Saharon Shelah- The Generalized Continuum Hypothesis Revisited

8/3/2019 Saharon Shelah- The Generalized Continuum Hypothesis Revisited

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THE GENERALIZED CONTINUUMHYPOTHESIS REVISITED

SH460

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat Ram

Jerusalem 91904, IsraelDepartment of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Partially supported by the Israeli Basic Research Fund and the BSF and I would like to thankAlice Leonhardt for the beautiful typing.Done - 8/91(1), 9/91(2,3),Latest Revision - 06/Oct/26Publ. No.460Revised according to the proofreading for the Journal

Typeset by AMS-TEX

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2 SAHARON SHELAH

Annotated Content

0 Introduction

[We explain why we consider the main theorem here a reasonable revisionof GCH (but provable in ZFC).]

1 The generic ultrapower proof

[We prove that for strong limit > 0 for every > for some < ,there is P []


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THE GENERALIZED CONTINUUM HYPOTHESIS REVISITED SH460 3

0 Introduction

I had a dream, quite a natural one for a mathematician in the twentieth century:to solve a Hilbert problem, preferably positively. This is quite hard for (at least)three reasons:

(a) those problems are almost always hard

(b) almost all have been solved

(c) my (lack of) knowledge excludes almost all.

Now (c) points out the first Hilbert problem as it is in set theory; also being thefirst it occupy a place of honor.

The problem asks is the continuum hypothesis true?, i.e.,

(1) is 20 = 1?

More generally, is the generalized continuum hypothesis true? Which means:

(2) is 2 = +1 for all ordinals ?

I think the meaning of the question is what are the laws of cardinal arithmetic;it was known that addition and multiplication of infinite cardinals is trivial, i.e.previous generations have not left us anything to solve:

+ = = max{, }.

This would have certainly made elementary school pupils happier than the usuallaws, but we have been left with exponentiation only. As there were two operationson infinite cardinals increasing them 2 and + it was most natural to assumethat those two operations are the same; in fact, in this case also exponentiationbecomes very simple; usually = max{, +}, the exception being that whencf() < we have = + where

cf() =: min{ : there are i < for i < such that =

i


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4 SAHARON SHELAH

Probably the interpretation of Hilberts first problem as find all laws of cardi-nal arithmetic is too broad1, still is cardinal arithmetic simple is a reasonableinterpretation.

Unfortunately, there are some difficulties. On the one hand, Godel had provedthat GCH may be true (specifically it holds in the universe of constructible sets,called L). On the other hand, Cohen had proved that CH may be false (by increas-ing the universe of sets by forcing), in fact, 20 can be anything reasonable, i.e.,cf(20) > 0.

Continuing Cohen, Solovay proved that 2n for n < can be anything reason-able: it should be non-decreasing and cf(2) > . Continuing this, Easton provedthat the function 2 for regular cardinals is arbitrary (except for the lawsabove). Well, we can still hope to salvage something by proving that (2) holds formost cardinals; unfortunately, Magidor had proved the consistency of 2 > +

for all in any pregiven initial segment of the cardinals and then Foreman andWoodin [FW] for all .

Such difficulties should not deter the truly dedicated ones; first note that weshould not identify exponentiation with the specific case of exponentiation 2, infact Eastons results indicate that on this (for regular) we cannot say anythingmore, but they do not rule out saying something on when < , and we canrephrase the GCH as

(3) for every regular < we have = .

Ahah, now that we have two parameters we can look again at for most pairsof cardinals (3) holds. However, this is a bad division, because, say, a failure for

= 1 implies a failure for = 0.To rectify this we suggest another division, we define to the revised power of,for regular < as

[] = Min

|P| :P a family of subsets of each of cardinality

such that any subset of of cardinality

is contained in the union of < members ofP

.

This answers the criticism above and is a better slicing because:

(A) for every > we have: = iff 2 and for every regular ,[] = .

1On this see [Sh:g] or [Sh 400a], note that under this interpretation of the problem there ismuch to say.


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(B) By Gitik, Shelah [GiSh 344], the values of, e.g., [0], . . . , [n] are essentiallyindependent.

Now we rephrase the generalized continuum hypothesis as:

(4) for most pairs (, ), [] =

Is such reformulation legitimate? As an argument, I can cite, from the book [Br]on Hilberts problems, Lorentzs article on the thirteenth problem. The problemwas

() Prove that the equation of the seventh degree x7 + ax3 + bx2 + cx + 1 = 0 isnot solvable with the help of any continuous functions of only two variables.

Lorentz does not even discuss the change from 7 to n and he shortly changes itto (see [Br, Ch.II,p.419])

() Prove that there are continuous functions of three variables not representedby continuous functions of two variables.

