Saharon Shelah- On Long EF-Equivalence in Non-Isomorphic Models

8/3/2019 Saharon Shelah- On Long EF-Equivalence in Non-Isomorphic Models

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ON LONG EF-EQUIVALENCE IN NON ISOMORPHIC

MODELS SH836

SAHARON SHELAH

Abstract. There has been a great deal of interest in constructing mod-els which are non-isomorphic, of cardinality , but are equivalent underthe Ehrefeuch-Fraisse game of length , even for every < . So underG.C.H. particularly for regular we know a lot. We deal here with con-structions of such pairs of models proven in ZFC, and get their existenceunder mild conditions.

The author would like to thank the ISF for partially supporting this research.First typed: 2 Sept 2003 and last revised in May 25, 2004. Publication 836.

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0. Introduction

There has been much work on constructing pairs of EF,

-equivalent non-isomorphic models of the same cardinality.

In Summer of 2003, Vaanenen has asked me whether we can provably inZFC construct a pair of non-isomorphic models of cardinality 1 which areEF-equivalent even for like

2. We try to shed light on the problem forgeneral cardinals. We construct such models for = cf() = 0 for every < simultaneously and then for singular = 0 . In subsequent work[HvSh 866] we shall investigate further: weaken the assumption = 0(e.g., = cf() > ) and we generalize the results for trees with no -branches and investigate the case of models of a first order complete T(mainly strongly dependent). We thank Chanoch Havlin and the referee fordetecting some inaccuracies.

Definition 0.1. (1) We say that M1, M2 are EF-equivalent if M1, M2are models (with same vocabulary) such that the isomorphism playerhas a winning strategy in the game 1 (M1, M2) defined below.

(1A) Replacing by < means: for every < ; similarly below.(2) We say that M1, M2 are EF,- equivalent when M2, M2 are models

with the same vocabulary such that the isomorphism player has awinning strategy in the game (M1, M2) defined below.

(3) For M1, M2, , as above and partial isomorphism f from M1 intoM2 we define the game (f, M1, M2) between the player ISO andAIS as follows:(a) the play lasts moves

(b) after moves a partial isomorphism f from M1 into M2 ischosen increasing continuous with

(c) in the + 1-th move, the player AIS chooses A,1 M1, A,2 M2 such that |A,1| + |A,2| < 1 + and then the player ISOchooses f+1 f such that

A,1 Dom(f+1) and A,2 Rang(f+1)

(d) if = 0, ISO chooses f0 = f; if is a limit ordinal ISO choosesf = {f : < }.

The ISO player loses if he had no legal move.(4) If f = we may write (M1, M2). If is 1 we may omit it. We

may write instead of +. The player ISO may be restricted tochoose f+1 such that(a)(a Dom(f+1) a / Dom(f) a A,1 f+1(a) A,2)

1. The Case of Regular = 0

Definition 1.1. (1) We say that x is a -parameter if x consists of(a) a cardinal and ordinal (b) a set I, and a set S I I (where we shall have compatibility

demand)


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ON LONG EF-EQUIVALENCE IN NON ISOMORPHIC MODELS SH836 3

(c) a function u : I P(); we let us = u(s) for s I(d) a set J and a function s : J I, we let st = s(t) for t J and

for s I we let Js = {t J : st = s}(e) a set T J J such that (t1, t2) T (st1 , st2) S

(1A) We say x is a full - parameter if in addition it consists of:(f) a function g with domain J such that gt = g(t) is a non-

decreasing function from us(t) to some <

(g) a function h with domain J such that ht = h(t) is a non-decreasing function from us(t) to such that

(h) if t1, t2 J and st1 = s = st2 , gt1 = g = gt2 and ht1 = h =ht2 ,

t1 = = t2 then t1 = t2 hence we write t = ts,g,h =

t(s,g,h).

(2) We may write

=

x , = x, I = Ix, J = Jx, Js = J

x

s , t

(s,g,h) =t,x(s,g,h), etc. Many times we omit x when clear from the context.

Definition 1.2. Let x be a -parameter.

(1) For s Ix, let Gxs be the group

1 generated freely by {xt : t Js}.(2) For (s1, s2) Sx let Gs1,s2 = G

xs1,s2 by the subgroup ofG

xs1 G

xs2

generated by

{(xt1 , xt2) : (t1, t2) Tx and t1 Jxs1, t2 J

xs2}

(3) We say x is (, )-parameter if s Ix |us| < .

