Saharon Shelah- Black Boxes

8/3/2019 Saharon Shelah- Black Boxes

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BLACK BOXES

SAHARON SHELAH

Abstract. We shall deal comprehensively with Black Boxes, the intentionbeing that provably in ZFC we have a sequence of guesses of extra structure onsmall subsets, the guesses are pairwise with quite little interaction, are far buttogether are dense. We first deal with the simplest case, were the existencecomes from winning a game by just writing down the opponents moves. Weshow how it help when instead orders we have trees with boundedly manylevels, having freedom in the last. After this we quite systematically look atexistence of black boxes, and make connection to non-saturation of naturalideals and diamonds on them.

Publication 309; Was supposed to be Chapter IV to the book Non-structure and probablywill be if it materialize.

1


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G

0. Introduction

The non-structure theorems we have discussed in [Sh:E59] rests usually on somefreedom on finite sequences and on a kind of order. When our freedom is relatedto infinite sequences, and to trees, our work is sometimes harder. In particular, wemay consider, for , regular, and = (x0, . . . , x, . . .)),where

[ a = a ], g(a) = g(xg()),

such that for we have:

MI |= (. . . , a, . . .)


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BLACK BOXES 3

(c) , For any I we have a model MI and sequences a (for

>), where

[ a = a ], g(a) = g(xg()),

such that for we have:

MI |= (. . . , a, . . .) I ,

and g(a) or just g(a) = .

Then (using> I ):

(1) There is no model M of cardinality into which every MI can be ()embedded (i.e., by a function preserving and ).

(2) For any Mi (for i < ), Mi , for some I satisfying > I ,

the model MI cannot be ()embedded into any Mi.Example 1.2. Consider the class of Boolean algebras and the formula

(. . . , xn, . . .) =: (n

xn) = 1

(i.e., there is no x = 0 such that x xn = 0 for each n).For > I , let MI be the Boolean algebra generated freely by x (for I) except the relations:

for I, if n < g() = then x xn = 0.

So MI = |I| [, 0 ] and in MI for we have: MI |= (n

xn) = 1 if and

only if / I (work a little in Boolean algebras).

So

Conclusion 1.3. If = 0 , then there is no Boolean algebra B of cardinality universal under embeddings (i.e., ones preserving countable unions).

Remark 1.4. This is from [Sh:a, VIII,Ex.2.2]).

Proof of the Theorem 1.1. First we note 1.5, 1.6 below:

Fact 1.5. There are functions f (for ) such that:

(i): Dom(f) = { : < },(ii): Rang(f) ,(iii): if f : > , then for some we have f f.

Proof: For let f be the function (with domain { : < }) such that:

f( ) = ().

So f : is well defined. Properties (i), (ii) are straightforward, so let usprove (iii). Let f : > . We define = i : i < by induction on .For = 0 or limit no problem.For + 1: let be the ordinal such that = f().

So =: i : i < is as required. 1.5

Fact 1.6. In 1.5:


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4 SAHARON SHELAH

(a): we can replace the range of f, f by any fixed set of cardinality ,(b): we can replace the domains of f, f by {a :

>}, {a : < },

respectively, as long as < < a = a.

Remark 1.7. We can present it as a game. (As in the book [Sh:a, VIII 2.5]).

Continuation of the Proof of Theorem 1.1. It suffices to prove 1.1(2). With-out loss of generality |Mi| : i < are pairwise disjoint. Now we use 1.6; for thedomain we use a : > from the assumption of 1.1, and for the range:i)

: for some i < , Rang(f) is a set of sequencesfrom |Mi| and Mi |= (. . . , f (a), . . .)


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BLACK BOXES 5

(c): In (b), there is a model M such that M = + and every model MI canbe ()embedded into M.

Remark:

(1) Essentially MI is (I+,), the addition of level predicates is immaterial,where I+ extends I nicely so that we can let a = for I.

(2) Clearly clause (c) also shows that weakening M = , even when + < 2

may make 1.1 false.

Proof: Let = {R : } { I let NI be the -model:

|NI| = I, RNI = I

,


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6 SAHARON SHELAH

This defines also fj : I(j,) Ij , which is forced to be a ()embedding and

also just an embedding.

So now we shall define for every I,>

I

, a model MI: clearly Ibelongs to some VPj . Let j = j(I) be the first such j, and let = (I) be suchthat I = I(j,). Let

MI(j,) = NIj ( and a = fj() for I(j,)).

We leave the details to the reader. 1.9

On the other hand, consistently we may easily have a better result.

Lemma 1.10. Suppose that, in the universe V,

= cf() = =

) areas required in clauses (a),(b),(c) of () of 1.1), () < , and N (for < ()) is a model in the relevant vocabulary,

Ii ,and MIi , a

i ( Ii) are as in () of 0), then for some i = j, MIi is not

embeddable into MIj .

