Saharon Shelah- Borel Sets with Large Squares

8/3/2019 Saharon Shelah- Borel Sets with Large Squares

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BOREL SETS WITH LARGE SQUARES

SAHARON SHELAH

Abstract. For a cardinal we give a sufficient condition (involvingranks measuring existence of independent sets) for:

: if a Borel set B R R contains a -square (i.e. a set of theform A A, |A| = ) then it contains a 20 -square and even aperfect square.

And also for: if L1, has a model of cardinality then it has a model of

cardinality continuum generated in a nice, absoluteway.Assuming MA+20 > for transparency, those three conditions (, and

) are equivalent, and by this we get e.g.

V


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2 SAHARON SHELAH

but no +-square. Lastly (in 1.15) assuming MA holds weprove exact results (e.g. equivalence of conditions).]

2 Some model theoretic problems.[When we restrict ourselves to models of cardinality up to thecontinuum, 1(0) is the Hanf number of L1, (see 2.1).

Also (in 2.4) if L1, has a model realizing many types(say in the countable set of formulas, many means 1(0))even after c.c.c. forcing, then

{{p : p a complete -type realized in M} : M |= }

has two to the continuum members.We then (2.5) assume L1, has a two cardinal model,

say for (, ) and we want to find a (, 0)-model, we need

1() . Next, more generally, we deal with -cardinalmodels (i.e. we demand that PM have cardinality ). We

define ranks (2.8), from them we can formulate sufficient con-ditions for transfer theorem and compactness. We can provethat the relevant ranks are (essentially) preserved under c.c.c.forcing as in 1, and the sufficient conditions hold for 1 un-der GCH.]

3 Finer analysis of square existence.

[We (3.1, 3.2) define for a sequence T = Tn : n < of trees(i.e. closed sets of the plane) a rank, degsq, whose value is abound for the size of the square it may contain. We then (3.3)

deal with analytic, or more generally -Souslin relations, ??patience incomplete-what has?? and use parallel degrees. Wethen prove that statements on the degrees are related to theexistence of squares in -Souslin relations in a way parallelto what we have on Borel, using (). We then (3.7 3.11)connect it to the existence of identities for 2-place colourings.In particular we get results of the form there is a Borel setB which contains a -square iff < (0) when MA +(0) < 20.]

4 Rectangles.

[We deal with the problem of the existence of rectangles

in Borel and -Souslin relations. The equivalence of therank (for models), the existence of perfect rectangles and themodel theoretic statements is more delicate, but is done.]

0. Introduction

We first review the old results (from 1, 2). The main one is


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BOREL SETS WITH LARGE SQUARES 3

()1 it is consistent, that for every successor ordinal < 1, there isa Borel subset of 2 2 containing an -square but no perfect

square.In fact

()+1 the result above follows from MA+20 > 1.

For this we define (Definition 1.1) for any ordinal a property Pr(; ) ofthe cardinals , . The maximal cardinal with the property of 1 (i.e. forevery small cardinal, c.c.c. forcing adds an example as in ()1) is charac-terized (as 1(0) where () = min{ : Pr(; )}); essentially it is notchanged by c.c.c. forcing; so in ()1:

()1 if in addition V = VP0 , where P is a c.c.c. forcing then 1(0)

(1)V0.

We will generally investigate Pr(; ), giving equivalent formulations(1.1 1.6), seeing how fast () increases, e.g.

+ < () ()(in 1.7, 1.8). For two variants we show: Pr2(;

+) ( +) is preservedby +-c.c. forcing, Prl(;

+) Pr(; +) and Pr(; +) is preservedby any extension of the universe of set theory. Now Pr1(; 0) implies thatthere is no Borel set as above (1.12) but if Pr1(; 0) fails then some c.c.c.forcing adds a Borel set as above (1.13). We cannot in ()1 omit some settheoretic assumption even for 2 - see 1.15, 1.16 (add many Cohen realsor many random reals to a universe satisfying e.g. 20 = 1, then, in thenew universe, every Borel set which contains an 2-square, also contains aperfect square). We can replace Borel by analytic or even -Souslin (usingPr

+()).

In 2 we deal with related model theoretic questions with less satisfactoryresults. By 2.1, 2.3, giving a kind of answer to a question from [Sh 49],

()2 essentially = 1(0) is the Hanf number for models of sentencesin L1, when we restrict ourselves to models of cardinality 2

0.

(What is the meaning of essentially? If 1(0) 20 this fails,

but if 1(0) < 20 it holds.)

In 2.4 we generalize it (the parallel of replacing Borel or analytic sets by-Souslin). We conclude (2.4(2)):

()3 if L1,(1), 0 1 are countable vocabularies, {(x) : L1,(0)} is countable and has a model which realizes

1(0) complete (, 1)-types then |{(M 0)/ =: M |= , M =}| min{2,2} (for any ), as we have models as in [Sh a, VII4]=[Sh c, VII 4].

If we allow parameters in the formulas of , and 20 < 21 then ()3 holdstoo. However even in the case 2 = 20 we prove some results in thisdirection, see [Sh 262] (better [Sh e, VII,5]. We then turn to three cardinal


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4 SAHARON SHELAH

theorems etc. trying to continue [Sh 49] (where e.g. (, 0) (20 , 0)

was proved).

We knew those results earlier than, or in 1980/1, but failed in effortsto prove the consistency of ZFC +1(0) > 1 (or proving ZFC1(0) = 1). By the mid seventies we knew how to get consistencyof results like those in 2 (forcing with P, adding many Cohen reals i.e. inVP getting ()3 for = (1)

V). This (older proof, not the one used) isclosely related to Silvers proof of every 11-relation with uncountably manyequivalence classes has a 20 ones (a deeper one is the proof of Harringtonof the Lauchli-Halpern theorem; see a generalization of the Lauchli-Halperntheorem, a partition theorem on >2, large by [Sh 288] 4).

In fact, about 88 I wrote down for W. Hodges proofs of (a) and (b) statedbelow.

(a) If, for simplicity, V satisfies GCH, and we add > 1 Cohen realsthen the Hanf number of L1, below the continuum is 1.

(b) If L1,(1) and some countable {(x) : L1,(0)} sat-isfies: in every forcing extension of V, has a model which realizes20 (or at least min{20, 1}) complete -types then the conclusionof ()3 above holds.

Hodges had intended to write it up. Later Hrushovski and Velickovicindependently proved the statement (a).

As indicated above, the results had seemed disappointing as the mainquestion is 1(0) = 1? is not answered. But Hjorth asked me about(essentially) ()1 which was mentioned in [HrSh 152] and urged me to write

this down.In 3 we define degree of Borel sets of the forms

n


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Set theory:BA = {f : f is a function from B to A}, the set of reals is 2.

S


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6 SAHARON SHELAH

Model theory:Vocabularies are denoted by , so languages are denoted by e.g L,(),

models are denoted by M, N. The universe of M is |M|, its cardinalityM. The vocabulary of M is (M) and the vocabulary of T (a theory ora sentence) is (T). RM is the interpretation of R in M (for R (M)).For a model M, and a set B M we have: a cl


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above for = 0 (by ()2) and limit (by ()1), so for = + 1 welet

()3 rkl(w, M; < ) +1 iff (letting n = |w|, w = {a0, . . . , an1})for every k < n and a quantifier free formula (x0, . . . , xn1)(in the vocabulary of M) for which M |= [a0, . . . , an1] wehave:

Case 1: l = 1. There are aim M for m < n, i < 2 such that :(a) rkl({aim : i < 2, m < n}, M; < ) ,(b) M |= [ai0, . . . , a

in1] (for i = 1, 2), so wlog there is no

repetition in ai0, . . . , ain1

(c) a0k = a1k but for m = k (such that m < n) we have

a0m = a1m.

