Saharon Shelah- Can the Order Determine the Addition for Models of PA

8/3/2019 Saharon Shelah- Can the Order Determine the Addition for Models of PA

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CAN THE ORDER DETERMINE THE ADDITION FOR MODELS

OF PA

SH924

SAHARON SHELAH

Abstract. We are interested in the question of how much the order of a non-standard model of PA can determine the model. In particular, for a modelM, we want to find a minimal set of pairs of non-standard elements (a, b)such that the linear orders {x : x < a} and {x : x < b} are isomorphic. It isproved that this set includes all pairs (a, b) satisfying the condition: there is anelement c such that for all standard n, cn < a,cn < b,a < bc and b < ac. We

prove that this is optimal, because if1 holds, then there is M of cardinality1 for which we get equality.

0. Introduction

Let M be a model of Peano Arithmetic (PA). For an a M, by M


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2 SAHARON SHELAH

We discussed those problems with Gregory Cherlin and he asked:{h.5}

Question 0.4. [Cherlin] Show that {M{


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CAN THE ORDER DETERMINE THE ADDITION FOR MODELS OF PA SH924 3

1. Somewhat rigid order

We define (in 1.2) some equivalence relations EM for models M (of PA). Weshall deal with their basic properties in 1.4, 1.12, the relations between them (in2.1), cofinalities of equivalence classes (in 1.11, 2.5), on order isomorphism/almost{


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4 SAHARON SHELAH

Remark 1.5. Note that 2.4 is better than 1.4(2) but the proof of 2.4 uses 1.4(2),also the proof of 1.4(2) applies to weaker versions of PA than the proof of 2.4, see

4.Proof. 1) Let = 3. If a1E3Ma2 and let c witness it, then c witnesses also a2E

3Ma2

and a2E3Ma1, so reflexivity and symmetry holds.

Lastly, assume M |= a1 < a2 < a3; if akE3Mak+1 and let ck witness thisfor k = 1, 2, then the product c1c2 witness a1E3Ma3 by part (1A) proved belowand if a1E

3Ma3 then the same witness gives a1E

3Ma2 a2E

3Ma3 so transitivity and

convexity holds.For = 1 the proof is similar (using part (1B) instead of part (1A)), also for

= 0, 2, 4 the proof is even easier and for = 5, 6 it holds by the definition.1A), 1B) Check.2) Without loss of generality, assume that a < (n 1)a < b < na, where 1 n N,and also there is a c such that (n 1)c = na b. Then c < a because otherwise

(n 1)a (n 1)c = na b hence b a, contradiction. Let X be a set ofrepresentatives for M/E0 and, without loss of generality, assume a, c X. Nowdefine f : M M by first defining it on X and then extending it to all of M inthe obvious way. (The obvious way is: if y = x + k, where x X and k Z, thenf(y) = f(x) + k and f is the identity on N.) Ifx X, then let

f(x) =

x if x c,

n(x a) + b otherwise.

Clearly, f(c) = c and f(a) = b. Now check that f is as required.3) First, if f exemplifies aE5Mb, i.e. is an automorphism of M mapping a to b thenfM


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{a15}Remark 1.8. 1) We have not said by the same isomorphism.

2) The assumption is too strong to be true but it makes sense for weaker versionsof PA, see 4.5 and part of the proof serves as proof to 2.7 so indirectly serves 2.6.{a17}

Question 1.9. Are M1, M2 isomorphic when (the main case is = 3):

(a) M1, M2 are isomorphic as linear orders

(b) M1 is -o.r.{a19}

Remark 1.10. In 1.9, it is less desirable but we may consider adding that also M2is o.r.

Proof. Proof of 1.7Easily by 1.4(1)

()0 each E2M -equivalence class is convex.

Without loss of generality

()1


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6 SAHARON SHELAH

[Why? First, assume aE2M2b; now without loss of generality a


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[Why? By ()8 + ()6.]

5 if a, b X1 then aE0

M

1

b h(a)E0

M

2

h(b).

[Why? By 3 and 4 and the definition of E0M

.]

6 if M |= a + b = c for = 1, 2 and h(a1) = a2 and h(b1) = b2 then(a) c1E

2M

c2 for = 1, 2

(b) c1E0M

c2 for = 1, 2.

[Why? If a1 N or b1 N the conclusion follows easily so we assume a1, b1 / N.For = 1, 2 by 2 we have M |= ab = c. Also by 4(b) we have x X1\N xE2M1h(x) recalling E

2M1

= E2M2 by ()6 we have a1E2M2

a2, b1E2M2b2 hence for somen N we have M2 |= a1 < n a2 a2 < n a1 b1 < n b2 b2 < n b1 henceM2 |= a1 b1 < n2 a2 b2 a2 b2 < n2 a1 b1 hence (a1 M2 b1), (a2 M2 b2)are E2M2-equivalent and also are E

2M2

-equivalent.

