ <
and
~.
So
N =
of M. We h a v e S I n ( N ) (v) ~ S m ( M ) (vIal) ~ B, SIn(N) (v) is a c l o s e d
unbounded
s u b s e t of
~ o ( X ) ( v ) in V, a n d t h e c l o s u r e SIn(N) v in V[G] is
S m (N) v[G] c B.
A consequence every stationary
of t h i s t h e o r e m subset
of
S
is t h a t f o r a c.c.c, e x t e n s i o n
D(X)(w) in V is a l s o a s t a t i o n a r y
V[G] of V subset
of
S B ( X ) (v[c]) in V[G]; in s h o r t , t h e e x t e n s i o n d o e s n o t d e s t r o y t h e s t a t i o n a r i t y of s t a t i o n a r y s u b s e t s of concept
of p r o p e r
forcing.
/) D(x)VlC] is g e n e r a t e d
S ~ o ( k ). We s h a l l u s e t h i s p r o p e r t y t o d e f i n e t h e While it is a c o n s e q u e n c e
b y t h e c l o s u r e s of t h e m e m b e r s
of t h e of
fact
that
D o ( A ) V is d o e s
n o t s e e m t o r e q u i r e as m u c h . 1.9 D e f i n i t i o n : A f o r c i n g n o t i o n P is c a l l e d proper if f o r e v e r y ( u n c o u n t abld) c a r d i n a l X, f o r c i n g with P p r e s e r v e s s t a t i o n a r i t y
modulo
D (~).
~0
We
s h a l l d e n o t e t h i s c o n d i t i o n f o r X with Conl(k ). N o t e t h a t Conl(So) is m e a n i n g less, o r t r i v i a l l y t r u e if y o u w a n t .
77 Note t h a t p r o p e r n e s s is p r e s e r v e d by e q u i v a l e n c e of f o r c i n g n o t i o n . 1.10 T h e o r e n ~ 1) P is p r o p e r iff t h e following c o n d i t i o n h o l d s f o r e a c h c a r d i n a l ~. Con2(A): F o r all iP~J : J < a~,i < a] ~ P s u c h t h a t a ~ A a n d a~ -: A f o r i < a,
~pr : j < a i ] is p r e d e n s e in P , a n d for all p 9
and suchthatforalli
0(h) 9 (~q 9 P)[q ~ p
S
aboveq for anyi 9
i
and
|p~j :j
is
predense
D 0(~t).
C~
2) Moreover, f o r a n y Am [P] §
j 9
is e q u i v a l e n t to Cona(h); a n d if P
is a c o m p l e t e B o o l e a n a l g e b r a w i t h o u t 1 t h e n Con l(h) is e q u i v l a e n t to Con2(~ ) f o r e v e r y u n c o u n t a b l e h. P r o o f : We a s s u m e first - Con2(k), i.e., t h e r e a r e a ": ~,
a , - h f o r i < a,
IPi,i"J < ai{ is p r e d e n s e in P a n d p 9 P s u c h t h a t t h e s e t T of all s 9
SB0(A ) f o r w h i c h
-(:~(q EP)[q >_p a n d [ p i j : j
is s t a t i o n a r y .
Let G C P
be a g e n e r i c s u b s e t of P s u c h t h a t
p c G. G m e e t s e v e r y p r e d e n s e s e t h e n c e t h e r e is in V[G] a f u n c t i o n f o n u such
that
p ~ , l ( ~ ) c G for
Srn(
is
i
For
the
algebra
/) 0(k)(vIa]). We shall show t h a t T A S r n ( < k , f > )
stationary
s e T N Sm((h,f)), we h a v e
all
in
V
is
no
longer
stationary
in
we
have
= ~b t h u s T
V[G].
Assume
t h e n , as T e V, c l e a r l y s e V. Since s is c l o s e d u n d e r f
V[G] ~ ( V i E s )
(i < a - * ( 3 j c s ) ( j < a i ^ p ~ J c G)), h e n c e s o m e
r E G f o r c e s this s t a t e m e n t . Since G is d i r e c t e d a n d p E G t h e r e is a q c G
q >-p,r hence q [ ~ ( V i c s ) ( i < a - * ( ~ j c s ) (j
Thus s satisfies e x a c t l y t h e
c o n d i t i o n of n o t b e l o n g i n g to T, which is a c o n t r a d i c t i o n .
7B
-Con2(A ) implies - C o n l ( A ) . We shall p r o v e t h a t
We h a v e p r o v e d t h a t
Con2(h ) implies Con l(h ) for h-> ] P ] , o r f o r all ~-~t~ z if P is a c o m p l e t e B o o l e a n a l g e b r a w i t h o u t 1. This suffices to p r o v e t h a t (Y~,) Con2(A) implies
(u
Conl(h ) since, as we shall now see if h > / ~ t ~ 1, Coni(h ) implies Cont(IZ ) Con2(h ) implies Con1(), ) (see 1.13 , 2.1 resp.)
a n d if ~ >/~,
F o r this p u r p o s e we shall p r o v e t h e following lemmas. 1.11 ~x ( j y
Lemma
" For
"x 9
unbounded
9
any
sets
( s t a t i o n a r y ) s u b s e t of
P r o o f : The p r o o f t h a t
Sbto(A ) iff W U
~'B0( A U B). (W U
there
S t % ( B ) is a c l o s e d
if W is c l o s e d u n b o u n d e d
in
S~0(A ) t h e n
St%(A LJ B ) i s trivial. Now a s s u m e t h a t
St%(A ), a n d s u p p o s e W U
Then
set
S So(A (3 B).
St%(B ) is c l o s e d u n b o u n d e d in
W is s t a t i o n a r y in
the
F o r all disjoint u n c o u n t a b l e s e t s A , B , Y i s a c l o s e d
u n b o u n d e d ( s t a t i o n a r y ) s u b s e t of
W -U
D,E we d e n o t e b y D ( J E
is a m o d e l
~t%(B) is n o t s t a t i o n a r y in
M = < A U B,F'~> n,g<~ s u c h
that
St%(B)) N Sm ( i ) = ~b. We c a n a s s u m e , w i t h o u t loss of g e n e r a l i t y , t h a t
t h e s e t of f u n c t i o n s |Fr~ : n , s < (o{ is c l o s e d u n d e r s u b s t i t u t i o n . We define a function ~
f o r n - t u p l e s of m e m b e r s of A as follows
(a0 . . . . .
[ F ~ ( a 0 . . . . . an_l) if F ~ ( a 0 . . . . . an_l) 9 n a n - l ) = {any member oJ' A otherurise
Let M = ( A , ~ ) n , g < f 9
o.
We shall
S t ~ ( B ) , s L) t 9
see t h a t
if s 9 SIn(M)
then
for some
Let t be t h e s u b a l g e b r a of M g e n e r a t e d b y s;
0
we h a v e
to
prove
that
t-s
~B.
Let
b 9
then
|/v~ : n , ~ < o ] is c l o s e d u n d e r s u b s t i t u t i o n b = F ~ ( a 0 . . . . . a 0 .....
a n _ ~ 9 s.
(a o . . . . . b 9
If
b 9 A
then
by
the
since
the
set
an_l) for some
definition
of
an_l) = b, and since s 9 Srn(M) , s is closed u n d e r T-~, c l e a r l y
which c a n n o t be t h e c a s e since b 9
b 9 icting o u r a s s u m p t i o n t h a t
~B.
-s.
Therefore b gA, hence
We c l a i m t h a t W N S m ( M ) = ~ b , c o n t r a d -
W is a s t a t i o n a r y s u b s e t of
~B0(A).
Suppose
79 s 9 W I~ SIn(M),
then,
as
we
have
shown,
for
some
t 9
~'
(B) 0
s 13 ~ 9 SIn(M).
However
(wO S o(B)) n sm( )
s 13 t 9 W U
S0(B)
contradicting
=r
Thus we have p r o v e d t h e "'only if "' p a r t . The "'if"' p a r t c a n be p r o v e d simil a r l y or by applying t h e "only if " p a r t to W1 =
1.12 Claim= (1) If f X c
S~
- If.
is a o n e - t o - o n e f u n c t i o n f r o m A into B, t h e n for
S S (A): X is a s t a t i o n a r y s u b s e t of is a s t a t i o n a r y s u b s e t of
S
(A) iff
~a 9
S0(B):f-l(a)
~S0(B).
(2) If f :A -*/? is o n e - t o - o n e onto, f 29(S#0(B)) onto ]9(~s(A)
t h e n t h e s t a t i o n a r i t y of s o m e
Proof:
Note
z e a < = > y e a] 9
induces a mapping from
) preserving Boolean operations and stationarity.
(3) If V C V? are m o d e l s of ZFC ,
t i o n a r i t y of some X c
9
X c
V k "IAI = IB] ':
A , B 9 V,
Sso(B)V is d e s t r o y e d in Vt if t h e sta-
SR0(A)V is d e s t r o y e d in Vt ( X 9 V of c o u r s e )
that
~a <
~s0(A tJ B):
if
y = $ (x)
then
D o(A U S ) .
The p r o o f is left to t h e r e a d e r .
1.13 Clain~ If A > ~ S l
then
Conl(A ) implies Conl(~).
P r o o f : Let W b e a s t a t i o n a r y s u b s e t of
S # (/~)v. Then, as we h a v e 0
p r o v e d in 1.11,
Con~(A) holds
S~0( ~ _ ~ ) v ~ W is a s t a t i o n a r y s u b s e t of
Ss0(A)V. Since
SS0(A - ~)v U w is also a stationary subset of SSo(X)v[~] in
80
V[G]. We c l a i m t h a t W is a s t a t i o n a r y s u b s e t of
S 0(~)v in V[G]. If this is
n o t t h e c a s e t h e n t h e r e is a c l o s e d u n b o u n d e d s u b s e t C of
V[G] s u c h t h a t C C~ W=~b. By L e m m a 1.11 unbounded
subset
of
SS0(A)v[a]
( S~0(A _ ~ ) ~ I ~ ] U C) N ( that
in
S~0(A-#)vU
SSo(#)v[g] in
~ S 0 ( A - ~ ) V l g ] U C is a c l o s e d
V[G]. Since
C N W=#
we
have
W ) = r c o n t r a d i c t i n g w h a t we g o t
S S ( A - ~ ) v U ;f is a s t a t i o n a r y s u b s e t of
~ S (h)vt~] in V[G].
C o n t i n u a t i o n of t h e P r o o f o f 1.10: We r e t u r n now to t h e p r o o f t h a t
Con2(A ) implies Con I(A) f o r a n y u n c o u n t a b l e ~,-> ]P] o r f o r all A ~ ~1 if P is a c o m p l e t e B o o l e a n a l g e b r a w i t h o u t 1. Let T be a s t a t i o n a r y s u b s e t of
. ~ ' (X) 0
in V . To p r o v e t h a t T is also a s t a t i o n a r y s u b s e t of to
prove
that
for
every
P-name
_.SSo(A)in V[G] we h a v e
M~= (A,F~)jZ,n
of
an
algebra,
~b II-e " T N SIn(M) # ~b". L e t p c P, we shall p r o v e t h a t t h e r e is a q ---p s u c h t h a t q ]~p "T C~ S r n ( ~ # ~".
Let h:~>A -* k be a o n e - t o - o n e f u n c t i o n .
We
d e n o t e t h e r e s t r i c t i o n of h to n - t u p l e s with h n. Let h~, f o r s < ~ be a f u n c tion such that = h(n,s
h~(h'Z(flo . . . . .
for n > s
.....
fl,~-l)) = fl~. F o r e a c h i < A if
fin-l) let .9, be a m a x i m a l a n t i c h a i n of P of c o n d i t i o n s
which f o r c e definite v a l u e s f o r F~(fl 0 . . . . .
fin-l)-
If ] P ] - - - A t h e n c l e a r l y
].9~]--~. If P i s a B o o l e a n a l g e b r a w i t h o u t 1, a n d with u <- v if u - > v
in t h e
B o o l e a n a l g e b r a , t h e n f o r e a c h fl < A we c a n p u t in -9i t h e s u m (i.e., t h e g.l.b. in P ) of all c o n d i t i o n s which f o r c e F~(fl 0 . . . . .
fin-l) = #, if t h e r e a r e s u c h
c o n d i t i o n s a n d t h e n .9, will be a m a x i m a l s e t of c o n d i t i o n s which f o r c e definite v a l u e s on F~(fl 0 . . . . . IP~,j : J < aiJ
so
that
fl,~-l) a n d ].9~] -< A. We t a k e f o r i < h
~p~,j : j < a~l = .9~.
Pi,i I~ " ~I~(flo . . . . .
fin-l) = 7 ( i , J ) " .
SIn(N) e
Let
DSo(A).
W= Is e
predense over q for allies]I,
We
Let
7(i,j) be s u c h N = ~h,hn,hn,7,n>n
a i and
Let
S~(~,) : ( ~ q - > p )
that then
[IP~,i : J e a ~ Cls}
is
D (k) b y Con2(k ). Since T i s a s t a t i o n R0
81
arysubset
of
$ 0 ( ~ , ) t h e r e is a n s
be s u c h t h a t [P~j : J c a ~ 80 . . . . .
fn-lcs.
Ns~
c T f~ S I n ( N ) C~ Ir
is p r e d e n s e
S i n c e [P~j : J c a i f~ s ] is p r e d e n s e q
contains
F~(flo . . . . . 7
we
for
,
~p
Assume ~cs.
i = hn+2(n,g,80 .....
Thus
Bin-i) c s.
o v e r q e v e r y g e n e r i c filter G w h i c h c o n -
P~.a"
for
j c a t ~ s,
and
therefore
f i n - i ) = 7 ( i , j ) in V[G] . S i n c e i , j c s a n d s i n c e s is c l o s e d u n d e r
have
80 . . . . .
some
E Wletq
above q for eachics.
By d e f i n i t i o n of N a n d s i n c e s c S m ( N )
n , s c s a n d s i n c e s is c l o s e d u n d e r h n+2 tains
Sinces
fl,~_l) c s .
Fr~(fl0 . . . . .
f n - 1 E s ) F'L~(fl0 . . . . .
every
G
which
q [~-"T ASm(M)#r
Thus
have
in
f n - 1 ) E s , h e n c e s c SIn(M).
contains (since
we
q
we
s ~ T).
So
have
V[G]
VnV~(V
Since this holds
q I[-"s c Sm(M)':
T is still
stationary
in
i.e., Vv, as
required.
1.1fi O b s e r v a t i o n s : It c a n be s e e n , b y m e a n s of Con2(h ) t h a t in o r d e r that
t h e f o r c i n g P b e p r o p e r it s u f f i c e s t o r e q u i r e
Cons(h ), o r Conl(X ) f o r
s o m e ~ -~ 2 IP] ( s e e Lerllrlla 2.2). We c a n a l s o r e p l a c e , e q u i v a l e n t l y , a , a i <- ~ in
Con2(A ) b y a = a t = A, a n d we c a n r e p l a c e
the predense
sets by maximal
a n t i c h a i n s a n d 21P] b y t h e n u m b e r of t h e m a x i m a l a n t i c h a i n s (see L e m m a 2.2 in t h e n e x t s e c t i o n ) . 1.6 L e r n m a : If P is a p r o p e r f o r c i n g t h e n in V[G] e v e r y c o u n t a b l e s e t of o r d i n a l s is i n c l u d e d in a c o u n t a b l e uncountable
in V[ G]).
Proof:
a E
L e t a be a c o u n t a b l e
o ,.~ via]., ~l%~^j
s u b s e t of
s e t of o r d i n a l s of V ( a n d h e n c e t~v is
S o(~,) v [ C ] : s ~a~ is o b v i o u s l y a c l o s e d u n b o u n d e d
Is c
St%(A)v[a]
s e t of o r d i n a l s in V[G], t h e n f o r s o m e
in V[G] .
S o(A)v is a s t a t i o n a r y s u b s e t of
in v h e n c e , s i n c e t h e f o r c i n g P is p r o p e r , it is also a s t a t i o n a r y S~o(~)g[c] in V[G]. T h u s
S~(A) g n Is c
a s u b s e t c o u n t a b l e in V w h i c h i n c l u d e s a .
S~0(~)v s u b s e t of
S~o(N)v[~] 9 s ] a I # ~ a n d N h a s
82 An
alternative
proof
is
that
if
s 9
~ ' (A) V[a],
then
0
W1 = ~t:s c t 9
) i n V [ G ] , b u t i n V WO=
S~,(A)I 9
SI%(A)V 9
DMo(A),
W0 is s t a t i o n a r y , h e n c e it is s t a t i o n a r y in V[G] h e n c e W0 N W1 r 0
so in V,
which is j u s t w h a t we need.
As a c o n s e q u e n c e , if a is a n ordinal s u c h t h a t cf (a) > co in V, we h a v e also cf (a) > ~ in V[G].
w More o n Properness.
2.1 Discussion: It is w o r t h noticing t h a t one c a n use for a s e t of g e n e r a t o r s of
/)s0(A) n o t only the s e t ~Sm(M) : universe of M = A a n d M is a n alge-
b r a with S0 o p e r a t i o n s ] b u t also t h e s o m e w h a t wider set [Sm(M) : u n i v e r s e of M = A a n d M is a p a r t i a l a l g e b r a with n 0 o p e r a t i o n s ]. This c a n be d o n e since if M is a p a r t i a l a l g e b r a , i.e., a n a l g e b r a whose o p e r a t i o n s are n o t n e c e s s a r i l y d e f i n e d f o r all a r g u m e n t s , t h e n Sm(M) is also a closed u n b o u n d e d s u b s e t of
m'~"o(A)"
It is easy to see that for ~ < A,
[PiJ : J < ar
To see that let
< a I be as r e q u i r e d by Cor~a(/z), i.e., a ~ / z and ar ~ / ~ for s < a.
Since
~ < A
D =Is 9
St%(A):2q[2 ~;q 9
above
Cona(h ) -~ Con2(~).
q]] 9
we
DS(A ).
can
apply es
Since
oq
Cow.2(A )
C~a) [ P i j "J 9 for
i
and
and Nar
obtain ispredense
a-----]~
we
have
0
D =(D N
S,(/~))U o
By T9
Lemma
1.11
D 0(A ).
since D f~
S~(A-/~)
(where A U B = I x
13Y : x 9
^y 9
o
for
Tc
T h e r e f o r e D /~
Sso(A),
if
.~So(/~) 9
TU
SN0(B) 9
D~o(A tJ B)
D0(/~ ) which e s t a b l i s h e s
SSo(]~ ) is e x a c t l y t h e set r e q u i r e d for Cona(l.~).
then
Con2(l.0
83
2.2
],emma:
Cone(21PI ) -* YX~I~ 0 Con2(X),
and
hence,
since
/z < X ^ Cone(h ) -* Cone(/z), t h e r e f o r e (3a =* 2 IPI) (Cone(a)) ==~ YX Cone(X)). Proof:
It
clearly
suffices
Cone(21Vl )-* Ce~e(X ). L e t p ,
(see
2.1)
to
prove
Cp~.j : j
d e n o t e t h e s u b s e t [p,,~. " j < a , I of P. Let ( ~ :
that
for
h>21PI
be as in Con2(k ). L e t 3 ~
i < 2]Vl>, be a listing w i t h o u t
r e c u r r e n c e of all s u b s e t s of P. Let r q~j : j < fi~> b e a listing of t h e m e m b e r s of ~ , t h e n we c a n h a v e fli ~; [ P [ . We define a p a r t i a l f u n c t i o n F:X -* 2 IPI b y F ( i ) = t h e 7 s u c h t h a t ~7 = - ~ '
f o r i < a.
We define also two p a r t i a l f u n c -
t i o n s G a n d H o n ~, • X, into X b y G ( i , j ) = t h e 7 s u c h t h a t Pi,i = qF(~),7, f o r i
j < a ~ , a n d H ( i , 3 ) = t h e l e a s t 7 s u c h t h a t P~,7 =gr(~),/' f o r i < a ,
J < flF(i)"
Since
Cone(21P] ) h o l d s t h e s e t A = Is c
S#0(21PI ) : (3q =-p)(V
i c s n P)([qi.j : J c s N ~ ] is p r e d e n s e a b o v e q)] is in
D o(21PI ). T h e r e -
f o r e t h e r e is a p a r t i a l a l g e b r a M with u n i v e r s e 21PI s u c h t h a t A :~ SIn(M). Let N be t h e p a r t i a l a l g e b r a whose u n i v e r s e is X a n d whose p a r t i a l o p e r a t i o n s a r e t h o s e of M t o g e t h e r with F,G a n d H. We shall show t h a t f o r e v e r y s : Srn(N) t h e r e is a q =*p s u c h t h a t f o r all i t s
Na,
IP~,j'J cs
na~
is p r e d e n s e
o v e r q. This will e s t a b l i s h Conf.(A) since t h e s e t which is r e q u i r e d by Cone(>~) t o b e in
D0(X ) h a s b e e n shown to i n c l u d e Srn(N) which is in
Let s c Srn(N);
since
N
contains
all t h e
partial
D#0(X).
operations
of M
s n 21PI c ~rn(M). S i n c e Srn(M) C A we h a v e s N 21Vl c A; t h e r e f o r e t h e r e is a q =*p s u c h t h a t |
~ s N fl)([q*d : J c s N fl, i is p r e d e n s e a b o v e q).
We shall show t h a t f o r this q
(Yi E s n a ) ( [ p i j : j c s N ar
is p r e d e n s e
a b o v e q), which is all w h a t is left to prove. Let i c s n a, since s is c l o s e d under
F
also
F ( i ) c s n 2lel
[qF(~)j : J C s n fl~]
is
(since
predense
above
|qY(~),i : J C s N Pi] = IP~j : J c s N a~] [p~j : s A a ~
is p r e d e n s e above q.
Rang (F) ~ 2 IP[) and
q. this
For j c s N Pi
We will
hence,
shall
by
@,
see
that
establish
that
qF(~)d =P~,H(iJ) b y t h e
d e f i n i t i o n of H. S i n c e i , j c s also H ( i , j ) c s a n d H ( i , j ) < ai b y t h e d e f i n i t i o n
84
of H.
qF(i)j = Pi.H(ij)
Thus
j 9 s CI a~ p i j
= qF(t),c(ij) 9
lPt,j J 9 s ~ nil. In the o t h e r d i r e c t i o n , f o r IqF(i) j e s (~ fltl, s i n c e s is c l o s e d u n d e r G.
9
"
m
2.3 Theorem:
IMI,''" >
Let M =<
be a model with countably
many
r e l a t i o n s a n d f u n c t i o n s , t h e n if M is u n c o u n t a b l e . [Igl
St%(IMI):N<MIc
e
D~0(IMI).
Proof: Let Mt be an algebra with universe M and with all the operations and
Skolem
functions
St%(IMI):N<MIe
e
M
as
operations.
S~o(IMI):N.~M].
Sm(Mt) cIge Ilgl
of
Since
as
Sm(Mt) e
is
well
known,
/)~0(IMI)
also
D~o(IMI).
2 . 4 D e f i n i t i o n : F o r a c a r d i n a l ;k we d e n o t e whose transitive
Then,
c l o s u r e is of c a r d i n a l i t y
with
H(A)
t h e s e t of a l l s e t s
< A. F o r a r e g u l a r u n c o u n t a b l e
k
is a m o d e l f o r all a x i o m s of ZFC e x c e p t m a y b e f o r t h e p o w e r s e t
axiom. 2.5 Definition: Let N be an elementary
substructure of < H ( X ) , E >
l e t P e N b e a f o r c i n g n o t i o n . F o r q e P we s a y t h a t q is ( N , P ) - g e n e r i c ,
and (or
N - g e n e r i c if i t is c l e a r w h i c h P we a r e d e a l i n g w i t h ) , if f o r e v e r y s u b s e t .9 of P w h i c h is p r e d e n s e
a n d is in N
.9 (~ N is p r e d e n s e
2 . 6 L e m m a : q is ( N , P ) - g e n e r i c nal in the P-forcing Nthen Te
iff
a b o v e q.
f o r e v e r y T w h i c h is a n a m e of a n o r d i -
n o t i o n , if T e N t h e n q ]~- " T e N " (i.e., if t h e n a m e is i n
q f o r c e s t h e v a l u e t o b e in
N) ifffor
every P-name
TeN,
q ]~-"if
VthenTeN" N
P r o o f : We p r o v e o n l y t h e f i r s t "iff" , t h e s e c o n d Assume
that
q is N - g e n e r i c
and
let T e N be a name
.9 = [ r E P : v ]~- " *r = a " , f o r s o m e o r d i n a l a ] . . 9
has the same proof. of a n o r d i n a l .
is o b v i o u s l y p r e d e n s e .
Let `9 is
85 d e f i n a b l e f r o m P a n d T in H(A)0 h e n c e .9 c N. S i n c e q is N - g e n e r i c , .9 (~ N is p r e d e n s e a b o v e q. Let f w h i c h r ]~- " T = a " (~ N
is
t h e n f is d e f i n a b l e in H(A) f r o m T, h e n c e f C N. Since
predense
above
q,
q 1 [ - " G C~ (-9 N N) # r
r E G'" T h e r e f o r e i f q c G , T = f ( r )
q I[-"(ar ~.gNN) r ~(,9(hN)
b e t h e f u n c t i o n o n ~ d e f i n e d b y ~f ( r ) = t h a t a f o r
V~ G, S i n c e r o N , a l s o f ( r )
cN
(asf
i.e.,
h o l d s i n V[G],where
V[G].
oN) thus rcgin
T h e r e f o r e q I[- " T c Y " .
Now a s s u m e
that
for e v e r y n a m e
T of a n
ordinal
, if T E N t h e n N
q I~-"T~N".
Let~NbepredenseinP.
I~1 o n t o ~, h e n c e t h e r e is s u c h a n f
T h e r e is a n f
cH(~)whichmaps
c G].
in N. We t a k e r = M i n l i ' f ( i )
N
Since~f,PCNand~is obviously a name q [t-"(3i cN)(f(i)
definable fromf of a n ordinal.
E G)"
Since
a n d P in H(k), also T e N ,
andTis
By o u r a s s u m p t i o n q I ~ - " ~ E N", h e n c e f
maps
the
members
of g ~ 1-91 to
m e m b e r s of N (~.9, b e i n g in N, we h a v e q [ [ - " ( 3 r a ~ V ~ N ) ( r
c G)". T h e r e -
f o r e ,9 V~ N is p r e d e n s e a b o v e q which is w h a t we h a d to prove.
2.7 R e m a r k : F o r A ~ IP
it d o e s n o t m a t t e r
in Con2(A) w h e t h e r we
r e q u i r e t h a t f o r e a c h i < a IP~j : J < a ~ b e p r e d e n s e o r w h e t h e r t h i s s e t is predense
above p.
On t h e f a c e of it t h e v e r s i o n w h e r e .we r e q u i r e
the
I P i j : J < a ~ to be p r e d e n s e is w e a k e r s i n c e it m a k e s a s t r o n g e r a s s u m p t i o n , b u t we n o w p r o v e f r o m it t h e s t r o n g e r v e r s i o n . S u p p o s e e a c h l p ~ j : j < a~] is predense above p.
Blow e a c h s u c h s e t b y a d d i n g to it all m e m b e r s of P
i n c o m p a t i b l e with p , t o g e t t h e s e t IP~,i : J < fl~l. S i n c e ] P I -< A I~il - ~
so
we c a n a p p l y t h e w e a k v e r s i o n of Con2(k) ( ~ m a y be > k b u t s i n c e o n l y t h e c a r d i n a l i t y f i g u r e s h e r e it is O.K. as long as fl~ < h+). We o b t a i n a s e t A in / 9 (k) s u c h t h a t f o r s c A we h a v e a q
~p
f o r which f o r e a c h i
cs
V~athe
D
s e t ~p~j : ] • s
C~fl~ is p r e d e n s e a b o v e q. F o r a ~ - < j < ~ , p ~ , j
is i n c o m p a t i -
ble with p a n d h e n c e also with q, t h e r e f o r e also t h e s e t [p~d : j c s (~ a~] is
86 p r e d e n s e a b o v e q which e s t a b l i s h e s t h e s t r o n g e r v e r s i o n of Con2(A). ~..8 T h e o r e r ~ (1) Let ~, > 2 IPI , ~, r e g u l a r
P is a p r o p e r f o r c i n g n o t i o n
iff f o r e v e r y c o u n t a b l e e l e m e n t a r y s u b s t r u c t u r e N of ( H ( A ) , c >
such that
P,p E N t h e r e is a p < q c P s u c h t h a t q is N - g e n e r i c . (2) F o r A - ~ 2 1 P I IN: N E
pert(A),
P is p r o p e r if[
~S"o(H(A))' a n d t h e r e is q ~ p
(g,P)-generic ] c
~D o(H(A))
f o r e v e r y p c P. P r o o f : (1) S u p p o s e t h e c o n d i t i o n of t h e t h e o r e m holds, a n d we shall
Conz(21PI ).
prove
Let
: i < a,j < hi>
(pij
be
as
in
Con2(21PI ).
Let
N<
< P ~ . i : i < a , j < a ~ > c N (we c a n a s s u m e
P E H(A)
an
since
H(~,) c o n t a i n s
isomorphic
copy
of P).
For
i oN,
~ i = IP~,j : J < a~] c N since it is definable in H(A) f r o m ( p i , j : i < a , j < a~> andi.
Let q ~ p
be N - g e n e r i c since IP~j : J < a ~ ] c N
a n d it is p r e d e n s e we
h a v e t h a t ~ i N N = [p~,j : j c N C~ a~] is p r e d e n s e a b o v e q. establish
Con2 (21PI )
suffices
IPI : N . ~ H ( ~ ) , c > , p , P ,
A-|NC~e D (21PI).
P,P,
it
The
to
that
the
set
set
A t = IN e S~,(H(A)): N ~ ~ H ( A ) , c > ,
( P i j : i < a , j < h i > c g] is in
implies t h a t also A c
prove
Therefore to
DM0(H(A)) b y T h e o r e m 2.3. This
D (21PI), by t h e t e c h n i q u e of using a m o d e l with ~0
o p e r a t i o n s c l o s e d u n d e r c o m p o s i t i o n which we h a v e a l r e a d y u s e d s e v e r a l times. Now we p r o v e t h e o t h e r d i r e c t i o n of t h e t h e o r e m .
Assume t h a t P is a
p r o p e r f o r c i n g , i.e., Con2(2]P]). Since A > 2 ]PI we c a n a s s u m e t h a t P c H(A). In H(~,) t h e r e is a f a m i l y < p ~ d :i < a , j
IF[,
a<-2 IPI s u c h t h a t
fo~ i < a ~ i = [p~j : j < a~] v a r i e s o v e r all t h e p r e d e n s e s u b s e t of P.
Con2(21PI ) t h e r e
is
a
partial
Srn(M) C Is e S ( 2 1 P I ) : (~q ~ p ) o v e r ql.
algebra
with
universe
21PI
such
By that
(u e s N a)IP~,i : J e s N a~| is p r e d e n s e
Since there are such~p~,i :i
a n d M i n H(A) t h e r e a r e
87 s u c h also in N. N o b v i o u s l y c o n t a i n s all t h e n a t u r a l n u m b e r s .
Since M is
g i v e n as a m a p p i n g of ~ on all p a r t i a l o p e r a t i o n s of M, all t h e s e o p e r a t i o n s b e l o n g t o N a n d h e n c e N is c l o s e d u n d e r t h e m . T h e r e f o r e N N 21PI 9 SIn(M) a n d t h e r e f o r e t h e r e is a q - > p p r e d e n s e o v e r q).
s u c h t h a t (Yi 9
Na)(lP~j
:J 9
Nazi
is
L e t `9 be a n y s u b s e t of P in N w h i c h is p r e d e n s e in P .
S i n c e in N it is t r u e t h a t " f o r e v e r y p r e d e n s e s u b s e t of P t h e r e is a n i < a s u c h t h a t ,9 =.9~ " ( s i n c e t h i s is t h e way we g e t
c l e a r l y t h i s is t r u e in N. T h e r e f o r e .9 = .9r f o r s o m e i 9 N N a. j 9
then alsop~j
one-to-one,
{pr
if
pij
9
9
(since
j 9
<ar NN=.gr N N = . g N N
:J 9 Na~-
in N)
F o r t h i s i, if
c a n be t a k e n to b e Thus
~pcj : j 9
a n d we k n o w t h a t
Nar
=
t h i s s e t is d e n s e
a b o v e q ( s i n c e i 9 N N a). T h u s we h a v e s h o w n q ->p to b e N - g e n e r i c . (2) Left to t h e r e a d e r .
2.9 Discussion: We shall n o w p r e s e n t a n o t h e r p r o o f of t h e f a c t t h a t if P s a t i s f i e s c.c.c, t h e n i t is p r o p e r . We shall p r o v e t h a t if T is a n a m e of a n ordi.q
nal, N ~ < H ( A ) , e >
and
TeN
then # I~-"TeN".
There is a maximal
a n t i c h a i n .9 s u c h t h a t f o r e a c h p 9 .9, p ~ "T = a " f o r a u n i q u e a. B e c a u s e of t h e c.c.c. 1`9] ":~0 so we c a n t a k e ` 9 = ~ p ~ ' i < a l ,
a<-c0.
The s e q u e n c e
< p i : i < a > is in H(A) a n d its p r o p e r t i e s c a n b e f o r m u l a t e d in H(A). T h e r e f o r e t h e r e is s u c h a s e q u e n c e in N. i 9 a,
and
if a i
is t h e
ordinal
Since ~cN
such
that
we h a v e p r
9
for every
p i I ] - " T 9 lar : i 9 ~ ] "
hence
r I~-"T~N". 2.10 T h e o r e m : If t h e f o r c i n g n o t i o n P is } I v c o m p l e t e t h e n it is p r o p e r . Proof:
Let
k be
p,PeN~
large
enough,
i.e.,
A regular
IN[ =~o. Let <.9~:i
and
k > 2 IPI,
let
be a list of all predense
sets which are in N. We define the sequence
NAP
by induction
on n:
po=p
and
T h e r e is s u c h a P ~ + I e N since P n + l ] [ - " P n is c o m p a t i b l e with s o m e m e m b e r s
88 of .gn";
.9~ b e i n g p r e d e n s e
of P t h e r e
is a q
s u c h t h a t q ~- P n f o r a l l n < co. q is N - g e n e r i c s i n c e f o r e v e r y p r e d e n s e
subset
.9i of P i n N,
`9n ( 3 N
s e t Q is p r e d e n s e
i n N. By t h e M1- c o m p l e t e n e s
is p r e d e n s e
a b o v e q s i n c e q ~ r n,
vn E . g n ( 3 N -
a b o v e q if f o r e v e r y p > q t h e r e is a m e m b e r
(A
of Q w h i c h is
compatible with p, but does not have to be > q). 2.11 Theorem= Let P E N <
L e t N[G] = I T [ G ] : r is a n a m e P r o o f : By r e p e a t i n g
, a n d let G b e a g e n e r i c s u b s e t of
^ T E N ] . T h e n we h a v e N[G] .< H(v[G])(~,).
the Forcing theorems
f o r N a n d .H(k) C l a i m I 5 . 1 7
i m p l i e s N[ G] C H(A) v [ c ] = H ( k ) [ G]. L e t ~o(x,y 1 . . . . .
Yn) b e a f i r s t o r d e r
/-/(v[c])(A) ~ (~x) 9 ( x , y 1 . . . . . a n z E N[G] s u c h Vaught criterion let
T1 .....
that
H(ViCl)(A) ~ ~o(x,y 1 . . . . .
be
names
of
II-" By
the
forcing
Ca* e H()Q)~o H(*)
the
above
names
(x,T 1 .....
lemma
(x,T 1 .....
T~)':
Y l .....
By t h e is
a
is
a
name
T n),
Then
"existential
complete-
a
such
T E H(?,)
r
where
Yn E N [ G ] ,
that
~oHC~')(a,T1 . . . . .
therefore
T2) -" 9H(A)(T,T1 . . . . .
T E H(A).
T~) ':
such
(~x
c
Forcing
that
H(~k))9H(I) statments
r e l a t i v i z e d t o HV[G](~) c a n b e d e f i n e d in H ( ~ ) , h e n c e H(A) ~ (3 a n a m e such
that
~b ] ~ - " ( ~ z g ( x , T 1 . . . . .
Tarski-Vaught
criterion
V[G] ~ ~OH(X)(T,T1 . . . . .
Tn))g( x)'"
-~ ~0(T,T I . . . . .
f o r N < H(A) t h e r e
if
Yn.
name
Tn) -* a E HCA) ^
there
~b I]- " a E H(A) -, a = T ' "
Since Yl .....
of
there
that
yn). T h e n , b y t h e T a r s k i -
w e s h a l l h a v e N[G] -
We s h a l l p r o v e
Yn E N[G] t h e n t h e r e is
yn) for some Yl .....
V[G] " ~ (~x E H ( A ) ) ~ (H(x)) (X,T 1 . . . . . ness"
formula.
is s u c h
Tn)H(X)].
a name
7) [
By
T E N.
the Thus
T~) a n d T [ G ] E N[G].
2.12 Theorem: Under the assumptions
of t h e l a s t t h e o r e m , t h e f o l l o w i n g
three conditions are equivalent. (a) G CI N is N - g e n e r i c ,
.gAN
N C #O. (b) N[ G ] N On = N N On.
i.e., f o r e v e r y .9 E N w h i c h is p r e d e n s e
in P ,
89 (c) N [ C ] ~ V = N P r o o f : (a)
x 9 N[G], ~q=|p
(c) . Let x 9
~
x = r[G]
9
=[p
~ V
:(3y)(p [~-r=y) 9
N V. We shall p r o v e x 9
for vp
some
-r 9 N.
Let
[[--T~ V]
I[-r=y) vp
9
Since
(by l e m m a 2.6 , if T E H ( h )
IFTCH(h)]
aqd p I~-"T=Y"
t h e n y E H ( k ) ) . A is obvi-
o u s l y p r e d e n s e in P . S i n c e ,9 is d e f i n a b l e in H(/Q f r o m T a n d T E N also ,9 9 N. ~ G. Since V[G] ]="T[G] 9 V " w e c a n n o t
By (a) t h e r e is a p 9 have p I}-"T~H(k)", x =r[c]
=y
9
hence
for
some
y 9
P I~-"T=Y",
hence
~ V.
(c) ==~ (b) is obvious. (b) ==e> (a). Let .9 9 N be p r e d e n s e in P . Let ~ t be a m a x i m a l a n t i c h a i n of P, s u c h t h a t f o r e v e r y p 9
p-~f(iv),f(p)
9
Since ~ q 9
t h e r e is s u c h a n .~t E N; since in Ht2Q t h e r e is a s e q u e n c e < q # : fl < a> c o u n t i n g t h e m e m b e r s of ~Ot t h e r e is s u c h in N. Let ~- be t h e c a n o n i c a l n a m e such that forfl
T[G] = f l i f q # 9
G. So T is a P - n a m e of a n ordinal, T 9 Since < q p : B < a >
L e t T[G] = 7 9
9
also q7 9
q7 9 G (since .~t is a m a x i m a l a n t i c h a i n , a n d if f o r s o m e
iTE GthenT[G]
=5),thusi
79
C~ G a n d . g N N
N G #~b.
2.13 C o r o n a r y : U n d e r t h e a s s u m p t i o n s of t h e t h e o r e m t h e following a r e equivalent: (a) q in N - g e n e r i c . (b) q I k N[G] C~ On = N VI On. (c) q II- N [ a ] n V = X n Y. P r o o f : E a c h of t h e p r e s e n t (a) - (c) is e q u i v a l e n t to t h e s t a t e m e n t t h a t t h e c o r r e s p o n d i n g c o n d i t i o n in t h e l a s t t h e o r e m h o l d s f o r all g e n e r i c s u b s e t s G of P which c o n t a i n g
90 w
P r e s e r v a t i o n of p r o p e r n e s s u n d e r c o u n t a b l e s u p p o r t i t e r a t i o n .
3.1 D e ~ U o , :
We call V = < P,,e. :~ < ~> (or < q.: ~ < ~>) a system of
countable support iterated forcing (or a CS iterated forcing system etc.) if the following holds. P~ = | f
: D o m ( f ) is a c o u n t a b l e s u b s e t of i a n d (Vj 9 Dora Pj-name
Q is a P f n a m e ~i
andf
tj
[~Pt " f ( J )
(f))[f(j)
is a
9 Q. '~]~" -3
of a p o s e t . We u s u a l l y r e s t r i c t t h e v a l u e s of f to b e c a n o n i c a l
n a m e s of o n e k i n d o r a n o t h e r to p r e v e n t P a f r o m b e i n g a p r o p e r class. The p a r t i a l o r d e r <~ o n P~ is d e f i n e d b y f ~g
r
(Vj e D o m ( f ) )
g tj I~-"f(J)':Qj g(J)"
F o r e v e r y j ~ D o m f we t a k e f ( j ) t o b e a n a m e ~bi of t h e m i n i m a l m e m b e r of Qj.
The
poser
defined
by
this
system
is
the
poser
Pa.
We
call
< a>
is a c o u n t a b l e - s u p p o r t i t e r a t e d f o r c i n g
P,, = Lira< Qj : j . 3.2 Theorem= If
< P~,Q.: i
s y s t e m a n d f o r e a c h i < a,
]~-p~" Q is p r o p e r "
then
Pa is
proper.
