Série Mathématiques
S AHARON S HELAH
Interpreting set theory in the endomorphism semi-group of a free algebra or in a category
Annales scientifiques de l’Université de Clermont-Ferrand 2, tome 60, série Mathéma- tiques, n
o13 (1976), p. 1-29
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INTERPRETING SET THEORY IN THE ENDOMORPHISM SEMI-GROUP OF A FREE ALGEBRA OR IN A CATEGORY
Saharon SHELAH
The Hebrew
University,
JERUSALEM, IsraelUniversite de LOUYAIN, LOUVAIN-LA-NEUVE,
Belgique
ABSTRACT. - We prove that in the
endomorphism semi-group
of a freealgebra Fh with h
-generators,we can interprete H(X+ ),
e > and get similar results forcategories.
The result was announced in[
S 3] .I would like to thank R. McKenzie, that
seeing
theproof, pointed
out theholding
of 3.3(A) (by
[ X ]).
I thank also M. Rubin for
stimulating
discussions on theproblem during
its solution, and forreading
theprevious version,
anddetecting
errors. The readers should thank him forurging
me to rewrite the paper,and in
particular
to stateexplicitely
that B is a Booleanalgebra.
§
0. INTRODUCTION.This paper has two lines of
thought
as motivation :comparing category theory
withset
theory,
andinvestigating
thecomplexity
of the theories of some natural structures.Lawvere
[
L] proved
that in thecategory
of all maps between sets, we caninterpret
set
theory (which
is notsurprising
in view of Rabin[ R] I interpreting
ageneral two-place
relation
by
twoone-place functions).
Eklof asked whether in Ab
(the category
of abeliangroup)
we can define the free group ofcardinality X .
Hegot
apositive
answerA
and h the firststrongly
compact
cardinal. Feferman asked whether for some p Ab Ab(Ab -
thecategory
of03BC 03BC
abelian group of
cardinality 03BC ).
It is natural to
replace
the class of abelian groupby
anyvariety,
and to concentrate on the free members.(In fact,
our results hold for moregeneral categories ; see § 5).
From another
direction,
there was interest in the first order theories ofpermutations
groups.
Mycielski [ My ]
asked it. ,This
problem
was dealt in Ershov[ L ],
McKenzie
[ M] ,
Pinus[ P ] ,
Shelah[ Sl ] ,
,[ S2 I
where it wastotally solved,
in fact.The
semi-group
ofendomorphism
of a freealgebra
is a naturalgeneralization.
For
simplicity
we shall restrict ourselves to X >L1,
L thelanguage
of V.Let for
regular X , H( ~)
be the set of sets ofhereditary
power X, and forsingular H( A)
=J.1 U
X
). Our
main result is that fora variety V,
inCatX (the category of
p X
members of V of
cardinality
X),
we caninterpret uniformly
a modelMh consisting
of somecopies
of(H( ~ ), e). For X
= 00 thisgeneralizes
Lauveretheorem,
and it also solves Fefermanproblem (i.e.
reduce it to aproblem
of settheory).
Incategories
in which the set of free,members is
definable,
Eklofproblem
is answered too(e.g.
forAb).
There is an
example showing
that notalways
we can define the set of free members.However,
we can characterize thealgebras
ofcardinality K p
if e.g.I
= ~(L-the language
of
V), and p
is definable inH( x )
and we can characterizealgebras
which are free sums ofsubalgebras
ofcardinality 2 LI .
REMARKS :
(1) By [ Sl ] , [ S2 ],
we cangive
a totalanalysis
of thecategory
of one-to-one maps ;by
which we cannotinterprete
settheory
in it.(2)
It is natural to ask what we caninterpret
in theautomorphism
group of a freealgebra [ or
thecategory
ofmonomorphism ]. Clearly
here the resultdepends
on thevariety.
This converges with the
question
of M. Rubin who asked on the classificated of first-ordertheories, by biinterpretability
of their saturated modelsautomorphism
groups.In
[
Ru] he
solved theproblem
for Booleanalgebras.