(d): In VP we can find a sequence I : < (so > I ) suchthat the MI s satisfy that no one is ()embeddable into another.

Proof: P is Q0 from the proof of 1.9. Let f be the generic function that is{f : f G

Q0}, clearly it is a function from to {0, 1}. Now clause (a) is trivial.

Next, concerning clause (b), we are given N : < (). Clearly for some A V ofsize smaller than , A , to compute the isomorphism types of N (for < ())it is enough to know f A. We can force by {f Q0 : Dom(f) A}, then f Bfor any B \ A of cardinality , (from V) gives us an I as required.

To prove clause (c) use system argument for the names of various MIs.The proof of (d) is like that of (c). 1.10

2. An Application for many models in

Discussion 2.1. Next we consider the following:Assume is regular, ( < )[ IA and

A B MIA = MIB .

We choose MiIA : i < : A with MIA =i


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BLACK BOXES 7

Of course, we have to strengthen the restrictions on MI. For IA , let

() =: {(i) + 1 : i < }, we are specially interested in such that is strictlyincreasing converging to some () UA; we shall put only such s in IA. Thedecision whether IA will be done by induction on () for all sets A. Arrivingto , we assume we know quite a lot on the isomorphism f : MIA MIB , speciallywe know

f


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8 SAHARON SHELAH

(i): for every model M with universe H { : < ()}.

For < (), let U = { : { i : i < } M}. We shall define byinduction on , for every A the set I[A] U so that on the one hand those

restrictions are compatible (so that we can define I[A] in the end, for each A ),and on the other hand they guarantee the non ()embeddability.

For each : (essentially we decide whether I[A] assuming M guessesrightly a function g : MI1 MI2 (I = I[A]), and A M

for = 1, 2, and wemake our decision to prevent this)

Case I: there are distinct subsets A1, A2 of and I1, I2 satisfying> I ,

and a ()embedding g of MI1 into MI2 and

M H


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Case II: Not I.No restriction is imposed.

The point is the two facts below which should be clear.

Fact 2.5. The choice of A1, A2, I1, I2, g is immaterial (any two candidates lead tothe same decision).

Proof: Use clause (d) of 2.3.

Fact 2.6. MI[A] (for A ) are pairwise non-isomorphic. Moreover, for A = B(subsets of ) there is no ()embedding of MI[A] into MI[B].

Proof: By the choice of the I[A]s and (i) of 2.4. 2.4

Still the assumption of 2.4 is too strong: it do not cover all the desirable cases,though it cover many of them. However, a statement weaker than the conclusionof 2.4 holds under weaker cardinality restrictions and the proof of 2.3 above worksusing it, thus we will finish the proof of 2.3.

Fact 2.7. Suppose = .Then there are {(M, A1 , A

2 ,

) : < ()} such that:

(): (i): for every model M with universe H


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10 SAHARON SHELAH

Discussion 2.9. What about Theorem 2.3 in the case we assume only = |T|.

Claim 2.10. In 2.3, we can restrict ourselves to I such that I0, I , where

I0, =>

: (i) = 0 for every i < large enough

.

Proof: By renaming. 2.10

3. Black Boxes

We try to give comprehensive treatment of black boxes, not few of them areuseful in some contexts and some parts are redone here, as explained in 0, 1.

Note that omitting countable types is a very useful device for building modelsof cardinality 0 and 1. The generalization to models of higher cardinality, or

+, usually requires us to increase the cardinality of the types to , and even sowe may encounter problems (see [Sh:E60] and background there). Note we do notlook mainly at the omitting type theorem per se, but its applications.

Jensen defined square and proved existence in L: in Facts 3.1 3.8, we dealwith related just weaker principles which can be proved in ZFC. E.g., for regular> 1, { <

+ : cf() < } is the union of sets, each has square (as defined there).You can skip them in first reading, particularly 3.1 (and later take references onbelief).

Then we deal with black boxes. In 3.11 we give the simplest case: regular> 0, =


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BLACK BOXES 11

It would be better if we can use for a strong limit > 0 = cf(),

0 = sup : for some n < and uniform ultrafilter D on ,cf n


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(h): if C is a closed unbounded subset of , and 0 < < then the set

{ S : C C}

is stationary.

Proof: (1) We can find {h1 , h2 : < } such that:

(a): for every we have h1 : and h2 : ,

(b): if A , |A| , and h1, h2 : A , then for some , h1 A = h1,

and h2 A = h2.