So by ()5 we have (a1 M2 b1), (a1 M1 b1) are E2M1-equivalent and also (a2 M1b2), (a2 M2 b2) are E

2M1

-equivalent hence together with the previous paragraph by

()6 they are E2M -equivalent, in particular c1, c2 are E2M

-equivalent as required inclause (a) of6. By 3 +1(b) also clause (b) of6 there follows.]

So by 4 + 6(b) we are done. 1.7

We have used{a19}

Observation 1.11. Assume a M\N1) a + n : n N is increasing and cofinal in a/E0M.2) a n : n N is decreasing and unbounded from below in a/E0M.3) n a : n N is increasing and cofinal in a/E2M.4) Moreover min{b : n M b a} : n N} is decreasing and unbounded from

below in a/E2M.5) Moreover for some b, 2b a < 2b+1 hence we can use in (3),(4) the sequence2b+n : n N, 2bn : n N.

{a21}Observation 1.12. 1) Assume Mk |= a b = ck for k = 1, 2 and M1{


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2. More for E3M

Here we say more on the equivalence relations EM. In 2.1 we deal with basicproperties: when E E+1 , when -o.r. implies ( + 1)-o.r., preservation under +and . We also prove one half of our answer to ?? that is in 2.4 we prove a1E3Mbimplies a1E

5a2. Concerning the weak form of uniqueness of the additive structure

in Theorem 2.6 we prove e.g. if M1, M2 are order isomorphic and M1 is 3-o.r. thenM1{


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of M mapping ak to bk. Combining there is an order-isomorphism g1 from M


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(a) the sequence a1+2n

: n N, that is max{b : b in M, a divides b and(b/a)2

n

a} : n N is a decreasing sequence from {b : a < b for every

a

a/E3M} unbounded from below in it

(b) the sequence a12n

: n N, that is max{b : (a/b)2n

a} : n Nis an increasing sequence included in {b : b < a for every a a/E3M} andunbounded from above in it.

2) For a M\N we have:

(a) the sequence ( 1 + 2n)a : n N, is a decreasing sequence in {b M : babove a/E1M} cofinal in it

(b) the sequence (1+2n)a : n N, is an increasing sequence in {b M : bbelow a/E1M} cofinal in it.

Proof. Straight. 2.5{b13}

Theorem 2.6. 1) If M1 is 3-o.r. and f is an order-isomorphism from M1 ontoM2 then f maps E

kM1

onto EkM2 for k = 3, 4.2) In part (1), moreover M1{


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[Why? Similarly to ()5, i.e. by 1.12(2).]

()7 E4

M1= E4

M2.

[Why? Let a, b M\N be given. For = 1, 2 and n N let a,n be such thatM |= an = a,n.

First, assume aE4M2b and without loss of generality a < b. So for some n Nwe have M2 |= a < b < an so M2 |= a < b < a2,n. Also a1,nE3M1a2,n by ()6 soby 2.1(1) we have a1,nE4M1a2,n hence for some m, M1 |= (a1,n)

m a2,n, in fact,even m = 2 is O.K. So M1 |= b < a1,mn but a


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{b21}Claim 2.8. Assume h is an order-isomorphism from M1 onto M2 andM1 is 4-o.r.

Then fora, b M we have h(a)E4

M2h(b) aE4

M1b.Proof. Without loss of generality h is the identity and let M1{


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3. Constructing somewhat rigid models{d0}

Hypothesis 3.1. is regular.{d1}

Definition 3.2. For any M (model of PA)

(a) let ZM = Z[M] be the ring M generates (so a ZM iff a = b a = b forsome b M, of course ZM is determined only up to isomorphism over M;similarly below); when, as usual, M is ordinary without loss of generalityZM Z

(b) let QM = Q[M] be the field of quotients of ZM; in fact, it is an orderedfield, if M is ordinary then without loss of generality QM Q

(c) let RM = R[M] be the closure of QM adding all definable cuts, so inparticular it is a real closed field, see below

(d) let SM = RbdM/R

infiM where

RbdM

= {a RM : RM |= n < a < n for some n N}

RinfiM = {a RbdM : RM |= 1/n < a < 1/n for every n N}

jM is the function from RbdM into R such that M |= n1/m1 < a (2(x)) = (2(x)).