If we allowquantification
over elements,we can
essentially
solved theproblem.
(3)
It will beinteresting,
tochange
somewhat our main theorem 5.5 toget biinterpretability.
Of course, if the set of identities ofvariety
is definable inl1 (À )
and thereare no non-trivial beautiful terms, this holds for /.
A
(4) Sabbagh
and the author note that if inCat, , Fl
is definable then eachautomorphism
of the skelton ofCat x (i.e.
the fullsubcategory
of a set ofrepresentatives
frnrneach
isomorphism
class ofobjects)
is inducedby
anautomorphism
of the(multi-sorted) algebras
of terms.NOT.ATION :
Let V be a fixed
variety,
and Cat : thecategory
of allalgebras
in V with allhomomorphisms.
Let K be a fixed
subcategory
ofCat, usually
we assume K is a fullsubcategory.
LetF A
be thefree
algebra (in V) generated by
X freegenerators {at :
t e I } . LetGX
be the endo-morphism semi-group
ofF~ .
Elements of I will be denotedby
t, s and at ,I,
t, s will appear in no other context. Severaltimes,
we deal with K =G x
*Two
endomorphisms (or elements,
orsubalgebras)
ofF~
are calledconjugate
if someautomorphism
ofF~
takes one to the other. We denote elements ofalgebras by
a,b,
c, d(a-usually
agenerator)
and alsoby
x, y, z which serve as individual constants too.For any f let Rnf be its range, and for a set A e
V,
Cl B is its closure inA,
b = ClI b l .
LetX,
denote finite sequences ofvariables, x = xo~
..., xn-l >usually.
For a term T we write T = T ...,
xn-1)
=T (x),
if every variableappearing
inTbelong
..., We can assume
w.l.o.g.
that if T(I,+)
= T is anidentity (of V),
X, are
pairwise disjoint,
then for some term a(x),
a(x) =
T(x,y)
is anidentity. A
term...,
xn-1)
is called reduced if for no i and a isan
identity.
Clearly
for every term...)
there is a reduced term a(xi(o), ...)
such that(i(0 ), i( 1 ), ...
are distinct andis an
identity.
Clearly
for every bE Cl B (B ç A e V)
there is a reduced T and distinctbi
E Bsuch that b =
T (b , bl, ...).
Also if T(D
isreduced, ti
e I aredistinct,
T(at ’
1..., at ) =
n o(as,, ..., as )
1 m(all
in someFh
A!) then {t1, ..., tn} c {s1,
...,sm} .
We say T
(at ,
0...)
is reduced ifT (x0, ...)
0 is reduced and theti
s are distinct. Notice thatany function h :
A A V,
has aunique extension h E Hom(F À
XA).
If h : I - I, h E
Gx
is definedby ah(t)’
We consider any K as a model : with two universes
(the algebras
and themorphisms,
and the relations f e
Hom(A,B),
gI
f o h.Naturally
a first-orderlanguage
L is associated with it. Forconvenience,
we can consider thislanguage
forGÀ
too.§ 1. ON DEEPNESS.
DEFINITION 1.1. : Let h be a function from B into B. For every x f
B,
we define itsdepth Dp(x) = Dp(x,h)
as an ordinal or -by defining
when it is ~ a :Dp(x) ~ 0 a
x c BDp(x) ~
6 -Dp(x) ~
a for every a 5(where
5 denote alimit
ordinal)
Dp(x) ~
a + 1 a for some y fB, f(y)
= x andDp(y) ~
a .LEMMA 1.1. : t f I }
freely generate F~ ,
and h is a function from I into I. Define(C)
If in(B),
T(xl,
...,xn)
is reduced and thet(e)
aredistinct,
thenequality
holds.PROOF.
(A)
Immediate.(B)
We proveby
induction on a thatand this suffices. For a = 0 or a
limit,
it is trivial. Fora= 0
+ 1,by
theassumption
anddefinition of
depth,
there ares(e)
EI, h(s(e))
=t(e)
and . Thenmin
Dp [ s(e), h] > B,
henceby
the inductionhypothesis Dp [
T(as(I)’
...,as(n)), 0
e
but ,
(C)
It suffices to proveby
induction on a that ,(for
alle).
