saharon shelah- the null ideal restricted to some non-null set may be ℵ1-saturated

8/3/2019 Saharon Shelah- The Null Ideal Restricted to Some Non-Null Set May Be 1-Saturated

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arXiv:math/9

705213v2

[math.L

O]16Aug2001

THE NULL IDEAL RESTRICTED TO SOMENON-NULL SET MAY BE 1-SATURATED

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

Abstract. Our main result is that possibly some non-null set of reals cannot bedivided to uncountably many non-null sets. We deal also with a non-null set of reals,the graph of any function from it is null and deal with our iterations somewhat moregenerally.

2000 Mathematics Subject Classification. 03E35, 03E55.Key words and phrases. set theory, forcing, FS iteration, null ideal, measure, 1-saturated

ideal, null functions, nep forcing.

I would like to thank Alice Leonhardt for the beautiful typing.This research was supported by The Israel Science Foundation founded by the Israel Academy ofSciences and Humanities. Publication 619

Typeset by AMS-TEX

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2 SAHARON SHELAH

Anotated Content

0 Introduction

[We review results and background, and give notation.]

1 The null ideal resricted to a non-null set may be 1-saturated[We explain the difficulty for the null case, solved by adding random reals+ in the end, measurable, but they are random over some subuniverses,so we add in the beginning Cohens r

i(i < ). The memory is devised

such that {A : {+ : A} null} will include a P()V\D, D a normalultrafilter on . Whereas in [Sh 592], the A (memory) were chosen closedenough, here we use automorphism of the memory structure.]

2 Non-null set with no non-null function[We show that consistently for some non-null set A of reals for every functionfrom A to the reals, its graph is a null subset of the plain.]

3 The L1,1-elmentary submodels and the forcing[We deal with general FS iterations of c.c.c. definable in subuniverses, andgive sufficient condition for PA P. As application we show consistency

of some values of 20 , add(meagre), cov(meagre), unif(meagre) b, d.]


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THE NULL IDEAL RESTRICTED, ETC. 3

0 Introduction

Note that the result stated in the abstract tells us that the positive result explained

below cannot be improved (to 1 sets). It is (Gitik, Shelah [GiSh 582])() given sets An of reals for n < , we can find Bn An, pairwise disjoint

such that An, Bn have the same outer Lebesgue measure.

Lately, we have proved ([Sh 592]):

0.1 Theorem. Con(cov(null) = + MAn) for each n < .

The idea of the proof was to use finite support Pi, Qi : i < , where say Qi has

generic real ri and Qi is random forcing in V[rj : j ai], ai i, ai closed enoughor r

i is Cohen real but the memory is not transitive, i.e. j ai aj ai.

As 0.1 was hard for me for long it seems reasonable to hope the solution willopen my eyes on other problems as well.

In this paper we deal mainly with can every non-null set be partitioned touncountably many non-null sets?, equivalently: can the ideal of null sets whichare subsets of a fixed non-null subset of R be 1-saturated?. P. Komjath [Ko],proved that it is consistent that there is a non-meager set A such that the ideal ofmeager subsets ofA is 1-saturated. The question whether a similar fact may holdfor measure dates back to Ulam, see also Prikrys thesis, Fremlin had asked both

versions since the seventies.

So we prove the following:

0.2 Theorem. It is consistent that there is a non-null setA R such that the idealof null subsets of A is 1-saturated (of course, provided that ZFC + measurableis consistent).

The proof of 0.1 was not directly applicable, but turning the tables make itrelevant as explained in 1.

The question appears on the current Fremlins list of problems, [Fe94] as problem

EL(a) respectively.We also have some further remarks, e.g. the exact cardinal assumption for 0.1.

We try to make the paper self-contained for readers with basic knowledge of forcingand of [Sh 592].

Also we answer the following problem which Komjath draws our attention to:


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4 SAHARON SHELAH

0.3 Theorem. It is consistent that:

there is a non-null A R such that: for every f : A R, the function fas a subset of the plane R

R is null

provided that ZFC + there is a measurable cardinal is consistent.

Lastly, in 3 we investigate when partial memory iteration behaves as in [Sh 592]and gives an example how to apply the method. This originally was a part of [Sh592], but as the main part of [Sh 592] was in final form and this was not and welike to add 3, we separate.

0.4 Notation. We denote:

(1) natural numbers by k,l ,m,n and also i, j

(2) ordinals by , , , , , ( always limit)

(3) cardinals by ,,,

(4) reals by a, b and positive reals (normally small) by

(5) subsets of or 2 or Ord by A , B, C , X , Y , Z (6) B is a Borel function

(7) finitely additive measures by

(8) sequences of natural numbers or ordinals by ,,

(9) s is used for various things.