This holds by Engelking-Karlowicz [EK65] (see for example [Sh:c, AP]).For < let C be a closed unbounded subset of of order type cf(). Now for

each < and a stationary a ask whether for every i < for some j < wehave

(),ai,j : the following subset of is stationary:

S,ai,j = { S : (i) if C, otp( C) / a then h1() = 0,

(ii) if C, otp( C) a then the h1()thmember of C belongs to [i, j),

(iii) if C then h2() = otp( C) }

Subfact 3.2. For some < and a stationary set a , for every i < for somej (i, ), the statement (),ai,j holds.

Proof: If not, then for every < and a stationary a , for some i = i(, a) < ,for every j < , j > i(, a), there is a closed unbounded subset C(,a,i,j) of

disjoint from S,ai,j .Let

i() =

{i(, a) + : < and a is stationary }.

Clearly i() < .For i() j < let

C(j) =

{C(,a,i(, a), j) : a is stationary and < } (i() + , ),

clearly it is a closed unbounded subset of . Let

C = { < : > i() and (j < )[ C(j)]}.

So C is a closed unbounded subset of , too. Let C+ be the set of accumulationpoints of C. Choose () C+ S, and we shall define

h1 : C() , h2 : C() .

For C() let h0() be:

min{ < : > 0 and the th member of C is > i()}

if = sup(C() ) > i(), and zero otherwise. Clearly the set

{ C() : h0() = 0}

is not stationary. Now we can define g : C() () by:

g() is the h0()th member of C.


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BLACK BOXES 13

Note that g is pressing down and { C() : g() i()} is not stationary. So(by the variant of Fodors Lemma speaking on an ordinal of uncountable cofinality)

for some j < sup(C()) = () the seta =: { C() C

: i() < g() < j }

is a stationary subset of (), and let h1 : C() be

h1() =

0 if otp( C ) / a,h0() if otp( C ) a.

.

Let h2 : C() be h2() = otp( C()). By the choice of (h

1 , h

2) : < ,

for some , we have h1 C() = h1 and h2 C() = h

2. Easily, () S,ai,j whichis disjoint to C(,a,i(), j), a contradiction to () C by the definition of C(j)and C.

So we have proved the subfact 3.2. 3.2

Having chosen , a we define by induction on < an ordinal i() < suchthat i() : < is increasing continuous, i(0) = 0, and (),ai(),i(+1) holds.

Now, for < we define ha() as follows: it is if h1() > 0 and the h

1()-th

member of C belongs to [i(1 + ), i(1 + + 1)), and it is zero otherwise. Lastly,let hb() =: h2() and W = a and

S =:

S : (i) for C, otp( C) = hb(),(ii) for C, hb(i) a ha() = ,(iii) for C , hb(i) / a ha(i) = 0

.

Now, it is easy to check that a, ha, hb, and S : 0 < < are as required.

(2) In the proof of 3.1(1) we shall now consider only sets a which satisfy thedemand in clause (e) of 3.1(2) on W [i.e., in the definition of C(j) during the proof

of Subfact 3.2 this makes a difference]. Also in (),ai,j in the definition of S,ai,j wechange (iii) to

(iii): if C, h2() codes the isomorphism type of, for example,C

C

C , < , , C , {i, : i C}

.

In the end, having chosen , a we can define C and S in the natural way. 3.1

Fact 3.3. (1) If is regular > 2, regular, S { < : cf() = } isstationary and for S, C0 , is a club of of order type (= cf()),

then we can find a club c of (see 3.4(1)) such that letting for S,C = C [c] =: { C0

: otp(C0

) c}, it is a club of and

(): for every club C we have:(a): if > 0, { S : C C} is stationary,(b): if = 0, then the set

{ S : (, )[ < C C (, ) C = ]}

is stationary.(2) If is a regular cardinal > 2, then we can find C : S : < 2

such that:

(a):

{S : < 2} = { < : 0 < cf() },


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14 SAHARON SHELAH

(b): C is a club of of order type cf(),(c): if S, cf() > > 0, then

{ C : cf() = , S and C C

}

is a stationary subset of .(3) If is regular, 2 > , then we can find C : S : < such

that:(a):

{S : < 2} = { < : 0 < cf() },

(b): C is a club of of order type cf(),

(c): if S, C, cf() > 0, then S and C C

,

(d): moreover, if , S , C, then

{(otp( C), otp( C)) : C}

depends on (otp( C), otp(C)) only.

(4) We can replace in (1)(a) and (b) of () stationary by / I for anynormal ideal I on .

Remark 3.4. (1) A club C of where cf() = 0 means here just an un-bounded subset of .

(2) In 3.3(1) instead of 2, the cardinal

min{|F| : F & (g )(f F)( < )[g() < f()]}

suffices.(3) In (b) above, it is equivalent to ask

{ S : (, )[ < C C otp((, ) C) > ]}

is stationary.

