saharon shelah- power set modulo small, the singular of uncountable cofinality

8/3/2019 Saharon Shelah- Power Set Modulo Small, The Singular of Uncountable Cofinality

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POWER SET MODULO SMALL, THESINGULAR OF UNCOUNTABLE COFINALITY

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat RamJerusalem 91904, Israel

Department of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Abstract. Let be singular of uncountable cofinality. If > 2cf(), we provethat in P = ([], ) as a forcing notion we have a natural complete embedding ofLevy(0, +) (so P collapses + to 0) and even Levy(0,UJbd

()). The natural

means that the forcing ({p [] : p closed}, ) is naturally embedded and isequivalent to the Levy algebra. Moreover we prove more than conjectured: if P failsthe -c.c. then it collapses to 0. We even prove the parallel results for the case > 0 is regular or of countable cofinality. We also prove: for regular uncountable, there is a family P of b partitions A = A : < of such that for anyA [] for some A : < P we have < |A A| = .

This research was supported by the United States-Israel Binational Science Foundation. Publica-tion 861.I would like to thank Alice Leonhardt for the beautiful typing.

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

0 Introduction

This work on the one hand continue the celebrated work of the Czech school on

the completion of the Boolean algebras P()/[] cf() = 0, (and () = h if = 0) consider the questions:

1 (a) is P isomorphic to the completion of the Levy collapse Levy((), 2

)?(b) which cardinals the forcing notion P collapse to () in particularis + collapsed

(c) is P (, )-nowhere distributive for = ()? This canbe phrased as: for some P-name f

of a function from () to ,

for every p P for some i < the set { < : p f

(i) = }

has cardinality .

The first, (a) is a full answer, the second, (b)seems central for set theories and

essentially give sufficient condition for the first, the last is sufficient if the densityis right, to get the first. The case of collapsing seems central (it also implies clause(c)) so we repeat the summary from Balcar, Simon [BaSi95] of what was known ofthe collapse of cardinals by P, i.e., 1(b). Let denote the fact that iscollapsed to by P

1 (i) for = 0, 20 h, (but P adds no new sequence of length < h sowe are done), Balcar, Pelant, Simon [BPS]

(ii) for uncountable and regular, b 0, (hence + 0),

Balcar, Simon [BaSi88]

(iii) for singular with cf() = 0, 20 1, Balcar, Simon [BaSi95]

(iv) for singular with cf() = 0, bcf() 0,Balcar, Simon [BaSi95];

under additional assumptions on cardinal arithmetic for singular cardinals more isknown

(v) for singular with cf() = 0 and 0 = 2, 0 1,


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POWER SET 3

Balcar, Simon [BaSi88]

(vi) for singular with cf() = 0 and 2 = +, 2 0, [BaSi88].

Now [BaSi95] finish with the following very reasonable conjecture.0.1 Conjecture: (Balcar and Simon) in ZFC: for a singular cardinal with countablecofinality, 0 1 and for a singular cardinal with an uncountable cofinality+ 0 (here we concentrate on the case cf() > 0, see below).

Concerning the other questions they prove

2 (i) Balcar, Franek [BaFr87]:if > cf() > 0, 2cf() = cf()+ then P is (0,

+)-nowheredistributive

(ii) Balcar, Simon [BaSi89, 5.20, pg.380]:

if 2

= +

and 2cf()

= cf()+

then P is equivalent:to Levy(0, +) if cf() > 0 and

to Levy(1, +) if cf() = 0

(iii) Balcar, Franek [BaFr87]:if 2 = +, = cf() > 0, J a -complete ideal on andJ is nowhere precipitous extending [] 0 the picture is reasonable, particularly if we recallthat by Baumgartner [Ba]

3 if = cf() < = cf() = 0; morever

4 (i) if > cf() = 0 then Levy(1, 0) can be completely embeddedinto the completion ofP. Moreover,

(ii) the embedding is natural: Levy(1, 0) is equivalent to Q whichis P where

Q = ({A : A a closed subset of of cardinality }, ).


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4 SAHARON SHELAH

Here we continue [KjSh 720] in 1, [BaSi89] in 2 but make it self contained. Bothsections use results on pcf (in addition to guessing clubs) Naturally we may add tothe questions (answered positively for the case cf() = 0 by [KjSh 720])

2 (a) can we strengthen P collapse to () to Levy((), ) is

completely embeddable into P (really P)

(b) can we find natural such embeddings.