T is as in Definition 2.9 of [Sh 592], t is a member ofT.We denote

(1) forcing notions by P, Q

(2) forcing conditions by p, q

(3) p q means p stronger than qand use r to denote members of Random (see below)

(1) Leb is Lebesgue measure (on {A : A 2})(2) Random will be the family

{r 2 :r is a subtree of (>2, ) (i.e. closed under initial segments, rwith no -maximal element (so lim(r) 2is closed)) and Leb(lim(r)) > 0 and moreover r lim(r[])is not null (on r[] see below)}


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ordered by inverse inclusion. We may sometimes use instead

{B : B is a Borel non-null subset of2}.

For >2, A 2 let

A[] = { A : }.We thank Tomek Bartoszynski and Mariusz Rabus and Heike Mildenberger for

reading and commenting and correcting.


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6 SAHARON SHELAH

1 The null ideal restricted to a non-null set may be 1-saturated

Let us first describe an outline of Komjaths solution to the problem for the meagre

ideal. Note that by a theorem of Solovay, the conclusion implies that there is ameasurable cardinal in some inner model. So Komjath starts with

V |= is measurable and D is a normal ultrafilter on .

He uses finite support iteration P, Q : 2, < 2 of c.c.c. forcing notions

such that for < the forcing notion Q adds a Cohen real 2 (so Q

=

(>2, )) and for each [, 2) for some A D, the forcing notion Q makes

the set

{

:

\A

}meagre (and every A

D appears). The point is that finite

support iterations tend to preserve non-meagreness, so in VP2 the set { : < }

remains non-meagre (and D is (i.e. generates) in VP2 an 1-saturated filter).For our aims this per se is doomed to failure: finite support iterations add Cohen

reals which make the set of old reals null, whereas countable support iterationstend to collapse 2. This seems to indicate that the solution should be delayedtill we have better other support iterations (the 20 = 3 problem; see [Sh:b,Ch.VII,Ch.VIII] and [Sh:f, Ch.VII,Ch.VIII]). But we start with a simpler remedy:we use finite support iteration P, Q

: , < of c.c.c. forcing notions

with cf() = , : < increasing with limit , Q is a partial random, say

RandomVPA()

where A() = A , adding the real .

Clearly we want that

|=P { : < } is not null.

For this it suffices to have: every countable A is included in some A().However, we also want that for B D the set {

: \B} will be null. For

this it is natural to demand that for some , Q is Cohen forcing and

( \B)( / A).

So we try to force a null set including { : \B} before we force the

s!

If is similar enough to being Cohen over

: \B we are done. So this

becomes similar to the problem in proving 0.1. But there we use 2 = , so that


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we need to carry only finitely additive measures

t : t T, i.e. have few

blueprints hence can make A (when |Q| ) closed enough. However, the

ts

come only to prove the existence of p which forces that many ps ( < ) arein the generic set (see the proof of [Sh 592, Claim 3.3]). So in fact we can definethe name of the finitely additive measure after we have the sequence p : < .Actually, we need this only for some specific cases and/or can embed our P intoanother iteration. So the problem boils down to having the As closed enough toenable us to produce finitely additive measures like the one in clause (i) of 2.11 orclause (d) of 2.16, [Sh 592]. The way we materialize the idea is by having enoughautomorphisms of the structure

P, Q , A, ,

: , < .

In fact we can replace below + by . If we would like to have e.g. {A :(A) = 1} to be a selective filter then the version is better: we can use{

+1+ : < }, which are Cohen, for ensuring this.If you do not like the use of automorphisms of Q and doing it through higher

s, later we analyze the method (of [Sh 592, 2,3]) more fully (using essentiallyL1,1 ), and then the proof is more direct (see 3.1 - 3.13).

1.1 Theorem. Let D be a -complete nonprincipal ultrafilter on . Then for somec.c.c. forcing notion P of cardinality 2, in VP we have

() for some A [2] we have(a) A is not null

(b) the ideal I = {B A : B is null} is 1-saturated.

Moreover,

() we can find pairwise distinct 2 (for < ) such that

A =

{ : <

}and Y

D

{ :

\Y

}is null.

1.2 Remark. 1) We can replace 2 by R.2) We can use as D any uniform 2-complete filter on (such that D is 1-saturatedin V and so in VP, see 1.11).3) In () of course D stands for the filter that D generated in VP.


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8 SAHARON SHELAH

1.3 Convention. For 2 (but we use only < (2)+) let g : D be suchthat

Y

D |{

< : g

() = Y}|

= .

For simplicity g is increasing with , and let h : D 2 be such that gh = idD.For < let

E = E = { < : g()}.

Finally we let

E = E = E : < .