Case 2: l = 0. As for l = 1 but in addition

(d)m am = a

0m

Case 3: l = 3. We give to an additional role and the definition is like case1 but i < ; i.e. there are aim M for m < n, i < such that

(a) for i < j < we have rkl({aim, ajm : m < n}, M; < )

(b) M |= [ai0, . . . , ain1] (for i < ; so wlog there are no

repetitions in ai0, . . . , ain1)

(c) for i < j < , aik = ajk but for m = k (such that m < n)

we have

i,j


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by l = 1, 3, 1, 0, 1, 4, 5, 1, 5 respectively (i.e. 2 4, 3 5 1,0 1, 2 3, 4 5, 3 1, 2 0, 2 1) and also if we decrease

,, or increase (the last two only when M is not a parameter).So the corresponding inequality on l(< , ) holds.

(2) Also rkl(w1, M; < ) rkl(w2, M; < ) for w1 w2 from [M]

.(3) Also if we expand M, the ranks (of w [M], of M) can only

decrease.(4) IfA M is defined by a quantifier free formula with parameters from

a finite subset w of M, M+ is M expanded by the relations definedby quantifier free formulas with parameters from w, M = M+A(for simplicity M has relations only) then for w [A] such thatw w we have rkl(w, M; < ) rkl(w w, M; < ). Hence ifw = , rkl(M; < k) rkl(M; < )

(5) In 1.1(3)()2 , if in the definition of cl


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Then for l = 2, 3, in (), () (of (a)) above equalities hold andthe inequality in () holds.

(c) Assume V1 = VP0 where P is 2-linked. Then for l = 4, 5

in clauses (), () (of (a)) above we have equality and the in-equality in () holds.

Proof 1) For < , n < , a quantifier free formula = (x0, . . . , xn1)and k < n let

Rn = {a0, . . . , an1 : am M for m < n and

= rkl({a0, . . . , an1}, M; < )},

Rn,k, = {a0, . . . , an1 Rn : M |= [a0, . . . , an1]

for no a1k |M| \ {a0, . . . , an1} we have() M |= [a0, . . . , ak1, a

1k, ak+1, . . . , an1]

() rkl({am : m < n} {a1k}, M; < ) },

M+ = (M , . . . Rn, Rn,k, . . . )


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10 SAHARON SHELAH

Convention 1.6. Writing Pr(; ) for + (omitting l) we mean l = 0.

Similarly (< , ) and so () etc.

Claim 1.7. Let l {0, 2, 4}.

(1) NPrl+1(+; )

(2) If is a limit ordinal < + (in fact 0 cf () < + suffice), and

NPr(; ) for <

then NPrl+1(


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then for some i < j < l we have rkl({i, j}, M+) and applying (b)

with {i}, j here standing for w, there we get rkl({fj (i)}, M)

hence + 1 rkl(M), contradiction.)Hence

()3 rkl(M+; ) rkl(M; ) + 1.

As rkl(M+; ) < clearly M+ witnesses NPr+1(+; ).

4) Like (3). 1.7 1.6

Conclusion 1.8. Remembering that () = min{ : Pr(; )} we have:

(1) for a limit ordinal () () and even 2() ().(2) for l even l() : 0 < < is strictly increasing, and for a limit

ordinal , () = sup


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12 SAHARON SHELAH

then < rkl(M; ) rkl(M ; ) hence rkl(M; ) . So

Pr(; ) hence l() so sup


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if w = {a0, . . . , an1} [M+], M+ |= Rn,,,k[a0, . . . , an1]

then rk({a0, . . . , an1}, M+; )

(note that by the choice of M

and Rn,,,k, if w [M+] then for somen , , , k we have M+ |= Rn,,,k[a0, . . . , an1]). This we prove by inductionon , so assume the conclusion fails; so

rk({a0, . . . , an1}, M+; ) + 1

(and eventually we shall get a contradiction). By the definition of rk3 appliedto = Rn,,,k, and k we know that there are a

im (for m < n,i <

+) asin Definition 1.1(3) case l = 3. In particular M+ |= Rn,,,k[a

i0, . . . , a

in1].

So for each i < + by the definition of Rn,,,k necessarily there is pi Psuch that

pi PM

|= Rn,[ai0, . . . , a

in1] and rk

({ai0, . . . , ain1}, M

; ) =

and [not rk({ai0, . . . , ain1}, M ; ) +1] is witnessed by = Rn, and k.

For part (1), as P satisfies the + cc, for some q P, q Y

= {i :pi G

P} has cardinality

+ (in fact pi forces it for every large enough i).Looking at the definition of the rank in VP we see that ai0, . . . , a

in1 : i

Y

cannot be a witness for the demand for rk3({ai00 , . . . , ai0n1}, M

; ) > for Rn,,k hold for any (or some) i0 Y

, so for part (1)

() q P for some i = j in Y

we have rk3({ai0, . . . , ain1, a

jk}, M

; ) <

(as the demand on equalities holds trivially).As we can increase q, wlog q forces a value to those i, j, hence wlog for

some n() = n + 1 < , () < and () < and for k() < n + 1 wehave

q P rk3({ai0, . . . , a

in1, a

jk}, M

; ) = (), and

rk3({ai0, . . . , ain1, a

jk}, M; ) () + 1 is witnessed by

= Rn(),()(x0, . . . , xn) and k() .

Hence by the definition of Rn(),(),(),k() we have

M+ |= Rn(),(),(),k()[ai0, . . . , a

in1, a

jk].

As () < by the induction hypothesis () holds hence

rk3({ai0, . . . , ain1, a

jk}, M

+; ) (),

but this contradicts the choice of aim(m < n, i < +) above (i.e. clause (a)of Definition 1.1(3) case l = 3). This contradiction finishes the inductionstep in the proof of hence the proof of 1.10(1).

For part (2), we have pi : i < + as above. In VP, if Y

= {i :pi GP} has cardinality

+, then ai0, . . . , ain1 : i Y

cannot wit-

ness rk5({a0, . . . , an1}, M; ) + 1 so there is a function F

0 : Y


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14 SAHARON SHELAH

witnessing it; i.e.

P if |Y

| = + theni Y

& j Y

& i = j & F

0(i) = F

0(j)

> rk5({ai0, . . . , ain1} {a

j0, . . . , a

jn1}, M; ).

If |Y

| , let F

0 : Y

be one to one. Let pi qi P, qi F

0(i) = i.As P is +-2-linked, for some function F1 : + we have (i , j 2 and >}, eachR0,1, a binary predicate, Q0, a unary predicate and

QM0, = { < : 0 & F(, )},


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RM0,1, = {(, ) : 0 & 1 & F(, )}.

By Pr1

() we know that rk0(M) 1.A pair (u, h) is called an n-approximation if u n2, h : u u n and

for every < 1 there is w [] such that:

(1) u = {n : w} and n = n for distinct , w(2) rk

0(w, M) (3) F(, ) n = h( n, n) for , w; hence

M |= Rn,n,h(n,n)[, ]

for , w.