Then, he discusses Kolmogorovs solution and improvements. He opens the secondsection with ([Br, p.421,16-22]): that having disproved the conjecture is not solvingit, we should reformulate the problem in the light of the counterexamples and proveit, which in his case: (due to Vituvskin) the fundamental theorem of the DifferentialCalculus: there are r-times continuously differential functions of n variables not

represented by superpositions of r times continuously times differential functionsof less than n variables.

Concerning the fifth problem, Gleason (who makes a major contribution to itssolution) says (in [AAC90]): Of course, many mathematicians are not aware thatthe problem as stated by Hilbert is not the problem that has been ultimately calledthe Fifth Problem. It was shown very, very early that what he was asking people toconsider was actually false. He asked to show that the action of a locally-euclideangroup on a manifold was always analytic, and thats false. Its only the group itselfthats analytic, the action on a manifold need not be. So you had to change thingsconsiderably before you could make the statement he was concerned with true.

Thats sort of interesting, I think. Its also part of the way a mathematical theorydevelops. People have ideas about what ought to be so and they propose this as agood question to work on, and then it turns out that part of it isnt so.

In our case, I feel that while the discovery of L (the constructible universe)by Godel and the discovery of forcing by Cohen are fundamental discoveries inset theory, things which are and will continue to be in its center, forming a basisfor flourishing research, and they provide for the first Hilbert problem a negative


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6 SAHARON SHELAH

solution which justifies our reinterpretation of it. Of course, it is very reasonableto include independence results in a reinterpretation.

Back to firmer grounds, how will we interpret for most? The simplest ways are

to say for each for most or for each for most . The second interpretationholds in a non-interesting way: for each for many s, = hence [] = (e.g. when 2). So the best we can hope for is: for every for most smalls (remember we have restricted ourselves to regular quite smaller than ). Tofix the difference we restrict ourselves to > > . Now what is a reasonableinterpretation of for most < ? The reader may well stop and reflect. Asall is forbidden (by [GiSh 344] even finitely many exceptions are possible), thesimplest offer I think is for all but boundedly many.So the best we can hope for is ( is for definiteness):

(5) if > , for every large enough regular < , [] = (and similarly replacing by any strong limit cardinal).

If the reader has agreed so far, he is trapped into admitting that here we solvedHilberts first problem positively (see 0.1 below). Now we turn from fun to business.

A consequence is

()6 for every for some n and2 P []


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In 1 we prove the theorem using a generic embedding based on [Sh:g, Ch.VI,1](hence using simple forcing) and give some applications, mainly, they are reformu-lations. For example, for for every regular < large enough, there is

no tree with nodes and > -branches. Also we explain that this is sufficient forproving that e.g. a topology (not necessarily even T0!) with a base of cardinality and > open sets has at least +1 open sets relying on [Sh 454a].

In 2.1 we give another proof (so not relying on 1), more inside pcf theory andsaying somewhat more. In 2.10 we show that a property of = which sufficesis: is a limit cardinal such that |a| < |pcf(a)| < giving a third proof.This is almost a converse to 2.9. Now 3 deals with applications: we show that for , 2 = + is equivalent to + (moreover = . Lastly in an appendix we prove there are notiny models for theories with a non-trivial type (see [LaPiRo]) of cardinality ,partially solving a problem from Laskowski, Pillay and Rothmaler [LaPiRo].

For other applications see [Sh 575, 8]. This work is continued in [Sh 513], forfurther discussion see [Sh 666]. For more on the Cantor discontinuum partitionproblem see [Sh 668].

We thank Todd Eisworth for many corrections and improving presentation.


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8 SAHARON SHELAH

1 The generic ultrapower proof

1.1 Theorem. Assume is strong limit singular and > . Then there are onlyboundedly many < such that for some (, ) we have pp(+,)() (so cf() < < ).

We list some conclusions, which are immediate by older works.

1.2 Conclusion. For every strong limit such that cf() = < < , for some < we have:

(1) for every a Reg (, ) of cardinality we havesup pcfcomplete (a) ,

(2) there is no familyP of> subsets of such that for some regular (, )

we have: A = B P |A B| < & |A| (3) cov(, +, +, ) (equivalently cov(,,,) as without loss of

generality cf() > ).

Hence

(4) there is P [] -branches when (, ) is regular.

Proof. By [Sh:g]; in detail (we repeat rather than quote immediate proofs).1) Let be as in 1.1. Without loss of generality cf() / [, ).Note that sup(pcf-complete(a)) sup{pp(|a|+,)(

) : = sup(a ) and cf() so cf() |a| < }, and easily the latter is by 1.1.2) By part (4) it is easy (let P4 []


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5) Follows by part (2): if the tree is T, without loss of generality its set of nodes is and the set of -branches cannot serve as a counterexample. 1.2

1.3 Remark. We can let be regular (strong limit > 0) if we restrict ourselvesin 1.2(1) to |a| < , and in 1.2(3),(4) to A [] 0, we need: forarbitrarily large < there is (2()+, ) such that cov(|B|, +, +, ) |B|,which holds by 1.1 (really in the proof there we use 1.4). 1.5

1.6 Proof of 1.1. Assume this fails. By Fodors Lemma (as in 1.3) without loss ofgenerality cf() = 0.

Without loss of generality for our given , is the minimal counterexample.