We may add that by [BaSi95] the Baire number ofU[], the space of all uniformultrafilters over uncountable is 1, except when > cf() = 0 and in that caseit is 2 under some reasonable assumptions. By [KjSh 720] the Baire number ofU[] is always = 2 when > cf() = 0.

Our original aim in this work has been to deal with > cf() > 0, proving the

conjecture of Balcar and Simon above (i.e., that + is collapsed to 0), first of allwhen 2cf() < answering 2(a)+(b) using pcf (and replacing + by ppJbd

cf()()). In

fact this seems, at least to me, the best we can reasonably expect. But a posterioriwe have more to say.

For = = cf() > 0, though by the above we know that some cardinal > is collapsed (that is b), we do not know what occurs up to 2

or when the c.c.fails. This leads to the following conjecture, (stronger than the Balcar, Simon onementioned above). Of course, it naturally breaks to cases according to .

0.2 Conjecture. If > 0 and P does not satisfy the -c.c., then forcing with Pcollapse to (), see Definition 0.6 below.

Note that

0.3 Observation. If conjecture 0.2 holds for > 0 then P is equivalent to a Levycollapse iff it fails the d(P)-c.c. where d(P) is the density ofP.

Lastly, we turn to the results; by 1.16(1):

0.4 Theorem. If > = cf() > 0 and > 2 then Q (a natural completesubforcing ofP, forcing with closed sets) is equivalent to Levy(0, UJbd ()).

By 1.17, 1.18 and 2.7 we have

0.5 Theorem. Conjecture 0.2 holds except possibly when 0 < cf() < < 2cf().

We shall in a subsequent paper prove the Balcar, Simon conjecture fully, i.e., in allcases.


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POWER SET 5

0.6 Definition. For > 0 we define () {0, 1} by() = 0 if cf() > 0() = 1 if > cf() = 0

and may add() = when = 0, h = .

We thank Menachem Kojman for discussions on earlier attempts, Shimoni Gartifor corrections and Bohuslav Bakar and Pek Simon for improving the presentation.


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1 Forcing with closed set is equivalent to the Levy algebra

1.1 Definition. 1) For f (Ord\{0}) and ideal I on let

UI(f) = Min{|P| :P [sup Rang(f)]

such that for every g ffor some u P

we have {i < : g(i) u} I+}.

2) Let UI() means UI(f) where f is the function with domain Dom(I) which isconstantly

1.2 Hypothesis.

(a) is a singular cardinal(b) = cf() > 0.

1.3 Definition. 1) P is the following forcing notion

p P iff p []

P |= p q iff p q.

2) P is the forcing notion with the same set of elements and with the partial order

P |= p q iff |q\p| < .

3) Q = Q0 is P {p P : p is closed in the order topology of }.

1.4 Choice/Definition. 1) Let i : i < be an increasing sequence of regularcardinals > with limit .2) Let i = {j : j < i}.3) For p P let a(p) = {i < : p [

i , i) = }.

4) Q1 = {p P : i < |p i| < i and for each i a(p) the set p i\i has

no last element, is closed in its supremum and has cardinality > |p i |}.

5) For p Q1 let chp

ia(p)

i be chp(i) = { + 1 : p [i , i)} and

cfp

ia(p)

i be cfp(i) = cf(chp(i)).

6) Q2 = {p Q1:cfp(i) > |p

i | for i a(p)}.


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1.5 Claim. 1) Q0,Q1,Q

2 are complete sub-forcings ofP.

2) For = 0, 1, 2 and p, q Q we have p Q q G iff |p\q| < and similarly

forP.

3) Q = Q0,Q1,Q2 are equivalent, in fact Q2 is a dense subset of Q1 and for = 0, 1, {p/ : p Q} does not depend on where is the equivalence relationofP, defined by p1 p2 iff (q P)(q P p1 G

q P p2 G

).

Proof. Easy.

Recall

1.6 Claim. 1) P can be completely embedded into P (naturally).

2) Q can be completely embedded into P (naturally).3) P is completely embeddable into Q (naturally).