Proof: (1) If 3.3(1) fails, for each club c of there is a club C[c] of exem-plifying its failure. So C+ =

{C[c] : c a club} is a club of . Choose S

which is an accumulation point of C+ and get contradiction easily.

(2) Let = cf() > 2, C be a club of of order type cf(), for each limit < . Without loss of generality

C & > sup( C) is a successor ordinal.

For any sequence c = c : 0 < = cf() such that each c is a club of , for S = { < : 0 < cf() } we let:

Cc =

C : otp(C ) ccf()

.

Now we define Sc, by defining by induction on < , the set Sc ; the only

problem is to define whether Sc knowing Sc ; now Sc if and only if (i) 0 < cf() ,

(ii) if 0 < = cf() < cf()then the set { Cc : cf() = , Sc }is stationary in .

Let c : < 2 list the possible sequences c, and let S = Sc and C = C

c

. Tofinish, note that for each < satisfying 0 < cf() , for some , S .

(3) Combine the proof of (2) and of 3.1.


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BLACK BOXES 15

(4) Similarly. 3.3

We may remark

Fact 3.5. Suppose that is a regular cardinal > 2, = cf() > 0, a set

S { < : cf() = }

is stationary, and I is a normal ideal on and S / I. If I is +saturated (i.e., inthe Boolean algebraP()/I, there is no family of + pairwise disjoint elements),then we can find C : S, C a club of of order type cf(), such that:

(): for every club C of we have { S : C \ C is unbounded in } I.

Proof: For S, let C be a club of of order type cf(). Call C = C : S

(where S stationary, S / I, C a club of ) I-large if:for every club C of the set

{ < : S

and C \ C is bounded in }does not belong to I.

We call C I-full if above { S : C \ C unbounded in } I.3.3(4) for every stationary S S, S / I there is a club c of such that C [c

] : S is Ilarge. Now note:

(): if C : S is I-large, S S, then for some S S, S / I,C : S is Ifull (hence S / I).

[Proof of (): Choose by induction on < +, a club C of such that:

(a): for < , C \ C is bounded in ,(b): if = + 1 then A \ A I+, where

A =: { S : C \ C

is unbounded in }.

As clearly

< A \ A is bounded in

(by (a) and the definition of A, A) and as I is +saturated, clearly for some we cannot define C. This cannot be true for = 0 or a limit , so necessarily = + 1. Now S \ A is not in I as C was assumed to be Ilarge. Check thatS =: S \ A is as required.]

Using repeatedly 3.3(4) and () we get the conclusion 3.5

Claim 3.6. Suppose = +, = , is a regular cardinal and S { < :cf() = } is stationary. Then we can find S, C : S and S : < suchthat:

(a):


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16 SAHARON SHELAH

Proof: Similar to the proof of 3.1 (or see [Sh:237e]).

We shall use in 3.25

Claim 3.7. Suppose = +, a limit ordinal of cofinality ,

h : { : = 1 or = cf() },

= ||, and S { < : cf() = } is stationary. Then we can find S,C : S

and S : < such that:

(a): 1, then { < + : cf() < } isa good stationary subset of + (i.e., it is in Igd[+], see [Sh:E62, 3.3] or[Sh:88r, 0.6,0.7] or 3.9(2) below).

(2): Suppose is regular > 1. Then we can find S : < such that:(a):


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Proof: Appears also in detail in [Sh:351] (originally proved for this work but as itsappearance was delayed we put it there, too). Of course,

(1) follows from (2).(2) Let S = { < + : cf() < }. For each S choose A such that:

(): A = Ai : i < is an increasing continuous sequence of subsets of of cardinality < , such that

i


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18 SAHARON SHELAH

Note that we have gotten

()2: E0 .

[Why? If cf() = 1 see above, if cf() = 1 this is trivial.]

Clause (ii): If B then B = B

.

We know that B = closure(A ) , by ()2 above. If A

then (by ()1) we

have B = closure(A ) , so we are done. So assume / A

. Then, by clause

() necessarily

< > sup(A ) and sup(A ) A

+1 A

.

But B = closure(A ) by ()2, hence together A

contains a club of and

cf() = cf(), but cf() = 1, so cf() = 1. Now, as in the proof of clause (i), we

get B = {B : A

}, so by the induction hypothesis we are done.

Clause (iii): If is limit then = sup(Ai ).

By clause () we know A is unbounded in , but A B

(by ()2) and we

are done.

So we have finished proving ()0 by induction on hence clause (a) of the claim.For proving (c) of 3.8(2), note that above, if is limit, C is a club of, C S,

and |C| < , then for every i large enough, C Ai , and even C Bi .