Let =n


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10 SAHARON SHELAH

n-complete ideal Jn Jbdan

we have n = sup(an) and an/Jn has true cofinalitywhich is . Let n = cf(n), so n n |an|.

Without loss of generality n > 0 hence without loss of generality |an| < hence

without loss of generality |an| < n+1 (and really even |pcf(an)| < n+1), hencethe ns are distinct hence the ns are distinct, and without loss of generality forn < we have n < n+1 and n < n+1 < , hence necessarily (by s minimality)

=n 3(1), for someg 1Ord from V (but depending on the generic subset G of Q), the set

{g/D

: g (1Ord)V , g


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We shall say in short g/D

is -like, note that for each there is at most

one such member in M (as the ordinals of M are linearly ordered).

However, we should remember V1

/D is, in general, not well-founded; still there isa canonical elementary embedding j of V into M = V1/D

(of course it depends

on G). Note that j maps the natural numbers onto {x M : M |= x j()},but this fails for 1; without loss of generality j ( +1) is the identity. IfM |= xan ordinal let cardM(x) be the cardinality in V

Q of {y : M |= y < x}. Note: alsoj() is -like and {j(n) : n < } is unbounded in j().

Without loss of generality for every n 1, n > |Q|, and Min(an+1) > n. Forevery regular (1,

+] there is x = g/D

which is -like. Note: g V (not

VQ\V), but we need the generic subset ofQ to know which member ofV it is. Let{g,i : i < i} V be a set such that Q for some i < i we have g,i/D

is -like

and i 3(1). For regular (in V) cardinal (, +), necessarily M |= x isregular > j() and g+/D

hence without loss of generality g+ = : < 1

(why? see (), by [Sh:g, Ch.V] for some normal filter D on 1 and we have but < 0 of cardinality 1 (as withoutloss of generality g,i() < for < 1 and recall < 0). For n 1 denotecn =: {Rang(g,i) : an, i < i} and dn =: j(cn) M; note V |= |cn|

|an| + |Q| = |an|. So M |= dn is a set of regular cardinals, each > j() but< j(0), of cardinality j(|an|) < j(n+1) < j(). Also for every an we haveM |= x dn as x = g,i/D

for some i < i and Rang(g,i) cn.

We can apply the theorem on the structure of pcf ([Sh:g, Ch.VIII,2.6]) in M(as M is elementarily equivalent to V) and get by[dn] : y pcf(dn) M and

fdn,yt : t < y : y pcf(dn)

M (this is not a real sequence, only M thinksso).

For y M such that M |= y a limit ordinal (e.g. a cardinal) let y be thecofinality (in VQ) of ({x : M |= x an ordinal < y}, |Q|

()4 assume |{a : a Mj(m)}| < n, then assuming for simplicity 1 < m < nM |= sup pcfj(m)-complete(dn g+/D

) g+/D

.

[Why? Assume not, so M |= sup pcfj(m)-complete(dn g+/D

) < g+/D

hence

M |= for every g (dn g+/D

) for some (y, a) : < j(m), y pcf(dn


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12 SAHARON SHELAH

g+/D

), a an ordinal < y we have g < sup


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But as j is an elementary embedding of V to M, the choice of (minimal) implies

M |= there is no < j() such that j(),

form a counterexample to the theorem.But as Rang [g+/D

] < j(0) < j(), clearly we have M |= g/D

< j().

By the last two sentences we get a contradiction to ()5. 1.1

1.7 Observation. Let Q, D

, G Q, VQ, M,j be as in the proof 1.6. Let for z M,

[z] = {t : M |= t y}. So

(1) IfY VQ, Y M, = Max

|Y|VQ

, |Q|V

then for some y V, |y|V

and x[x Y M |= x j(y)].

(2) Assume M |= d is a set of regular cardinals > |d|, > j

|Q|V

and y(when M |= y limit ordinal) is as in 1.6 (its cofinality in VQ).

(a) If M |= y pcf(d), J is (in VQ) the ideal on [d] generated by{[b[d]] : M |= pcf(d) & < y} {[d \ by[d]]}

then (in VQ)x[d]

x/J has true cofinality y

(b) cf({y : y [d]}) = max{y : y [pcfd]}.

Proof. Straightforward (and we use only part (1)). For (2)(b) rememberM |= y is finite [y] finite.