Proof. 1) Known: just a [] can be mapped to {[i , i) : i a}.2) By [KjSh 720, 2.2].3) Should be clear (map A [] to {[i , i] : i A}). 1.6

1.7 Choice/Definition. = UJbd ().

Recall1.8 Claim. Assume > 2.

1) = sup{ppJbd () : < , cf() = } = sup{tcf(

i


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8 SAHARON SHELAH

2) By [Sh 589, 1.1] we actually get the stronger conclusion.3) It follows easily from the definitions 1.1 and 1.7, and from the inequalities 2 < < . 1.8

1.9 Claim/Definition. Fix a set P [] exemplifying = UJbd ().

1) There is C = C : < such that:

(a) C is a subset of [i , i) closed in its supremum when (

i , i]

(b) if i < , < i, is a regular cardinal and C is a closed subset of [i , i)

of order type ++, then for some (i , i), C C and otp(C

) = .

2) Q3 = {p Q2 : if i a(p) then p [

i , i) {C

: (

i , i)} and for some

u P, { < : for some i < , p [i , i) = C} u} is a dense subset of

Q

1

,Q2

, hence ofQ.3) For p Q3 let cdp

ia(p)

i be such that cdp(i) (i , i) is the minimal

[i , i) such that p [i , i) = C

. Notice that for every p Q

3, there is

some u P with Rang(cdp) u.

Proof. 1) It is enough, for any limit (i , i) and regular , + < cf(), to find

a family P, of closed subsets of (i , ) of order type such that any club of

contains (at least) one of them. This holds by guessing clubs, see [Sh:g, III, 2].

2), 3) By the definitions. 1.9

1.10 Claim. 1) If > 2 (or just 2) thenQ2 (hence Q

1) has a dense subset

of cardinality .2) If > 2 (or just 2) thenQ

3 is a dense subset ofQ

1 and has cardinality

.

Proof. 1) By part (2).2) By 1.9(2) it suffices to deal with Q3. The cardinality of the set P from 1.9 is

. Whenever p Q3, then the function cdp is uniquely determined by its range,because i Dom(cdp) iff Rang(cdp) [

i , i) = and the value cdp(i) = iff

[i , i)Rang(cdp). Also, the function cdp uniquely determines p by p ={Ccdp(i) : i Dom(p)}. Since Rang(cdp) u, u P, we get |Q

3| 2

= .1.10

From now on (till the end of this section)


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POWER SET 9

1.11 Hypothesis. 2 < (in addition to 1.2).Recall (Claim 1.13(1) is Balcar, Simon [BaSi89, 1.15] and 1.13(2) is a variant).

1.12 Definition. A forcing notion P is (, )-nowhere distributive when there aremaximal antichains p = p : < ofP for < such that for every p P forsome < , we have |{ < : p, p are compatible}|.

1.13 Claim. 1) If

(a) P is a forcing notion, (, )-nowhere distributive

(b) P has density

(c) > 0 P has a -complete dense subset

thenP is equivalent to Levy(, ).2) IfP is a forcing notion of density collapsing to 0 then P is equivalent toLevy(0, ).3) IfP is a forcing notion of density and is (, )-nowhere distributive then Pcollapses to (and may or may not collapse ). 1.13

1.14 Claim. Assume b : < is a sequence of pairwise disjoint members of[] with union b. Then we can find an antichainI ofQ3 such that:

() if q Q3 and ( < )(a(q) b []), then q is compatible with =:

UJbd

() of the members ofI

.

Proof. Let

I = {p Q3 : we can find an increasing sequence i : <

such that i b\, a(p) {i : < } and

i a(p) p [i

, i) has order type }.

Let J = {p Q3: for every < we have a(p) b []}.Clearly

(a) |I| = UJbd ()

[Why? As I Q3.]

(b) ifI I, |I| < and q J then there is r such that q r I andr is incompatible with every p I.


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[Why? Let = |I|+, it is < , hence we can find an increasing sequence : i() > (1 < )(i(1) < i()). Asq Q3 it follows that (q [

i(), i())) has order type otp(a(q)) > . Let

Cq, = { : q, [i(), i()) and otp(q [

i(), i()) ) is < }. Now

for every p I the set p [i(), i()) {[

i , i) : i b} if non-empty has

cardinality which is < hence p Cq, is a bounded subset of Cq,, call the

lub p,. As = |I| + < tcf(


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POWER SET 11

Hence q fits the demand in 1.14 with A here standing for b : < . Hence it iscompatible with members ofI which, of course, shows that we are done.