Case 2: d ap,n.Let 1(x) = 2(x) ((x) > bp,n+1) and

2(x) = 2(x). Now

= (1, 2) is

as required noting (by ()4.2) that

Ma |= |1(M)| |{c 1(M) : (c) d}|

1

2|1(M)|


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()5.4 (a) k N\{0} be large enough such that ((1) (2))/(2) > 2/k

(b) let n(1) N be large enough such that:

(1) (2) > 1/n(1) (2) > (k + 1)/n(1)

(c) let n(2) N be > n(1)

()5.5 let

(a) 1(x1, x2; y1, y2) = x1 < x2 1(x1) 1(x2) y1 < y2 < b

(b) 2(x1, x2; y1, y2) is the conjunction of: 1(x1, x2, y1, y2)

|{x : 1(x)x1 x < x2}|n(1) |{y : 2(y)y1 y < y2}|n(1)a(c) 2(x1, x2) = (y1, y2)(2(x1, x2; y1, y2)

(d) 3(x1, x2; y1, y2) is the conjunction of 2(x1, x2; y1, y2)

|{x : 1(x) x1 x < x2 (x) / [y1, y2)}|n(2) < |{y : 2(a) y1 y < y2}|n(2) a

(e) 3(x1, x2) = (y1, y2)3(x1, x2, y1, y2).

So by our assumptions (for clause (b) use ()5.3 does not apply) we have

()5.6 (a) ifa1 < a2 are from 1(Ma) then Ma |= 1[a1, a2; F[2](a1), F[2](a2)]

(b) if Ma |= 2[a1, a2; F[2](a1), F

[2](a2)] then

Ma |= 3[a1, a2; F[2](a1), F[2](a2)]

Ma |= 3[a1, a2].

Clearly

()5.7 if Ma |= 3[a1, a2; b1, b2] for = 1, 2 then [b

11, b

12)Ma [b

21, b

22)Ma = .

It is well known that for a linear order, for any finite family of intervals, theirintersection is non-empty iff the intersection of any two is non-empty.

Now a version of this can be proved in PA hence

()5.8 for some (x1, x2) 2a, we have: if Ma |= 3[a1, a2] then (a1, a2) 2(M) and for every b1, b2 2(M) we have Ma |= 3[a1, a2; b1, b2] implies(a1, a2) [b1, b2)Ma.

Now

()5.9 (a) let QM RM be a true rational such that(2) > > (2)k/(k + 1) + 1/n(1)

(b) let d = (a) Ma computed in Ra and c = (a)1/n(1)

()5.10 in Ma we can define an increasing sequence a1,i : i < i(), so i() Masuch that

a1,0 = 0, a1,i() = a

a1,i+1 = min{a : 1(a) and a1,i < a and |1(Ma) [a1,i, a)| is d}

|{a : 1(a) and a1,i()1 a < a}| is d but 2d


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()5.11 (a) in Ma we can defineu = {i < i() : Ma |= 3[a1,i, a1,i+1]}

v = {i < i() : i / u}(b) let 1,i(x) := 1(x) a1,i x < a1,i+1.

So (will be used in Case 1 below)

()5.12 (a) (1,i(x)) = for i < i()

(b) if i < i() and i v then

Ma |= |2(Ma) (F(a1,i), F(a1,i+1))Ma| |1,i(Ma)| a1/n(1)

= b a1/n(1) .

[Why? Clause (a) is obvious by the definition of() and a1,i+1. For clause (b) notethat by the definition of3 in ()5.5(e) we have Ma |= 3[a1,i, a2,i; F[2](a1,i), F[2](a2,i)],

but by ()5.6(a) we have Ma |= 3[a1,i, a2,i; F(a1,i), F(a1,i)]. By the definition of3 in ()5.5(d) we are done.]Now towards Case 2 note

()5.13 if i1 < i2 are from u then F(a1,i1) < M(a1,i2 , a1,i2+1).

[Why? Obvious by ()5.8.]

()5.14 we define terms 1(x1, x2), 2(x1, x2) 2a such that if i < i() then:(a) Ma |= 1(a1,i, a1,i+1) < (a1,i, a1,i+1) < 2(a1,i, a1,i+1)

(b) 2(Ma) [Ma(a1,i, a1,i+1), Ma2 (a1,i, a1,i+1))Ma has c elements, in

Mas-sense

(c) 2(Ma) [Ma1 (a1,i, a1,i+1),

Ma((a1,i, a1,i+1))Ma has c elements in

Mas-sense(d) if i < j are from u then Ma |= 2(a1,i, a1,i+1) < (a1,j , a1,j+1)

(e) Ma |= 1(a1,i, a1,i+1) < (a1,i, a1,i+1) < 2(a1,i, a1,i+1)

(f) if i u then Ma |= 1(a1,i, a1,i+1) < F(a1,i) < (a1,i, a1,i+1) (2) is a real substitute, see clause (d) in part (2).

4) Why clause () in 2(a), defining P? Otherwise 2() may be irrelevant to thetype we like to omit, so impossible.5) By such approximations, i.e. member ofP,

(A) why can we arrive to a complete type?

Answer: As if we divide 1 to two sets at least one has the same ():

(B) why can we continue to omit p(x) a?