For a = 0 or a a limitordinal,
this is trivial. For a =J3+ 1, by
the definition ofdepth,
there is a reduced term a(xl,
...,xm)
and distincts(e) (e
=l,m)
such thatBy (ii)
and the inductionhypothesis Dp [
for e =l,m.
By (i)
and the definition ofh,
As T
(x I, ---, xn)
is reduced and thet(e)
aredistinct,
.{t(1), ...,
{h(s(1)), ..., h(s(m»} .
So foreach e * l,n
there iske’ 1 ~ ke 4
msuch that
t(e) - h(s(ke))
henceDp[ t(e), h] ~ Dp[ s(k.), h]
+ I ~{3
+ 1 = a .LEMMA 1.2. : Let
hl, h2
be functions from B into B which commute i.e.hi
oh2 = h2
ohl.
Then for any x e BPROOF We prove
by
induction on a thatFor a =
0,
or a a limitordinal,
it is immediate.For a = B + 1.
If
Dp [x, hI] ~ {3 +1,
then for some y eB, hi(y) =
x andDp [
y,So
hl(h2(y))
=h,
oh2(y)
=h2
ohl(y)
=h2(hl(y)) = h2(x),
andby
the inductionhypothesis {3 j3 )
henceLEMMA 1.3. :
freely generates
and let B = t e
J I -
Then we can find f EG~ ,
so that(A) if t 6 J
then a(B)
if g EG ,
g and f commute, and g maps B intoB,
then for every a a(0),
(C)
every function gfrom {at :
tE J}
into Bsatisfying
the condition from(B),
can beextended to an
endomorphism
ofG~ , commuting
with f andmapping
B into B(D)
f is the form mentioned in 1.1.PROOF :
By renaming
we can assumeI-J
=10 u { 0, t, 11 > :
t eJa ,
aa(0)
R, ( r~ )
> 0, 11 adecreasing
sequenceordinals,
11(U ) ~c a} U { 1, t,
n> : 0 nm I
and we
identify ( 0,
t,> )
and1,
t, 0 > with t. Let us define a function h on I :Let f be h as defined in lemma I ,I . It is easy to prove that if t E
J, t ( r~ ) =
n ,then
Dp [ 0, t , q
> , h] =
q(n-1).
Hence(A)
followsimmediately by
1.1(C),
andf e
G by
1.1(A).
As forpart (B),
if g e commutes withf,
g maps B intoB,
t e
J a
9 a a(0) ,
letg(at)
= a(as(1)’
...,as(n))
be reduced. As g maps B intoB,
e
B,
hencenecessarily s(e)
EJ
for e =1,
n ; so lets(e)
EJa (e). By
1.1(C)
and aremark above
On the other hand
by
lemma1.2,
as g commutes withf,
Combining
both weget a ~
a(e)
for e =l,n.
Henceg(at)
E C1{ at :
t Ej9
1a (0) 1 ,
so weproved (B).
As for
(C),
extend g to a function g, from{ at :
intoF. ) by :
It is easy to check that
g,
is well defined(because
t eJa , t(e)
eJ (3
=>a ~ (3)
and it has a
unique
extension tog~
eG .
In order to check thatg2
and f commute, it suffices to prove that for every s eI,
f og2(as) - g2
of(as)
which isquite
easy.Now
(D)
holdsby
the definition of f.§ 2. SIMPLE PROPERTIES EXPRESSIBLE IN FIRST.ORDER LOGIC.
LEMMA 2.1. : Each of the
following properties (in
a fullsubcategory K)
isexpressible by
formulas
(of L) :
(A) proj (f) -
which means f is aprojection (B)
f is anautomorphism, aut(f)
in short(C)
g is aprojection,
andRn(f) C Rn(g) (D)
g is aprojection, Rn(f trng) c Rn(g)
(E)
g is aprojection, Rn(fe) c Rn(g)
for e =1, 2,
3 and(F)
g, f areprojections
andRn(g)
=Rn(f).