Now assume that the conclusion of (c) fails (for fixed () and ). Then for each < we have a club E0 exemplifying it. Now, E

0 =:


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BLACK BOXES 19

We define C (for S) as follows:if / E,

C =: C

\ (max( E) + 1),

if E, otp(C E) e {} then

C = { C E

: otp( C E) e},

and if E, otp(C E) / e {} let

C = C \

max{ : otp(C E

) e {}} + 1

.

One easily checks that (d) and square hold for C : S. So, we just have to

add C : S to {C : S : < } for any , (), (for which we choose

and E). 3.8

Claim 3.10. (1): Assume that 0 < = cf(), +

< = cf(), S {


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20 SAHARON SHELAH

Lemma 3.11. Suppose that , and () are regular cardinals and =


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BLACK BOXES 21

(2): Remember, (< ) =:


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22 SAHARON SHELAH

The case <


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BLACK BOXES 23

(1): Straightforward (play the game).(2): Similar to the proof of 3.1.

(3): Obvious.(4): Easy.(5): Easy.(6): Follows.

6.7

Claim 3.16. Assume that ,, are as in 3.13 with || . Then

(1): the sequence { < i : cf() = } : i < is good for ( ,,).(2): If 1 < and Si : i < 1 is good for ( 1, , ) then

Si : i < 1{ < i : cf() = } : 1 i <

is good for ( ,,).(3): If Si : i <

1 is good for ( ,,) and i() < , then we can partition

Si() to pairwise disjoint sets Si(), : < i such that for each < i, thesequence

Si : i < i()Si(),

{ : < i, cf() = } : i() < i <

is good for ( ,,),(4): S good for ( ,,) implies that Si is a stationary subset of i for each

i < g().

Proof: Like 3.15 [in 3.16(3) we choose for Si(), a club C of of order typecf(); for j < , < i(), let

Sji(),

= { Si() : is the jth member of C};

for some j and unbounded A i(), S

j

i(), : A are as required]. 3.16Now we remove from 3.11 (and subsequently 3.18) the hypothesis is regular

when cf() ().

Lemma 3.17. Suppose =


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24 SAHARON SHELAH

sup(|Mi | cf()) < () = sup(|M | cf()),

(a2): as in 3.11.(b0): , (b1), (b2) as in 3.11 but in clause (b3) we demand i = 2 mod 3.

Remark: To make it similar to 3.11, we can fix Sa, Sai , Sbi , S

bi,a,

i as in the first

paragraph of the proof below.

Proof: First, by 3.15 [(1) + (2)], we can find pairwise disjoint Sai cf() fori < cf(), each good for (cf(), , ()) (and Sai > i & cf() = ), andlet Sa =

i


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BLACK BOXES 25

Note that () = + + 1 Mi and M,i is definable in M

,i+1 as

M,i M,i+1 (by the definition ofW

0 in the proof of 3.11). Similarly, M

,j : j i

is definable in Mi+1. It is easy to check that the pair (M,, ,) satisfies condition(a1) of 3.11.Next we choose by induction on U , () < () as follows:

(C): () is the first < () such that: if < but + () > then:

(): {, j : j < } M,() .

This is possible and easy, as for () it suffices to have for each suitable , / M,() ,

so each disqualifies < () ordinals as candidates for (), and there are < ()such s, and () is by the assumptions (see 3.11) regular.

Now

W = {(N,(), ,()) : U}, U , h U

are as required except that we should replace U by an ordinal (and adjust , h

accordingly). In the end replace N

i by N

i H


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26 SAHARON SHELAH

(iv): if () = = () and =


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BLACK BOXES 27

We proceed as in the proof of 3.11 after W0 was defined (only () = S1,1,1 h() = 1).

Suppose G is a winning strategy for player I. So suppose that if player II haschosen (0), (1), . . . , (n 1), player I will choose M. So |M| is a subset ofH, T is closed under initial segments, T,

T ()[ T]

,

cn(T Tn) is constant.

It follows that fixing any lim(T) we can find h; T such that h isan isomorphism from M g() onto M increasing with .

Note that above all those isomorphisms are unique as the interpretation of satisfies comprehension. Also clause (c1) follows from the use of R.The rest should be clear. 3.20

Lemma 3.22. LetS, , , be as in (*) of 3.20 and in addition:

0 = cf() < (), ( < ())[


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28 SAHARON SHELAH

(): Fx = {Fx,n : n < }(): Fx,n = {f : f is an elementary embedding of Mx,n into A}

(): n(f) = k if and only if f Fx,k(): rank(f) = {+1 : for every < there is g F

x,n(f) extending f, such that =rankx(g) and Rang(g) = Rang(f) }

NowCase 1: for no x W and f Fx,0 do we have rankx(f) =

For every x W and f Fx let (f, x) be the first ordinal < such thatif rankx(f) = then there is no g Fx,n(f)+1 extending f with rankx(g) = andRang(g) = Rang(f) .