Note that ({}, {((, ), )}) is a 0-approximation. Moreover

()0 if (u, h) is an n-approximation and u

then there are m > n and an m-approximation (u+

, h+

) such that:(i) u \ {} (!+)( + u+),

(ii) (!2+)( + u+) (where !2x means there are exactly 2xs)

(iii) u+ n u and(iv) if 1, 2 u

+ then

[h(1n, 2n) h+(1, 2) or (1n = 2n =

& 1 = 2)]

[Why? For each < 1 choose w satisfying (1), (2) and (3) for + 1,now apply the definition of rk0 (if w = {

l : l < |w|},

k, k < |w|

we apply it to k) to get w+ = w {} satisfying (1), (2) and (3)

for , then choose m (n, ) such that m : w+

is withno repetitions. Lastly as there are only countably many possibilities form, { m : w

+}, {( m, m, F(, ) m) : , w

+}

for < 1, so one value is gotten for uncountably many , let be one of

them. Choose m = m, u+ = { m : w+} and define h

+ to satisfy(3).]

Repeating |u|-times the procedure of ()0 we get

()1 ifu = {l : l < k} n2 (no repetition), (u, h) is an n-approximation

then there are m, u+ = {+l : l < 2k} and h+ such that (u+, h+) is

an m-approximation for some m > n and(i) l

+2l , l

+2l+1,

+2l =

+2l+1,

(ii) if l < k, i < 2 then h(l, l) h+

(

+

2l+i,

+

2l+i) and(iii) ifl1 = l2, l1, l2 < k and i, j < 2 then h(l1 , l2)h+(+2l1+i,

+2l2+j

).

Consequently we have:

()2 there are sequences ni : i < and (ui, hi) : i such thatni < ni+1, (ui, hi) is an ni-approximation and (ui, hi), (ui+1, hi+1)are like (u, h), (u+, h+) of ()1.


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Now, let ni : i < and (ui, hi) : i be as in ()2. Define

P = { 2 : (i )(ni ui)}.

By ()1 for (ui+1, hi+1) we know that P is a perfect set. We claim thatP P B. Suppose that , P and = . Then ni() =

ni()for some i() < and the sequence hi(ni, ni) : i() i < is -increasing and (as (ui, hi) are approximations) (

ni, ni, hi(

ni, ni))

T is increasing for i [i(), ). The case = P is easier. The claimis proved. 1.121.9

Theorem 1.13. Assume NPr1() and = 0 . Then for some c.c.c.

forcing notionP, |P| = andP20 = and in VP we have:

(*) there is a Borel set B 2 2 such that :(a) it contains a -square. i.e. there are pairwise distinct

2

for < such that (, ) B for , < (b) let V |= 0 = 1; B contains no

+1 -square, i.e. there are

no 2 (for < +1 ) such that [ = = ] and

(, ) B for , < +

(c) B contains no perfect square.Actually B is a countable union of closed sets.

Proof Stage A: Clearly for some () < 1 we have NPr1()(). Let M

be a model with universe and a countable vocabulary such that rk1(M) 0 then

tqm = tp0

m {(qal1

, qal2

) : {0, 1} and distinct l1, l2 < |u[p0]|

satisfying gp0

(a0l1, a0l2

) = m}

and if m [m[p0], m[q]), m = gq(, ) and = then

tqm

= {(ql, q

l) : l nq}

(vii) if m [m[p0], m[q]) then for one and only one pair (, ) we havem = gq(, ) and for this pair (, ) we have = , {, } u[p0]and {, } u[p1]

(viii) The function fq is determined by the function gp and clauses 8, 9 ofstage A.


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20 SAHARON SHELAH

Of course we have to check that no contradiction appears when we define fq

(i.e. we have to check q of the Explanation inside stage A for q). So suppose

that w0, w1 u[q], l n[q], u, h are as in q. If w0 u[pi] (for somei < 2) then gq(, ) < m[p0] for , w0 and hence g

q[w1 w1] m[p0].

Consequently either w1 u[p0] or w1 u[p1]. If l = nq then necessarily

w0 = w1 so we have nothing to prove. If l < nq then (u, h) Dom (fp

0)

(and fp0

= fp1

) and clause 8 of stage A applies.If w0 is contained neither in u[p

0] nor in u[p1] then the function gq satisfiesgq(, ) [m[p0], m[q]) for some , w0 hence l = n

q and so as {q l : w0} = {

q l : w1} clearly w0 = w1, so we are done.

Next we have to check condition 7. As we remarked (in the Explanationinside Stage A) we have to consider cases of (u, h) such that fq(u, h) = 1only. Suppose that u, l < l() nq, ei, h,

u, u and h are as

in 7 (and fq

(u, h) = 1). Let w u[q] , 0, 1 w be such thatu = {ql() : w}, ei() = qil() (for i = 0, 1). If w u[p

i]for some i < 2 then we can apply clause 7 for pi and get a contradiction(if l() = nq then note that {qnp : w} are already distinct). Since w\{0, 1} implies g

q(, 0) = gq(, 1) (by the relation between h and

h) we are left with the case w \ {0, 1} u[p0] u[p1], 0 u[p

0] \ u[p1],1 u[p

1] \ u[p0] (or conversely). Then necessarily 0 = a0k0

, 1 = a1k1

for

some k0, k1 [0, |u[p0]|). Now k1 = k

M(w \ {0}) = kM(w \ {1}) = k0 by

the requirements in condition 7. Now we see that for each i < 1

M |= M(w \ {0})[w \ {0, 1} {aik1

}]

and this contradicts the fact that M(w\{0}), 1 witness rk1(w\{0}, M) =1.

Stage C: |P| = hence P20 . We shall get the equality by

clause () at stage E below.

Stage D: The following subsets ofP?? are dense (for m,n < , < ):

I1m = {p P : m[p] m}

I2n = {p P : np n}

I3 = {p P : u[p]}

Proof Let p P, 0 \ u[p] be given, we shall find q, p q I1m[p]+1 I

2n[p]+1 I

30

: this clearly suffices. We may assume that u[p] =

and 0 < . Let

(a) nq = np+1, mq = mp+2|(u[p])|, mq = mpmq, uq = up{0},(b) for up we let q =

p 0,

q0

(np+1)2 is the sequence con-stantly equal 1,


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(c) gq is any two place function from uq to mq extending gp suchthat gq(, ) = 0, gq(, ) = 0 for = and

gq(, ) = gq(, ) & (, ) = (, ) (, ) up up & (, ) up up

(d) tqm is defined as follows:() if m < mp, m = 0 then tqm = t

pm {(0 0, 1 0) : (0, 1)

tpm (n[p]2 n[p]2)}() if m [mp, mq), m = gq(, ), = then tqm = {(

ql,

ql) :

l nq}(e) fq extends fp and satisfies 7,8 and 9 of stage A (note that fq is

determined by gq)

Now check [similarly as at stage B].

