Jakob Kellner and Saharon Shelah- Preserving Preservation

8/3/2019 Jakob Kellner and Saharon Shelah- Preserving Preservation

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PRESERVING PRESERVATION

JAKOB KELLNER AND SAHARON SHELAH

Abstract. We prove that the property P doesnt make the old reals Lebesgue null is preserved undercountable support iterations of proper forcings, under the additional assumption that the forcings are nep (ageneralization of Suslin proper) in an absolute way. We also give some results for general Suslin ccc ideals.

1. Introduction. Let us consider the following

Hypothesis 1: Let ( P ,Q ) < be a countable support iteration of proper

forcings ( a limit) such that each P ( < ) forces that the set of old reals X V 2 remains Lebesgue positive. Then P forces this as well.

The main result of this paper (9.4) is that hypothesis 1 is true under some additional(relatively mild) requirements on the P . It seems that such requirements are needed(this is argued in section 4).

Preservation theorems of this kind have proven to be extremely useful in indepen-dence proofs. Hypothesis 1 specically is used in the proof of the following two theo-rems of [9]:

It cannot be decided in ZFC whether every superposition-measurable func-tion is measurable. (A function f : R 2 R is superposition-measurable,if for every measurable g : R R the superposition function f g : R R , x f ( x, g( x)) is measurable.)

and (von Weizs ackers problem):

It cannot be decided in ZFC whether for every f : R R there is a contin-uous function g : R R such that { x R : f ( x) = g( x)} is of positive outermeasure.

Forcing is a very general method for proving independence results, i.e., results of the form formula is neither provable nor refutable in ZFC. Forcing gives a methodfor modifying a given set-theoretical universe V to a new universe V in which someformula is guaranteed to hold. In a forcing argument (for example, to violate CH) onetypically has to

Received by the editors April 2004.1991 Mathematics Subject Classication. 03E40, 03E17. partially supported by FWF grant P17627-N12. supported by the United States-Israel Binational Science Foundation (Grant no. 2002323), publication

828.The authors thank a referee for proposing numerous enhancements, including a substantial simplication

of lemma 5.11.

1


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PRESERVING PRESERVATION 3

Hypothesis 2: Assume X has positive outer measure. Then the following property is

preserved by countable support iterations of proper forcings:Forcing with P leaves the set X positive.So P1 does not seem to be iterable. The Lebesgue version of tools-preserving (9.2)

is an iterable property that implies P1. In this paper we show (in sections 6 and 9) thatthe following property P2 implies Lebesgue-tools-preservation:

ZFC proves that P is nep and satises P1.(P2)

So the iteration of forcings satisfying P2 satises Lebesgue-tools-preservationand there-fore P1.

Non-elementary proper forcing (nep) has been introduced in [13]. It is a generaliza-tion of Suslin + (introduced in [2]), which in turn is a generalization of Suslin proper. Forexample, Cohen, random, amoeba and Hechler forcing are Suslin ccc, Mathias forcing

is Suslin proper, Laver, Miller and Sacks forcing are Suslin+

. An introduction to transi-tive nep forcing and Suslin ccc ideals can be found in [7].We will investigate not only the Lebesgue ideal, but general Suslin ccc ideals (such

as the meager ideal) as well. The case of the meager ideal has already been solved byGoldstern and Shelah in [12, Lem XVIII.3.11, p.920].

Annotated contents.

Section 2, p. 3: We recall the denition and basic properties of Suslin ccc ideals,the corresponding notions of positivity and outer measure, and the Cohen andrandom algebras on 2 .

Section 3, p. 8: We dene preservation of positivity and of outer measure, and listsome basic properties.

Section 4, p. 11: We give a partial counterexample to hypothesis 1. To be more

exact: We show that Hypothesis 2 is consistently false.Section 5, p. 13: We introduce true preservation (of positivity and of outer measure),a notion using the stationary ideal on [ ] 0 . We show that these notions are re-lated to (strong) preservation of generics. Apart from denition 5.9, this sectionis not required for the main result 9.4.

Section 6, p. 18: We prove that under certain assumptions, preservation of positivityimplies strong preservation.

Section 7, p. 24: We recall the Case A or tools preservation theorem for count-able support iterations of proper forcings.

Section 8, p. 25: We review the case of the meager ideal.Section 9, p. 26: We deal with the case of the Lebesgue ideal and show that strong

preservation is equivalent to Lebesgue-tools-preservation, and that thereforestrong preservation is preserved in countable support iterations.

Diagrams of implications (for the general case, as well as for meager and Lebesguenull) can be found on pages 29 and 30.

2. Notation and Basic Results. In this paper, the notion N H ( ) always meansthat N is a countable elementary submodel.

Forcings are written downwards, i.e. q < p means q is a stronger condition than p.Usually, the symbols for stronger conditions will be chosen lexicographically biggerthan those for weaker conditions.


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4 JAKOB KELLNER AND SAHARON SHELAH

Names for objects in the forcing extension are usually written with a tilde below, such

as . The standard name for an object x V is denoted by x. The name of the genericlter however (as well as a generic lter itself) will usually be called G (or e.g. G P if wewant to stress the forcing P ).

ro( Q) denotes the complete Boolean algebra of regular open sets of Q.We will x a Suslin ccc ideal I (Suslin ccc ideals are dened in 2.2). We will use the

phrases null, measure 1 or outer measure1 for every such I , even if I is not relatedto a measure. This seems more intuitive than terminology such as having outer Borelapproximation 2 . Note that our notation does not mention the ideal I as parameter:we will say null instead of e.g. I -null (although the notion does of course dependon the ideal I used).

We will mainly be interested in the case that I is the set of Lebesgue null sets.C denotes the Cohen algebra and B the random algebra.

Suslin ccc Ideals. We assume that Q I is a Suslin ccc forcing:

D 2.1. A (denition for a) forcing Q is Suslin ccc, if Q 2 , x Q and x Q y are

1

1 statements, x and y are compatible is Borel, and Q is ccc.

So Q I is dened using a real parameter r Q . A candidate is a countabletransitive modelof some ZFC ZFC containing r Q (see denition 6.3 for more details on ZFC ).

In addition, we assume that I is a hereditarily countable name for a new real (i.e. Q I

I \ V )

such that in all candidates { I (n) = m , n, m } generates ro( Q I ). (Such a real is

sometimes called generic real.) Note that e.g. for Cohen forcing the canonical namefor the Cohen real has this property; analogously for random forcing.

A Suslin ccc ideal I is an ideal dened from a pair ( Q I , I ) as above in the following

way:

D 2.2. A BC means A is a Borel code. For A BC, AV denotes the evaluation of A in V

(i.e. AV is the Borel set corresponding to the code A). A Borel code A is null, or: A I BC , if Q I

I AV [GQ I ].

A is positive, or: A I +BC , if A is not null. A has measure 1 if the code for 2 \ A is null.

A subset X of 2 is null, or: X I , if for some A I BC , X AV . X 2 is positive, or: X I + , if it is not null. X is of measure 1 if 2 \ X is null.

For an arbitrary set N and a real r 2 , r is called I -generic over N ,or: r Gen( N ), if r AV for all A I BC N .So Gen( N ) = 2 \ { AV : A I

BC N }.

For example, if Q I is the random algebra B , then I is the ideal of Lebesgue null sets,and if Q I is Cohen forcing C , then I is the ideal of meager sets.

Note that the notation we use assumes that the ideal I is understood, e.g. we saypositive instead of positive with respect to I .

The following can be found e.g. in [7]:

L 2.3. I is a -complete ccc ideal containing all singletons, andro( Q I ) Borel / I (as a complete Boolean algebra).


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For a Borel code A, the sentences q Q I

I AV [GQ I ] and A I BC are 1

2 .

(So in particular they are absolute.) If N is countable, then Gen( N ) is a Borel set of measure 1. If N is a countable elementary submodel of H ( ) and M the transitive collapse of

N , then r is I -generic over N iff r is I -generic over M . Let M be a candidate. r is I -generic over M iff there is (in V ) a Q I -generic lter G

over M such that I [G ] = r .

If M is a candidate and q Q I M , then there is a positive Borel code Bq M such that M

I B M [G]q ro( Q I ) = q. Such a Bq satises

{ I [G] : G V is Q-generic over M and contains q} =

= \ { AV : A M , q I A} = Gen( M ) BV q .