Remark 1.3. (1) We may use Sa set ofn-tuples from I (or (< )-tuples)

then we have to change Definitions 1.2(2) accordingly.Definition 1.4. For a -parameter x we define a model M = Mx as follows(where below I = Ix, etc.).

(A) its vocabulary consist of() Ps, a unary predicate, for s Ix() Qs1,s2, a binary predicate for (s1, s2) Sx() Fs,a, a unary function for s Ix, a G

xs

(B) the universe of M is {(s, x) : s Ix, x Gxs}

(C) for s Ix let PMs = {(s, x) : x G

xs}

(D) QMs1,s2 = {((s1, x1), (s2, x2)) : (x1, x2) Gxs1,s2)} for (s1, s2) Sx

(E) ifs Ix and a Gxs then FMs,a is the unary function from P

Ms to P

Ms

defined by FMs,a(y) = ay, multiplication in Gxs (for y M \ PMs wecan let FMs,a(y) be y or undefined).

Remark 1.5. We can expand Mx by the following linear order: let


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4 SAHARON SHELAH

is an ordered group, exists as ??Fxs is free and let


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Claim 1.10. Let x be a full -parameter s Ix and c1, c2 PMs , c

Gsand F

Mxs,c(c1) = c2. A sufficient condition for (Mx, c1), (Mx, c2) are EF,-

equivalent where x, is the existence of R, I, c such that:

(a) R is a partial order,(b) I = Ir : r R such that Ir Ix and r2 R r2 Ir1 Ir2(c) R is the disjoint union of R : < , R0 = (d) c = cr : r R where cr CIr and r1 r2 = c

r1 = cr2 Ir1and crs = c

so s {Ir : r R}(e) if r : <

is R-increasing, < r R and < then it has an R-ub from R

(f) if r1 R, + 1 < and I I, |I| < then (r2)(r1 r2

R+1 I Ir2).

Proof: Easy. Using 1.9(1),(5). 1.10

Claim 1.11. (1) Let x be a -parameter and I Ix. A necessary andsufficient condition for MxI is Ps-rigid is:

1 there is no c CxI with cs = eGs .

(2) Letx be a full -parameter and assume that s() Ix, < x,

for notational simplicity andt Jxs(). The models M1 = (M, (s, eGs)), M2 =

(M, (s, xt)) are EF,-equivalent when:2, (i) is regular, s Ix |u

xs| <

(ii) if s Ix and g Gx and uxs Dom (g) then t

,xs,gus,hgus

is well defined

(iii) if (s1, s2) Sx and t1 = ts1,g1,h1, t2 = ts2,g2,h2 are welldefined then (t1, t2) Tx when for some g Gx we havegt1 gt2 g and h1 h2 hg

(iv) t = t,xs(),g,hg

where g : us() {0} and hg is constantly

= {+ 1 : us()}.

Proof

(1) Toward contradiction assume that f is an automorphism ofMxI such

that f PMxs is not the identity. By 1.9(4) for some c C

xI we have

f = fc. So fc PMxs = f P

Mxs = id hence by 1.9(1) we have

cs = eGs , contradicting the assumption 1.(2) We apply 1.10. For every i < and non-decreasing function g

Gx from some ordinal = g into i we define cg = c

g,s : s

Igp, cg,s = (s, xtg,s), t

g,s = t

s,gus,hgus

. Let Ri = {g : g a non-

decreasing function from some < to 1+i such that , g

is constantly zero, < g() = 1} and let R = {Ri : i < }ordered by inclusion. Let I = Ig : g R and c = c

g : g R. It

is easy to check that (R, I, c) is as required. 1.11


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Claim 1.12. (1) Assume = cf() = 0 . Then for some full(, 1)-parameter x we have |I| = = |J|,

x =

and condition

1 of 1.11(1) holds and for every s() Ix\{} condition2, of1.11(2) holds whenever < .

(2) Moreover, ifs Ix\{} then for some c1 = c2 PMxs and(M, c1), (M, c2)

are EF,-equivalent for every < x but not EFx ,-equivalent.

Claim 1.12(1) clearly implies

Conclusion 1.13. (1) If = cf() = 0 , then for some modelM of cardinality we have:(a) M has no non-trivial automorphism(b) for every < for some c1 = c2 M, the model (M, c1), (M, c2)

are EF-equivalent and even EF,-equivalent.