1.15

1.16 Conclusion. 1) If 2 < (and 0 < = cf() < , of course) then Q isequivalent to Levy(0, ), i.e., they have isomorphic completions (recalling Q isnaturally completely embeddable into the completion ofP = ([]

, )).2) If ( < )(|| < ) then Q is equivalent to Levy(0, ).3) If is strong limit (singular of uncountable cofinality ), then P is equivalentto Levy(0, ) = Levy(0, 2).

Proof. 1) By 1.10(1), Q3 has density (even cardinality) and by 1.15 it is (b, )-

nowhere distributive hence by 1.13(3), we know that Q3

collapses to b. But

P is completely embeddable into Q2 (see 1.6(3)) and P collapses b to 0 (e.g.

see 2) and Q3 is dense in Q2. Together forcing with Q

3 collapses to 0. As Q

3

has density , by 1.13(2) we get that Q2 is equivalent to Levy(0, ).

Lastly Q,Q3 are equivalent by 1.5(3) + 1.9(2) so we are done.

2) Recalling 1.8, by [Sh:g, VIII] we have = (alternatively directly as in [Sh

506, 3]). Now apply part (1).3) By easy cardinal arithmetic = 2. Enough to check the demands in 1.13(2).Now as Q collapses to 0 by part (1) and Q can be completely embeddableinto P (see 1.6(2)) clearly P collapses to 0. But |P| |[]| = 2, so P hasdensity 2.

Lastly = 2 by [Sh:g, VIII]. So we are done. 1.16

1.17 Claim. Assume that P does not satisfy the -c.c. Then forcing withPcollapses to 0.

Proof. By the nature of the conclusion without loss of generality is regular. Nowwe can find X such that

()1 (a) X = X : <

(b) X P

(c) X X []


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12 SAHARON SHELAH

[Why? For each i < fix some partition Wi, : < i of i into i (pairwisedisjoint) sets each of cardinality i. Now for each p P we shall choose A = A

p P as required in ()2 such that P := {A

p : p P} has cardinality this

suffice; so fix p P. By induction on < we can find < of cofinality ++

such that p is unbounded in and > { : < }. There is a club C1

of of order type ++ with min(C

1 ) > { : < }. Let C

2 = { C

1 : is

a limit ordinal such that C1 p is unbounded in and has order type divisible by+ }, it is a club of . But by the club guessing (see 1.9) there is C

3 such that:

C3 C2 ( C

1 ) and otp(C

3 ) = .

By the definition ofQ3, there is some a [] such that

{C3 : a} Q

3.

Lastly, let us define A = A : < by

A = {[, min(C3\(+ 1)) : a satisfies

< and C3 and

otp(C3 ) W,}.

Easily A : < is as required in ()2, and since A is determined by an elementofQ3 (and the constant Wi, : < i : i < ), the cardinality |P| |Q

3| .]

Now for A P we define a P-name A as follows: for G P generic over V,

()3 A[G] = iff is minimal such that {A : X} G

clearly

()4 A[G] is defined in at most one way;

()5 for every p P for some A P for every < we have p A = .

[Why? Let A P be such that ( < )( = |p A|), it exists by ()2. Now wecan find q satisfying p q P such that ( < )(q A is a singleton) and foreach < let q = {A q : X}. Clearly < |X X| < {A : X} q {A q : X} = {A q : X X} []


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1.18 Claim. If > cf() = 0 andP fails the -c.c., thenP collapses to1; note that in this case Q is equivalent to Levy(1,

0 ) by [KjSh 720].

Proof. Let = 0 .

By Kojman, Shelah [KjSh 720], P collapses to 1 hence it suffices to provethat P collapse to assuming > (otherwise the conclusion is known). Letn : n < be a sequence of regular uncountable cardinals with limit . Nowrepeat the proof of 1.17 1.18


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2 The regular uncountable case

We prove that (for regular uncountable),P collapse to 0 iff

P fail the-c.c. This continues Balcar, Simon [BaSi88, 2.8] so we first re-represent what they

do; the proof of 2.6 is made to help later. In the present notation they let = b(rather that bspc as below, let f : < b be a sequence exemplifying it;let C = { < : ( < )(f() < ), a limit ordinal} and let B = \C, soB : < is a (, )-sequence (see 2.5(1)), derive a good (, >)-sequence fromit (see 2.5(2)), define n(A), n(A) and used the A,,is to define the P-names

n

and prove P {g(

n) : n < } = b (see 2.6). We then prove the new result: if

P fail the -c.c. then it collapses to 0.