For example, if we we chose Q I to be Cohen forcing, then we get the followingwell known facts: The meager ideal is a -complete ccc ideal, ro( Q I ) is Borel modulomeager, for a Borel code A the statement A is meager is absolute, a real c is Q I -genericover a model M iff it is I -generic (i.e. if it avoids all meager Borel sets of M ), etc.

For any Suslin ccc ideal I there is a notion analogous to the Lebesgue outer measure.Note however that this generalized outer measure will be a Borel set, not a real number:

D 2.4. Let X be a subset of 2 . A Borel set B is (a representant of) the outer measure of X if B is (modulo I ) the

smallest Borel superset of X . I.e. B X , and for every other Borel set B X , B \ B is null.

X has outer measure 1, if 2 is outer measure of X .

Instead of B X we could use X \ B I in the denition of outer measure. 1

Clearly, every X has an outer measure (unique modulo I ); the outer measure of aBorel set A is A itself; the outer measure of a countable union is the union of the outermeasures; etc.

If I is the Lebesgue ideal, then the outer measure of X (according to our denition) isa Borel set B containing X such that Leb( B) = Leb ( X ), where Leb ( X ) R is the outermeasure according to the usual denition.

If I is the ideal of meager sets, then the outer measure of a set X is 2 minus the unionof all clopen sets C such that C X is meager. (This follows from the fact that everypositive Borel set contains (modulo I ) a clopen set and that there are only countablemany clopen sets).

The Random and Cohen Algebras on 2 . We can add many Cohen (or random)reals simultaneously using the Cohen algebra C (or random algebra B ) on 2 . Wewill need these forcings only for the counterexamples 3.3 and 4.1. We briey recallsome well known facts.

Let J be any set. For i J and a { 0, 1} dene [ i a ] { x 2 J : x(i) = a}. Abasic clopen set is a nite intersection of such sets [ i a ]. These sets form a basis of the topology, and the clopen subsets of 2 J are exactly the nite unions of basic clopensets.

L 2.5. Let B J be the -algebra on 2 J generated by the (basic) clopen sets.

1 That makes no di ff erence modulo I , since every null set is contained in a Borel null set.


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The requirement that A is a maximal antichain is upwards absolute is very strong

and will usually only be satised by ccc forcings. For example every Suslin ccc forcingis strongly absolute, but Mathias forcing is not (although it is nicely denable and inparticular Suslin proper). Note that for strongly absolute forcings p and q are compat-ible is absolute.

L 2.9. For any J (suitable denitions of forcings equivalent to) B J and C J arestrongly absolute.

Quotient forcings. The following is a basic fact of forcing theory: If f : P Q is acomplete embedding, then

Q is equivalent to P

R, where

R contains all q Q that are compatible with f ( p)for all p G P . So in particular

for every Q-generic lter G Q over V there is a P -generic lter G P over V and an

R[G P ]-generic lter G R over V [G P ] such that V [G Q ] = V [G P ][G R], and

for every P -generic lter G P over V and every

R[G P ]-generic lter G R over V [G P ]there is a Q-generic lter G Q over V such that V [G Q ] = V [G P ][G R].

Sometimes it is more convenient to use the following analogon that doesnt mentioncomplete embeddings (which is folklore, but we do not have a reference):

L 2.10. Let P and Q be arbitrary partial orders.1. If G Q is Q-generic over V , and

if in V [G Q ] there is a P -generic lter G P over V ,then there is a forcing R V [G P ] and an R-generic lter G R over V [G P ] such thatV [G Q ] = V [G P ][G R]. R can be chosen to be a subset of ro( Q)V (and G R is essentially the same as G Q ).

2. Assume that Q forces that for all p P there is a P -generic lter over V containing

p. Then there is a P -name R for a subset of ro( Q)V

such that the following holds:If G P is P -generic over V and G R is

R[G P ]-generic over V [G P ], then G R is ro( Q)-generic over V and V [G P ][G R]

R[GP ] = V [G R]ro( Q) .

3. Q forces: If (2 P )V is countable, then for all p P there is a P -generic lter G Pover V containing p.

P . (1) Assume towards a contradiction that q ro( Q) forces thatG is P -generic

but there is no such R in V [G ]P . There is a p0 P such that q ( p

G) for all p p0 .

(Otherwise the set D { p P : q ( pG)} is dense, so q forces that there is a

pG D .) In particular the truth value q0 p0

G q is positive. There is a

complete embedding f from P p0 to ro( Q) q0 (just set f ( p) p G ).

So ro( Q) q0 can be factorized as P p0 R, where

R is the P -name for the set of all

q ro( Q) q0 such that p G q 0 for all p G P . If G P is P p0 -generic over V

and G R is R[G P ]-generic over V [G P ], then G R is ro( Q) q0 -generic over V , and thereforeV [G R]Q is a Q-generic extension by a lter containing q.In V [G R],

G[G R] = G P :

G R

R[G P ], so pG q 0 for every q G R.

If pG[G R] then p

G q = 0 for some q G R, so p G P .

If pG[G R] then p

G[G R] for all p p, so p G P for all p p, and p G P .

So in V [G] there is an R as required after all and we get a contradiction.

(2) For p P pick a ro( Q)-nameG for a P -generic lter containing p. Then there is

a p p such that ( pG ) for all p p (as in the proof of (1)). Choose a maximal


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antichain A P of such p and call the associated names for ltersG p . For a A set

qa

a Ga

. So qa

( p Ga) for all p a .Let

R be the following P -name: If G P A = {a} then let

R consist of those q ro( Q) qa

such that pG a q 0 for all p G P .

Assume that G P is P -generic and that G P A = {a}. Then V [G P ] is an extensionby P a , and

R[G P ] is the quotient of the complete embedding f : P a ro( Q) qa . So

every

R[G P ]-generic G R over V [G P ] is ro( Q) qa -generic (and therefore Q-generic) overV .

(3) If only countably many subsets of P are in V , then we can start with any p andcan construct a decreasing sequence of length meeting all these dense sets.

3. Preservation. Recall that we have xed a Suslin ccc ideal I and the correspond-ing notions of positivity.

D 3.1. Let X 2

be positive with outer measure B, and P a forcing. P preserves positivity of X if P X I

+ . P preserves Borel positivity if P preserves the positivity of AV for all positive

Borel codes A (i.e. P AV I + ).

P preserves positivity if P preserves the positivity of X for all positive X . P preserves outer measure of X if P ( BV [G] is outer measure of X ). P preserves Borel outer measure if P preserves the outer measure of AV for all

Borel codes A (i.e. P AV [G] is outer measure of AV ). P preserves outer measure if P preserves the outer measure of X for all X .

Of special interest is preservation of positivity (or outer measure) of 2 (we will alsosay: of V ), i.e. of the set of all old reals.

We have already mentioned the following: If I is the ideal of Lebesgue null sets, then

the random algebra B preserves positivity, and the Cohen algebra C does not preservepositivity. Dually, if I is the ideal of meager sets, then the Cohen algebra C preservespositivity, and the random algebra B does not preserve positivity.

It is clear that preserving outer measure of X implies preserving positivity of X (sincebeing null is absolute for Borel codes, and the outer measure of X is a null set i ff X isnull).

Preserving the outer measure of V is equivalent to preserving Borel outer measure:Let A be a Borel set in V . Then in V [G], the outer measure of X 2 V is the disjointunion of the outer measure of X AV [G] = AV and the outer measure of X \ AV [G] =(2 \ A)V . So if the outer measureof A decreases, then the outer measure of V decreases.

So another way to characterize Borel outer measure preserving is:No positive Borel set disjoint to V is added.

If in every forcing extension of V the set of old reals 2 V has either outer measure0 or 1 then clearly preservation of positivity of V implies preservation of Borel outermeasure. Note that this is the case (for any P ) if I is either the Lebesgue null or themeager ideal.

Other than for outer measure, positivity preservation of V and of all Borel sets is notequivalent. A trivial counterexample is the following:Set B0 { x 2 : x(0) = 0}, B1 2 \ B0. Let Q I add a

I 2 such that either

I B0 and

I is random or

I B1 and

I is Cohen. Q I

I B iff Q I

I B B0 and

Q I I B B1 , i.e. iff B B0 is Lebesgue null and B B1 is meager. In particular, B0


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and B1 are positive Borel sets. So C forces that BV 0 is null and that BV 1 remains positive.