Stage E: We define some P-names

(a) =

{p : p G

P} for <

(b) Tm =

{tpm : p G

P} for m <

(c) g

=

{gp : p GP}

Clearly it is forced (P) that

() g

is a function from {(, ) : , < } to

(Why? Because I3 are dense subsets ofP and by clause 5 of stageA)

()

2

(Why? Because both I2n

and I3

are dense subsets of P)()

=

for = (< )

(Why? By clause 2. of the definition of p P.)() T

m

l


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22 SAHARON SHELAH

Stage F: We want to show (*)(c) of the conclusion of 1.13. Let P ={p P : u[p] }. Clearly P 2 ti ni, m[pi] > max Rang(mi).

Since pi P (0, 1) Tmi(0,1) for 0, 1 ti

ni2 we easily get (byclause 8 of the definition of P, stage A) that (ti

ni2, mi) Dom(fpi). By

clause 7 (of the definition of P) (and 1.2(2)+ clause 8 of the definition ofP) we deduce that

fpi+10 (ti+1

ni+12, mi+1) < fpi0 (ti

ni2, mi)

for each i < and this gives a contradiction (to the ordinals being well

ordered).Stage G: To prove ()(b) of theorem 1.13 we may assume that V |=

0 = 1 < . Let P1 = {p P : u[p] 1}


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F we have VP1 |=the Borel set B

does not contain a perfect square.Suppose that after adding Cohen reals (over VP1 ) we have a +1 -square

contained in B. We have +1 -branches ( < +1 ), each is a Pv -name for

some countable v \ 1. By the -system lemma wlog we assume that = v v = v

. Working in VP1v we see that Pv\v is really theCohen forcing notion and

is a Pv\v -name. Without loss of generality

v = [1, 1 + ), v = v {1 + + } and all names

are the same

(under the natural isomorphism). So we have found a Cohen forcing name

VP1+ such that:

if c0, c1 are (mutually) Cohen reals over VP1+

then VP1 [c0, c1] |= (c0,

c1) B

&

c0 =

c1.

But the Cohen forcing adds a perfect set of (mutually) Cohen reals. Byabsoluteness this produces a perfect set (in VP1 ) whose square is containedin B

. Once again by absoluteness we conclude that B

contains a perfect

square in VP already, a contradiction. 1.13 1.10

Remark 1.14. Note that if B is a subset of the plane (, ) which is G(i.e.

n n,large enough, we have ({ : u} { }, n) K.

The following depends on 3:

Theorem 1.15. Assume MA and 20 1(0) or 20 > . Then: there

is a Borel subset of the plane with a -square but with no perfect square iff < 1(0).

Proof The first clause implies the second clause by 1.12. If the secondclause holds, let (0) and < 1, by 3.2(6) letting i

2 for i <


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24 SAHARON SHELAH

be pairwise distinct we can find an -sequence of (2, 2)-trees T such that(i, j)

n

(lim Tn) for i, j < and degsq(T) = (just use A = {(i, j) :

i, j < } there). By 3.2(3) the setn

(lim Tn) contains no +1(0)-square.

1.15th1.12

Fact 1.16. Assume P is adding > Cohen reals or random reals and > 20 . Then in VP we have

() there is no Borel set (or analytic) B 2 2 such that(a) there are

2 for < such that [ = = ], and(, ) B for , < .

(b) B contains no perfect square.

Proof Straight as in the (last) stage G of the proof of theorem 1.13

(except that no relevance of (7) of Stage A there).Let P be adding r : < , assume p P forces that: B

a Borel

set, : < are as in clause (a), (b) above. Let

be names in

Pv = P {r : v}, and B

be a name in Pv = P {r

: v} where

v, v are countable subsets of. Wlog v : < (20)+ is a -system with

heart v and otp(v \ v) = otp(v0 \ v). In VPv we have B

and 20 = (20)V,

so wlog v = and otp(v) does not depend on .Without loss of generality the order preserving function f, from v onto

v maps to

. So for Q=Cohen in the Cohen case we have a name

such that Cohen

(r

) 2 is new, CohenCohen (

(r

1);

(r2

)) B,and we can finish easily. The random case is similar. 1.161.13

Conclusion 1.17. (1) For (1, 1) the statement () of 1.16 is notdecided by ZFC +20 > 1 (i.e. it and its negation are consistentwith ZFC).

(2) 1.16 applies to the forcing notion of 1.13 (with instead of 20)

Proof 1) Starting with universe V satisfying CH, fact 1.16 shows theconsistency of yes. As by 1.7(1) we know that 1(0) 1 and 1 > (by assumption), Theorem 1.15 (with the classical consistency of MA+20 > 1) gives the consistency of no (in fact in both cases it works forall simultaneously).

2) Left to the reader.

2. Some model theoretic related problemsWe turn to the model theoretic aspect: getting Hanf numbers below the

continuum i.e. if L1, has a model of cardinality 1(0) then ithas a model of cardinality continuum. We get that Pr1() is equivalentto a statement of the form if L1, has a model of cardinality thenit has a model generated by an indiscernible set indexed by 2 (the


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indiscernibility is with respect to the tree (2, , ,


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26 SAHARON SHELAH

is a model of (just for each subformulan2 : v is finite nonempty and its members are pairwise -

incomparable and for some n, v n2 n+12}.Z = {(v, (. . . , x, . . . )v) : v Y, a formula in T1 with the set of

free variables included in {x : v} and for every < 1 thereare a R

M for v such that : [ = from v a = a] and

rk0({a : v}, M) and M |= [. . . , a , . . . ]v}.

We say for (vl, l) Z (l = 1, 2) that (v2, 2) succ(v1, 1) if for some v1 (called (v1, v2)) we have v2 = (v1\{}) { 0, 1} and lettingfor i < 2 the function hi : v1 v2 be hi() is if = and it is i if = , we demand for i = 0, 1:

2 1(. . . , xhi(), . . . )v1 .

Choose inductively (vl, l) : l < such that (vl+1, l+1) succ(vl, l) isgeneric enough, i.e.:

()1 if = (x0, . . . , xk1) L(T1) then for some l < for everym [l, ) and 0, . . . , k1 vm we have:

m (x0 , . . . , xk1) or m (x0, . . . , xk1)

()2 for every p(x) and for every function symbol f = f(x0, . . . , xn1)(note: in T1 definable function is equivalent to some function sym-bol), for some l < for every m [l, ), for every 0, . . . , n1 vm

for some (x) p(x) we have

m (f(x0), . . . , f (xn1))

It is straightforward to carry the induction (to simplify you may demandin ()1, ()2 just for arbitrarily large m [, ), this does not matterand the stronger version of ()1, ()2 can be gotten (replacing the

>2 bya perfect subtree T and then renaming a for lim(T) as a for

2)).Then define the model by the compactness.

2. 1.If not, then NPr1() hence for some model M with vocabulary , ||

0, cardinality we have ()def= rk0(M) < 1. Let () L1,() be

as in 2.3 below, so necessarily M |= (). Apply to it clause (2) whichholds by our present assumption (with RM = ), so () has a model M1as there, (so M1 |= ()). But {a :

2} easily witnesses rk0(M1) = ,moreover, for every nonempty finite w {a :

2} and an ordinal wehave rk0(w, M) . This can be easily proved by induction on (using()2(b) of (2) (and =

2 a = a of ()2(a))). 2.12.1


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Fact 2.3. (1) For every < + and vocabulary , || , there isa sentence L+,[] (of quantifier depth ) such that for any

-model M:M |= iff rk

0(M; < 0) = .