Next let Ai : i < be an increasing continuous sequence of elementary sub-models ofA, each of cardinality < such that Aj : j i Ai+1.

Easily the set E = {i < : Ai = i > } is a club of .

Choose by induction on n < an ordinal in increasing with n such that in Eis of cofinality , possible as 2


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BLACK BOXES 29

subset of of order type , of course). Then we can find N : suchthat

(a): = { : S} where { : in an increasing -sequence ofordinals < with limit } and () = when , S.

(b): N is N,n : n in -increasing, and we let N = N,(c): each N is a model of cardinality with vocabulary H(+) for

notational simplicity, and universe := () and N,n = Nnwhere n is the n-the member of C

(d): for every distinct , where S, for some n < we haveN N = N,n = N,n

(e): for every, the models N, N are isomorphic moreover thereis such isomorphism f which preserve the order of the ordinals andmaps N,n onto N,n

(f): ifA is a model with universe and vocabulary H(+) then forstationarily many S for some we have N A .Moreover, if = and h is a one to one function from into then we can add: if (N,n) then h() N,n

Proof: (1) Let g0, g1 be two place functions from to such that for [, ] : g0(, i) : i < enumerate {j : j < } without repetitions, andg1(, g0(, i)) = i for i < .

Now we can restrict ourselves to M such that each Mi (for i ) is closedunder g0, g1. Then (c3) follows immediately from

[() = () |M | = |M | ]

(required in (c1)).(2) Should be clear.(3) This just rephrase what we have proved above.

3.24

Lemma 3.25. Suppose that = +, = 0 = 2 > 20, cf() = 0 andS { < : cf() = 0} is stationary, = 0, 0 < () = cf(()) < . Thenwe can find W = {(M, ) : < ()} and functions

: () S, h : ()

and C : S with n : n < listing C in increasing order such that:

(a0)-(a1): as in 3.11,(a2) : as in 3.22,

(b0)-(b2): as in 3.11 and even(b1): = , { n : n < } M implies < and even () n0

and 1 2 3 satisfying:

1 |M | [

n,

n+1),

2 |M | [

n,

n+1),

3 |M | [

n,

n+1),


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30 SAHARON SHELAH

i.e., either sup

[n, n+1) |M

|

< min

[n, n+1) |M

|

or sup [n, n+1) |M | < min [n, n+1) |M |;(c5): if < and there is B , |B| = 0 which contains no perfect set

with density (holds trivially if is strong limit), then also { : < ()}does not contain such a set. (See 3.26).

Proof: We repeat the proof of 3.20 with some changes.Let S,, : < , < , < be pairwise disjoint stationary subsets of S.

Let g0, g1 be as in the proof of 3.24. By 3.7 there is a sequence C : S suchthat:

(i): C is a club of of order type , not !, 0 / C,(ii): for < , < , < , for every club C of , the set

{ S,, : C C}

is stationary.We then define W, (j , Mj,n : n < , j , Cj) for j < ,A for < , and R asin the proof of 3.20.Now, for S,, let W1 be the collection of all systems M, :

> suchthat

(i): is an increasing sequence of ordinals of length g(),(ii): otp(C ()) = 1 + () for < g(),(iii): there are isomorphisms h : > such that h maps M onto

M,g() preserving , R, cd(x) = y, g0(x1, x2) = y, g

1(x1, x2) = y (andtheir negations),

(iv): if then h h , M M, M M ,(v): MC = , and M

[(), ()+1), where is the -th member

of C ,(vi): if >, < g(), is the (1 + ())-th member of C then M

depends only on , and M M,(vii): N = M.

Now clearly |W1 | , so let W1 = {(M

j ,

j) :,

> : j < }. Letj : j < be a list of distinct members of; for (c5) choose as there.

Let

Mj =


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BLACK BOXES 31

for = 0; by [Sh:E12, Part D] the requirement holds, e.g., for for a club of < 1

Moreover, under this assumption on we can demand (essentially, this is ex-panded in 3.30)

(c4) : we strengthen clause (c4) to:if () = () = but = then for some n0 1, we haveeither for everyn [n1, ) we have sup

[n,

n+1) |M

|

< min

[n, n+1) |M

|

or for everyn [n1, ) we have sup

[n, n+1) |M

|

< min

[n, n+1) |M

|

;

Lemma 3.27. We can combine 3.25 with 3.22.

Proof: Left to the reader.