Therefore C preserves positivity of V , but not of Borel sets.However, preservation of positivity of V does imply Borel positivity preservation if additional requirements are satised, for example once again if we know that the outermeasure of V in V [G P ] is either 0 or 1. Another su ffi cient condition is the following(which also is satised in case that I is Lebesgue null or meager, for any P ):

L 3.2. Assume that P preserves positivity of V , and thatfor every A, B I +BC there is

an A I +BC and a Borel (denition of a) function f : A Bsuch that

A A and P forces that for all null sets X B, f 1 ( X ) is null.Then P preserves positivity of Borel sets.

P . (from [13]) Assume that G P is P -generic over V and that in V [G P ], BV is null.In V , let X be a maximal family of positive Borel sets such that for every A X thereis a f A : A B as in the assumption and such that for A A X , A A I . X is countable and its union is 2 (modulo I ). In V [G P ], A V f

1 A ( B V ) is null for

each A X . So 2 V = A X ( A V ) is null.Borel positivity (or outer measure) preserving generally (consistently) does not imply

positivity preserving, not even for Cohen or random.The standard counterexample is the following:

E 3.3. Assume I is the Lebesgue null ideal and R is B 1 .(Or I is the meager ideal and R is C 1 .)Let G R be R-generic over V . Then in V [G R], X V 2 is positive and there is a cccforcing P that preserves Borel outer measure but destroys the positivity of X .

P . We assume that I is meager (the Lebesgue case is analog). Note that in bothcases, it is enough to show that P preserves positivity of (2 )V [G R ] (this implies preser-vation of Borel outer measure).Assume r is B -generic over V , and (c i)i 1 is C

V [r ]1 -generic over V [r ]. Then ( ci)i 1 is

C V 1 -generic over V as well. So B C 1 can be factored as C 1

P , where P is a C 1 -namefor a ccc forcing. Set V V [(ci)i 1 ], and let G P be the corresponding P -generic lterover V .In V , X = V 2 is not meager, but in V [G P ] = V [r ][(ci)i 1 ] it clearly is (cf. 2.6 and2.7). So in V , P does not preserve positivity.On the other hand, in V [G P ] = V [r ][(ci)i 1 ] the set Y {ci : i 1} V is a Luzinset, in particular non-meager (cf. 2.7). So in V , P preserves positivity of (2 )V .

However, if P is (absolutely) Borel positivity preserving and nep (for example Suslin

proper), then positivity preserving does follow, see theorem 6.1.Note that in any case, preservation of positivity (or outer measure) is trivially pre-served by composition of forcings (or equivalently: in successor steps of iterations). Inthis paper we investigate what happens at limit stages.

We will restrict ourselves to countable support iterations. Note that for example fornite support iterations, in all limit steps of countable conalities Cohen reals are added,so preservation of Lebesgue positivity is never preserved in nite support iterations.

Preservation of positivity is connected to preservation of generics (e.g. random reals)over models:


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L 3.4. If P is proper and X is positive, then the following are equivalent:

1. P preserves the positivity of X .2. for all N H ( ) and p P N there is an X and an N -generic q p forcing

that is I -generic over N [G].3. for all p P there are unbounded (in 2 ) many N H ( ) containing p such that

there is an X and an N -generic q p forcing that is I -generic over N [G ].

Here, A { N H ( )} is called unbounded in 2 , if for every x 2 there is a N Asuch that x N (or equivalently, if for all y 2 countable there is an N A such that y N ).

P . (1) (2): Assume that N H ( ), that q0 p is N -generic, and that G P isP -generic over V and contains q0. In V [G P ], Gen( N [G P ]) is a measure 1 set, and X ispositive, so Gen( N [G P ]) X is nonempty. This is forced by some q q0 in G P .

(2)

(3) is clear.(3) (1): Assume p forces that X is null, i.e. that X

AV [GP ] for some Borel nullcode

A. According to (3), there is an N H ( ) containing p and

A, and there are X

and an N -generic q p forcing that is I -generic over N [G ].If G P is P -generic over V and contains q, then G P is P -generic over N as well, and

A[G P ]

is a Borel null code in N [G P ]. In V [G P ], is I -generic over N [G P ], so

A[G P ]V [GP ]

X , a contradiction.

L 3.5. If P is proper, then the following are equivalent:

1. P preserves positivity.2. For all N H ( ), there is a set A of measure 1 such that for all p N and A

there is an N -generic q p forcing that is I -generic over N [G ].3. For all p there are unbounded (in 2 ) many N H ( ) containing p such that for

some measure 1 set A and for all A there is an N -generic q p forcing that is I -generic over N [G].

P . (1) (2): Since there are only countable many ps in N , it is enough to showthat for all N H ( ) and all p P N there is a set A as in (2). Let X be the set of exceptions, i.e. X iff every N -generic q p forces that is not I -generic over N [G ].We have to show that X is a null set. Otherwise (according to lemma 3.4) there is an X and an N -generic q p forcing that Gen( N [G]), a contradiction.

(2) (3) is clear, and (3) (1) follows from lemma 3.4.

Why are we interested in preservation of I -generics over models instead of preserva-tion of positivity? It is not clear how the iterability of preservation of positivity can beshown directly. On the other hand, in some important cases it turns out that preservation

of generics is iterable (e.g. if I is meager, see section 8, or if I is Lebesgue null underadditional assumptions, see section 9). However, to be able to apply the according it-eration theorems, we will generally need that all I -generics are preserved, not just ameasure 1 set of them (as in lemma 3.5).

It seems that preservation of all I -generics really is necessary, more specically thatthe statement

preservation of Lebesgue positivity is preserved in proper countable support iterations(and the analog statement for meager) is (consistently) false. A counterexample seemsto be diffi cult, but we can give a counterexample to the following (stronger) statement:


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Let X be a positive set. Then

preservation of positivity of X is preserved under proper countable support iterations

4. A Counterexample.

E 4.1. Assume that I is the Lebesgue null ideal and that R is B 1 .(Or I is the meager ideal and R is C 1 .)Then R forces the following: There is a positive set X and a forcing iteration ( P n ,

Qn)n


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Generally, we dene

P n B J 0 C B J n 1 C B J n .

We dene the P n-namer n 2

J to be the concatenation of the random sequences for allthe random algebras used in P n .

L 4.2. There is a complete embedding f n,n+ 1 from P n to P n+ 1 which leaves therandom sequence invariant. 5

Assuming this lemma, the rest of the proof is straightforward:As usual, we interpret

r J n as a sequence of ( 1 many) random reals. Let

X n be the

set of these reals. Set

X n X n ;

X < n

X 1

X 2

X n 1 ; and

X n

X \

X


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is compatible with [G P ] = f 0

[G Q]. Now pick a q q Q forcing that [G Q ]

SV [GQ ]

is a witness for this compatibility. Then ( q , ) f 1(( p, )), (q, ), and we get acontradiction.

P 4.2. First note that B J n is equivalent to B J n B J n+ 1 . So we have tond a complete embedding

from P n =P

B J 0 C C B J n 1 C B J n B J n+ 1to P n+ 1 = B J 0 C C B J n 1 C B J n C

QB J n+ 1

It is clear that the identity (lets call it f 0 ) is a complete embedding between P and Q (thetwo blocks marked above). (Generally, for all R,

S the identity is a complete embedding

from R into RS .) Therefore we can apply lemma 4.3 to get a complete embedding

f 1 : P n P n+ 1 . It is clear that f 1 leaves the random J -sequence invariant.

5. True Preservation. Preservation of all generics (not just a measure-1-set of them) is closely related to preserving true positivity, a notion using the stationaryideal on [ ] 0 .

From this section only denition 5.9 is needed for the proof of main result 9.4. 8

D 5.1. Let I be arbitrary and C [I ] 0 a family of countable subsets of I . C is called unbounded, if for all A [I ] 0 there is a B C such that B A. C is a club set (or: club), if C is unbounded and closed under increasing countable

unions. The club lter is the family of subsets of [ I ] 0 containing a club set. A set S [I ] 0 is stationary if every club set C meets S , i.e. C S .

(Or equivalently, if the complement of S , [ I ] 0 \ S , is not in the club lter.)

First we recall some basic facts:

L 5.2. Let I and H 1 H 2 be arbitrary.1. (Jech) The club lter on [ I ] 0 is closed under countable intersections.2. (Menas) C [I ] 0 contains a club i ff there is an f : [I ]2 [I ] 0 such that

C ( f ) C , where C ( f ) { x [I ] 0 : ( i j x) f ({i, j}) x}.3. If C [H 1] 0 is club, then C

H 2 { B [H 2] 0 : B H 1 C } is club.4. If C [H 2] 0 is club, then C

H 1 { B H 1 : B C } contains a club.5. If C 0 [I ] 0 is club and P an arbitrary forcing, then there is a C 1 C 0 club and

a nameC such that P forces that

C [I ] 0 is club and that

C V = C 1 .