1.8 Remark. Of course, the proof of 1.1 gives somewhat more than stated (say afterfixing 0 = 1). E.g.,

the cardinal satisfies the conclusion of 1.1 for if

> cf() = 0 (as before this suffices) and = sup{ < : is regularuncountable and there is a forcing notion Q satisfying the -c.c. of cardinality

0 < } such that Q for every 1-complete filter D on from Vcontaining the co-countable sets there is an ultrafilter D

on P()V extend-

ing D as in [Sh:g, Ch.VI,1] for regular cardinal > + which is complete forpartitions of from V to countably many parts.

Alternatively, we can phrase the theorem after fixing D.


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14 SAHARON SHELAH

2 The Main Theorem Revisited

We give another proof and get more refined information. Note that in 2.1 if is

strong limit, we can choose R such that: if < are in R then 2 < and then0R,,1 is immediate.

2.1 Theorem. Suppose is a limit singular cardinal satisfying:

0 for any R Reg unbounded, for some R, > cf() and 1, cf() we have:

1, for some < we have:

1,, for everya (, ) Reg of cardinality < , pcfcomplete(a) .

Before we prove it, note:

2.2 Observation. Assume:

(a) wni : i < is a sequence of pairwise disjoint sets, wn =

i


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Then for some i there are xn wni such thatn

hn(xn+1) = xn.

2.3 Remark. Hence for the Jm-majority of y wm there is xn : n < as abovesuch that y = xm.

Proof. Without loss of generality wni : n < , i < are pairwise disjoint. Now

we define by induction on the ordinal for each i < a set ui wi =:n


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16 SAHARON SHELAH

2.4 Subclaim. In [Sh 430, 6.7A] we can add:

(j) max pcf(b[a]) = (when defined).

Also in [Sh 430, 6.7] we can add

() max pcf(b) = .

Proof. This is proved during the proof of [Sh 430, 6.7] (see ()4 in that proof,p.103). Actually we have to state it earlier in ()2 there, i.e. add

() max pcf(bi,j ) .

We then quote [Sh:g, Ch.VIII,1.3,p.316], but there this is stated.Lastly, concerning [Sh 430, 6.7A] the addition is inherited from [Sh 430, 6.7].2.4

2.5 Subclaim. In [Sh 430, 6.7A] we can deduce:

() if a +1

(ii) a n

b() [a]

(iii) ((), ) b [a] a = b

()

[a] a

(iv) = max pcf(a\k


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[0, ), 0 (0, ). For + 1 apply the case = 0 to a\k

bkk

[a] and get

+1, +1.

Clauses (), (). Easier. 2.5

2.6 Claim. 1) Assume 0 is regular, a cardinal, J the -complete idealgenerated by J |a|, a / J, ai J for i < cf()), so it is enough to deal separately

with each un, = u(n, ) =: {x wn : [x] = and x > +} where R. If

un, Jn we have nothing to do. Otherwise choose R, > , and I ,

, : < witnessing R. By [Sh:g, Ch.IX,4.1] applied to , < + =


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tcf

xu(n,)

x/Jn, for each < we can find a sequence n,,x : x un,, n,,x

regular < x but + and

xu(n,)

n,,x /Jn has true cofinality ,.

Now apply 2.6(3) with ,,I,J,, i : i < , ij : j < , j : j <

there standing for , u(n, ), I, Jn u(n, ), +, , : < ,

n,,x : x

u(n, ), : x u(n, ). This gives us objects bn,,x : x u(n, ), < x andJn, as there. We could have changed some values ofn,,x to

+ to guarantee that+ = max pcf{n,,x : x u(n, ), < }, so without loss of generality {

n,,x :

< } = |{f

n, (x) : A

n,, x Bn, }|.

Let us define


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wn+1i, = {(x,,) : ( An,)[x Bn, & = f

n, (x) & x w

ni ]}

hni, : wn+1i, w

ni is h

ni,((x,,)) = x when x >

+

x un, (x,,) = n,,x .

Recall we are assuming un, J+n , if i un, Jn we let w

n+1i, = . Now we

switch integrating on all R:

wn+1i =

R

wn+1i,

We let

wn+1 =R

i


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(x,,) wn+1 }.For An,,

Bn, {x u(n, ) : ( < x)[n,,x bn,,x (x, ) v(n, )]},

and so v(n, ) / In,. Thus v(n, ) is not in the -complete ideal generatedby Jn,, and the definitions of Jn, and Jn+1 imply w

n+1 / Jn+1.]

3 For every A Jn+1 the set B =: {x wn : (y wn+1)[hn(y) = x y A]} belongs to Jn.[Why? Suppose toward contradiction that B J+n , and choose R

such that B u(n, ) J+n . Let A = {(x,,) wn+1 : x B}, and let

A = {(x, ) : (x,,) A}. For An, as Bn, = u(n, ) mod Jn clearly

B Bn, J+n ; also

B Bn, {x u(n, ) : ( < x)[n,,x b

n,,x (x, ) A

]},

and since B Bn, J+n by the definition ofI

n,, we know A / In, henceA1 / Jn+1 but by the definition of B, A clearly A1 A hence A / Jn+1,contradiction.]