(2) We can strengthen clause (b) to: for some c1 = c2 for every < the models (M, c1), (M, c2) are EF,-equivalent.

Proof of 1.12: 1) Assume > for notational simplicity. We define x by(x = and):

(a) () I = {u : u []0}() the function u is the identity on I() S = {(u1, u2) : u1 u2 I}() x =

(b) () J is the set of quadruple (u,,g,h) satisfying(i) u I, <

(ii) h is a non-decreasing function from u to

(iii) g is a non-decreasing function from u to (iv) if 1, 2 u and g(1) = g(2) then h(1) = h(2)(v) h() >

() let t = (ut, t, gt, ht) for t J so naturally st = u,gt = g

t, ht = ht

() T = {(t1, t2) J J : t1 = t2 , ut1 ut2, ht1 ht2

and gt1 gt2}.

Now

()0 x is a full (, 1)-parameter[Why? Just read Definition 1.1 and 1.2(3).]

()1 for any s() I\{}, x satisfies the demands for 2,(i), (ii), (iii), (iv)

from 1.11(2) for every <

[Why? just check]()2 if u1 u2 I, we define the function u1,u2 : Ju2 Ju1 by

u1,u2(t) = (u1, t, gt u1, h

t u1) for t Ju2,[Why is u1,u2 a function from Ju2 into Ju1? Just check]

()3 for u1 u2 we have() T (Ju1 Ju2) = {(u1,u2(t2), t2) : t2 Ju2} hence() Gu1,u2 = {(u1,u2(c2), c2) : c2 Gu2} where u1,u2 Hom(G

xu2 ,G

xu1)

is the unique homomorphism from Gxu2 into Gxu1 mapping xt2


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to xt1 whenever u1,u2(t2) = t1[Why? Check.]

()4 if u1 u2 u3 I, t3 Ju3 and t = u,u3(t3) for = 1, 2 thengt1 , gt2 are compatible functions as well as ht1 , ht2 and

t1 = t2

moreover gt1 gt2 is non-decreasing, ht1 ht2 is non-decreasing[Why? just check]

()5 clause 1 of 1.11(1) holds for I = I, s() I \ {}

[Why? Assume c CxI is such that cs() = eGs(). For each u I, cu is a

word in the generators {xt : t Ju} ofGu and let n(u) be the length of thisword and m(u) the number of generators appearing in it.Now by ()3 we have u1 u2 n(u1) n(u2) m(u1) m(u2). As (I, )is 1-directed, for some u I we have u u I n(u) = nm(u) = mand let cu = (. . . , x

i()t(u,)

, . . .) 0.

As {u : u u I} is directed, by ()4 above, for each < n any twoof the functions {gt(u,) : u u I} are compatible so g =: {g

t(u,) :u I} is a non-decreasing function from = {u : u I} to andh =: {h

t(u,) : u u I} is similarly a non-decreasing function from to . It also follows that for some we have

=:

t(u,) whenever

u u I in fact =

t(u,) is O.K. For each i Rang(g) choose

,i < such that g(,i) = i and let E = { < : a limit ordinal> sup(u) such that i <

& < n & i Rang(g) ,i < and

< & < n h() < }, it is a club of . Choose u such that u uand Min(u\u) =

E.Now what can g( Min (u\u)) be?

It has to be i for some i < < hence i Rang(g) so for some

u1, u u1 and ,i u1 so h(,i) < hence considering u u1

and recalling clause ()(vi) of (b) from definition of x in the beginning ofthe proof we have h(,i) < h(

) hence by (clause (b)()(v)) we havei = g(,i) < g(

), contradiction.]2) A minor change is needed in the choice of Tx

Tx = {(t1, t2) : (t1, t2) J J and ut1 ut2 , ht1 ht2 , gt1 gt2 ,

t1 t2 and if Rang(gt1) {0} then t1 = t2}.

1.12


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2. The singular case

We deal here with singular = 0 and our aim is the parallel of 1.13constructing a pair of EF-equivalent for every < non-isomorphic modelsof cardinality . But it is natural to try to construct a stronger example:This is done here:

for each < = cf(), in the following game the ISO player wins.