2.1 Context. is a fixed regular uncountable cardinal.

2.2 Definition. 1) Let bspc be the set of regular > such that there is a

(b) B []

(c) if 1 2 > then B2 B1 which means B2\B1 []

and A [B] then for some < we have A B []

(f) if > and < < then B B and B\B [] and B\B [].


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3) For a (, >)-sequence B and A [] we try to define an ordinal k(A, B) byinduction on k < . If = (A, B) : < k is well defined (holds for k = 0) andthere is an < such that A B ( < )(A B []

)-parameter when:

(a) B = B : > is a (, >)-sequence

(b) is an S -ladder which means that = : S, is an increasing

sequence of ordinals of length with limit , where S = { < : cf() = }.

5) We say (B, ) is a good (, >)-parameter when (a)+(b) of part (4) holds and

(c) if A [] then for some n < , n and S and A [A] we have

() (A, B) = () for < n

() for many ordinals < we have ( < )(AB\Bbelongs to []).

6) B is a good (, >)-sequence if clause (a) of (4) and clause (c) of (5) holds forsome S-ladder (see above). We say B is a weakly good sequence if clause (a) of (4)and clause (c) of (5) which means that we ignore subclause () there. Similarly(B, ) is a weakly good (, >)-parameter.

2.4 Observation. 1) In 2.3(5)(c)(), the for many ordinal < implies forclub many ordinals < 0.2) In 2.3(6) it doesnt matter which S -ladder you choose.

Proof. If 1, 2 are increasing and sup(1) = sup(2) = , then {i < :

j


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16 SAHARON SHELAH

defined and the pair (B, ) is a good (, >)-parameter where we define B and fas follows:

(a)

B = B :

>

(b) f = f : >

(c) B = , f = id

(d) B [], f is a function from B onto , non-decreasing,

and not eventually constant

(e) if the pair (B, f) is defined and < then we let

B = { B : f() \C}

(f) if = and B, f and B are defined then we letf : B be defined by f(i) = otp(C f(i))

for eachi < ,

hence

(g) if > then B B and i B f(i) > 0 f(i) > f(i).

Proof. 1) Recall S := { < : cf() = }.By the definition of bspc there is an


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Clearly C : < is as required.Lastly, let B = \C, it is easy to check that B : < is a (, )-sequence.

2) Clearly BC, fC are well defined and (B, ) is a (,>)-parameter and clauses

(a)-(g) of holds. Why is it good? Toward contradiction assume that it is not, sochoose A [] which exemplify the failure of clause (c) of Definition 2.3(5) anddefine

T0 = T0A =

> : there is A [A] such that

(A, B) : < g() is well defined and equal to

.

and define

T1 = T1A :=

T0A : for every k < g() there are <

ordinals < (k) such that ( k) T0

.

Clearly

()1 T0 T1 are non-empty subsets of> (in fact T1 T0)

()2 T0,T1 are closed under initial segments.

For T let SucT() = { T : g() = g() + 1 and }.We define A [B]

for T1 by induction on g():

()3 (a) A = A

(b) if A is defined and T1 then we let

A = A B\

{B : < and T1}.

Now

()4 if T1 then

(a) if B [A] and (B, B) : < g() is well defined and equal to then B A

(b) if SucTj () has cardinality < then A\ {A : SucTj()} hascardinality < for j = 1 (actually j = 0 is O.K., too).

(c) If SucT1() has cardinality < then SucT0() = SucT1()


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18 SAHARON SHELAH

[Why? First we can prove clause (a) by induction on g() using the definition ofT1 and clause (c) of 2.3(2). Second, we can prove clause (b) from it. Third whyclause (c) holds?

Otherwise, as T1 T0, there is an with n

SucT0(n)\ SucT1(n). Henceby the definition ofT1 the set u := { < : n T0} has cardinality but then u | u| < n T1 which implies that |SucT1(n)| ,contradiction to the assumption of clause (c).]