6. (Shelah) A forcing P is proper i ff for arbitrary I and S [I ] 0 stationary, P forcesthat S remains stationary.

7. The set of countable elementary submodels of H ( ) contains a club of [ H ( )] 0 .8. Assume [ I ] 0 H ( ). Then the following are equivalent:

C [I ] 0 contains a club. For all N H ( ) containing I and C , N I C . For club many N H ( ), N I C .

8 This proof only needs implication (3) (1) of lemma 9.3, i.e. the fact that a strongly preserving forcingis tools-preserving. However we do use the notion of true preservation to show implication (3) (1) of 5.11,which in turn is used in the proof that Lebesgue-tools-preservation is equivalent to strong preservation.


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9. Assume [ I ] 0 H ( ). Then the following are equivalent:

S [I ]0

is stationary. There is an N H ( ) containing I and S such that N I S . For stationary many N H ( ), N I S .

Note that if C is club in V , then generally C will not be club any more in a forcingextension V [G P ], even if P is proper.

P . We refer to Kanamoris Higher Innite [6] or Jechs Millennium Edition [5].The proof of (1) is straightforward (see 25.2 or 8.22). (2) is proven in [6, 25.3] or[5, 8.26]. (3) is trivial. (4) and (5) follow from (2) (for the latter, set C 1 C ( f )V and

C C ( f )V [GP ]).For (6) see e.g. Proper and improper forcing [12].For (7), consider the family of countable subsets of H ( ) closed under some xedSkolem function.

(8): If C N is club, then clearly I N C . If C { N H ( )} [ H ( )] 0 is club,then C I = { N I : N C } [I ] 0 contains a club according to (4).(9) follows directly from(8), since S is stationary i ff [I ] 0 \ S does not contain a club.Assume that I is an arbitrary index-set, S [I ] 0 is stationary and = (s : s S ) is asequence of reals. Pick any H I 2 (think of H to be a H ( )). For C [H ] 0 , wedene 9

S (C ) {s S : N C : N I = s & s Gen( N )}, and

(C ) {s : s S (C )}.

So we get S (C ) the following way: Take an N C (which will be a countable elemen-tary submodel of H ( )), and let s be the intersection of N with I (so s is a countablesubset of I ). If s is an element of S , and if s is I -generic over N , then put s into S (C ).

D 5.3. Assume H I 2 . is truly positive, if (C ) is positive for every club set C [H ]0 . B is the true outer measure of , if it is the smallest Borel set containing any of the

(C ), i.e. if the following holds: B is Borel, there is a C [H ] 0 club such that(C ) B, and for no club C [H ] 0 there is a Borel B such that (C ) B and

B \ B I .

L 5.4. 1. The above notions do not dependon H (provided that H I 2 ).2. The true outer measure always exists.3. The following are equivalent:

is truly positive. (C ) for every club set C [H ] 0 . for all x H ( ) there is an N H ( ) containing x, I , S and such that

N I = s S and s Gen( N ).

P . (1) Assume that I 2 H 1 H 2 and that C [H 1] 0 is club.By denition s S (C ) iff for some N C , s = N I S and s Gen( N ).In particular s S (C H 2 ) iff for some N [H 2] 0 , N N H 1 is in C ,

s = N I is in S and s Gen( N ).So since N and N contain the same elements of I and 2 , S (C ) = S (C H 2 ).

9 Gen( N ) was dened in 2.2.


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The same argument works with C [H 2] 0 and C H 1 .

For general H 1 , H 2 , apply the argument to the pairs H 1 , H 1 H 2 and H 2 , H 1 H 2 .(2) The family { (C ) : C club } is semi-closed under countable intersections:If (C i)i is a countable sequence of club sets, and C C i its intersection, then C is club, and (C ) (C i).Let X be the family of Borel sets B such that for some club set C , B (C ). So X isclosed under countable intersections. Therefore X contains a minimal element (modulo I ), since I is a ccc-ideal.(3) Without loss of generality H = H ( ). Assume is not truly positive. Then for someclub set C and Borel null set B, (C ) B. Set

C { N H ( ) : N C , B N }.

C is club. For any N C and any I -generic over N , is not in B. So (C ) 2 \ B.But (C ) (C ) B, so (C ) = . The rest is similar.

D 5.5. Let P be a forcing. P preserves true positivity if for all truly positive, P forces that remains truly

positive. P preserves true outer measure, if for any with true outer measure AV , P forces

that AV [G P ] remains the true outer measure of .

L 5.6. 1. If P is true outer measure preserving, then it is true positivity pre-serving.

2. If P is true positivity preserving, then it is proper and positivity preserving.3. If P is true outer measure preserving, then it is outer measure preserving.

It seems that true positivity preserving generally does not imply true outer measure

preserving. (But the equivalence holds if I is the ideal of meager sets, see lemma 8.1;or if I is the ideal of Lebesgue null sets and P is weakly homogeneous, see lemma 9.1).

P . (1) is clear since a sequence is truly positive i ff its true outer measure is not0.True positivity preservation implies properness because of 5.2(6).So for (2) and (3) it is enough to show the following: If X is positive (or: has true outermeasure B) then there is a truly positive (or: an with true outer measure B) such that{s : s S } X . Let I be 2 .For (2), pick for each N H ( ) an X Gen( N ). (Recall that Gen( N ) is a measure1 set.) Then is truly nonempty (cf. 5.4(3)).For (3), set 2 0 . As cited in Kanamori [6, 25.6(a)] or Jech [5, 38.10(i)], [ I ] 0 can bepartitioned into 2 0 many stationary sets, i.e. [ I ] 0 = S . Enumerate all positive

Borel subsets of B as ( B : ). For each N H ( ) let be such that N S andpick an B Gen( N ). Assume towards a contradiction that the true outer measureof is B B and that B = B \ B is positive. Since C is club and S stationary, thereis an N C S . So N B (C ), a contradiction.

As announced, the true notions are closely related to preservation of generics:

D 5.7. P preserves generics, if for all N H ( ), p N and Gen( N )there is a q p N -generic forcing that Gen( N [G P ]).


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N . Instead of for all N , we can equivalently say for club many N . (This

follows from the proof of lemma 5.8.) Of course the notion does not depend on , provided is regular and large enough(in relation to |P |).

It is clear that preservation of generics is preserved under composition and impliesproperness.

L 5.8. P preserves generics i ff P is true positivity preserving.

P . : Assume otherwise, i.e. assume that is truly positive, and p (C ) =

for some nameC of a club set in H ( )V [G] .

In V , the set

S { N H ( ) : N I = s S, s Gen( N )}

is stationary. (Otherwise, the complement of S would witness that is truly empty.)Pick some . According to 5.2(9), there is an N H ( ) containing , S , , p, P, C such that N N H ( ) S (and such that P preserves generics for N , if we assumepreservation for club many N only). So N I = N I = s S , and there is an N -generic q p forcing that s Gen( N [G]) (since N and N contain the same subsets of P ). Let G be a P -generic lter over V containing q. In V [G], N [G] I = N [G] I = s(since G is N [G]-generic), and N

H [G]

C [G ] (since

C N [G] is club). So

s (C ), a contradiction.

: Assume towards a contradiction that N H ( ), p, is a counterexample.Without loss of generality there is a N such that |P | . Set

S { N H ( ) : N is counterexample for p and some }.

This set is stationary, since S N and N H ( ) S .For each N S , pick an N witnessing the counterexample. Then is truly positive:

If N C S , then N (C ).Let G be a P -generic lter over V containing p. In V [G], set

C gen { N H V ( ) : G is N -generic }.

(Note that the elements of C gen are generally not in V , only subsets of V .)C gen contains a club: N H V ( ) is guaranteed if N is closed under a Skolem functions of H V ( ).G is N -generic means that for every dense subset D P in N , G N D is nonempty.So C gen contains the set of N closed under countably many operations.Therefore (still in V [G])

C 1 C H V [G P ] ( )gen = { N H

V [GP ]( ) countable : N V C gen }

contains a club as well, as does the setC { N H V [GP ]( ) : G N and N C 1}.

By the assumption (C ) , i.e. for some N H ( ), we get: N N V S (note thatS V ), and N Gen( N ), and G is N -generic and element of N . Therefore N [G ] N ,and N Gen( N [G]). G contains some q p forcing this all. But we assumed that N is a counterexample, therefore no N -generic q p can force that N Gen( N ).