REMARK :
Clearly
for any f EHom(A, B), Rn(f)
is asubalgebra
of B.PRnOF° : We
give
theexpression
or an indication of it in each case(A)fof
= fgot = lA) [1 A -
theidentity
of A defineby ( Yf,B)(f e Hom(B,A)-~
h o f =f) I
(C) proj (g) A
g o f = f.As g is a
projection,
x eRn(g) 0* g(x)
= x. Hence whenproj (g),
and f : B -A, g:A-~
A.(D ) proj (g) A g o f o g = f o g
(the proof
similar to theprevious one)
(F)
Immediateby (C).
CLAIM 2.2.
(A)
If B is the range of someprojection
f EG~ ,
then anyhomomorphism
h : B ~ B(B
considered as asubalgebra)
can’be extended to an h’ EG~
(B)
If B =where J c
Ifreely generates F~ ,
then ahomomorphism
h : B ~ B is onto B iff for somehomomorphism
g : B-~ Bh o g
isthe
identity.
PROOF.
(A) Clearly
h o f EG~
extend h and its domain isF~ ,
(B) Clearly
ifh o g
is theidentity (on B),
then for any y eB, g(y) e
B andh(g(y))
= y, hence h is into B.Let h be onto
B,
so for every t fJ, at
=h(T t(as 1 t , ...))
for some termT t
andJ.
There is aunique homomorphism
g : B +B, g(at) =
Tw). Clearly
for anyt e
J,
h o =’at
henceh o g
is theidentity.
REMARK.
Sabbagh
hadproved
that of is one-to-one» and of is onto » are(first-order)
definable in
Cat~ .
*§ 3. BEAUTIFUL TERMS.
DEFINITION 3.1.: The term T
(xl’
...,xn)
is called beautiful if(A)
For any term a(xl, ..., xm)
is an
identity (of F~ )
is an
identity
(C)
T(x, ..., x)
= x is anidentity.
REMARK : The beautiful terms for n > 1 cause us much trouble. For the free abelian group,
only
x is beautiful and reduced.However,
if F is a freealgebra
of the identitiesT(TI(x), T~(x))
= x,Tl(T(x,y))
= x andT2(T(x,y))=
y andT3(x,y) = T(Tl(x), T2(y»,
-then the identities which hold in
() ; T3)
define avariety
for whichT3(x,y)
is abeautiful term.
LEMMA 3.1.
(A)
The set of beautiful terms is closed undersubstitution,
i.e. if ,a
(xl,
...,xrJ,
Ti(xI’
...,xn(i)) (i =1,
...,n)
are beautiful terms, then so is(B)
x is a beautiful term, and there is no other beautiful term T(x) ;
and T
(xl, ..., xn)
=xi
is beautiful. ’(C)
Thetwo-place
beautiful terms( T(x,y)) generate by
substitution all the beautiful terms.PROOF.
(A)
Thechecking
has noproblems.
(B) By
the third demand in Def.3.1 ;
the secondphrase - by checking.
So we can prove our assertion
by
induction on n.DEFINITION 3.2. : The beautiful Boolean
algebra
of avariety
is the Booleanalgebra B
such that(1)
Its elements are beautiful terms of the form T(x,y) (x,y
here arefixed).
(2)
Its zero is y, its unit is x.(3)
The intersection is definedby
(4)
Thecomplement
is defineT’(x,y)
whereT (x , y)
=T ’(y,x).
DEFINITION 3.3. : For any filter T
of B,
letw. T
be the relation(defined
on elements ofmembers of
Cat)
and it is
similarly
defined on eachHom(A,B), A,
BE Cat(see
Def. 3.4(A)).