If not said otherwise we assume that = 0

2

(for simplicity).Note: our intention is that

will exemplify {

+ : \g()} is null, for which

it is enough that will be (at least somewhat) like Cohen over V[

+ :

\g()], so it is reasonable to ask that is not in the subuniverse over which

+ is random when \g(), i.e. / E.

1.4 Definition. K is the family of P, Q, A, ,

: < such that

(A)

P, Q

: <

is a FS iteration with direct limit P

(B) A , = 0, and either |A| < , Q is Cohen forcing, the Cohen

real or |A|

, Q is Random

V[ :A]

(defined as in [Sh 592, 2.2])

and the random real. This is a particular case of [Sh 592, 2.2] except

replacing there by here so we can use [Sh 592].

Note that Definition 1.5(1), Claim 1.6(2) are vacuous when = 1, the case we usehere but they are natural for less specific cases. Let

=

: < , note that

more accurately we should write not Q

= RandomV[

A]

but RandomV[

A],

A

or RandomV,

A

as Q may be a proper subset of RandomV[

A]

depending on A, too. For the good cases equality holds. In shortness we shall write

RandomA

.


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1.5 Definition. For Q K :1) A g(Q) is called Q-closed if

( < g(Q) &|A|

< &

A)

A

A.

2) Let1

PAUT(Q) = {f :f is a one to one partial function from g(Q) tog(Q) such that dom(f) and rang(f) are Q-closed and for

dom(f) we have |A | < |Af()| < and if, dom(f) then A f() Af()}.

3) We define the following by induction on g(Q). For f PAUT(Q) let f be thepartial function from Pdom(f) to P

rang(f), (see [Sh 592, Definition 2.2(3)]) defined

by:

p1 = f(p0) when p0 Pdom(f), dom(p1) = {f() : dom(p0)}

p1(f()) = B(. . . , truth value( f()), . . . )wp0

when p0() = B(. . . truth value( ) . . . )wp0

;

(so wp1f() = {f() : wp0 }).

For such Q, f for any Pdom(f)-name we define f(

), a Prang(f)-name, naturally.

4) For Q K let

EAUT(Q) = {f PAUT(Q) : dom(f) = rang(f) = g(Q)}.

1.6 Fact. Let Q K .1) If < g(Q), Q = Q then Q

K and PAUT(Q

)

PAUT(Q).

2) If g(Q), then { : < } is Q-closed; and the family of Q-closed sets isclosed under intersection and union (of any family).3) If f PAUT(Q) and A g(Q) is Q-closed, then f (A dom(f)) belongs to

1this suffices when we iterate just partial random forcing (when |A | ) and Cohen forcing(when |Q | < ), which is enough here. For a more complicated situation, see 3.


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10 SAHARON SHELAH

PAUT(Q).

4) If2 f EAUT(Q), then f is an automorphism of Pg(Q)

.

5) IfA is Q-closed and Q,A below holds, then PA Pg(Q). Moreover, ifq Pg(Q)then:

(a) q A PA(b) P

g(Q)|= q A q

(c) if q A p PA then q = p (q (g(Q)\A)) belongs to Pg(Q) and is alub of p, q

where

Q,A if

A,

|A

| and B

is countable then for some

f EAUT(Q ) we have:(i) f (B A) = the identity

(ii) f(B) A(iii) f(B A) A A.

Proof. Straight. For part (5) we prove by induction on (Q) that replacingPg(Q)

, A by P , A = A

so q

P , clauses (a), (b), (c) in (5) hold.

In successor stages = + 1, if / A or |A| < it is trivial, so assume A & |A| . So PA P etc., and it is enough to show:

() if in VPA,I is a maximal antichain in RandomVPAA then in VP

,

the set I is a maximal antichain of RandomVPA .

By the c.c.c. this is equivalent to

() if < 1, {p : < } PA(+1), p PA and p PA {p() :

< and p

GP

A} is a predense subset of RandomAA

then p P {p() : < and p GP} is a predense subset ofRandomV

PA .

2does f PAUT(Q) imply f is an isomorphism from Pdom(f)

onto Prang(f)

? The problem is

that the order is inherited from P so is not necessarily the same.


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Assume () fails, so we can find q such that

p q P

q P {p() : < and p GP} is not a predense subset of RandomA

.

So for some GP-name r

q P r RandomA(= Q

) and is incompatible with every p() Q

.

Possibly increasing q, without loss of generality r is a B(. . . , truth value( ), . . . ))w where w A is countable. Let us define (supp for support is from

[Sh 592, Def.2.2(1)(F)], i.e. above supp(r

) = w)

B = w dom(q)


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12 SAHARON SHELAH

p f(q) P,

f(r) is B(. . . , f(), . . . )w, and f(w) f(B A) A A

hence P f(r) Random(AA)

andf(q) P

Ain Q

, the conditions f(r

) and p() are incompatible for <

,

getting a contradiction.