For the analog equivalence to true outer measure preservation we need the notion of interpretation:


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D 5.9. Let p be a condition in P .

T is an interpretation of T with respect to p, if: T is a positive Borel set,

T a P -name for a positive Borel set, and

for all positive Borel sets AV T , p does not force that AV [G] T is null.

P strongly preserves generics if the following holds:For all N H ( ), p, T ,

T N and 2 such that

T is an interpretation of T with respect to p and

T Gen( N )there is an N -generic q p forcing that

T Gen( N [G P ]).

N . If T is an interpretation of T with respect to p, p

T

T , and

T T is positive, then then T is an interpretation of T .

Again, instead of for all N , we can equivalently say for club many N , and thenotion does not depend on .

L 5.10. For every p P and every nameT for a positive Borel set there is an

interpretation T of T with respect to p.

P . Set

X { B I +BC : p BV [G]

T I }.

Let Y be a maximal family of pairwise disjoint members of X . Then Y is countable(since I is a ccc ideal) and Y is not of measure 1 (since p forces that

T is positive

and that Y V [G] T I ). Set T \ Y . Then T is an interpretation of

T with

respect to p.

L 5.11. The following are equivalent:1. P preserves true outer measure.2. P strongly preserves generics.3. If p

T I +BC , then there is a T I

+

BC and a p p such that:T is an interpretation of

T with respect to p , and

if N H ( ), p , T ,T N , and T Gen( N ),

then there is an N -generic q p forcing that T Gen( N [G]).

P . This is similar to the proof of 5.8.(1) (2) Assume that N H ( ), p, T ,

T , is a counterexample. Without loss of

generality there is a N such that |P | . Set

S { N H ( ) : N is counterexample for p, T ,T and some }

This set is stationary, since S N and N H ( ) S .For each N S , pick an N witnessing the counterexample. So in particular

N T Gen( N ).

Let B T be a true outer measure of . B is positive (which just means that is trulypositive). So there is a p p forcing that B

T is positive (since T is an interpretation

of T with respect to p).Let G be a P -generic lter over V containing p . In V [G ], set

C { N H V [G P ]( ) : G, p, T ,T N and G is N V -generic }.


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B

B

T

T

(C )

(C )

F 1. T is an interpretation of T with respect to p .

C contains a club (as in the proof of 5.8). Assume N (C ). Then for some N C N N H V ( ) S , G is N -generic, and N is I -generic over N .

Note that N [G] N (since G N ). So N is I -generic over N [G] as well. Since N Sand N , p, T , T , N is a counterexample, we know that N cannot be in

T [G]. Therefore

(C ) B \T [G], i.e. the true outer measure of decreases ( p forces that B

T is

positive).(2) (3) This follows from fact 5.10 (for every p,

T there is an interpretation T of

T with respect to p).

(3) (1) Assume that B (C ) is an outer measure of , and p P forces thatthere is a Borel code

B and a club set

C of H ( )V [G]

such that (C ) (

C ),

B

B,

T B \

B I + and

B (

C ).

Without loss of generality, s B for every s S . Now choose a p p and aninterpretation T of

T with respect to p according to (3). Without loss of generality

T B (cf. gure 1). In V , the set

S { N H ( ) : p , P, T ,T N , N I = s S ,

sGen( N ) T }

is stationary. (Otherwise, the complement of S contains a club C . If (C ), then T . So (C ) B \ T , and B cannot be true outer measure of .)

Pick some . There is an N H ( ) containing , S , p , T ,T ,

C etc such that

N N H ( ) S . So N I = s S , and there is an N -generic q p forcingthat s Gen( N [G])

T . Let G be a P -generic lter over V containing q. In V [G],

N [G ] I = N I = s (since G is N -generic), and N H ( )V [G]C [G] (since

C [G] N [G] is club). So s

T [G] (

C [G]), a contradiction to (

C [G])

B[G ]

and

B[G] T [G] = .

6. StrongPreservationof Generics fornep Forcings. We already know that preser-vation of Borel outer measure generally does not imply preservation of positivity. An

example was presented in 3.3. Note that the P of this example is very undenable.In this section we will show that under some additional assumptions on P , we even getstrong preservation:

T 6.1. If in all forcing extensions of V , P is nep and Borel outer measurepreserving then P strongly preserves generics.

In particular this requirement will be satised if there is a proof in ZFC that (a def-inition of the forcing) P is nep and Borel outer measure preserving. We formulate thisas a corollary:


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C 6.2. If P is provably nep and provably Borel outer measure preserving

then P strongly preserves generics.The notion of nep forcing (a generalization of Suslin proper) was introduced in [13].

The most important instances of nep forcings are

1. nicely denable, proper (but not necessarily Suslin proper) forcings P that arecoded as set of reals. We will call such forcings transitive nep.

2. nicely denable (long) iterations of forcings as in (1).

Examples for transitive nep forcings are Sacks and Laver. Such forcings typically areSuslin + . In this paper, we will not dene the general notion of nep which would allowus to deal with (2) since the denition requires technicalities such as non-transitivecandidates. 10 Instead, in the following section we will recall the denition of transitivenep. 11

About transitive nep forcing. In the denition of nep (or Suslin proper) we usecandidates, i.e. models of some xed -theory ZFC . Intuitively, we would like to useZFC (just as we would like to use N V in the denition of proper forcing), butfor obvious technical reasons this is not possible. So we will restrict ourselves to areasonable choice of ZFC :

D 6.3. A recursive theory ZFC ZFC is called strongly normal, if thefollowing is provable in ZFC:

H ( ) ZFC for all su ffi ciently large regular .

You can think of ZFC as ZFC minus the power set axiom plus something like exists. 12

We assume that the forcing P is dened by formulas ( x) and ( x, y), using a realparameter r P . Fixing a strongly normal ZFC , we call M a candidate if it is a countabletransitive ZFC model and r P M . So in any candidate ( P M , M ) is dened (but thisforcing is generally not equal to P M , since the denitions do not have to be absolute).It is important that the transitive collapse of an N H ( ) (containing r P ) is a candidate(for su ffi ciently large regular ).

If M is a candidate, then G is P -generic over M if for all A M such that

M A P is a maximal antichain ,

|G A| = 1. (Note that in this case it is not enough that G meets all dense sets, sinceincompatibility is generally not absolute).q is called M -generic is q forces that G is P -generic over M .

10 Note that including forcings of type (2) is not needed for the main result 9.4 of this paper: We will showthat forcings of type (1) satisfy a strong, iterable condition, therefore this condition is satised by forcings of type (2) anyway.

11 Usually transitive nep forcings are in fact Suslin + . Nevertheless we dene transitive nep here insteadof Suslin + since the denition is actually simpler and better isolates the property needed for the proof. Also,there are examples of transitive nep forcings P that are not Suslin + , to be more exact: whose natural denitionsare not Suslin + , e.g. because p P is

11 and not

11 . (It is a diff erent question whether for these examplesthere are equivalent forcings P that do have (possibly less natural) Suslin + denitions.)

12 For the usual transitive nep forcings we could actually x this ZFC . Generally however we should for technical reasons not do that, just as we should not x e.g. H ( ) = H ( + ) in the denition of properforcing.


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D 6.4. A (denition of a) forcing P is transitive nep (with respect to ZFC ),

if p P and q p are upwards absolute between candidates and between

candidates and V as well. 13

P H ( 1) (in V and all candidates), and p P and q p are absolutebetween the universe and H ( ) (for large regular ).

If M is a candidate and p P M then there is an M -generic q p.

So transitive nep is a direct generalization of Suslin proper. Since (transitive collapsesof) elementary submodels (containing r P ) are candidates, every transitive nep forcingis proper. There are popular forcings that are transitive nep and not Suslin proper,for example Laver, Miller or Sacks (all these forcings are Suslin + and even satisfy aneff ective version of Axiom A, see [7] for a proof).

If P is nep and M a candidate, then

M p ( ) iff M [G] (

[G ]) for every lter G containing p which is P -generic over M and V .14

When we say P is nep we mean(a suitable denition of) P is nep with respect to some strongly normal ZFC .

In practice the choice of ZFC is immaterial (similar to the choice of in the denitionof proper forcing). If you believe this you can skip the following explanation, continueat the proof of theorem 6.1, let S ( ) be some regular and ignore the argumentswhy certain models are in fact candidates.

Why do we use strongly normal here, and not just normal, i.e. H ( ) ZFC for large regular ?