DEFINITION 3.4. :
(A)
Iffi :
A -~ B(in Cat), T
a beautiful term, T(f 1, "-, fn) :
A y B is definedby
(B)
IfAi (i
=1,
...,n) belong
toCat,
T a beautiful term, thealgebra
T ...,An)
is defined as follows :
its elements are a,, ...,
a,>
where ai cAi
and ...,an, = bl,
...,bn>
iff for each
i, ai ~ T.bi
whereTi
is the filtergenerated by
T(y,
..., y, x, y, ...,y)
’
1
(the
x-in the ithplace) ;
for a term a ,(C)
Iff i : Ai
-~Bi (in Cat)
T a beautiful term, thenis defined
naturally.
REMARK : There is an
ambiguity
in the definition of T ...,f~) ;
so when both definitionsn
are
meaningful
weprefer (A) ;
and even better : in(B) if A
1 or T(x~,..., xn)
=xi
is an
identity,
we makeT (AI, ..., An)
=Ai.
In factT (A1, ...)
isessentially
defined up toisomorphism.
NOT ATION: If x e =
x ,
..., x~ > then0 m-1
LEMM.B :3.2. :
(A)
The beautiful Booleanalgebra
isreally
a Booleanalgebra.
(B)
For any filter T of B and A ECat, ~
is anequivalence
relation overA,
and even acongruence
relation,
soA/~T
is defined and e Cat and T(x,y) E
Timplies
theidentity I (x,y)
= x holds inA/ -T-
(C)
For anyfi :
A - B(in Cat)
and any beautiful T , T(fi , ..., fn)
is ahomomorphism
from A into B. If A
= B, fi
aprojection
then T ...,fn)
is aprojection.
(D)
IfAi
ECat,
T isbeautiful
then T(Al,
..., Cat.(E)
Iffl : Ai ~ Bi
inCat,
T beautiful then(F)
If T " B is anultrafilter,
for thecategory
A cCat} ,
there are no beautifulterms. It is the
category
of thevariety,
whose identities are those of V and {T(x,y)
= x : T e T} . This holds for filters T too.PROOF :
Easy.
I,EMMA 3.3. :
( ~1)
For every Ti(x,y)
= T2(y,x),
letTe
be the filtergenerated by
Te(x,y).
Then anyA e Cat is the direct
product
of I and A/ In A/ = x is ane e
identity.
(B)
If ourvariety
V is the modules over aring R,
then the beautifultwo-place
terms areax +
(1-a)y
where a is in the center of R and isidempotent (i.e. a2
=a).
PROOF :
Easy.
LEMM 1B 3.~. : Let T be a beautiful term.
(A)
Let K 6 Cat be asubcategory,
such that forfi :
A -~ B inK,
T(f 1,
...,fn)
is in K.In the proper
language L* (i.e.
with variables forobjects,
and formorphisms,
and apartial
operation
ofcomposition,
and the relation f eHom(A,B) plus
ourT )
for every first order formula o0 (f l,
... ;A1, ,..), there
is a finite set o of formulas ofL *
with the same free variables such that :totally determine o* =
hence in
particular
the truth of oo(g 1, - - - ; B1, ...)
(iii)
ifin (ii) the 4l j ’s
areequal then = 4l j .
(B)
We canexpand L* by letting
in variables forelements,
and more terms of ourvariety ;
and also let in
(ii)
B,1i(l)
-~
; and
the same conclusionholds, provided
that we arecareful for the
question
whenREMARK : This is
just
a varient of FefermanVaught [ FV ] ,
as refinedby
Weinstein[W]
and G alvin [ G J .
PROOF:
(A)
We define 4lby
induction on the structure of ~ 0.(if T (fg)
is notdefined,
h = T(f,g)
isfalse)
It is now easy to prove
(ii) by
induction and(iii)
followstrivially by choosing
g~
= for alli, j .
i i
(B)
Left to the reader.§ 4. DEFINING ARBITRARY SETS WITH PARAMETERS.
THEOREM 4.1. : There is a formula
q,b
such that thefollowing
holds. Let K be a full. m
subcategory
ofCat,
andF Jl f K. Suppose A, B
eK, )B) ~ lu ,
andfi (i lu + )
is an
m-tuple
of members of Hom(A,B).
Then we canfind g (from K) such
that :for some beautiful term T and
i(1) ~ .., ~ i(n).