Note that we use: in VP, two conditions in Random

B

are compatible in RandomB

iff they are compatible in RandomVP

, for every B ; in particular for B = AAand B = f(A A). 1.6

1.7 Remark. 1) Instead automorphisms (f EAUT(Q )) we can usef PAUT(Q ) but having more explicit assumptions and more to carry byinduction, see 3.2) The use of Random is not essential for the last claim, we just need enoughabsoluteness.

Proof of 1.1. We shall define

Q = P, Q , A, ,

: + and < +

as an iteration fromK . More accurately, P = P = P,, Q

= Q

= Q

,, A =

A = A,, = = ,

=

=

, are such that

for < we have: A = and Q is as in [Sh 592, Def.2.5], i.e. the

Cohen forcing notion (and is the Cohen real),

for [, + ) we let:A = E

[, )

Q = Random

A


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(and is the (partial) random real)

(as in clause (F) of Definition 2.2 in [Sh 592].Let P = P = P+. We define P

(= P

,) as in [Sh 592, Definition 2.2(3)], i.e.

p P :for each dom(p) the condition p() is either a Cohen

condition or has the form B(. . . , truth value( ), . . . )wp

,

where wp A is a countable set,B is a Borel function withdomain and range of the right form (and B, wp are not

P-names but actual objects).

More generally, P,A for A + denote PA for a (this to help when we dealwith more than one ).

1.8 Definition/Fact. 1) We define E (for < , = 0 2), an equivalence

relation on by: E iff g() = g() .2) E (for < , =

0 2) is an equivalence relation on with 2||equivalence classes, each of cardinality .3) If < then E refines E and |E (/E )| = for every < .

1.9 Fact. 1) If < then (in Q)

(a) PA+ = P(A+)[,+) = P

E[,+) P

+.

Moreover

(b) if q P+ and q (E [, + )) p PE[,+) then

p (q (\E ) [, + )) P+is least upper bound of p, q.

2) If then P,[,+) P

,+ and P

,[,+) is isomorphic to

P,+ say by h : P,+ P,[,+) where h = h,

is the canonical map-

ping, i.e. let h : + + be < h() = , < h( + ) = + ,


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14 SAHARON SHELAH

now if h(p) = p then dom(h(p)) = {h() : dom(p)}, etc.3) If < + , |A| (equivalently, [, + )) then Q

, = V

A

= VPA .

Proof. 1) By 1.6(5) (really this is a particularly simple case). I.e. let A = A+ =E [, + ), so by 1.6(5) it is enough to check Q(+)\A there. So let Abe such that |A| and B is countable, and we should find f as there. As|A| necessarily so for some < , = + . By 1.8(3) the existence off is immediate.2) Straight.3) By [Sh 592, 2.3(7)] and clause (a) of part (1). 1.9

1.10 Fact. X = {+ : < } is not null (in VP+).

Proof. It if is null, it is included in a Borel null set X which is coded by a real swhich is determined by truth value(p GP+) : < for some p : < where p P+. Let w =

0 and)(i) (

)( +

w

< )

(ii) w E ( A+).

This is possible as we have 0 demands; for each one the set of < satisfying itis in D. Now

+ is random over V

PA+ , and X is (definable) in VPA+ (see [Sh

592, 2.3]). Hence + is not in the Borel set X, a contradiction. (Alternatively

follows the proof of [Sh 592, 2.3(2)]). 1.10

1.11 Fact. If D is a -complete ultrafilter then in VP, D is a -complete

1-saturated filter.

Proof. Well known (see [J]), as P is a c.c.c. forcing notion.

So the only point left:


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1.12 Fact. If Y D then P {+ : \Y} is null.

This follows from

1.13 Main Fact. If < , < , / E then P,+

+ N(

), where

N() is as in [Sh 592, Definition 2.4].

Proof. Note that

() for <

EAUT(Q ( + )) {f :f a permutation of + ,f [, + ) is identity and

f maps E onto E for < }.

Assume toward a contradiction that the desired conclusion fails, hence for somep P+ we have p

+ / N(

). In fact, possibly increasing p, without loss

of generality for some

p

+ n

a

where n

, a

: < is as in [Sh 592, Definition 2.4].We may assume that p P++1. Let G P+ be generic over V such that

p ( + ) G. So in V[G]:

p( + ) Q+ + T

where T

= T(a

)

=: { >2 : if [, ) and n

g()

then n

a

}.