Normal would denitely be enough to imply proper. The point in using strongly normalis that we can assume without loss of generality that not only the candidate M satisesZFC , but also e.g. all forcing extensions M [G] (for forcings that are small in M ). (Thisis of course not possible with a ZFC that is just normal. For example if V = L thenthere is a normal ZFC containing V = L, but ZFC fails to be normal in any nontrivialforcing extension.) 15

Lets explain that in more detail:First note that we are dealing with two forcings, Q I and P . Q I is Suslin ccc. This

implies that Q is Suslin proper (and therefore transitive nep) with respect to any ZFCthat contains a certain strongly normal sentence .16 Each two strongly normal theoriesare compatible (i.e. the union is strongly normal as well), and a forcing remains nep if we strengthen ZFC (since then there are fewer candidates). So we can assume withoutloss of generality that Q and P are nep with respect to the same ZFC .

13 I.e. if M 2 , M 1 are candidates, M 2 M 1 is a candidate, and M 1 p P , then M 2 p P andV p P , and the same for .

14 More formally this reads: p forces: If G is M -generic, then M [G ] ( [G ]).

15 We can still formulate the theorem for forcings that are nep with respect to not necessarily stronglynormal theories, but then we have to use two theories ZFC and ZFC and have to assume something likethe following:A forcing extension of a ZFC -candidate is a ZFC -candidate; P is nep with respect to ZFC ; ZFC impliesthat every small forcing R forces that P is nep with respect to ZFC .So the formulation of the theorem gets messy, while there is no gain in practice, where ZFC is stronglynormal anyway.

16 is the completeness theorem for Keisler logic. This follows from the proof that Suslin ccc impliesSuslin proper in [4], see [7] for a discussion.


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If 2 R H ( ) then R H V ( )[G] = H V [G]( ) and r ( H V [G] ( ) ) iff H ( ) (r ).

So if P is nep (or Borel outer measure preserving) in V , then it is nep (or preserving) in H ( ) (for su ffi ciently large regular ).Let R be any forcing notion. We assumed that R forces that P is nep and preserving.Also ZFC is strongly normal, so R forces that H V [G R ]( ) satises

ZFC plus P is nep and Borel outer measure preserving( )

for every regular

R. Clearly we can nd a R such that R forces that

R < R, andwe can do that for all R H ( ). So for all there is a regular S ( ) such that

H ( ) thinks that R forces ( ) for all regular S ( ) and R H ( ).( 2)

Proof of theorem 6.1. The proof is very similar to the proof of preserving a littleimplies preserving much in [13] (or its version in [7]). The point of the proof is thatwe mix internal forcing extensions (i.e. by M -generic lters in V ) for Q I and R (acollapse) with external forcing extensions for P , and use absoluteness to compare whatP forces in the di ff erent internal models.

We will prove the theorem for the case that P is transitive nep. If you know thegeneral denition of nep, you will see that the same proof works for general nep aswell. 17

Recall that the Suslin ccc ideal I was dened by a Suslin ccc forcing Q I and a genericreal

I (see 2.2).

We have to show the following: if T is an interpretation of T with respect to p,

then for all (or just: conally many) N H ( ) containing p, T ,T and for all

Gen( N ) T the following holds:

there is an N -generic q p forcing that T Gen( N [G P ]).( )

P is nep with respect to a strongly normal ZFC . Set 0 1 , 1 S ( 0), 2S ( 1), and 3 S ( 2).There are conally many such N 0 H ( 3) containing P, p, T ,

T . We x such an

N 0 . So it is enough to show ( ) for N 0 . Let i : N 0 M 0 be the transitive collapse.For i { 1, 2}, set i i( i ), and set

T i(T ). Note that i doesnt change p, T or

I , since these objects are hereditarily countable ( T is a Borel code, i.e. a real number).Not surprisingly, ( ) for N 0 is equivalent to the following:

there is an M 0 -generic q p forcing that T Gen( M 0[G P ]).( 2)

This is straightforward: A lter G is M 0-generic i ff it is N 0 -generic (since i doesntchange the elements of P ). Also, the evaluation of a name

s of a real number is absolute:

If G is M 0-generic, thens[G ] = i(

s)[G]. To see this, pick in N 0 maximal antichains An

deciding

s(n). Fix n. If G is N 0-generic, then G chooses an element of p An N 0forcing that s[G ](n) = m. M 0 thinks that pn forces that i( s)(n) = m, so i( s)[G](n) = m.So in particular

T [G] =

T [G ], and N 0 [G ] and M 0 [G] see the same Borel null sets, i.e.

Gen( N 0[G]) = Gen( M 0[G ]).From now on for the rest of the proof we x the above M 0 , p, T,

T .

We formulate ( 2) as a property of :

17 You just have to use the ord-collapse instead of the transitive collapse, and keep in mind that the evalu-ation of names

[G ] has to be redened for non-transitive candidates. And you have to formulate awkward

requirements on ZFC if you allow ZFC to be a ( , P )-theory.


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D 6.5. Let M be a candidate containing p, T andT . is absolutely generic

over M , or: Genabs

( M ), if T and there is an M -generic q P p forcing thatT Gen( M [G P ]).

(Note that the denition of absolutely generic does not only depend on M , but on p,T and

T as well. However, these parameters are xed.)

Using this notion, ( 2) reads as follows:Gen( M 0) T = Gen abs ( M 0).

For any candidate M , Gen( M ) is a measure 1 set. So if Gen( M ) T = Gen abs ( M ) thenGen abs ( M ) is a measure 1 set in T . As the rst step in our proof we show that Gen abs ( M )is at least nonempty:

L 6.6. Assume that P is Borel outer measure preserving, M is a candidate, and M thinks that

T is an interpretation of T with respect to p, and

A is a positive Borel (code for a) subset of T .Then Gen abs ( M ) A I + .

P . In M , A is an interpretation of T with respect to p, since A T is positive.

So without loss of generality A = T , i.e. we just have to show that Gen abs ( M ) is positive.Pick (in M ) a p p forcing that T

T is positive. Let q p be M -generic, and

G a P -generic lter over V containing q. So in M [G] (and therefore by absoluteness inV [G]) T V [G]

T [G] is positive. T V [G] is the outer measure of T V (since P preserves

outer measure). So T V T [G] is positive. Also, Gen( M [G ]) is of measure 1. Therefore

X Gen( M [G]) T V T [G ]

is positive in V [G]. Clearly X Gen abs ( M )V . So in V , Gen abs ( M ) has to be positive.Let (in M 0 , for i { 1, 2}) Ri be the collapse of H M 0 ( i) to 0 , i.e. the set of nite

partial functions from to H M 0 ( i). Fix an Gen( M 0) T . We have to show that Gen abs ( M 0). Gen( M 0) means that there is (in V ) a Q I -generic lter G Q over M 0 such that

I [G Q ] = . Pick (again in V ) an R2-generic lter G R2 over M 0[G Q ]. Set

M M 0[G Q ][G R2 ]. So we get (in V ) the following forcing extensions:

M 0 M 0 [G Q ] M M 0 [G Q ][G R2 ]

M sees all relevant information about H 1 H ( 1) M 0 (in particular M knows that H 1is a candidate). So it is enough to show that M thinks that is absolutely generic for H 1 :

L 6.7. M Gen abs ( H 1)

If we assume this, theorem 6.1 follows immediately: Q I R2 H M 0 ( 2). M 0 is the

transitive collapse of N 0 H (S ( 2)), so according to ( 2) M is a candidate and P istransitive nep in M . Clearly M knows that H 1 is a candidate. So according to thelemma there is a p M such that

M p p is H 1-generic, and p P T Gen( H 1[G P ]) .

Let (in V ) q P p be M -generic. We claim that q witnesses ( 2), i.e. thatq is M 0-generic and forces that Gen( M 0[G P ])

T .

So let G P be a P -generic lter over V containing q. Then G P is M -generic. M [G P ] thinks that G P (i.e. G P P M ) is H 1-generic, since this is forced by p G P .


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M 0 R1 / /

Q

GQ c c

c c c c c c

c c M 1 Gen abs ( H 1) B

M 1q

R

G2

c c c

c c c c

c c c c c

M 0 [ ]

R1 / Q

G1

? ? M 2 Gen abs ( H 1)

R2

G1 G2 / /

F 2. The models used in the proof of lemma 6.7.