Answer: As if () is a definable (in Ma) function with domain 1 let d bemaximal such that |{a 1(M) : (a) < d}|

12

|2(M)|, i.e. is in the middle inthe right sense.

If 1{d} is large enough we easily finish; otherwise for some n we have d /(ap,n

, bp,n

), so 1

(M)

(x) /

(ap,n

, bp,n

) has

1

2(

2(M)) elements

(C) why can guarantee that such (x) does not realize the forbidden new type?

Answer: This is a major point. If(1) > 2(2) this is easy (as in the case we useaE4Mb) and if for some a1 < a2 we have (

2) > (

2) we let

2(x) = (2(x) a1 x < a2 (x) / (F(a1), F(a2))

and we let

2 (x) := (2 F(a1) x < F(a2))

we are done, so assume there are no such a1, a2.We consider two possible reasons for the failure of a suggested pair (a1, a2).

One reason is that maybe the length of the interval [F(a1), F(a2)) of2(M1) is toolarge. The second is that it is small enough but () maps the large majority of1(M) [a1, a2) into [F(a1), F(a2)). In the second version we can define a versionof its property satisfied by (a1, a2, F(a1), F(a2)). So we have enough intervals ofpseudo second kind (pseudo means using the definable version of the property). So

dividing 1(M) to convex subsets of equal (suitable) size (essentially a(2) , R>0

small enough) by ai : i < i() we have: for some such interval [ai, ai+1) there areb1, b2 as above. For those for which we cannot define (F(a1), F(a2)) we can defineit up to a good approximation. If there are enough, (this may include pseudocases in respect to F) we can replace 1(M) by

1(M) = {ai : i < i()} and

2() defined by the function above.So |(M)| is significantly smaller than |1(M)|, essentially (1) = (1)

(ai+1 ai) (1) (2) + , quite small. But we are over-compensating sowe decrease 2(x) to

2(x) which is quite closed to {F(ai) : [ai, ai+1) is of the

pseudo second kind} and (2) is essentially (2) (2) + . So both losesimilarly in the () measure but now, if we have arranged the numbers correctly(2) > 2(

2), a case we know to solve.

If there are not enough is of the pseudo second kind, the function essentiallyinflates the image getting a finite cardinality arithmetic contradiction.

{d35}Theorem 3.9. Assume 1. If M is a countable model of PA, then M has anelementary extension N of cardinality 1 such that E

5N = E

3N, i.e. is 3-o.r.


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22 SAHARON SHELAH

Proof. Without loss of generality M has universe a countable ordinal. As weare assuming 1 , we choose F a partial function from to for < 1, i.e.

F = F, < 1 such that for every partial function F : 1 1, for stationarilymany countable limit ordinals we have F = F.

We now choose a AP0 by induction on < 1 such that

(a) Ma0 = M

a0 =

(b) a : is AP-increasing continuous

(c) if = + 1, is a countable limit ordinal, Ma has universe and forsome a, b the tuple (a, a, b, F) satisfies the assumptions of 3.6 on(a, a, b, F), they are necessarily unique (see 3.6(A)(c)), then a+1 sat-isfies its conclusion (for some c).

Why can we carry the induction?

For = 0 recall clause (a).For = 1, as a0 = let Ma1 be a countable model such that M = Ma0

Ma1, M = Ma1 and without loss of generality the universe of Ma1 is a countableordinal.

Lastly, let a1 = .For a limit ordinal use 3.5(2), i.e. choose the union, this is obvious.For = + 1, if clause (c) apply use Claim 3.6.For = + 1 > 1 when clause (c) does not apply, this is easier than 3.6 (or

choose (a, h, F) such that (a , a, b, F) are as in the assumption 3.6, this ispossible because Ma is non-standard, see the case = 1, and note that a, b Ma\N aE

5Ma

b because Ma is countable; so we can use 3.6).Having carried the induction let N = {Ma : < 1}.

Clearly N is a model of T of cardinality 1. We know that E3

N E5

N by2.4. Toward contradiction assume aE5Nb but (aE3Nb) where a, b N\M.

Without loss of generality b < a and let F be an order-isomorphism from N


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4. Weaker version of PA

We may wonder what is the weakest version of PA needed in the various resultsso below we define some variants and then remark when they suffice. But say whenwe add the function 2x, we prefer to add to the vocabulary a new function symboland the relevant axioms (rather than an axiom stating that some definition of ithas those properties). So we shall comment what version of PA is needed in theresults of 1,2.

{k0}Convention 4.1. A model is a model of PA4 (see below) of vocabulary PA ifnot said otherwise.

{k3}

Definition 4.2. We define the first order theories PA for {1, . . . , 4} andlet PAcom be the set of completions of PA.