Now for some q we have:

p ( + ) q G and

q p( + ) = B(. . . , truth value( ), . . . )


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16 SAHARON SHELAH

where B V is a Borel function, A+ and < for < . So (see [Sh592, Definition 2.2(1)(F)()]

t

=:B

(. . . , truth value(

), . . . )

is a perfect subtree of (>2, ) of positive measure above every node, i.e. t

0 < Leb({ 2 : and n < n t

}).

So by the choice of p and q we have q t

T

. Without loss of generality

q P+. Let w = dom(q) {supp(q()) : dom(q)} { : < } {}.

We can choose p P+, f , (by induction on < (20)+) such that(a) f (w

\{}) = the identity(b) f [, + + 1) is the identity

(c) f EAUT(Q ( + + 1)),(d) = f(),

(e) / { : < },(f) q = f(q) and

= f() for < .

Hence

(A) q B(. . . , truth value(

), . . . )2, ). Hence3 Y

contains a perfect subtree, which was exactly our problem in proving theorem 0.1

in [Sh 592]. So we would like to continue as in the proof of [Sh 592, 0.1], but forthis we need a suitable

.

First Proof:

3In fact, we could have demanded

= for < , < (20)+ so t hence Y includes t

itself


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We start repeating the proof of [Sh 592, 3.3]. As there we choose = : < a sequence of positive reals satisfying + and let R V be {f : f a partial function from to {0, 1} with

countable domain}. The iterations Q+k and the properties we are interested in,are the same in V and in VR, so we can work in VR.

So we can choose : < , forced to be pairwise distinct members of2.

Possibility B: (Definition) The set T of weak blueprints as in [Sh 592, Definition2.9] replacing wt and the t-s by t so replacing clauses (a), (c), (i), (j) by

(a) t = tn,k : n < n

t, k < ,(c)

tn,k

an ordinal increasing with n

(i) tn1,k1

= tn2,k2

n1 = n2 and for k1(j) for each n nt the sequence t

n,k : k < is constant or is strictlyincreasing; if it is constant and n Dom(ht0) then ht0(n) is constant

(k) for k2 < n1 < n2 < nt one of the following occurs as follows:

() nt

, mt

, ht

0 , ht

1 , ht

2 , nt is as in the proof of [Sh 592, 3.4]

() t

i, is i, Dom(p) above (for i < i = nt, < ).

Nothing relevant to us changes.The actual difference between the two possibilities for the rest of the proof is

small and we shall use the second. We define the weak blueprint t = t() =(t

, nt

, mt

, ht

0 , ht

1 , ht

2 , nt).

Clearly t T; we would like to proceed and choose

(but first choose

B,n for n < such that: if Y {A,+ , \A,+} for < , X n


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18 SAHARON SHELAH

{B,n, \B,n} and


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(so ak(r, r, n) [0, 1] is well defined for k < , t = r : < , {r, r}

Random, n =

n : <

, n < n+1 < ).

Let us carry the induction.

Case 1: n = 0.Necessarily jn =

+j . Note that < jn A, = and the sets {tn,k : t T,0} for n < nt , k < are pairwise disjoint. Hence the proof if easy (similarto [Sh 592, Lemma 2.14]).

Case 2: n + 1, jn+1 > jn + 1.

Similar to Case 1.

Case 3: n + 1 and jn+1 = jn + 1.Let n be such that j

n,k = jn for k < . Ifn Dom(h0) nothing to do, so assume

n Dom(h1). We would like to repeat the proof of [Sh 592, Lemma 2.16(1)], butthis proof needs, in our notation, not just P

+j ,Ajn

P+j ,

jn

(which we have

proved) but also

jnP()

P+j,A

jn is a P

+j ,Ajn

-name, which is not part of our

induction hypothesis here. But we have in our induction hypothesis

j+1n . We let

be large enough and choose M0, M1 elementary submodels of (H(), ,


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20 SAHARON SHELAH

n < n} and { < +n : / E+n

} [+n

, ) if = + n + n.It suffices to prove that for some we do not have +n + : < n, , p : < as above for the Q defined above. We take k = n.

[Why? Choose >> , and an elementary submodel M of (H(), ,


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2 Non-null set with no non-null function

Peter Komjath referred me to the following problem, answered below.

2.1 Theorem. It is consistent that:

there is a non-null A R such that: for every f : A R, the function fas a subset of the plane RR is null

provided that ZFC + there is a measurable cardinal is consistent.

Proof. We can use 2 instead ofR. Let be measurable, D a normal ultrafilter on. Let = 0 , 2 for simplicity, and we use the same forcing as in 1. Nowwe interpret the Cohen forcing Q

(for < ) as

(n, a) : < : < , n < , a is a subset of(n2) (n2), n2 |{ n2 : (, ) a}|/2n 1 4

.