()5 |T1|

[Why? Otherwise by ()4 the set A := {A\ {A : SucT0()} : T1} isa subset of of cardinality < and by clause (d) of of the present claim alsoA = {f1 {0} : T1} is a subset of of cardinality < . So we can choosej A\(A A). Now we try to choose n T1 by induction on n such that

g(n) = n, n+1 SucT1(n) and j An .So 0 = belongs to T1 by ()1 + ()3(a). Now assume n is well defined, thenSucT0(n) = SucT1(n) by ()4(2) and our present assumption toward contradict-ing |T1| < .

Now j / A, A An\ {A : SucT1(n)}, but j An hence clearlyj {A : SucT1(n)}, so we can choose n+1 as required. So we have carriedthe definition of n : n < .

As j An Bn by ()3(b) above, clearly fn(j) is well defined (for eachn < ). As j / A and f1n {0} A

, so j / f1n {0}, necessarily fn(j) = 0 and sofn(j) > fn+1(j) by the choice of fn+1 in clauses (g) of. Hence fn(j) : n < is decreasing (sequence of ordinals), contradiction. So ()5 holds.]

Let n < be maximal such that |T1n| < , it exists as |T1| = cf() > 0and n = 0 |T1n| = 1 < , and let T1n be such that SucT1() has members; it exists as is regular. We can choose an increasing sequence i : i < of ordinals such that i is the i-th member of the set { < : T1} andlet Ai [A] be such that (Ai, B) : n = i and let = {i : i < },so S . Let

A = {Ai : i < } B\ {A() : < g()

and < () and ( ) T1}

(note that that number of pairs (, ) as mentioned above is < ).Clearly (A, B) = () for < g() hence (A Ai, B) = () for i )-parameter and 1 bspc then the forcing

notionP collapses 1 to 0.

Proof. Let (B, ) be a good (, >)-parameter.Note

1 if A1 A2 are from [] and (A2, B) is well defined then (A1, B) iswell defined and equal to (A2, B), recalling Definition 2.3(3).

Let h = h : < 1 exemplify 1 bspc , i.e., is as in Definition 2.2 and

without loss of generality [i < j < i < h(i) < h(j)]. For each Sand > let A,,i = B\ {B : j < i} for i < soA,,i : i < are pairwise disjoint subsets of (each of cardinality ). For n <

and A []

we try to define an ordinal n(A,

B, ,h) as follows:

2 n(A, B, , h) = iff for some n and S we have (A, B) :

n = so in particular is well defined and A {A,,ih(i) : i < }but for every < we have A {A,,i h(i) : i < } []


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Let S : < 1 be pairwise disjoint stationary subsets of S1 and define

g : 1 1 by g() = if S ( 1\

+, this suffice3) We could have seperated the different roles of in the proof of case 1. Say

(a) (B, 1) will be a good (, >(1))-parameter,

(b) h : < 2 exemplify 2 bspc and : S2 is an S

2 -ladder system

(so S2 in the proof)

4) Actually, we can revise case 2 to cover Case 1, too: for S+ choose C

a club of of order type +. Now for each we can repeat the construction of

names from the proof of Case 2, for each p P for some we succeed to show below.

Proof. The proof is divided to two cases.

Case 1: bspc , > +, e.g. = b.

So is regular > + and a good (, >) sequence B exists (by 2.5).Let = : S

be such that

is increasing continuous with limit and guesses clubs (i.e. for every club C of , for stationarily many S wehave Rang() C); exists by [Sh:g, III,2] because = cf() >

+. As B is agood (, >)-sequence, (B, ) is a good (, >)-parameter by 2.4 (or use 2.5).

Let h : < exemplify bspc without loss of generality i < j < i , S and i < , recall that A,,i = B\{B :j < i} and let n(A, B, , h),

n =

n(B, , h) be defined as in the proof of 2.6. For

>, S

and < let B,,

:= {A,,i h(i) : i < }. So clearly (for

each >, S) the sequence B,, : < is

-increasing. For Sand i < let A,,,i := B

,,(i+1)

\ {B,,(j+1) : j < i}. So A,,,i : i <

are pairwise disjoint subsets of . Note that (by the proof of 2.6 but not used)for each pair (, ) as above for some club E, of , for every

S E,and i < , A,,,i has cardinality . We shall show during the proof of (1) that

{A,,,i : i < : >, S ,

S} is as required in part (2), so this will

prove part (2) when b > +.