Lemma 3.28. Suppose 0 = < () = cf(()) and: 0 = = 0,

and let S consist of strong limit singular cardinals of cofinality 0 and bestationary. Then we can find W = {(M, ) : < ()} and functions :() S and h : () such that:

(a0)-(a2): of 3.11 (except that h() depends not only on ()),(b0),(b3): of 3.11,(b1)+: of 3.18,(c3): if () = = () then |M | |M

| is a bounded subset of .

Remark:

(1): See [Sh:45] for essentially a use of a weaker version.(2): We can generalize 3.22.

Proof: See the proof of [Sh:331, 1.10(3)] but there sup(N ) < .

Lemma 3.30. (1): Suppose that = +, = = 2, < cf(()) =() < , is strong limit, > cf() = > 0, S { < : cf() = }is stationary. Then we can find W = {(M, ) : < ()} (actually, a

sequence), functions : () S and h : () and C : Ssuch that:

(a1)-(a2): as in 3.11,(b0): = for = ,


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32 SAHARON SHELAH

(b1): if { i : i < } M and = then < and even

() < (),

(b2): if (j + 1) M then Mj M ,(c2): C = C : S, C a club of of order type , and every club of

contains C for stationarily many S,(c3): if S, C = {,i : i < } is the increasing enumeration, <

satisfies () = , then there is ,i, +,i : i < odd such that

,i Mi , M

i

+,i, ,i <

,i <

+,i < ,i+1

and(): if () = (), < then for every large enough odd i < ,

+,i < ,i (hence [

,i,

+,i) [

,i ,

+,i) = ) and

[,i , +,i) M

= .

(2): In part (1), assume = 0 andpp() =+ 2. Then the conclusion holds;moreover, (c3) (from 3.24).

Remark: The assumption pp() = 2 holds, for example, for = for a club of < (see [Sh400, 5]).

Proof:(1) By 3.6 we can find C = C : S, C a club of, of order type such thatfor any club C of for stationarily many S, we have: C C.

First case: assume (= 2) is regular.By [Sh:g, II, 5.9], we can find an increasing sequence i : i < of regular

cardinals > () such that =

i


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BLACK BOXES 33

On the other hand, if cf() = (2)+, without loss of generality, cf

f(i)

= cf()for every i < (see [Sh:g, II, 1]), so there is < such that

h < f mod Jbd where h(i) = sup

f(i)

j H


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34 SAHARON SHELAH

Then we can find

W = {(M, ) : < ()}

and functions

: () S and h : ()

such that:

(a0): like 3.11,(a1): M = Mi : i

2 is an increasing continuous elementary chain((Mi ), the vocabulary, may be increasing too and belongs to H


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BLACK BOXES 35

(iii): if E is a club of + then for stationarily many < +, cf() = ,C E and the set

{i < : for every C , cf() = i+1 E}

is unbounded in .

Now copying the black box of on each < +, cf() = , we can finish easily.

Lemma 3.33. If , , , , (), S are as in 3.30, and

< () || < ()

then there is a stationary S {A : |A| < ()} and a one-to-one function cdfrom S to such that:

A S & B S & A = B & A B cd(A) B.

Remark: This gives another positive instance to a problem of Zwicker. (See [Sh:247].)

Proof: Similar to the proof of 3.30 only choose

cd : {A : A and |A| < ()}

one-to-one, and then define

S {A : A , |A| < ()}

by induction on . 3.33

Problem 3.34. (1): Can we prove in ZFC that for some regular > (),,(): we can define for S

= { < : 0 cf() = } a

model M with a countable vocabulary and universe an unboundedsubset of of cardinality < (), M() is an ordinal such that: forevery model M with countable vocabulary and universe , for some(equivalently: stationarily many) S , M M.

(2): The same dealing with relational vocabularies only (we call it ()rel,,).

Remark 3.35. Note that by 3.8 if (),,, = cf() > then ()+,,.

* * *

Now (3.363.40) we return to black boxes for singular , i.e., we deal with thecase cf() .

Lemma 3.36. Suppose that = . Assume further

(): cf() ,(): =

iw

i, |w| , w + (usually w = cf()) and [i < j i < j ],

and each i is regular < and

cf() > 0 cf() = w = cf(),

(): > , is a regular cardinal, D is a uniform filter on w (so { w : > } D for each w), is the true cofinality of

iw

(i,


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36 SAHARON SHELAH

(): f = fi/D : i < exemplifies the true cofinality ofi

(i, cf(), S, then for some A D and unbounded B we

have

B B < i A f(i) < f(i),

i.e., f A : B is


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BLACK BOXES 37

Fact 3.38. If i : i w, f , D are as in 3.36, then

S = { < : cf() = and there are A D, and unbounded B

such that [ B B < i Af(i) < f(i)] }

is good for (,,()).