Let PA consist of the following first order sentences in the vocabulary {0, 1, 0M = {s SM : 0 < s}

let jM : RbdM SM be jM(b) = b/RinfiM

jM(b) = ifRbdM < b

jM(b) = if b < RbdM(f) for a closed interval I = [a, b]M of M let g(I) = b a M

(g) for interval I, J of M let = jM(log2(g(J))/log2(g(I)).{c37}

Claim 6.2. E4M refines E5M when M is 1-saturated.

Proof. Let N = M{0M = It[a] : t S

>0M

(b) SM R, which is a pure additive subgroup of (R, +) with 1 S

(c) {It : t S>0} list {b/E3M : b a/M}

(d) ifR |= 0 < s < t then Is < It, i.e. b Is c It b 0 and a Is for = 1, 2, 3 thena1 M a2 Is3

(f) moreover

S does not depend on a,

S is a real closed subfield ofR

if M is 1-saturated then S = R

2) If b It and t S>0M then t = sup{

mn : m N, n N\{0} and b

mn < It[a]} =

sup{mn : m N, n N\{0} and bm < an}, (we can start with this).

3) We say (J1, J2) is an s-pair or really s a-pair when

(a) J1, J2 are closed intervals of M of length a/E4M(b) s S>0M


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28 SAHARON SHELAH

(c) inRM we compute (log2(g(J2))/(log2(g(J1)) RM, it belongs to RbdM and

jM maps it to s.

4) a-pair means s-pair for some s S>0M .5) Let (J1, J2) = s which means (J1, J2) is an s-pair.

Discussion 6.4. So by 2.4 it suffices to prove

if s1 < s2 are from S>0M then NIs1

= MIs2 .

Toward proving {c41}

Definition 6.5. Let s S>0M . We say x is an s-approximation or x APs = APsM

when x consists of the following:

(a) an s-pair (Ix,1, Ix,2)

(b) a set Ix, of sub-intervals ofIx, (may be a singleton) which form a partition

of Ix,, we let ex, be the equivalence relation this set defines(c) if J Ix, then (Ix,1, J) is a pair or J is a singleton

(d) an order-preserving function fx from Ix,1 onto Ix,2(e) if fx(J1) = J2 then (J1, J2) is an s-pair or both are singletons.

{c43}Definition 6.6. Let S>0M , we define a two-place relation on the set of-approximations:

x < y iff

(a) x, y are -intervals

(b) ey refines ex if I Ix,, J Iy, and J I and J is not a singleton, then(Ix,, J) < (Iy,, I).

{c45}Observation 6.7. 1) 0M .

3) If xi : i < 1 is a


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(c) aEI1b g(a)EI2g(b).

Let() X1 I1 is a set of representatives for I1/EI1 such that min(I1), max(I1)

X1

() X2 = {b I1 : a X1\{max(I1)}, b = g(a) or a = max(I2)}.

Let g be the natural one-to-one order preserving function from X1 onto X2:

g|max(I1)) = max(I2)

g(a) = g(a) if a X1\{max(I1)}.

We now define X+ as

S+ = X {a + g(I)11/n : 1 < n N and a X\{max(I)} {a

g(I)11/n : 1 < n N and a X\{min(I)}} g+ is the unique order preserving function from X+1 onto X

+2 extending

X+2 .

{c50}

Claim 6.9. If the sequence xn : n < is


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30 SAHARON SHELAH

Moved 2010.8.24 from after ()4.8 in the proof of 3.6, pgs. 15,16:Moreover

()5.9 there are (x1, x2) a for = 1, 2, 3, 4 we have: ifMa |= 2[a1, a2] then : Ma |= 1(a1, a2) < 2(a1, a2) < (a1, a2) < 3(a1, a2) < 4(a2, a4)

b all are from 2(Ma)

if Ma |= [a1, a2, b1, b2] then Ma |= 1(a1, a2) < b1 < (a1, a2) andMa |= 3(a1, a2) < b2 < 4(a1)

Ma |= |2(M) [(a1, a2); +1(a1, a2)| a1(n(2)) for = 1, 3

Ma |= |2(M) (a1, a2)| |2(M) [1(a1, a2), 3(a1, a2))| a1/n(2) .

[Why? 1(a1, a2) = min{b1 : Ma |= (y)22(a1, a2, b1, y)}2(a1, a2) = max{b2 : Ma |= 2(a1, a2, 1(a1, a2), b2)}4(a1, a2) = max{b2 : Ma |= (y1)2(a1, a2, y1, b2]}

3(a1, a2) = min{b1 : Ma |= 2[a1, a2, b1, 4(a1, a4)}. Now it is easy to checkthat those terms are as required.]