We interpret the generic sequence a = (n , a ) : < as the following nullsubset of the plane:

null(a) =: {(, ) : 2 and for infinitely many < we have ( n , n

) / a }.

Let X

= {+i : i < }, as in 1, it is not null (in fact everywhere). Now

suppose p g

: X

2; of course, this g

is unrelated to the gs from 1. Sofor each i < , g

(

+i) is a P+-name involving the conditions pi, P+ (for

< ). Let for < , M (H((+)+), ,


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22 SAHARON SHELAH

(c) w [, + ) is constant, w { + : A} = { + }, moreoversup(w) < + min(A\(+ 1)) and w [, + ) w,

(d) for , A, the models M , M are isomorphic, with the isomorphismmapping to , and Q, + : < , g to themselves, in fact is theidentity on M.

So for all A we have an isomorphic situation.For A let r = r,m : m < be the


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We do it by cases (note that in each case for some Am is equivalent to for

every Am).

Case 1: r,m g

(+) is in V[

: M+].

Clearly ()2,m holds as the graph of a Borel function is null and apply this tothe function B(. . . ,

, . . . , . . . ; )M.

Case 2: r,m (+ , g

(

+)) (2 2\


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24 SAHARON SHELAH

p forces that (+, g

(

+)) is not in (

2 2)VP++1 as the second coordinateis not in VP++1.

Case 4: Neither Case 1 nor Case 2, nor Case 3.As in the Case 3, but we can have f (( + 1) M) is the identity. We let

hence f (P+1 M) is the identity. We let B1 = { : < + 1 and /M( + )\M}, using f for f EAUT(Q) we can easily show that PB1 P+,and GB1 = GP++1 PB1 (or see 3). Now we have in V[GP++1]:

E

[GB1 ] = { : p ( + + 1) GB1} is unbounded in (20)+.

We can assume that p g

(+) / V[G

B1 ], p I for some (as in 1), as

if not without loss of generalityp forces g(+) = such that = B(. . . ,truthvalue( ), . . . )


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3 The L1,1-elementary submodels and the forcing

We may wonder what is really needed in

1,

2 (and [Sh 592],

2,

3). Here we

generalize one feature: iterating with partial memory but without transitivity inthe memory: not restricting ourselves to Cohen and random.

3.1 Definition. Ks is the family of sequences of Q of this form

Q = P, Q, A, B, , , Y

,

: <

(we write = g(Q)) such that:

(a) P, Q : < is an FS iteration of c.c.c. forcing notions (so P denotes

the limit)

(b) < (see clause (d) below), is a Q

-name (i.e. a P-name of one),

P+1 and

is the generic of Q

(c) B A and B B B and |B| < .

First simple version:

(d) = 0, 1, 2), = 0, 1, 0 = 0(x, Y),

1 =

1(x,y,Y

), let

= 2

(e) the set of elements ofQ is a non empty subset of Q

pos

= { :

(

0

) inthe universe V[

: A]

(f) Y is a P-name of a subset of

1, moreover we have sequences B =

B, : < 1, = ,,n : < 1, n < and , = ,,n : () if = 0, 3, a 4-place relation on

>() if

= 1, 2, 4 is binary, 5 is a 5-place relation, and for i < 6 we have iis a subset4 of { : = (0, . . . , j1), for some n each is a sequenceof ordinals of length n, and (i, j) {(0, 3), (1, 4), (2, 4), (3, 3), (4, 2), (5, 5)}}(can read more natural restrictions), i is closed under initial segments (i.e.letting i = 0, = (0, 1, 2) 0 n = (0 n, 1 n, 2 n) 0, ofcourse n is an abuse of notation),

lim() = {(,,) :,, are sequences of length such that n < ( n, n, n) }

(depend on the universe) and similarly for other i

(e) the set of elements of Q is a (nonempty) subset of Qpos = { :

(0) V[ : A]}

(f) =

, : <

, < 6,

, =

,,m : m < , and

,,m is a

P-name of a natural number (or of a member of {0, 1}), it has the formB,,m(. . . ,truth value(

,,m,k ,,m,k), . . . ))k


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(k) in V[ : A] we have {pn : n < : pn Q

for n < , {pn : n < }

is predense in Q} = {pn : n < : for some (Ord) and < 3 we

have (p,,

3

,) lim(3

) [where p = pn[n]2

([n]) : n < ]}, and wewrite 2(p) for the minimal such (l) in V[

: A] we have {(p,) : p Q

, <

4 and p Q

} = {(p,): for some (4) and < 4 we have p, 4}

(m) if G


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28 SAHARON SHELAH

P =

p P :for every dom(p), p() is computed by someBorel function B (so B V, and is an object not a name),B is a function to V, from some

truth value( ) : < where

< and < and we say in this case that

the truth value of ( ) appear in p() for <

and p forcing a value to 0

(p())

.