Thus we have carried out the induction and hence get by 2.2 the contradiction andfinish the proof. 2.1

2.8 Remark. 1) We can be more specific phrasing 2.1: let R be unbounded, = : R

, a set of ideals on ; the desired conclusion is: for every > for some < we have: if R\, i (, ) Reg for i < , J,

J then pcf

i > 1, an uncountable limit cardinal and we have:

1.5, for every (, ], we have 1, (from the conclusion of 2.1).

Then

2, () a (, ), a Reg, |a| < | pcf(a)|

() if is regular then (for a Reg):


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a (, ) & |a| < | pcf(a)| < .

Proof. Let () =

+ if is singular if is regular .

So assume a (, ) Reg, |a| < , and pcf(a) has cardinality ().Let 0 = Min(a) and i+1 : i () list the first (() + 1)-members of(pcf(a))\{0} (remember pcf(a) has a last member), and for limit (), let

=i sup pcf-complete({i() : < }).Why is this possible? We know pcf-complete({i() : < }) cannot have a member (), (hence > () being regular), by the choice of . Alsopcf-complete({i() : < }) cannot be unbounded in () (because cf(()) =() (remember () is regular) as then it will have a member > (), see[Sh:g, Ch.I,1.11]). So it is bounded below () hence i() exists.

Now we get contradiction to [Sh 410, 3.5], version (b) of (iv) there(use e.g. i() : < ( + |a|)

+4). (Alternatively to [Sh 430, 6.7F](5)). 2.9

2.10 Theorem. Let be a limit uncountable singular cardinal, < and[|a| < & a Reg (, ) | pcf(a)| < ]or at least:

, for every large enough Reg , we have:

, if a Reg (, ), |a| < then | pcf-complete(a)| <

Then for every large enough < we have 1

, of 2.1, hence cov(,,,) = .

Remark. This proof relies on [Sh 420, 5].

Proof. Without loss of generality cf() = 0 (e.g. force by Levy(0,cf()) as nothingrelevant is changed, or argue as in 1.3, as , implies that for each [cf(), ),


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the cardinal sup{ pcf-complete(a) : a Reg (, ) and |a| }, however, wecan just repeat the proof).

Assume this fails. Without loss of generality is minimal, so cf() = 0. Failure

means (by 2.7) that = sup(R) when

R =

: Reg and for some Reg (, ) for < ,

and -complete ideal I on

we have + = tcf( . This makes a minor restriction say for one we may get +

instead of < + (which is equivalent to < ).So by [Sh 420, 5], for some uncountable regular < from R\ cf()+, , fromthe assumption of the theorem holds and for some family E of ideals on normal

by a function : and J E and i = cf(i) (, ), + = tcf

i


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28 SAHARON SHELAH

I Jbd a -complete ideal on such that (, ) and tcf


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So there is some \X, and for i < choose i ci such that

i bi [a](well defined by the choice of ci). So by smoothness of the representation

d {i < : i b[a]} {i < :

i b[a]} J.

Now by the pcf theorem for some A J+ we haveiA

i /J has true cofinality

which we call , so necessarily pcf-complete({

i : i A}) d (see thedefinition of d) but this contradicts the previous sentence (recall d by theminimality of ()). 2.10


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3 Applications

Of course

3.1 Claim. If is as in 2.1, then the conclusions of 1.2 and 1.1 hold.

3.2 Claim. If then:

(a) 2 = + +

(b) =


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= 0 regular and stationary S = S1 S2 we have(D)+S (D)

S (D)S and (D)S1 (D)S2 but (D)

S (D)

S1

, (D)+S2

(D)+S1 .

3.5 Proof of 3.2. By 3.4(1) it suffices to prove clause (b). Trivially (D) = , for each < , R there is P, a family of < subsets of of cardinality such that if A , |A| = then A is included in the union of< members ofP.Let P = {B : for some R ( +1) and A P

we have B A} so P

is a

family of< subsets of. For each A we define hA : by defining hA()

by induction on : for non-limit hA() is the first ordinal i >


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32 SAHARON SHELAH

(remember A : < lists the bounded subsets of each appearing unboundedlyoften).

Now for any A we have E =: EA =: < : limit and


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= 3()+, let Mi : i < be such that: Mi (H(), , subsets of each ofcardinality with the intersection of any two having cardinality < .