P r o o f : In T h e o r e m 2.8 we s h o w e d t h a t P is a p r o p e r f o r c i n g iff f o r s o m e A > 81PI e v e r y c o u n t a b l e e l e m e n t a r y s u b s t r u c t u r e P~
N of < g ( A ) , e >
such that
9 N h a s a q , p -: q 9 P s u c h t h a t q is N - g e n e r i c . As e a s i l y s e e n f r o m t h e
p r o o f it s u f f i c e s to r e q u i r e t h i s o n l y f o r all s u c h N which c o n t a i n s s o m e fixed m e m b e r y of H(A).
F o r o u r p r e s e n t p r o o f we c h o o s e a r e g u l a r c a r d i n a l A which is v e r y l a r g e with r e s p e c t to I P a l , a n d we shall show t h a t P a is p r o p e r b y showing t h a t f o r every
countable
elementary
Pa 9
substructure
and for allp
9
N
of
H(A)
such
C~N t h e r e is a p
that 9
91
w h i c h is N - g e n e r i c . We shall show b y i n d u c t i o n on j ~ a t h a t e v e r y Pj 9 N h a s a somewhat stronger property: (*) F o r all i < j ,
i 9
a n d f o r all p 9
,
and q 9
if q is
a n d q : ~ p r i t h e n t h e r e is a n r 9 Pj s u c h t h a t r is (N,Pj)-
(N,Pr
g e n e r i c , r ~ - p a n d r ~- q a n d r F i = q. F o r j = 0(*) is v a c u o u s l y t r u e . Since j+l
Now we a s s u m e (*) f o r j a n d p r o v e it f o r j + l .
9 N also j 9 N. T h e r e f o r e , s i n c e (*) h o l d s f o r j we m a y a s s u m e ,
w i t h o u t loss of g e n e r a l i t y t h a t i = j . Let Gj b e a g e n e r i c s u b s e t of Pj w h i c h contains
By T h e o r e m 2.11 we h a v e N[Gi] ~ H ( A ) [ G j ] = H ( ~ ) v[G~], s i n c e
q.
Pj 9 N ( b e c a u s e j , ~ P C , Q : i < a ~ 9 N a n d Pj is d e f i n a b l e in H(~,) f r o m j a n d ~Q
Q 9 a n d h e n c e Q [Gj] 9 ~J ~i forcing in V[Gi]. Since p,j
R e m e m b e r t h a t Q.[Gj] is a "~ also P ( J ) = P l 9 and
:i
proper
9
pj[Gj] 9 Q.[Gj]; s i n c e ~, is still s u f f i c i e n t l y l a r g e a n d Qj[Gj] is p r o p e r t h e r e is a n r i 9 Qj[Gj] s u c h t h a t r i ~pi[G] a n d vj is ( N [ G j ] , Q j [ G j ] ) - g e n e r i c .
Since
t h e only r e q u i r e m e n t we h a d a b o u t t h e g e n e r i c s u b s e t G~. of Pj was t h a t is c o n t a i n s q , q f o r c e s t h e e x i s t e n c e of a n r t a s a b o v e . completeness
lemma
q ]~-j"r 9 Q ^r.;~pj ~j ~j ~j r =q
tJ|(J,ri)].
I
3.1
^r ~j
is
0bviouslyr
there
is
a
name
(N[G.] , Q . ) - g e n e r i c . ~~ ~j 9
By t h e e x i s t e n t i a l r ~j "
such We
set
= q . Also s i n c e q ~ p
andrrj
that now
rj and
q [~- "'r ~PJ "we h a v e r ~ p . We still h a v e to p r o v e t h a t r is ( N , P j + l ) - g e n e r i c . ~i
By t h e c o r o l l a r y in 2.13 in o r d e r to p r o v e t h a t r is ( N , P j + l ) - g e n e r i c it s u f f i c e s to p r o v e t h a t f o r all g e n e r i c s u b s e t s
N[G] (~ On = N V~ On. r e G , q e Gj.
Let
G/
be
the
G of P j + I w h i c h c o n t a i n r ,
part
of
G
up
to
j.
Since
S i n c e q is (g,Pi)-generic we h a v e N[Gj] N On = g F ~ On.
L e t G c ~Qj[Gj], b e t h e " j - t h
component
" of G.
S i n c e r e G , r j [ G ] e G.
S i n c e r j is (N[Gj], Q [ G j ] ) - g e n e r i c we h a v e N [ G j ] [ G ] N On = N[Gi] F~ On, a n d using
the
equality
observe that
a b o v e we g e t N [ G j ] [ G ] V~ On = N F~ On.
We h a v e
N[G] c N [ G j ] [ G ] , t h e n we h a v e N[G] V~ On = N F~ On.
to For
92 every Pj+vname definable
from
T in N t h e r e is a n a m e Y e N T and
Pj+I,
and
is
hence
as in l e m m a II. 1.5 (Y is in
N).
By T h e o r e m
2.11,
T[a] = v [ a j ] [ ~ ] c N [ a j ] [ ~ ] , Now we c o m e t o deal with t h e c a s e w h e r e j ('i" n : n < ~ >
is a limit ordinal. Let
be a s e q u e n c e of all P~.-names of o r d i n a l s w h i c h a r e in N.
N N J is a c o u n t a b l e set, so let i = i 0 < i 1 < - - .
< ~), be a
s e q u e n c e cofinal in it. For
p 1 9 P1
and
q U(plt(j-in)).
q 9 P%
such
that
Since q > - p l t i n ,
q >- P 1 t ~
q-UplePj
let and
q U P 1 denote q-Upl~'pl
.
For
P l , P 2 9 Pj we write p 1 m P2 f o r p l < P 2 ^ P2 ~ P l .
We define now two s e q u e n c e s < q ~ : n < o > a n d < p n ' n qo = q , P o = P a n d f o r a l l n
< o> such that
(1) qn e P~ a n d q~ is ( N , P ~ ) - g e n e r i c
(z) q~+l t in (3) p n 9
= q~
andDom(pn)
=g
nJ
(4) q~ _>p,~ t~, (5) Pn+1 t in = P n ~ in a n d qn+l U P n + I >-Pn (6) qn ]l-p~, "(as 9 P1 n N) (aqt ~ q n ) ( s t in < q t ^ q t e Gt~% ^ q f U s ~ q t u p n ) ' "
(7) qr~ U Pn+l II"ej '"rn 9 g "
Let us a s s u m e t h a t qn a n d Pn a r e defined a n d t h a t t h e y s a t i s f y (1) - (7). We shall now define P n + l and qn+l- Let G b e a g e n e r i c s u b s e t of P~ s u c h t h a t qn E G. We shall see t h a t t h e r e a r e q t c (a) q t - ~ q n a n d f o r s o m e s (b) s ~ [ i n -= q*, By (6) t h e r e is a n s
cPj
s ~ c Fj AN, 9
G ands ~
AN,
s tin aq*,
qt-Upn
mqt-~s.
s ~ >- s a n d s ~
NNandaqt:*qn
, s tin
n.
93 qt ~ s m qt -(JPn. The s e t -9o = Is * : s * is i n c o m p a t i b l e with s, o r s ~ -~ s a n d s ~ d e c i d e s t h e value of Tn{ is obviously a p r e d e n s e s u b s e t of Pj.
This s e t
b e l o n g s t o N s i n c e it is definable f r o m t h e p a r a m e t e r s s a n d Tn w h i c h a r e in N. Let .9 d e n o t e t h e s e t w h i c h c o n s i s t s of t h e r e s t r i c t i o n s of t h e m e m b e r s of -9o to/-n. ~q is a p r e d e n s e s u b s e t of P~. Since qn is ( N , P ~ ) - g e n e r i c .9 (~ N is p r e d e n s e a b o v e qn. T h e r e f o r e . g f ~ N CI G ~ b .
Let r E . g f ~ N
(~ G, t h e n " r
is a r e s t r i c t i o n t o / n of a s* E Pj s u c h t h a t s* _~ s a n d s* d e c i d e s Tn ' ' i s t r u e in H(h).
By Tarski V a u g h t ' s c r i t e r i o n t h e r e
r =s*r/n
E G, a n d w e c a n t a k e q t t o b e - ~ s ' r / n .
is s u c h a n s* in N.
Thus
Thus qt,s * are as r e q u i r e d
b y (a) a n d (b) . T h e r e f o r e V[G] ~ s * ( 3 q t c
G)(s* a n d q t a r e as in (a) a n d
(b) , a n d s * is t h e first s u c h e l e m e n t in s o m e fixed well o r d e r i n g of P i ' ) By t h e existential
completeness
qn [}- (3qt c G)
lemma
there
is
a
P~-name
g
such
that
[~,q+ a r e as in (a) a n d (b) a n d g is t h e l e a s t s u c h ] .
Since
e a c h s * is in N a n d it satisfies I Dom (s *) ] < ~0 in H(~) it, satisfies it also in N, h e n c e D a m i s ' ) c N (since a n e n u m e r a t i o n of Dora (s*) is in N).
Pn+l as follows. P n + l r i n =Pn t / n . F o r y c j
~N-/n
let
We define
Pn+l(7) be t h e P T -
n a m e of t h e m e m b e r of Q7 d e t e r m i n e d b y g (i.e., if ~ is a set of p a i r s of m e m b e r s of P~ a n d m e m b e r s of PJ N g t h e n Pn+l(7) = I ( r , t >
: ('Jrt
Now let us define q~+l- F o r e a c h s c Pj (~ N s u c h t h a t s ~/n ": qn t h e r e is , b y t h e i n d u c t i o n h y p o t h e s i s an q,~+l(s) E P~,.I s u c h t h a t q n + l ( s ) t / n = qn,
qn+l(s) >- s t/n+l a n d q n + l ( s ) is (N,Pc,+l)-generic. We define qn+l as follows. The d o m a i n of qn+l is t h e u n i o n of all t h e d o m a i n s of t h e q n + l ( s ) ' s f o r s E N as above, a n d s i n c e N is c o u n t a b l e t h e d o m a i n of qn+l is c o u n t a b l e . qn+l[/n =qn. generic
For/~--<7
of
P~,,
then
Let
q n + l i f q n c G, and G i s a V[ G] ~:~u(3s E Pj ~ N)
([qn U s m qn U P n + I ] ^ u = q~+l(s)). By t h e e x i s t e n t i a l c o m p l e t e n e s s leroxna t h e r e is a P ~ - n a m e ~
s u c h t h a t qn [}-"(as c Pj 0 N)
of a P~r
(qn -U s ~ q,~ U P n + l ^ ~ = q,~+l(s))'"
Now
~ d e t e r m i n e s c a n o n i c a l l y a QT-
c o n d i t i o n , which is t a k e n to be t h e v a l u e of qn+l(7).
94 We s h a l l n o t p r e s e n t h e r e t h e p r o o f t h a t Pn+l a n d qn+l t h u s d e f i n e d s a t i s f i e s (1) - (7),
Now we define r = (3 qn- Clearly
r b e l o n g s to Pj.
We c l a i m t h a t f o r
n ~
every n
, r ~Pn-
To p r o v e t h a t we h a v e to show t h a t f o r e v e r y ? e Dora (pn), Since7 9
r r7 I~-"i~ such
that
9' < i~ , n -: k
then,
we h a v e , b y ( 3 ) , 7 9 by
(4)
Letlc be
qk -> P~ r i~,
hence
q~ t 7 ]~ "Pk(7) <- q~(7)': By t h e d e f i n i t i o n o f t , q~ t 7 = r ~ 7 , q~(7) = r(7). Also, b y (5) p ~ ( T ) = P . ( 7 ) , h e n c e r t 7 I [ - " p n ( 7 ) ~ r ( 7 ) " Thus, b y (7) , r ]~ " T . e N " a n d t h e r e f o r e r is ( N , P j ) - g e n e r i c w h i c h f i n i s h e s o u r p r o o f .
3.3 T h e g e n e r a l a s s o c i a t i v i t y T h e o r e n ~ S u p p o s e
Proof: Left to t h e r e a d e r .
3.4 Theorem= S u p p o s e < Q j : j < a > is a (
Pj=IAm
<j>.
If I ] - p j " Q r Qt , Q a d e n s e s u b s e t of Q t " a n d P ] = L i m < ~ ~ Q t : i < j > , ~j ~j ~j ~j ~-
then P j r
Pj* is a dense s u b s e t of --P2"
R e m a r k : By L e m m a l 5.1 (a), we c a n r e p l a c e Q. b y a n y e q u i v a l e n t Q? -3 ~J ( j u s t u s e 3.4 few t i m e s ) . P r o o f : Left to t h e r e a d e r .
95 w l~artin Axiom Revisited.
Why is c.c.c, f o r c i n g so p o p u l a r ?
I t h i n k t h e m a i n r e a s o n is t h a t s u c h
f o r c i n g n o t i o n s p r e s e r v e c a r d i n a l i t i e s a n d cofinalities, so why shall we n o t be interested
in t h e p r o p e r t y "'P d o e s n o t c o l l a p s e c a r d i n a l s
" instead
"P
satisfies t h e c.c.c. " ? In p a r t i c u l a r Magidor a n d Stavi h a d w o n d e r e d o n t h e role of t h e c.c.c, m a i n l y in MA a n d a s k e d :
"Is it c o n s i s t e n t t h a t f o r a n y f o r c i n g n o t i o n P of p o w e r 81 n o t c o l l a p s i n g cardinals G~P,
(i.e.,
81)
and
dense
D r ~ P (i <81)
there
is
a
directed
G nDc~bfori<sl?
In p a r t i c u l a r B a u r n g a r t n e r , H a r r i n g t o n a n d Kleinberg [BHK] p r o v e d t h a t if S C ~l is s t a t i o n a r y c o - s t a t i o n a r y , a n d CH holds, t h e n t h e r e is a f o r c i n g notion
P s = [C : C
a
countable
closed
subset
of
S!
C1 ": C2
iff
C1 = C2 N (Sup C1§ ) w h i c h d o e s n o t c h a n g e c a r d i n a l i t i e s a n d cofinalities a n d w h i c h c o l l a p s e S ( i.e., c o l l a p s e its s t a t i o n a r i t y , i.e., I~-?s " S C co1 is n o t stationary "
So why n o t i n c l u d e s u c h f o r c i n g in MA? B e c a u s e we c a n find pairwise disj o i n t s t a t i o n a r y s e t s Sr~ r ~ 1 , ~ 1 = (J S n . s t a t i o n a r y , ~o1 m u s t be collapsed. ings Ps,, a f t e r r s t e p s not matter ~/)a~ = I C c P s ~
If we m a k e e a c h Sn in t u r n n o t
More e x a c t l y , if we t r y to i t e r a t e t h e f o r o -
81 collapses, no m a t t e r how t h e limit is t a k e n . It d o e s
if we l o o k a t t h e d e s i r e d v e r s i o n of MA, in s o m e :Max C~-a~.
Thus if G ( ~ ) a ~ r
forn
V a n d let
t h e n in
V, e a c h S n is n o t s t a t i o n a r y .
You c a n still a r g u e t h a t CH is t h e c a u s e of t h e p r o b l e m b u t we shall p r o v e in t h e o r e m 4.4 t h a t e v e n ~ 0 > 81 , S ~ a i s t a t i o n a r y c o - s t a t i o n a r y t h e r e is a f o r c i n g n o t i o n P of p o w e r 81, n o t c h a n g i n g c a r d i n a l i t i e s a n d cofinalities b u t still c o l l a p s i n g S.
96 S o it is n a t u r a l stationary
to change
t h e q u e s t i o n t o " P of p o w e r tl l, n o t c o l l a p s i n g
s u b s e t s of tl i "', a n d we s h a l l a n s w e r i t p o s i t i v e l y , a s s u m i n g t h e r e is
a m o d e l V of ZFC w i t h a s t r o n g l y i n a c c e s s i b l e c a r d i n a l .
The natural
s c h e m e is t o i t e r a t e
1~l, i n a n i t e r a t i o n
( b y CS i t e r a t i o n ) p r o p e r f o r c i n g of p o w e r
of l e n g t h co2. H o w e v e r t o p r o v e t h e c o n s i s t e n c y
anything
by iterating
condition
is s a t i s f i e d , w h e r e g will b e t h e n e w 1~2 a n d t h e l e n g t h of t h e i t e r a -
tion.
proper
of a l m o s t
We h a v e a p r o b l e m
large power because using g strongly
Another
of t h e m a n y
, I~-P "IQI =l~l "'
names.
We c a n o v e r c o m e
or showing that
P * Q have a this either
t h e s e t of n a m e s
by
which are
c o u n t a b l e is d e n s e .
p r o b l e m is t h a t " n o t d e s t r o y i n g
the same as "proper". ]]-c~(~,,2tpl)
e v e n if I P] = ~ l
inaccessible,
essentially hereditarily
f o r c i n g we u s u a l l y h a v e t o p r o v e t h e g - c h a i n
s u b s e t s of co1'' is n o t
H o w e v e r we s h a l l p r o v e t h a t if P is n o t p r o p e r , t h e n
"P destroys a stationary
dealing with a candidate
stationary
s u b s e t of col''. S o i n s t e a d of " h o n e s t l y "
P i.e., a f o r c i n g n o t i o n w h i c h d o e s n o t d e s t r o y
sta-
t i o n a r y s u b s e t s of o l , b u t is n o t p r o p e r we c h e a t a n d m a k e i t t o d e s t r o y a s t a t i o n a r y s u b s e t of co1.
4.1 T h e o r e i ~ power
S u p p o s e Q =
"
i < g > is a CS i t e r a t i o n
II-p~
"
~Q
h a s
gisregularand(Y/z
T h e n P ~ = Lira Q s a t i s f i e s t h e g - c . c . , a n d e a c h Pi(i < ~) e v e n h a s a d e n s e s u b s e t of p o w e r
< g. H e n c e f o r i < g,
I]-P~ "28~ < ~ " .
P r o o f : By L e m m a I. 5.4 , w.l.o.g, t h e s e t of e l e m e n t s /z. < ~; ( i . e . , / z . is a P ~ - n a m e of a c a r d i n a l For a forcing notion P and P-name ments/z of Q:
of Q. is a c a r d i n a l
< ~). Q of a f o r c i n g n o t i o n w i t h s e t of e l e -
( a c a r d i n a l ) we d e f i n e a h e r e d i t a r y
countable
P-name
of a m e m b e r
97 it is t h e c l o s u r e of t h e s e t of o r d i n a l s b y t h e following two o p e r a t i o n s : (a) if ~'n(n < co) a r e s u c h n a m e s , p ~ ~ P ( n < co) t h e n l e t T b e t h e T~ f o r the firstn
, P n c G , a n d ~b if t h e r e is n o s u c h n . ~p
(b) if T n , m ( n < ~ T <~
, m <~)
such that for every n
a r e s u c h n a m e , let T b e t h e f i r s t o r d i n a l < co] is p r e d e n s e
, |'rr,,m'm
o v e r T; a n d ~b if
t h e r e is n o s u c h We s h a l l p r o v e b y i n d u c t i o n o n ~ ~ g t h a t P I s a t i s f i e s t h e g - c h a i n c o n d i tion. S u p p o s e t h i s h o l d s f o r e v e r y r < 6, so f o r r < ~ we h a v e b y L e m m a I. 3.6 < g p o s s i b l e v a l u e s of ~
< g (as ~ is r e g u l a r ) .
(we c a n a d d t o ~r t h e o r d i n a l i , d e f i n e b y P~ = I f : f
/~r ,z i < / ~
s u c h t h a t i is m m i n i m a l ) .
of a n o r d i n a l
We n o w c a n p r o v e b y i n d u c t i o n o n r
PC, u s i n g t h e p r o o f t h a t p r o p e r n e s s that
~ w.l.o.g. ~r =/~r Let us
a f u n c t i o n w i t h d o m a i n a c o u n t a b l e s u b s e t of ~ , je (i)
is a h e r e d i t a r i l y c o u n t a b l e P c n a m e order.
Induction on ~-
I P ~ l <- I~l ~~ so f o r ~ < g ,
P ~ C P~ i n h e r i t its
$, t h a t P~ is a d e n s e s u b s e t of
is p r e s e r v e d b y CS i t e r a t i o n .
P~has
a d e n s e s u b s e t of p o w e r
It is c l e a r
So we
finish. F o r $ = g, if p~ c P ~ f o r i < g, c l e a r l y S = li < g : cf i = I~l] is s t a t i o n a r y , f (i) = S u p [i f-I D o m p i ] < i is a p r e s s i n g d o w n f u n c t i o n , h e n c e o n s o m e s t a tionary
S 1 c S , h has a constant
C~gsuchthatiffl~ ary, hence
C , a
v a l u e 7-
T h e r e is a c l o s e d u n b o u n d e d
ThenDompa~fl.
h a s p o w e r ~ a n d f o r a , B c C f"l $1
Pa F7 , P~ r7 are compatible
S o S 1 f~ C is still s t a t i o n , Pa,P~
are
(in P~ o r P~, d o e s n o t m a t t e r ) .
compatible
iff
B u t we h a v e
p r o v e d t h a t Q7 s a t i s f i e s t h e g - c h a i n c o n d i t i o n , so we finish.
4.2 T h e o r e l i x S u p p o s e P is n o t p r o p e r , t h e n t h e r e is a n t ~ l - c o m p l e t e forcing notion Q such that S ~ ~ l "-
] Q ] -- 2 IPI a n d
I~-Q " P c o l l a p s e s o m e s t a t i o n a r y
98 P r o o f : As P is n o t p r o p e r , t h e r e is a s t a t i o n a r y S C
_~' (~) w h i c h P 0
d e s t r o y s , S0 < ~ ~ 2 IFI. So t h e r e a r e P - n a m e s F ~ of n - p l a c e f u n c t i o n s f r o m ~n
Let Q = Col(Ml,~) = I f : f
t o ~, s u c h t h a t I~-p "Srn(<~,F~0,""" > ) C~ S = r a f u n c t i o n f r o m s o m e a < e l into/~]. F a c t A: {~-<; "' S is a s t a t i o n a r y s u b s e t of
SS0(~ ) ' '
This is b e c a u s e Q is
s, -complete hence proper. F a c t B: The s t a t e m e n t {~-p " S c
S s (/~) is n o t s t a t i o n a r y " i s a b s o l u t e , 0
i.e., if it h o l d s in V it h o l d s in VQ. We j u s t h a v e to c h e c k t h a t t h e P - n a m e s #~ c o n t i n u e to s a t i s f y t h e suitable r e q u i r e m e n t . F a c t C: J~-Q " t h e o r d i n a l ~ h a s p o w e r S 1 '" Which is trivial. F a c t D: If f o r c i n g by P d e s t r o y s a s t a t i o n a r y s u b s e t of
SMo(A)(A = ~ in
o u r c a s e ), A of p o w e r Sl, t h e n f o r c i n g b y P d e s t r o y s s o m e s t a t i o n a r y s u b s e t of r
(follows f r o m Claim 1.12. (3)).
4.3 T h e o r e m : S u p p o s e ZFC h a s a m o d e l with a s t r o n g l y i n a c c e s s i b l e c a r dinal ~. T h e n ZFC h a s a m o d e l in w h i c h 2s~ = S 2 a n d (*) If P is a f o r c i n g n o t i o n of p o w e r S 1 n o t d e s t r o y i n g s t a t i o n a r y s u b s e t s of~l
, ~)i C P d e n s e f o r i < ~ l
t h e n t h e r e is a d i r e c t e d G c P
, G ~)r
#
f o r i < ~t-
Proof: Notice t h a t / 3 " V ~ > S A-c.c.] the~..
VP ~
~<" =
0^~
{p{ _<~^ ( ~ A - - z ) [ P h a s t h e
~ ^ ~, > S 0.
This is p r o v e d e x a c t l y as t h e parallel f a c t in T h e o r e m II 3.4.
99 Now l e t [ S a : a < ~] b e a p a r t i o n of ~ to ~ s e t s s u c h t h a t f l e S a ==> fl :~ a , a n d ]S=[ = ~.
Let < <
Define
:e 9
by
induction
be a l i s t
on
i < ~
a
CS i t e r a t e d
forcing
system
of t h e c a n o n i c a l P a - n a m e s of p o s e t s on r
The
inr u c t i o n h y p o t h e s i s f o r i < ~ are: (1) Q1 is p r o p e r f o r j < i.
(~)
As~,uming< Pj,Q/
IPil <";. :j
< i> is a l r e a d y
defined,
let
I <~l' if [~-p,<~i,<> /s p r o p e r
-q,= [col (.,,e"')
o,her'..' e
C l e a r l y Pi is p r o p e r (by T h e o r e m 3.2 a n d r e m e m b e r i n g t h a t Col (St,2 #') is S 1c o m p l e t e h e n c e b y T h e o r e m 2.10 p r o p e r ) . We still h a v e to c h e c k t h a t ]Pi[ < b u t it is e a s y , n o t e t h a t we u s e ~<~ = ~ which we h a v e b y t h e first o b s e r v a t i o n in t h i s p r o o f a n d T h e o r e m 4.1. T h e o r m e 4.1 ) s a t i s f y t h e ~-c.c.
Finally also P z is p r o p e r
and (again by
which m a k e s it p o s s i b l e to p r o v e V/" ~ (*)
e x a c t l y a s in T h e o r e m II 3.4. Note t h a t in view of T h e o r e m 4.1 we h a v e a p a r a l l e l of MA f o r p r o p e r f o r c ing w i t h o u t a s s u m i n g a n i n a c c e s s i b l e . 4.4 T h e o r e i ~ S u p p o s e S c r stationary unbounded
too).
Then there
is s t a t i o n a r y c o - s t a t i o n a r y (i.e., ~ , - S
is
is a f o r c i n g n o t i o n P s w h i c h s h o t a c l o s e d
C c S (i.e., a d d s u c h a s e t ) w i t h o u t c o l l a p s i n g c a r d i n a l s
(or
c h a n g i n g cofinalities). R e m a r k : So we c a n n o t a n s w e r Magidor S t a v i ' s q u e s t i o n p o s i t i v e l y in t h e original version. R e m a r k : A s s u m i n g CH t h i s was d o n e b y B a u m g a r t n e r
, Harrington,
K l e i n b e r g [BHK] Without CH, A v r a h a m [A] a n d B a u m g a r t n e r
[B] i n t r o d u c e
forcings notions
which
add
a new closed unbounded
subset
of
~1 ( f o r
700 d i f f e r e n t p u r p o s e s ) . We c a n a d o p t e a c h f o r p r o v i n g 4.4., a n d will use a f o r c i n g similar to A v r a h a m ' s . P r o o f : Let P = Col(~o, < ~l)- So P is e s s e n t i a l l y a d d i n g ~l C o h e n reals. Gp ~ P is ( d i r e c t e d a n d ) g e n e r i c over V, t h e n Gp is also g e n e r i c o v e r L ( t h e c o n s t r u c t i b l e u n i v e r s e ) as P 9 L. By T h e o r e m I. 6.7 Gp satisfies t h e c o u n t a b l e
If
chain condition and ~[c~] = ~=
l ~ G p ] a n d V , V[Gp] h a v e t h e s a m e c a r d i -
nals a n d cofinalities. Let Q = ~C : C a c l o s e d b o u n d e d s u b s e t of S w h i c h b e l o n g to C 1 < C2 i f f C l = C 2 N ( M a x
L[Gp]]
Ci+l )
Clearly Q is a f o r c i n g n o t i o n of p o w e r Sl, so it c a n n o t c o l l a p s e c a r d i n a l s o r r e g u l a r i t y of c a r d i n a l s e x c e p t p o s s i b l y Si- So we shall p r o v e t h a t P * Q d o e s not
collapse
(p,q) 9
* Q9
Si.
So
let
, (p,q) 9
(in
V)
N ~
N (h~l 9
g
and suppose
countable,
G p c P is g e n e r i c
Gp. As P satisfies t h e c o u n t a b l e c h a i n c o n d i t i o n , p is (N,P)g e n e r i c , h e n c e N[Gp] N V = N . In L[Gp] we h a v e a n e n u m e r a t i o n of Q[Gp] N N[Gp] (of l e n g t h w) ( b u t n o t of t h e s e t of d e n s e s u b s e t s ); in f a c t we over V , p 9
h a v e it e v e n in
L[Gpr(6+l)],
~=NNwl.
Now in L[Gp] t h e r e is a C o h e n
g e n e r i c r e a l o v e r V [ G p [ ( 6 + I ) ] a n d we use it to c o n s t r u c t
.L[Gp].
V C N[Gp] N Q[Gp]
is
generic
qt = uC n 9 Q[Gp] is (N[ Gp],~Q[Gp])-generic.
over
Cn 9
Q[Gp] in
V [ G p r (6+1)].
So
Going to n a m e s we finish.
w On Aronszajn trees. 5.1 Definition: A c a r d i n a l z is said to h a v e t h e t r e e p r o p e r t y if e v e r y t r e e of h e i g h t z in w h i c h e v e r y level h a s <:z m e m b e r s h a s a b r a n c h of l e n g t h z. A t r e e w h i c h is a c o u n t e r e x a m p l e to t h e t r e e p r o p e r t y of ~ is called a ~-
Aronszajn tree. By t h e Konig infinity l e m m a ~ h a s t h e t r e e p r o p e r t y .
101
A ~-Aronszajn tree in which every antichain
a , ; - S o u s l i n t r e e . An ~ - A r o n s z a j n Aronszajn
tree , and an
and a Souslin tree, respectively.
is of c a r d i n a l i t y < ~ is c a l l e d ~l-Souslin tree are called an
An A r o n s z a j n
t r e e is s a i d t o b e
s p e c i a l if i t is t h e u n i o n of l~0 a n t i c h a i n s . A s p e c i a l A r o n s z a j n t r e e c a n n o t Soustin, since in a Souslin tree every antichain being uncountable,
is c o u n t a b l e ,
be
hence the tree,
c a n n o t b e t h e u n i o n of l~0 a n t i c h a i n s .
R e m a r k : I t is e a s y t o s h o w t h a t a n A r o n s z a j n t r e e T is s p e c i a l iff t h e r e is a function f
: T -* ~ w h i c h is o r d e r p r e s e r v i n g .
5 . 2 T h e o r e m = T h e r e is a s p e c i a l A r o n s z a j n t r e e T. of t h e a - t h l e v e l T a of T will b e i n c r e a s i n g b o u n d e d
Proof: The members sequences define
of r a t i o n a l m e m b e r s
Ta we a s s u m e
that
q >Sup(x)
there
rational 17#1 = n 0 .
If
a=O
numbers.
is
a
qc~^q
and
Ta = [ < >].
>Sup(z)~,
ing sequence
x such that
<~]
we introduced
~
is
the
a sequence
=a.
Let ~qn :n
<~>
=q,
set
of
all
hypothesis for every
y of l e n g t h a sequence
be an increas-
hypothesis
S u p ( x n + l ) = q,~+l- T a k e y =
U xn, then
for each x c
t T= 1 "~ t~0-
by the
Assume
induction
is I J i g n = a a n d S u p ( y )
o n l y ~0 s u c h p a i r s t h e n
and take
x n c T~n s u c h t h a t S u p ( x n ) = q,~ a s f o l l o w s : x 0 = x .
one suchy
every
s u c h t h a t q0 ~- S u p x a n d S u p n < o q n = q" We d e f i n e
is a n 2,,+ 1 c Ta**l s u c h t h a t
the length ofy
a=7+l
L e t ign b e a n i n c r e a s i n g
x n E T#~ is d e f i n e d ; S u p (an) = qn < qn+l t h e n there
If
If a is a l i m i t o r d i n a l t h e n
S u p ( y ) = q.
a n d S u p [Bin : n
of r a t i o n a l s
now the member
Sup(y)
where
x ~ U T# a n d e v e r y q > S u p ( x ) we s h a l l c o n s t r u c t lg
and
O b v i o u s l y ]T a] = ] Ta] . n o = ~o. T h e i n d u c t i o n
h o l d s a l s o f o r fl < a , a s e a s i l y s e e n .
which extends
W h e n we c o m e t o
f o r a l l 9 c T# a n d
y c T7 , x C y
take
T,.=|z^:zcTv^ rational
with ~ as the tree relation.
for all fl<9'
=Supn
=Supn
U T# a n d q > S u p ( x ) p
=q"
Since
and there are
102
T h a s n o b r a n c h of l e n g t h ~ l s i n c e if S is s u c h a b r a n c h t h e n t J S is a n i n c r e a s i n g s e q u e n c e of r a t i o n a l s of l e n g t h ~1, w h i c h is i m p o s s i b l e . By o u r c o n s t r u c t i o n Therefore
T=
of T, f o r e v e r y x c T
k)~x E T:Sup(x)
clearly an antichain.
=q],
a n d e a c h s e t ~x c T : S u p ( x ) = q ]
is
T h u s t h e t r e e T is a s p e c i a l A r o n s z a j n t r e e .
W h e n we w a n t t o c o n s t r u c t
a g + - A r o n s z a j n t r e e we u s e , i n s t e a d of t h e
r a t i o n a l s , t h e s e t I1= of all f i n i t e s e q u e n c e s sequences
S u p x is a r a t i o n a l n u m b e r .
of l e n g t h eo w i t h b l a n k s p a c e s
of o r d i n a l s
considered
identified with 0 ordered
g r a p h i c a l l y . We c a n p r o c e e d a s i n t h e c o n s t r u c t i o n b u t w h e n we c o n s t r u c t
as
lexico-
of t h e ~ l - A r o n s z a j n t r e e ,
Ta, f o r a l i m i t o r d i n a l a s u c h t h a t cf ( a ) < g, we h a v e
t o p u t i n T a all t h e i n c r e a s i n g s e q u e n c e s y of m e m b e r s of I1= of l e n g t h a s u c h t h a t y r fi c Ta f o r e v e r y fi < a .
O t h e r w i s e we h a v e n o a s s u r a n c e
t h a t we c a n
c a r r y o u t t h e c o n s t r u c t i o n of Ta f o r a l i m i t o r d i n a l a s u c h t h a t cf ( a ) = ,z. I n order to be sure that
I 7'=1 "< J;, f o r e v e r y a < ~+ we n e e d t h a t ~<~ = ~ ~
= ~,
s i n c e t h i s will e n a b l e u s t o p r o v e t h a t if f o r a l i m i t o r d i n a l a w i t h cf ( a ) < r we construct
T= a s m e n t i o n e ~ l a b o v e we s t i l l h a v e ] T=] ,z ~. So we h a v e p r o v e d :
5.3 T h e o r e m = If ~ = ~<= t h e n t h e r e is a i r
tree.
If t h e c o n t i n u u m h y p o t h e s i s h o l d s t h e n ~70 = R1 a n d t h e r e f o r e Aronszajn tree. the continuum
is a n
~2-
T h e r e f o r e , if we l o o k f o r a m o d e l w i t h n o ~ 2 - A r o n s z a j n t r e e , h y p o t h e s i s s h o u l d fail to h o l d i n s u c h a m o d e l .
theorem which says that in such a model r
T h e r e is a
is a w e a k l y c o m p a c t c a r d i n a l i n
L, h e n c e t h e c o n s i s t e n c y of t h e i n e x i s t e n c e of c 0 2 - A r o n s z a j n t r e e s is a t l e a s t a s s t r o n g a s t h e c o n s i s t e n c y of t h e e x i s t e n c e of a w e a k l y c o m p a c t c a r d i n a l ; we s h a l l s e e t h a t t h e s e two c o n s i s t e n c y a s s u m p t i o n s a r e e q u i v a l e n t . M i t c h e l l had proved this theorem, and Baumgartner forcing.
gave a simpler proof by proper
103 T h e f o l l o w i n g t h e o r e m is d u e t o B a u m g a r t n e r , Malitz a n d R e i n h a r t [BMR].
5 . 4 T h e o r e m : F o r e v e r y t r e e T of h e i g h t ~1 w i t h n o b r a n c h of l e n g t h co1 t h e r e is a c.c.c, f o r c i n g n o t i o n P s u c h t h a t i n t h e g e n e r i c e x t e n s i o n of V b y P
, T is s p e c i a l . If I T I < 21% t h e n b y M a r t i n ' s a x i o m i t f o l l o w s t h a t
T is s p e -
cial.
As a c o n s e q u e n c e ,
if we a s s u m e
M a r t i n ' s a x i o m a n d 21%> l~l, t h e n
all
Aronszajn trees are special and hence there are no Souslin trees.
Proof:
Let P b e t h e s e t of all f i n i t e f u n c t i o n s p f r o m T i n t o ~ s u c h t h a t
=p(y)
if p ( z )
t h e n x a n d y a r e i n c o m p a r a b l e . F o r e v e r y x c T t h e s e t Dx of
all m e m b e r s of P w h o s e d o m a i n c o n t a i n s x is o b v i o u s l y d e n s e i n P, h e n c e if G is a g e n e r i c s u b s e t of P
U G is d e f i n e d o n all of T a n d if F ( x ) = F ( y )
, F=
If we h a v e M a r t i n ' s a x i o m a n d ] T] < 2 t~~ t h e n
then x and y are incomparable. there are sect
< L:~v d e n s e s e t s D= a n d t h e d i r e c t e d s e t G c a n b e t a k e n t o i n t e r -
all of
them,
and
F=
U G is a s
above,
i.e.,
it
specializes
T since
T = U~= : r ( x ) =n~. n
We s t i l l h a v e t o p r o v e t h a t uncountable
subset
If of P
P s a t i s f i e s t h e c.c.e. whose
members
W i t h o u t l o s s of g e n e r a l i t y we c a n a s s u m e t h a t
are
Suppose there pairwise
all m e m b e r s
is a n
incompatible. of If h a v e t h e
same cardinality, that their domains form a A-system with the same kernel s a n d t h a t f o r a l l / ) e If , p t s and let Dom (Pa-s)
is t h e s a m e f u n c t i o n .
= ~xa, 1 . . . . .
x=,•].
Denote W= |Pa:a
L e t a,fl < ~1, P a a n d p p a r e i n c o m -
p a t i b l e , h e n c e P a U P p t~ P . S i n c e P a a n d p ~ c o i n c i d e o n s a n d t h e r e s t of their
pa(Xa]r
domains
are
= pH(xpl)
Ya,k.s = ~fl < r l. As we s a w
disjoint while
we
must
Xa, ~
and
have
zp,~.
for are
some
1 <- k , ~ ~ n ,
comparable.
Let
: ~ # a, p a ( Z a , k ) =p~(xp.lZ ) a n d Xa, ~ a n d xfl,~ a r e c o m p a r a b l e U
Y~.~,~ = ~1 - | a ] .
Let E b e a u n i f o r m u l t r a f i l t e r o n ~1,
l
t h e n f o r e v e r y a t h e r e a r e k a n d s s u c h t h a t Ya,~,~ e E, l e t / c ( a ) a n d ~ ( a ) b e such.
Therefore for an uncountable
s u b s e t A of r
, k(a) =/c and ~(a) = s
104
for a 9
Let a,fl 9
therefore
then
Ya,~,~,
Y~,~,~ 9
hence
Ya,~,~ (~ Y~,k,~ 9
and
] Ya,k,s N Yfl,/c,g] =bll- L e t 7 9 Ya,]c,~ /~ Y~Jr163t h e n za, ~ a n d x~,e a r e
comparable
with
x7, ~.
Now
XT,~'s w i t h
7 9 Ya,~.~ (3 YI3,k,e. a r e
different
different, and since there are only countably many members
of T b e l o w xa, e
o r b e l o w xp,~ t h e r e m u s t be s o m e 7 9 Y,~,~,e. A Yl3,k,e. s u c h t h a t
x7, ~ is g r e a t e r
t h a n b o t h x a k " a n d x~, e a n d s i n c e T is a t r e e , x a k " is c o m p a r a b l e
with xp,~.
This h o l d s f o r all a,jff 9 A h e n c e T h a s a l i n e a r l y o r d e r e d s u b s e t of c a r d i n a l i t y ~1 a n d t h e r e f o r e a b r a n c h of l e n g t h ~1, c o n t r a d i c t i n g o u r a s s u m p t i o n .
w
M a y b e t h e r e is n o t~2-Aronszajn t r e e .
6.1 L e r a m a : A s s u m e V [= " ~ ~ > Sl ^ T is a n c02-Aronszajn t r e e . " an ~l-complete
poser.
Then
V[P] ~ T h a s n o c o f i n a l b r a n c h e s .
Let P be (co2 m a y
b e c o m e of c a r d i n a l i t y coI in V[P] so it d o e s n o t h a v e t o s t a y a n ~ 2 - A r o n s z a j n tree.) P r o o f : A s s u m e that P 0 []-"' E is a c o f i n a l b r a n c h in T ' : We s h a l l d e f i n e in V t w o f~anctions F : 2 <~ -* T t a f o r s o m e a < ~2 a n d S : 2 <~ -. P s u c h t h a t : (i) F ( < > ) =
t h e r o o t of T , S ( < > )
=Po
S(x) [~"F(x) eE'"
(ii) f o r a l l x e 2 <~
(iii) x .6, y --. S ( x )
F(x)
S(z)
and
are
S(z) ~PPo , S(z) member
defined
has,
by
induction
f o r e v e r y fl < ~2, a n
of Tp (i.e., t h e s e t of v e r t i c e s
If > F ( x ) : (~p >-p S ( z ) ) p I~-" t e E"] members
of T t h e n
hypothesis.
on
the
length
extension
which
of
x.
Since
puts
some
of h e i g h t fl in t h e t r e e )
were
t h e y w o u l d be a b r a n c h
a
set
of p a i r w i s e
in B.
of T in V, c o n t r a d i c t i n g
T h e r e f o r e t h e r e a r e two i n c o m p a r a b l e
If
comparable our
f ' s in t h i s set, t a k e o n e t o
105 b e F ( z ^ <0>) a n d t h e o t h e r to b e F ( z S(x
^ <1>)
a s c o n d i t i o n s -> S ( x )
for i 9 lo,ll.