2) For p P , dom(p) let

supp(p()) = { : for some the truth value of ( ) appear in p()}

and let supp(p) = dom(p)

dom(p)supp(p()).

3) For A let PA = {p P : dom(p) A and dom(p) supp(p()) A}, i.e. PA = {p P : supp(p) A} with the order inherited from P which isinherited from PA (recall: only for some As, P

A P

).

4) A is called Q-closed if A

g(Q), and

A

B

A. We call A strongly

Q-closed if A g(Q), A A A. We let cQ(A) be the Q-closure of A andscQ(A) be the strong Q-closure of A.5) Let

Simple version:


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PAUT(Q) =

f :(i) f is a one-to-one function,

(ii) domain and range off are Q-closed

(in particular are

g(Q))

(iii) for 1, 2 dom(f) we have:1 A2 f(1) Af(2)

(iv) for 1, 2 dom(f) we have1 B2 (f1) Bf(2)

(v) if f(1) = 2 then

f maps Y1 to Y

2 , i.e. 1 = 2 ,

1 = 2 ,B1, = B2, ,

1,,n = 2,,n, 1,,n = 2,,n

.

Non simple version:


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30 SAHARON SHELAH

PAUT(Q) =


(ii) domain and range off are Q-closed

(in particular are g(Q))(iii) for 1, 2 dom(f) we have :

1

A2

f(1)

Af(2

)

(iv) for 1, 2 dom(f) we have1 B2 (f1) Bf(2)


1 = 2

, 1 = 2

, 1 = 2

and f maps

1,

to

2,

(i.e. B1,,m = B2,,m

for < 6, < 1 = 2

, m <

and 1,,m,k = 2,,m,k

for < 6, < 1 = 2

, m < , k < but, of course

is replaced by

f()

that is 2,,m,k = f(1,,m,k

)

.

6) For f PAUT(Q), f is the natural map which f induces from Pdom(f) ontoPrang(f); similarly for part (7).7) For Q1, Q2 Ks let PAUT(Q1, Q2) be defined similarly:


31/44


PAUT(Q1, Q2) =


(ii) domain off is Q1-closed

(in particular is g(Q1))and range of f is Q2-closed

(in particular is g(Q2))(iii) for 1, 2 dom(f) we have :

1 A12 f(1) A2f(2)(iv) for 1, 2 dom(f) we have

1

B12

f(1

)

B2f(2)


simple version :11 =

21 ,11 =

22 ,

and f maps Y

11

to Y

22

,

i.e. B11, = B22,

, 11,,m = 22,,m

, 11,,m = 22,,m

non simple version : (for < 6)

11 =22 ,

11 =22 ,

11 =22

and f maps 11, to 22,

(i.e. B2,,m = B1,,m

2,,m,k = 1,,m,k

2,,m,k = f(1,,m,k

) for < 1 = 2

, m < ,k < )

.

3.4 Claim. Let Q Ks be of length .1) P

is a dense subset of P .

2) In VP, from [GQ ] we can reconstruct GQ and vice versa. From

: 0)(b) F = F : < 1 and F is decreasing with (for notational simplicity)


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(c) F PAUT(Q1, Q2) for < 1(d) if f F then dom(f) g(Q1) and rang(f) g(Q2)(e) if f1

F1 , 2 < 1 < 1, C1

g(Q1), C2

g(Q2),

|C1

|< ,

|C2

|<

then for some f2 F2 we havef1 f2, C1 dom(f2), C2 rang(f2).

2) We say F is an explicit -witness for (Q1, Q2) if (a) - (d) above hold and

(e) if f1 F1 , 2 < 1 < 1, C1 g(Q1) scl(Q1)(dom(f1)), C2 g(Q2) scl(Q2)(dom(f2)), |C1| < , |C2| < , then for some f2 F2 we have f1

F2, C1 dom(f2), C2 rang(f2), dom(f2) sclQ1(dom(f1)), rang(f2) sclQ2(rang(f1))

(f) for f F , g f g F(g) if f F and , dom(f), then

sclQ1({}) f() sclQ2({f()})

(h) if C1 g(Q1), C2 g(Q2) are countable and < 1, then for somef F we have C1 dom(f), C2 rang(f).

3.11 Claim. Assume F is a -witness for (Q1, Q2).1) If

(0, 1) and f

F and , < g(Q

1), then

sclQ1({}) f() sclQ2({f()}).