Let X : < be an antichain ofP, it exists by the assumption. We now for,, as above define P-names

,, : for G P generic over V we let:

0 ,, [G] = iff for some A G, n < and n and ,

S we have:

(a) (A, B) : < n = so in particular is well defined

(b) n(A, B) = S

(c) n(A, B, , h) = S

(d) A A,,,i has at most one member for each i <

(e) A {A,,,i : i X }

Note that demands (a),(b),(c) are natural but actually not being used; with themwe could have defined the P-names

n which is

,, when defined. Now clearly

1 ,, is a P-name of an ordinal < (may have no value)

2 for every p P for some > and , S , for every < there is

q such that p q P and q P ,, = .

[Why? We start as in the proof of 2.6. First there are n < , n and S

such that p A,,i []

for many ordinals i < .Second, there is a club Cp of such that:if < < are from Cp then p B

,,\B

,, []

. Indeed, Cp = { < : forevery < the set p B,, \ B

,, is from []

} is as required.

Now by the choice of , i.e., club guessing, there is acc(Cp) S such that(i < )((i) Cp). So (as we have used (i + 1), (j + 1) in the definition ofA,,,i)


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22 SAHARON SHELAH

i < p A,,,i [].

This fulfills the promise needed for proving part (2) in the present case 1. Choosei p A,,,i for i < . Now for every < let q = {i : i X

}. Recall that

X : < is an antichain in P. Clearly for < we have P |= p q andq

,, = ; so we have finished proving 2.]

This is enough for proving

3 forcing with P collapse to 0.[Why? By 1 + 2 we know that P = {

,, :

>, S and

S}, so it is forced that || ||. As we already have by 2.6 thatP || = 0, we are done.]

Case 2: b = +.

Let = + and B be a good (, >)-sequence. Let S : < be a partition

of S+

to (pairwise disjoint) stationary sets. For < + let ui : i < be

an increasing continuous sequence of subsets of each of cardinality < with

union and without loss of generality < (i < )(ui = ui ). Let

h = h : < + exemplifying + bspc be such that each h is strictly increasing,

(i)h(i) > i and let C = { < : is a limit ordinal and for every i < we haveh(i) < } and let (B, ) be a good (, >)-parameter; exists by 2.5(2). Now for > and S we define A,,i(i < ), B

,,( < ) as in Case 1. Now for

>, S , < + and < + we define the sequence Y,,,, : < by

Y,,,, := {B,,[i, Min(C\(i+1))\{B

,,1

: 1 ui } : i C satisfy u

i }.

So Y,,,, : < is a sequence of pairwise disjoint subsets of and for < let

Z,,,, := {Y,,,, : S }.

Clearly

1 Z,,, = Z,,,, : < is a sequence of pairwise disjoint subsets of .

We shall show during the proof of (1) that

Z,,,, : < :

>, S , < , <


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POWER SET 23

exemplify part (2); you may wonder: possibly for some quadruple (,,, ) we donot have ( < )[|Z,,,,| = ], so? However the quadruple (,,,) for whichthis fails, cannot satisfy the desired property in part (2), so we can just omit them.

Let X : < be a family of sets from []

such that the intersection of any twohave cardinality < , it exists as P fail the -c.c.. For each

>, S , < +

and < + we define a P-name ,,, as follows:

2 for G P generic over V, ,,,[G] = iff

() for some A G we have

(a) < A Z,,,, has at most one member

(b) A {Z,,,, : X }

() if for no A G does (a)+(b) hold and = 0.

Clearly

3 ,,, is a well defined (P-name) (by 2).

Now

4 for every p P, for some >, S , < +, < + we have:

for every < for some q P above p we have q ,,, = and

< |Z,,, p| = .

As in Case 1, this is enough for proving that P collapse to = +. But by 2.6

we already know that forcing with P collapses + to 0 and so we are done.

Note: we can eliminate from the ,,,, but not worth it. So we are left with

proving 4.Why does 4 hold? First, as in the earlier cases, find > and S such

that p A,,i [] for ordinals i < . Second, for some club Cp of we have < Cp p B,,\B

,, []

. As S (for < ) is a stationary subsetof and Cp a club of for each < we can choose

S Cp. Hence there

is < + large enough such that < < Cp. Now define a function

h : by induction on i, as follows:

h(i) = Min{j :j (i, ) and i1 < i h(i1) < j and

if the pair (, ) is such that u

i S then

p (i, j) B,,\ {B,,1

: 1 u

i } is not empty}.