^ <1>) a n d c h o o s e S ( x
^ <0>) a n d
such that S(x ^ ) II-"'F(x ^ ) 9 E"
Since t h e r a n g e of F is c o u n t a b l e it is i n c l u d e d in s o m e T t a f o r
s o m e a < ~2- S i n c e P is ~1 - c o m p l e t e , f o r e v e r y ~ 9 2 ~ , P c o n t a i n s a condition p n w h i c h is a n u p p e r b o u n d of [S(17 t n ) : n 9 ~ . Since p n ->P0 t h e r e is a
q,1-~P.landat n 9
Tasuchthatq.1 ][-"tn 9
Letv #77, ~,~7 9
L e t n be
the least such that u r 7% ~ ~7 ~ n, then by requirement (iv) above we have that F(y ~ ~%) and F(~7 r n) are incomparable in T. q,1 ~P,I -> S(~7 F n), hence also q. I[-"F(~7 try) 9 E': t.,F(~7 t~%) 9 ~clearly of F ( r I t n )
is < a
also T~ > F ( v t n ) ,
Since q. forces that E is a branch
of T and that
t.1 and F(~7 t r~) are comparable in T. Since the height
a n d t h e h e i g h t of t,1 is a we h a v e t,~>2"(7/~%).
Similarly
a n d since F(~7 t n ) a n d F ( u t n ) a r e i n c o m p a r a b l e also t,~
a n d t~ a r e i n c o m p a r a b l e
and hence different.
Thus
Ta contains
2 hI" >R1
d i f f e r e n t m e m b e r s tn' c o n t r a d i c t i n g t h e a s s u m p t i o n t h a t T is a n ~ 2 - A r o n s z a j n tree. 6.2 Theorem= If ZFC is c o n s i s t e n t with t h e e x i s t e n c e of a w e a k l y c o m p a c t c a r d i n a l t h e n ZFC is c o n s i s t e n t with 2 ~ ~
~2 a n d t h e n o n - e x i s t e n c e of
c02-Aronszajn t r e e s . R e m a r k : By w h a t was m e n t i o n e d theorem.
l a s t s e c t i o n we h a v e "iff" in t h i s
Mitchell h a d p r o v e d t h e t h e o r e m
and Baumgartner
thai gave a
s i m p l e r p r o o f b y p r o p e r forcing. Proof: L e t g b e a w e a k l y c o m p a c t
c a r d i n a l . We shall u s e a s y s t e m
< Q : i < ~> of i t e r a t e d f o r c i n g with c o u n t a b l e s u p p o r t .
Q~ will b e t h e c o m p o -
sition of two f o r c i n g n o t i o n s Q~,0 a n d Q~,I . Now Q~,0 will b e t h e p o s e r of c o u n t a b l e f u n c t i o n s o n ~1 i n t o c02 (in V[Pi]) w h i c h c o l l a p s e s ~o2. Q~,0 is o b v i o u s l y ~ l - c o m p l e t e . cardinality
~
with
no
iffa=a
cofinal
branch
t^~
V[Pt][Q~,o] c o n t a i n s w l - t r e e s of
(e.g.,
I
with
L e t W be t h e disjoint u n i o n of all s u c h
t r e e s , as a single t r e e with a t m o s t 2 ~t r o o t s ( we t a k e all t h e t r e e s to b e
106
~
x li{ with s o m e p a r t i a l o r d e r i n g s ) .
By T h e o r e m 5.4 t h e r e is a c.c.c, f o r c i n g
Q~,~ w h i c h m a k e s t h i s t r e e s p e c i a l , a n d h e n c e m a k e s e v e r y ~ -
t r e e of c a r d i -
n a l i t y t~l s p e c i a l , p r o v i d e d it h a s n o ~ v b r a n c h . Let
us
notice
that
these
descriptions
of
Q~.0 in
V[P~] a n d
Q~,I in
V[P~][Q~,o] = V[Pi * Q~.0] r e a l l y y i e l d c o r r e s p o n d i n g n a m e s Q a n d Q b y t h e ~.0 ~i. I L e m m a of the existential completeness which we proved. S i n c e Q~,0 is m v c o m p l e t e
o v e r V[P~] a n d s i n c e Qi,1 s a t i s f i e s t h e c.c.c, o v e r
V[P~][Q~,o] b o t h a r e p r o p e r , h e n c e e a c h P~ , i < g, is p r o p e r .
Q~,0 * Q~,I is p r o p e r
a n i n a c e s s i b l e c a r d i n a l ~ <- g. O u r c o n s t r u c t i o n i
h a v e IQ~I < ~ ,
and therefore
also
T h u s ml is n o t c o l l a p s e d e v e n in V[P~]. L e t ~ b e of Pi a n d Q~ a r e s u c h t h a t f o r
h e n c e e a c h Q~ h a s c a r d i n a l i t y
p r o v e d , a l s o P~ s a t i s f i e s t h e h - c . c , a n d t h e r e f o r e
<~.
T h e r e f o r e , a s we
~ is n o t c o l l a p s e d in V[P~]
a n d t h u s V[Px] ~ " h =* Oz" ( s i n c e e~1 t o o is n o t c o l l a p s e d ) . 6.3 L e m m a : V[Ph] ~ " t h e r e
P r o o f : It s u f f i c e s t o p r o v e V [ P ~ + I ] - V[P~][Qi,0].
a r e a t l e a s t )~ r e a l n u m b e r s
that
for every i < ~ there
We s h a l l s e e t h a t
Vt f o r V[P~][Q~,o] , Q f o r Q~,I a n d
makes special. j
<~-,aj
Let (aj:
a,b 9 T a n d F ( a ) = F ( b ) t h e n a ifF(aj)
T f o r t h e t r e e in Vt w h i c h Q in T in Vt (i.e.,
a f u n c t i o n F o n T i n t o ~ s u c h t h a t if
andb
is e v e n a n d tj = 1 i f F ( a i )
are incomparable.
isodd.
Forj
We s h a l l s e e t h a t t = ( t j
a C o h e n r e a l o v e r Vt, i.e., f o r e v e r y d e n s e s u b s e t ~ )
of 2 < ~ i n
:j
n
(Is = O ^ p ( a i )
For
p 9 Q
let
is e v e n ] o r [s = t ^ p ( a i )
p ~
i =0 :j <~>
for
some
in
Let u s s i m p l i f y t h e n o t a t i o n
j < ~0) b e a n a s c e n d i n g s e q u e n c e Q introduces
is a r e a l
a f o r c i n g n o t i o n s u c h a s Q~,I i n t r o -
duces a Cohen real over the previous universe. by writing
" for every
V~ , t r n
is 9
is o d d ] }. L e t Q~b=|p 9 Q : p ~149
we s h a l l s e e t h a t Q~bis a d e n s e s u b s e t of Q in Vt. Q ~ 9 V t s i n c e it is d e f i n e d in Vt. F o r q 9 Q l e t r
~q ~ , r 9
Let
strict
n
be
a
upper
1a =q U l(aj,n+2J +r(j)> :j 9
t h e r e is s u c h a n t bound
of
the
s i n c e ~) is d e n s e in 2 <~, range
Aa:. ~Dom(q)I.
of
q.
Let
Obviously p 9 Q,
107 andp
~r
eric
subset
9
hencep of
Q
9 Q~b S i n c e p
then
there
is
mq
a
, Q2bis d e n s e .
Let G be the gen-
G ~ Q2b, P* e~.),
and
for
some
"n, < co , p " 9 2n. S i n c e p C F , t r n = p ", h e n c e t t n 9 ~/), w h i c h e s t a b l i s h e s t h a t t is a C o h e n r e a l .
6.4 L e m m a : F o r e v e r y i n a c c e s s i b l e A < r Let
Proof:
G
FIG] ~ " A > B 2 " . F 9 FIG] b e IF(a,fl) :fl
be
a
generic
Suppose a
now
function
subset
that
on
, V[P~,] ~="~ = B~" of
PA.
We s a w
FIG] ~ " B 2 = l z " ,
/~xo l
such
that
already
that
/z
Let
0
where for
all
=a, i.e., F ( a , - ) is a m a p p i n g of r I o n a. L e t F b e a n a m e of
F a n d l e t p 0 9 G f o r c e t h a t F is as we d e s c r i b e d .
F o r e a c h a < / z a n d fl < / z let
-ga,p b e a m a x i m a l a n t i c h a i n of m e m b e r s of Px w h i c h a r e r a P 0 a n d w h i c h give d e f i n i t e v a l u e s t o 1~a,fl). I-ga,~l < A . member .gU
Let _9=
l.J
S i n c e a s we saw, P x s a t i s f i e s t h e A-c.c. c o n d i t i o n _ga.~-
Since
h is r e g u l a r
1_91 < A .
Since
each
of .9 [J [p0] is a countable function on A there is a 7 < A such that
IP0] O P T
Let G 7 = G (~ PT' then clearly F 9
FIGT] (since w e
F((x,~) in FIGT] to be that 7 for which there is a q 9
q I~-"F(a,~6)=T'9-
V[GT] ~ "/~ = B e ' :
~ G 7 such that
T h e n V[GT] ~ = / z ~ B 2 , b u t s i n c e F I G ] ~ " ] z = B 2 " w e B u t s i n c e we f o r c e a b o v e
define
have
FIGT] w i t h QT,0[GT] w h i c h col-
l a p s e s t h e B2 of FIGT], we h a v e FIGT][GT,0] ~ "/z < S 2 " h e n c e
F I G ] ~= "/~ < B2 ' ,
w h i c h is a c o n t r a d i c t i o n . Continuation
of t h e P r o o f of T h e o r e m 6.2: Now let u s go o n with t h e
proof of the theorem.
Assume
that there is a P 0 e P ~ such that P 0 ]~-p~ "
there is an o~2-Aronszajn tree," i.e.,P 0 l~-P." there is a/c-Aronszajn tree T on r a n d a function r on g such that for a < r by the l e m m a
FIPz]
, F(a) is the rank of {xin T, (since
[= "~2 = ~'9, i.e.,P 0 [~-P." there is a transitive relation T
on r such that for all a,~,7 6 r if a T 7 a n d ~ T 7 then a = ~ v aT~ A ~Ta a n d there is a function F
on ~ into r such that for all a,j~ 6 ~ if a Tfl then
F(a) < Y(~), for all a e m and 7 < F(a) there is a ~Ta such that F(~) = T, a n d
108
f o r a l l 7 e:_ g t h e r e is a f l e g s u c h t h a t f o r all a e ,: if F ( a ) for all B C g there
= ' 7 t h e n a -- fl, a n d
a r e a , f l e B s u c h t h a t a ~ fl ^ - a T f l ^ - f l T a , o r e l s e t h e r e
is a fl < ~ s u c h t h a t B C fl". T h i s i m p l i e s , b y t h e e x i s t e n t i a l there are canonical
names
(*) P o 1]-- " T is
of T a n d F of r e l a t i o n s
a transitive
relation
completeness
on ~ such that:
on
g^ (Va,fl,g e g)(aT'l^ ~
0, = ~ ^ a T ~ ^ ~ T a ) ^ (~F: ~ ~ ~) ^ (Va,fl c ~)
(~ <
that
fiT7 +
~
~
[(aT~ ~ F ( a ) < F(#)) ^
F ( a ) -, ('~7 r ,:)(3,T~ ^ F ( 7 ) = / ~ ) ) ] ^ ~
~
(v7 r ~)(a# r ~)(va
and
r ~ ) ( p ( = ) = 7 -" =
for every
standard
n a m e B of a s u b s e t
(*)po I1-,~ " ( a = , ~ We
have
e
B)(,~
# # ^ - =T~^
to use a different formal
for a limit ordinal
~ > p(p)
- # T = ) ~ (a~ r ~ ) ( B c
representation
B)".
for the names
if UT i is a canonical
name
LIT, so that
of a subset
of ~ or
4,e[
x ~ then
U%
E R(A)
. This follows
that ~ is large enough. statement
mentioned
is a II~ statement
therefore
about
this structure, about
II~ indescribable
and
I. 5.6. and
that (**) holds
this structure.
there
that
Theorem
from
Consider now the structure < H(~),c,,, T,E>
first order
and
from
is an
Since
inaccessible
is an
cardinal
B as
compact A < ~ such
substructure
P z and [[-z are definable in
definitions give PA and
fact
(*) is a
for every
~ is weakly
elementary
the
of
and the same
]~-A in
Therefore
(T N A x A)[GA] is a A-Aronszajn tree in V[GA] (where G is the generic subset of Pz over V a n d
G A = G /'~PA). We claim that ( T N A x A ) [ G A ]
is the part of
N
the tree T[G] up to l e v e l A . Let
p IFpa
hence p ]~-p "
the relation between the two above mentioned
(by
structures), hence, since
109
p c G ,
since
and
(*) h o l d s f o r t h e s t r u c t u r e
an
ordinal
(FNXxX)(a)=/~a
fi < ~
such
I~p,,(Vae~)
thereforep
> p
there
[~-v~(Va E X)
(F(a) =/~-,a
s i n c e p C G we h a v e in V[G] t h a t all t h e o r d i n a l s in t h e level ~ a r e
V[Px]
T h u s in
T N (/~ • •)
is a
and
< fl < k.
is a ~ - A r o n s z a j n t r e e , i.e., a n o 2 - A r o n s z a j n
N
tree.
As we s a w in V[Px] t h e r e a r e a t l e a s t ~ r e a l n u m b e r s ,
V[Px]-
Qx,0 is a n
T N ~" x ~
still
V[Px][Qx,o] ,
col-complete has
no
poser
cofinal
in
V[Px], t h e r e f o r e ,
branch
in
i.e., 2 I% ~ l~a in by lemma
V[Px][Qx,0].
Since
6.1 in
[~,[ =l~ 1 t h e r e is a s u b s e t a of X of o r d e r - t y p e ~i. L e t b be t h e
s e t of all t h e o r d i n a l s in T N k • X w h o s e level is in a , T N b x b is a n e o s t r e e with no cofinal branch
in
V[Px][Qx,o].
Qx.1 m a k e s t h i s t r e e s p e c i a l .
T N (b x b) is a s p e c i a l Col-tree in V[Px+I] a n d t h e r e f o r e
Thus
it s t a y s so a l s o in
V[Px] s i n c e t h e f u n c t i o n w h i c h m a k e s it s p e c i a l is in V [ P x + I ] a n d h e n c e also in
V[P,].
T h u s in V [ P , ]
T N (k x ~) t o o h a s
T C~ (b x b) h a s n o c o f i n a l b r a n c h
no cofinal branch,
but
and therefore
t h i s is a c o n t r a d i c t i o n
since
T N (~ x k) is t h e p a r t u p t o level k of t h e ~ - t r e e T a n d a s s u c h it m u s t h a v e b r a n c h e s of l e n g t h X.
We u s e d a w e a k l y c o m p a c t ~ t o o b t a i n a g e n e r i c e x t e n s i o n in w h i c h t h e r e are no r
t r e e s . If all we w a n t is t o o b t a i n a n e x t e n s i o n of V in
w h i c h t h e r e a r e n o s p e c i a l w 2 - A r o n s z a j n t r e e s ( i.e., t r e e s w h i c h a r e t h e u n i o n of R i a n t i c h a i n s ) t h e n it s u f f i c e s t o u s e a Mahlo c a r d i n a l ~.
w
Closed u n b o u n d e d subsets of c01 c a n run a w a y f r o m m a n y sets.
110
Baumgartner ZFC + 2~~
[B3] h a s if A~ ~ 1
proved
the
is infinite
c o n s i s t e n c y of t h e countable
then
following with
there
is
a
closed
u n b o u n d e d C C ~1 s u c h t h a t A~ ~' C f o r e v e r y i < ~1. We p r o v e a s o m e w h a t s t r o n g e r a s s e r t i o n . 7.1 Theorem= Z F C + 2~~ = ~2 is c o n s i s t e n t with: (*) if A~ c ~1 h a s n o last e l e m e n t a n d is n o n - e m p t y a n d h a s o r d e r t y p e < Sup A~, f o r i < ~1, t h e n t h e r e is a c l o s e d u n b o u n d e d subs e t C of ~1, s u c h t h a t C C~ A~ is b o u n d e d in A~ for e v e r y i.
R e m a r k : L a t e l y A v r a h a m i m p r o v e d B a u m g a r t n e r ' s r e s u l t to Z F C + '2~~ = anything + (**) t h e r e are B2 c l o s e d u n b o u n d e d s u b s e t s of ~1, the i n t e r s e c t i o n of a n y t~1 of t h e m is finite. Galvin p r o v e d p r e v i o u s l y t h a t CH implies (*) fail.
[Sh80]
Our p r o o f is similar to
w
P r o o f : We s t a r t with a m o d e l V s a t i s f y i n g CH, a n d use CS i t e r a t e d f o r c ing of l e n g t h I~2, s u c h t h a t in t h e i n t e r m e d i a t e s t a g e s CH still holds.
So b y
4.1., it suffices to prove. 7.2 L e m m a : S u p p o s e V satisfies CH. T h e r e is a p r o p e r f o r c i n g n o t i o n of power
~1
which
adds
a
closed
unbounded
subset
of
C
of
~1
and
Sup [C ~ A] < Sup A f o r a n y infinite A C ~1 with no last e l e m e n t a n d o r d e r t y p e < Sup A, which b e l o n g s to V.
D i s c u s s i o n : A plausible f o r c i n g to e x e m p l i f y 7.2 is: Q = | C : C a c l o s e d c o u n t a b l e s u b s e t of ~1 so t h a t if A E V , ~=SupA
limit, A h a s o r d e r t y p e < ~ ,
t h e n Sup [A C~ C] < ~1 with t h e o r d e r C 1 < C2 iff C2 N [(Max C1) + 1] = C 1.
111
So t h e e l e m e n t s
of Q a r e a p p r o x i m a t i o n s
to the required
that a generic subset of Q gives a C as required,
p r o v i d e d t h a t ~z is n o t c o l -
l a p s e d ; h e n c e t h e m a i n p o i n t is t o p r o v e t h e p r o p e r n e s s it seemed
Q is n o t p r o p e r .
of Q. U n f o r t u n a t e l y
H o w e v e r if we w a n t t o a d d a C c Q w h i c h is ( N , Q ) -
generic for some N < ~H(X),c) force with some P,
C. I t is c l e a r
a c . c . c , f o r c i n g is e n o u g h .
I P I =1~2 , P s a t i s f i e s t h e c . c . c ,
So we c o u l d f i r s t
such that
[I-p "2~~ =1~2
a n d MA h o l d s " T h e n w e d e f i n e Q i n V[G] f o r G ~ P g e n e r i c o v e r V; s i m i l a r t o the definition above, Q = I C c V[G]: C a c l o s e d c o u n t a b l e with
no
last
element
and
s u b s e t of co1, a n d if A c V is i n f i n i t e
(order
type
A ) < S u p A < M a x C,
then
Sup (C Fl A) < Sup A] Now Q is p r o p e r , a n d a d d a C a s r e q u i r e d , s o P * Q a d d a C a s r e q u i r e d . N
Unfortunately inaccessible
P * Q a l s o c o l l a p s e s I~2, so if we a r e willing t o u s e s o m e s t r o n g l y g > I~0, t h e r e
is n o p r o b l e m .
s i o n of MA , w h i c h is c o n s i s t e n t
O t h e r w i s e , we u s e a r e s t r i c t e d
ver-
w i t h 2 I% =l~ 1, s o 7.3, 7.4, 7.5 b e l o w p r o v e
L e m m a 7.2 h e n c e T h e o r e m 7.1.
7.3 Claim" S u p p o s e (a) V s a t i s f i e s CH, P is a f o r c i n g n o t i o n of p o w e r I~I s a t i s f y i n g t h e c . c . c . G ~ P g e n e r i c o v e r V. (b) 6 < ~ I is a l i m i t o r d i n a l , ( V a , ~ < 6)a § fl < 6, able
family
of
closed
bounded
subsets
of
6,
R c V[G] is a c o u n t -
ordered
by
CI--< C 2
iff
for eachi
, A~ E V i s
C i = C 2 (h(MaxC x§ (c) D e f i n e (in V[G]): Q= QR =l(C,{(A't,a~):i < h i ) : a s u b s e t of 6 o f o r d e r t y p e (CI,I(A~I,~{ 1 ) : i • n i l )
for every/
CcR
, n <e,
< 6 , a n d C KIA~ C a ~ a n d a i is a n o r d i n a l < 6] ---~ (C2,{(Ai2,/li 2) : i < ~l.2J) i j ~ C 1 < C 2 , ~r~1 ~ n 2 a n d
< n 1 for some j < n 2
(A~l,a 1) = (A~,a]).
Then 1) QR s a t i s f i e s t h e c . c . c .
112
8) if f o r e v e r y C c R ;
ifl<~:
C U |~] c Q] h a s o r d e r t y p e 6 t h e n
" U C is an u n b o u n d e d s u b s e t of 6 ' : CEG
P r o o f : 1) Trivial, as a n y two c o n d i t i o n s with t h e s a m e first c o o r d i n a t e a r e c o m p a t i b l e , a n d t h e r e a r e only c o u n t a b l y m a n y possibilities f o r t h e first coordinate. 8) Trivial, b e c a u s e if A 1 . . . . .
An a r e s u b s e t s of 6 of o r d e r t y p e
< 6, t h e r e u n i o n h a s o r d e r t y p e < 6 . See D u s h n i k Milner [DM] . 7.4 ClHim: S u p p o s e V satisfies CH. T h e r e is a f o r c i n g n o t i o n P of p o w e r ~1 s a t i s f y i n g t h e c.c.c., s u c h t h a t t h e following s t a t e m e n t is f o r c e d : (*) S u p p o s e 6 < r
is limit, ( V a , ~ < 6) a + ~ < 6 ( e q u i v a l e n t l y 6 is a n ordi-
n a l p o w e r of co) a n d R is a c o u n t a b l e f a m i l y of c l o s e d b o u n d e d s u b s e t of 6 such
that
(YCcR)(V/~)
(Max C < p < 6 - *
C U {~| ~ R )
a n d ~0n c Q R
is a
d e n s e o p e n s u b s e t s o[ R ( o r d e r e d as in 7.3). Then t h e r e a r e Ca c R , Ca < Cn+ l,
Sup Max Ca = 6 , Ca+z c~O n a n d f o r
e v e r y A E V , A ~ 6 of o r d e r t y p e < 5 ,
S u p [ A / ~ ( u C n ) ] < 5 ; m o r e o v e r we
n
c a n c h o o s e Co c R a r b i t r a r i l y . P r o o f : We c a n u s e a n FS i t e r a t i o n ~ P ~ , Q . : i < ~ l )
of f o r c i n g n o t i o n s
s a t i s f y i n g t h e c.c.c., s u c h t h a t f o r e v e r y possible R a n d 6. (which a r e in V or a p p e a r in V P~ f o r s o m e i < B1) f o r u n c o u n t a b l y m a n y j < co1 , Qj = QR7.5 Clain~ S u p p o s e V satisfies CH, P is as in 7.4, a n d Q = i C : C a c l o s e d b o u n d e d s u b s e t of co1, s u c h t h a t f o r e v e r y infinite c o u n t a b l e A E V
A ~ r
with n o l a s t e l e m e n t , if Sup A < Max C t h e n Sup (A C~ C) < Sup A i. T h e n Q is N
p r o p e r (i.e., I]-p "Q is p r o p e r ':)
P r o o f : Let G c P be g e n e r i c o v e r V , J~ be r e g u l a r big e n o u g h a n d let
113
S = [N"
N~
such t h a t N i < N , < N a. : j - - i >
9
and t h e r e is a sequence < N r N= uN, and~=N
<~>
f~l]
It is easy to check the following facts, and by 2.8 t h e y imply Q is proper, so we finish the proof of 7.2, hence 7.1. F a c t A: S c Dt40 ((H(A)) (remember we do not distinguish strictly between N its sets of elements). FactB:If ~S P, t h e n
I(C,l(A,,a~)
, ~=N
f~l,
R=P
AN
, .gcNis
a d e n s e subset of
: i < ~l) e QR C 9 o0] is a dense subset of QR-
IV. On o r a c l e - c . c , a n d "p(c0) / finite h a s n o n o n - t r i v i a l a u t o m o r p h i s m A b s t r a c t : H e r e we p r e s e n t h o w to i t e r a t e f o r c i n g s a t i s f y i n g t h e M-c.c. f o r a n o r a c l e M. This e n a b l e s us, in s o m e s e n s e , to " c o n s t r u c t " t h e s e t of r e a l s in co~. s t e p s with o m i t t i n g t y p e s a l o n g t h e way (i.e., i m i t a t i n g c o n s t r u c t i o n s of m o d e l t h e o r y ) . We u s e it to p r o v e t h e c o n s i s t e n c y of "D(c0)/finite h a s n o n o n - t r i v i a l a u t o m o r p h i s m " , w h i c h is e q u i v a l e n t to " I ~ ( N ) - N h a s o n l y trivial h o m e o m o r p h i s m s " .
w
Introduction.
It is well k n o w n t h a t if CH h o l d s t h e n f)(c0)/finite is a s a t u r a t e d ( m o d e l t h e o r e t i c a l l y ) a t o m l e s s B o o l e a n a l g e b r a of p o w e r I~i, h e n c e h a s m a n y a u t o m o r p h i s m s , as a n y i s o m o r p h i s m f r o m o n e c o u n t a b l y s u b a l g e b r a to a n o t h e r , c a n b e e x t e n d e d to a n a u t o m o r p h i s m .
It was n o t c l e a r w h a t t h e s i t u a t i o n is if
CH fails.
On t h e o t h e r h a n d a n y o n e - t o - o n e f u n c t i o n f f r o m o n e c o - f i n i t e s u b s e t of c0 to a n o t h e r c o - f i n i t e s u b s e t of c0, i n d u c e s a n a u t o m o r p h i s m of ~(c0) / finite : A / f i n i t e is m a p p e d to f (A) / f i n i t e . We call s u c h a n a u t o m o r p h i s m trivial. A p r i o r i it is n o t c l e a r w h e t h e r f)(c0)/finite h a s n o n - t r i v i a l a u t o m o r p h i s m s . . Our m a i n c o n c l u s i o n is t h a t n o t n e c e s s a r i l y m u s t t h e r e b e a n y n o n - t r i v i a l a u t o morphism.
Our m a i n tool is " o r a c l e c h a i n c o n d i t i o n f o r c i n g " . I t e r a t i n g f o r c i n g s a t i s fying t h e i l l - c h a i n c o n d i t i o n , i n t r o d u c e d b y S o l o v a y a n d T e n n e n b a u m is well known. We u s e m u c h of t h e s a m e f r a m e w o r k : we s t a r t , e.g., with t h e c o n s t r u c t i b l e u n i v e r s e V = L, a n d u s e (finite s u p p o r t ) i t e r a t i o n of l e n g t h ca2 of f o r c i n g notions,
< c0~>, e a c h Qi s a t i s f y i n g t h e t~;-c.c, a n d of p o w e r t~1. At
s t a g e a we g u e s s a n a u t o m o r p h i s m f a of p(c0) in
VP', m o r e
exactly a P=-name
115
of s u c h a n a u t o m o r p h i s m s u c h t h a t f o r a n y P ~ - n a m e f of a n a u t o m o r p h i s m of p ( ~ ) / f i n i t e f o r s t a t i o n a r i l y m a n y a < oa, we have: cf a = ~1, a n d f
c f
(this is p o s s i b l e if we h a v e a d i a m o n d s e q u e n c e on 16 < ~2:cf (~) = otJ, b y s o m e simple c o n s i d e r a t i o n s ) .
Our m e a n s of killing t h e a u t o m o r p h i s m , i.e.,
g u a r a n t e e i n g f = c a n n o t be e x t e n d e d to an a u t o m o r p h i s m of p ( ~ ) / f i n i t e in VP ' , is to a d d a s e t X ~ ~ so t h a t f o r n o Y ~ : (*) X C~ A is finite <=> Y N f a ( A ) is finite
f o r e v e r y A e -'^ -[/2(~)]IF~ The Pm
d e m a n d (*) h e l p s u s s i n c e if f D f = is a n a u t o m o r p h i s m of [/9(~)] v 8/finite t h e n Y / f i n i t e = f ( X / finite ) satisfies (*).
This l o o k s v e r y r e a s o n a b l e , b u t e v e n if we s u c c e e d to show t h a t w h e n we a d d X (by t h e f o r c i n g Q=) we do n o t a d d a Y s a t i s f y i n g (*), h o w c a n we k n o w t h a t s u c h a Y is n o t a d d e d l a t e r o n in t h e i t e r a t i o n ?
This is t h e role of t h e o r a c l e c h a i n c o n d i t i o n .
The b e s t w a y we c a n
explain t h e c o n s t r u c t i o n is as follows. We c a n look at i t e r a t e d f o r c i n g as a c o n s t r u c t i o n of t h e c o n t i n u u m in ~2 steps, i.e., a t e a c h s u c c e s s o r s t e p we a d d m o r e reals, a n d even a t limit s t e p s of c o f i n a l i t y ~ we do so ( b u t n o t a t limit s t e p s of c o f i n a l i t y ~1). Now p r o m i s i n g n o t to a d d a Y as in (*) is a n o m i t t i n g t y p e obligation, a n d building a m o d e l of p o w e r S2 by a c h a i n of ~2 a p p r o x i m a tions, p r o m i s i n g to o m i t t y p e s a l o n g t h e way, is a widely u s e d m e t h o d in m o d e l t h e o r y ( m a i n l y f o r St i n s t e a d of ~ ) .
See e.g., Keisler's work on L(Q), f o r ~1;
t h e H2 c a s e like o u r s is s o m e w h a t m o r e difficult, see [Sh 80c ].
The o r a c l e c h a i n c o n d i t i o n itself is in a s e n s e an effective v e r s i o n of t h e H1-c,c. One of t h e u s u a l ways to c o n s t r u c t a n ~i-c.c. f o r c i n g P of p o w e r t~I is
to m a k e its set of elements be ~I; let M = < M 8 : 6 < r~1> (the COl -oracle ) be a sequence of countable sets which is a diamond sequence ( i.e.,such that for every A c COl,Ir :A f~ ~ e Mr
is stationary) and demand:
if A 6 M~ is p r e d e n s e in P t ~ t h e n it is p r e d e n s e in P.
116
We call (a v a r i a n t of) this t h e M -c.c. , ( a s s u m i n g e a c h ~/8 is c l o s e d enough).
The c o n n e c t i o n b e t w e e n it a n d t h e o m i t t i n g t y p e a r g u m e n t dis-
c u s s e d a b o v e is t h e following. S u p p o s e B~ a r e Borel s e t s of r e a l s (i < 111) s u c h t h a t t h e i r i n t e r s e c t i o n is empty e v e n if we a d d a C o h e n real. T h e n (if a Q ,
- s e q u e n c e exists a t all ) 1
there is an ~I- oracle M such that in any generic extension of V by a forcing s a t i s f y i n g t h e M - c h a i n c o n d i t i o n no real is in all t h e B~'s ( n o t e t h a t in a b i g g e r u n i v e r s e we r e i n t e r p r e t e a c h H~ u s i n g t h e s a m e definition).
Clearly in o r d e r t o a p p l y o r a c l e c h a i n c o n d i t i o n f o r c i n g s we h a v e to p r o v e r e l e v a n t l e m m a s on c o m p o s i t i o n , p r e s e r v a t i o n b y d i r e c t limit, a n d it is v e r y h e l p f u l to m a k e 111 c o n d i t i o n s on s a t i s f y i n g t h e o r a c l e c h a i n c o n d i t i o n i n t o one. F o r this we h a v e to c h o o s e t h e r i g h t v a r i a n t of t h e definition. The n e t r e s u l t is t h a t at s t e p a we a r e given an c01-oraele w h i c h
Qa h a s to satisfy, a n d
a r e allowed to d e m a n d , for 1t1 Borel t y p e s , t h a t I~'~ will o m i t t h e m p r o v i d e d t h a t n o t o n l y o u r u n i v e r s e VP" o m i t s t h e m , b u t even f o r c i n g b y C o h e n f o r c i n g d o e s n o t c h a n g e this.
This is d o n e in w
a n d h a s little to do with
~co)/finite. A f t e r all this we still have to do t h e i n d u c t i o n step, i.e., w o r k i n g in VPa a n d b e i n g given a n a u t o m o r p h i s m f = , a n d a n c01-oracle M we h a v e to find Q , a d d i n g a real X, a n d s a t i s f y i n g t h e M-c.c., so t h a t (*) h o l d s f o r no Y e v e n if we N
a d d a C o h e n real.
Of c o u r s e if
fa
is trivial this c a n n o t be done.
We shall c o n s t r u c t Oa,
a s s u m e t h e r e is a n a m e Y (for Q=- x Cohen) f o r c e d to s a t i s f y (*), a n d a n a l y s ing this we shall a t l a s t p r o v e / = name
of a
unbounded
non-trivial
is trivial. We have to n o t e t h a t if / is a P~,-
automorphism
of p(c0)/finite,
then
for a
closed
C c c 0 2 , f o r e v e r y 5 c C with cf (6) =c01, f r [ p ( ~ 0 ) ] Cs h a s a P6-
n a m e f o r c e d (for P~) t o be a n o n - t r i v i a l a u t o m o r p h i s m of p(c0)/finite, h e n c e
117
t h e d i a m o n d o n 16 < ~2:cf ( 6 ) = c011 g u e s s e d it s t a t i o n a r i l y o f t e n , h e n c e we h a v e "killed" it s o m e w h e r e in t h e p r o c e s s .
A n a t u r a l q u e s t i o n is: Q u e s t i o n : Find a p a r a l l e l of M-c.c., r e p l a c i n g C o h e n f o r c i n g (in t h e a s s u m p t i o n s of t h e o m i t t i n g t y p e t h e o r e m ) b y r a n d o m r e a l f o r c i n g , o r a n y o t h e r r e a s o n a b l e f o r c i n g a d d d i n g a real.
N o t e t h a t if P ~ Q t h e n P
t h e n J ' ~ is a m a x i m a l a n t i c h a i n of P ,
h e n c e of Q, t h e r e f o r e it is p r e d e n s e in Q a n d so is .9).
w i O n oracle chain conditions 1.1. Definition= An ~1- o r a c l e is a s e q u e n c e ~ = ( M ~ : 6 a limit o r d i n a l <: ~ 1 > , w h e r e M~ is a c o u n t a b l e t r a n s i t i v e m o d e l of Z F C - ( o r a l a r g e e n o u g h p o r t i o n of ZFC ) s u c h t h a t 6 ~ Ms, M~ ~ " 5 is c o u n t a b l e " a n d ~ satisfies: (V A C e01)[16:A N 6 E M ~ is s t a t i o n a r y ]. (ZFC - is ZFC w i t h o u t t h e p o w e r s e t a x i o m ) . R e m a r k : N o t e t h a t t h e e x i s t e n c e of a n r h o l d i n g of ~
is e q u i v a l e n t to t h e
(by a t h e o r e m of Kunen), a n d t h a t ~ 1
i m p l i e s CH. 1
In o r d e r t h a t t h e r e a d e r c a n s e e how we i n t e n d to a s s o c i a t e a c o n d i t i o n o n f o r c i n g n o t i o n s with e a c h ~ z - o r a c l e M, we give h e r e a t e n t a t i v e d e f i n i t i o n (which we will.modify l a t e r ) : 1.2 F i r s t t r y : A f o r c i n g P with u n i v e r s e ~1 s a t i s f i e s t h e M-c.c. if (V6)(VA c M~) [A C 6 ^ A is p r e d e n s e in P r 6 -* A is p r e d e n s e in P ] .
118
B u t by this it is possible t h a t P, P? a r e i s o m o r p h i c ; o n e satisfies this a n d one doesn't.
The s o l u t i o n will be to r e q u i r e t h e p r o p e r t y n o t f o r all 6, b u t
only f o r a l a r g e e n o u g h s e t of 6's: 1.3 Definition: With e a c h o r a c l e M we a s s o c i a t e a filter D~ o v e r col g e n e r a t e d b y t h e s e t s I~I(A ) = [6 < col : A (~ 6 9 M~{ for e v e r y A C: COl" Let p(6) b e
~A "A C
6i.
1.4 Claim: i) If < M~ ~ p(6):6 < col> is a O
"-sequence then every gen-
erator of Dj/, I~(A), contains a closed unbounded set; 2)
For
every
A , B c col
there
is
C C col
such
that
I~(C) = I~(A) N I~(B). 3)
D~
is
a
proper
normal
filter,
containing
every
closed
u n b o u n d e d s e t of limit o r d i n a l s < c0i.
Remark: Recall that ~Sa: 6 < col> is a O
"-sequence if IS~I ---S0 and
for every A C caI, 16 :A (~ 6 9 S~I contains a closed unbounded subset of r and if V = L then a 0
"-sequence exists.
P r o o f : i) By t h e definitions of a O * - s e q u e n c e a n d of I~(A). 2) Let g:(o 1 -* col be the m a p p i n g defined by g ( a ) = 2a, a n d let f : c o l - ' c o l be t h e m a p p i n g defined b y f ( a )
=2a§
1. If 6 < c o 1 is a limit ordi-
nal, t h e n 6 is c l o s e d u n d e r g , f , a n d g r 6, f t 6 e M$. Now, t a k i n g C = g (A) (J f (B) we o b t a i n w h a t we n e e d ( r e m e m b e r i n g M~ is a m o d e l of ZFC-). 3) a) DZ1 contains every closed u n b o u n d e d set of l i m i t ordinals
< cop F o r this, we h a v e to c o n s t r u c t , f o r given M a n d i n c r e a s i n g c o n t i n u o u s s e q u e n c e <6~ : i < c o l > , a s u b s e t A of col s u c h t h a t if 6 < c o l , & # 6 ~ f o r all < (a l, t h e n A N 6 ~( Ma. We c o n s t r u c t s u c h A p i e c e b y piece, n a m e l y d e t e r m i n i n g A N[6~,6~ +co) by i n d u c t i o n on i
( we b e g i n with i = - 1
taking
119
~-l = 0; o u t s i d e t h e s e i n t e r v a l s all o r d i n a l s a r e in A).
Having d e t e r m i n e d
A ~ 6i, we h a v e 2 ~~ possibilities f o r A /'] ( ~ + c0); we c h o o s e one which d o e s n o t b e l o n g to t h e c o u n t a b l e s e t I S (h (~r
c M~, ~r < 6 < ~r
It is e a s y t o c h e c k t h a t A satisfies o u r r e q u i r e m e n t . b) D~ is a p r o p e r fllter. By 2) e v e r y s e t in D~/ c o n t a i n s some IF(A ) which is s t a t i o n a r y , h e n c e nonempty. c) D~I is n o r m a l . It suffices t o show t h a t if A r A N 6 ~ M ~ implies A~ ( h ~ E M ~
r
< ~1) t h e n t h e r e is A ~ ~1 s u c h t h a t
for alli<~.
By a) a n d 2) it suffices if t h e
i m p l i c a t i o n h o l d s f o r c l o s e d u n b o u n d e d l y m a n y ~'s. Let
<-,->
be
a
nicely
defined
ordinals, ( p r e s e r v i n g c o u n t a b i l i t y ) <-,->
pairing
function
from
ordinals
to
, a n d let C = 16 < ~1:~ is c l o s e d u n d e r
a n d t h e r e s p e c t i v e p r o j e c t i o n s J. F o r ~ E C, t h e r e s t r i c t i o n s of < - - , - >
a n d t h e r e s p e c t i v e p r o j e c t i o n s to 6 b e l o n g to M8. C is c l o s e d u n b o u n d e d in ~1"
LetA = |
C a n d A (] ~ c M S, a n d l e t i < ~ . T h e n
also Ar N ~ c Ms, which c o m p l e t e s t h e p r o o f .
1.5 Definition ( r e a l one): A f o r c i n g n o t i o n P satisfies the M-c,c.:
a) If IPI ":~ob) If ]P] = 8 1 a n d f o r s o m e (every) f : P - , a ~ 1 which is o n e - t o - o n e , 16 < w l : A c M6,A C ~,
f-l(A)
is
predense
in f - 1 ( 6 )
implies
f-l(A)
is
p r e d e n s e in P | c DB. ( t h e e q u i v a l e n c e of t h e " s o m e " a n d " e v e r y " v e r s i o n s is s e e n to be t r u e b y c o n s i d e r i n g I~(B), w h e r e B C wl c o d e s a p a r t i a l o n e - t o - o n e f u n c t i o n f r o m ~1 into e~1 i n d u c e d by two r e p r e s e n t a t i o n s f ,g of P). c) If IPI >81 a n d f o r e v e r y p t c p (P" ~ P ) a n d f :P" -, ~1, as in b).
if ]P?I -<81 t h e n t h e r e a r e P" ~ P t
120
R e m a r k : In a n y c a s e , u s u a l l y we use i t e r a t i o n s of l e n g t h r
s t a r t i n g with
V = L, so we w o r k e x p l i c i t l y only with ] P ] = ~ l1.6 Claimc 1) If P satisfies t h e M-c.c. f o r s o m e r
~ then P
s a t i s f i e s t h e ~l-c.c. 2) L e t M b e a n r
P
s a t i s f i e s t h e M-c.c. 3) In Definition 1.5 f o r I P ] >~1 we c a n d e m a n d P"
satisfies the
_~-c.c., t h e r e
is ~ e C V~ I~(o~ with t h e
property:
[A e M6, A is a p r e d e n s e s u b s e t of P t ~] - - > A is p r e d e n s e in P ( a c t u a l l y , t h e s e t of t h e s e ~'s is in D~). B u t ~ N ~ is a c o u n t e r e x a m p l e , a s a n y p e ~ - ~ is i n c o m p a t i b l e with e v e r y q e ~ C~ ~8) A s s u m e first t h a t I P I = ]Q] = s 1. W.l.o.g. t h e u n i v e r s e of Q is r
We h a v e to show t h a t f o r a s e t i n D ~ of ~'s, if A is p r e d e n s e i n P t ~ a n d
A e Ms t h e n A is p r e d e n s e in P.