Definition 2.1. (1) For models M1, M2, and partial isomorphism ffrom M1 to M2 and < cf() we define a game ,(f, M1, M2). Aplay lasts moves, in the < move a partial isomorphism f wasformed increasing with , extending f, satisfying |Dom(f)| < . Inthe -th move if = 0, the player ISO choose f0 = f, if is a limitordinal the ISO player chooses f = {f : < }. In the + 1 < move the player AIS chooses < and then they play a sub-

game 1 (f, M1, M2) from 0.1(3) producing an increasing sequence

of partial isomorphisms fi : i < and let their union be f+1.ISO wins if he always has a legal move.

(2) If ISO wins the game (i.e. has a winning strategy) then we sayM1, M2 are EF

,-equivalent, we omit if clear from the context. If

f = we may write ,(M1, M2)

Remark: For (M, c1), (M, c2) to be EF 0, = i : i < is increasingcontinuous with limit , 0 = 0, 1 = (= cf()), i+1 is regular >

+i and

let = and for < let i() = Min{i : i < i+1}.

Definition 2.3. Under the Hypothesis 2.2 we define a -parameter x = xj,as follows:

(a) () I is the set of u [ \ ]0

() u : I P( \ ) is the identity,() S = {(u1, u2) : u1 u2 I}() x = j

(b) J is the set of tuples t = (u,j,g,h) = (ut, jt, gt, ht) such that

() u I() j < j() (i) g is a non-decreasing function from ug = uvg to where

vg = {i() : u and g() = +i()}

(ii) u g() [i(), +i()

]

(iii) if i vg then g(i) < j t(< = 1)(iv) vg is an initial segment of {i() : u}

() (i) h is a non-decreasing function with domain ug vg


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(ii) u h() [i(), i()+1] and ifi vg then h(i) < (iii) if1 < 2 are from ug vg and i(1) = i(2) then g(1) =

g(2) h(1) = h(2)(iv) < h() for ug vg and g() =

+i()

h() =

i()+1 for u(c) T is the set of pairs (t1, t2) J J satisfying

(i) ut1 ut2 I and(ii) gt1 gt2 , ht1 ht2 , jt1 = jt2

Observation 2.4. x = xj, is a full -parameter.

Proof: Read the Definition 1.1(1)+1.1(1A)

Claim 2.5. Assume s Ix, c1 = (s, eGs), c2 = (s, xt), t Js, and for simplic-ity Rang(gt[1+i, 1+i+1)) {1+i}, Rang(g

t) = {0} and < jt < j.Then (Mx, c1), (Mx, c2) are EF,jt-equivalent.

Proof: So t, jt are fixed. For i < , j < j let

(a) Bi = { : = i : i < and i i i+1 and 0 = i and(1+i = 1+i+1 1 + i < i)}

(b) for Bi let A = {[i, i) : i < } which by our conventions isequal to i

{[j , j+1) : 1 j < i}

{[i, i) : i [i, )}

(c) for Bi let Gj,i, = {g : g is a function from A to , non-decreasing and the function g is into j and the function g[1+i, 1+i+1)is into [i,

+i ] and 1 i < i ()(i < i+1 g() =

+i )}

(d) for g Gj,i, Bi we define hg : A as follows: if A

then h() = Min{ i(): if i() > 0 g() = +i() then =i()+1, otherwise

[i(), i()] and = i() g() < g(

)}

(e) Gj,i = {Gj,i, : Bi} and Gj = {Gj,i : i < }

Let R = Gjt and for g R let i(g) be the unique i < such that g Gjt,iand g the unique Bi such that g Gjt,i(g), and = i(g) : i <

On R we define a partial order g1 g2 g1 g2 hg1 hg2For g R we define Ig, cg as follows

(a) Ig = {u I : u Dom(g) \ }(b) cg = cg,s : s Ig(c) cg,s = xtg(s) where tg(s) = (s,j,gug,s, hgug,s) where ug,s =

u {i() : u and g() =

+

i()}Let g G1 be chosen such that for i > 0, i(g) = sup({g

t() : ut [i, i+1)} {i}) and 0(g) = {i() + 1 : u

t and gt() = +i()

} {1}.