Moved 2010.8.24, a proof of 3.11, FILL?, pg. 18:

Hypothesis 6.10. Assume is strong limit singular of cofinality 0, =n


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{d15}Claim 6.16. There are b, a such that a AP b, a Mb realizing (A0, A1), so

when :(a) a AP

(b) (A0, A0) cut(Ma)

(c) if X M is definable catching (A0, A1) and F : X M is definable thenF converge to no p which means: for every < , for some (a0, a1) A0xA1 we have M |= (x)[a0 < x < a1 x X ap, < F(x) < bp,.

{d17}Claim 6.17. Assume a AP, Ma |= n < a an < b for n < and F is anorder-preserving function from (Ma)


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32 SAHARON SHELAH

7. Private AppendixExistence of order-rigid models

Saharon: check againThe previous sections stress the value of proving the existence of some version of

order-rigid models. From the general treatment in [Sh:472], [Sh:800] we have ZFCexistence results and see [Sh:384] on the history of such results. But as they are notyet ready we deduce the results under extra set theoretic assumptions by [Sh:107]for T a completion of PA, so that we can define internally the cardinality of abounded set. On bigness notions, alternatively, see [Sh:509].

{c3}Theorem 7.1. Assume = + and werestrict ourselves to M C (and to A,B, C, a , b , . . . , C.3) E = EC for 6, see Definition 1.2.

{c7}Definition 7.3. 1) For M C we say M |= |(M, a)| = b when in M there is adefinable one-to-one function from {c : M |= [c, a]} onto [0, b)M = {c : M |= c (d2) define P2e2 with the parallel property invertingthe orders. Clearly P1e1 ,P

2e2 are not empty! Also q P

1e1 q(C) [0, r , d2] by

clause (e)1 and q P1e1 q(C) has no last member, so q(C) [0, , d2]. Similarlyq P2 a(C) [a, b].

For q P1 let Aq1 = {c ((d1), (d2)): there is e, e e and e realizes q},

easily q1, q2 P1 Aq1 = Aq2 and call it A1. Similarly there is A2 [(d1), (d2)]such that q P2 A2 = {e [(d1), (d2)]: there is e, e e and e realizes q}.Clearly A1 is an initial segment of [(d1), (d2)] and by clause (e)2 clearly A2 is an

end-segment of it.

Case 1: there is e A1 A2.Let q = tp(e, A {d1, d2}).So we can find q1 P1 and c1 > e realizing q1 and q2 P2 and c2 < e realizing

q2. Easily if e q(C) then there is c1 > e realizing q1 hence e < c1 (d2) and

there is c2 < e realizing q2 hence e > c1 (d1), so q(C) ((d1), (d2)) as

promised in clause (f).

Case 2: A1 A2 = .


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Let q1 P1, q2 P2, let q0(x, y1, y2) be q1(y1) q2(y2) {y2 y1}, thistype cannot be realized hence there are 1(y1) q1(y), 2(y2) q2(y2) such that

{1(y1), 2(y2), y2 y1} is not realized.Without loss of generality 1(y1) y1 < d2. Clearly 1(C) is non-empty

and [0, , d2) so max(1(C)) is well defined. If c1 = max(1(C)) belongs to((d1), (d2)) then q := tp(c1, A {d1, d2}) is required, so we are left with thecase max(1(C)) = (d1) 1 so q = tp(max1(C) + 2, A {d1, d2} is as required(note: that (d1) + 1 < (d2) as a1 + 1 < d2 as p(x, y) {d1 < x < d2} is -bigq S(A {d1, d2}) it is as promised in clause (f).

Clause (g): as q(y) d1 < y < d2 there is a formula (y, z1, z2) with parameters

from A such that (y, d1, d2) q0(y) and (C, d1, d2) (d1, d2). let F(d1, d2) =max((C, d1, d2) {0}).

Clause (h): obvious.

For this F

3 there are 0(x), F and c:(a) c < b and x c is -small

(b) 0(x) x < b

(c) F(z1, z2) is a function definable in C over A

(d) Dom(F) (a){(a1, d2) : d1 < d2 < b and ((C) (d1, d2)) c(C))}

(e) if (d1, d2) Dom(F) then FC(e, d1, d2) < a

(f) if d0 < d1 d2 < d3 < b

(g) Fc(d2, d2+1) = e for = 0, 1 then e

0 < e

1.

In C define the unary F1 , F2 , F3

4 (a) F1 : {d < b : ((C) (d, b)) c + 1}

(b) F2 (d) = min{d : d < d < b and |(C) (d, b)| c}

(c) F2 : Dom(F1) [0, a)

(d) F2 (d) = {F(d, Fi(d)}

(e) F3 : Dom(F2) [0, a)

(f) F3 (d) = min{F2(d) : d d Dom(F)}

(g) 2(x) = x Dom(F1) F1(x) Dom(F2).