S i n c e Q s a t i s f i e s t h e M-c.c., it s u f f i c e s to
s h o w t h a t A is p r e d e n s e in Q t ~. F o r q c Q define ~)q = | r c P : v is i n c o m p a t i b l e with q, or ( u
~ r ) ( r t c P -+ r t is c o m p a t i b l e
with q)]. L e t ~q be a m a x i m a l s e t of p a i r w i s e i n c o m p a t i b l e e l e m e n t s in ~ q .
As ~)r
is d e n s e in P, ~q is a m a x i m a l a n t i c h a i n in P, h e n c e in Q (since P
c P t6
121
c o m p a t i b l e in P a r e c o m p a t i b l e in P r 6, a n d s i m i l a r l y f o r Q.
Now, l e t A b e p r e d e n s e in P r 6 a n d let q c Q r 6. Since rq c P r 6, t h e r e is lo E A c o m p a t i b l e with rq. Let r t E P b e a b o v e p , rq. By t h e c h o i c e of rq, v t is c o m p a t i b l e with q, h e n c e p is c o m p a t i b l e with q, so we finished. B e f o r e p r o v i n g t h e c l a i m f o r all P , Q we e s t a b l i s h t h e following s i m p l e f a c t on f o r c i n g n o t i o n s : 1.6A F a c t : (CH). If P i C P2, IP1] --t~l, a n d P2 s a t i s f i e s t h e tti-c.c., t h e n t h e r e is P3 s u c h t h a t ]P3 ] < t~1 a n d P1 C P3 <~ P2P r o o f of F a c t : We define b y i n d u c t i o n a s e q u e n c e < P ( a ) : a < cot> as follows: p(0) = P i . p(a) = U P(P) f o r limit a. te
F o r a = g + 1, f o r e v e r y c o u n t a b l e s u b s e t ~9 of P(#) w h i c h is n o t p r e d e n s e in P2, p i c k P3 c P2 e x e m p l i f y i n g this. Let P~=) be t h e s e t o b t a i n e d b y a d d i n g the pjs
to P(#) . Let p(a) be t h e s e t o b t a i n e d f r o m p~=) b y a d d i n g a c o m m o n
u p p e r b o u n d f o r e v e r y p a i r in p~a) h a v i n g o n e in P2Now, Pa =
U P(=) is as r e q u i r e d , p r o v i n g t h e F a c t . a,C~ i
R e t u r n i n g to o u r claim, we d i s t i n g u i s h t h e following c a s e s : (i) ] P I --
s u c h t h a t p t s a t i s f i e s t h e M-c.c.
] p t ] =t~ 1 (this follows f r o m t h e d e f i n i t i o n of Q s a t i s f y i n g t h e M-c.c.).
S i n c e P
122
(ii) P" satisfies t h e M-c.c. H e n c e P satisfies t h e M-c.c. 3) Let P satisfy t h e M-c.c. (by Definition 1.5), IP] > 141' a n d let p t c p , IP?I ~ 1
and choose P",
]P"I ~ 8 1 s u c h t h a t P? c P'~ <~ P . F r o m 2)
we k n o w t h a t P" satisfies t h e M-c.c., h e n c e is as r e q u i r e d .
w T h e o m i t t i n g t y p e theoremu 2.1 L e m m a : ( ~ ) .
S u p p o s e ~bi(x)(i < c01) a r e II 1 (i.e., V r e a l 3 r e a l ... )
f o r m u l a s o n r e a l s (with a r e a l p a r a m e t e r , possibly). S u p p o s e f u r t h e r t h a t t h e r e is,no s o l u t i o n t o ./~ @~(x) in V; a n d m o r e o v e r e v e n if we a d d a C o h e n to "b qE~a)I
V r e a l t h e r e will be n o n e .
Then t h e r e is a n ~1 - o r a c l e M s u c h t h a t P ~ M-c.c.
implies in VP t h e r e is n o s o l u t i o n t o / ~ . Ip~(x).
P r o o f : Let n < ~ b e large e n o u g h so t h a t t h e f o r c i n g t h e o r e m s c a n b e p r o v e d f r o m ~,,~ s e n t e n c e s in ZFC , a n d t h e a s s u m p t i o n of t h e l e m m a c a n b e f o r m u l a t e d as a ~,,,~ s t a t e m e n t . Now, f o r a given c o u n t a b l e f o r c i n g n o t i o n P a n d r e a l ~, let M(P,~) be a c o u n t a b l e ~ - e l e m e n t a r y a n d ~@~: i < ~ 1 ~ .
a given P - n a m e f o r a
s u b m o d e l of V c o n t a i n i n g P,
Let D ( P , ~ ) be t h e c o l l e c t i o n of all p r e d e n s e s u b s e t s of
P lying i n / g ( P , T ) .
2.1A Clain~ If P c p ? a n d e v e r y A c D ( P , T ) is p r e d e n s e in P?, t h e n V/n ~ ~ A
~(r)
( n o t e t h a t ~ is also a P L n a m e ) .
Proof of Claim: Let G be P L g e n e r i c o v e r V. S i n c e e v e r y A C ~}(P,T) is p r e d e n s e in p t a n d t h e r e f o r e i n t e r s e c t s G, G y~ P is P - g e n e r i c o v e r M(P,'r). Also ~[G] = T[G C~ P]-
Since M(P,~) is a ~ n - e l e m e n t a r y
s u b m o d e l of V, it
satisfies t h e a s s u m p t i o n of t h e l e m m a , h e n c e M(P,~)[G ~ P] ]= ~ Ipi(~[G])
123 .U(P,.~)
f o r s o m e i < ~1
" ( r e m e m b e r t h a t P is c o u n t a b l e a n d t h e r e f o r e a d d i n g a
g e n e r i c s u b s e t of it is e q u i v a l e n t to adding a C o h e n real). ments
pass
upward
from
M(P,~)[G N P]
V[G]
to
Since ~@ s t a t e -
, it
follows
that
V[ G] ~ ~ "tl,~(T), p r o v i n g t h e claim.
Now, using
%
a n d a closing p r o c e s s , we c a n o b t a i n a n r 1
= < M ~ : ~ < ~1>
satisfying: if P h a s u n i v e r s e 6 < ~1 a n d P , [ 9 M~ t h e n
~ ( P , T ) c M a ( r e m e m b e r t h a t P,D(P,T) a r e c o u n t a b l e ) .
We will show t h a t
satisfies t h e r e q u i r e m e n t of t h e lemma.
Assume, on t h e c o n t r a r y , t h a t p t ~ M-c.c. b u t T is a P t - n a m e of a r e a l so N
t h a t V/~ ~ / ~
~bg(~). We m a y a s s u m e t h a t IPt I = S l since:
a) if I p t I --<S0 t h e n this is impossible b y t h e a s s u m p t i o n of t h e l e m m a . b) if IPt I > S l t h e n b y !.6(1) a n d 1.6(3) we c a n find P " s u c h t h a t T is a
P"-name, ]P"] =I~1, P"
). So P" can replace pt in the
sequel. W.l.o.g. t h e u n i v e r s e of p t is ~t-
We c a n find 6 < ~1 s u c h t h a t l e t t i n g
P = P t F 6 t h e following will hold: (i) T is a P - n a m e . (ii) P , T e M~. (iii) P c P t. (iv) A e M,~, A is a p r e d e n s e s u b s e t of P ------>A is p r e d e n s e in pt. m
F r o m t h e s e f a c t s , t h e claim a n d t h e c o n s t r u c t i o n of M, it follows t h a t
V i~ ~ ~ /k. ~(T), a c o n t r a d i c t i o n .
2.2 E x a m p l e : If V ~ [ OM ^ A c s ^ A I
is of t h e s e c o n d c a t e g o r y ] t h e n
124
m
t h e r e is a n ~ l - o r a e l e M s u c h t h a t : P ~ _~-c.c. ==~ VP ~ "A is of t h e s e c o n d c a t e g o r y "
Proof: Let A =~v~ : i < ~ l l , closed nowhere dense, where ~
~(x)
=(r~ cBz)
where
Bz = L) B~x , B~x
is "simply" d e s c r i b e d by x so as t o i n c l u d e
e v e r y s e t of t h e first c a t e g o r y as Hx f o r s o m e x. E a c h ~p~ is trivially H~. We still h a v e to show: if P is c o u n t a b l e , t h e n not: ]{-p" f o r s o m e r e a l x ,
for every i < ~ 1
r i c B x ^ / X / ~ z is n o w h e r e d e n s e " . n S u p p o s e , on t h e c o n t r a r y , t h a t t h e r e are r c P a n d a P - n a m e x, so t h a t N
r I~-y"x is a c o u n t e r e x a m p l e ". T h e n for e v e r y i, f o r s o m e p c P a n d n , p H - p " r ~ e t ~ z "' L e t / ~ =
lr~'p
l~-p"r~ c B zn ' ' ] ; we h a v e s h o w n A C u A ~ .
We shall s h o w t h a t e v e r y A~ is n o w h e r e d e n s e a n d g e t a c o n t r a d i c t i o n .
Let •
< ~, p E P , ( q l , q 2 ) a r a t i o n a l i n t e r v a l .
It s u f f i c e s t o f i n d a r a t i o n a l
s u b i n t e r v a l disjoint f r o m A~. But I~-p"B~ is n o w h e r e d e n s e ". So t h e r e a r e
ptmp,(q~,q~)
c(ql,q2) suchthatpt
]~-p"B~xc~(q~,q~) = r
Clearly:
N
r~ e (q~,q~) ~------>p t ]]-p,,ri Ig ~
"~---------->
w Iterations of M-c.c. forcings. 3.1 Claim: If ~/i-(i < Wl) a r e Wl-oracles t h e n t h e r e is a n ~ l - o r a e l e M s u c h that:
125 (P ~ [~-c.c. ) ===>( / k P ]= ~_~-c.c. ).
Proof: Let M, =/kM~ :6 <~l>C h o o s e M~ s u c h t h a t h) J ~ c M~. It s u f f i c e s to p r o v e t h a t : ~<~
M1,M2 a r e ~ l - o r a c l e s , I & : M ~ ~ M~{ is c o - b o u n d e d (i.e., with a b o u n d e d c o m p l e m e n t ) t h e n M2-c.c. ------>Ml-C.c.. (*) if
S i n c e f o r A c ~1, e v e r y s u f f i c i e n t l y l a r g e ~ 9
I~?~(A) is in I~?,(A), it suffices
to show t h a t if 6 h a s t h e r e q u i r e d p r o p e r t y with r e s p e c t to M 2 ( a n d ~ is s u f f i c i e n t l y l a r g e ) t h e n it h a s t h e r e q u i r e d p r o p e r t y with r e s p e c t to Mz. But t h i s follows t r i v i a l l y f r o m M~ c M~. 8.2 C]a!m: If
P~(i < a) is t h e r e s u l t of finite s u p p o r t i t e r a t i o n , e a c h P~
s a t i s f y i n g t h e J~-c.c., P = t3 P~, f h e n P s a t i s f i e s t h e M-c.c.
P r o o f : We d e m a n d s u p e r f i c i a l l y less on t h e Pi's: P~
Clearly
P 9 [P~:i < a ] (it is P ~ - x ) so b y a s s u m p t i o n P s a t i s f i e s t h e M-e.c.
C a s e a: ef ( a ) > ~ 1.
We a r e g i v e n p t c P ,
[Ptl ~t~ 1 a n d h a v e t o find a p p r o p r i a t e P " , 3'- By
a s s u m p t i o n f o r s o m e i, P I c Pi.
F o r it we c a n find P "
cPt,f
As Pi c P
we
are through. C a s e m : af ( a ) = c01. Obviously it is s u f f i c i e n t to d e a l with t h e c a s e a = ~1- We will n o w show
126
t h a t it is sufficient to deal with t h e c a s e I P I = t~.
I n d e e d , if I P I <~1 t h e n t h e c o n c l u s i o n of o u r c l a i m holds.
So a s s u m e
I P I > ~ I , a n d P d o e s n o t s a t i s f y t h e M-c.c. T h e n t h e r e is P t C P of p o w e r t~1 so t h a t P? C P" C P, I P~ ] =t~ 1 imply t h a t P" d o e s n o t s a t i s f y t h e M-c.c. We n o t e t h a t P satisfies t h e t~Vc.c. (See Claim 1.6(1) a n d L e m m a II 2.8). We now r e f e r t h e r e a d e r to t h e p r o o f of F a c t 1.6A (in t h e p r o o f of Claim 1.6(2)). We r e p e a t h e r e , f o r p t ~ p , t h e s a m e c o n s t r u c t i o n , with t h e a d d i t i o n a l p r o v i s i o n t h a t p i is always c h o s e n to b e l o n g to a P~ with t h e l e a s t possible i ( a n d t h e s a m e is d o n e w h e n a d d i n g c o m m o n u p p e r b o u n d s ) .
The effect of this is t h a t
we o b t a i n P" (P3 in t h e n o t a t i o n u s e d t h e r e ) so t h a t n o t o n l y P"
H e n c e for e a c h i
M-c.c. Also, as P" f3 P~ <" P" (h P~+I (since P" (3 P~ <~ P~ <~ P ) a n d f o r limit P" f h P ~ =
r
u(P"
NP~),
it
follows t h a t
(hP~ : i < ~ >
satisfies
our
h y p o t h e s i s . But ] P " I =t~l a n d it d o e s n o t s a t i s f y t h e /~-c.c., so we h a v e r e d u c e d o u r c o u n t e r e x a m p l e t o o n e of p o w e r t~. We a s s u m e now, w.l.o.g., t h a t Pi - CJPj C r i
X tiJ. We let #4 = 16:C E M8 is
a p r e d e n s e s u b s e t of P i r (6 x 6) implies C is p r e d e n s e in P~{.
F o r e a c h i
We let pri be a f u n c t i o n f r o m P x ~
into P~
s u c h t h a t : p c P~ is c o m p a t i b l e with q c P in P, iff f o r s o m e n , p r , ( q , n ) , p a r e compatible.
(existing
as
Pi <~ P,
Pi ~ ~Vc.c.) -
We
let
B~ = 16:pr~((6 x 6) x ~) ~ 6 x 6 a n d p r ~ t ((6 x 6) x ~) c M~]. F o r e a c h i < ~1 , B~ c D~. We let A = .K~ = | 6 " every i <6
, 6cB~]
for e v e r y i < 6 Then A f ~ B c D ] / .
, ~cA~
a n d B=A.B~ = l ~ :
A s s u m e now t h a t
6cA
for f~B,
C c Ms is a p r e d e n s e s u b s e t of P t (6 x 6). Notice t h a t (*) 1o E P, is c o m p a t i b l e with some q c C iff it is c o m p a t i b l e with some
127
We k n o w t h a t f o r all i < ~ p r c ( C x ca) 9 M,~ a n d is a p r e d e n s e s u b s e t of Pct(~x~)-
since ~ 9
T h e r e f o r e , as 6 9
pr~(Cx~)
ispredenseinPc.
Using (*) again, (as P$ = U P~ , P r (~ x 6) c P~) C is p r e d e n s e in P$, a n d s i n c e P8 <~ P, C is p r e d e n s e in P. Thus we h a v e p r o v e d t h a t P satisfies t h e M-c.c. C a s e IV: cf ( a ) = t~.
One c a n deal with this c a s e as with Case III ( t h e d i a g o n a l i n t e r s e c t i o n s a r e replaced by countable intersections).
We shall deal n o w with t h e " r e a l " s u c c e s s o r c a s e : 8.3 Clnira: A s s u m e V ~ ~
M is a n ~ l - o r a c l e , P is a f o r c i n g n o t i o n of f
p o w e r ~1 s a t i s f y i n g t h e M-c.c. T h e n in V P t h e r e is a n c01-oracle M ~ s u c h t h a t : if Q 9 VP satisfies t h e M~
(in VP) t h e n P * Q (which is in V) satisfies t h e
M-c.c.
P r o o f : F i r s t we o b s e r v e t h a t if M~ in VP is g o o d f o r all Q of p o w e r ~ B1 t h e n it is g o o d f o r all Q. H e n c e we shall t r e a t Q as a P - n a m e of a b i n a r y relat i o n on coz ( b u t we shall define M* w i t h o u t d e p e n d i n g on a given Q). We also a s s u m e t h a t P h a s u n i v e r s e ~1, a n d we fix f o r t h e r e s t of the p r o o f a g e n e r i c G ~ P (which h a s a c a n o n i c a l P - n a m e in I/).
F o r A ~ wl x wl define A c = [a:(~p 9 G ) ( p , a ) 9
AJ.
Let ~ < col be a limit o r d i n a l . We let M; be a c o u n t a b l e t r a n s i t i v e m o d e l of ZFC - c o n t a i n i n g 6 , M6 a n d ~A a " A c ~ x ~,A 9 M,~J. We will first show t h a t M" is a ~ - s e q u e n c e
in V[G].
So we a r e given a P - n a m e S of a s u b s e t of ~1, a n d we h a v e to show t h a t N
[~:S[G] N ~ 9 M~] is s t a t i o n a r y in V[G].
Since P satisfies t h e ~l-C.C., e v e r y
c l o s e d u n b o u n d e d s u b s e t of wl in V[G] c o n t a i n s one in V, so it suffices to
128
show t h a t if C c V is a c l o s e d u n b o u n d e d s u b s e t of ~1 t h e n t h e r e is 5 a C s u c h t h a t S[G] N ~ 9 M~. Let, f o r e a c h a < ~l,.ga ~ P be a m a x i m a l a n t i c h a i n of c o n d i t i o n s d e c i d i n g t h e t r u t h v a l u e of a c S.
As P satisfies t h e t~l-c.c., e a c h ,9a is c o u n t a b l e , so
w.l.o.g, f o r all ~ 9 C, a < 5 implies ~ga C 5, a n d P t 5 ~ P , (so if p , q 9 P t 5 h a v e a c o m m o n u p p e r b o u n d in P, t h e y h a v e one in P t 5.) Now, let T = I ( p , a ) : p ][-p " a 9 S"] . As M is a n ~ l - o r a c l e , t h e r e a r e 5 c C N
and A EM$ such t h a t A = T f ~ ( h x h ) .
By o u r c o n s t r u c t i o n of M~ A a 9
We will show t h a t S[ G ] N 5 = A g. Assume
first
p ]]-p"aeS",
that
hence
a9 aeA
Nh. a.
Let
p 9 G N~ga,
Assume now t h a t
aeA
then g.
10 < 5
and
Then for some
N
p 9 G,p I [ - p " a 9
and a<6,
hence a 9
f~5. T h u s IS[G] N h = A
~ ,
a n d we h a v e p r o v e d t h a t M" is a n ~ t - o r a c l e in V[G].
We
now
( P *Q) t ( h x h ) .
assume
that
A ~5x5,
We w a n t to s h o w t h a t A
A cM~
and
A
is
predense
in
is p r e d e n s e in P *Q. The p r o o f will
be b r o k e n into t h r e e steps, a n d as we p r o c e e d we will i m p o s e r e s t r i c t i o n s on 5, t h e r e a d e r will v e r i f y t h a t t h e s e r e s t r i c t i o n s a r e allowable (in t h e s e n s e of D~) a n d do n o t d e p e n d on A a n d G. F i r s t S t e p : A G ~ 5, A a c M~ a n d A ~ is p r e d e n s e in ~ G ] t 5. N
P r o o f : The first two f a c t s a r e i m m e d i a t e .
Let (io,q),(a,fl) c P * Q. We define J'(p,r
w h e n e v e r ( p , q ) a n d (at,p)
a r e eolifpatible , as t h e l e a s t possible P - p a r t of a c o n d i t i o n a b o v e ( p , q ) a n d (a,~) in P * Q. We m a y a s s u m e t h a t f o r e a c h p , q
<5
f(p,q)(hxh) C5 and
f o r e a c h q < 5 t h e m a p p i n g p -~ f@,q) t (5 x 5) f o r p < 5 b e l o n g s t o M~.
129
L e t q c Q [ G ] r 6 , a n d define D q = u f o ~ , q ) ( A ) 9 T h e n ~ ) q ~ , D q
9
~)q is d e n s e in P t 6 (as A is p r e d e n s e in ( P * Q) r (6 x 6) a n d b y t h e d e f i n i t i o n of f (p,q)). H e n c e we m a y a s s u m e t h a t ~)q is p r e d e n s e in .P (by o u r a s s u m p t i o n t h a t P s a t s i f i e s t h e M-c.c.).
Let r r
Then for somep
<6, s eQ
a n d (a,fl) c A
we h a v e in P * Q N
( r , S ) >-- ( p , q ),(a,~).
G As r ~ a (in P), r I[-p "~x 9 G " so r I ~ p " # 9 A ' P " ; as Np
r ]~-p"fl <-9 s a n d q <--9s " i t follows t h a t r ]~-p"q is c o m p a t i b l e with s o m e elem e n t of A c ' '
As t h i s h o l d s f o r a l l r
9
andDq
ispredenseinP,
itistrue
in V[G], a n d a s t h i s h o l d s in V[G] f o r all q 9 Q[G] t 6 , we o b t a i n t h a t A a is p r e d e n s e in Q[G] r 6.
S e c o n d Step: A a is p r e d e n s e in Q[G].
P r o o f : This follows f r o m t h e a s s u m p t i o n t h a t Q s a t i s f i e s t h e M*-c.c. We o n l y h a v e to o b s e r v e h e r e t h a t t h e r e s t r i c t i o n to a s e t in D~" c a n b e s u b s t i t u t e d b y a r e s t r i c t i o n to a s e t in D~. The p r o o f of this p o i n t follows closely t h e p r o o f of t h e f a c t t h a t M~ is a O - s e q u e n c e " T h i r d S t e p : A is p r e d e n s e in P * Q. P r o o f : L e t ( p , q ) c P * Q. S i n c e in V[G] existr->p
, (a,fl) c A a n d s
9 Qsuchthatr
A a is p r e d e n s e in Q[G], t h e r e ][-p"a c G andfl---•s N p
As r I I p " a
e G ", r - ~ a ,
so ( r , s ) > - ( p , q ) , ( a , f l )
~
,q---Qs" m
, in P *Q, p r o v i n g w h a t we
need.
w R e d u c t i o n of t h e Main T h e o r e m t o t h e Main I ~ m m a
130
General topologists have been interested
in w h a t
AUT(~CO)/finite) c a n b e
(p(co) as a B o o l e a n a l g e b r a of s e t s , p ( c o ) / f i n i t e is t h e q u o t i e n t w h e n we divide b y t h e i d e a l of finite s u b s e t s of co). Or e q u i v a l e n t l y : w h a t c a n b e t h e g r o u p of homeomorphisms
of rico - ~0?
N o t e t h a t AUT~CO)) is i s o m o r p h i c (Per(~o)). It is well k n o w n t h a t :
to the
4.1 Claim: (CH) ig~co)/finite is a s a t u r a t e d bra)
of p o w e r
1t1 so it h a s
set
of p e r m u t a t i o n s
atomless
L#1 a u t o m o r p h i s m s
of co
BA ( = B o o l e a n a l g e -
(saturated
in t h e
model
theoretic sense). R e m a r k : This c l a i m will n o t b e u s e d h e r e .
I t is a l s o e a s y to n o t e t h a t :
f EPer(co) t h e n
If
-*f(A)/finite is a n
A/finite
automorphism
of
~g~co)/finite. Moreover: 4.2 Claim: If 3" is a o n e - t o - o n e
s u b s e t s of co t h e n A / f i n i t e (Such an automorphism
function, Dom ()'),Rang (f) are cofinite
~ f ( A ) / f i n i t e is a n a u t o m o r p h i s m
of ~ c o /
finite.
is c a l l e d t r i v i a l , a n d we s a y it is i n d u c e d b y f ).
4.3 Clain~ If we s t a r t with, e.g., V = L, a n d d e f i n e finite s u p p o r t i t e r a t e d forcing
Pi (i~co2) s u c h t h a t
]Pil
then
we c a n Pm2
d e f i n e F~ C [p(co) x p(co)] ve~, F e V p'~, t h e s e t
[i:i < ~o2, cf (i) = col, F t [p(co)] ve~ = F~J is s t a t i o n a r y in co2.
Proof: Translate (remember 4.4
Fi c VP~ s u c h t h a t f o r a n y F C [p(co) x ~ e 0 ) ] v ,
everything
to P ~ 2 - n a m e s a n d a p p l y ~
|6 < ~8:cf (6) =,.,l!
the ~tl-chain condition). Claim:
Suppose
Pi
(i --< co2)
are
as
in
Claim
4.3,
131
F C [ A U T ( ] ~ ) / f i n i t e ] ~'a a n d li < co2:cf (i) = ~1, F t [p(~0)/finite ] d ' is (i) n o t in Vel o r (ii) n o t a n a u t o m o r p h i s m of []9(~0)/finite ] ve~ o r (iii) i n d u c e d b y some p e r m u t a t i o n f
of ~ ( f c Vg~, of c o u r s e ) l .
is
s t a t i o n a r y then F is trivial. P r o o f : By F o d o r ' s lemma. We w a n t to show: 4.5 Main theorem= CON (ZFC) implies CON (ZFC + 2 ~~ = ~2 § e v e r y a u t o m o r p h i s m of ]9(~)/finiLe is trivial).
We
postpone
the
work
concerning
specifically
automorphisms
of
p ( c 0 ) / f i n i t e to t h e n e x t s e c t i o n . To b e p r e c i s e we will p r o v e t h e r e t h e following. 4.6
Main
Lemma
: Suppose
V ~"2~~
1, M~ is
an
~1-oracle,
F c AUT(p(~o)/finite ) is n o t trivial. ': Then t h e r e is a f o r c i n g n o t i o n P s u c h t h a t 1) [PI =~1, 2) P satisfies t h e M~ 3) P i n t r o d u c e s a s u b s e t X of ca, s u c h t h a t if Q is Cohen forcing, P*q t h e n in V - t h e r e is n o Y ~ co s a t i s f y i n g I Y (~ F(C) =aeF(B) <=~ X N C =a,B: C,B E [ p ( ~ ) ] v I (where A =aeA ? m e a n s A ~aeA ? a n d At caeA, w h e r e A Ca, At iff A - A t is finite). R e m a r k : Note we r e a l l y r e p l a c e F by s o m e F? f r o m p(r t h a t F(A/finite) = F t ( A ) / f i n i t e . P r o o f of 4.5. ( a s s u m i n g 4.6):
to ~(r
such
132
O u r g r o u n d m o d e l is a m o d e l of V = L. We i n t e n d to d e f i n e P i ( i "~ ~ ) as in Claim 4 . 3 , so we h a v e F~ p r e d i c t i n g a n y a u t o m o r p h i s m of D(e~)/finite in VP'* ( F~ is a P , - n a m e . ~
So it b e c o m e s m e a n i n g f u l w h e n we h a v e d e f i n e d P,, b u t u s i n g
la
d e p e n d on t h e a c t u a l c o n s t r u c t i o n of t h e P,'s).
So h o w do we define P~ f o r i -< ~27 As we a r e g o i n g to i t e r a t e with finite s u p p o r t , P , ( i ": ~2) will b e d e t e r m i n e d as s o o n as we define < Q.:i < a~2> b y t h e r e l a t i o n s P0 = [~b], Pi+l = Pi * Q. a n d P$ = t3 Pi f o r limit ~. Actually, we will define b y i n d u c t i o n Q. a n d a n w l - o r a c l e M~ c VP~*~ f o r i < ~2 ( r e m e m b e r ~t
t h a t h a v i n g d e f i n e d Q we h a v e d e t e r m i n e d P~§
so we c a n go a b o u t o u r n e x t
t a s k ).
We a s s u m e t h a t F~ c Vp~ is a n o n - t r i v i a l a u t o m o r p h i s m of ] 9 ( ~ ) / f i n i t e ; o t h e r w i s e we l e t Q be a P i - n a m e of C o h e n f o r c i n g a n d ~r~ b e a n y ~ l - o r a c l e in V~§ ( r e m e m b e r t h a t ~ t ~ l h o l d s in t h e g r o u n d m o d e l a n d t h e P~'s s a t i s f y t h e t~l-c.c, a n d h a v e p o w e r < t~2). F o r e a c h j < i, Pj,i C VPj is t h e f o r c i n g n o t i o n (of p o w e r t~l) s u c h t h a t P~ = Pj * P j , , a n d l e t / ~ j + l c VP' ( f o r j < i ) b e a n ~1" o r a c l e with t h e p r o p e r t y : if Pj+I,~ ~ Mj-c.c. in VPJ*' a n d
t h e n Pj+l,i * Q ~ Mj-c.c. in VP'§
Vv'
j§
Q ]= Mi+l-c.c. in
e x i s t s b y Claim 3.3.).
Now, let
M~ E VP~ b e a n ~01-oracle t h e s a t i s f a c t i o n of w h o s e c.c. implies s a t i s f a c t i o n of all MJ§
f o r j < i (existing b y Claim 3.1 ). Apply now 4.6 in Vy~ to o b t a i n
P = Q~ c VP* s a t i s f y i n g VP~*' = VP ~
1), 2),
3)
of 4.6.
Finally
apply
Lemma
2.1
in
f o r t h e t y p e a p p e a r i n g in 3) of 4.6, t o o b t a i n a n ~ l - o r a c l e
~ . c VP~*~ s u c h t h a t t h e t y p e is o m i t t e d in g e n e r i c e x t e n s i o n s of VP~§ with f o r c i n g n o t i o n s s a t i s f y i n g t h e Mi-c.c.
As
P~2
s a t i s f i e s t h e t~l-c.c, a n d h a s p o w e r N2, t h e r e a r e a t m o s t t~" = t~2
P ~ s - n a m e s of r e a l s ; o n t h e o t h e r h a n d we h a v e a d d e d r e a l s t~2 t i m e s , so
133
Ve'2 I=''2~~ = ~2" Pw
L e t F 9 [AUT(]9(co)/finite)] v 2; we w a n t t o s h o w t h a t F is trivial. n o t , t h e n b y C l a i m 4.4. we k n o w t h a t t h e r e is a c l o s e d u n b o u n d e d ~2
with
the
property:
i c C
implies
If it is
s u b s e t C of
cf (i) # o 1
or
F t [ p ( o ) / f i n i t e ] ~2"~9 [AVTO~(~)/finite ] ve~ b u t is n o t i n d u c e d b y a n y f in VP~. But
by
Claim
4.3
we
can
find
i E C
with
cofinality
F r []9(~)] re* = F i ( r e f e r t o t h e r e m a r k p r e c e d i n g t h i s p r o o f ) . h a d Qi i n t r o d u c i n g X ~ ~ a n d a n ~ l - o r a c l e / ~ i
~l
satisfying
So a t s t a g e i we
9 VP**I s u c h t h a t F ( X ) c a n n o t
e x i s t in g e n e r i c e x t e n s i o n s of V P~*' w i t h f o r c i n g n o t i o n s s a t i s f y i n g t h e Mi-c.c. Examining our construction
and remembering
Claim 3.2 it is e a s y t o p r o v e b y
i n d u c t i o n o n a < ~2 t h a t f o r a < i in V P~*I, Pi+l,a s a t i s f e s t h e ~ - c . c . .
Hence
V P'2 = (VP~"~)p'+*''2 is s u c h a n e x t e n s i o n of V P~§ - a c o n t r a d i c t i o n . 4.7 C o n c l u s i o n : In t h i s s c h e m e :
l) for any second category A c l a < ~ e : c f ( a ) = ~1 ~
A A [ ~ o ) ] #~' c V P~ a n d is of 2 n d c a t e g o r y J
contains a closed unbounded
set.
2) if A C []9(~)] v~*, A c VP~, A is of 2 n d c a t e g o r y , t h e n t h e r e is a n o r a c l e e n s u r i n g it r e m a i n s of t h e 2 n d c a t e g o r y . P r o o f : F o r 1) s e e 4 . 4 , f o r 2) s e e 2.2. 4.8 Conclusion: 1) We c a n a s s u m e that. in VP'~ a n y s e c o n d c a t e g o r y s e t c o n t a i n s s u c h s u b s e t of c a r d i n a l i t y t~1. 2) We c a n find t~1 - d e n s e s e t s of r e a l s , s o m e of t h e f i r s t c a t e g o r y , s o m e of t h e s e c o n d c a t e g o r y . 3) B a u m g a r t n e r ' s a s his r e s u l t fails.
construction
[B4] c a n n o t be c a r r i e d o u t h e r e
134 w P r o o f o f Main L e m m a
4.6.
F o r a sequence A =
of ~, define
p ( ~ ) = ~f :jr is a p a r t i a l f u n c t i o n f r o m ~ to 2 = 10,1] , f is a finite u n i o n of f u n c t i o n s of t h e f o r m : 1) finite f u n c t i o n s 2) 1~ - finite 3) 0A, - finite ] w h e r e g n is t h e f u n c t i o n f ( a ) = JZ if a e A, u n d e f i n e d o t h e r w i s e , a n d f - f i n i t e is f r ( d o m ( f ) - A) f o r s o m e finite A. T h e i d e a of t h e p r o o f is as follows: we t r y t o d e f i n e A =
function
of t h e l e r n m a f o r X w h o s e c h a r a c -
is L I [ P " P e Gp(2i)~ . We will h a v e a g e n e r a l
pattern
for
d e f i n i n g A. O u r w o r k will b e t o s h o w t h a t if all i n s t a n c e s of t h i s p a t t e r n
fail
t h e n F m u s t b e trivial. R e m a r k : If t h e p r e s e n t a t i o n
does not look rigorous enough to you, look
a t a r e m a r k a t t h e end.
O u r f i r s t w o r r y is t o m a k e P ( A ) s a t i s f y t h e M*-c.c.
T h e following (*1)
t a k e s c a r e of t h i s while l e a v i n g u s w i t h a s u f f i c i e n t d e g r e e of f r e e d o m in t h e construction
of A.
(*i) If A =
A c Cg,
[AI = [C~ --AI =M0,
-A>).
every
.gEM a
is
predense
in
P r o o f of (*1): We d e f i n e b y i n d u c t i o n o n n < ca d i s j o i n t finite s u b s e t s C,~1 ,C~2 of B s u c h t h a t C~ U Cn2 D B (5 n . I n e a c h s t a g e we d e a l w i t h a n o b l i g a t i o n of t h e following k i n d :
135
.9 9 M., T is a t e r m for a m e m b e r
of P(A ^
(or C a in the other
c a s e ); o u r t a s k is to e n s u r e W will b e c o m p a t i b l e with s o m e m e m b e r of .9 n o m a t t e r h o w A c C 1 is defined.
has the form f
U i A - ~ U J c , - . 4 - ~ w h e r e i,j 9
k <e,
f eP(A)
( m a y b e t h e r e will b e a c o n t r a d i c t i o n ; t h i s is OK ). We d o n ' t k n o w w h a t A C~ C.* is. T h e r e a r e 2 Ic41 s u b s e t s of C~ :A l . . . . .
Ae . . . . .
for 1 ~s
~ 21c41; l o o k a t
f e = f U i.4~-~ U Jc;-Ae-~. S u p p o s e t h i s is a f u n c t i o n , t h e n it b e l o n g s t o P ( A ) , h e n c e is c o m p a t i b l e with s o m e gs e . 9
Now Dora (ge) N B is finite ( b y
B b e i n g a l m o s t d i s j o i n t f r o m e v e r y A~. a n d g e 9 .9 C P()I-)). Let : n s = Max ( D a m (gs
/'1 B) f o r 1 -< s -< 2 Ic~l, a n d n o = n ;
n " = ~ a x I(n e + i):0-< r <- 21&'l{: CI+I = C1;
T a k i n g a d d i t i o n a l c a r e t o m a k e C 1 = g C~, 02 = U C~ infinite, we o b t a i n what we need.
In view of (*i) ,it is convenient to introduce the following notation:
Po, wiu denote the forcing notion corresponding to < A t :j < Q will denote C o h e n forcing. P = will denote the forcing notion corresponding to A =
Ps
will d e n o t e P(A -
f o r g 6 l l , 2 I , A c C~ a s in ( ' 1 ) .
(we
136
P~[A] will d e n o t e Pr If
* Q.
p,q e P o l X Q, p = ((2)0,(2)1) a n d q = ((q )0,(q )l), we will write p U q
w h e n we m e a n ((2)0 U (q)0,(2)1 U (q)l). Forp
For
eP,.,t,
q 9
Q, p U q
means(2
U ( q ) 0 , ( q ) 0 , etc.
A C B, Ch~ will d e n o t e 1A U 0B-A-
Note t h a t (*l) e n a b l e s u s to c o n s t r u c t A as follows: We let
What a b o u t Ma?
U
< co>,
Bin).
It is t a k e n to i n c l u d e all t h a t is n e e d e d in o r d e r to
p r e p a r e t h e s a t i s f a c t i o n of t h e ~ ' - c . c . , as well as a d d i t i o n a l p r e d e n s e s u b s e t s w h i c h will be specified in t h e sequel, when t h e i r n e c e s s i t y b e c o m e s a p p a r e n t ; a n d of c o u r s e t h e y i n c r e a s e with a. S o m e t i m e s , t h e s e will be s u b s e t s of Pax
Q which a r e to r e m a i n p r e d e n s e in
P~[A], b u t we c a n deal with s u c h
s u b s e t s in t h e f r a m e w o r k of ( ' 1 ) w i t h o u t difficulty . Also, we c a n p r e s e r v e p r e d e n s i t y a b o v e a given c o n d i t i o n .
We c a n define t h e
Ma's t o g e t h e r with t h e
Aa's.
Now, in using (*1) we c a n c h o o s e b e t w e e n v a r i o u s c a n d i d a t e s for B , a n d f o r e a c h of t h e m we c a n c h o o s e b e t w e e n v a r i o u s c o n t i n u a t i o n s of A (namely, we c a n c h o o s e ~ e | 1 , 3 | a n d A c C~). How do we c h o o s e ?
In t h e v e r i f i c a t i o n of 3) of t h e lemma, we h a v e to c o n s i d e r all possible Y w h i c h a r e P~, x Q - n a m e s of s u b s e t s of to. By CH a n d t h e ~l-c.c., we c a n find P,, x Q - n a m e s Y , w h i c h a r e d e t e r m i n e d s c h e m a t i c a l l y b e f o r e t h e a c t u a l c o n ~a
s t r u c t i o n of A, so t h a t t h e s e q u e n c e < Y :co ~ a < col, a a n e v e n o r d i n a l > will
137
c o n t a i n all p o s s i b l e Y e a c h a p p e a r i n g t l l - t i m e s . [More explicitly, we c a n i d e n tify aFfix truth
Q - n a m e of a Y c c 0 w i t h C q ~ d , l g j
value,
for
each
q~j [~-"[~ ~ ~ =- l g j " .
: i , j < w ~ , q~j c P ~ x
i <~a C qi,a" : J
is
a
maximal
Q,
1gd a
antiehain,
We c o n s i d e r Y o n l y if M a a l r e a d y e n s u r e s u s e a c h
C q ~ j : j < c0~ will c o n t i n u e to b e a m a x i m a l a n t i c h a i n , a n d it h a s t h e r i g h t form.
As t h e r e a r e
~ tl 1 c a n d i d a t e s , a n d e a c h c a n d i d a t e a p p e a r s wl t i m e s
e v e r y t h i n g is c l e a r ] . Our c h o i c e of B,E,A will b e as follows: w h e n e v e r p o s s i b l e , we c h o o s e t h e m so t h a t [Ch~,~ ]l-p~[A]"Ya N F(Ce) = ~ F(A)"] is false, ( a n d i n c l u d e s o m e s e t s in Ma+ 2 as e x p l a i n e d below).
A s s u m e now t h a t P~t o t a i n e d b y s u c h a p r o c e d u r e
fails to s a t i s f y t h e
r e q u i r e m e n t s of t h e l e m m a . As 4.6 (1) is c l e a r l y s a t i s f i e d a n d we e n s u r e d t h e satisfaction
of 4.6 (2) , only 4.6 (3) m a y b e false.
X = UIf-l(ll]):f
We u s e t h i s f a c t f o r
c G] w h e r e G c P~I is g e n e r i c . If t h e e x i s t e n c e of a p p r o p r i -
a t e Y is n o t f o r c e d a l r e a d y b y ~b, t h e n s o m e r c P~I x Q f o r c e s t h a t t h e r e is n o s u c h Y a n d in t h i s c a s e P ~ r l P "P -~ (r)0] s a t i s f i e s t h e r e q u i r e m e n t s
of
t h e l e m m a (as t h e Ma's m a y t a k e c a r e of t h e s a t i s f a c t i o n of t h e M ' - c . c . b y s u c h r e s t r i c t i o n s of P~, as well). H e n c e we m a y a s s u m e t h a t r f o r c e s this. So, t h e r e is Y = Y (w ~ a < ~al,a e v e n ) which r e a l i z e s t h e r e l e v a n t t y p e , a n d N
N~
w.l.o.g, a is l a r g e r e l a t i v e to Y ( r e m e m b e r e a c h Y a p p e a r s tl 1 t i m e s . ) S i n c e
Ch~.~ It-p.~ x Q "'X CI C~g = A '" ( r e f e r r i n g t o C~ a n d A with w h i c h we a c t u a l l y c o n t i n u e d a t s t a g e a), it follows t h a t C h ~ It- eol x Q "Y~ N F(C~g) = ~ F ( A ) ':
Now, w h e n we a r e a t s t a g e a w e k n o w Y
, P= b u t n o t P ~ . S o w e w a n t t o
N~
s h o w t h a t t h i s is a l r e a d y t r u e if we c o n s i d e r f o r c i n g in P~[A].