Being H 1-generic is absolute (it just says that |G P A| = 1 for all maximal antichainsin H 1). So G P really is H 1 -generic. Analogously is I -generic over H 1[G P ].Recall that H 1 = H ( 1) M 0 , the denition of P is absolute between the universe and H ( ) and 1 is suffi ciently large. Therefore P H 1 = P M 0 and H 1 contains exactly the

same subsets of P as M 0 does. So G P is M 0 -generic as well. Also, H 1[G P ] containsexactly the same reals (in particular Borel codes) as M 0 [G P ]. (This can again be seenby deciding

r (n) by a maximal antichain.) Therefore is I -generic over M 0 [G P ]. This

is all forced by some q p, so we are nished.

P 6.7. We already know that is I -generic over M 0 . Using the factsthat H 1 and M 0 see the same subsets of P and that H 1 is countable in M we get M Gen( H 1) T .

Assume towards a contradiction that M Gen abs ( H 1). Since M = M 0 [G Q ][G R]this is forced by some q G Q and r R2 . However since R2 is homogeneous and thesentence Gen abs ( H 1) only contains parameters in M 0[G Q ] we can assume thatr = 1, i.e.

M 0 q Q ( I T , R

2 I Gen( H 1) \ Gen abs ( H 1)) .( )

Fix a positive Borel code Bq M 0 such that

{ I [G] : G V is an M 0 -generic lter containing q} = Gen( M 0) BV q

(see 2.3). Without loss of generality Bq T , since (in M 0) q I T .

Choose in V an R1-generic lter G R1 over M 0 , and set M 1 M 0[G R1 ]. M 1 knowsthat H 1 is a candidate, and that B

M 1q T is positive. M 1 knows that P preserves Borel

outer measure (because of ( 2)), and H 1 thinks that T is an interpretation of T with

respect to p (since H 1 is the collapse of H V ( 1)). So we can apply lemma 6.6 in M 1 (i.e. M 1 is the universe V and H 1 the candidate M ) and nd a Gen abs ( H 1) B

M 1q . In

particular is I -generic over H 1 and therefore over M 0 . So in M 1 there is a M 0 -genericlter G Q such that [G Q ] = B

V q . Therefore q G Q , and we can factorize R1 as

R1 = Q R1 / Q such that G R1 = G Q G1 . (See gure 2 for a diagram of the forcingextensions we are going to construct).

In M 0[G Q ] = M 0 [ ] we look at the forcing R2 = R M 02 (the nite partial functions

from to H M 0 ( 1) = H 1). 2 1 and R1 / Q is a subforcing of ro( R1), where R1 H 1 .So according to 2.10 (2) and (3), R2 can be factorized as R2 = ( R1/ Q) R . We alreadyhave the ( R1 / Q)-generic lter G 1 (over M 0[G Q ]), now choose (as always in V ) a R -generic lter G 2 over M 1 . Set G R2 = G 1 G2 . So G R2 is R2-generic over M 0[G Q ]. Set M 2 M 0 [ ][G R2 ].


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Now set H 2 H ( 2) M 1 . H 2 thinks that P is nep and that Gen abs ( H 1) (by

absoluteness in M 1). Also, H 2 is a candidate (by ( 2), since H 2 = H M 0

[G R1 ]). So forsome p1 H 2 ,

H 2 p1 p is H 1 -generic , p1 Gen( H 1[G P ]) T .

In M 2 , there is an H 2-generic p2 p1 (since M 2 thinks that P is nep and that H 2is a candidate.) Let G P be a P -generic lter over M 2 containing p2 . From now on,we work in M 2[G P ]. G P is H 2 -generic and contains p1 , so by absoluteness G P is H 1-generic as well and Gen( H 1[G P ])

T . On the other hand, according to ( ),

Gen( H 1[G P ]) T , a contradiction.

Note that if we set T =T = 2 this proof gives us preserving a little implies

preserving much of [13]:

T 6.8. If in all forcing extensions of V , P is nep and preserves Borel positiv-

ity, then P preserves generics.In particular, if P is provably nep and provably preserves Borel positivity, then P pre-serves generics.

7. A general preservation theorem. For proving the main result 9.4 we will use ageneral iteration theorem for countable support iterations of proper forcings. It appearedas Case A in Proper and improper forcing [12, XVIII, 3]. The proof there is not easilydigestible, though. A simplied version appeared in Section 5 of Goldsterns Tools [2].This version uses the additional requirement that every forcing of the iteration adds anew real. Note that this requirement is met in many applications anyway (e.g. in theforcings of [9] cited in the introduction).

A proof of the iteration theorem without this additional requirement appeared in [8]

and was copied into Set Theory of the Reals [1] (as rst preservation theorem 6.1.B),but Schlindwein pointed out a problem in this proof. 18 Another proof (building on theone in [2]) will appear in [3].

The general preservation theorem uses the following setting: Fix a sequence of in-creasing arithmetical two-place relations Rn on . Let R be the union of the Rn . As-sume

C { f : f R for some } is closed, { f : f Rn } is closed for all n , , and for every countable N there is an such that f R for all f N C

(in this case we say covers N ).

D 7.1. Let P be a forcing notion, p P . f f 1 . . . f k is a tools-interpretation of

f

f 1 , . . . ,

f k under p, if each

f i is a

P -name for an element of C, and there is an decreasing chain p = p0 p1 . . .of conditions in P such that p i (

f 1 i = f 1 i & . . . &

f k i = f k i).

A forcing notion P is tools-preserving, if for all N H ( ), covering N ,

18 See [11]. In this paper Schlindwein wrote a simple proof for the special case of -bounding, howeverhe later found a problem in his own proof [C. Schlindwein, personal commuinication, April 2005]. He ispreparing a new version [10].


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p, ni, f ,

f N such that f is a tools-interpretation of

f under p and f i Rni

there is an N -generic q p, forcing that covers N [G P ], and

f i Rn i for all i k .

Note that if f is a tools-interpretation, then f l C .Tools-interpretations di ff er from the interpretations of denition 5.9. They obviously

deal with functions from to instead of Borel sets modulo I . But there is anothertechnical di ff erence: For tools-interpretations, we require that there is a decreasing se-quence of conditions p p1 p2 . . . , not just that for all n, the truth value of ( m n)

f (m) = f (m) is positive. 19

Now we can formulate the rst preservation theorem [1, 6.1.B] already mentioned:

T 7.2. Assume ( P i,Q i)i


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P . This is proven as application 3 in [2] or as theorem 6.3.20 of Set theory of the

reals [1]. It is an easy application of tools-preservation (7.2):Let be the set of clopen sets of 2 . Set

C = { f : ( U ) f (U ) U }.

We dene f Rn by

f C and 2 and for some k n, f (U k ).

Then for any N H ( ), covers N iff is Cohen over N . Also { f : f Rn } is clopen, so

f i Rni can be forced by determining

f i m for some m. Therefore P preserves Cohens

iff P is tools-preserving. This nishes the proof of 8.2.Note that in this simple case tools-preservation isnt really needed. It is enough to

trivially modify the proof that a countable support iteration of proper forcings is proper(see e.g. [1, 6.1.3]). In the following we point out the changes that have to be made to

this specic proof:The lemma now reads:

Suppose ( P ,Q )


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L 9.1. If P is weakly homogeneous 21 and preserves (true) positivity then P pre-

serves (true) outer measure.P . For the untrue version, this is [1, Lem 6.3.10]. The same proof works for

true outer measure as well: Assume that B is a true outer measure of , that Leb( B) = r 1and that p forces that

B (

C ) and Leb(

B ) < r 2 < r 1 , r 2 rational. We have to show

that there is a truly positive that fails to be truly positive after forcing with P .So p forces that there is a sequence

I n of clopen sets such that

I n (

C ) and

Leb( I n) < r 2 . Let pn , h(n), I n be such that for all m h(n),

pn Leb(m> h(n)

I m) < 1/ n & ( k < m) I k = I k .

So Leb( I m) r 2 , and B \ I m is not null. Therefore

S {s S : s I m}

is is stationary (otherwise, the complement of S would witness that B is not the trueouter measure or ). Dene S . So is truly positive.

pn Leb( I m \ I m) < 1/ n, and

pn (C )

I m \ I m, i.e.

pn Leb ( (C )) < 1/ n. So

pn T-Leb ( ) 1/ n.

Since the last statement does not contain any names except standard-names, and since Pis weakly homogeneous, we get 1 P T-Leb ( ) 1/ n for all n, i.e. T-Leb ( ) = 0.So the truly positive becomes null after forcing with P .

Now we are going to show that strong preservation is equivalent to the Lebesgueversion of tools-preservation (see denition 7.1).