Let c = cg and f = fxc is the partial automorphism of M with domain

{PMxu : u Ig} from Definition 1.7. We prove that the player ISO wins

in the game ,j(f, M1, M1), as f(c1) = c2( PMxut

) this is enough. Recallthat a play last j moves; now the player ISO commit himself to choose inthe < j move on the side a function g G1+, increasing with , g0 = g


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and his actual move f is fxc

where c = cg . For the -th move if = 0 or

limit let g = {g : < }g G1+. In the (+1)-th move let the AIS

player choose < . Now the player ISO, on the side, first choose i < such that i(g) < i, and i > , second he chooses g

+ G1++1,i

satisfying:

(a) g+ extends g,

(b) Dom(g+ ) = i(c) g+(i \ Dom(g)) is constantly 1 +

(d) if 0 < i Dom(g) then g+[i, i+1) = g[i, i+1)

(e) ifi / (Dom(g)) and i Dom(g+ ) then Dom(g

+[i, i+1)) =

[i, i+1) and [i, i+1) \ Dom(g) g+ () =

+i

(f) ifi < ,i / Dom(g+ ) then g+[i, i+1) = g[i, i+1)

Now ISO and AIS has to play the sub-game 1 (f, M1, M2). The playerISO has to play f, in the -th move for and on the side hechooses g, G1++1 with large enough domain and range, to make it alegal move, increasing with , and g,0 = g

+ and g,i = g

+i . Now

obviously {g : g G1++1, g+ g} is closed under increasing union of length

< i , it is enough to show that he can make the ( + 1)-th move which istrivial so we are done. 2.5

Claim 2.6. Mx is Ps-rigid for s I.

Proof: We imitate the proof of 1.12.

()0 x is a full (, 1)-parameter()1 if u1 u2 I, we define the function u1,u2 : Ju2 Ju1 by

Fu1,u2(t) = (u1, jt, gt u1, h

t u1) for t Ju2,()2 if u1 u2 u3 are from I then u1,u3 = u1,u2 u2,u3 that is

u1,u2(t) = u1,u2(u2,u3(t))()3 for u1 u2 we have

() T (Ju1 Ju2) = {(u1,u2(t2), t2) : t2 Ju2}() Gu1,u2 = {(u1,u2(c2), c2) : c2 Gu2} where u1,u2 Hom(G

xu2 ,G

xu1)

is the unique homomorphism from Gxu2 into Gxu1 mapping xt2

to xt1 whenever u1,u2(t2) = t1[Why? Check.]

()4 if u1 u2 u3 I, t3 Ju3

and t = u,u3

(t3) for = 1, 2 then,recalling Definition 1.1(1A)(h), gt1 , gt2 are compatible functions aswell as ht1 , ht2 and jt1 = jt2 moreover gt1 gt2 is non-decreasing,ht1 ht2 is non-decreasing[Why? just check]

()5 clause 1 of 1.11(1) holds for I = I(= Ix)

Why? Assume c CxI is such that cs() = eGs() for some s() I. For each

u I, cu is a word in the generators {xt : t Ju} ofGu and let n(u) be thelength of this word and m(u) the number of generators appearing in it.


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Now by clause () of ()3 we have u1 u2 n(u1) n(u2) m(u1) m(u2). As (I, ) is 1-directed, for some u I, n < and m < we

have u u I n(u) = nm(u) = m and let cu = (. . . , xk(u,)t(u,) , . . .) 0 and letk() = k(u, ).

As {u : u u I} is directed, by ()4 above, for each < n any two ofthe functions {gt(u,) : u u I} are compatible so g =: {g

t(u,) : u I}is a non-decreasing function from Yi() to where Yi() = ( \ ) i()

for some i() and h =: {ht(u,) : u u I} is similarly a non-

decreasing function from Yi() to . Also g maps [i, i+1) into [i, +i ] for

i < and maps to .Case 1: i() = .

It also follows that for some j we have j =: j

t(u,) whenever u u I

in fact j = jt(u,) is O.K. and j < j . For each i Rang(g)

choose ,i < such that g(,i) = i and let E = { < : a limit ordinal> sup(u ) such that i < j

& < n & i Rang(g) ,i < and

< & < n h() < }, it is a club of . Choose u such that u uand Min(u \u) = E.

Now what can gt(u,)( Min (u\u)) be?It has to be i for some i < j < j

hence i Rang(g) so for some u1, u u1

and ,i u1 so h(,i) < hence considering u u1 and recalling

clause ()(iv) of (b) from definition 2.3 of x we have h(,i) < h() henceby (clause (b)()(iii)) we have i = g(,i) < g(

), contradiction.