REM,4RK :
(1)
Thef ~
i are added as individual constantduring
theproof,
in order not to mentionthem
explicitely during
theproof.
In theend,
we include them ing.
PROOF : Let
{at :
t eI*} freely generate F , and
letJ
C1*, P .. jj I = j I 1* .
. Fornotational
simplicity
letJ = t a, ~i> : a -
=
,andi(0)=
p(as
we can allowrepeations).
MAIN LEMMA 4.2. : There is a formula o
(f),
such that[ f iff
there is a beautifulT = T
..., and ordinals a
(1),...,
a(n),
so that forevery B
p,f(a§ ) "
n r~
PROOF OF 4.1 FROM 4.2 :
Let
f* :
0 F - J1.A,
be such that itmaps {aa :
0 ap I
ontoA,
andPROOF OF 4.2 :
We
partition
ourproof
to three cases[ it
suffice to prove each oneseparately,
and thenby comining
the formulas andchoosing
theparameters
for each case, we caneasily
find aunique formula ] .
We shall
use §
2freely :
and restrict ourselves to F and G .Jl P
Clearly
there are first orderformulas ~
such that :( 1 ) ~ 1 (f )
says that ft B I
is intoB 1
and it commutes withB 1 (by 2.1, D, as f;
is aprojection,
andRnf* = B1)
’
’
7 7
(2) o2 (f)
saysthat o1 (f)
andf t B2
is intoB2
and it commutes withB2
4
(3) o
3(f)
saysthat o2 (f)
and for any g,o2 (g) implies
that fl B2and
(f
5 o g of5*)
5l B2
commute.(f; 0 f) t B
5 0 6 0
(5) o5 (f )
says that o 4(f)
and(f*
of) ) B = f* I B
2 0 2 0
(6) ~
6(f)
says that for somef’, f’ ~ B3 = f ~ B3 and ~ 5 (f’)
whereB3=
cl{ am : 0 m w lNow we notice ’
[ Clearly
the «if»part holds ;
for the otherdirection,
as 0 eB 1 ’
I=
T (aO
, ...,a09 )
for some T , k(i) ; apply f ~
n times and weget
our resultI ,
0
Q-(1) . (k)
3the «if»
part
isimmediate,
and for the«only
if» use(1 *),
andapply f ~
m times on the 4right
side ~T
(xl,
...,xk)
satisfies condition(A)
forbeing
beautiful.for every n,m Now for every term So
clearly
by § 3 ’s definition,
3
computing
weget
As this holds for any o , we prove that T satisfies condition
(A)
from Definition ~.1, The other direction in(3#)
should be easynow ],
, and the 1" from
(3*)
satisfies condition(B)
fromto the
equality
ofThis
clearly
isequivalent
to condition(B)
from Def. 3.1 on T, The other direction is easy.] (the P, (i)
aredistinct)
where[Because, assuming
f or some beautiful T , and 9,
(i)
w , for every n cj.[ Immediate, by (5), noticing f ~
5 7 5of ~
of ~
is aprojection
ontoB3] -
So we
clearly
finish case I.Case - II : u =
I J is regular >
N0. -o .
11
0
1
= p. , and let us denote = aR
> . where = u. But forusing
in caseIII,
° a
a,B,n>’
-i "we from now up to the end of the
proof
of claim4.3,
assumeonly 03BC1 +l p
, isregular.
For each limit choose anincreasing
sequence 5(n) (n w )
ofordinals 5 , whose limit is 6 ; such that for
6 ~.:j3= S(n)}
is a
stationary
subset of(see
e.g.Solovay [ So] ).
Let us define somefe’s , by defining
(t
e1*) understanding
that whenf* (at)
is notexplicitely defined,
it isao.
Solet,
e e 0
Clearly f*
areprojections
ontoB2 , B3
resp. and letf * , f * , f*
3 4 9 10 11 12
be
projections
ontoB4 , B5, B6
resp.’