Lemma 3.39. Let(1) = () + (< ()).In 3.36, if = (1), we can strengthen (b1) to (b1)+ (of 3.18).

Proof: Combine proofs of 3.36, 3.18.

Lemma 3.40. 3.173.11

3.29 and 3.193.11

3.37 hold (we need also the parallel to 3.30).

Proof: Left to the reader.

* * *Now we draw some conclusions.

The first, 3.41, gives what we need in 2.7 (so 2.3 ).

Conclusion 3.41. Suppose =


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38 SAHARON SHELAH

Claim 3.42. (1): In 3.11, if =


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BLACK BOXES 39

Exercise 3.47. Let , , , and Y []


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40 SAHARON SHELAH

4. On Partitions to stationary sets

We present some results on the club filter on [ ]0 and [] and some relatives,and (see Def. [Sh:E62, 3] or 4.4(2) below). There are overlaps of the claimshence redundant parts which still have some interest.

Claim 4.1. Assume is a cardinal > 1, then []0 can be partitioned to 0

(pairwise disjoint) stationary sets.

Proof: Follows by 4.2 below, [in details, let be the vocabulary {cn : n < } whereeach cn is an individual constant. By 4.2 below there is a sequence M = Mu : u []0 of -models, with Mu having universe u such that M is a diamond sequence.

For each let S be the set u []0 such that for every n < we have

cMun = (n)}.By the choice of M necessarily each set S is a stationary subset of []0 , and

trivially those set s are pairwise disjoint.]

Claim 4.2. Let > 1. Then we have diamond on []0 (modulo the filter of clubs

on it, see 4.4(2) or [Sh:E62]), and we can find A []0 for < def= 2

0 suchthat each is stationary but the intersection of any two is not.

Proof: The existence of the A-s for < follows from the other result. Let bea countable vocabulary, 1 = { n + 3 when n > 0. Let C : S20 be club guessing, whereS20 = { < 2 : cf() = 0}, such that C = sup(C) has order type .

Let (A , ) : < 20 list without repetitions the pairs (A, ), A a model withvocabulary 1 and universe a limit countable ordinal and = n : n < an

increasing sequence of ordinals with limit sup(A

) andAn

A. Let En be thefollowing equivalence relation relation on 20 : En iff (An,

n) is isomorphicto (An,

n) which means: there is an isomorphism f from A n onto A

n which maps A k onto A

k for k < n and is an order preserving function(for the ordinals, alternatively we restrict ourselves to the case < is interpreted asa well ordering.

We can find subsets t of such that

(): for , < 20 we have t sn = t sn iffA n = A n and

k =

k

for k n. Also t sn is infinite and = 0 > |t t | for simplicity

( so t sn depend just on /En, in fact code it).

For < 20 let

Sdef= a []0 : t = {|Csup(a) | : a},

and let

S = {a St : otp(a) = otp(A

)},

and for a S let Na be the model isomorphic to A by the function fa, where

Dom(fa) = a, fa() = otp( a).Let S be the union ofS for < 2

0 . Clearly = S S = , and soS S

= . Hence Na is well defined for a S

Let Kn be the set of pairs (A, ) such that A is a 1-model with universe acountable subset of with no last member, and is an increasing sequence of


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BLACK BOXES 41

ordinals < of length n such that k < sup(A) and [k, k+1) A = andA k A. So clearly there is a function cdn : Kn P(sn) such that: if < 20

then cdn(A, ) = t

sn iff the pairs (A, ), ((A

,

n)) are isomorphic.Let M be a 1model with universe . Now (see [Sh:E62], or history in the

introduction of3, and the proof of 3.22) we can find a full subtree T of>(2) (i.e.,it is non-empty, closed under initial segments and each member has 2 immediatesuccessors) and elementary submodels N of M for T such that:

(a): Rang() N,(b): if is an initial segment of then N is a submodel N, moreover N2

is an initial segment of N 2.

Now let E be the set of < 2 satisfying: if T and > then N 2 is a

bounded subset of, and is a limit ordinal. Let E1 be the set of E such thatif T> then for every < there is such that < < and T.So by the choice of C : S for some S we have C E1

Let ,k : k < list C in increasing order.Now we choose by induction on n a triple (n, sn, n, kn) such that

(*) (a) n T has length n (so 0 is necessarily ).(b) ifn = m + 1 then n is a successor of m(c) sn is cdn((Nn , : < n)) if the pair (Nn , : < n) belongs to

Kn and is sn otherwise; actually it is so,(d) n = sup(Nn) + 1(e) kn = min{k : Nn ,k} and k0 = 0 and n[0, kn]

,kn1() (km, kn) is disjoint to 2. For each c []0 , if otp(c) = otp(c

2,

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