So

5 (a) F1 , F

2 , F

3 are well defined and definable in C

(b) F3 is non-decreasing

(c) F3 (d) < F3 (F2(d)) when d, F

2 (d) Dom(F), i.e. d 2(C).

The contradiction should be clear; we get contradiction to pigeon-hull principle(which is provable from PA1) 7.7

The theorem 7.7 suffice because


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36 SAHARON SHELAH

{c19}

Claim 7.8. Assume A B C, |A| = and a1i : i < is increasing, a2i : i <

is decreasing, a1i < a

2i , a

ci A, A1 = {a A : a < a

1i for some i < }, A2 = A\A1.

If for not (x, y) L(T) and c g(y)B do we have A (M, c) = A1, then thetype r(x) = {a1i < x < a

2i : i < } has no support over B (i.e. forTh(C, b)bB in

the sense of [Sh:82].{c21}

Discussion 7.9. [2010.7.08 Referee - cryptic?] We may like to generalize Theorem1.7, 2.6 (and their complement)?

For this the following may help:{c23}

Definition 7.10. Let M be a model of PA (or as in 7.7 and for simlicity adduniform coding of bounded subsets).1) We say M1 is v.r. (very rigid) when: for any definable increasing functionf : {0, . . . , b} M and e, d such that dn < b

for every n N and a definable

norm N with d-additivity.{c25}Question 7.11. PA3 enough?

{c27}Remark 7.12. We shall try to code in M1 a set such that in M2 we can recode it.We may strengthen M is o.r. hoping for 1.9.

We may consider{c29}

Definition 7.13. M1 is strongly o.r. when :for any M2 satisfying M = M1{


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Part II

8.Introduction to the second part

Compared to 0 - 3, from now on we note the necessary version of PA neededand called PA [Saharon say M |= PA4 if not said otherwise]

Convention 8.1. M, N are ordinary models of PA4 if not said otherwise.

Let us review the present work. First, in 1 we introduce and deal with therelevant equivalence relations, and as a warm up it is proved that if M is a modelof a very weak version of PA (called PA1), the so called aE2Mb implies ()M,a,b soaE2Mb is a sufficient condition for a positive answer to Question to 0.1(1), while forthe so called 2-order rigid model M, we prove that the isomorphic type of M{


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38 SAHARON SHELAH

{h.19}Definition 8.5. 1) We say M1, M2 are order-isomorphic if some is an order

isomorphism form M1 onto M2, see below.2) is an order-isomorphism from M1 onto M2 iff (< belongs to M1 M2 , ofcourse and) is an isomorphism from M1{


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9. Somewhat rigid order

We define (in 9.2) some equivalence relation EM for models M (of version).We shall deal with their basic properties in 9.3, 9.11, the relations between them(in 10.1) cofinalities of equivalence classes (in 10.5, 9.10, 10.5), on order isomor-phic/almost isomorphic in 9.1, 9.4, 10.4 (including proving for -o.r. models provingEmM E

5).

Lastly, for inverses of those see Theorem 9.6, 10.6.

{1a.1}Definition 9.1. We say M, N are almost {


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40 SAHARON SHELAH

4) aE6Mb iff there is c M min{a, b} and an order-automorphism f of M such thatmax{a, b} M f(c) iff this holds for c = min{a, b}.

Proof. 1) Let = 3. If a1E3Ma2 and let c witness it, then c witnesses also a2E3Ma2

and a2E3Ma1, so reflexivity and symmetry holds.Lastly, assume M |= a1 < a2 < a3; if akE3Mak+1 and let ck witness this for

k = 1, 2, then the product c1c2 witness a1EM3 a3 by part (1A) and if a1E

3Ma3 then

the same witness give a1E3Ma2 a2E3Ma3 so transitivity and convexity holds.

For = 1 the proof is similar (using part (1B) instead of part (1A)), also for = 0, 2, 4 the proof is even easier and for = 5, 6 it holds by the definition.1A), 1B) Check.2) Without loss of generality assume a


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{1a.8d}Claim 9.4. 1) If f is an order isomorphism from M1 onto M2 then f maps E

0M1

onto E0

M2 .2) If f is as almost {


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42 SAHARON SHELAH

[Why? For = 1, 2 as M |= PA2 it follows that (M


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(b) aE1Mb iff f(a)E3M

f(b)

(c) aE2Mb iff f(a)E4M

f(b).

[Why? Look at the definitions and do basic arithmetic.]

4 there is an order isomorphism h from X1 onto X2 such that(a) hN = idN(b) if a M\N then h maps (X1 (a/E2M1) onto X2 (a/E

2M1

).

[Why? By ()8 + ()6.]

5 if a, b X1 then aE0M1 b h(a)E0M2

h(b).

[Why? By 3 and 4 and the definition of E0M

.]