Now we
apply
lemma1.3,
with f a, ~i > : a J.l.,~i
for,J,
and { a,B
> : a P i forJ ,
and1*
forI,
andget
eG
as mentioned there.0 13
,
Let the first order
formula (f,g)
1 says thatf,g
areconjugate
tof2,Rnf,
2Rng c B3
and there is h E G
commuting
withf*12,
andmapping B3
intoitself,
such that h o f = g.p 13
We shall
write (f g)
also in the form f ~ g. Soby 1. 3,
iff,g
areconjugate
tof*
2 andLet the first order
formula cp 2(f)
says thatf t B2 , f t B5 , f ~ BO’ f r B6
are intoB3, B~,B~,B~
resp.and for any g
conjugate
tof; ,
ifRng
C7B~ ,
theng 6 f o g ;
andf t B5
commutewith
f , f , f
and fB
commute withf .
7 8 9 3 3
*
CLAIM 4.3 :
[f ]
iff foreach 0 p. 1
for some T , a
(e) (which
do notdepend
on(3 !)..
PROOF: Assume
[ f ] ,
and leta
( (3 ,e)
increase withe,w.l.o.g. By part of ~2 saying :
gconjugate
tofRng 5; B 2 implies g ’ f o ;
it follows that(choose
g such thatg(at) = ap
g g t 0
As 03BC1
isregular, 03BC1
> N0 for anyj8
p001I
sup {y
(13 ,e) :
e =1, ..., k(13 ), 13 j3 }
henceS
= { ~i 0
:00 J.l 1
; and(3 ~i 0 ,1 ~ {3 0 }
is an unbounded subset
of p ;
andby
its definition it is closed.Now we shall prove that for 6 e
S,
cf 6= N 0 implies y ( 8 ,e) =
6 . Forsuppose
’Y = ’Y (8 ,e ) :/= 8
then as said above 6 y(6 ,e ). As 8 (n)
isincreasing (as
a function ofn)
0 0 1
and its limit
is 5 , (or 6(n)
is notdefined)
for some n E wbig enough,
5 7(b ,e ) (n),
and 110y
(8,
7e2) y (6 ,e ) (n) 7 ( 5 e2) (n) (when they
aredefined).
1 2 1 2
As
f l
B is into Bnecessarily
y(6 ,e)
hascofinality N0,
and asf lB5
commutes withf*
67 . 05 7
As
f ~
B commutes withf*
5 9
and T
(a -t( 6 1) (n)
...)
is reduced(by the
choice of T and ofn).
5
a(S, 1) )
6But
f(a0 (n)
isequal
also to T5 (n) a~ (S (n),l) ...)
which is reduced too.Hence y e 0) (n)
e{y ( 6 (n),e)
1 ek(8 (n))}
0
but on the one hand 6 7
(6 e 0) (n) by
the choice ofe0 and
n,0 0
and on the other hand 8
(n) d = y ( 6 (n),e)
8 as 8 cS,
contradiction.Hence y (ð ,e)
= 5 for every 6 c S.Now for every
{3 p 11
n we know { 6 p : cfð = NO’
ðn) " {3 }
is
stationary,
so there is 8 ES,
cf 8 = ~0 such
that 6(n) = (3
.As before we can show that
and the last is reduced.
Hence y (b (n),e)
e f 6(n)}
for each e, hence y( b (n),e) =
5(n),
thatis y (B ,e)
for
each
and e. ’Hence,
’ as T(a «(3 1 ) ...)
isreduced,
the ordinals a(o e 0 (
a 1 ’ )
’1 ~ e ~
k( (3 )
are distinct.As
f t
B commutes withf* ,
for everyj3
- -
3 3
As the a
«(3 ,e)
aredistinct, necessarily
{ a(j3 ,e) :
1 ~ e -’k(e)} _
{a
(o,e)
1 e 4k(O) },
but as a(j8 ,e)
isincreasing
with e(for each B, by
the choice ofT0( , ...,)) necessarily
a({3 ,e)
=a (o,e), k({3 )
=k(O)
and we can assume thatT ~
=T O.
Now
clearly
as beforeSo we finish one direction of claim 3.3 where as the other is immediate.