We list the clopen subsets of 2 as ( I i)i , and interpret a function f as asequence of clopen sets. We set

C { f : i Leb( I f (i)) < 2 i}, and

f Rn iff f C , 2 , and for all l > n, I f (l).

For f C , the set N f n i> n I f (i) is a null set. In fact, every null set is containedin a N f for some f C (see e.g. [1, 2.3.10] or [2]). f R just means N f .

is random over a model N iff is not element of any null set coded by a real in N .So is random over N iff covers N .

For reference we baptize this version of tools-preservation:

D 9.2. P is called Lebesgue-tools-preserving if it is tools-preserving for the

Rn dened as above.It is clear that Lebesgue-tools-preserving implies preservation of generics and there-

fore preservation of positivity. Lebesgue-tools-preservation is preservation of genericsplus side functions. It turns out that this is equivalent to strong preservation:

L 9.3. The following are equivalent:1. P is Lebesgue-tools-preserving.

21 So if only contains standard-names, then ( p P ) implies (1 P P ).


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2. P is Lebesgue-tools-preserving for k = 1 and n1 = 0.

3. P strongly preserves randoms.P . (2) (1): Assume N H ( ), p, , f 1 , . . . , f k ,

f 1 , . . . ,

f k and n1, . . . , nk are

as in the denition of Lebesgue-tools-preserving.Set n max( k , n1, . . . , nk ). pn p forces that f i n =

f i n . Let g be such

that I g (m) = i= 1...k I f i (n + m) , andg the name of a function in such that p forces that

I g(m) = i= 1...k I

f i (n + m). So for all m, p Leb( I g (m)) < k 2

(n + m) < 2 m, i.e. pg C .

g is a tools-interpretation of g under pn (this is clear if we assume that the list I m

contains no repetitions; otherwise we just have to chooseg(m) accordingly).

I g (m) for all m; i.e. g R0 .Since we assume (2) we can nd an N -generic q pn forcing that is random over N [G] and that

g R0 .

This means that q forces that I

f i (m) for all i k and m > n . And for n i m n ,

pn forces that I f i (m) = I f i (m) and therefore that I f i(m) . So q forces that f i Rni .

(2) (3): We show the equivalent property (3) of lemma 5.11. So x p and N andassume p

T I +BC . We want to show that there is a T and a p p such that T is an

interpretation of T with respect to p , and for every T N there is an N -generic

q p forcing that Gen( N [G]) T .

Since every positive set contains a positive closed set we can assume without lossof generality that p forces that

T is closed and that the measure of

T is at least some

rational number r .Any measurable A 2 can be approximated from the outside by countable unions

of clopen sets. If A is closed (i.e. compact), then any open cover of A has a nite sub-cover. So for any > 0 there is a clopen set C A such that Leb( C \ A) < . Inparticular there is a sequence of clopen sets

2 = A0 A1 Asuch that Leb( An \ A) = Leb( An) Leb( A) < 2

n and A = An . Set Bn := An \ An+ 1 .Then the Bn are a disjoint sequence of clopen sets, A = 2 \ Bn and

Leb( Bn) = Leb( An) Leb( An+ 1) Leb( An) Leb( A) < 2 n .

So the sequence ( Bn) is coded by an f C . Also Leb( Bn) = 1 Leb( A).Applying this to

T we get that p forces that there is the according

f for

T . Pick an

N H ( ) containing p and f , and let G V be an N -generic lter. Then f

f [G ]

is a tools-interpretation of f , witnessed by a decreasing sequence pn of elements of G .

Let ( Bn)n be the sequence of clopen sets corresponding to f . Bn is disjoint to Bmfor m n (since p forces this for

B). Also p forces that Leb(

Bn) 1 Leb(

T ) < 1 r ,

and therefore Leb( Bn) 1 r . So T 2 \ Bn is positive. T is an interpretation

of T with respect to p: Assume A T has measure s > 2 n

. Then pn p forces thatLeb(T \ m< n

Bm) = < 2 n and that m< n

Bm = m< n Bm. A is disjoint to n


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Set T 2 \ m>1 I f (m) and T 2 \ m>1 I

f (m) .

Then T is an interpretation of T with respect to p2 :Assume A T is a positive Borel set. Pick N such that n N 2

n < Leb( A). p N p2forces that f (i) =

f (i) for all i < N . So p N forces that A m< N I

f (m) is empty, and that

Leb( m N I f (m)) < Leb( A), and therefore that A

T is positive.

So by (3) we know that there is an N -generic q p2 forcing that is random over N [G ]and that

T .

T means that for all m > 1, I

f (m). Since q p2 , q forces that

is not in I f (0) = I f (0) or I

f (1) = I f (1) either. So q forces that

f R0 .

Using this lemma, theorem 6.1 and the fact that strong preservation implies preserva-tion we get: 22

C 9.4. Assume that ( P i,Q i)i


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preserving trueouter measure

O O5.11

/ /preserving true

positivity O O

5.8

strongly

preserving generics / /

5.6

preservinggenerics

preserving many

generics O O

3.5

preserving

outer measure

/ / preservingpositivity

preserving Borel

outer measure O O

/ /

P nep: 6.1

V R

H C

9 5 1 ' #

B B

preservingBorel positivity

P is Borelhomogeneous: 3.2

O O 1 1 1

P nep: 6.8

# ' 1

\ \

9 C

H R

V

preserving outermeasure of V

preservingpositivity of V

V has outer measure0 or 1 o o

Preservation of (Borel) positivity and outer measure is dened in 3.1, the true notionsin 5.5, and (strong) preservation of generics in 5.7 and 5.9. For P nep see section 6.

REFERENCES

[1] T B and H J , Set theory: On the structure of the real line , A K Peters,Wellesley, MA, 1995.

[2] M G , Tools for Your Forcing Construction , Set theory of the reals (Haim Judah, editor),Israel Mathematical Conference Proceedings, vol. 6, American Mathematical Society, 1993, pp. 305360.

[3] M G and J K , New reals: Can live with them, can live without them , preprint,see h ttp: // arxiv.org / math.LO / 0505471.

[4] J I (H J ) and S S , Souslin forcing , The Journal of Symbolic Logic ,vol. 53 (1988), pp. 11881207.

[5] T J , Set theory , Monographs in mathematics, Springer, 2002, 3. millennium ed.[6] A K , The higher innite , Perspectives in Mathematical Logic, Springer, 1994.[7] J K , Preserving non-null with suslin + forcings , preprint, see

h ttp: // arxiv.org / math.LO / 0211385.[8] M R , Goldstern-Judah-Shelah preservation theorem for countable support iterations ,

Fundamenta Mathematicae , vol. 144 (1994), pp. 5572.


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[9] A R and S S , Measured creatures , Israel Journal of Mathematics ,

vol. accepted, math.LO / 0010070.[10] C S , Understanding preservation theorems: omega-omega bounding , preprint, see

h ttp: // arxiv.org / math.LO / 0505645 .[11] , A short proof of the preservation of the -bounding property , MLQ. Mathematical Logic

Quarterly , vol. 50 (2004), no. 1, pp. 2932.[12] S S , Proper and improper forcing , Perspectives in Mathematical Logic, Springer,

1998.[13] , Properness Without Elementaricity , Journal of Applied Analysis , vol. 10 (2004), pp. 168

289, math.LO / 9712283.[14] R. M. S and S. T , Iterated Cohen extensions and Souslins problem , Annals of

Mathematics. Second Series , vol. 94 (1971), pp. 201245.[15] J Z , Descriptive set theory and denable forcing , Memoirs of the American Mathe-

matical Society , vol. 167,3 (2004), no. 793, pp. vii + 141.

JAKOB KELLNERINSTITUT F UR DISKRETE MATHEMATIK UND GEOMETRIE

TECHNISCHE UNIVERSIT AT WIEN1050 WIEN, AUSTRIA

E-mail : [email protected] : http: // www.logic.univie.ac.at / kellner

SAHARON SHELAHEINSTEIN INSTITUTE OF MATHEMATICS

EDMOND J. SAFRA CAMPUS, GIVAT RAMTHE HEBREW UNIVERSITY OF JERUSALEM

JERUSALEM, 91904, ISRAELand

DEPARTMENT OF MATHEMATICSRUTGERS UNIVERSITY

NEW BRUNSWICK, NJ 08854, USA E-mail : [email protected] : http: // www.math.rutgers.edu / shelah

Documents

Jakob Kellner and Saharon Shelah- Preserving Preservation