(2) For every < +, l {0, 1} and vocabulary , || there is asentence L+,(

)[] ( is the quantifier there are

many xs) such that for any -model M: M |= l iff rkl(M; 0 the following are equivalent(a) Pr+(; )(b) If M is a model, (M) countable, R, R0 (M) unary predi-

cates, |RM0 | , |RM| then we can findM0, M1, a(

2)

such that(i) M1 is a model of the (first order) universal theory of M

(and is a (M)-model)(ii) a RM1 for 2 are pairwise distinct

(iii) M1 is the closure of{a : 2}M0 under the functions

of M1 (so


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28 SAHARON SHELAH

() M1 also includes the individual constants of M; ingeneral M1 = 2

0

() if (M) has predicates only then |M1| = {a : 2} |M0|)

(iv) M0 is countable, M0 M, M0 M1, M0 = clM(M0

RM0 ), RM10 = R

M00 ( R

M0 ); in fact we can have:

(*) (M1, c)cM0 is a model of the universal theory of(M, c)cM0

(v) for every n < and a quantifier free first order formula = (x0, . . . , xn1) L((M)) there is n

< suchthat: for everyk (n, ) and0, . . . , n1

2, 0, . . . , n1 2 satisfying

m


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We need here the for every < + because we want to fix elements ofRM0 , and there are possibly choices.

(a) (b) Like the proof of 2.3; assume NPr(; ), < +, let M wit-ness it, choose R0 = +1, R = , w.l.o.g (M) = {Rn, : n < , < }, Rn,is n-place, in M every quantifier free formula is equivalent to some Rn,, letRn,k =: {(i0, . . . , in1, , ) : M |= Rn,(i0, . . . , in1), {i0, . . . , in1} is withno repetition, increasing for simplicity, and rk({i0, . . . , in1}, M; ) = ,with rk({i0, . . . , in1}, M; ) +1 being witnessed by ({i0, . . . , in1}) =Rn,, k({i0, . . . , in1}) = k} where the functions , k are as in the proofof 1.13. Let M be (,


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30 SAHARON SHELAH

For a RM let (a) be the such that a RM (e.g. RM = \


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m = m : < (), m (0, ) for some w A we have

|w RM | =

m and rk(w; M,) + k. Then for = + 1, choose distinct

ai A RM (i < +m+m ) and use polarized partition (see Erdos,

Hajnal, Mate, Rado [EHMR]) on ai : i < : < (). For ()infinite use A RM such that wA = { : A RM = } is finite non-empty,

wA |A RM | and proceed as above. 2.10 2.8

Claim 2.11. Let () < , 0 < < (),

= (+1, ) : < ()

(for ).

(1) IfPrn(n) forn < , and for some 0 there is a tree T

>(0 )of cardinality 0 with

() -branches then:

every first order sentence which has a n-model for each n, also

has a -model moreover, if T is a first order theory of cardinality 0 and

every finite T T has a n-model for each n then T has a-model

(2) So if = for are as above then we have 0 -compactnessfor the class of -models. Where() a class K of models is -compact when for every set T of

first order sentences, if every finite subset of T has a model inK then T has a model inK.

(3) In part (1) we can use n with domain n, if n n+1 and () ={wn : n < }.

Proof Straight if you have read [Sh 8], [Sh 18], [Sh 37] or read the proofof 2.12 below (only that now the theory is not necessary countable, no typesomitted, and by compactness it is enough to deal with the case () is finite).2.11 2.9

Claim 2.12. Let () < 1, = (+1,

) : < () for each 1

(and strictly increasing with). IfPr() for every < 1 and

1 2

0

and L1, and for each < 1 there is a -model satisfying then

there is a 1-model satisfying .

Proof For simplicity, again like [Sh 8], [Sh 18], [Sh 37]. Let M be a

model of for < 1. By expanding the Ms, by a pairing function and

giving names of subformulas of we have a countable first order theory Twith Skolem functions, a countable set of 1-types and M+ such that

(a) M+ is a -model of T omitting each p

(b) if M is a model of T omitting every p then M is a model of .

Now as in the proof of 2.1 we can find a model M+ and a for < (), 2 such that


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32 SAHARON SHELAH

() M+ a model of T

() a RM and = a = a

() for every first order formula (x0, . . . , xn1) L,((T)) and (0), . . . ,(n 1) < () ordinals, there is k < such that: if 0, . . . , n1 2, 0, . . . , n1

2 and k : < n is with no repeti-

tions and k = k then M+ |= (a

(0)0 , . . . , a

(n1)n1 )

(a(0)0 , . . . , a

(n1)n1 )

() if (x0, . . . , xn1) is a term of (T) and (0), . . . , (n 1) < (),and p then for some k < for any 0, . . . , (n 1) k2 pairwisedistinct there is (x) p(x) such that:

() 2 M+ |= (a

(0)0 , . . . , a

(n1)n1 ).

(i.e. this is our way to omit the types in )() If (0), . . . , (n 1) < (), (x0, . . . , xn1) is a term of (T), and

m < n, then for some k < , we have() if 0, . . . , n1

2, and 0, . . . , n1 2 and k = k

and k : < is without repetitions and () < (m) = then

M+ |= Q(m)((a(0)0 , . . . , a

(n1)n1 )) & Q(m)((a

(0)0 , . . . , a

(n1)n1 )

(a(0)0 , . . . , a(n1)n1 ) = (a

(0)0 , . . . , a

(n1)n1 ).

Now choose Y 2 of cardinality 1 and let M

be the (M)-

reduct of the Skolem hull in M+ of{a : < () and Y}. Thisis a model as required.

2.122.10

Conclusion 2.13. If V0 |= GCH, V = VP0 for some c.c.c. forcing notion

P then e.g.

() if3 < 20 , 0, , + , +, ++ (see [Sh 8]) i.e.,

letting

0 = (0, ), (, +), and 1 = (, +), (+ , ++),

for any countable first order T, if every finite T T has a 0-modelthen T has a 1-model.

() : () : () if

+ +1 and 0

1

() 2

0 (and versions like 2.11(1)).

Proof Why? By 2.10 if = ( , ) : < (), () thenPrn() (really k() for k depending on n only suffices, see [EHMR]).Now ccc forcing preserves this and now apply 2.11. Similarly we can use+-cc forcing P and deal with cardinals in the interval (, 2) in VP. 2.132.11


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Remark 2.14. We can say parallel things for the compactness of (), for singular 20 (or + |T | < number of -branches of T), e.g. we get

the parallel of 2.13.In more details, if V0 = V

P, P satisfies the +-c.c. then

() in VP0 , for any singular (, 2) such that V0 |= is strong

limit we have() the class

{(,


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34 SAHARON SHELAH

( u)[h0() = h1() = ]

u = Rang(h0) Rang(h1)

g is a 2-place function from u to g(hi(), hi()) = g(, ) for i < 2 and ,

u

degsqT(u, g) (so (u, g) pfap).

(2) We define degsqT(u, g) = iff

for every ordinal , degsq(u, g)

(so = 1, = are legal values).(3) We define degsq(T) =

{degsqT(u, g) + 1 : (u, g) pfapT}.

Claim 3.2. Assume T is an -sequence of (2,2)-trees.