As t h e f o r c i n g
r e l a t i o n f o r e a c h f o r m u l a n c Yis n o t a f f e c t e d b y t h i s shift, o u r only p r o b l e m a r i s e s f r o m t h e f a c t t h a t e q u a l i t y is f o r c e d to h o l d only he. This difficulty is o v e r c o m e as follows: if, on t h e c o n t r a r y , t h e r e is p >- Ch~ in P~[A] f o r c i n g
" Y N F(Ce) #a~ F ( A ) ' ; t h e n f o r e a c h k < ~ t h e s e t of t h o s e c o n d i t i o n s "knowing" s o m e c o u n t e r e x a m p l e
to t h e e q u a l i t y a b o v e k is p r e d e n s e a b o v e p in
138
P ~ [ A ] ; so, a s we r e m a r k e d in P~, x O , b u t ( a s p
->
a b o v e , we c a n a s s u m e t h i s p r e d e n s i t y is still t r u e
Ch~)
P ]~-P,,xO " Y N F(C~) =aeF(A)", a c o n t r a d i c -
tion.
=ae F(A)" B u t we c h o s e B,~ a n d A d o i n g o u r
So, ChcA~ I ~ / ~ [ A ] " Y N F ( C s
b e s t t o a v o i d this, so we m u s t h a v e h a d n o b e t t e r c h o i c e ; t h u s : (*2) L e t B b e i n f i n i t e a n d a l m o s t d i s j o i n t f r o m e a c h B = C 1 U C2 be t h e p a r t i t i o n o b t a i n e d in ( ' 1 ) . e v e r y A ~ Cs s u c h t h a t I A I = I C s
Ch~, II-~[A]
T h e n f o r e v e r y s ~ ~1,21 a n d
I =e0, " Y C~ F ( C~) = , . F ( A ) '"
We n o w a s s u m e B is as in ( ' 2 ) a n d c o n c e n t r a t e s i m i l a r l y . So n o w we s h a l l u s e Let Pax
P a x Q, ~
Q = [ps
<~J,
= |q~:i < ~ ] ,
Aj(j < a), a n d l e t
o n C1; C 2 c a n b e t r e a t e d
PQ][A]. a n d let f o r n < ~ ,
such that
~n
be a d e n s e
subset
q~ I~-P~xQ"[n e Y] --- t~~ ' ' ( t~~ d e n o t e s
of a
t r u t h v a l u e , 1 is t r u e , 0 is false).
We s h a l l n o w d e f i n e b y i n d u c t i o n o n /c < co a n a t r u a l function
~ f ~ : a ~ - , 10,1] w h e r e
a k = [ n ~ , n ~ + l ) ( 3 C1.
number
We w a n t
n~ a n d a
o u r n~,~f k t o
s a t i s f y (*3) a n d (*4) w h i c h follow:
Let us denote by
IF(t:) t h e s e t of t h e f u n c t i o n s f , D o m ( f ) a p r o p e r initial
s e g m e n t of C 1 - [0,n~)
, R a n g ( f ) ~ 10,1].
(*3) (a) n0 = 0 (b) h e + 1 > h e , R a n g ( r e ) = |O, lJ
(e) F o r any{~
, m
a f u n c t i o n f r o m C1 (][O,n;~) to IO,lJ
a n d t d e n o t i n g a t r u t h v a l u e , t h e following h o l d s :
i f for some i, t ~ m = t a n d f ufe UPe Uq~cPaxQ then f o r s o m e j , t~n = t a n d f o r e v e r y h c I F ( k §
139 .f U $ ~ u h
UP~ U q ~ ' e P , , x Q.
(*4) F o r e v e r y s < ]c a n d g0,g 1 f u n c t i o n s f r o m [0,n~) CI Cl t o 10,11, o n e of t h e following holds: ( F is f r o m t h e h y p o t h e s i s of t h e m a i n L e m m a ) Case(da):
for
no
i ( 1 ) < ca , tt~0) # tr
m E F(Cl)-[O,nk+l)
and
h e/F(/c+l),
ht , h ~ ht ~ IF(/c+l), gl Uf~
uhl
UPg
are
such
go U f k U h l U P ~
Uq~l)
are both inPax
there that
i ( 0 ) < ca, for
every
U qr
and
Q.
C a s e (alp): f o r e v e r y lc(*) > k a n d f u n c t i o n h f r o m [ n ~ + l , n ~ ( , ) ) CI Cl t o 10,1]
there
are
m E F(C1) N [n~(,),n~(*)+l),
tt~0) # till),
and
for
i ( 0 ) < o~ , i ( 1 ) < co s u c h every
that
h i c IF(lc (*)+1),
go U f ~ U h U f ~ ( , ) U h i UP~ U q~0) gl Uf~
U h U f/c(,) U h I U P ~
U qr
and are bothin
Pax
q.
We h a v e n o w t o c o n v i n c e t h e r e a d e r t h a t we c a n d e f i n e n ~ , f ~ t h a t s a t i s f y (*3) , (*4). A s s u m e t h a t we h a v e d e f i n e d n 0 . . . . . a n d we w a n t t o d e f i n e n ~ + l , f ~ . (i)
instances
J' :Ct N [0,•k)
of
n~_l,n ~ and f0 .....
fk-1
We h a v e f i n i t e l y m a n y t a s k s , of t h r e e t y p e s :
(*3),
c)
namely:
given
s < k,
m
of (*4) ( d a ) , n a m e l y : g i v e n s < / c , go,g1 c 10,1] [~
A c~, we
-* |0,1], t, we h a v e t o s a t i s f y " i f f o r s o m e i . . . . .
(ii) i n s t a n c e s
t r y to s a t i s f y " f o r no m 9 9 9 " ( b u t s o m e t i m e s we fail - t h e n we r e m e m b e r
the
f a i l u r e f o r (iii)). (iii) f a i l u r e s of ( * 4 ) ( d ~ ) in t h e p a s t : g i v e n a t a s k of t y p e (ii) w h i c h we d i d n o t fulfil a t a p r e v i o u s s t e p , we h a v e to s a t i s f y (da) f o r k (*) = o u r p r e s e n t k.
We will d e f i n e 3"~ b y t a k i n g a p p r o x i m a t i o n s intended
of it in I F ( k ) , e a c h o n e of t h e m
t o fulfil o n e t a s k a n d c o n t a i n i n g its p r e d e c e s s o r
o n c e we h a v e c o n s t r u c t e d
an approximation
- this ensures that
fulfilling a g i v e n t a s k , t h e f ~
t h a t we will e v e n t u a l l y o b t a i n will a l s o fulfil it. O u r initial a p p r o x i m a t i o n arbitrary
e l e m e n t of IF(It) a s s u m i n g t h e v a l u e s 0 a n d 1 .
is a n
140
So now we d e a l s e p a r a t e l y with e a c h type: (i) A s s u m i n g s u c h i e x i s t s (with f k i n t e r p r e t e d as o u r c u r r e n t a p p r o x i m a t i o n of it), l e t j be t h e l e a s t s u c h i.
The a d d i t i o n of h c a n c r e a t e a c o n -
t r a d i c t i o n o n l y inside C 1 (by t h e d e f i n i t i o n of
IF(k +1)), a n d s i n c e C 1 is a l m o s t
disjoint f r o m t h e d o m a i n of a n y c o n d i t i o n in P a (in t h i s c a s e (ps U q~)0), b y defining f k in t h e d a n g e r o u s finite p o r t i o n a c c o r d i n g to (p~ U q~n)0 we fulfil our task. (ii) L e t
IFc(k ) d e n o t e t h e s u b s e t of IF(k) c o n s i s t i n g of t h o s e f u n c t i o n s
c o n t a i n i n g o u r c u r r e n t a p p r o x i m a t i o n of f k . A s s u m e t h a t :
There
exists
f e IFr
such
that
for
every
m c
F(C1),
f t e I F c ( k ), f t D f , and for every i(O),i(1)
and
for
every
A s s u m i n g this, we t a k e s u c h f as o u r n e x t a p p r o x i m a t i o n of f ~ , which will s a t i s f y t h i s i n s t a n c e of
(Cla).
So if we fail, we k n o w t h a t :
f eIFr t h e r e a r e m eF(C1), m > S u p ( D o m ( f ) ) and f t e IFc(~:), f t ~ f a n d i(O),i(1) < a) such that t~To)~ t~l ) a n d f o r e v e r y f " elFc(k), f " ~ f t b o t h g o U f " U P ~ u q ~ 0 ) a n d g I u f " U P s U q~l) For
every
a r e in P a • Q(iii) A d a p t i n g o u r n o t a t i o n to t h a t of (ii), we d e n o t e o u r p r e s e n t k a s k (*) a n d a s s u m e t h a t t h e p r o p e r t y w r i t t e n a t t h e e n d of (ii) h e l d f o r s < k and
go,g1 e [0,1J [~
n c,
We a r e g i v e n h c 10,1] [n*§
r cl; let f o ( , ) b e o u r c u r r e n t a p p r o x i m a -
f~(,) a n d u s e t h e p r o p e r t y f o r f = f k U h U f o ( , ) to o b t a i n re,f t, i(O),i(1) a s d e s c r i b e d t h e r e . L e t t i n g f~(o) = f t - ( f ~ U h) a n d e x t e n d i n g it if n e c e s s a r y to e n s u r e t h a t m < n~(,)+l, we o b t a i n f~(,) fulfilling o u r t a s k . t i o n of
141
So t h e d e f i n i t i o n of n k , f k c a n be c a r r i e d o u t . on concentrates
o n A * = U a3k+l.
Our treatment
from here
The o t h e r t w o t h i r d s of C 1 c a n b e t r e a t e d
k <e~
similarly.
L e t f ~ = U ( f 3 ~ U f 3 k + 2 ) , A1 = f
*-1(11|).
N o t i c e t h a t b y (*3) b),
b o t h A 1 a n d its c o m p l e m e n t in D o m ( f ~ a r e infinite.
T h u s , f r o m ('21 it follows t h a t : (*5) F o r e v e r y A ~ A*, Ch~. O f ~ ]bp~[AUAt]"YC~ F(C1) = ~ F(A O A1)". H e n c e , f o r e v e r y A C A ~ , Ch~, U f ~ c a n b e e x t e n d e d
to a condition forcing
e q u a l i t y a b o v e a c e r t a i n n a n d in p a r t i c u l a r d e c i d i n g f o r e a c h m m n in F(C1) whether
it b e l o n g s t o Y (as n o o t h e r n a m e s
a r e i n v o l v e d in t h e e q u a l i t y ) .
L o o k i n g a t p ( A ~ a s a t o p o l o g i c a l s p a c e , we c o n c l u d e (as I P a x Q I = ~0):
(*6)
There
exist
Dora(p*) NA*CDom(h~
a
condition
p * E Pax
Q,
a
finite
h ",
CA*,andn'<~suchthat
. ~ = IA C A " : h ~ c C h ~ , a n d f o r e a c h m -> n ~ in F(C1)
[p" u ch~. u f" IFp[[AuA,]"~ c Y"orp" u Ch2. u f" II-~[AuA,] '''~ r ~'11 is of t h e s e c o n d c a t e g o r y e v e r y w h e r e " a b o v e " h * (i.e., its i n t e r s e c t i o n with any open set determined
b y a finite e x t e n s i o n
of h * is of t h e s e c o n d
category). We n o w define, f o r A c _~,
FI(A ) : | n e d F ( C1) : p ~ U ChA~ U f ~ II--.P~'[AU,Ad ' ' m C Y"] It follows f r o m (*5) t h a t
(*6A) F ( A U A1) =a, F I ( A ) f o r all A c _~. We will v e r i f y n e x t t h a t _3 is a B o r e l set.
T h e d e m a n d h ~ c Ch~. d e f i n e s
a n o p e n set, a n d t h e n we h a v e a c o n j u n c t i o n o v e r all m ~ n * in F(C1) of c o n -
142 d i t i o n s e a c h r e q u i r i n g t h a t p ~ (3 Ch~, U f ~ d e c i d e w h e t h e r m b e l o n g s t o Y. Denoting
_,4q = ~A c A ~
is
incompatible
= ~q~ : i < r
with
p ~ U Ch~, U f ~
is d e n s e in P a x
and
remembering
that s
Q a n d we c a n p r e s e r v e
its p r e d e n s i t y
in P ~ [ A UA1], we s e e t h a t A c A " s a t i s f i e s t h e r e q u i r e m e n t
for
m if[ it b e l o n g s t o [C~l_~qt~ : i < ~ , t~m is t r u e ]] U [ ( ~ l - ~ q ~ : i < ~ , t~m is f a l s e ~]. As e a c h _~q is o p e n , we h a v e s h o w n t h a t _~4 is Borel.
As _,4 is a B o r e l set, we k n o w b y a t h e o r e m
of B a i r e t h a t (by e x t e n d i n g h *)
we m a y a s s u m e t h a t |A c A" : A r _ 2 , h " c Ch~, ] is of t h e f i r s t c a t e g o r y . L e t k b e l a r g e e n o u g h so t h a t n3~ m ~z ~ a n d JZ < 3k - 1 w h e r e p ~ = p g . For A 9
let t~ denote the truth
m 9 [n3~: , n3~+s ) C~ F(C1),
v a l u e of m 9 F I ( A ). T h e n f o r e v e r y
P ~ U ChAA9 t ) f ~ is c o m p a t i b l e w i t h s o m e q~n b u t
is n o t c o m p a t i b l e w i t h a n y q~n s u c h t h a t t~m # t~n ( r e m e m b e r t h e d e f i n i t i o n s of 9F I ( A ) a n d _,4 a n d o b s e r v e t h a t m ~ n * , m 9 F ( C 1 ) ). Hence
by
(c)
of
(*3)
m 9 [n3~ , na~+3 ) N F(C1), t~ = t
iff
for
some
t~
by
is
determined,
for
A 9 _,4
Ch~, r [0,n3~+4 ) = Ch~. r [0,n3~+2);
i,
t~m = t
and
for
every
that
and is:
h 9 IF(3Ir +4)
Ch~, t [0,n3~+2 ) U / " ~ r [0,n3~+4) U h U P ~ U q m 9 p= x Q. Now we w a n t Ch~. r a3~+i
to show that
alone.
So
let
t~
is d e t e r m i n e d
us
assume
(for A,m
that
as a b o v e ) b y
A (~ , A (0 e _~4
and
Ch~- F a 3 ~ + l = Ch~. r a3~+1, a n d f o r s o m e m 9 [~%3k,~3~+3) C% F ( C 1 ) we h a v e t~
# t~,~. L e t f o r t = 0,1
t~0~ , t~l) a r e d e t e r m i n e d tinct,
it
follows
p" = ps , (da)
that
fails
gt(ChAC2 U f ~ r ( [ 0 , n 3 k _ l ) N Cl)- F r o m t h e w a y (see e n d of f o r m e r p a r a g r a p h )
at
( take
stage
w i t h t h o s e of A(~ from there
B(~
of
(*4),
for
the
above
h = f " r (a3k U a3~+2 U a 3 ~ + 3 ) U
w h i c h d o e s n o t d e p e n d o n t). want to construct
3k-1
a n d t h e i r b e i n g dis-
H e n c e (dS) h o l d s in t h i s c a s e .
(1) e _,4 s u c h t h a t t h e i r c h a r a c t e r i s t i c
go,g1
and
Ch~ r176t a3~+l, Using this, we functions agree
(1) r e s p e c t i v e l y u p t o n 3 ~ _ 1, a n d a g r e e with e a c h o t h e r
on, while f o r i n f i n i t e l y m a n y m 9 F(C1), ChAm2 a n d ChAs~,*) d e t e r -
143
m i n e t]~o),t~l) r e s p e c t i v e l y to be distinct.
=~ F(B(~ U A1) =a. F(B(O U A1) =ae FI(B(1)), so this will be a c o n -
FI(B(~
tradiction.
B(~
S i n c e B (~ = ~ B (1), also by (*6A)
Thus, it r e m a i n s to show how to c a r r y o u t t h e c o n s t r u c t i o n of
(1). We do it by a n m - s e q u e n c e of finite a p p r o x i m a t i o n s . We s t a r t
with
the
A(~
(1) ~ . ~ ,
both
sections
B (~
(1)
IA ~ A ~
(finitely
respective
and
are
sections
of A(~
compatible
identically)
so
(1) up
with h ' ,
that
they
to
r~ak_ 1.
so we first contain
As
extend h'.
As
r ..~ , h ~ ~ CA~. ~ is of t h e first c a t e g o r y , we h a v e a c o u n t a b l e f a m -
ily of n o w h e r e d e n s e s u b s e t s of p(A ~ which a r e to be a v o i d e d b y B(~ o r d e r t h a t B(~
(1), in
(1) E_~. So we a r r a n g e all t h e s e t a s k s ( e a c h t i m e d e a l i n g
with e i t h e r B (~ or B (1) a n d one n o w h e r e d e n s e s u b s e t ) in a s e q u e n c e , a n d b y t h e d e f i n i t i o n of " n o w h e r e d e n s e " we o b t a i n e a c h time a finite e x t e n s i o n of o u r f o r m e r a p p r o x i m a t i o n s w h i c h e n s u r e s t h e i m p l e m e n t a t i o n of o u r t a s k .
We will show t h a t o u r c o n s t r u c t i o n is all r i g h t by proving: if k (*) > 3 k - 1 a n d k ( * ) --- 2 (mod 3 ) t h e n t h e r e is m E F(C1) A [n~(,.),n~(*)+l) s u c h t h a t f o r t = 0,1 t h e r e a r e i ( t ) < co s u c h t h a t t~0) ~ t~r~l) a n d f o r e v e r y h E IF(It(*)§
CAA~2) t[O,n~(.)) u f
~
uh
UP*
uq~t)
cPax
Q. To p r o v e this,
u s e (d~) f o r s u c h /c(*) a n d h = (Ch~ ') U f ") t [n3~,n~(.)). We t h u s g e t a c o n tradiction.
So, b y t h e last f o u r p a r a g r a p h s a n d (*6A), as F is a n a u t o m o r p h i s m a n d A 1 is fixed, we h a v e p r o v e n . (*7) T h e r e a r e f u n c t i o n s F~ :]9(a3~+1 ) -* ]")(F(A *) N [n3~ ,n3~+3)) such
that
for
F(A) =ae F ( A u A 1 ) - F(A1) =ae FI(A) - F(A1) =~
every
AEA
U Fk(A N as~.+l). Ir
Now, (*7) would be g o o d e n o u g h (as we shall see l a t e r ) if it were n o t r e s t r i c t e d to .~. So we w a n t to a c h i e v e a similar r e s u l t w i t h o u t this r e s t r i c t i o n .
144
We define by i n d u c t i o n a s e q u e n c e of p a i r s
is
a
partition
A*-Dom(h')
of
into
< co) s u c h t h a t finite
subsets
and
ge:urZ --, [0,11, as follows: for r < co, let -~2r = --d2r+l be t h e r - t h e l e m e n t in a sequence
of
nowhere
dense
subsets
of
]D(A*)
showing
IA C A*:A r ...d , h" C Ch~.] is of t h e first c a t e g o r y . E a c h
that
is d e f i n e d
so t h a t :
(i) u g is disjoint f r o m Urn f o r all m < g. (ii) i f s e ( A * - D o m ( h * ) ) - (iii) i f A c A ~ , g~ c ChAA9
U um t h e n g e u ~ . m
To see t h a t this is possible, u s e t h e f a c t t h a t _.de is n o w h e r e d e n s e t a k i n g into account
one
v = ( U u,n)
after
the
other
all
the
elements
of
[0,11~
where
uDom(h*)-
m
Now, f o r t = 0 , 1 let At*= O U 2 r + t a n d g t ' = and
is a p a r t i t i o n of A*.
Og2r+t.
T h e n D o m ( g 0 =At*
F r o m (iii) it follows t h a t if
Ch~. t h e n A c -~.
F r o m (*7) we c a n t h u s derive a definition f o r F(A CI A~) working unif o r m l y f o r all A c A * we c a n
(using (*7) f o r (A V/A~) LJ ( g ; LJ f ")-1(|1]))- Similarly
d e r i v e a d e f i n i t i o n f o r F(A VIAl).
From these
two definitions,
r e m e m b e r i n g t h a t Dom (h *) is finite, we obtain: ( ' 8 ) T h e r e a r e f u n c t i o n s Gk:p(ask+l ) -* ]')(F(A") 0 [na~,?zs~+3)) s u c h that f o r everyA
C
A* F(A) :v~ D Gk(A (~ a3k§
With t h i s r e s u l t in h a n d , we h a v e . g o o d c o n t r o l of F r ]9(A *). We p r o v e first: (*9) F o r e v e r y l a r g e e n o u g h / c a n d disjoint b 1,b2 C a3~+1,
145
Otherwise b~ n by =~b,
there
is a n
infinite
S c~o,
and
CT~(b~) N C%(b~) #~b.
Then
let
b~,by c a a k + l f o r k e S, A1 = ub~
, A2=
ubY. /ceS Clearly A I N A2 = ~, AI.A2 C A ~ and F(AI) n F(A2) # ~ ~. This contradicts /~eS
F/finite c o m m u t i n g with N. S i m i l a r l y as F c o m m u t e s with U ( ' 1 0 ) F o r e v e r y l a r g e e n o u g h k a n d d i s j o i n t b 1,b2 c a3k+t,
C~(bl U 52) = C~(bl) U C,~(b2) a n d as F is o n t o ( a n d m o n o t o n i c ) ( * l l ) F o r e v e r y l a r g e e n o u g h k, f o r a n y s i n g l e t o n b c (Z3/r 1 C%(b ) is a s i n g l e t o n .
Define g , Dora (g) = A ' , G~(IIZ]) = g ( g ) f o r g 9 as~+1,k l a r g e e n o u g h .
So
c l e a r l y b y (*8) - (*I1): ('12) For every
A c A ", F(A) =he g(A) ~Ylg(i):i 9 AI a n d g is o n e - t o - o n e .
R e m e m b e r t h a t A ~ is o n e t h i r d of C 1. Doing t h i s s e p a r a t e l y f o r e a c h t h i r d a n d t h e n t h e s a m e f o r Ca, we o b t a i n f o r e v e r y B w h i c h is a l m o s t d i s j o i n t f r o m each
Aj(j < ix): ( t h a t we c a n g e t a o n e - t o - o n e g is p r o v e d like (*9))
('13) There is a one-to-one function g, Dora (g) =a, B, such that (VA C
B ) [ g ( A ) = ~ F(A)].
`9= ~A C o : F t f ~ A ) / =he is i n d u c e d b y s o m e f u n c t i o n g as a b o v e ]; c l e a r l y .9 is a n ideal. Let J be t h e ideal g e n e r a t e d b y IAj:j < a] a n d t h e finite
Let
s u b s e t s of co. L e t J1 = ~B c c0:(YA e ~ ) I B N A ] < ~0|. Clearly, ~1 is also a n ideal.
146
In o u r n e w t e r m i n o l o g y , ( ' 1 3 ) s t a t e s t h a t ~1 c ~ . I t f o l l o w s t h a t e v e r y i n f i n i t e A c co c o n t a i n s a n i n f i n i t e s u b s e t in ~ - s i n c e we can construct
A as before with A0 = ~-A
(w.l.o.g. , A 0 is i n f i n i t e ) , a n d t h e n
take any infinite B E ~l and discard its finite intersection
w i t h A 0.
Notice also: ( ' 1 4 ) E v e r y A w h i c h is n o t i n ~ c o n t a i n s a n i n f i n i t e s u b s e t in ~ l -
To see this, let < B n:n < a~> be an r We t r y t o d e f i n e a s e q u e n c e an c A -
U Bn. If we s u c c e e d
define a n , then
A c(
U Bin)U
of the generators of ~. such that am #an
for m <~
we o b t a i n a s u b s e t a s r e q u i r e d .
lain :m < n ] ,
hence
A E~,
and
If we c a n n o t
contrary
to our
r n ~:~z
assumption.
As e v e r y i n f i n i t e s u b s e t have constructed
of ~ c o n t a i n s
an infinite subset
A w i t h e a c h Aj c ~ . So we a s s u m e
fail (to satisfy the requirements course, to this construction
of t h e l e m m a ) .
i n .~, w e c o u l d
n o w t h a t we d o t h i s a n d
O u r f o r m e r r e s u l t s a p p l y , of
{for s o m e a ) .
L e t B E ~ i b e i n f i n i t e , B = C i U C2 a s i n (*1). i n d u c e F r / ' ) ( B ) (in t h e s e n s e of (*13)), a n d l e t I ~ c 11,2].
(*15) L e t g Then for everyp ( V m c Cs
E Pa x Q t h e r e a r e g , p ~ q c P a x
~_n :::::::>[c1 U I<m,O>J
(v,,~ e c~)(~ =-~ -----> [q u t < ~ , i > l Proof
of ('15):
Suppose
Ibe.xr
Q and n such that:
r r'])
IFP.,,q"g(m) c ~"])
it fails for g,p
, and
, w.l.o.g., s = i .
Let
igi~:iZ < ~] b e t h e s e t of all c o n d i t i o n s i n P ~ • Q a b o v e p . L e t n -* ( ( n ) 0 , ( ~ ) 1 ) b e a m a p p i n g of a o n t o ~ • ~.
147 We
will
define
functions
h n(~z < ca),
hn
increasing
and
finite,
Dom (h~) c Ci, Rang ( h ~ + l - h ~ ) = ~0,1], a n d if n r163C~ t h e n n E Dora (h~+~); we will also define rn~ ~ C l ( n < ca), as follows:
We l e t h 0 = (P)o r (Dom ((P)o) r~ Cl). A s s u m e t h a t we have defined h~, a n d we will see h o w we define ~
and
hn+ 1. If h n is n o t c o m p a t i b l e with q ( ~ ) 0 , ~ is c h o s e n a r b i t r a r i l y a n d hn+ 1 ~ h~ is c h o s e n in a c c o r d a n c e with o u r r e q u i r e m e n t s above. If h~ is c o m p a t i b l e with q(,~)o, we u s e o u r a s s u m p t i o n f o r q = h~ U q(n)0" Notice t h a t if m is l a r g e enough
then
m if Dom (q) f~ C 1
and
[m E C1 ~
g (m)
is
defined,
g ( r n ) ~_ F(C1) , g ( m ) ~- ( n ) l ]. So we c a n find m = rn n E C i s a t i s f y i n g all t h e s e ,
and, e.g., ~ , U q(~)o U ~ ( ' ~ , 0 > ! I/-~.~Q"g("~) if r". Let
qJ ~ h~ u q(~)o u ~ ( , ~ , , o > i
h~+l = (qJ)0 r (Dom ((qnt)0) (3 C1), Rang (hn+ 1 - h n )
and
let
.,g(.~) ~ r: h~+ 1 D h~+ I
Let satisfy
= ~0,1{ a n d n 9 Dom (hn+l) if n e C 1.
Now, Dora ( U hn) = Ci. U hn.
function
force
A
Let A c C1 be t h e s u b s e t with c h a r a c t e r i s t i c
and
C1-A
are
infinite,
so
by
(*2)
ChAc, I~e[[A] ' ' Y (~ F(C1) =a~ F(A)"" By o u r c h o i c e of h o, Ch~., = U h~ is comp a t i b l e with p ,
so s o m e e x t e n s i o n of p U ChoAI in P~[A]
forces for some
i n t e g e r e q u a l i t y a b o v e it. Let n be s u c h t h a t :
q(~)0 U ChcA, II-pflA] ''Y N
F(Cl)
- (n)l
= F(A) - (n)~"
a n d F(A ) - ( n ) l = g (A ) - ( n ) l .
We r e f e r now t o t h e definition of ran, hn+ 1. Since h n ~ Ch~, is c o m p a t i b l e with
q (~)o,
we
had
h~ U q(n)0 U ~ m n , 0 ) ~m~,l>;
qtnll-p, x Q " g ( m n ) E Y ' "
where
qt
was
above
{ (or it c o u l d be t h e o t h e r c a s e , with g ( m ~ ) if Y a n d
this is d e a l t with similarly). Now, m n if A since h~+ 1 s a y s so, h e n c e
148
g(mn) r g(A). As g(rn~) ~- (n)l, g(mn) C F ( A ) ; b a t g(mn) oF(C1), h e n c e q(n)0 U Ch~, I~-p[[A]"g(rnn) ~ }'". But q t f o r c e s t h e o p p o s i t e (we k n o w it in - P a x Q, b u t it d o e s n o t m a t t e r ) , a n d t h e y a r e c o m p a t i b l e (since q(n)0 C qnt a n d a c o n t r a d i c t i o n b e t w e e n Ch~, a n d
qJ is
i m p o s s i b l e by o u r c h o i c e of h t + l ) - a
c o n t r a d i c t i o n . So we h a v e f i n i s h e d t h e p r o o f of ( ' 1 5 ) .
We define now, f o r e v e r y
p c Pa x Q, a p a r t i a l f u n c t i o n f r o m ~ to ~ as fol-
lows:
gp(,..n) = k P U ~{m,1)
iff
{ I]-P.•
[p u I/xm,O> I II-,,:',,,,q ''k r ~"
and
c Y " a n d / c is t h e o n l y o n e s a t i s f y i n g this].
Our i n t e n t i o n in defining gp is to h a v e a f u n c t i o n (for t h e Lime b e i n g c o u n t a b l y m a n y f u n c t i o n s ) w h i c h i n d u c e s F r ]:)(B) u n i f o r m l y f o r all B c ill. ( ' 1 6 ) F o r e v e r y B c J1 a n d e v e r y p c P a x Q t h e r e is q, p -~ q E P a x s u c h t h a t / ? Cae Dora (gq) a n d (VA C
Q,
B)F(A) =ae gq(A).
P r o o f of (*16): IF H is finite t h e n t h e r e is n o t h i n g to p r o v e , so we a s s u m e t h a t H is infinite. Let g e x e m p l i f y H < .9 ( r e m e m b e r fil C ~).
Let u s call k a c a n d i d a t e Lo b e
gp(m), if it s a t i s f i e s t h e f i r s t two p a r t s of
the definition ( but maybe not the uniqueness requirement). /c is a c a n d i d a t e to b e then
We n o t i c e t h a t if
gp(m) a n d /ct is a c a n d i d a t e to b e g p ( m t ) , m # r o t ,
k # let ( c o n s i d e r p U I ~ r n , 0 ~ , ( m t ,
l)],
it f o r c e s b o t h
k ~ Y and
k t E Y). N o t i c e also t h a t in ( ' 1 5 ) we h a v e s h o w n t h a t f o r l a r g e e n o u g h m c C~ N
( a n d a p p r o p r i a t e q) g ( m ) is a c a n d i d a t e to be gq ( m ) .
Thus, g i v e n B a n d p , we a p p l y ( ' 1 5 ) f i r s t f o r s = 1 a n d p , a n d t h e n f o r = 2 a n d t h e c o n d i t i o n o b t a i n e d f o r ~ = 1, o b t a i n i n g q = ' p s u c h t h a t f o r large enough
m c.B
g(m) is a c a n d i d a t e to b e gq(m). Now, it suffices to gq(m) is d e f i n e d t h e r e is o n l y o n e
show t h a t B Ea~ D o m (gq), s i n c e w h e n e v e r
149
c a n d i d a t e , b u t g ( m ) is one, so f o r a l m o s t all m c B
gq (rn) = g ( m ) a n d we
k n o w t h a t g i n d u c e s F r ]9(B).
A s s u m e t h a t B ~a, Dora (gq).
T h e n w.l.o.g.
B is disjoint f r o m Dom (gq)
a n d we h a v e k m ( m E B) s u c h t h a t / c m a n d g ( r n ) a r e d i s t i n c t c a n d i d a t e s to be C/~!. Then C is infinite a n d disjoint f r o m g(B) (as n o
g a ( m ) . Let C = I/r
k is a c a n d i d a t e for two re's).
We let D = F - I ( C ) ; t h e n as g(B) =a, F(B),, D is
a l m o s t disjoint f r o m B.
Case I:/) ~ 81" In this case, we a p p l y ( ' 1 5 ) f o r D a n d q twice (as we did a b o v e f o r B a n d p ) to o b t a i n q t _~ q s u c h t h a t f o r l a r g e e n o u g h m t c D
g t ( m t ) is a c a n d i d a t e
to be gq~(mt) (where g t i n d u c e s F r/9(D)).
But f o r l a r g e e n o u g h m t c D
mt~B
gqr
and
gt(mt)
is
a
candidate
to
be
for some
mcB
(as
gt(D) =he F(D), F(D) = |lcm:m e B], a n d q < qt e Pa x Q) - a c o n t r a d i c t i o n (as in t h e b e g i n n i n g of t h e p r o o f of (*16) we o b s e r v e n o k is a c a n d i d a t e to be
gq~(n) f o r two n ' s ) . Case II: D ~ ill. T h e n w.l.o.g. D c Aj f o r s o m e j < a.
By s t r e n g t h e n i n g q we c a n a s s u m e
w.l.o.g, t h a t tAj -finite ~ q o r 0Aj - finite -< q. Assume t h a t t h e f o r m e r is t h e c a s e (the s e c o n d c a s e is d e a l t with similarly). h e n c e q I~p,•
Then q ]~-p, lxq"Aj Ca* X",
~Y". As we have s e e n in t h e p r o o f of ( ' 2 ) , b y
p r e s e r v i n g t h e p r e d e n s i t y of ( c o u n t a b l y m a n y ) a p p r o p r i a t e s u b s e t s a b o v e corresponding
conditions,
we
can
conclude
that
c1 ]l-p,xr
Y".
H e n c e t h e r e a r e n a n d qt k q in P a x Q s u c h t h a t qt ][-p, xq,,nt c Y'" for all
~zt ~ F(Aj), n t ~ n. B u t C Ca* _~'(Aj) , so t a k i n g n t =/c~n l a r g e e n o u g h we k n o w t h a t n t is a c a n d i d a t e to be gqt(m) - a c o n t r a d i c t i o n . So we h a v e finished t h e p r o o f of ( ' 1 6 ) .
150
('17) F o r p 1,P2 9 Pa x Q,
B = ~n:gp,(n),g~,,(n) a r e d e f i n e d a n d d i s t i n c t J 9 ~. P r o o f of (*17): If not, t h e n b y ( ' 1 4 ) t h e r e e x i s t s a n infinite tYl 9 ~1 s u c h t h a t gp,,gp, a r e d e f i n e d e v e r y w h e r e in B 1 b u t t h e y n e v e r a g r e e t h e r e . By ( ' 1 6 ) we c a n find f o r i = 1,2 q~ ~ p~ s u c h t h a t t71 ~ae D o m (gq,) a n d (VA ~ BOF(A ) =ae gq,(A).
It follows t h a t gql t t71 =ae gq~ r t71; o t h e r w i s e , we
h a v e a n infinite /72 c / 7 1 on which t h e f u n c t i o n s n e v e r a g r e e , a n d we c a n divide /?2 i n t o t h r e e p a r t s e a c h h a v i n g d i s j o i n t i m a g e s u n d e r t h e two f u n c t i o n s ( d e c i d e i n d u c t i v e l y for n 9
to w h i c h p a r t is belongs, a g o o d d e c i s i o n
a l w a y s e x i s t s as t h e f u n c t i o n s a r e o n e - t o - o n e ) ; o n e p a r t a t l e a s t is infinite, call it /?3, t h e n F(B3) is a l m o s t e q u a l t o b o t h gql(B3) a n d gqe(/?3) w h i c h a r e disjoint.
But, w h e n e v e r gp,,gq, a r e b o t h
ql ~ P l ) ,
a n d t h e s a m e h o l d s f o r p 2 , q 2 , so we o b t a i n t h a t gp~ t/71 =ae gp t/71,
defined they must
agree
(since
c o n t r a d i c t i n g o u r a s s u m p t i o n . So ( ' 1 7 ) h o l d s
By a r r a n g i n g t h e c o n d i t i o n s in P a x
Q in a n co- s e q u e n c e a n d u s i n g ( ' 1 7 )
we o b t a i n t h a t t h e r e is a p a r t i a l f u n c t i o n gO f r o m co to ~ s u c h t h a t f o r e v e r y
p epax
Q
~ z : g p ( n ) is d e f i n e d b u t g ~
is n o t d e f i n e d or is # g p ( n ) ] c ~ .
H e n c e , b y (*16), f o r e v e r y / ? 9 ~1, B ~ae D o m (gO) a n d
(Va c P)F(A) = ~ g~ We h a v e a l m o s t a c h i e v e d o u r goal of i n d u c i n g F t ]9(/t) u n i f o r m l y f o r all B e ~ 1 - t h e m i s s i n g p o i n t is t h a t we w a n t go to be o n e - t o - o n e . B u t t h i s m u s t b e t r u e a f t e r d i s c a r d i n g f r o m Dora (gO) a s e t in ~, b e c a u s e o t h e r w i s e we c a n construct
bn # b m f o r n # m , g~176
) and
B = ~bn:n < c0~ 9 fll (see t h e p r o o f of ('14)); t h e n f o r l a r g e e n o u g h n a n d a n a p p r o p r i a t e gq ( i n d u c i n g F tD(B)) g~ ) = gq(bn), so gq is n o t o n e - t o - o n e , a contradiction.
S i n c e d i s c a r d i n g a s e t in ~ d o e s n o t a f f e c t t h e o t h e r p r o p e r -
t i e s of gO we h a v e a o n e - t o - o n e gO i n d u c i n g F r ]9(/?) u n i f o r m l y f o r a l l / ? 9 ill-
As (Vj < a)Aj 9 ..9, a n d t h e y a r e a l m o s t disjoint, a < ~1, t h e r e is g 1 , o n e -
151
t o - o n e p a r t i a l f u n c t i o n f r o m r to r
so t h a t f o r e v e r y 17 c ~, F ( B ) =a* g t ( B )
a n d B ~a, D o m (gl). C l e a r l y D o m ( g l ) (j Dora (gO) is c o - f i n i t e (use ( ' 1 4 ) ) .
Let fl* be the i d e a l fl U f l , g e n e r a t e s .
Let D = i n : g l ( n )
#- g ~
and both
a r e d e f i n e d J. If D fs ~ " we c a n find (as we f o u n d B 3 C B 2 in t h e p r o o f of ('1.7)) D i c D such that D 1 ~*
and
gl(D1) ~ g~
= ~b.
Dt C Dr, F(Dt) == F(D1) N g~ T h e n D t c fit, o t h e r w i s e it h a s a n infinite s u b s e t D~ c J, so r(D"t)=azgl(D't) h e n c e F(D'~) c ~ gl(D1), b u t F(D~) Cae F(D~) co, g~ a c o n t r a d i c t i o n . Similarly, As F is onto, f o r s o m e
Dt - Dr c fl; so D1 c ~*, a c o n t r a d i c t i o n . So
D c ~ * , h e n c e b y trivial c h a n g e s in
gO,gt D =~b. Let g =gl ugO.
F(A) =~ g (A), so as fl* is d e n s e , F is a n a u t o m o r p h i s m ,
F o r e v e r y A E ~*
t h i s h o l d s f o r a n y A. As F(c0) =a~ g ( D o m (g)), Rang (g)
is co-finite.
Why c a n we a s s u m e g is o n e - t o - o n e ? gin s e t A, s i n c e t h e n
No i n t e g e r c a n h a v e a n infinite ori-
F(A) is infinite while g(A) is a s i n g l e t o n . 0 n l y finitely
m a n y i n t e g e r s c a n h a v e a n o n - s i n g l e t o n origin set, o t h e r w i s e we would h a v e two d i s j o i n t infinite s e t s with t h e s a m e infinite i m a g e u n d e r g.
So we c a n
t h r o w o u t t h e p r o b l e m a t i c finite p a r t of g.
T h u s F is trivial.
R e m a r k : If y o u w e r e n o t s a t i s f i e d with t h e
proof, because
of t h e
r e t r e a t s (i.e., "we c o u l d h a v e .... ") h e r e is a s h o r t e x p l a n a t i o n . S u p p o s e _~r~
a r e g i v e n in a u n i v e r s e V ( a n d M* is a n col-oracle, h e n c e
V ~=CH), a n d F is a f u n c t i o n f r o m ]9(o) to ]9(o) s u c h t h a t F i n d u c e s a n a u t o m o r p h i s m of ] 9 ( o ) / f i n i t e .
C h o o s e k big e n o u g h (e.g., (2~~ +) a n d let < " b e a
152
well-ordering of H(k). Now let M4(i < wl) be s u c h t h a t : Mi is a c o u n t a b l e elem e n t a r y submodel of
after
('13)).