2) Let F1 = F1 : < 1, F1 = {f1 : f F}. Then F1is a -witness for(Q2, Q1).3) If A g(Q) and: A = sclQ(A) for = 1, 2 and we let F = {f X : f F1+ , X A1 A2}, then F : < 1 is a -witness for (Q1 A1, Q2 A2) andan explicit -witness for it, note that by renaming A can become an ordinal.4) If p (P1

g(Q1)), q (P1

g(Q1)) andI (P1

g(Q1)) is countable, then for some

< 1 we have: for every f

F we have:

() if supp(p) supp(q) dom(f) then (P1g(Q1)

) |= p q iff (P2g(Q2)

) |=f(p) f(q)

() if supp(p) supp(q) dom(f) then p, q are compatible in (P1g(Q1)

) iff

f(p), f(q) are incompatible in (P2g(Q1)

)


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40 SAHARON SHELAH

() ifrI

supp(r) dom(f) then I is predense in (P1g(Q1)

) iff f(I) =

{f(r) : r I} is predense in (P2g(Q2)

).

5) For every Borel functionB = B(. . . , tn, . . . )n (for tn a truth value, valuesin V) and (n, n) : n < , n < g(Q1), and p (P1g(Q1)) there is < 1 suchthat: if f F , {n : n < } supp(p) dom(f) and x V then

p P1g(Q1)

B(. . . , truth value(n n), . . . )n


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()2 p, q are compatible in P11 iff f(p), f(q) are compatible in (P2g(Q2))

()3 I is predense in (P11) iff{f(r) : r I} is predense in (P2g(Q2)).

As in all three cases we can replace (P11) by (P1g(Q1)) and f by f we are done.

Case 3: cf(g(Q1)) = 0.Concerning clauses (), () (of 3.11(4)) the proof is just as in case 2, so we

deal with clause (). Let I (P1g(Q1)

) be countable. We choose n < g(Q1)

such that n < n+1 andn


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42 SAHARON SHELAH

hence for p I, f(p n) = f(p A1n) = f(p) A2n. As p I n

(p In)

(as dom(p) is bounded in g(Q1)) clearly f(I) =

n


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3.15 Remark. We can use other three cardinals, and use very little cardinal arith-metic assumption.

Proof. Let Q =Pi, Q

j, a

j,

i: i

3+

2, j <

3+

2be FS iteration, with

(a) i the generic real of Qj

(b) if j < 3 then Qj is Cohen say (

>, ) so j is undominated

(c) if j = 3 + , < 2 then Qj is RandomV[i:iaj]

(d) if b a [3 + 2]1 then for arbitrarily large i (3, 3 + 2) we havea ai = b.

Let P = P3+2 . Trivially |P| = 3. So in V we know:

Clause (a): 20 = 3.As |P| 3 the inequality holds, the other inequality, , holds as

i : i < 3

is a sequence of3 distinct reals.

Clause (b): add(meagre) = 1.As in [Sh 592] we know that b is not increased by P so b = 1 but add(meagre)

b.

Clause (c): cov(meagre) = 2.If for i <

1we have A

i 2 are Borel sets in VP then for some <

2,Ai

:i < 1 VP3+i , the forcing P3+i+/P3+i add a Cohen real over VP3+i (aswe are using FS iteration).

So cov(meagre) > 1.On the other hand

3+i : i < 2 is a non-null set of reals hence unif(null)

2, but cov(meagre) unif(null) so cov(meagre) 2. So together cov(meagre)= 2.

Clause (d): d = 3.Now d 20 = 3, on the other hand, as in [Sh 592] for no A [3]2 from V,

is there


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44 SAHARON SHELAH

REFERENCES.

[Fe94] David Fremlin. Problem list. Circulated notes; available fromhttp://www.essex.ac.uk/maths/staff/fremlin/measur.htm .

[GiSh 582] Moti Gitik and Saharon Shelah. More on real-valued measurable cardi-nals and forcing with ideals. Israel Journal of Mathematics, submitted.math.LO/95072086

[J] Thomas Jech. Set theory. Academic Press, New York, 1978.

[Ko] Peter Komjath. On second-category sets. Proc. Amer. Math. Soc.,107:653654, 1989.

[Sh:b] Saharon Shelah. Proper forcing, volume 940 of Lecture Notes in Math-ematics. Springer-Verlag, Berlin-New York, xxix+496 pp, 1982.

[Sh:f] Saharon Shelah. Proper and improper forcing. Perspectives in Mathe-matical Logic. Springer, 1998.

[Sh 592] Saharon Shelah. Covering of the null ideal may have countable cofinal-ity. Fundamenta Mathematicae, 166:109136, 2000. math.LO/9810181

6References of the form math.XX/ refer to the xxx.lanl.gov archive

saharon shelah- the null ideal restricted to some non-null set may be ℵ1-saturated

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