Case 2: i() = so i() < .Clearly if i (i(), ) and [i, i+1) then g() =

+i (see clause

(b)()(iii) of Definition 2.3) hence g[i, i+1) is a non-decreasing functionfrom [i, i+1) to

+i , but i+1 is regular >

+i (see Hypothesis 2.2) hence

g[i, i+1) is eventually constant say i [i, i+1) and g[i, i+1) isconstantly i [i,

+i ). So also h[i,

+i+1) is constant and its value is

< i+1, and we get contradiction as in case 1.2.6

Conclusion 2.7. If = 0 > cf() > 0 then for every < cf() there arenon-isomorphic models M1, M2 of cardinality which are EF

,-equivalent.

Proof: By 2.5+2.6 as the cardinality of Mx is . 2.7


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Remark 2.8. By minor changes, for some t PMu , u = letting c1 =eGu, c2 = xt we have: (Mx, c1), (Mx, c2) are non-isomorphism but EF

,j -

equivalent for every j < = cf(). This is similar to the parallel remark inthe end of 1.


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Private Appendix

3. For every large enough

Naturally we would like to prove this for all are at least in some sense formost . Naturally, for me at least we do it by using the RGCH (the

revised G.C.H., see [Sh 460] or [Sh 829, 1]). Specifically, this holds forevery , moreover we phrase a weaker condition which conceivably??is provable in every 20. So instead every countable u and function g

from u . . . we shall try to use for density means?? So this leads to thefollowing.

Conclusion 3.1. Like 1.12 (hence also 1.13) assuming just = cf() > or at least

there is P []0 of cardinality such that (A [])(u P)(u A).

Proof: We define y = y as in the proof of 1.12 see there except that[]


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14 SAHARON SHELAH

|B| < [why? otherwise we can find B for < , pairwise distinct.

So for < there are v I, t z+v and be 1,, 2, v such thatht(1,) = and 1, < < 2,. As is regular without loss ofgenerality (ht(1,), h

t(2,)) = (1 ,

2 ) and h

t(1,) = .Let (w, t

) be such that v w I, t

zw and v,w(t

) = t.

By the assumption we know that for some , || = 0 andw = {w : } I. Now for each there is s z

+v

such that w,w(s) = t. But = s = s, so we get a

contradiction.]

So we can find < such that

2 if 1 [, ) then for no , 2 and u I, t z+u do we have 1, 2

u, 1 ht(1) < 2

We can find u1 I such that u1 u u1 hence zu1 = and lets zu1 , = h

t() and let u2 I be such that u1 {+ 1} u2 I, sothere is t Zu2 such that u1,u2(t) = s hence

ht() = hs() = < + 1 u2 so (u2, , + 1) witness then

Bht(),ht(+1) B, contradiction. 3.1

Conclusion 3.2. Like 2.7 assuming only cf() > 0 and > cf() > 0or just: there is P []

0 of cardinality such that

(a) if for every A of cardinality there is u A, u P(b) for every A cf () of cardinality there is u A, u P

TO BE FILLED : singular.


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4. Havning trees instead <

When <


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16 SAHARON SHELAH

5. On 0-independent theories

Our aim is to prove

ifT T1 are complete first order theorem T with the 0-independenceproperty, = cf() > |T| then(a) there are M1, M1 P C(T1, T) of cardinality which are EF,-

equivalent for every < but not isomorphism.(b) the singular.(c) Karp complexity.

Program:

We use EM(I, ), I Korgr = class of ordered graphs of cardinality .From a nice -parameter p, we drive a model N Korgr as follows: for

each Gps we attached Nps and the action of x G

ps and define the graph of

Np {Nps

: s S} such that the partial automorphism of Mp i.e.e = cs : s set induce a partial automorphism of the ordered graph.

So the problem will be to make M1 M2. Better: from one -parameter pwe define two ordered graphs Nps,1N

ps,2 and partial automorphism of each+

partial isomorphism from one to the other- those are the really interestingobjects.


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Remark: Note that J Koi we can use PJ only in particular definingEM(J, )

Definition 5.1. 1)Koi is the class of structures J of the form (A,Q ,P

Documents

Saharon Shelah- On Long EF-Equivalence in Non-Isomorphic Models