(1) For every (u, g) pfapT, degsqT(u, g) is an ordinal, or 1. Anyautomorphism F of (2, ) preserves this (it acts on T too, i.e.

degsqT(u, g) = degsqF(Tn):n


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f((1) n, (2) n) then R(u,f)(0, 1, . . . , k1) and we can then prove

rk({0, . . . , k1}, M) degsqT(u, f)

(by induction on the left ordinal). But M is a model with countable vocab-ulary and cardinality = ()+1(0). Hence by the definition of()+1 we

have rk(M) ()+1, so ()+ 1 rk(M) degsq(T) () (by previoussentence, earlier sentence and a hypothesis respectively). Contradiction.

4) Let

W = { : is a (strictly) decreasing sequence of ordinals, possibly empty}.

We choose by induction on i < , ni and an indexed set (uix, f

ix,

ix) : x

Xi such that

(a) ni < , n0 = 0, ni < ni+1(b) Xi finite including

j


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36 SAHARON SHELAH

We then let

Tn = {(, ) : for some i < and x Xi and , uix we have

fix(, ) = nand for some ni we have (, ) = (

, )}

and T = Tn : n < . Now it is straight to compute the rank.5) By the completeness theorem for L1,(Q) (see Keisler [Ke71])6) By the proof of 1.13. 3.23.2

* * *Now we turn to -Souslin sets.

Definition 3.3. Let T be a (2, 2, )-tree. Let set(T) be the set of all pairs(u, f) such that

(n = n(u, f))[u n2 & f : u u n & , u (,,f (, )) T].

We want to define degsqT(x) for x set(T). For this we define by inductionon the ordinal when degsqT(x) :

Case 1 = 1degsqT(u, f) iff (u, f) set(T)

Case 2 limitdegsqT(u, f) iff degsqT(u, f) for every <

Case 3 = + 1degsqT(u, f) iff for every

u, for some m > n(u, f) thereare (u, f) set(T) and functions h0, h1 such that degsqT(u

, f) and

(i) n(u, f) = m(ii) hi is a function from u to m2

(iii) hi() for i < 2(iv) for u we have h0() = h1() =

(v) for 1 = 2 u, i < 2 we have f(1, 2) f(hi(1), hi(2))

(vi) for u we have f(n, n) f(, )

Lastly, degsqT(u, f) = iff for every ordinal we have [ degsqT(u, f) ]

Also let degsq(T) = degsqT({}, {< , >}).

Claim 3.4. (1) For a (2, 2, )-tree T, for (u, f) set(T), degsqT(u, f)is an ordinal or infinity or = 1. And similarly degsq(T). All are

absolute. Also degsq(T) +

implies degsq(T) = and similarlyfor degsqT(u, f).(2) degsq(T) = iff P prjlim T (= {(, )

2 2 : for some ,

n


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is adding Cohen reals for = ??1() some iff for some P,Pprjlim(T) contains ?? +(0)-square. .

(3) If() = degsq(T) < +, thenprj lim(T) does not contain a()+1()-square.

Proof 1) Easy.2) Assume degsq(T) = , and note that = {degsqT(u, f) : (u, f)

set(T) and degsqT(u, f) < }\{} is an ordinal so (u, f) set(T) & degsqT(u, f) degsqT(u, f) = (in fact any ordinal sup{degsqT(u, f) + 1 :(u, f) set(T)} will do). Let set(T) = {(u, f) set(T) : degsqT(u, f) =}. Now

()1 there is (u, f) set(T)()2 for every (u, f) set

(T) and u we can find (u+, f+)

set(T) and for e = 1, 2 he : u u+ such that ( u)( he()),(, u)(f(, ) f+(he(), he()), ( u)[h1() = h2() =].

[Why? As degsqT(u, f) = it is + 1 so by the definition we can

find (u+, f+), h1, h2 as above but only with degsqT(u+, f+) , but this

implies degsqT(u+, f+) = .]

()3 for every (u, f) set(T) with u = { : n2} (n1)2 (no

repetition) we can find n2 > n1 and (u+, f+) set(T) with u+ =

{ : n+12} (n2)2 (no repetitions) such that(i) n+12 n

(ii) for 1, 2 n+1

2, 1 n = 2 n f(1n, 2n)f+

(1 , 2)(iii) for n+12, f(n, n) f(, )

[Why? Repeat ()2 2n times.]

So we can find : n2, fn by induction on n such that :

n2 iswith no repetition, degsqT({ :

n2}, fn) = , and for each n clauses (i),(ii), (iii) of ()3 hold, i.e. 1, 2

n+12, 1 n = 2 n fn(1n, 2n) fn+1(1 , 2) and for

n+12 we have fn(n, n) fn+1(, ) and of

course { : m2} (kn)2, kn < kn+1 < .

So we have proved that the first clause implies the second (about theforcing: the degsq(T) = is absolute so holds also in VP for any forcingnotion P). Trivially the second clause implies the third and fourth. So

assume the third clause and we shall prove the first. By 1.8 2+() is welldefined (e.g. +), but

3+

() = 2+

() by 1.5(3), let P be the forcing

notion adding 2+

() Cohen reals. By 1.1(2) in VP, +() 2+

()

20 , and so there are pairwise disjoint i 2 for i < +() such that

(i, j) prjlim(T) for i, j < +(). Lastly we prove forcing implies firstin VP. By part (3) of the claim proved below we get for every < +, as


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38 SAHARON SHELAH

() +() that [ = degsq(T)]. Hence degsq(T) +, but by part

(1) this implies degsq(T) = .

3) Just like the proof of 3.2(3). 3.43.4

We shall prove in [Sh 532]

Claim 3.5. Assume () < + and < ()().

(1) For some ccc forcing notion P, in VP there is a -Souslin subsetA = prj lim(T) (where T is a (2, 2, )-tree) such that() A contains a -square but degsq(T) ()

(2) For given B (2 2)V of cardinality we can replace () by() A (2 2)V = B but degsq(T) ().

Remark 3.6. The following says in fact that colouring of pairs is enough:say for the Hanf number of L1, below the continuum, for clarification see3.8.

Claim 3.7 (MA). Assume < 20 and () < 1 is a limit ordinal, 1, 0


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() for every n < and pairwise distinct 0, . . . , n1 < thereare < and pairwise distinct 0, . . . , n1 < such that

k


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(3) For a model M, and

w [M]def

= {u : u = (u1, u2), ui RMi are finite nonempty and(c)(a, b)[a u1 & b u2 F(a, b) = c]}

we shall define the truth value of rkrcl(w, M; < ) by inductionon the ordinal (for l = 0, 1, can be omitted). If we write winstead of w1, w2 we mean w1 = w R

M1 , w2 = w R

M2 (here

RM1 RM2 = helps). Then we can note

()0 & rkrcl(w, M; < ) rkrcl(w, M; < )

()1 rkrcl(w, M; < ) ( limit ) iff


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(b) M |= [ai0, . . . , ain1] (for i < ; so there are no repeti-

tions in ai0, . . . , ain1)

(c) for i < j < , aik = ajk but if m < n and (am RM1 ak / R

M1 ) then a

im = a

jm

(d) ifam1 RM1 , am2 R

M2 then for any i, j, F

M(aim1 , ajm2) =

FM(am1 , am2).Case 4: l = 2. Like case 3 but in addition

(e) am = a0m for m < n

Case 5: l = 5. Like case 3 except that we replace clause (a) by(a) for every function H, Dom (H) = , |Rang(H)| < for

some i < j < we have H(i) = H(j) and

rkrcl({aim, ajm : m < n}, M; < ) .