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34 SAHARON SHELAH

3.9 Remark. 1) The holding of ()2 is characterized in [Sh 410, 6].2) On earlier results concerning such problems and earlier history see Hajnal,Juhasz, Shelah [HJSh 249]. In particular, the following is quite a well known

problem:the Cantor discontinuum partition problem: Can every topological spacebe divided into two pieces, such that no part contains a closed homeomor-phic copy of 2 (or any topological space Y such that every scattered setis countable, and the closure of a non-scattered set has cardinality contin-uum)?

3) Note that the condition in ()2 holds if = 20 > , = 0, = 1 and 1

(from 2.1) (which holds e.g. if V = VP0 , P a c.c.c. forcing making the continuum

> V0 ). So in this case the answer to is positive.4) Also if = 20 > 1, and ()[ 2

0 ,1 ] then the answer to in

(2) is yes; now on ,1 see [Sh 410, 6].

Proof. We prove by induction on [, |X|] that:

() if Z, Y are disjoint subsets of X, |Y| , then there are h, Y+ such that

(a) Y Y+ X\Z

(b) |Y+|

(c) h is a function from h : Y+ to

(d) if A P, < , , |A Y+| and |A Z| < then

h (A Y+) is onto .

Case 1. = , so |Y| .Without loss of generality [B Y & |B| & |c(B)| = c(B)\Z Y].

Now just note that PY =: {c(B) Y : B Y, |B| , |c(B) Y| = } hascardinality = , and by the definition ofP (using the demand |A Z| < in ()), it suffices that h satisfies: [A PY h Z is onto ], which is easilyaccomplished.

Case 2. > .


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Let =

2+

, Ni : i an increasing continuous sequence of elementarysubmodels of (H(), ,


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36 SAHARON SHELAH

Subcase 2. = i + 1.So |A Ni| < , hence |A

j


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P = P

I=: A X :|A| = and for every pairwise distinct

x A for < we have

{u : |c{x : u}| < }

is included in some I I

.

and replace ()2 by

()3 For every assumeF {( , I , f ) : I I, = Dom(I), f : is one to one}

and if (, I, f) F for = 1, 2 are distinct then { < 2 : f2() Rang f1} I2.Then |F| .Note that the present P fits for dealing with of 3.9(2) and repeating the proofof 3.8.

3.11 Discussion of Consistency of no: There are some restrictions on such theo-rems. Suppose

() GCH and there is a stationary S { < +1 : cf() = 1} andA : S such that: A = sup A, otp(A) = 1 and1 = 2 |A1 A2 | < 0.

(This statement is consistent by [HJSh 249, 4.6,p.384] which continues [Sh 108].)Now on 1 we define a closure operation:

c(u) ( S)[ A & (u A) 0].

This certainly falls under the statement of 3.8(2) with = = 0, = 1 exceptthe pcf assumptions ()1 and ()2 fail. However, this is not a case of our theorem.


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38 SAHARON SHELAH

Appendix: Existence of tiny models

We deal now with a model theoretic problem, the existence of tiny models; we

continue Laskowski, Pillay, and Rothmaler [LaPiRo]; our main result is in 3.17.

3.12 Context. Assume T is a complete first order theory. Let |T| be the number offirst order formulas (x), x = x : < n, n < , up to equivalence modulo T.Assume T is categorical in all cardinals > =: |T| and call a model M of T tinyif M < (= |T|). It is known that a T with a tiny model satisfies exactly one ofthe following:

(a) T is totally transcendental, trivial (i.e. any regular type is trivial)

(b) T is not totally transcendental.

3.13 Question. For which < are there T, |T| = (which is categorical in +

and) with a tiny model of cardinality ?

3.14 Discussion. By [LaPiRo] we can deal with just the following two cases(see [LaPiRo], 0.3,p.386 and 387121 and 1.7,p.390).

Case A. x = x is a minimal formula and its prime model consists of individualconstants.

Case B. T is superstable not totally transcendental and is uni-dimensional, theformula x = x is weakly minimal, regular types are trivial and its prime modelconsists of individual constants.

They proved: ()[0 + in case A, = 0], (see [LaPiRo,2.1,p.341]).Actually more is true by continuing their argument.

3.15 Lemma. If ,,T are as above, in Case A, then:

(i) < ,

(ii) we can find n : n < such that: 0 = , n n+1,

=n


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Proof. By 1.2(2),(ii) (i), so let us prove (ii). Let M be a tiny model of T,M = .

For n 0, let Bn be the family of definable (with parameters) subsets ofn+1M.

Clearly |T| n n = |Bn| by clauses (), ()) and{fR : R } is a set of functions as required. 3.15

3.16 Lemma. Suppose (),,, < . Then

(a) there is a group G of permutations of such that |G| = andf = g G { < : f() = g()} is finite

(b) there is a theory T as in 1.1, |T| = , with a tiny model of cardinality ofCase A.