M" c [vii
( Igj : j < i > ~ M~, ~
are
c U M~ a n d M~*c M~. Let ~ be d e f i n e d as in t h e proof We
now
define
by
induction
on
a , ~ < a < ~1,
< A~ : i < 2 a > s u c h that: (i) [A4 : i < 2hi is a family of a l m o s t disjoint s u b s e t of ~. (ii) If (hA C ~ ) [ [ A [ = n 0 ^ - ( a B C A ) ( I B ] = ~ 0 ^ B e . J ~ ]
then A =e-A 0
satisfies - (2B c A) ( ]B [ = n 0 ^ B c .9). (iii) If .9 is d e n s e (i.e., t h e h y p o t h e s i s of (ii) fails ) t h e n A4 E .9 f o r every i <2a. (iv) Let 7= = Min IT:
P[(a:
i <
of P [ < A ~ : i < e f t > I ,
~
D is still a p r e d e n s e s u b s e t of
].
(v) We c h o o s e { Y : a < ~ , >
B,g,A as in t h e proof (a page b e f o r e
N a
(*2))and this i n s t r u c t s us how to c h o o s e A2a,A2a+I.
V a - PROPERNESS AND NOT ADDING REA[~ w1 I t e r a t i o n of f o r c i n g n o t i o n s w h i c h d o e s n o t a d d reals.
N e x t t o n o t c o l l a p s i n g 111 n o t a d d i n g r e a l s s e e m s t h e m o s t n a t u r a l r e q u i r e m e n t on f o r c i n g n o t i o n .
There are many works deducing various assertions
f r o m CH a n d m a n y o t h e r s who did it f r o m d i a m o n d of 111- If we w a n t to show t h a t t h e u s e of d i a m o n d is n e c e s s a r y , we u s u a l l y h a v e to build a m o d e l of ZFC in which CH h o l d s b u t t h e a s s e r t i o n fails, b y i t e r a t i n g s u i t a b l e forcing. A c r u cial p a r t in s u c h a p r o o f is showing t h a t t h e f o r c i n g n o t i o n s do n o t a d d r e a l s e v e n w h e n we i t e r a t e t h e m . So we w a n t a r e a s o n a b l e c o n d i t i o n on Q (in VP~) w h i c h e n s u r e s t h a t f o r c i n g with
Pa
does not add reals when
~Pi,O:i < coa is
a
CS i t e r a t e d f o r c i n g s y s t e m . A n o t h e r r e p r e s e n t a t i o n of t h e p r o b l e m is "find a p a r a l l e l of MA c o n s i s t e n t with C.C.H.".
The s p e c i f i c q u e s t i o n w h i c h drew m y a t t e n t i o n to t h e a b o v e w a s w h e t h e r t h e r e m a y b e a n o n - f r e e W h i t e h e a d g r o u p of p o w e r 111 ( f r o m [Sh 74] we k n e w t h a t t h e r e is n o s u c h g r o u p if V = L o r e v e n if ~.r h o l d s f o r a n y s t a t i o n a r y S C co1, a n d t h a t t h e r e is s u c h a g r o u p if MA + 21% > 111holds)- This is e s s e n tially e q u i v a l e n t to: is t h e r e
a stationary
S c co1, a n d f o r e a c h 3 E S a n
u n b o u n d e d s u b s e t A S of o r d e r - t y p e ~0, s u c h t h a t A = ~ A s : ~ c S t formization property.
It is e a s y to see t h a t ~
h a s t h e uni-
implies (Ai:i c S t
does not
h a v e t h e u n i f o r m i z a t i o n p r o p e r t y ( a n d in II 4.3 , we p r o v e d , f r o m ZFC + MA + 2 t% > 111, t h a t A h a s t h e u n i f o r m i z a t i o n p r o p e r t y ) .
The s o l u t i o n was s u r p r i s i n g . By Devlin a n d S h e l a h [DS] ( s e e XIV w h e r e ), t h e r e is,a v a r i a n t ~ 1 of ~
which n o t o n l y follows f r o m CH b u t is e q u i v a l e n t 1
t o 211~< 2111. This v a r i a n t implies m a n y c o n s e q u e n c e s of t h e d i a m o n d (see [Sh 1987], a n d in t h e s y s t e m a t i c d e v e l o p m e n t in A v r a h a m a n d S h e l a h [ASh] ere).
154 In p a r t i c u l a r
S>
does not have the uniformization property when
S c Din, i.e. S c o n t a i n s a c l o s e d u n b o u n d e d s u b s e t of t01. This still l e a v e s o p e n t h e q u e s t i o n for S a s t a t i o n a r y c o s t a t i o n a r y s u b s e t of ml. Now f o r s u c h sets it was p r o v e d in [Sh 77] t h a t t h e u n i f o r m i z a t i o n p r o p e r t y m a y h o l d f o r a fixed stationary
costationary
subset
S of ~1, (i.e. f o r all
u n b o u n d e d of o r d e r t y p e ~). However
~,-S ( a n d e v e n
r
S>, A~ C6
r
~,_~) m a y
still hold:) N o t a t i o n : In this c h a p t e r ~,/~ will s t a n d f o r u n c o u n t a b l e c a r d i n a l s , if n o t
e x p l i c i t l y s t a t e d otherwise. 1.1 Definition: Let E b e a family of s u b s e t s of always
S~Q(/~) (we a s s u m e
S~0(/~ 1 E E, so /~ is r e c o n s t r u c t i b l e f r o m E b u t using specific E we
m a y f o r g e t to write
~t%(/~), p a r t i c u l a r l y using a s t a t i o n a r y S ~ ~1 m o r e
e x a c t l y IS] 1(1) We s a y t h a t E is n o n trivial if f o r e v e r y k l a r g e e n o u g h , t h e r e is a c o u n t a b l e N < < H ( k ) , ~ > , E c N a n d N n ~ E A for e v e r y A c E CI N. We s a y in s u c h c a s e s t h a t N is s u i t a b l e for E. (21 We say, f o r a n o n - t r i v i a l E , t h a t a f o r c i n g n o t i o n P is E c o m p l e t e if f o r e v e r y ~, large e n o u g h , a n d N ~ < Y ( A ) , e > c o u n t a b l e , s u i t a b l e f o r E, t o which P belongs, t h e p a i r (31 The p a i r < p n : n < ~> f o r (4)
(N,P)
(N,P 1
(N,P) is
c o m p l e t e (see below).
is c o m p l e t e
h a s a n u p p e r b o u n d in
P,
if e v e r y g e n e r i c
sequence
where:
a generic sequence for
(N,P)
if
Pn e P n N,
P ]=Pn "=Pn+l, a n d f o r e v e r y d e n s e o p e n s u b s e t .9 of P which belongs to N,
.9 n Ip,,:n < ~l ~ r 1.2 Claim: In t h e definition, in (1) f o r all k s u c h t h a t E,
SB (/~) c H(k) 0
we g e t t h e s a m e a n s w e r (so we m a y r e p l a c e t h e u n i v e r s a l q u a n t i f i e r on ~. b y
155
existencial). P r o o f : Easy, b y a n a r g u m e n t s i m i l a r to t h e p r o o f of II 2.2. E x p l a n a t i o n of (4) k is big e n o u g h to e n s u r e e v e r y t h i n g a b o u t P, f o r c ing e t c ... is e x p r e s s i b l e in H(~,), now as N is a s u b m o d e l , it is l e g i t i m a t e to a s k w h a t g o e s on w h e n y o u f o r c e w i t h P s t a r t i n g in N, of c o u r s e a g e n e r i c s e q u e n c e is n o t f a r f r o m b e i n g a g e n e r i c s u b s e t , so w h a t (4) s a y s is t h a t f o r a n y g e n e r i c e x t e n s i o n N[G], t h e r e is p 9 P who k n o w s e v e r y t h i n g a b o u t it. (G c P N N g e n e r i c o v e r N). 1.3 T h e o r e m :
(1) If P is E - c o m p l e t e (so E n o n trivial) t h e n [~-p"E is n o t trivial
(2) If Q = < P i , Q :i < a >
is a c o u n t a b l e
support
iteration
,
I~-p~ "Q. is E - c o m p l e t e " t h e n P= = Lim Q is E - c o m p l e t e .
R e m a r k : So in (2) it is e n o u g h to a s s u m e E is n o t trivial a n d
IbP,
" if E is n o t trivial t h e n Q is E - c o m p l e t e . " ~~. P r o o f : (1) L e t N ~ < H ( k ) , C >
complete.
C h o o s e
be s u i t a b l e f o r E, P c N
hence (N,P)is
a g e n e r i c s e q u e n c e f o r (N,P), Po =P a n d
c h o o s e p * -~ p,~ f o r all n < ~. S i n c e p ~ is ( N , P ) - g e n e r i c b y c o r o l l a r y III 2.13 p * [~-" N[ G] .~ < H ( k ) , c > ; E 9 N[ G ] a n d N[ G ] N tz = g N I~': h e n c e NN~ 9
N
A=
AEE
AEN
n
A
A~E A~N[G]
So p ~ f o r c e N[G] to e x e m p l i f y t h a t E is n o t trivial. (2) Let k be big e n o u g h ,
g ~
s u i t a b l e f o r E. And
< p n : n < co> be a generic sequence f o r (N,P=). (note Pn ~ N, D o m p n countable hence
c N n a,p~
Dompn
~ N).
is a m e m b e r
Define p * a P=: its domain
is N N a, and
for
of Q. which is an upper bound for Ipn(i):~ < ~]
156 if t h e r e is s u c h u p p e r b o u n d (in Q.). We now p r o v e b y i n d u c t i o n o n i 6 N (~ a t h a t p * r i ->pn t i f o r e v e r y n . T h e r e a r e no s p e c i a l p r o b l e m s .
1.4 Claim: (1) The m i n i m a l n o r m a l fine filter o n
.~' 0(/~) which i n c l u d e s
E is D = D(E): (Where fine m e a n s : f o r a n y o r d i n a l a < / z l A c / z : a e A ] b e l o n g s t o t h e filter .) A eD
if a n d o n l y if t h e r e is C e
s u c h that l a 9 C:(Vi e a)a
9
/)0(/z) a n d At e E
U [ S#0(/z)t f o r i
At] C A.
(2) E is n o n trivial if a n d o n l y if r tg D(E). (3) P is E - c o m p l e t e if a n d o n l y if P is D - c o m p l e t e . P r o o f : Easy.
1.5 ].emma: A s s u m e 3"" = , , . port i t e r a t i o n
I[-P, " I Q
=•1",
If Q =
E a family of
n o n - t r i v i a l , a n d e a c h O is E - c o m p l e t e . ~i
subsets
of
S , (/z) which is
T h e n Pa = Lim Q s a t i s f i e s t h e ~
-chain condition. P r o o f : Let [Pie Pa:i < I~a] be g i v e n we shall find two c o m p a t i b l e c o n d i tions
among
them.
Pick
k
regular
large
enough
for
every
i
Ni < < H ( k ) , e > c o u n t a b l e s u c h t h a t I Q,P~,i,E] r Nr a n d Nt s u i t a b l e f o r E. F a c t : We c a n find i < j < co2 a n d a n i s o m o r p h i s m h:N, -* Nj s u c h t h a t
h(p,) =pj a n d h t (Ni N Nj) =ida, r~Nt" P r o o f of t h e F a c t : D e n o t e S~ = l a < t~2:cfa = ~ l J , c l e a r l y it is a s t a t i o n a r y set; define f ( a )
= Min lfl:Na N ( u N i ) = Na N ( u N , ) { 9 Since I[Na[[ =t~0 ~
a n d a e S12 ==> c f a = I~1 f is a r e g r e s s i v e f u n c t i o n o n S~ , h e n c e b y F o d o r ' s l e m m a t h e r e e x i s t s S c S~ s t a t i o n a r y a n d fl < t~2 s u c h t h a t f [S] = lfl]. T h e number
of
countable
subsets
of
U Nr
is
157
II ulv II
(I 'I ".o)"~
= (2"~ "~ = 2 "~ =.,,
t h e r e f o r e ~ e m a y c h o o s e f ~ S of e a r d i n a l i t y ~a a n d a m o d e l N* s u c h t h a t ( W ~ T) iV, n (U,<~N,) = N ' . F o r e v e r y a ~ r define ~ The n u m b e r of i s o m o r p h i s m t y p e s is ~ 2 ~ = ~ IS] =~a such that i ~ j e S ~
= {N~x~,c>o
~'.
h e n c e we m a y c h o o s e S t c T,
N~? -~ 5/] . P i c k s u c h
i,j let h:N~ -~ N~ be t h e
i s o m o r p h i s m a s e a c h c a N ~ is a n i n d i v i d u a l c o n s t a n t , h is a n i s o m o r p h i s m o v e r N ~ ( i.e., it is t h e i d e n t i t y o n N ~ a n d
h~)
=pj . This is t h e i s o m o r -
p h i s m we p r o m i s e d in t h e F a c t so we h a v e p r o v e d t h e f a c t .
IP~ c Pa ('/N~:s < ~ s u c h thatpi =p~ ... and for every~)cNdense s u b s e t of Pathere exists n such that p~c~) a n d let IP~ ~ P a A N j :~ < ~ ] b e d e f i n e d b y C o n t i n u a t i o n of t h e p r o o f of 1.6: Now c h o o s e
p~ = h(pfi).
Define a c o n d i t i o n r a s follows: D o m r = ( a CI N,) U ( a f l Nj),
f o r $ c a CI N, - N j
r (~) will be a n a m e of a n u p p e r b o u n d of lp~(~):n < ~]
if t h e r e is s u c h a b o u n d , a n d o t h e r w i s e ~b, f o r ~ c a ~ Nj
r ( ~ ) will be a n a m e
of a n u p p e r b o u n d of lp~(~):n < ~] if t h e r e is s u c h a n e l e m e n t , a n d o t h e r w i s e ~b. It s u f f i c e s to p r o v e t h a t f o r e v e r y 7 < a
poty,
pOry_r
r7.
We shall
p r o v e b y i n d u c t i o n on 7 < a t h a t f o r e v e r y n <
p~r T, p~F T<<-r iT. F o r 7 limit, it follows f r o m t h e i n d u c t i o n h y p o t h e s i s . F o r 7 = ( + 1, n o t i c e t h a t D o m p ~ C N~ CI a, D o m p ~ C Nj N a, a n d Divide t o 4 c a s e s : 1. ~ r N~ a n d ~ r
trivial.
2. ~ E Ni - Nd, it suffices to p r o v e (*) r t ~
It-p,"P~(~) < r ( ~ ) ' : If
Ip~'(~):n < r I h a s a n u p p e r b o u n d in Qt t h e n r ( ~ ) is s u c h t h a t a n d of c o u r s e r ( ~ ) >--P~(t).
b u t r is (N~,Pt)-generic.
Clearly ( P i r
~: n < ~
is a g e n e r i c
(Ni,P$) , h e n c e b y t h e i n d u c t i o n h y p o t h e s i s r t $ is (Ni,Pt)- g e n eric. So r ~ ~ I ] - P , " N ~ [ G ] is E - s u i t a b l e a n d ( p ~ ( ~ ) ' n < ~ > is a g e n e r i c sequence for
sequence b o u n d in
f o r (N~[G ],Q )".
Q "so we finish.
~f
Hence
rt~i[-"ip~(~):n<~)
has an upper
158
3. ~ E Nj - N~, s y m m e t r i c p r o o f to 2. 4. $ E Ni CI Nj; r e m e m b e r r
t h a t w.l.o.g. Q~ ( s e t of e l e m e n t s ) is
r - is (N~,P~)-generic a n d ( N j , P ~ ) - g e n e r i c .
m e n t of ~1 N~ N r
S i n c e N~ CI o i is a n initial seg-
= N1 N ~i. Also r d e t e r m i n e s G~ CI N~ a n d G~ N Nj h e n c e
t h e r e e x i s t s a ~ c N~ s u c h t h a t (an < ~) p~ I~- "' p~n(~) = a ~ " T h e r e is s u c h
a~ENj
a s a b o v e b y t h e d e f i n i t i o n of h, h ( p ~ ) = p ?
a~ = a~,
h(a~)=a?
hence
t h e r e f o r e r t ~ ]~-" p ~ = p ~ " . Now c o n t i n u e a s in p r e v i o u s c a s e s .
1.6 T h e o r e m : S u p p o s e in V
s u b s e t s of
SH0(/z),
~'
CH holds, a n d E is a n o n - t r i v i a l f a m i l y of
= X-
T h e n V h a s a g e n e r i c e x t e n s i o n V 1 b y p r o p e r f o r c i n g in which:
*)
a) CH holds, E is n o t trivial, 2 ~' = X, a n d b) If P is E - c o m p l e t e p r o p e r f o r c i n g n o t i o n , I P I = ~l a n d ~)~ ~ P is d e n s e f o r i < i 0 < c f x
then
t h e r e is a d i r e c t e d G ~ P ,
G CI~)~ # ~ b f o r i
R e m a r k : (1) We c a n r e p l a c e i < cf X b y i < X1 w h e r e X1 < X is r e g u l a r . (2) P r o p e r n e s s is n o t e s s e n t i a l in t h e p r o o f of t h e t h e o r e m , i t ' s u s e will a p p e a r in 1.7. (3) Also t h e r e a d e r s h o u l d b e a w a r e of t h e f a c t t h a t Dao(/z )- c o m p l e t e n e s s is
more t h a n p r o p e r n e s s , o t h e r w i s e 1.6 would say: MA~I ^ CH w h i c h is of c o u r s e impossible. Proof:
We
use
countable
that a < n + a n d
support
iterated
forcing
systems
I[-e, " IQ.] =sl, Q. is p r o p e r a n d is E -
c o m p l e t e ": F o r a n y s u c h s y s t e m Q, Lira Q is a n E - c o m p l e t e f o r c i n g n o t i o n w h i c h s a t i s f i e s t h e B2-chain c o n d i t i o n .
For a n y s u c h ~l we c a n find ~2 l e n g t h e n i n g it s u c h
t h a t (let Ps = Lim Q~): A)
V~
has
subsets
of ~1 n o t
in
Vel (if Q~ =
let
159
a 2 = a 1 + 1,Q=t = ~f : f 9 ~
o r d e r e d by inclusion).
or, if we like. B) If R 9 V Pt is a f o r c i n g n o t i o n e i t h e r in V P~ t h e r e is a d i r e c t e d s u b s e t g e n e r i c o v e r V P', o r V P' ~ " I R I
>t~l " o r R is n o t p r o p e r , or f o r no 9 2
E - c o m p l e t e in V Pz''.
( l e n g t h e n i n g 91) is R
Using A a n d B a n d a n y list of tasks, t h e r e is no p r o b l e m to find Q as above, s u c h t h a t f o r c i n g with Lim Q gives a g e n e r i c e x t e n s i o n as r e q u i r e d .
1.7 C o n c l u s i o n : In t h e m o d e l f r o m 1.6, if E = t ~ 1 - S ] , S C o 1, s t a t i o n a r y c o s t a t i o n a r y t h e following holds: a) f o r a n y As c 5 u n b o u n d e d of o r d e r t y p e o, f o r 5 e S, ( ( A ~ : 8 9 S > ,t%) has the uniformization property. b) S is still s t a t i o n a r y (by p r o p e r n e s s ) . P r o o f : Let < A $ : 6 c S > from
some
a <wi
to
~,
be as above, Is:A~-, ~, such
that
for
every
Q = |f:f a f u n c t i o n
&---a, 6 9
(6
limit)
3" tA~=a, f s t A ~ ] We h a v e to c h e c k . F a c t A: Q is E - c o m p l e t e . F a c t B : I f p c Q, D o m p -< a < r
t h e n t h e r e is q , p - q c Q, Dora q = a.
F a c t C: If p c Q, A c ~1 - D o m p is finite, f a f u n c t i o n f r o m A to ~ t h e n there isq,p
<-q E Q , f
cq.
F a c t D: Q is p r o p e r .
Proof of F a c t A: Trivial.
160 P r o o f of F a c t B: Let ~6~:g < j -< ~1 be a list of all limit o r d i n a l s 6 9 S, Dora p < 6 -< a. As A~, h a s o r d e r t y p e ~, a n d Sup (A~) = 6~. A$~ (~ A ~ is finite for m <s
h e n c e we c a n define f o r s < ~ ,
fie < 6 e
such that fig->Domp.
fig > Max ( A ~ C~ A e , ) f o r m < g. Now define q , a f u n c t i o n f r o m a to ~: p (i)
q(~) =
i 9 Domp
j'~(i)
~ 9 (A~-~e)
0
otherudse (but
(i < a)
Now q is well d e f i n e d as t h e fl~ w e r e d e f i n e d s u c h t h a t t h e A $ , - fie a r e p a i r w i s e d i s j o i n t a n d disjoint to D o m p .
It is trivial to c h e c k q 9 Q.
P r o o f of F a c t C: Let A = lae:~Z < m {, a n d we define b y i n d u c t i o n p g ( s <_ 2rn),
P0 = P , D o m p 2 e + l = at~,
Domp2e+l = ag+l,
P2e+l(ae) = f ( a e ) , P ~ <--Pe+l. Now t h e d e f i n i t i o n o f p 2 e + I is b y F a c t B, a n d t h e definition
of Pae+2
6 - Domp2e+ 1 ~
is
direct
[it
belongs
to
Q as
for
every
limit
6,
6 ~ D o m p 2 e ] . Now q = P2m is as r e q u i r e d .
P r o o f of F a c t D: Let h be big e n o u g h , ]~ > (2x) +. < a well o r d e r i n g of H ( ~ ) , f o r w h i c h h is t h e first e l e m e n t .
It suffices to p r o v e t h a t f o r a n y g i v e n
c o u n t a b l e N < < H ( i ~ ) , e , < > , Q 9 N, p e g (~ Q t h e r e is q 9 G, ( N , Q ) - g e n e r i c , p -- q. So let 6 = N C% ~I a n d choose a n < 6, such that a n < ~xn+ 1 Let Ibg :JZ < ~
hull
of
be a list of all m e m b e r s
L.J an.
N~ e N
in
<(H(/~), 9
b e i n g t h e first, h e n c e < H ( A ) , e , < > b e l o n g s to N). It is also c l e a r
N o w define by induction o n ~%,IDn such that:
a ) P o = P , P n <--Pn+l
b) p,~ c N,~ n Q thenpn,
f~agreeonA~
is
definable
in
b e l o n g s to An h e n c e to
some N k .
c) i f 6 6 S ,
A
model
Nt=Ntg(h)
t l N k = N t, so e v e r y p r e d e n s e ~ ) C Q , D 9
(as
the
clearly
=
of N /~ H(A) a n d N~ be the S k o l e m
iQ,pg{ t ] l b g : g < k ! [J li:i < an] So
, 6
c~Nn - D o m p 0
161 d) if s < n , a t a n o p e n d e n s e s u b s e t o f Q t h e n l a n ~ a t e) a n < D o m p n + l F o r n = 0, p 0 = P s a t i s f i e s a l l t h e r e q u i r e m e n t s .
If Pn is d e f i n e d a n d s a t i s f i e s t h e r e q u i r e m e n t s , f i n i t e a s Nn+ 1 c N, S u p (Nn+ 1 FI o l ) c N h e n c e A~ h a s o r d e r
type ~ with Sup(Aa) =~.
f i r s t n o t e t h a t AS f~ Nn+ 1 is S u p (Nn+ 1 f~ r
By F a c t C t h e r e
< ~, w h e r e a s
ispn t->j0 n ,p~c
Q,
p t ~ f ~ r (Nn+ 1 f~ A~ - D o m P 0 ) a n d b y F a c t B. w.l.o.g, a n C D o m p n ? . As Nn+ 1 ~ ( H ( A ) , c , < ) , Pn,Q c Nn+l, w e satisfies all the requirement
can
find
such
p n? c Nn+ 1.
o n Pn+l e x c e p t m a y b e d) f o r s = n .
Now
p~*
So if a n is a n
o p e n d e n s e s u b s e t of Q, we c h o o s e Pn+l c a~, Pn+l ->PIn,Pn+l ~ Nn+~, a n d if a n is n o t a n o p e n d e n s e s u b s e t of Q, w e c h o o s e Pn+~ = P n?-
S o we h a v e c a r r i e d t h e d e f i n i t i o n b y i n d u c t i o n of t h e P n ' S - Now q = is a m e m b e r
of Q: b y a) , q is a f u n c t i o n
Domp n C~,
and
by
e)
a n CDompn
from
hence
an
ordinal
Domq
belongs to every
a t w h i c h is a n
t o ~; b y b)
=~;
q rA~,=~e f~, as q tA,II~P~ c Q for some n and q rA~=a,
U Pn
for
f~by
~l<~,
c). A l s o q
o p e n d e n s e s u b s e t of Q ( b y d ) a s p e + l
-- q ) .
B u t e v e r y o p e n d e n s e s u b s e t of Q w h i c h b e l o n g s t o N, b e l o n g s t o H ( h ) ( a s ~, is b i g e n o u g h ) , h e n c e t o N? h e n c e is a t f o r s o m e g, s o q is ( N , Q ) - g e n e r i c .
We c a n e a s i l y g e t s i m i l a r l y ( m o r e e x a c t l y , c o m b i n i n g
1.8 C o n c l u s i o n : I n t h e m o d e l f r o m 1.6, if E = Ir costationary
, the following holds:
1.7 a n d [ S h 74]):
- S ] , S ~ wl s t a t i o n a r y
If G is a n a b e l i a n g r o u p ,
G =
U Gi , G~
increasing
countable,
f r e e , G i / G 1 f r e e w h e n j r S, j < i t h e n
G is a W h i t e -
head group.
A l s o n o t e t h a t if E is a n o r m a l f i l t e r o n co1, b y E - c o m p l e t e "shoot"
through
all A ~ E c l o s e d u n b o u n d e d
subsets
filters we can
of o 1. We c a n i n 1.7
162
(hence
1.8) r e p l a c e ~ 1 -- S~, b y a n o r m a l n o n t r i v i a l i d e a l o n c01; w h i c h we c a n
assume
is d e n s e in t h e s e n s e t h a t e v e r y s t a t i o n a r y
S C co1 c o n t a i n s
a subset
in the ideal.
w Generalizations
We s h a l l r e p e a t
of properness.
m o s t t h i s s e c t i o n in t h e n e x t o n e w i t h m o r e d e t a i l s a n d
less generality. 2.1 D e f i n i t i o n : F o r a n u n c o u n t a b l e
c a r d i n a l k, c o u n t a b l e
ordinal a and
s162 (I) Let SQSa~ (~) is the set of sequence ~Ni:i ~ a t such that: a) N i a countable submodel of ~ H(k),c~ b) ioN,
and < N j : j ~ i ~
c N~+,+~ or at least ~ Nj:j ~= i>
is definable in N~+j+g. c) If ~ first order, = c N~, ~H(X),c>
~ 3z~(x,a) then for
some b ~ N~+~,
is (a,s
if for ~ big enough,
= ~N~:i---a> E SQS~(k), such that P c N O , and for every p c N 0. p c P, there is r E P, v ~ p which is (N,P,E)-generic (or (P,~)-generic for ~/, or ~generic for N), which means: f o r e v e r y i , r [~-p " N~.[fp] ("l V c N~+2~ "' (3) If P c N ~ H ( h ) , G C P g e n e r i c t h e n N [ G ] = N U [ T [ G ] : T E N a P-name
]. We d e f i n e N [ G ] s i m i l a r l y
R e m a r k : N o t e t h a t f o r lg = 0, N i ~ ( H ( ~ ) , E ) .
163 r
2.2
Definition:
(1)
_ ~ ( , ) -- I< a~:i ___a > :
a~.6
.~ ( , )
o
and
the
o
s e q u e n c e is i n c r e a s i n g c o n t i n u o u s I(2) We call E c
( a , g ) - n o n trivial, if f o r e v e r y
UP(S~o(P')) ~,'
enough A, SQSs (A,E) < Ni:i ~- a> e SQS~ (A) such
large
# r
SQS~a(h,E) E 9 NO a n d
is
where
that:
the
if
set
a
of
= 16 + 7,
T
In s u c h a c a s e we call N s u i t a b l e for E. (3) So ( E , a , g ) - p r o p e r is defined as in 2.1(2) for N 9
SQS~a(A,E).
(4) In Definition 2.1 , 2.2 we m a y s u p p r e s s g w h e n it is zero. (5) A f o r c i n g n o t i o n which is ( E , 0 , 0 ) - p r o p e r will be called Eproper. (6)
A forcing
notion
will
be
called
(~,l)-proper
if
it
is
(~ ~t%(/~)I,~,l)-proper. 2.3 T h e o r e m : C o u n t a b l e s u p p o r t i t e r a t i o n p r e s e r v e s
(E,a,g)-properness
p r o v i d e d t h a t g = 0 or a is a limit ordinal. P r o o f : F o r g = 0, e x a c t l y as t h e p r o o f of T h e o r e m IIl 3.2, by i n d u c t i o n on a. F o r g > 0 use L e m m a 3.10 ( f r o m t h e n e x t s e c t i o n ) a n d Claim 2.4 below. R e m a r k : T h e r e a r e e x a m p l e s t h a t t h e n o t i o n s a r e distinct, p r o p e r is
([ SI4 (/~)I,0,0) -proper. Note that by Lemma 3.10 it follows that (E,a,g)0
properness is equivalent to (E,a,Ic)-properness for k ,g > 0.
SQS~a(A),P e N 0, " N [ G ] e SQS~a(A)"'
2.4 Claim: 1) If N e eric, p e G t h e n VP ~ 2) If ~ / e
SQS~ (A),
G c P is g e n e r i c p is (N,P,IZ) - g e n -
Q = < P ~ , Q :g < n >
e N o is i t e r a t e d forcing,
164
r c, a P c - n a m e of a m e m b e r of ~QC' is (N[Gc]:QC,g) - g e n e r i c (Gc ~ Pr t h e g e n e r i c set), t h e n r = < r 0 , r 1 . . . . .
r n _ l > is ( N , P n , 2 n g ) - g e n e r i c .
w a - p r o p e r n e s s a n d (E,a)- p r o p e r n e s s r e v i s i t e d
In w we g a v e s o m e s o l u t i o n to "which f o r c i n g d o e s n o t a d d r e a l s " . What o c c u r s is t h a t we m a y h a v e a s m a l l s t a t i o n a r y s u b s e t of cop on which e.g. unif o r r n i z a t i o n p r o p e r t i e s hold. But we w a n t e.g. to b e a b l e to p u t a p r o o f on t h e c o n s i s t e n c y of CH + S H , w h i c h is i m p o s s i b l e b y w
m e t h o d , b e c a u s e it is p o s -
s i b l e that the model V, from Theorem 1 6 satisfies also 0
o,, and even 0 ; , - s
( s e e w or [Sh 80a]).
H e r e we m a k e a n i n v e s t m e n t f o r t h i s g o a l b y d e v e l o p i n g a - p r o p e r n e s s (and (E,a)-properness)
which is a g e n e r a l i z a t i o n of p r o p e r n e s s , w h e n t h e
g e n e r i c i t y is o b t a i n e d f o r s o m e t o w e r of m o d e l s s i m u l t a n e o u s l y .
In a l m o s t all
c a s e s t h e p r o o f t h a t p r o p e r n e s s h o l d s gives a - p r o p e r n e s s . The p o i n t is t h a t f o r m a n y p r o p e r t i e s X, " X + a - p r o p e r n e s s
To a l a r g e d e g r e e we r e d o h e r e w
" i s p r e s e r v e d b y CS i t e r a t i o n .
with m o r e e x p l a n a t i o n , b u t f o r ~ = 0
only, f o r n o t a t i o n a l s i m p l i c i t y only. 3.1 Definition: F o r a < ~1 t h e f o r c i n g n o t i o n P is said to be a-proper if f o r e v e r y s u f f i c i e n t l y l a r g e ~ a n d f o r e v e r y s e q u e n c e
165
conditions
and
s i n c e P is a - p r o p e r
there
is a p <- q e P w h i c h is (N~,P)-
g e n e r i c f o r e v e r y i <-- a). T h e r e f o r e , in p a r t i c u l a r , e v e r y a - p r o p e r P is p r o p e r . (3) If P is a - p r o p e r
I i _- a + : + a > , p
p ~ q e P
it is also a +
he as requ:red.
w h i c h is ( N ~ , P ) - g e n e r i c
l+a
Si.oe
proper.
To s e e t h i s let
P is a - p r o p e r
for every i---a.
t h e r e is a q,
S i n c e Na+ 1 ~ H(A) a n d
p,P,
p is (Ni,P)-generic for i < a (and P 6 No) then p is (N,P)-generic where N=uN~ ~
As a c o n s e q u e n c e ,
if P is p r o p e r it is a - p r o p e r
f o r all n < o a n d
if P is ~ - p r o p e r it is a - p r o p e r f o r all o ~ a < r 2. a
all i --- a a n d < ~ ' i -- a> is c o n t i n u o u s l y i n c r e a s i n g ]. Let
F
be
a
function
into
.~
(A)
and
let
1
G(F) = l < a ~ : i - - < a >
e
_~
(A)'(Vi
(V
finite
b Ca~+l)
1
F(< a~ I j < i>,b l c a~+ll ^ (V finite b Cao) F(b ) c aot, L e t F~,n <~, b e functions
into
._~I(A)
and
let. F
be
given
by
F(x)=
U F~(x),
then
a ( F ) c r3 a(F,~), h e n c e t h e s e t of all G ( F ) ' s g e n e r a t e s a n ~ t - e o m p l e t e filter on
166
3.4 Theorem= The f o r c i n g n o t i o n P is a - p r o p e r if a n d only if it p r e s e r v e s t h e property
of b e i n g a s t a t i o n a r y
s e t (i.e., a s e t of p o s i t i v e m e a s u r e
) with
a
r e s p e c t t o t h e filter
2~)
The p r o o f of this t h e o r e m is like t h e p r o o f of t h e c o r r e s p o n d i n g f a c t f o r proper forcing. 3.5 T h e o r e m : F o r e a c h a < o 1, a - p r o p e r n e s s is p r e s e r v e d b y c o u n t a b l e support iteration. Again t h e p r o o f is s i m i l a r to t h e o n e on p r o p e r n e s s . Now we a d d E as p a r a m e t e r , w h e r e E is s i m i l a r to w h a t we do in w a
3.6 Definition: A f a m i l y E of s u b s e t s of
U
SI%(/~ ) is a - n o n - t r i v i a l if:
enough, t h e r e is a s e q u e n c e N = ~ N~:i .c a t ' of c o u n t a b l e e l e m e n t a r y s u b m o d e l s of ~ H(X),c>, ~ Nj:j ~--i> c N~§ 1, E c N O s u c h t h a t : if F o r e v e r y k big
If C ] 9 ( S ; 0 ( / ~ ) ) ' If c E N N~, t h e n < N~+j N / 2 : j ---a - i >
9 If: a n d K/is c a l l e d
a
suitable for E and for
(E,a). Let ,,F~a(E) = IS c
a
~o(/~):E U | So(/*) - SI is
a - t r i v i a l 1. Remark:
We
can
if
we
want
use
a
filter
DaA(E )
(for
3.7 Definition: A f o r c i n g n o t i o n P is ( E , a ) - p r o p e r ( E as a b o v e a - n o n trivial) i f f o r e v e r y /V which is s u i t a b l e f o r ( E , a ) a n d p c N 0, p c N 0 ( ~ P t h e r e is q >- p (in P ) q
(Nr
f o r e v e r y i --< a.
1.7 3.8 T h e o r e r ~ S u p p o s e E is a - n o n - t r i v i a l , E C U J~ ~'~ (/~))7<~1
(1) If P is ( E , a ) - p r o p e r , t h e n [~-p " E is a - n o n - t r i v i a l "
167
(2) I]-p~ " Qr is
If
Q
:
is
a
countable
(E,a)-proper ": PD = ]Am Q, then Pp is
support
iteration,
(E,a)-proper.
R e m a r k : So in (2) (by (1)) it s u f f i c e s to a s s u m e
]l-p~ " if E is a - n o n -
trivial (in VP0 t h e n t,*?+ Q. is ( E , a ) - p r o p e r "
P r o o f : No new p o i n t . 3.9 Theorem= If P is ( E , a ) - p r o p e r
and
@ is n o t ( E , a ) - p r o p e r
then
IJ-P " Q is n o t ( E , a ) - p r o p e r . " Proof: Easy. 3.10 L e m m a : The following p r o p e r t i e s of a f o r c i n g n o t i o n P, p C P , c o u n t a b l e limit o r d i n a l a, a n d A r e g u l a r big e n o u g h a n d < a well o r d e r i n g of H(A) a r e e q u i v a l e n t :
(~ is a n a - s e q u e n c e
of o r d i n a l s t r i c t l y i n c r e a s i n g
,
9'(i) < i + ~, ~ a n a - s e q u e n c e of n a t u r a l n u m b e r s , k (i) -~ 2): (1
)
There
is
a
function
F,
Rang F ~
S
Dora F =
S<~ ( ~ P ) )
s u c h t h a t f o r e v e r y i n c r e a s i n g s e q u e n c e of c o u n t a b l e
1
subsets q ~p,
of p ( P ) ,
Ar
< a)
satisfying:
such that for every , i
.9 (~ F(< Ai:j <-{> is
F(
and JcAr
there
is q c P,
a m a x i m a l a n t i c h a i n of P ,
p r e d e n s e a b o v e q.
(2) F o r e v e r y N~ C < H(A),E,<> f o r i < a s a t i s f y i n g a),b),c) listed below a n d p
c N O t h e r e is q c P, q ->p s u c h t h a t
(*) f o r e v e r y i < a a n d P - n a m e ~ of a n o r d i n a l
where N~ i n c r e a s i n g . increasing.
is c o u n t a b l e ,
i c N~,
Nr
168 b)
For
~
every
(first-order)
(]x)~(x,~) implies
formula
~a(x,e.),~ 9 Ni,
i
< H ( A ) , e , < > ~ (~x 6 N,+,) ~a(x,~)
e) In b) we c a n allow a c N~ kJ ~N~]. (3) ~ The same as 2) o m i t t i n g c), replacing N~ by N~(i) in (*) a n d a s s u m i n g ~ r ~ e N~ for every ~ < a. (4)~ The same as in 2), r e p l a c i n g Ni+ 1 by N~+~(r in (*), o m i t t i n g c) a n d a s s u m i n g ~ r/~ 9 Ng for ~ < a , / c ( i ) _> 2. (5) Like 4) for/c c o n s t a n t l y 2 . R e m a r k : (1) Note t h a t (2) is j u s t (a,1) -proper. (2) " k big e n o u g h " j u s t m e a n s D(P) 9 H(A); we c a n r e p l a c e /9(p) by t h e family of m a x i m a l a n t i c h a i n s of P. (3) As (1) does n o t d e p e n d on A, we get t h e equivalence of t h e o t h e r s for all suitable k. Proof: (1) ~
(2): We c a n a s s u m e t h a t t h e F exemplifying 1) is definable
in < H ( A ) , e , < > (By a f o r m u l a with the p a r a m e t e r s first
F
satisfying
(1).
f = < f 0,f 1,...> n <~ 9 H(A)
Moreover
we
similarly
p,P
can
only); j u s t t a k e t h e assume
definable But
for
there
such
eve
is
that
together
Set < a + ~0).
Now
also We
Nc=Nc(
the
--i>)
Skolem
for
r162
hull
(in
F(
< H(A),e,<>)
of:
[ < g 7 : 7-< fl>:fl < {'] U IAT0")l:J -ci, 7(J) < 2{'1 for s u c c e s s o r a, a n d u N ~ for
169
limit
Nr
F(
Now
r
<---/>)depends on < A j : 7 ( j ) < 2 ~ ' >
only. If
and
note
that
sequence N~(
as required in 3) ~. it is easy to check that it is as required in (i). (1) -------> (5): The s a m e p r o o f as (1) ------> (2). (5) ==> (4) E "Trivial. (4) E
~(o) =/r
~
(3) ~: f o r some y (i.e. ~ d e p e n d s on E):
~,(~) = (u~,(J) + k ( u ~ ( ; ) ) ~'
+ a
(2) ==~ (4) ~: f o r /c c o n s t a n t l y 3. Let Et~ 9
for 6
Given ~ define
be as in (4) t~, i.e.
a n d a) , b) a r e satisfies, a n d p 9
o . Let us d e f i n e
N,t : N2,, Now
(2))
<--i> =
d <-i> 9
cN2,+ihenceby
I.
So
2) a). a n d trivially it satisfies (2) b). as in f a c t (2) c) (as
s u c h t h a t f o r e v e r y P - n a m e 18 of a n o r d i n a l , ~ 9 N,t, q [[-p "' fl 9 N,t+l ': h e n c e q I~P " if i < r ~ 9 N2~. ~ an ordinal then ~ 9 N2~+I[G].
Hence q IbP " if ~ < ~'E 9 g~,~ an ordinal t h e n ~ E
w
Ni+3[G].
P r e s e r v a t i o n of ~- p r o p e r n e s s + t h e co~ - b o u n d i n g p r o p e r t y 4.1 Definition: A f o r c i n g n o t i o n P h a s t h e ~0~-bounding p r o p e r t y if f o r
a n y f e (o~) rIG] (G c P g e n e r i c ) t h e r e is g 9 (r162 s u c h t h a t ~ -< g (i.e.