Case 6: l = 4. Like case 4 using clause (a) instead (a).(4) For M as above and c QM we define rkrc(M, c; < ) as

sup{rkrc(w, M; < ) + 1 : w [M] and(a w1)(b w2)[F(a, b) = c]}.

(5) Let Prrd(,, ) mean rkrc(M, c; < , ) for every M for some

c M when M is such that |RM1 | = 1, |RM2 | = 2, |(M)| 0,

FM : RM1 RM2 Q

M, |QM| = 1. Let NPrrd(,, ) mean its

negation and rd(, ) be the minimal such that Prrd(,,, ).

Remark 4.2. The reader may wonder why in addition to Prrc we use the

variant Prrd. The point is that for the existence of the rectangle X1 X2with F (X1 X2) constantly c

, this constant plays a special role. So inour main claim 4.6, to get a model as there, we need to choose it, one out of1, but the other choices are out of . So though the difference between thetwo variants is small (see 4.5 below) we actually prefer the Prrd version.

Claim 4.3. The parallels of 1.2 (+statements in 1.1), also 1.3, 1.5(2), 1.6,1.10 hold.

Claim 4.4. (1) If wi [RMi ]

for i = 1, 2 then

rkrcl(w1, w2, M; ) rkl(w1 w2, M; ).

(2) IfRM

1 = RM

2 (abuse of the notation) then rkrcl

(M; < ) rkl

(M; ).(3) If 1 = 2 = then Pr(, ) Prrc(1, 2; , ).

Claim 4.5. rd(, ) = rc(, ) if is a successor ordinal or cf () > .

Claim 4.6. Assume < 1, 2. Then the following are equivalent:

A. Prrd1+(1, 2; , )


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44 SAHARON SHELAH

B. Assume M is a model with a countable vocabulary, |RMl | = l forl = 1, 2, PM = , QM = , and FM a two-place function (really

just F(RM1 RM2 ) interests us) and the range of F(RM1 RM2 ) isincluded in QM and G is a function from [RM1 ]

[RM2 ] to PM.

Then we can find (M)-models M0, N and elements, c, a, b (for

2) such that :(i) N is a model with the vocabulary of M (but functions may be

interpreted as partial ones, i.e. as relations)(ii) a R

N1 , b R

N2 are pairwise distinct and F

N(a, b) = c(

N)(iii) M0 countable, M0 M, c

QM, M0 is the closure of (M0 PM) {c} in M, in fact for some M0 M we have M0 =closure of PM

0 {c}, c M0,

(iv) M0 N, PM0

= PN

,(v) |N| = {(a, b, d) : is a (M)term , 2, 2 and d

M0}(vi) for {l : l < l()}, {m : m < m()}

2 (both without repeti-tions non-empty) there is d1 PM0 such that if d PM0 andquantifier free formulas l,m are such that

N |=

l


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Proof (B) (A) Toward contradiction assume NPrrd+(1, 2; )hence there is a model M witnessing it, so |(M)| . So c QM

rkrc1(M, c; ) < + (note that Prrd was defined by cases of rkrc(M,c,)).Let {i(x, y) : i < } list the quantifier free formulas in L,((M)) with

free variables x, y. Let {ui : i < } list the finite subsets of . For c QM

and a0, . . . , a()1 RM

1 , b0, . . . , bm()1 RM

2 (a = a0, . . . , a()1,

b = b0, . . . , bm()1 and for notation let an+1+ = b) let

c,a,b = rkrc1(({a0, . . . , a()1}, {b0, . . . , bm()1}), M

, c; ),

and kc,a,b, c,a,b be witnesses for rkrc1((a, b), M, c; , ) c,a,b + 1. Let

i(c, a, b) < be such that c,a,b is a conjunction of formulas of the form

j(xl, ym) for j ui(c,a,b). We define M:

the universe is |M

|,the function FM

, relations RM

1 , RM

2 , QM , PM

, the pairingfunction on ordinals,

Rn =

(i,a,b) : a RM1 , b RM2 and

if |ui| > n then M |= j [a, b]where j is the n-th member of ui

and let Hc be one to one from rkrc(M, c; , ) into

; we define the function G:

Gc(a, b) = H(Gc,0(a, b), Gc,1(a, b), Gc,2(a, b)) = H(kc,a,b, c,a,b, ic,a,b).

Now we can apply statement (B)-of 4.6 which we are assuming and get

M0, N , c, a, b (for 2) satisfying clauses (i)(vii) there. So c M0 M

N, so = rkrc(M, c; ) satisfies < , even < +. Clearlyrkrc1(N, c; ) .

Consider all sequences

: < (), m : m < m(), d1, d, ,m : < (), m < m(),a : < (), bm : m < m()

which are as in clause (vi) of (B).Among those tuples choose one with = rkrc1({a : < ()}, {bm : m f(2). For each < +, this property is-Borel. 4.9 4.7

Conclusion 4.10. If is an n-Souslin subset of2 2 containing a

(rd1(n), rd1(n))-rectangle thenit contains a perfect rectangle. (Note:n can replaced by if cf (S0(), ) = e.g. 4, by [Sh g, IX, 4].)

Conclusion 4.11. (1) For l < 6, Prrcl(+, (2

+)+; ).

(2) IfV = VP0 , P |= c.c.c. and V0 |=GCH then Prrc1(1, 3).

Proof 1) For a model M, letting (1, 2) = (+, (2

+)+) choose (for

m = 1, 2) ami a nonempty sequence from RMm for i < m, {ami : i < m}pairwise disjoint. For (i, j) 1 2 let i,j = rkrc(a

1i , a

2j , M) with wit-

nesses kM(a1i , a2j ),

M(a1i , a2j ) for rkrc

1(a1i , a2j , M) > i,j . As 2 = (2

1)+,

|(M)| , for some B2 2, |B2| = 2 and for every i < 1 the followingdoes not depend on j B2:

kM(a1i , a2j ),

M(a1i , a2j ).

Similarly, there is B1 1, |B1| = 1(= +) such that for j = min(B2) thevalues

kM(a1i , a2j ),

M(a1i , a2j )

are the same for all i B1; but they do not depend on j B2 either. So for

(i, j) B1 B2 we have kM(a1i , a2j ) = k, M(a1i , a2j ) = . Let k speakon a1i , for definiteness only. Choose distinct i in B1 (for <

+). Wlog

rkrc1(a1i0 , a1j , M) rkrc

1(a1i , a1j , M).

Now a1 give contradiction to rkrcl(a1i0 , a

2j , M) > i0,j .

2) This can be proved directly (or see [Sh 532] through preservation byc.c.c. forcing notion of rank which are relations of rkrc similarly to 1.10.

Remark 4.12. 1) IfT is an (, )-tree and A B lim(T), with A, B uncountable (or just not scattered) then lim(T) contains a perfect rectangle.Instead lim(T) (i.e. a closed set) we can use countable intersection of opensets. The proof is just like 1.17.

2) We can define a rank for (2, 2, )-trees measuring whether prj lim(T) 22 contains a perfect rectangle, and similarly for (, )-tree T measuringwhether lim(T) contains a perfect rectangle. We then have theorems parallelto those of 1. See below and in [Sh 532].

The use of below is just notational change.