Let 6 (f)
say that f mapsB 2
intoB 3
and ifthere . is f1,f1 lB2 f lB2
It is easy to check that for
every 9 ;
for some T , 7(e) :
Let o4(f)
say that f mapsB2 into B3, o 3 (f ),
andif o3 (g),
thenf*
of t a-0
=f* [ a-0 . (note
that7°
= Rnf* ).
As in case I we can check that2 0 2 0 0 2
and T is beautiful.
0
Case
III : p
asingular
cardinal.We
give
this case with less details.We let p I p
and
B , B ,
B be as in caseII,
J.lregular ,
>NO. By
case II1 1 2 3 1 1 0
we can define
f*’s
eproperly,
sothat,
forsome o0 r- f= #°
~À-(f)
iff there are T anddistinct a , ..., a so that for each J.l ,
f(J3 ) ..., a13 ).
1 n 1 0
aI
a nLet
/> 1 (f) ( )
say Y thatf(a0
!: B and for everyg, o
°(g) implies
that2 Y
g, o0 (g)
P(f
0g) t lio = (g
of) t a-0 .
It is easy to check thatG. t-- /> 1 [ f iff
there are and
0 0 A -
distinct
13 1 , ( )
...,13 (m) ( )
such that for every Y a ,(
= g(a13 ( 1 ) , ..., a13 (m) ,
a
(
a a
)
a
satisfying (A)
of D ef . 3.1.As JJ. 1 is
regular,
we can use case II. So thereis # 2
such thatG
2[ f I
iff there area beautiful term a and distinct
0 i )
such that for every Y a , f(ao
(a) a(i (I),
a ,..a0
a(m) )~
Let
Y- p(i),
J.l(i)
p , ~(i) increasing.
°
i
We
just
prove that for every 7cfp thereisi*
as constructed in case II andabove,
T such that
(i) Gx
1BF w 2 [f, f*] ]
iff there are a beautiful term a anddistinct 0 (i)
p(1) ,
T -
such that for every a
f(a0a)
a = g(a 9
a(’)1
..., a(ii) f* (0)
is aprojection
onto C1 : a ii,B 03BC (y)+}
~ ~ a
Checking
the constructioncarefully,
we see that :if T, g are beautiful terms,
j3 (i) p (’y R, )j- (i = 1, ..., m, R, = l, ..., n)
Now
checking
theproof
of 4.1 from4.2,
the existence of thef*
mentionedabove,
’Y is sufficient to prove 4.1 when
I(0)
03BC.Hence there are
p3
andg *
such that[f ,g~ ] if f
f = T(f 7 (1)
, ...,f* 7(n) )
f or some beautiful T . So let the formula o4 [ f,g *] ] say «there is f 1 ’
suchthat o3 [f 1 ’ ]
and for everyf2
which satisfies~ 3 [f2 ’ [* j A
Rn Rnf2 (0)
satisfies also1>2 [f, f2 ]
».Clearly G ~ ~4 [f, g*] ]
iff for some beautiful Tand (3 (i) ~ , for
every a ,f(aO) =
T(a (3 (1), ...) (f*
o f of* , g*)
is our desired formula.(a
(
a
~
)
55
5~g )
THEOREM
4.4:
We can find ’(in
thelanguage
ofG )
A such thatfor any ultrafilter and free set of
generators {at :
t EF
andft
e F X definedby ft(as)
=at ;
there isf
eG
such thatf U = ft
for some t andx X 0
(A) G F= 1>b
X[ f, f* ]
iff f = T(ft, ...)
for some beautiful T . .1
(B) for
anyf,g, (see
Def.3.3)
-(D)
For anytwo-place
relation R on the classof ~ eq -equivalence classes,
thereis g
e Gsuch that for any
f 1, f 2
eG ~
Gx
is aprojection
onto some Cl h...) :
A A /
(for
some n w, beautifulT e ,
andt(e,1 ), ...
c1*.
P ROO F : The set
R = {
f,g> : f,
g c...) :
Lbeautiful, t(e)
e1*1
and for some T