Proof. As (a) (b) is proved in [LaPiRo], p.3922331 we concentrate on (a). Letpr(, ) be a pairing function on i.e. pr is one-to-one from onto . Solet {A : < } [] be such that = 0 > |A A|. Clearly

0 hence there is a list = : < of distinct members of . By renamingwe can have the family {A,n : < , n < }, such that (A,n [], [(, n) =


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40 SAHARON SHELAH

(, m) |A,n A,m| < 0] and)).

For n + 1, = < y > n+1{1, +1} we let

() fx (, ) = (, ) when

x = y, fy

(, ) = (, ) (by the previous stage)() fx (, ) = (g

1,n+1(), < x >) when () does not apply.

Easily f is a well-defined permutation of A.Now {f : < } generates a group G of permutations of A. We shall prove itgenerates G freely, moreover:

ifn < , t = ((), x()) : n is such that () < , x() {1, 1}, andfor no < n do we have () = ( + 1) & x() = x( + 1)

(i.e.

nfx()() is a non-trivial group term) then At = {a A : (

nfx()() )(a) =

a} is finite.

As |A| = , this clearly suffices.

As this property ofn

fx()() is preserved by conjugation without loss of generality

()0 n () = ( + 1) x() = x( + 1) where n + 1 is interpreted as zero.

For any a At let

()1 btm[a] = (n

=mfx()()

)(a) for m n + 1

(so btn+1[a] = a = bt0[a] and for m = 0, . . . , n we have b

tm[a] = f

x(m)(m)

(btm+1[a]))

()2 btm[a] = (tm[a],

tm[a])

Choose m < large enough such that:

()3 if m m and 0 1 < 2 n and (1) = (2) then

A(1),m A(2),m = .


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For a At let m = m[a] n + 1 be such that lg(tm[a]) is maximal and call thelength k = k[a]. As f(, ) = , implies lg() {lg() 1, lg() + 1}, clearly

()4 lg(t

m1[a]) = lg(t

m+1[a]) = lg(t

m[a]) 1 (where m 1, m + 1 meansmod n + 1).

Clearly

()5(a) btm[a] = fx(m)(m) (b

tm+1[a])

(b) btm1[a] = fx(m1)(m1) (b

tm[a]) hence (as (f

x(m1)(m1) )

1 = fx(m1)(m1) ) we have

(b) btm[a] = fx(m1)(m1) (b

tm1[a]).

Looking at the definition of fx(m1)(m1) (b

tm1[a]), as m = m[a] by ()4 clause () in

the definition of f applies so

()6(a) fx(m1)(m1) (b

tm1[a]) = (g

1(m1),k[a](

tm1[a]), (

tm1[a])x(m 1)).

Similarly looking at the definition fx(m)(m)

(btm+1[a]), by ()4 clause () applies so

()6(b) fx(m)(m)

(btm+1[a]) = (g1(m),k[a](

tm+1[a]), (

tm+1[a])x(m)).

By ()5(b) + ()6(a) we have

()7(a) btm[a] = (g1(m1),k[a](

tm1[a]), (

tm1[a])x(m 1)).

By ()5(a) + ()6(b) we have

()7(b) btm[a] = (g1(m),k[a](

tm+1[a]), (

tm+1[a])x(m))).

We can conclude by ()7(a) + ()7(b) that

()8 x(m) = x(m 1) hence x(m) = x(m 1).

So by ()0 applied to m 1 we get

()9 (m) = (m 1).

Clearly by ()7(a) + ()7(b) we have

()10 g1(m),k[a](

tm+1[a]) = g

1(m1),k[a](

tm1[a]).


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42 SAHARON SHELAH

Now by the choice of the g1 s (and the pairing function) and ()10

()11 tm+1[a] =

tm1[a] and

g

0

(m),k[a](t

m+1[a]) = g

0

(m1),k[a](t

m1[a]).

So by ()11 and the choice of the g0 s

()12 g0(m),k[a](tm+1[a]) = g

0(m1),k[a](

tm1) A(m),k[a] A(m1),k[a].

Ifk[a] > m we get a contradiction (by ()3), so remembering m = m[a] necessarily

lg(tm[a][a]) m + 1, hence by the choice of m[a] we have

lg(t[a]) m.

So {t[a] : < n + 1 : a At} is finite, hence it suffices to prove for each n+1{1, 1} the finiteness of

At, = {a At : t[a] : < n + 1 = }.

Let us fix .As for a At, we have g(tm[a]) m

for n + 1, it is enough to prove thatfor each k = k : n the following set is finite:

At,,k =: {a At, : g(t[a]) = k for < n + 1}.

Let K(k) = { n + 1 : k is k1, k+1} (i.e. a local maximum).