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48 SAHARON SHELAH

Definition 4.13. For T a (, )-tree we define a function degrcT (rectangle

degree). Its domain is rcpr(T)def= {(u1, u2) : for some l < , u1, u2 are

finite nonempty subsets of l and g a function from u1 u2 to such that(0, 1) T for i ui}. Its value is an ordinal degrcT(u1, u2) (or 1 or). For this we define the truth value of degrcT(u1, u2) by inductionon the ordinal .

case 1 = 1degrcT(u1, u2) 1 iff (u1, u2) is in rcpr(T)

case 2 limitdegrcT(u1, u2) iff degrcT(u1, u2) for every <

case 3 = + 1degrcT(u1, u2) iff for k {1, 2},

uk we can findl() < , and functions h0, h1, such that :

Dom (hi) = u1u2, [ u1u2 hi() l()]such that h0(

) = h1(), u1k h0() =

h1() and letting u1i = Rang(h0ui) Rang(h1ui)

we have degrcT(u10, u

11) .

Lastly define: degrcT(u0, u1) = iff

[degrcT(u0, u1) ] (

an ordinal or ).Also degrc(T) = degrcT({}, {}).

Claim 4.14. Assume T is in an (, )-tree.

(1) For every (u0, u1) rcpr(T), degrcT(u0, u1) is an ordinal or or

1; if f is an automorphism of (>, )then degrcT(u0, u1) = degrcf(T)(f(u0), f(u1)).

(2) degrc(T) = iff there is a perfect rectangle in lim(T) iffdegrcT(u0, u1) 1 for some (u0, u1) (so those statements are ab-solute)

(3) If degrc(T) = () < 1 then lim(T) contains no

rc()+1(0), rc()+1(0)

-rectangle.

(4) If T = Tn : n < is a sequence of (, )-trees and degrc(Tn) (), and A =

n


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5) Let = rc()+1(), and let T = Ti : i < , degrc(Ti) () andA =

i


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50 SAHARON SHELAH

(2) np = n[p] < and p = [p] n[p] for up such that =

up p =

p,

(3) 0 < mp < and tpm

{l l : l np} closed under initialsegments and such that the -maximal elements have the length np

and tp,(4) the domain of fp is {u = (u1, u2) : for some l = l(u1, u2) n

p

and m = m(u1, u2) < mp we have u t

p l and if 1 u1,

2 u2, p1 l u1,

p2 l u2 then g

p(1, 2) = m} andfp(u) = (fp0 (u), f

p1 (u), f

p2 (u)) () (u1 u2) L1,((M)),

(5) a function gp : up1 up2 {0, . . . , m

p 1}, mp < ,

(6) tpm (np) = {(p,

p) : u

p0, u

p1 and m = g

p(, )}

(7) if = u tl, fp(u1, u2) = (, , ), l < l() np, for i = 0, 1

a function ei, has the domain u, [(l)( u ei,() tp

l())], [ / u & u e0,() = e1,()], [ u e0,() =e1,(

)] and fp(e0,1(u1) e1,1(u1), e0,2(u2) e1,2(u2)) = (, , )(so well defined)then < ,

(8) if l np, for = 1, 2 we have u up are non empty, the sequence

pl : u is with no repetition, and u = {p l : u} and

fp(u1, u2) is well defined,

thenfp2 (u

1, u

2) =

M(u1, u2),fp1 (u

1, u

2) =

pl

where is kM(u1, u2) and

f

p

0 (u

1, u

2, h) = rkrc(u1, u2, M),(9) if (u1, u2) Dom (f

p) then there are l, u1, u2 as above,(10) if (1, 2) t

pm (n2 n2) then for some 1 u

p1, 2 u

p2 we have

gp(1, 2) = m and 1 p1, 2

p2 .

4.144.11

Remark 4.16. We can generalize 4.13, 4.14, 4.15 to Souslin relations.

References

[CK] Chen C. Chang and Jerome H. Keisler. Model Theory, volume 73 ofStudies in

Logic and the Foundation of Math. North Holland Publishing Co., Amsterdam,1973.

[EHMR] Paul Erdos, Andras Ha jnal, A. Mate, and Richard Rado. Combinatorial settheory: Partition Relations for Cardinals, volume 106 of Studies in Logic andthe Foundation of Math. North Holland Publ. Co, Amsterdam, 1984.

[GcSh 491] Martin Gilchrist and Saharon Shelah. Identities on cardinals less than .Journal of Symbolic Logic, 61:780787, 1996.


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[HrSh 152] Leo Harrington and Saharon Shelah. Counting equivalence classes for co--Souslin equivalence relations. In Logic Col loquium 80 (Prague, 1980), vol-

ume 108 of Stud. Logic Foundations Math, pages 147152. North-Holland,Amsterdam-New York, 1982. eds. van Dalen, D., Lascar, D. and Smiley, T.J.[Ke71] H. Jerome Keisler. Model theory for infinitary logic. Logic with countable con-

junctions and finite quantifiers, volume 62 of Studies in Logic and the Foun-dations of Mathematics. North-Holland Publishing Co., Amsterdam-London,1971.

[Mo] Michael Morley. Omitting classes of elements. In The theory of models, page265. North Holland Publ. Co., 1965.

[Sh a] Saharon Shelah. Classification theory and the number of nonisomorphic mod-els, volume 92 of Studies in Logic and the Foundations of Mathematics. North-Holland Publishing Co., Amsterdam-New York, xvi+544 pp, $62.25, 1978.

[Sh c] Saharon Shelah. Classification theory and the number of nonisomorphic mod-els, volume 92 of Studies in Logic and the Foundations of Mathematics. North-Holland Publishing Co., Amsterdam, xxxiv+705 pp, 1990.

[Sh e] Saharon Shelah. Nonstructure theory. Oxford University Press, in prepara-tion.

[Sh g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford Logic Guides. Ox-ford University Press, 1994.

[Sh 8] Saharon Shelah. Two cardinal compactness. Israel Journal of Mathematics,9:193198, 1971 ,Notice ??18 (197).

[Sh 18] Saharon Shelah. On models with power-like orderings. Journal of SymbolicLogic, 37:247267, 1972.

[Sh 37] Saharon Shelah. A two-cardinal theorem. Proceedings of the American Math-ematical Society, 48:207213, 1975.

[Sh 49] Saharon Shelah. A two-cardinal theorem and a combinatorial theorem. Pro-ceedings of the American Mathematical Society, 62:134136, 1976.

[Sh 202] Saharon Shelah. On co--Souslin relations. Israel Journal of Mathematics,47

:139153, 1984.[Sh 262] Saharon Shelah. The number of pairwise non-elementarily-embeddable mod-els. The Journal of Symbolic Logic, 54:14311455, 1989.

[Sh 288] Saharon Shelah. Strong Partition Relations Below the Power Set: Consistency,Was Sierpinski Right, II? In Proceedings of the Conference on Set Theory andits Applications in honor of A.Hajnal and V.T.Sos, Budapest, 1/91, volume 60ofColloquia Mathematica Societatis Janos Bolyai. Sets, Graphs, and Numbers,pages 637638. 1991.

[Sh 532] Saharon Shelah. More on co--Souslin equivalence relations. in preparation.

Institute of Mathematics, The Hebrew University, Jerusalem 91904, Israel,

and Rutgers University, Mathematics Department, New Brunswick, NJ 08854,

USA

E-mail address: [email protected]

Documents

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