6 if M |= a + b = c for = 1, 2, h(a1) = a2 and h(b2) = b2 then

(a) c1E2M

c2 for = 1, 2

(b) c1E0M

c2 for = 1, 2.

If a1 N or b1 N the conclusion follows easily so we assume a1, b1 / N. For = 1, 2 by 2 we have M |= a b = c. Also x X1\N xE2Mh(x)recalling E2M1 = E

2M2

, so a1E2M2

a2, b1E2M2

b2 hence for some n N we have M2 |=a1 < n a2 a2 < n a1 b1 < n b2 b2 < n b1 hence M2 |= a1 b1 ci and density in I is 1

2nfor some n N.

3) If h : M1 M2 is order isomorphic (onto) then f maps E4M1 = {(a, b)} :nN

log[2](a) log[1](b) [n, n)} onto E4M where log[2] = log2 log2; similarly for

log[2k].

4) Old part of the introduction, before ??... and for a 3-order rigid model M, theorder determines M{+, 0}. Weknow that on each A there are order automorphisms, quite many. But is A = Afor < R2>0. We have to look again at ()1 of version B.2) A variant of the 7.7 is when in clause (d): use = 1a . We replace clauses (), ()by:

() d1, d2, (d1), (d2) B

() q(x) includes {d1 < x < d2 < bx} {(x, (d1), a) = (x, (d2), a) : a fromA}.


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13. Private Appendix

Moved from proof of 10.6 in Oct.2009:

Let c,n be such that M |= cn = c,n for = 1, 2 and n < so M2 |= c1,n() d2, (y > d1)} is -big

hence

p(x, y) {x > d1, y d2} is -big

which means (d1, d2) / R, and by ()2 we get (d1, d2) / R and ()3 holds.]

Recall p0(x) = p(x, y)x


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50 SAHARON SHELAH

()4 if d p0(M) then both p(x, y) {x < d}, p(x, y) {x > d} are -big.

[Why? If not, there is (x, y) p(x, y) so if d (M) not both {(x, y) x d} are -big and (x, y) x < b. Let (x) = (y)(x, y)so (M) is a subset of [0, b) and choose d which is in the middle of (M), i.e.((M) 2|{e (M) : e d}| and |(M)| |{e (M) : e d}|. For d we geta contradiction.]

()5 if d2 < d2 < b, d2 = (d2), d1 = (d

2) hence d1 < d

1 and p(x, y) {d2 .

We shall use 2a , so we arrive to

()4 (a) X is a definable suset of [0, b)


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52 SAHARON SHELAH

(b) |X| (a)k for every k N; older: k + 1 bk, k N

(c) f is a definable weakly increasing function from X into [0, a)

(d) for every b, |{x X : x < b f(x) (b)}| is 2a-small.]

References

[GT06] Rudiger Gobel and Jan Trlifaj, Approximations and endomorphism algebras of modules,de Gruyter Expositions in Mathematics, vol. 41, Walter de Gruyter, Berlin, 2006.

[KS93] H. Kaiser and N. Sauer, On order polynomially complete lattices, Algebra Universalis 30(1993), 171176.

[KS06] R. Kossak and J. Schmerl, The structure of models of Peano arithmetic, Oxford UniversityPress, 2006.

[Sh:a] Saharon Shelah, Classification theory and the number of nonisomorphic models, Studies inLogic and the Foundations of Mathematics, vol. 92, North-Holland Publishing Co., Amsterdam-New York, xvi+544 pp, $62.25, 1978.

[Sh:82] , Models with second order properties. III. Omitting types for L(Q), Archiv fur

Mathematische Logik und Grundlagenforschung21

(1981), 111.[Sh:107] , Models with second order properties. IV. A general method and eliminatingdiamonds, Annals of Pure and Applied Logic 25 (1983), 183212.

[Sh:384] , Compact logics in ZFC : Complete embeddings of atomless Boolean rings, Nonstructure theory, Ch X.

[Sh:472] , Categoricity of Theories in L, when is a measurable cardinal. Part II,

Fundamenta Mathematicae 170 (2001), 165196, math.LO/9604241.[Sh:482] , Compactness in ZFC of the Quantifier on Complete embedding of BAs, .[Sh:509] , Vive la difference III, Israel Journal of Mathematics 166 (2008), 6196,

math.LO/0112237.[Sh:800] , On complicated models, Preprint.[Sh:924] , Can the order determine the addition for models of PA, Mathematical Logic

Quarterly submitted.[Sh:950] , A dependent dream and recounting types.

Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The He-brew University of Jerusalem, Jerusalem, 91904, Israel, and, Department of Mathe-

matics, Hill Center - Busch Campus, Rutgers, The State University of New Jersey, 110

Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

E-mail address: [email protected]: http://shelah.logic.at

Saharon Shelah- Can the Order Determine the Addition for Models of PA

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