( v n < ~ ) f (~) ~ g ( M ) . 4.2 Discussion: Clearly t h e ~J~-bounding p r o p e r t y c a n b e c o n s i d e r e d as
170
an a p p r o x i m a t i o n to t h e p r o p e r t y " n o t a d d i n g reals". Also this p r o p e r t y , a n d similar p r o p e r t i e s play c r u c i a l p a r t s in m a n y i n d e p e n d e n c e p r o o f s . T h a t is, m a n y t i m e s we w a n t o n t h e o n e h a n d to a d d m a n y reals, b u t o n t h e o t h e r h a n d t o p r e s e r v e s o m e t h i n g , e.g. to p r e s e r v e : t h e s e t of old (or c o n s t r u c t i b l e ) r e a l s is of t h e s e c o n d c a t e g o r y or d o e s n o t h a v e m e a s u r e zero, o r e v e r y n e w r e a l b e l o n g s to an old Borel s e t of s p e c i a l kinds, etc. In t h e n e x t c h a p t e r we shall deal with v a r i o u s s u c h p r o p e r t i e s .
B u t h e r e we c h o o s e to deal with w~-
b o u n d i n g , as it is v e r y n a t u r a l , a n d as t h e p r o o f of its p r e s e r v a t i o n is a p r o t o t y p e to m a n y o t h e r s u c h proofs. To be m o r e e x a c t we do n o t p r o v e t h a t it is p r e s e r v e d , o n l y t h a t t o g e t h e r with w - p r o p e r n e s s it is p r e s e r v e d . 4.3 T h e o r e m : The p r o p e r t y " w - p r o p e r n e s s + t h e wV-bounding p r o p e r t y
"'is p r e s e r v e d b y c o u n t a b l e s u p p o r t i t e r a t i o n . Proof: Let < P ~ , Q
I i < a > be an iterated forcing system. We prove that
it has the w~-bounding property by induction on a (the preservation of wproperness follows from T h e o r e m 3.5 ). For a = 0 there is nothing to prove. F o r a + 1 we h a v e Vp a § t i o n g 9 VPa s u c h t h a t g > f
pa)@~. I f f
9 Vp=§
(VP=~I)Q ~ t h e n t h e r e i s a f u n c -
(since Q= h a s t h e w~-bounding p r o p e r t y ) a n d b y
t h e i n d u c t i o n h y p o t h e s i s t h e r e is a n h 9 V s u c h t h a t h -> g, so h -> f .
Now a s s u m e t h a t a is a limit o r d i n a l a n d c f ( a ) > ~0- Let f 6 VP" a n d let f be a P a - n a m e of f , a n d let q e Pa. Let N be a c o u n t a b l e e l e m e n t a r y s u b m o del of H(A) s u c h t h a t q , f , P a e N, a n d let q ~ p 9 P a be (N,Pa)-generic.
For
e a c h ~z let ~gn be t h e s e t of all c o n d i t i o n s in P a w h i c h d e c i d e t h e v a l u e of
.gn is d e n s e in Pa a n d since it is definable f r o m Pa it is in N. T h e r e f o r e ~gn C~ N is p r e d e n s e a b o v e p .
The u n i o n of all t h e d o m a i n s of t h e m e m b e r s of
tJ (Sin FI N) (J ~p ~ is i n c l u d e d in Pj for s o m e j < a. Let f t be t h e P j - n a m e of a m e m b e r of w~ c o n s t r u c t e d so t h a t if r E ~9~ h o l d s (i.e. b e l o n g to G.) t h e n ~3
171
f t ( n ) h a s t h e v a l u e w h i c h r f o r c e s on f ( n ) . p we h a v e p II.f?andap
= ft. T h e r e f o r e f o r e v e r y q c Pa t h e r e is a j < a, a P f n a m e
~qssuchthatp
f t E: V[Py]. g~_.ft=f
f
S i n c e ~qn D N is p r e d e n s e a b o v e
By t h e
[~-f =ft
induction
T h u s in V[Pa] f = f t f o r s o m e , j
hypothesis there
is a g c ( ~ ) v
such that
.
We h a v e r e a l l y p r o v e d . 4.4 Claim= If Q =
< a t is a countable support i t e r a t e d f o r c i n g > 1to, t h e n a n y r e a l
n o t i o n , P a = Lim Q, e a c h Q. p r o p e r a n d a h a s c o f i n a l i t y
(or e v e n co-sequence) a d d e d by Pa, is a d d e d b y Pi f o r s o m e "/ < a.
So we a r e l e f t with t h e c a s e a is a limit o r d i n a l of c o f i n a l i t y H0, t h e following l e m m a is t h e m a i n point. 4.5 L e m m a : S u p p o s e a)
P
(p,q) c P * Q
a
forcing
P,Q
have
notion, the
N O
Q
a
P-name
~0~-bounding
of
a
property,
forcing ~
big
notion, enough,
P * Q,(p,q) b e l o n g t o g 0.
b) N O a N1,N 1 c N 2 , a n d e a c h Ng is c o u n t a b l e . c) r c P is ( N g , P ) - g e n e r i c f o r ~ = 0, 1, 2 a n d r - > p . d)<.~:E < ~
No, A~ D N~ C &
is a list of all m a x i m a l a n t i c h a i n s of P which b e l o n g s t o is f i n i t e . . ~ , *e p r e d e n s e
above r
and
<.~jg:s < ~ > ~ N 1,
<
e)<Jg:Jg < ~ > E N I is a list of the maximal antichains of P * Q which b e l o n g s to N O.
m~
there is q~ c Q n ~,, ~" c }~ n No finite for (~ < ~), suoh that
172
(r,q)>-{p,q),
and
each
a~* is
predense
above
(r%ql)
(Hence
(r,ql)
is
(No,P * Q)- g e n e r i c . ) ~
R e m a r k s : (1) I n s t e a d m a x i m a l a n t i c h a i n s we c a n look a t n a m e s of a n ordinal, or d e n s e s u b s e t s . (2) The s i t u a t i o n f o r P,No,NI,r in t h e a s s u m p t i o n is similar to t h e s i t u a t i o n of P * Q, No,N 2, (r,ql) in t h e c o n c l u s i o n w h e n ql > ~ql is (N2,Q) generic.
So we p r e s e r v e t h e s i t u a t i o n n o t i n c r e a s i n g t h e c o n d i t i o n in P. So
we a d v a n c e one s t e p in the i t e r a t i o n a n d lose g e n e r i c i t y f o r one of t h e m o d e l s (N1) a n d this will give u s t h e i n d u c t i o n s t e p in t h e p r o o f of 4.3 f o r cf a = t~0. P r o o f of Lemma:
F o r helping us in u n d e r s t a n d i n g let G c P be g e n e r i c o v e r V, and we shall work s o m e t i m e s in V[G] s o m e t i m e s in V. Note t h a t if r c G (which is t h e i n t e r e s t i n g c a s e for us ) t h e n Nr~[G] N H(A) V = gr a n d N~[G] ~ ~ H ( A ) [ G ] , e ) , which is always t r u e . F i r s t try: As ~o is a m a x i m a l anLichain of P
*Q,
t h e r e is 090,q0 ) E~o, P o E G,
V[G] ~ " q o [ G ] , / [ G ] a r e c o m p a t i b l e in g G ] " H e n c e in V, t h e r e is a m a x i m a l antichain ~
~ P, ~o+ = ~P,7:W c Yll (T1 C 1~/~), s u c h t h a t for e a c h 7 / c T 1 t h e r e I~P " Iq,qO~~_~is a c o m p a t i b l e s e t (in ~Q) " (a
is (5oO,!:) C J0, P ~ p O ~ p ~ , p .
c o m p a t i b l e s e t m e a n s it has an u p p e r b o u n d ) .
Similarly for e a c h 7 / c T1; t h e r e a r e p l _>p~, a n d (pl,ql) c ~ 1 , p l ->Pl, p l i~_p ..iq:,ql I is c o m p a t i b l e
"
So t h e r e
is
T2, T2 C212 f o r s o m e /~,
~7 c T2 ===>7/t 1 E T 1 a n d f o r e v e r y ~71 c T 1 f o r some 7/2 , ~11 =772t 1, a n d ~
= Ip~:7/ C T21 is a m a x i m a l a n t i c h a i n , p ~ - ~ p , rl, P ~ P I ,
P ~ I ~ - " l q : ~ 1,q~7 r 21] is c o m p a t i b l e in Q".
( p l , q ~ ) c fit,
173
So we c a n
Tn, Pn(rl 9 Tn),~+-1 a n d
easily define i n d u c t i v e l y on n (for
9
%).
Looking a t t h e way we h a v e d e f i n e d this, c l e a r l y we c a n a s s u m e
(i.e.
the
function
n -* P n
inN0
)
T~, and
< ( 9 ~ - 1 , q"W " - 1 ) : ~ 9 T.> 9 N0, But as <3n:~%
<<
T=,
Tr,:n < ca> 9 No, etc., b u t
T~>>:I---n
O is a m a x i m a l a n t i c h a i n of P, for s o m e n ( s
hence 3~(g) C 3n(~) N NO = J~ ('~NO is predense above r and is
=.~(s
finite. Let T n = ~ e
Tn :pn e~(s
So it is natural to look for ql 9 Q[G]
(where r 9 G) such that for each s < ca, lqs149
T~+ I I is predense above
ql. This will be sufficient - it implies (in V[G]) ql is (N0,Q[G])-generic, ql ~ q, and in V, ~(P~,~):~7 9 T~+II is a finite subset of ~
f~ N O predense above
(r,ql); moreover clearly we can choose ql 9 N2, in fact ql 9 NI.
U n f o r t u n a t e l y t h e r e is no r e a s o n to a s s s u m e extreme
case
G ~N0). ~P
Tr~ = I~Tn]- (e.g. when P
So
in
V[G] we
know
q l exists. Look a t t h e
is t~1- c o m p l e t e a n d r q,
q~+1
(s
and
determine we
know
s
~q'
q~+l :s < s
assume
lq,q~+,:s<
is compatible for every s < ca;this is not a good reason to cal is compatible, except when Q is Ml-complete and any
two compatible members have a least upper bound.
Second Try: Let Js N N 0 = ~(Pm~,qs
1,
"IT/,
<~:s <~>
6NI, we can a s s u m e
~gS = I r a
that
<<(pm~,q~m):m < m >
this i s a P - n a m e , S e 9
1. In
:s
V[G] t h e r e i s a f u n c t i o n
174
f : 6 ) - * 6) s u c h t h a t q : = q ^/~_ \/ q ~ [ G ] is c o n s i s t e n t ( b e c a u s e Q h a s t h e g m < f (~) ~ m m eS~[6:]
co~
property) (equivalently, there is q l -> q such that for every
s < 6)' ~qs
< f (g),m 9 S[G]t
is p r e d e n s e a b o v e ql in ~Q[G]" And also
e q u i v a l e n t l y ql ] ~ - Q " q 9 G a n d f o r e v e r y s <6) f o r s o m e m < f ( s ~Q q e [ C ] 9 C '?.
~ e G,
B u t as P h a s t h e 6)~-bounding p r o p e r t y , we c a n a s s u m e t h a t f as N I [ G ] ~ < H ( A ) [ G ] , 9
9 V. Also
a n d t h e p a r a m e t e r s a p p e a r i n g in t h e r e q u i r e m e n t s
on f b e l o n g to N I [ G ], we c a n a s s u m e f
9 Y N NI[G]
= gl.
Now in V we h a v e a P - n a m e of it f 9 N 1 w h i c h r f o r c e s to b e as above. As r
is
(N1,P)-generic
we
have
just
countably
many
candidates
for
f 9 (6)w)v N Ni, c l e a r l y t h e r e is in V a f u n c t i o n f " e to~ s u c h t h a t f o r e v e r y f
9
\co~
f "~a, f "
(i.e. | n
s o n a b l e to t r y ql = q ~
\/ q~ [G], (i.e. it is c o n s i s t e n t . rn.cf "(~)~rn.
So it is r e a -
See in t h e begin-
rn E S~[ a J
ning of t h e s e c o n d t r y f o r a n e x a c t definition.) So t h e q u e s t i o n is w h e t h e r
r I~-p "q ^ /~
\/
q~ [G ] is consistent in Q[Gp] "
m,.r "(~) -m. ~P
-- --
(If so we c a n use a s u i t a b l e P - n a m e for q 1)-
U n f o r t u n a t e l y t h o u g h f ' is a v e r y plausible c a n d i d a t e , t h e f a c t is t h a t if G C P is g e n e r i c o v e r V, r 9 G, m a y b e f [ G ] --< f ~ fail.
Third try:
Here we shall r e s c u e s o m e t h i n g f r o m the first t r y to c o r r e c t t h e s e c o n d try. Using t h e first t r y n o t a t i o n ( a n d w h a t we prove). R e m e m b e r ~ " = .9~(g), and
.9,~(~) is
a
finite
subset
of .gn(~)f~ No ( p r e d e n s e
above
r).
'~Let
-gn(~) = |P~:W 6 T~+i|, T~+ I a finite subset of T~+ I and so for s o m e k(g) < ~,
175
for every n ~ rA, ~ Clearly
,g~) 9 ~C~,qe~~):~ < ~(e)l. < ~ > 9 N1, a n d so in t h e s e c o n d t r y w.l.o.g. ] c ( g ) < f *(g)
f o r e v e r y g < co.
N o w return to the beginning of the argument in the second try. We k n o w that
for
every
qt e Q[G] Fl N0[G]
there
is
f 9 (coo)v such
that
~
qt ^ / k
\/
q~ [G]
is
consistent.
In
particular
for
every
=%s[z] w% 0 ..... w%n_ I < ~ there is an f =
.--<m f ..... m._t> 9 (~.)I' such that if
{q 'q-~v o ,qm, i ..... q~.~, | is compatible (in Q[G]) then
is consistent. m0~k(0
Let for i < ~ f+(i) = M a x
) ..... ~ _ l ~ k ( ~ - - l ) ] .
~
<~0
The m a x i m u m
..... .~_t>(i)-1% ~;i,
is taken over a finite set,
hence, it is a well defined natural number, so f + 9 (~o)Y[G]. So there is in V
It e_ (COo)V f + ~ f t . For each 7/e~ such
that
if f ~
t h e r e is a P - n a m e as
above
exists,
f then
of a f u n c t i o n ; f r o m co to ~ f r o m V, f is s u c h -@
a
function;
w.l.o.g.
< I : ~ 9 1 7 6 9 "i, and remember that
N o t e t h a t r ]~- "
N
the f
are as above "
N o w we shall prove that I" ]~-p" q ^ / ~
qI~[G]m ~P is consistent. And as remarked in the end
t
g<~ m
So f t : f t [ G ]
e_N I ClaW(c V) h e n c e f t ~ a ,
f~
So f o r s o m e i, f o r
176
e v e r y j , i <_j < w, f t ( j ) < - f ' ( j ) . o r d i n a l s s u c h t h a t ~? r g e Tg,
Also t h e r e is a u n i q u u e ~ - s e q u e n c e 7/ of ~-1 Pre e G f o r 1 - - g < w ( f r o m t h e first t r y
remember
maximal
[P~:V e T~ 1,
s ptg,q~i~l) = (p~-l,~ra~q~-l)
a for
some
antichain
m g < a~.
By t h e
P t m < p ~)" definition
So
of k ( g ) ,
( g ) , m (g) _
It(J) a n d t h e r e s t is trivial. C o n t i n u a t i o n of t h e p r o o f of t h e t h e o r e m 4.3.
We w e r e p r o v i n g b y i n d u c t i o n on a t h a t if [~p," Q is w - p r o p e r a n d h a s t h e p r o p e r t y ': Q =
w~
is w - p r o p e r a n d h a s t h e w ~
property.
The ~ - p r o p e r n e s s follows b y
3.5, a n d f o r t h e w ~ - b o u n d i n g p r o p e r t y only t h e c a s e of c f a = ~o w a s left. Now b y lI 2.4(c) w.l.o.g, a = ~. Let f be a t='=- n a m e p e P=, P ]~- " f 9 w~" a n d we h a v e to f i n d g e ( w ~ v, q , p - - q
9
q [~-"f--g"
We c a n a s s u m e
w.l.o.g, t h a t
(*) f ( n ) is a P n - n a m e (this
is i n c l u d e d
N~ < < H ( A ) , e >
by
(X big
the
proof
enough
that ) be
P~ an
is p r o p e r increasing
see chain
III 3.2 such
).
Let
that, p,
We w a n t to find q e P=, q _~ p , r ]l-p= " f ~ g " f o r s o m e g
e ( w o) v we n o w
d e f i n e d b y i n d u c t i o n o n n, qn e Pn 1) qn+l r n = qn, p r n <_ qn
2) qn is ( N k , P n ) - g e n e r i c f o r k = 0, a n d n + l -~ k < w
177
3) t h e r e is a f u n c t i o n F n E Nn+ 1, w h o s e d o m a i n is t h e s e t of m a x imal a n t i c h a i n s of P~ w h i c h b e l o n g t o N 0, a n d f o r e v e r y .9 c Dom Fn, F~(~) is a finite s u b s e t of .~ V~ No p r e d e n s e a b o v e r n.
Clearly if we s u c c e e d t h e n q = U qn E P... is as r e q u i r e d .
g(n)
as t h e minimal
F o r ~z = 0
g(n), q,~ [[--" f(n)< g(n)"'g (n)
We c a n define
e x i s t s b y 3) a n d (*).
use t h e c 0 - p r o p e r n e s s of Q , a n d f o r n + 1 we use first t h e *'0
lemma and then o-properness.
w What f o r c i n g c a n we i t e r a t e w i t h o u t a d d d i n g reals.
In s e c t i o n 1 we h a v e p r o v e d t h a t we c a n i t e r a t e f o r c i n g n o t i o n s of special k i n d ( E - c o m p l e t e ) w i t h o u t a d d i n g reals. As a r e s u l t we g e t a parallel of MA for s u c h f o r c i n g s a n d g e t t h e c o n s i s t e n c y of s o m e u n i f o r m i z a t i o n p r o p e r t y . Howe v e r this union, quite s t r o n g in s o m e r e s p e c t s , is c o n s i s t e n t with d i a m o n d o n S,: (see [Sh 77 ] [Sh 80a] . ) On s o m e s t a t i o n a r y s u b s e t s of col it c a n s a y m u c h , b u t on o t h e r s n o t h i n g .
So we shall t r y h e r e to find a p r o p e r t y of f o r c i n g n o t i o n s , so t h a t f o r c i n g withLimQ
, Q=
a CS i t e r a t i o n of s u c h forcing, d o e s n o t a d d
reals. 5.1 E x a m p l e : Assume 2t% = ~1 ( o r even 2t% < L~t' suffice),
Let A~ ~ ~ be u n b o u n d e d of o r d e r t y p e co, f o r B < o i limit, so b y [DS] (or see XV w
~A~:8 < col> d o e s n o t h a v e t h e u n i f o r m i z a t i o n p r o p e r t y , h e n c e
t h e r e a r e ff6:A~-* 10,iI s u c h t h a t f o r no f:r 1 -~ [ 0 , i ] , is f t A~ =a. ff~ f o r
Let f = ( I
<
178
P)r = I f : D o m f is a n o r d i n a l a < ~1, ~ ~ a ~
[ f r A S =ae f # ] l
order by induction, D~ : I f :i "C Dom f , f 9 So c l e a r l y t h e r e is no d i r e c t e d G c Py, G (~ D e ~ ~ for e v e r y i.
R e m a r k : P r e v i o u s l y J e n s e n a n d J o h n s b r a t e n show, t h a t t h o u g h f o r c i n g with Souslin t r e e s d o e s n o t a d d reals, b u t s t a r t i n g with V = L (at l e a s t with V=
~
) t h e r e is a CS i t e r a t i o n of s u c h f o r c i n g of l e n g t h ~, s u c h t h a t f o r c -
ing b y t h e limit a d d reals. This h o w e v e r d o e s n o t e x c l u d e a s u i t a b l e MA as t h e example above.
Now P is v e r y nice f o r c i n g - e.g. it is a - p r o p e r f o r e v e r y a < ~1, b u t o u r d e s i r e d p r o p e r t y s h o u l d e x c l u d e it. 5.2 Definition: (1) We call [~ a c o m p l e t e n e s s s y s t e m if f o r s o m e ~, ~ is a function
defined
N < (H(h),c)
on
the
set
of
triples
( N,P,.p ) ,
p c N C~ P,P c N,
N c o u n t a b l e s u c h t h a t ( P is m e a n t h e r e as a p r e d i c a t e o n N,
i.e., P • N):
[~(iv,p,p) = I~(N,P,p) is a filter, o r even a family of n o n - e m p t y s u b s e t of G e n ( N , P ) = IG:G ~ N A P, G d i r e c t e d a n d G N .9 # ~b f o r a n y d e n s e s u b s e t .9 of P w h i c h b e l o n g to N] s u c h t h a t if G E G e n ( N , P ) b e l o n g to a n y m e m b e r of
[~(N,P,~) t h e n p E G. (2) We call ~ a k - c o m p l e t e n e s s (h m a y also be finite or ~0) s y s t e m
[~(N,PX) h a s t h e p r o p e r t y t h a t t h e i n t e r s e c t i o n of a n y i elem e n t s is n o n e m p t y f o r i < 1 + h (so f o r k--> ~0, ~ ( N , P x ) g e n e r a t e s a filter ). Now s u c h [~ c a n be n a t u r a l l y e x t e n d e d to i n c l u d e g ~ ( H ( ~ t ) , e ) ~ c N, /z < izt b y [~(N,P@ ) = [~(N r~ H((~),P,p ). We do n o t d i s t i n g u i s h strictly. if e a c h f a m i l y
(3) We s a y [~ is on/~. 5.3 Definition: (I) S u p p o s e P is a f o r c i n g n o t i o n , E a n o n trivial family
179
of s u b s e t s of SH0(/z) a n d ~ a c o m p l e t e n e s s s y s t e m . We s a y P is ( E , [ ~ ) - c o m p l e t e if f o r e v e r y big e n o u g h ~k, i f P , E,[~ c N , P ~ P C~ N, g < ( H ( A ) , E ) ,
g countable, A c E N N ~
g N Iz c A, t h e n t h e
following s e t b e l o n g s t o [~(;v,P,p') (i.e., [ ~ N (~ H(~),P,:~)): | G c G e n ( N , P ) : p c G a n d t:here is a n u p p e r b o u n d f o r G in P t (2) If E = I SS0(/~)] we write j u s t [ ~ - c o m p l e t e .
5.4 Remark:
(1) We c a n t h i n k of [~(~,p,p> as a filter on d i r e c t e d s u b s e t s G of P N N g e n e r i c o v e r N , to which p b e l o n g s a n d t h e d e m a n d " ( E , ~ ) - c o m p l e t e " m e a n s t h a t ( f o r D ( E ) - m a j o r i t y of s u c h N ' s ) t h e " m a j o r i t y " of s u c h G's h a v e a n u p p e r b o u n d in P h e n c e t h e n a m e ( E , ~ ) ) - c o m p l e t e n e s s . (2) In s o m e s e n s e t h e d e f i n i t i o n s a b o v e a r e trivial: if P is E p r o p e r a n d d o e s n o t a d d r e a l s t h e n t h e r e is a ~ - c o m p l e t e n e s s s y s t e m [~ s u c h t h a t P is ( E , [ ~ ) - c o m p l e t e f o r all z s i m u l t a n e o u s l y . B e c a u s e , g i v e n ( N , P , p ~ , we e x t e n d p to q c P which is ( N , P ) - g e n e r i c . d e n s e s u b s e t s of P w h i c h b e l o n g to N, ~
If ~.~n:n < ~] is a list of t h e
N N = IPn0~ :0
from which
we c a n d e f i n e a P - n a m e x: x = | ( n ,/r ~ :n < ~ a n d / r is !minimal s u c h t h a t :Pn,~ ~ G ] Clearly q I~-p"X c ~ " , r,
q < r E P,
a n d s i n c e P d o e s n o t a d d r e a l s t h e r e is x " c (v~)v, a n d
r I~-p "X = x " .
Let
Gr = ~Pt C P v~ N : p t < r l .
Clearly
Gr c G e n ( N , P ) a n d let
So w h a t is t h e p o i n t of s u c h a d e f i n i t i o n ? We shall u s e a l m o s t a l w a y s c o m p l e t e n e s s s y s t e m s r e s t r i c t e d in s o m e sense: [~(N,PaD> is d e f i n e d in a r e a s o n a b l y s i m p l e way. T h e p o i n t is t h a t u s u a l l y w h e n we w a n t to d e c i d e w h e t h e r s o m e G E G e n ( N , P ) h a s a n u p p e r b o u n d , we do n o t n e e d t o k n o w t h e whole P,
180
b u t r a t h e r s o m e s u b s e t of
N,
e.g. a f u n c t i o n f r o m N to itself.
Check the
e x a m p l e we d i s c u s s e d b e f o r e : if ~ = N N ~1, t h e n we j u s t n e e d to k n o w f $ . B u t two f ~ ' s
m a y give i n c o m p a t i b l e
demands,
so f o r it t h e
s y s t e m is a
1-
c o m p l e t e n e s s s y s t e m . So if we deal with ~ 0 - c o m p l e t e n e s s s y s t e m , we e x c l u d e it (in f a c t l a t e r we shall d i s c u s s e v e n 3 - c o m p l e t e n e s s s y s t e m ).
The e x p l i c a t i o n of " d e f i n e d in a r e a s o n a b l y simple way" is: 5.5 Definition: (1) A c o m p l e t e n e s s s y s t e m [~ is called simple if t h e r e is a first o r d e r f o r m u l a 1# s u c h t h a t
[~(N,P,p) = ~Az:x
a f i n i t o r y r e l a t i o n o n N!
where
A~ = IG e Gen(N,P):< N tJ P(INI),e,.P,P,N> ~ lb[G,x]] (2) A c o m p l e t e n e s s s y s t e m li~ is called a l m o s t simple o v e r V0 (V 0 a class, u s u a l l y a s u b u n i v e r s e ) i f t h e r e is a first o r d e r f o r m u l a @ s u c h t h a t
~(N,P,p) = ~Ax,z:x
a r e l a t i o n on N, z e V01
where A,, z = | G c G e n ( g , P ) : < V0 LJ g t J / 9 ( N ) , c , p , P , V0,N>
~ @[G,x,z]]
(3) If in (2) we o m i t z we call [~ simple o v e r V0. 5.6 Claim-" (1) A k - c o m p l e t e n e s s s y s t e m is a k ' - c o m p l e t e n e s s s y s t e m f o r e v e r y k ~ < k. (2) P is ( E , ~ ) - c o m p l e t e for s o m e [~ if a n d only if P is E - p r o p e r a n d ( f o r c i n g wiLh P ) d o e s n o t a d d new reals.
Proof: (1) Trivial. (2) The d i r e c t i o n <== (i.e. "if" ) was p r o v e d in R e m a r k 5.4(2) above. So let us p r o v e t h e " o n l y if" p a r t . So P is (E,[~)-complete.
181
N ~ r H(k),e), p c P, (5 A. So B = I G 9 A~ENN
p ,P,E,[~, 9 N,
Suppose N (5~e
p 9 G~ 9
hence B ~
has
an
N
upper
(by Definition 5 . 2 ) a n d l e t G 9
countable bound
and
So G h a s
a n u p p e r b o u n d q (by t h e d e f i n i t i o n of ( E , [ ~ ) - c o m p l e t e n e s s ) , G 9 G e n ( N , P ) , and
by
Definition
p ]~-p"f 9
f 9
5.3
too,
p 9 G.
So
q ~p
is
(N,P)-generic.
As
] ~ - " f ( n ) = k '" f o r
t h e n f o r e v e r y n , -gn = i r 9
s o m e ]r < c01 is a d e n s e s u b s e t of P which b e l o n g to N, h e n c e .gn N G ~ #. H e n c e q d e t e r m i n e s t h e v a l u e of f ( n ) . So q d e t e r m i n e s f ( n )
for every n.
H e n c e it d e t e r m i n e s f , i.e. f is n o t a n e w real.
5.7 E x a m p l e : F o r c i n g b y a Souslin t r e e T is n o t [ ~ - c o m p l e t e f o r a n y s i m ple 2 - c o m p l e t e n e s s s y s t e m ~ .
Let N ~ C H ( } , ) , E ~ b e c o u n t a b l e , consist
of
the
branches
of
T E N, 6 = N ('1 w 1. Note t h a t G e n ( N , P )
T ~ N
which
has
As = tY e T:y < xl, where x c T a = t h e & t h l e v e l o f
an
upper
bound,
i.e.
T. N o w N " d o n o t k n o w "
w h a t is t h e s e t of s u c h b r a n c h e s of T (5 N, a n d two disjoint s e t s a r e p o s s i b l e .
The a b o v e is a n a r g u m e n t , n o t a proof.
To be e x a c t , we can, a s s u m i n g
d i a m o n d of Sl , build a S o u s l i n t r e e , s u c h t h a t no first o r d e r f o r m u l a lh define a s i m p l e 2 - c o m p l e t e n e s s s y s t e m f o r w h i c h T is G - c o m p l e t e .
w Speciali~ag an Aronszajn tree without adding reals.
The t r a d i t i o n a l t e s t s f o r g e n e r a l i z i n g MA h a s b e e n Souslin H y p o t h e s i s . J e n s e n h a s p r o v e d t h e c o n s i s t e n c y of t h e Souslin H y p o t h e s i s with G.C.H. ( s e e Devlin a n d J o h n s b r a t e n [DJ]) He i t e r a t e s f o r c i n g n o t i o n s of Souslin t r e e s , in limit p o i n t s of c o f i n a l i t y ~0 he u s e s d i a m o n d to r e f i n e t h e i n v e r s e limit of t h e t r e e s , in limit p o i n t s of c o f i n a l i t y ~t h e u s e s t h e s q u a r e on ~2 ( a n d p r e p a r a t o r y m e a s u r e s in p r e v i o u s steps). In s u c c e s s o r s t a g e s h e s p e c i a l i z e s a s p e c i f i c t r e e , b y first f o r c i n g c l o s e d u n b o u n d e d
s e t a n d t h e n building Souslin t r e e
182
using ~
( m o r e e x a c t l y h e a d d s 82 c l o s e d u n b o u n d e d s u b s e t s in t h e begin1
ning).
We s h a l l p r o v e t h a t t h e r e is a Q - c o m p l e t e f o r c i n g n o t i o n P r s p e c i a l i z i n g a n A r o n s z a j n t r e e T, f o r Q a s i m p l e 8 1 - c o m p l e t e n e s s s y s t e m .
The p r o o f is
c l o s e to J e n s e n ' s s u c c e s s o r s t a g e . Notation: For an Sl-tree z 9 Tl~, a ~ f l ,
T, T~ is t h e i - t h level, T t i = U Ti, a n d f o r
x raistheuniquey
9 T=,y ~z.
6.1 T h e o r e r m T h e r e is a n ~ l - c o m p l e t e n e s s s y s t e m Q, s u c h t h a t f o r e v e r y A r o n s z a j n t r e e T, t h e r e is a Q - c o m p l e t e f o r c i n g n o t i o n P r , s p e c i a l i z i n g it, i.e. I~-er " T is a s p e c i a l A r o n s z a j n t r e e , '" also PT is a - p r o p e r f o r e v e r y c o u n t a b l e o r d i n a l a. Proof:
F i r s t Approximation: 0
Let P = If :f a function from T r (atl)
to • ( s e t of r a t i o n a l n u m b e r s ) s u c h
T 0
that
a < c o 1, x < y = = ~ f ( x )
The
order-inclusion.
If
f c
P, 7
Domf
= T r ( a + l ) we s a y f
has height a + 1 , ht(f)
= a + 1.
0
Clearly
P r s p e c i a l i z e T, b u t we h a v e to p r o v e t h a t it is p r o p e r a n d d o e s 0
n o t a d d reals. Let N < < H ( ~ ) , e > , T,
e N, N ~ co1 = ~, ( N c o u n t a b l e ) . Let
P r
( t h e $ - t h level of T be)
T~ = ~x~:n < col. It is trivial t h a t we c a n e x t e n d a n y
c o n d i t i o n to a c o n d i t i o n of a r b i t r a r i l y l a r g e h e i g h t . So we h a v e to define a n 0 i n c r e a s i n g s e q u e n c e of c o n d i t i o n s p ~ c P A N, w h i c h will b e g e n e r i c f o r N ( h e n e e t h e i r h e i g h t s c o n v e r g e to ~) a n d h a s a n u p p e r b o u n d .
Now in o r d e r
183
t h a t ~Pn:~ < ~
h a s a n u p p e r b o u n d , it is n e c e s s a r y t h a t f o r e a c h s < co, the
s e q u e n c e (of r a t i o n a l s ) ~ p n ( z c z t h t ( p n ) ) : n
< co> is b o u n d e d . So a n a t u r a l
c o n d i t i o n t o e n s u r e it is e.g. (*) f o r s < n , pn(zfZ t ht (pn) ) + 1 / 2 n > pn+l(zs t ht(pn+t) )
This is n o t difficult by itself, b u t we h a v e also to e n s u r e t h e g e n e r i c i t y of :n < co>. So it c l e a r l y suffices to p r o v e 0
(**) if p e Nand~<0,
0
P r A N, n < c0, ~) an o p e n d e n s e s u b s e t of P r w h i c h b e l o n g to 0 PrNN'p'=q'q
t h e n t h e r e is q e
eDand
p(zr~ t h t ( p ) ) + r > q(zg t h t ( q ) ) f o r s < n U n f o r t u n a t e l y we see no r e a s o n f o r (**) t o hold. 0
E x p l a n a t i o n : So we shall c h a n g e s o m e w h a t
P
by adding "promises" T s u c h t h a t if (the p a r a l l e l to) (**) fail, t h e n we c a n a d d o n e m o r e p r o m i s e to p g u a r a n t e e i n g t h a t p h a s no e x t e n s i o n in D , c o n t r a d i c t i o n to D b e i n g o p e n and dense. Actual Proof 6.2 Definition: We call F a p r o m i s e ( m o r e e x a c t l y a T - p r o m i s e ) if t h e r e a r e a c l o s e d u n b o u n d e d s u b s e t C of co1 a n d n < r
( d e n o t e d by C(F),n(F))
such that: a) t h e m e m b e r s of F a r e n - t u p l e s ~ x o . . . . . from
Ta f o r some a E C .
xO <-- YO . . . . .
We say r
0 .....
xn_l> of disjoint e l e m e n t s
Xn_l> ~ r
0 .....
y._,>
if
xn-1 ~ Yn-1.
b) if a < / / a r e in C, z e F N n ( T a ) t h e n t h e r e a r e infinitely m a n y •'s,
which
s e q u e n c e s a r e disjoint).
are
pairwise
disjoint
(i.e.
the
ranges
of
the
184 6.3 Definition: We let
P1 = [ ( f , C ) : C is a c h a r a c t e r i s t i c f u n c t i o n of a c l o s e d b o u n d e d s u b s e t of ~x, with t h e l a s t e l e m e n t r
, and f
is a m o n o t o n i c i n c r e a s i n g f u n c t i o n
f r o m U Tr t o 0].
(fl,C1) < (fz,C2)
if
C 1 = C 2 t ( g t ( C 1 ) + 1)) a n d f l
and
only
if
C 1 ~ C2
(equivalently
C f2.
6.4 Definition: We s a y t h a t ( f , C ) E P1 fulfills or s a t i s f i e s a p r o m i s e r if: c(I') ~ C-MinC(F)
and for every a
in C(F) N C a n d
g CFNn(Ta)
( w h e r e n = n ( F ) ) t h e following holds: (]) f o r e v e r y v > 0 t h e r e a r e infinitely m a n y p a i r w i s e d i s j o i n t ~ ~: F FI '~(T0) s u c h t h a t f (zg) < f (yg) < f (z~.) + ~ f o r g < ~z. 6.5 T h e Main Definition: P = PT = I(f,C,~I,): ( f , C ) c P 1 ,
and
~I" is a
c o u n t a b l e s e t of p r o m i s e s which ( f , C) fulfills { ( f 1,Cl,ql) -- (f2,CZfl12) if: ( f l, C1) ~; ( f l, C2) ( i n P 1 ) and~I" 1 c a / 2 a n d a
E Ce - C
limpliesaE
(~ C(F).
N o t a t i o n : If p = ( f , C,@) we write f = f p , C = Cp ,x~ = ~zp, s tp = g t (Cp).
6.6 ? a c t : If p c P, p < e01 t h e n (1) t h e r e is q ~ P, q ~ p , (2) m o r e o v e r if t i c
a n d gtq _> p
N C(F),fl>gtp,thenwecanhavegfq FE~,p
(3) m o r e o v e r if m < co, Yo . . . . . t i o n to (2) d e m a n d f q ( y r P r o o f : (1) C l e a r l y
=ft.
Ym-1 c T a, e > 0 we c a n in addi-
g tp) < f g (yr < f g (yr
g tp) + ~ f o r i < m .
N C(F) is a c l o s e d u n b o u n d e d s u b s e t of ~1 (as 9 is Fc#~
c o u n t a b l e a n d e a c h C(F) is a c l o s e d u n b o u n d e d s u b s e t of c01). H e n c e t h e r e is
185 f t , ~ t > / g , f t > t t p a n d fit e
(~ C(F), a n d a p p l y (2). re,I,,
(2) L e t a = s
We d e f i n e
Cq = Cp I.)[fl~ (with d o m a i n #g+ 1)
~I,q = ~I'~, so we still h a v e to d e f i n e fq, b u t a s we w a n t t o h a v e f ~ c f q , we h a v e t o d e f i n e j u s t f q r Tp. We h a v e o n it two d e m a n d s in o r d e r t h a t q c P : (i) m o n o t o n i c i t y : f p(x r a) = f q(x r a) < f q(x) f o r x c T~ (ii)
$
from
Definition
6.4
for
ex1 < a a
s E 11 A n(r)(T,1) w h e n a e = fl ( f o r a e < fl u s e p
If we s u c c e e d
to d e f i n e
fq
in
C(~,
F c 4,p =,I,q,
c P)
t T#~ s u c h t h a t if s a t i s f i e s i), ii) t h e n q is well
d e f i n e d , a n d t r i v i a l l y b e l o n g s to P a n d is > p .
Now ii) c o n s i s t s of c o u n t a b l y m a n y d e m a n d s
o n t h e e x i s t e n c e of i n f i n i t e l y
m a n y y e ~ (T#).
Let
[(F m , 7 m , ~ m ) : m
< ~]
be
a
list
of
the
triples
F c ~I,p, ~ ~ r CI n(r)(TT), 7 < fl, T c Cp N c ( r ) , e a c h a p p e a r i n g
(F,y,~),
infinitely often
(if t h i s f a m i l y is e m p t y , we h a v e n o w o r k a t all). We n o w d e f i n e b y i n d u c t i o n o n m , a f u n c t i o n f m s u c h t h a t :
f m is a f u n c t i o n f r o m a finite s u b s e t f ~(x r a) < fro(x) f o r x e Dora Ff m. a)
of
Tp t o
0 such
that
b) f m C fm+l
< ~m a n d f o r e v e r y g < n (F m) ( w h i c h is t h e l e n g t h of era) fp (x~) < f m (Y ~) < f p (x~) + 1 / m . e)
There
is
gm C Dom fm§
-- D o m tim,
This will b e e n o u g h , a s a n y t r i p l e a p p e a r s
~m
infinitely often and 1/rn
con-
v e r g e to z e r o , so a n y c o m p l e t i o n of tJfm t o a f u n c t i o n f r o m Ta t o 0 s a t i s f y i n g m
i) is a s r e q u i r e d .
186
We let f 0 b e a r b i t r a r y s a t i s f y i n g a), e.g. t h e e m p t y f u n c t i o n .
If f m 7m c C(F),
is defined, let n = n ( F ) , (part
of
if Tm = a
requirements
h y p o t h e s i s of F a c t 6.6(2)).
of
we k n o w t h a t F is a p r o m i s e ,
(Fro,Tin,era)) a n d
/~
C(F)
by
a
H e n c e (by t h e d e f i n i t i o n of a p r o m i s e ) t h e r e a r e
infinitely m a n y p a i r w i s e disjoint if's, ern < if, ?7 E F N n(Tp) . As t h e d o m a i n of
f m is finite t h e r e is s u c h ff d i s j o i n t to D o m fro" So we let: D o m f m+l = Dora f m U lYo . . . . . fm+l(~/s
: f p (Ys t a) ..I- l / Z m
If 7m < a, we u s e t h e f a c t t h a t Tm C C(F)
and
a r C(I~
(by
a c Cp, a > 7m ~ Min C(F)).
So
f~(ze) < f~o(x~) + 1/3rn.
Yr~-ll
( f p , C ~ ) c P 1, s a t i s f i e s t h e p r o m i s e F,
Definition there
6.4
is
C(F) ~ Cp-Min C(F)
and
~ c F (~ n(Ta) 2 m < ~,
and
Now we a p p l y t h e a r g u m e n t a b o v e r e p l a c i n g ~rn b y
(3) The s a m e p r o o f as t h a t of (2), u s i n g o u r f r e e d o m to c h o o s e
$0So we finish t h e p r o o f of F a c t 6.6.
Now we shall p r o v e t h e c r u x of t h e m a t t e r : t h e p a r a l l e l of (**). 6.7 F a c t : ( 1 ) If N ~ ( H ( h ) , c ) able Nf~ca 1 =~, open
dense
~>0
subset
(hbig
and x 0 ..... of
P
then
enough) P,p cN, p cP,
there
is
(2) In (1) we c a n i n s t e a d of = 0 . . . . . branches
Beta
as
of T N N the
(i.e.
unique
fq(Bet~tq) < fp(Be t s
q cD
x e B e N T~, a n d
in t h e
~N,
q ~p,
xn_ 1 h a v e B 0
B e = | x ~ e : i <~], xie e Tr x~e < x f
T:V < z e | , a n d ~ / 2 .
.....
for i <j)
conclusion
+~.