For each m K(k), the arguments in ()3

()12

apply, so by ()11

, if a At,,k then the value g(

tm[a]) is determined and g

0(m),km

(tm+1[a]) A(m),km

A(m1),km , but the latter is finite so we can fix g0(m),km

(tm+1[a]) = m but

g1(m),km

(tm+1[a]) can be computed from = g0(m),km

(tm+1[a]) and ((m), km)

i.e. as pr(otp(A(m),km ), m).But by ()7(b) the latter is tm[a] and as

tm[a] = m the value of b

tm[a] is uniquely

determined. Similarly by induction we can compute the other btm [a] for every m,

in particular bt0[a] = a, so we are done. 3.17

3.17 Conclusion. For a cardinal , the following are equivalent:

(a) there is a T as in 3.16(b) (i.e. T categorical in |T|+, |T| > ), with a tinymodel M, M = as in Case A above

(b) (),,+

(c) there is a group G of permutations of , |G| = + such that for g G,{ < : g() = } is finite or is .


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REFERENCES.

[AAC90] D.L. Alben, G.L. Alexanderson, and C. Reid (editors). More Mathe-matical People. Harcourt Brace Jovanovich, 1990.

[Br] F.E. Browder (editor). Mathematical developments arising fromHilberts Problems. Proc. of Symposium in Pure Math, 28:421, 1974.

[FW] Matthew Foreman and Hugh Woodin. The generalized continuum hy-pothesis can fail everywhere. Annals Math., 133:136, 1991.

[GiSh 344] Moti Gitik and Saharon Shelah. On certain indestructibility of strongcardinals and a question of Hajnal. Archive for Mathematical Logic,28:3542, 1989.

[Gre] John Gregory. Higher Souslin trees and the generalized continuum

hypothesis. Journal of Symbolic Logic, 41(3):663671, 1976.

[HJSh 249] Andras Hajnal, Istvan Juhasz, and Saharon Shelah. Splitting stronglyalmost disjoint families. Transactions of the American MathematicalSociety, 295:369387, 1986.

[HLSh 162] Bradd Hart, Claude Laflamme, and Saharon Shelah. Models with sec-ond order properties, V: A General principle. Annals of Pure andApplied Logic, 64:169194, 1993. math.LO/9311211.

[LaPiRo] Michael C. Laskowski, Anand Pillay, and Philipp Rothmaler. Tiny

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[Sh:E12] Saharon Shelah. Analytical Guide and Corrections to [Sh:g].math.LO/9906022.

[Sh 108] Saharon Shelah. On successors of singular cardinals. In Logic Collo-quium 78 (Mons, 1978), volume 97 of Stud. Logic Foundations Math,pages 357380. North-Holland, Amsterdam-New York, 1979.

[Sh 82] Saharon Shelah. Models with second order properties. III. Omittingtypes for L(Q). Archiv fur Mathematische Logik und Grundlagen-

forschung, 21:111, 1981.

[Sh 351] Saharon Shelah. Reflecting stationary sets and successors of singularcardinals. Archive for Mathematical Logic, 31:2553, 1991.

[Sh 400a] Saharon Shelah. Cardinal arithmetic for skeptics. Ameri-can Mathematical Society. Bulletin. New Series, 26:197210, 1992.math.LO/9201251.


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[Sh 420] Saharon Shelah. Advances in Cardinal Arithmetic. In Finite and Infi-nite Combinatorics in Sets and Logic, pages 355383. Kluwer AcademicPublishers, 1993. N.W. Sauer et al (eds.). 0708.1979.

[Sh 410] Saharon Shelah. More on Cardinal Arithmetic. Archive for Mathemat-ical Logic, 32:399428, 1993. math.LO/0406550.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Sh 454a] Saharon Shelah. Cardinalities of topologies with small base. Annals ofPure and Applied Logic, 68:95113, 1994. math.LO/9403219.

[Sh 430] Saharon Shelah. Further cardinal arithmetic. Israel Journal of Math-ematics, 95:61114, 1996. math.LO/9610226.

[Sh 589] Saharon Shelah. Applications of PCF theory. Journal of SymbolicLogic, 65:16241674, 2000.

[Sh 575] Saharon Shelah. Cellularity of free products of Boolean alge-bras (or topologies). Fundamenta Mathematica, 166:153208, 2000.math.LO/9508221.

[Sh 666] Saharon Shelah. On what I do not understand (and have some-thing to say:) Part I. Fundamenta Mathematicae, 166:182, 2000.math.LO/9906113.

[Sh 513] Saharon Shelah. PCF and infinite free subsets in an algebra. Archivefor Mathematical Logic, 41:321359, 2002. math.LO/9807177.

[Sh 668] Saharon Shelah. Antihomogeneous Partitions of a Topological Space.Scientiae Mathematicae Japonicae, 59, No. 2; (special issue:e9,449501):203255, 2004. math.LO/9906025.

Documents

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