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24 SAHARON SHELAH

it is well defined as for a given i < the number of pairs (, ) such that u

i Sis |u

i | < and is increasing; next we define

C = {j < : j is a limit ordinal such that i < j h(i) < j}.

Clearly C is a club of and let h be defined by h(i) = h(Min(C\(i + 1)).By the choice of h : < there is < such that for many ordinalsi < , h(i) < h(i). Recall that C = { < : is a limit ordinal and for everyi < we have h(i) < }.

So W1 = {i < : h(i) < h(i)}, by the choice of clearly W1 []

. Leti0j : j < be an enumeration of the club C acc(C) of in an increasing order,

so clearly U := {j < : W1 [i0j , i

0j+1) = } is unbounded in . For each j U

let i2j be the first member of W1 [i0j , i

0j+1), then let i

1j = sup(C (i

2j + 1)), it is

well defined as i0j C acc(C), and so i0j i1j and let i3j = min (C \ (i2j + 1)) soi2j < i

3j and

() i0j i1j i

2j < h(i

2j ) < h

(i2j ) < h(i2j ) < i

3j .

[why? as said above by the choice of i1j by the choice of h, by the choice of the

pair (C, h), by i2j W1, by the choice of i3j and C respectively.]

(1) () i1j < i3j are successive members of C

[Why? both are members of C by their choices hence it is enough toprove that C (i

1

j

, i3j

) = . But C (i1

j

, i2j

] = by the choice of i1j

and

(i2j , i3j) = by the choice of i

3j ]

Now for each < we know that S Cp = {u

i : i < } and

u

i : i < is -increasing hence for some j() < if j U \j() then u

i0j

hence by the choice of h(i1j ) and () we have p (i1j , i

3j) B

,,

\ {B,,1 : 1

u

i0j} is not empty; but i1j < i

3j are successive members of C by ()

so the

definition of Y,,,, implies that p Y,,,, (i1j , i

3j) = .

As this holds for every large enough j U i.e., for every j U\j() andU [] it follows that p Y,,,, []

. By the definition of Z,,,, it

follows that p Z,,,, [].We have proved this for every < . Choose p Z,,,, for every < .

Now for each < let

q = { : X }.

So clearly:


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< P |= p q and q P ,,, = .

2.7

2.8 Conclusion. If is regular uncountable and P fail the 2-c.c. then comp(P)

is isomorphic to the completion of Levy(0, 2).


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REFERENCES.

[BaFr87] Bohuslav Balcar and Frantisek Franek. Completion of factor algebras ofideals. Proceedings of the American Mathematical Society, 100:205212,1987.

[BPS] Bohuslav Balcar, Jan Pelant, and Petr Simon. The space of ultrafil-ters on N covered by nowhere dense sets. Fundamenta Mathematicae,CX:1124, 1980.

[BaSi88] Bohuslav Balcar and Petr Simon. On collections of almost disjoint fam-ilies. Commentationes Mathematicae Universitatis Carolinae, 29:631646, 1988.

[BaSi89] Bohuslav Balcar and Petr Simon. Disjoint refinement. In Handbook

of Boolean Algebras, volume 2, pages 333388. NorthHolland, 1989.Monk D., Bonnet R. eds.

[BaSi95] Bohuslav Balcar and Petr Simon. Baire number of the spaces of uniformultrafilters. Israel Journal of Mathematics, 92:263272, 1995.

[Ba] James E. Baumgartner. Almost disjoint sets, the dense set problem andpartition calculus. Annals of Math Logic, 9:401439, 1976.

[KjSh 720] Menachem Kojman and Saharon Shelah. Fallen Cardinals. Annals ofPure and Applied Logic, 109:117129, 2001. math.LO/0009079.

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

[Sh 506] Saharon Shelah. The pcf-theorem revisited. In The Mathematics of PaulErdos, II, volume 14 of Algorithms and Combinatorics, pages 420459.Springer, 1997. Graham, Nesetril, eds.. math.LO/9502233.

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

saharon shelah- power set modulo small, the singular of uncountable cofinality

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