Proof: (1) By (2), u s i n g B e = l y c
Ncount-
xn_ 1 c Te ( a r e d i s t i n c t ) a n d D e N
of (1)
is and
Bn_ 1
define have
187
(2) S u p p o s e p,N,~,.8 0 . . . . . a =s
Bn_ 1 f o r m s a c o u n t e r e x a m p l e , let
and zs_ = BEI" tx, .~" = < Z 0 ..... Zn_l>.
Let r I = I•: f o r s o m e T ~ : p m a ,
a n d ~ e n(TT) y e n(T#), g ~ y
~;~ and
= T, q e ~) a n d f q ( z ~ ) < fq(xcl ) + v~.
t h e r e is n o q e P, g _~p s
So I"1 is, in a sense, t h e s e t of " b a d " ~]'s; t h e p l a c e s t o w h i c h we c a n n o t e x t e n d p suitably. More explicitly: Subfact:
If
T e
(~ C(F) - (s re#p
+ 1),
T < ~
then
Br~_lr7> e f t .
Otherwise t h e r e is a q e P w i t n e s s i n g < B 0 r T . . . . . t h e n q exemplifies Io, N, ~, B 0 . . . . .
B,~_t r T > r F 1 b u t
Hn_ 1 do n o t f o r m a c o u n t e r e x a m p l e , (for
P = 7, z~ = B~ r p) e x c e p t t h a t m a y b e q ~ N.
Now F 1 is definable in H(A) using p a r a m e t e r s
which a r e in N, h e n c e
F 1 9 N. Now t h e r e q u i r e m e n t s on q a r e also first o r d e r with p a r a m e t e r s in N, so w.l.o.g, q 9 N a n d we g e t t h e c o n t r a d i c t i o n .
Let C 1 =
(3 C ( F ) - s
19
By t h e s u b f a c t
N ~ " f o r e v e r y T e C l t h e r e is y e r 1 N ~ ( T T ) ' ' b u t N 9 < H ( A ) , E > h e n c e also t h e u n i v e r s e V satisfies t h e s t a t e m e n t .
Our
plan
is
to
get
a
promise
F C F 1,
and
T = (fp ,C; ,~l'p U ~F~) 6 P, p ~ T, and above ~" there is no m e m b e r getting a contradiction to " ~) is open dense subset of P "
Let
show
that
of ~Z), thus
188
I'2 = ~7 9 F l : t h e r e a r e u n c o u n t a b l y
m a n y ~ 9 F1,7t < z l
By t h e a b o v e g 9 F i. We s h a l l p r o v e l a t e r :
B. S u b f a c t :
C" ~ ~0~, a = Min C~
T h e r e is a c l o s e d u n b o u n d e d
~ ~ C1
s u c h t h a t F = l!/ ~ F2: f o r s o m e i 9 C ~ y 9 n ( T , ) { is a p r o m i s e . Let us show that because
t h i s will b e e n o u g h
t o p r o v e 6.7, h e n c e
Theorem
6.1;
a s b e f o r e we c a n a s s u m e C* E N: a n d a s Min C(F) = Min C* = a = gtl0, = (fp,@,'~l~ U | F ] ) c P F~ N p
clearlypt of P t h e r e
As q 9
is q > p t i n D .
<---.pt. A s ~ ) is o p e n a n d d e n s e s u b s e t
(fq,Cq) s a t i s f i e s t h e p r o m i s e F, - s o a s
a 9 C(F) N Cq, a l s o ~ = g tq 9 C(Y) N Cq, h e n c e b y t h e d e f i n i t i o n of " f u l f i l l i n g a p r o m i s e " , t h e r e is y 9 ), ~ . < y , fq(y~.) < f p ( x s § So b y F l ' s d e f i n i t i o n y ~ F 1 ( a s q 9 D ) b u t y 9 F c F 2 E F1; s o we a r r i v e t o a c o n t r a d i c t i o n t h u s p r o v i n g P a c t 6.7, e x c e p t t h a t we n e e d :
P r o o f o f S u b f a c t B: N o t e t h a t a) if ~ 9 U n ( T i ) , g < y 9 F i ; so c l e a r l y a l s o F a h a s t h i s p r o p e r t y .
Next note that b ) f o r a n y i / 9 Fe | z E F2:Y -< ~{ is u n c o u n t a b l e .
(If n o t , f o r s o m e 7 > ~ are distinct
~
there
just
are
T ~ i < ~1), ~i: 7 -
t h e r e is n o ~ 9
E F 1, ~ i
(as
f o r i
countably T7
z, y - ~ ,
many
is
possible
countable
~
(z~ 9 t 7 .....
)
hence
i < ~ l , z ~ t 7 : zi~ f o r E < n J is u n c o u n t a b l e , c) f o r a n y ! /
cF 2 N n(T~),i <j
T h i s is j u s t a c o m b i n a t i o n d)
for
any
Y 9
~ ff u n ( T i ) . But there i
for
(for
some
h e n c e ~ 9 F2).
< ~01 t h e r e is ~ c a ( T j ) N F 2 , y < ~
of a) a n d b). Nn(T~)
~1'z2 e r 2 C~ n(T~), Y < ~ l y < ~2.
there
is
j,
i <j
1
and
disjoint
189
Otherwise let f o r i < j i <~<~<~l,
forsomes
< ~ 1 let ~J c F 2 ~ n ( T j ) , z
<
z
5~2J
(by c)).
So f o r
(otherwise u s e c ) o n Z ~ a n d ~ , ~ f o r i , j
,to
g e t c o n t r a d i c t i o n ) this c o n t r a d i c t s T being A r o n s z a j n b y t h e p r o o f of t h e o r e m III 5.4. e) f o r a n y 77 c I"2 A n(T~) a n d m < r t h e r e a r e j < ~1, J > i a n d pairwise disjoint ~,1 . . . .
, ~m ~ F ~ (~ n(Tj) ~7 < ~
forJg = 1 . . . . .
m.
J u s t b y i n d u c t i o n on n , d) a n d c).
Now we p r o v e t h e s u b f a c t itself.
F o r a n y 77 c F2 t h e r e a r e (by e)) Jm(Y)
( m < co) s u c h t h a t t h e r e a r e m pairwise disjoint m e m b e r s of F 2 N m(Ti,(y )) this h o l d s f o r a n y j >-jm(y).
w h i c h a r e >77 . B y c )
Now i f j ~- U Jm(?7) t h e n r~ ,{o
we c a n find f o r e v e r y m , m pairwise disjoint m e m b e r s of Y2 U n(Tj) w h i c h a r e > y . Let |~r members
are
be a m a x i m a l s u b s e t of ~
pairwise
disjoint.
If i 0 < w
c F 2 N n(Yi):Y < z l w h o s e
, choose
another
such
set
lyg:s < ni o + 1} (exists a s j -~ UJm(Y) : now a t l e a s t one 77~ s h o u l d be disjoint to all s i ' s . C o n t r a d i c t i o n to t h e m a x i m a l i t y of |Z~:i < i0I. H e n c e i 0 is infinite.
Let >
C" = li:i
is
a
or
i > a,
i c C1
implies
u Y~(y)l. It is e a s y to c h e c k C* is as r e q u i r e d . So we finish t h e p r o o f of 6.7.
Continuation of the proof of 6.1: Triv~ al.
w Iteration of (E,[~)- complete forcing notions
?7 ci12 U u n ( T j ) ,
and
190
T h e d i s c u s s i o n in t h e two p r e v i o u s s e c t i o n s l a c k a c r u c i a l p o i n t t h a t we can iterate such forcing notions without adding reals.
In order to get a rea-
s o n a b l e a x i o m we n e e d t o i t e r a t e u p t o s o m e r e g u l a r ~ > 1~1 a n d h a v e t h e ~c.c. F o r ~ = l~2, L e m m a 1.5 d o e s n o t s u f f i c e a s [ P r ] = 1~2 ( f r o m T h e o r e m 6.2 ) b u t m e a n w h i l e ~ s t r o n g l y i n a c c e s s i b l e will suffice ( s e e VII w aesthetic
o r VII w
An
d r a w b a c k of t h e p r o o f is t h a t we do n o t p r o v e t h a t t h e f o r c i n g we
g e t b y t h e i t e r a t i o n e n j o y t h e s a m e p r o p e r t y we r e q u i r e f r o m t h e i n d i v i d u a l forcing notions.
7.1 Theorem: L e t Q = Pa = Lim 0
a > b e a countable support i t e r a t i o n ,
; E a n o n t r i v i a l f a m i l y of s u b s e t s of (1) If e a c h Q. is a - p r o p e r
for some simple Ill-completeness system (2) We c a n r e p l a c e
Sbl0(N).
for every a < ol, and (E,~)-complete
then Pa d o e s
not add reals.
in (1) " s i m p l e " b y " a l m o s t s i m p l e o v e r V"
( n o t e V a n d n o t VP').
C o m b i n i n g t h e i d e a s of t h e p r o o f s of 7.1 a n d of 4.3 we c a n p r o v e 7.2 Theorem: In T h e o r e m 7.1 we c a n w e a k e n ' ~ l - c o m p l e t e n e s s
system "
to "I~0-completeness system "
H o w e v e r we s h a l l n o t p r o v e it now. P r o o f o f T h e o r e m 7.1: F o r c l a r i t y of p r e s e n t a t i o n
we f i r s t d e a l with t h e c a s e a = ~ ( f o r a <
t h e r e is n o t h i n g t o p r o v e ) .
L e t N~ < < H ( A ) , e >
be c o u n t a b l e ( f o r i < r
[~, Q e N 0, N~ e N(+ l ( h e n c e
N( < N(+I) e a c h N( is s u i t a b l e f o r E a n d p e P ~ N No a n d f e N o b e a P o - n a m e of a real.
191
Now we shall define b y i n d u c t i o n on n < ~ c o n d i t i o n s r~,pn s u c h t h a t :
A)
(1) r~ ~ P~, r~ : r ~ + l r n (2) r~ is (N~,Pn)-generic f o r i = 0 a n d n + l < i < ~0.
B) (1 / T h e r e is G~ 9 Gen (No,Pn) which is b o u n d e d by r~ a n d b e l o n g to Nn+ 1.
(2)
pn<-p,~+l,
pn E P ~ N o ,
po=p
and
pr, tn~--~'~
equivalently (Pn t n ) E G,~. (3) Let I.~m:n < ml be a list of the open dense subsets of P~ which belong to No; then Pn+l c .~. Let u s c a r r y t h e i n d u c t i o n . n = 0 Trivial ( h e r e we n e e d only m - p r o p e r n e s s a n d t h a t
Q0 is ( E , ~ ) -
complete) r~ + 1 We shall first define P n + I t h e n G~+ 1 a n d a t last rn+ 1. First step : We w a n t to find P n § .~n c P~ r 9
is
dense,
E P~ N No, P n + l -~Pn, Pn+l r n c G~ Pn+l c .~n. As . ~ E N 0,
above
every
C~N0 N - ~ n , r - - ~ q . L e t f i n = 1 r 9
v ~ t r~ = r ] ,
clearly
it
is
dense
q E P~ N No ~149
above
Pn t n,
there
is
~149 (in
Pn).
But
p n r n e G~ 9 Gen ( g 0 , P n ) , a n d -~n 9 No, h e n c e t h e r e is r 9 G~ N fin, a n d so t h e r e is r * E P ~ , r~
r~
r~
=r.
So c l e a r l y r * is t h e p n + l
we
need.
Second step: Let G~ C Pn be g e n e r i c o v e r V, rn E Gn ( h e n c e G~ ~ Gn) a n d we shall t r y to see w h a t a r e t h e d e m a n d s on G~+ 1. Really we w a n t in V[Gn], to find a m e m b e r of Gen (No, Q [Gn] ) which h a s a n u p p e r b o u n d . N~
192
N o t e t h a t ~ n is a P n - n a m e name #
w h i c h b e l o n g s t o No. So a l s o t h e r e is a p n -
f o r t h e ~b a p p e a r i n g in t h e d e f i n i t i o n of s i m p l i c i t y , a n d it b e l o n g s t o
[Gn]. So we k n o w t h a t f o r
N 0. As we k n o w Gn N NO = G,[, we " k n o w " @n = #
s o m e Ax,y (x a r e l a t i o n o n No, y 9 V, s e e D e f i n i t i o n 5.5) e v e r y G e Az, y h a s a n u p p e r b o u n d (in Q [Gn] ), w h e r e
Az,y = IG 6 G e n (No[G.], Q [G,~]):
u N0[C~] u P(N0[C~]) V(c~'')
So we h a v e P n - n a m e s x , y f o r z a n d y . Now z,y a r e q u i t e u n l i k e l y t o be in N o ( a s t h e i r d e f i n i t i o n s u s e d N o a s a parameter
)
but
they
Nn+l[Gn] ~
can
be
chosen
r,z e a n,
as
rn
in is
Nn+l[Gn]
(remember
(Nn+l,Pn)-generic,
and
N 0 [ G n ] 9 Nn+l[an]).
M o r e o v e r , t h o u g h we n e e d t o k n o w a n t o b e a b l e t o find t h e i r e x a c t v a l u e s , we k n o w t h a t t h e y a r e in V ( r e m e m b e r p ( N [ a n]) e V a n d Pn d o e s n o t a d d reals).
So a s r n is (Nn+l,Pn)-generic
r,~ [b "" z , y 9 Nn+ 1 ", so we h a v e j u s t c o u n t N
ably many
possible
pairs
(those
in Nn+l).
Now Nn+ 1 " t h i n k "
m a n y p o s s i b i l i t i e s , b u t as Nn+ 1 9 Nn+ 2, Nn+ a ~ "
uncountably
a b l e , ' ; in Nn+ 2 we " k n o w " t h a t
there
are
Nn+ 1 is c o u n t -
N [ A z , y : x , y e Nn+l] is n o n e m p t y ( r e m e m b e r
G~ 9 N n + I h e n c e g 0 [ G ~~ e Nn+l).
So in Nn+ z there is G n C ~Qn [Gn] N N0[G,~]
which belong to the intersection.
So t h o u g h w e do not k n o w exactly w h o will
be x a n d y , w e k n o w (as long a s r 9 G )) that G ~ 9
easily
compute -- I
Third
.....
step:
G~+ 1 9 G e n (No,Pn+I), .....
. From
G
,Gi w e can
G~+I N P ~ =Gn, q
9 NO
and
193
We n o w h a v e t o d e f i n e r n + 1 6 Pn§
so a s we r e q u i r e rn+ 1 t n = rn, we j u s t
h a v e t o d e f i n e r,~+l(n ). What a r e t h e r e q u i r e m e n t s induction demand,
+r
looking at the
just:
r IF . " r.+ICn) is a b o v e all m e m b e r s
eric f o r n
f r o m it?
of G[ and is
gen-
<_i < c 0 "
a n d t h e r e is n o p r o b l e m in this.
We h a v e f i n i s h e d t h e p r o o f of 7.1 f o r a = co.
Now we h a v e t o t u r n to t h e g e n e r a l c a s e i.e. w i t h n o r e s t r i c t i o n o n a.
Let
p,Q 9 N O <
No C ~ ( a +
1) = [ f l ~ : i _ 7 J ,
7)-
( a + 1) we
Now
i<j
can
,N O
be
countable,
===>fli
(so N O C ~ ( a +
find
N~ < < H ( A ) , e >
for
P 9 Pa.
Let
1) h a s o r d e r t y p e
i-<7,
N~ c o u n t a b l e
,
As N 0 9
1,7+
1 cNlhencei
e N i.
7.3 D e f i n i t i o n : A p a i r (r,G) is c a l l e d a n ith a p p r o x i m a t i o n if: 1) r 9 P#~ a n d r is ( N j , P s , ) g e n e r i c f o r j = 0 a n d i + 1 -< 2" <- 72) G 9 G e n (No,P~,), a n d G is b o u n d e d b y r a n d G 9 N~+ 1.
Now it s u f f i c e s t o p r o v e 7 . 4 Claim= If 0 ~ i
<j--<7,
p rfli 9 G t h e n
there
is a j t h
rtFfl~ =r
NP~,
= G.
andGt
(r,G) an ith approximation, approximation
(r?,G t)
such
W h y is the claim sufficient? Just use i =0, 2" = 7 (so ~ = a )
p 9 Pp~ (~ No, that
p 9 G t,
and we get
194
what we need.
Now the proof of the claim is by induction on j (for all i), Then for successor this is just like the induction step for a = ~ and for limit j we diagonalize using the induction hypothesis (taking care of also 1) of Definition 7.3.
VI P-POINTS AND PRESERVATION THEOREMS w1 A general preservation theoremu
An i m p o r t a n t p a r t of m a n y i n d e p e n d e n c e p r o o f s u s i n g i t e r a t e d f o r c i n g , is t o show t h a t s o m e p r o p e r t y X is p r e s e r v e d (if s a t i s f i e d b y e a c h i t e r a n d ) . h a v e d e a l t with s u c h a p r o b l e m ( t h e ~ - b o u n d i n g
We
property) and similar proofs
c a n b e w o r k e d o u t f o r s e v e r a l o t h e r e x a m p l e s . We give h e r e a g e n e r a l c o n t e x t which serves for many reals").
examples (but unfortunately
not for "adding no
In f a c t t h e r e is m o r e in c o m m o n b e t w e e n t h e e x a m p l e s d i s c u s s e d
l a t e r e v e n t h a n e x p r e s s e d by t h e s t r i c t e r c o n t e x t s u g g e s t e d h e r e (fine c o v e r ing m o d e l ) (i.e., t h e u s e of t r e e s T, T (~ no0 finite ) b u t t h e saving will b e minimal.
F o r a p p l i c a t i o n s it suffices to r e a d Definition 1.1, 1.2 ( t h e fine c o v e r i n g m o d e l ) a n d C o r o l l a r y 1.7 (on p r e s e r v a t i o n ) . 1.1. Definition: We call (D,R) a w e a k c o v e r i n g m o d e l (in If) if a)D a set, R a two p l a c e r e l a t i o n on it, z R T implies T is a c l o s e d s u b t r e e of ~>~(i.e., < > c T, T c l o s e d u n d e r initial s e g m e n t s , a n d a b o v e a n y 7 / c T t h e r e a r e a r b i t r a r i l y l a r g e m e m b e r s of T ), l e t Lim T = [~7 E ~ : r / r k c T f o r every k < ~| b) f o r e v e r y ~ / ~ ~eo a n d z c D o m R (=|x:(:IT)zRT]) t h e r e is T c D s u c h t h a t xRT a n d T / c Lim T Re~rn~rk: The i n t u i t i v e m e a n i n g is: z R T m e a n s " s i z e " a t m o s t x.
T is a c l o s e d t r e e of
In Definition 1.2, w h i c h e x p l o i t m o r e of o u r i n t u i t i o n , we
h a v e a n o r d e r on t h e s e t of p o s s i b l e x ' s , x < y with t h e i n t u i t i v e m e a n i n g " x is a s m a l l e r size t h a n y % So it would be n a t u r a l t o d e m a n d :
zRT, x < y ~
yRT
zRT, T t c T ==~ zRTt
196
However, no n e e d arises. Note also t h a t s o m e t i m e s x a p p e a r t r i v i a l l y (like
UF(F) in w a n d t h e ~ O - b o u n d i n g m o d e l in w 1.2 Definition: (1) A fine c o v e r i n g m o d e l is (D,R,<) s u c h t h a t
(a) (D,R) is a w e a k c o v e r i n g m o d e l ( ~ ) < is a p a r t i a l o r d e r on D o m R, s u c h t h a t e Dora R)(.~x e D o m R)(x < y)
(i) ( u (ii)(vy,=
e Dora
< y
>yn f o r n < ~ a n d
(7) I f x > x t > y n + l t h e r e is T * c D , xRT ~
= <
Tn c D ,
< y)
y n R Tn ( f o r n
c~)then
o
Lim T * ~ : f o r e v e r y i
< w , ~Trn/ c U T/]
( 6 ) c o n d i t i o n (7) c o n t i n u e s to h o l d in a n y g e n e r i c e x t e n s i o n in w h i c h a ) holds, (2) F o r a p r o p e r t y X of f o r c i n g n o t i o n s , (D,R,<) is a fine c o v e r i n g m o d e l f o r X - f o r c i n g if Definition 1.2(1) h o l d s w h e n we r e s t r i c t o u r s e l v e s in (6) t o X - f o r c i n g n o t i o n s only. 1.2J~ R e m a r k : 1) In a n a b u s e of n o t a t i o n we do n o t a l w a y s d i s t i n g u i s h b e t w e e n (D,R,<) a n d (D,R). (2) Look c a r e f u l l y a t (6), it is in a sense, m e t a - m a t h e m a t i c a l . (3) Note t h a t if y <x, yRT then for some T ~ T C T ~ (4) So if (D,R,<) is a fine c o v e r i n g model. P a (D,R) - p r e s e r v i n g f o r c i n g n o t i o n [ s e e Definition 1.4(1)] t h e n in VP (D,R,<) is still a fine c o v e r i n g model. 1.2B C o n v e n t i o n :
If t h e
order
< is n o t
specified
it is x < y
iff
x,y E ~co, x ( n ) <- y ( n ) f o r e v e r y n a n d y ( n )l x(n ) d i v e r g e to ~. 1.3
Definition:
(1)
For
a
weak
covering
model
(D,R)
and
x , y E Dora R,(D,R) ~ ~4~s(x,y) i f for every T s u c h t h a t xRT a n d f u n c t i o n F f r o m D x w to R a n g R = lz :(3t c D)tRz ~ s u c h t h a t
( Y n ) ( Y x E DomR)zRF(z,n) n 0
1
there is
T*,yRT* a n d
of n a t u r a l n u m b e r s , a n d z ~ c D o m R f o r s
increasing
sequence
<wsuchthat:
197 T" D |~/:there is {~ < o such that,? t ~%~ 9 T, and ~7 9 F(z~,~)]
Note t h a t t h e t r u t h value of (D,R) ~ ~d~s(x,y) d e p e n d s o n V. {2) We call (D,R) a c o v e r i n g m o d e l if it is a w e a k c o v e r i n g m o d e l and (c) f o r e v e r y y e Dom R t h e r e is x e Dom R, (D,R) ~ ~a.a (x,y)
(D,R), a n d z 9 D o m R , a n d = < z n :n < co>, z~ 9 D o m R we say that (D,R) ~ ~a~s(~,z) if for every Tn, x ~ R Tn(~%
T" D TO t3 |~/: f o r e v e r y i, ~7 r ~
9 13 [Tj:j <-i]]
(4) (D,R) is a s t r o n g c o v e r i n g m o d e l if it is a c o v e r i n g m o d e l and: d) F o r e v e r y z 9 Dom R t h e r e a r e a n (n < co) s u c h t h a t (D,R) ~ ~a~s(<:ZO,X,, . . . >,z) ^ ^ wA, ( [ ~ 9,a~s(xn,zn+,) ^ ,,
A
:,,,:,,+,,:,,+,,...
1.4 Definition: 1) Let (D,R) be a w e a k c o v e r i n g model.
We s a y t h a t a
f o r c i n g n o t i o n P p r e s e r v e s (D,R) o r is ( D , R ) - p r e s e r v i n g if ]]-p"(D,R) is a w e a k c o v e r i n g m o d e l ". 2) Let K be a p r o p e r t y of c o v e r i n g models. We say t h a t a f o r c i n g n o t i o n P is K - p r e s e r v i n g if for a n y (D,R) e V s a t i s f y i n g K, P p r e s e r v e s (D,R). 3) We call a c o v e r i n g m o d e l (D,R) s m o o t h if for a n y (D,R)p r e s e r v i n g f o r c i n g n o t i o n P, ] [ - p " ( D , R ) is a c o v e r i n g m o d e l ". 4) We call a s t r o n g c o v e r i n g m o d e l (D,R) s t r o n g l y s m o o t h if f o r a n y ( D , R ) - p r e s e r v i n g f o r c i n g n o t i o n P , ] ] - p " ( D , R ) is a s t r o n g c o v e r i n g m o d e l ,,.
1.5 Claim= If (D,R,<) is a fine c o v e r i n g m o d e l t h e n (D,R) is a s t r o n g l y s m o o t h s t r o n g c o v e r i n g model. P r o o f : We h a v e to c h e c k c o n d i t i o n s c),d) of Definition 1.3 in o r d e r to p r o v e (D,R) is a s t r o n g c o v e r i n g m o d e l c) It suffices to p r o v e t h a t x < y ----->(D,R) ~ ~ais(z,y). So s u p p o s e T e D, xRT a n d F is a f u n c t i o n f r o m D • co to Rang R, s u c h t h a t
198
~ D o m R)[z R F ( x , n ) ]
(Yn)(Yx So
let,
by
y >y?>xn+~>xn
(fl)
of
>..->x
yt,xr,(n < to) b e s u c h t h a t z~=xg a n d To = T, T~+l=F(xn,r~).
Definition 0=x,
1.2
let
Now a p p l y c o n d i t i o n (~,) of D e f i n i t i o n 1.2 t o g e t T ~ n o < n t < -. - s u c h t h a t
T* D [~/: f o r e v e r y / < to, ~ / t n ~ ~ U Tj~. Now T" is a s r e q u i r e d in 1.3(1):
We h a v e t o p r o v e t h a t (i~ T c
T'and
(ii) 7 / t g ~ T ^ ~ c F ( z ~ , s
~7c
T~
C l e a r l y (i) holds:
~ E T,g < t a z : : ~ v l t n s E T = ToE u T j . j .c.~ F o r (ii) s u p p o s e ~/t~/e E T, 7 / c F ( z e , s 9/ ~
(so ~ / E Te+I), a n d we h a v e to p r o v e
T ~ so it s u f f i c e s to p r o v e , f o r a n y /
rItrt~ e F ( z e , g ) = T~§ l c
U
Tj E u T j ,
uTj. j-*:i
and
If g + 1 < i if
then
i < g + 1,
~/r rtr ~ ~/t rtg E T = T O C U Tj, so Y/E T" is a s r e q u i r e d so we h a v e p r o v e d c). j
-
d)
Choose
xt
and
then
by
induction
on
r~
xn
such
that
xn < i n + 1 < - - - < x t < z ( t h e y e x i s t b y (B) of D e f i n i t i o n 1.2.) By t h e d e f i n i t i o n of @a~s a n d (7) of D e f i n i t i o n 1.2 (D,R) ~ "iba,:s(<Xo,,Xl,''' >,z).
By t h e p r o o f
(D,R) ~ ~bdis(x~,mn§ (D,R) ~ d ~ s ( < X r t , x n + l , x n + l , ' ' ' ~ , x n + 2 ) ,
prove
of
c)
above
ran+ 1 R TI§
By (~)
xrt+l
of D e f i n i t i o n
So so
1.2 t h e r e
we let are
have Tg
to
be
given,
only
x n R T O,
y'f,yn(n < to) s u c h t h a t By r e m a r k
1.2A (3) t h e r e
T~. So b y (9') of D e f i n i t i o n 1.2 t h e r e a r e n 0 < n
1 <---
a n d T*,zRT ~ a n d T* D Iv/: f o r e v e r y g,~7 t n g e: U T]}, h e n c e c l e a r l y T" 3 117: for
every
g,r/tn
c U 7"i}.
(D,R) ~ q/ais(<X,~,xn§
So
T
is
as
required,
we
. . . > ,z) h e n c e finish t o p r o v e d).
have
proved
199
(D,R) strongly smooth? By r e m a r k 1.2A (4). Suppose P is (D,R)-preserving t h e n is VP (D,R) is still a weak covering model as P is (D,R)-preserving, hence (a) of Definition 1.2 holds in VP, (~) is trivial, and (T), (6) hold by (6). So (D,R,<) is a fine covering model in VP. Now why is
1.6 Theorem: Suppose
(D,R) is a smooth strong covering model,
Q=
(D,R) -preserving for i < ~, and 6 is a limit ordinal. (Recall
(~, 1) -properness from V. 2.1). Then P8 = Lim Q is (D,R) -preserving. 1.7 Corollary: if ~ = < P~Q :i < a >
is a countable s u p p o r t iteration,
(D,R,<) a fine covering model and each q. is ( ~ , l ) - p r o p e r and (D,R,<)preserving, t h e n so is Lim Q. Proof: By 1.6, 1.5 . (1.5 t a k e s care of the induction for successor ordinals). Proof of Theorem 1.6: As each Q. is p r o p e r (by V 2.1 as it is (~,I)N$
proper) by III 3.2, w.l.o.g. 6 = cf 6; and by V 4.4. if cf 6 > ~0 t h e n each real in YP* belong to some VP~,i < 6, so the conclusion is trivial. So we can assume
So let z c D o m R $
EP~, ~ a P~-name. p ]~-p,"f e ~ w "
p?, p ~ p ? ~ P~, T c D such t h a t zRT a n d p t
iI-p,~ r
We have to find
T'" By the proof of
t h e preservation of properness, we can assume w.l.o.g, t h a t
f ( n ) is a Pn-name N
(we have to increase p ).
We can write the members of P• as < q : s larly
for
P,.
Let
p = < q : s < ~>,
< n>,
]~p,"qs ~ Q~" and simi-
and
let
(in
VP")
200 Pnl Pm = Q
* Q
Now
p"eP~
*...
* Q
"r~+l
','m.
we
define
....
p ~ =
.
~n--1
"
by
induction
n < o,
on
a
condition
q_:_ 1>, and for each ~ - " ~ a ~ . - n a m e
~.~
such
that: a ) p r n ~ p . , P n <--P n + l r n , m o r e o v e r
~)
If
v[c,.],
is
Gm C P m
q#-i>
.....
Equivalently,
<~,,
.....
generic
ll-a.~."L(,~) q~_,>
qg .....
[~-P~"q~z--< q ; <- ~JZqn+l,, (over
V)
m ~ n,
then
in
= t.,..[~.~],,. ll-e"g(~)
= t,~,,."
This is e a s i l y d o n e ( f o r e a c h 'n,, we i n c r e a s e p " n t i m e s ) .
Let f
7)
be the Pn-name.
r
r
~) I k o , " /
2' ( m )
= tin,n; SO
II-v.j'/
~~+t
tn
rn =Irn"
By D e f i n i t i o n 1.3(4)(d) t h e r e a r e xn (n < ~) s u c h t h a t
(~.R) ~ ,,~(<~0.
Now
- 9 - > .~) ^ ~
let
be
~(~,~+,1^
such
that
z,
~ ~ ~0, ~nd also L'< ~:~ < ~ < ~>' < ~",~:~ "- ~ < 0
b e l o n g to N o .
We n o w d e f i n e b y i n d u c t i o n
on n conditions r ~ =
, rn_l>
e P.
(so t r i v i a l l y r ~ = ~.n+l [, W,) s u c h t h a t :
a) p r ,-,. ~ r~, ~ b)
For
is < ~ : 8 ~ ~ ~ < ~ > - ~ e n e r i c ~or (P~,2~). some
Tn e D CI N s . , r n
l t- " fN?% e Lira Tn ",
sa~ R T~, T~ c :r~+l c) F o r e a c h n < o, E < ~, I[-p " r n , q ~ a r e c o m p a t i b l e in Q "
and
201
If we succeed t h e n
for any g:~-,
~, t h e r e
II-e."r~ 9 r
a r e rnt,
and
~n
rt~ ~ rn, qgn(n)'' ( b y e) a b o v e ) . C l e a r l y r t = < r ~ , r t . . . . .
rnt, . - . > is ~ p ,
also
if g is i n c r e a s i n g , m < n , g ( m ) > n , t h e n
( r t t n ) IkPj'Y ~n =Ira r,~". hence
(rttn)
II-p~"Y r,~ 9 T~"
H e n c e b y t h e d e f i n i t i o n of lkd/s ( s e e D e f i n i t i o n 1.3 (3)) if g " g r o w s f a s t enough"
rt I~p."f
r/t (g(m)--l)
9
U Tr
9 Lim T " " and
where
T" = ~/:
for
every
m
xRT*
i-:rn
T h e r e is n o p r o b l e m f o r n = 0,1.
So we h a v e j u s t t o do t h e i n d u c t i o n s t e p . A s s u m e r n, Tn a r e d e f i n e d , a n d we s h a l l d e f i n e r n+t, Tn+ 1.
L e t Gn E Pn b e g e n e r i c o v e r V, r n 9 Gn, so je f
/ Gn of a m e m b e r of ~
Nn-I- 1
q 9 Q [Gr~], z 9
there
but
, becomes a Q [Gn]-name
(D,R) is p r e s e r v e d b y Pn+l, h e n c e f o r e v e r y
is a c o n d i t i o n
q t ~ q ( q t 9 Q [Gn]), s u c h t h a t
Nn
~n
qt I~-"fn+i/Gn 9 Lira Tt " f o r s o m e Tt 9 D, z R T t.
Hence for everyq
E Q [Gn], z E D o m
R t h e r e a r e qt, q t c
Q , q'zqt
and
Tt,xnRT t, qt {FQ [c.]..f + I 9 Lim T t'' Let Tt
Fo(q,z), qt = F2(q,z). By o u r h y p o t h e s i s ( s m o o t h n e s s )
in
V[Gn]
(D,R) is still a c o v e r i n g m o d e l . So n o w we a p p l y c o n d i t i o n c) of D e f i n i t i o n 1.3(2) ( t h e d e f i n i t i o n of a c o v e r i n g m o d e l ) a s ~a~(x2m,X2m+1 ) f o r t h e t r e e Tn a n d t h e f u n c t i o n F, ( w h i c h is
D) F ( z , m ) =Fo(qnm,z), so we g e t a t r e e T ~ Tn C T* a s d e s c r i b e d t h e r e . W o r k i n g in V w e h a v e a P n - n a m e T* , l e t f o r s c Pn, T*(s) b e
from DomRxca
to
202
t h e u n i q u e T s u c h t h a t s []- "'T* = T'" if it exists. Hence for a n open d e n s e s u b s e t .9 of Pn, T ~
is d e f i n e d for t e .~. Clearly t h e f u n c t i o n T~
belong
to N 1 (more e x a c t l y s u c h f u n c t i o n s belong to N1) a n d also ~ e N 1. If r ~ belong to t h e g e n e r i c s u b s e t of P~, t h e n some t e .9 N Ns-+2- belong to it by c o n d i t i o n a) above.
Let
-9 Yl N8-+2- = | ~ : 0
and
by
( D , R ) I [ - ~ b ~ ( < z a n , x 2 ~ + l , x 2 n + v . . . >,zan+a);
the
choice
of
apply Definition
the
xj's,
1.3(3)
with
<X2w.,X2.+I,Z2n+l,2.2.+I,''" >, Z2.+2, < T~t,T'(tl),T'(t2),T'(t3)'""> ingfor<xo,Xl,ZmZm'">,z,
stand-
Wegetatree T . + l , m o r e o v e r
we c a n c h o o s e it in N8.+2.**. Now Tn+ 1 c l e a r l y i n c l u d e Tn, a n d (*)
if
G. C P .
is g e n e r i c
qt e q~[c~],]~[c~] -< qt, qt
over
IF/+/a~ e Lira T*"
As we can assume w.l.o.g,that Q
can
find
rt
V, r EG~,
compatible
r n _ l , r ~ > is (< Nr
then
there
are
is closed under countable disjunction we
with
q~
for
< W 0 ..... v/~_i,v}~> [[-i%.,"f +i e Lim Tn+i ~ Ns~+2,.8. N o w use the (r
s <~,
every
g
such
that
Moreover we can find it in
properness of Qn to find rn ~ r~t such that
+l) ~ s ~ ~>,P~§
-generic.
So we finish t h e i n d u c t i o n step, h e n c e t h e proof.
N o w let 1.8 Definition: TTR = I T D ~co: T C a>co a closed tree, T f3 ~'w finite ~. TTR has a natural tree structure: t < s h t (t ) = rain | ~ : t r ~r
1.9. F a c t
if t = s C t ~ e
for s o m e n; let
TTR n = | t : h t ( t ) = rt ~.
: A closed tree T c ~
induce a branch [T D n~e~
< r
such that (VTt)
I
T f ~ l
and is its union. N o w TTR is isomorphic to
203
w Three known properties In t h i s s e c t i o n we u s e t h e m a c h i n e r y f r o m t h e p r e v i o u s s e c t i o n . We d e a l with t h r e e p r o p e r t i e s which we call: ~ - b o u n d i n g
property, Sacks property
a n d L a v e r p r o p e r t y . All h a v e b e e n used, a n d we show t h a t t h e i r p r e s e r v a t i o n b y c o u n t a b l e s u p p o r t i t e r a t i o n follows f r o m 1.7. We t h e n p r e s e n t t h e " c l a s s i c a l " e x a m p l e s of s u c h f o r c i n g .
In t h e n e x t s e c t i o n we shall d e a l with o n e
e x a m p l e in m u c h m o r e detail. N o t a t i o n : DP(~ diverge
to
infinity,
= Ix E ~ : z ( n ) i.e.
for
every
~ 1 for every n m <~
for
some
and (z(n):n ]c < ~,
for
< aJ) every
n >h:,z(n)~.m.| 2.1 Definition: 1) We call (D,R) a n w ~ - b o u n d i n g m o d e l if a) F o r s o m e s u b u n i v e r s e Vt (with t h e s a m e ~1) ]9 = H(R1) ~ b) xRT iff z , T ~_ D,z c DP(~e~), T ~ ~>~ a c l o s e d t r e e a n d ( V n ) ( T N n ~ is finite) (so z h a s r e a l l y n o role) c) it c o v e r s se~ i.e:,
(V~? e =~)(Yz c DP(WaO N D)(:~T e D)[zRT ^ ~? e Lira T] 2) The f a m i l y of e~~ - b o u n d i n g
K~,(D,R) is t h e full m o d e l if D = H(lr
models
will be d e n o t e d
by
(so it is u n i q u e ) a n d its R is d e n o t e d
by R~. 2.2 Claim: Any coS-bounding m o d e l is a fine c o v e r i n g m o d e l . P r o o f : Let us c h e c k t h e c o n d i t i o n s in Definition 1.2. a ) a) Trivial b y a),b) of Definition 2.1. b) P a r t c) of Definition 2. I is a r e s t a t e m e n t of b) ~) Trivial ( r e m e m b e r C o n v e n t i o n 1.2B) T) Follows f r o m ~ ) ( f o r t h e trivial e x t e n s i o n ) ~) L e t z > z t > y~ > Yn,Y~ R Tn, b u t t h e s e q u e n c e (yn,Tr,:n < ~
is in
204
V[G] in w h i c h (D,R) is still a w e a k c o v e r i n g m o d e l . L e t n k = k, T ~ = | y : f o r e v e r y k < =, ~/r/c c U Tj.J j~k
T" is a s r e q u i r e d e x c e p t t h a t t h e r e is n o r e a s o n why T* 9 D.
Now we u s e F a c t 1.9, a p p l i e d t o T ' , (i.e t o t h e b r a n c h of TTR w h i c h T" induced.)
So t h e r e
is a c l o s e d
t r e e Cc TTR,c 9
and
for every ~,
T" V~ ~>~ ~ C Let T " = [I/ 9 =>e: for s o m e t c O , Clearly T ~176 c D
z/ 9
( as C 9 D,D = H(t~1) v~ for s o m e
Tit), T'" is a closed tree
c =>e, a n d T ~ c T**. in addition, for every n
T** C~~
= LJ It Ut ~co:t 9
TTt~§
so, b e i n g a finite u n i o n of finite s e t s , T'* ~ n e is finite. 2.3 C l a i m : I ) A f o r c i n g is ( D , R ) - p r e s e r v i n g f o r s o m e r
model
iff it is (D,R) p r e s e r v i n g f o r all c0=-bounding m o d e l s 2) A f o r c i n g n o t i o n is (D,R)-preserving f o r s o m e c ~ - b o u n d i n g m o d e l ifJ' it h a s t h e wW-bounding p r o p e r t y (see Definition V 4.1 ). P r o o f : Easy. 2.4 C o n c l u s i o n : The " e e - b o u n d i n g
property
+ (~,l)-properness
" is
preserved by countable support iteration. Z.~ Definition: 1) We call (D,R) a S a c k s m o d e l if: a) F o r s o m e s u b u n i v e r s e Vt with t h e s a m e #I,D = H(Rt) ~ b)
z R T
n < ~ , T (~ ~
~ff
x,T 9
and
z cDP(~'~),Tc~>r
and
for
every
has at most z(n) elements.
c) It c o v e r s , i.e.,
(v~ 9 ~ ) ( v = 9 D P ( ~ ) n D)(~7 c D)[= R T ^ n e ~ m
T]
2) The class of Sacks models is Ksc,(D,R) is the full Sacks m o d e l
205
if D = H(l~z), a n d t h e n we d e n o t e R b y Rsc. We s a y P h a s t h e S a c k s p r o p e r t y if it is K~ - p r e s e r v i n g . 2.{} Clairo= Any S a c k s m o d e l is a fine c o v e r i n g m o d e l . P r o o f : Let u s c h e c k t h e c o n d i t i o n s in Definition 1.2. ( a ) a) Trivial b y a), b) of Definition 2.5. b) P a r t c) of Definition 2.1 is a r e s t a t e m e n t of it. (~) Trivial, b y t h e d e f i n i t i o n of t h e p a r t i a l o r d e r . (7) L e t z > x t > Y~+2 > Y~,Y~ R T~ ( f o r n < ~).
Define nk
as t h e first m < ~
~ k , y ~ ( E ) / x t ( ~ ) is ~ 1 / ( / c + 2 ) , (~) L e t
such that
f o r e v e r y ~, m - E