moti gitik and saharon shelah- more on real-valued measurable cardinals and forcing with ideals

8/3/2019 Moti Gitik and Saharon Shelah- More on Real-Valued Measurable Cardinals and Forcing with Ideals

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MORE ON REAL-VALUED MEASURABLE CARDINALS

AND FORCING WITH IDEALS

by

Moti Gitik Saharon Shelah

School of Mathematical Sciences Department of MathematicsSackler Faculty of Exact Sciences Hebrew University of Jerusalem

Tel Aviv University Givat Ram, JerusalemRamat Aviv 69978 Israel

[email protected] Department of MathematicsRutgers University

New Brunswick, NJ, [email protected]

Abstract.

(1) It is shown that if c is real-valued measurable then the Maharam type of

(c, P(c), ) is 2c. This answers a question of D. Fremlin [Fr,(P2f)].

(2) A different construction of a model with a real-valued measurable cardinal

is given from that of R. Solovay [So]. This answers a question of D. Fremlin

[Fr,(P1)].

(3) The forcing with a -complete ideal over a set X, |X| cannot be iso-

morphic to RandomCohen or CohenRandom. The result for X = was

proved in [Gi-Sh1] but as was pointed out to us by M. Burke the application

of it in [Gi-Sh2] requires dealing with any X. The application is: if An is a

set of reals for n < then for some pairwise disjoint Bn (for n < ) we have

Bn An but they have the same outer Lebesgue measure.

Partially supported by the Israeli Basic Research Fund. Publ. Number 582.

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In Section 1 we deal with the Maharam types of real-valued measurable

cardinals. The result (1) stated in the abstract and its stronger version are

proved. The proofs are based on Shelahs strong covering lemmas and his revised

power set operation.In Section 2 a model with a real-valued measurable which is not obtained as

the Solovay one by forcing random reals over a model with a measurable.

In Section 3, the result (3) stated in the abstract is proved.

Theorem 1.1 and the construction of Section 2 are due to the first author.

Theorem 1.2 is joint and the result of Section 3 is due to the second author.

We are grateful to David Fremlin for bringing the questions on real-valued

measurability to our attention. His excellent survey article [Fr] gave the inspira-

tion for the present paper. We wish to thank the Max Burke for pointing out a

missing stage in the argument of [Gi-Sh2].

1. On Number of Cohen or Random Reals

D. Fremlin asked the following in [Fr,(P2f)]:

If c is a real-valued measurable with witnessing probability , does it follow that the

Maharam type of (c, P(c), ) is 2c? or in equivalent formulation:

If c is a real-valued measurable does the forcing with witnessing ideal isomorphic to the

forcing for adding 2c random reals?

The next theorem provides the affirmative answer; see also 1.2.

Theorem 1.1. Suppose that I is a 20-complete ideal over 20 and the forcing with it

(i.e. P

20

/I) is isomorphic to the adding of-Cohen or-random reals. Then = 22

0.

Proof: Suppose otherwise. Denote 20 by . Let j : V N be a generic elementary

embedding.

Claim 1. j() > (+)V.

Proof: By a theorem of Prikry [Pr] (see also [Gi-Sh2] for a generalization) for every

< 2 = 20 = . Then, in N, 2 = j(). But (P())V N, so j() (2)V. By

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[Gi-Sh2 Lemma 2.2.], then (2)V = cov(,, 1, 2). So cov(,, 1, 2) +. Clearly,

cov(,, 1, 2) cov(, 1, 1, 2) (cov(, 1, 1, 2))N .

The last inequality holds since N is obtained by a c.c.c. forcing and so every countable

set of ordinals in N can be covered by a countable set of V. By Shelah [Sh430,3.2(2)], in

N cov(, 1, 1, 2) < j(). Hence + cov(,, 1, 2) (cov(, 1, 1, 2))N < j().

of the claim.

By Shelah [Sh 430,3.2(1),(2)] for every i < 3 there is Si = si | < []i

unbounded of cardinality . We can assume by shrinking Si that for every regular <

if si | < is unbounded in []i then for all , < si . For < and

i < 3 let Si = {P SiP }. Fix a function f representing in a genericultrapower and restrict everything to a condition forcing this. Without loss of generality

f() for every < .

Claim 2. Let i < 3 then { < Si f() is unbounded in [f()]0 and |Si f()| =

f()} I where I is the filter dual to I.

Proof: We drop the index i for a while. |S| = in V, so S is in a generic ultrapower.

Suppose that in a generic ultrapower S is bounded. There will be some t countable

such that for every s S s t. Using c.c.c. of the forcing we find a countable subset of

in V, t t. Since S is unbounded in V some s S contains t. Contradiction. Now,

j(S) = S, since is regular, S = s | < is unbounded in []i and hence no

s for every , < j(). of the claim.

Let N be a generic ultrapower. By [Gi-Sh1] there are in N at least Cohen (or

random) reals over V.

Claim 3. There exists a sequence r | < of reals in V so that

(1) every real of V appears in r | < .

(2) for almost all (mod I) r+i | i < f() are f()-Cohen (random) generic over

L[S f() | < 3, r | < f()].

Proof: Construct r | < by induction. On nonlimit stages add reals in order to

satisfy (1). For limit s with 0 f() unbounded in [f()]0 , for every < 3, add

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f()-Cohen (or random) reals. It is possible since there are at least candidates in a

generic ultrapower by [Gi-Sh1].

of the claim.

Now work in N. rngf A is unbounded in , for every A I. Let j(r | < ) =r | < j() where r | < is a sequence given by Claim 3.

Then, using Claim 3 in N we can find some < j() satisfying (2) of Claim 3 such

that j(S) f() is unbounded in [f()]0 , for every < 3 and j(f)() (+)V. It

is possible since by Claim 1, (+)V < j() and, in V the range of f restricted to a set not

in I is unbounded in .

The following will provide the contradiction and complete the proof of the theorem.

Claim 4. r+i | i < j(f)() is a sequence of Cohen (random) reals over V.

Proof: r+i | i < j(f)() is Cohen (random)-generic over L[j(S) j(f)(

) | 1, some final segment of

s | < will be generic over L[s]. Then, in V, for I-almost every regular < there

is a sequence t

= t | < such that for every s 1 and of cofinality > 1,

some final segment of t is Cohen generic over L[s]. Back in N, we use this for some

+, for = + which is still below j().

of the claim.

Let us fix such a sequence r | < + in N. We split it into blocks each of the

length 1. Denote such changed sequence by ri | < +, i < 1. Now back in V,

let us use the fact that cov(, 2, 2, 1) = . We know that for every <

+

the blockri | i < 1 is added by using only 1 Cohen reals from the Cohen reals, s | < .

Work in V. For every < + and i < 1 pick a condition pi in the Cohen forcing for

adding -Cohen reals which decides the value of r,i (0). Let ,i 2 be such a value ,

i.e. pi r,i(0) = ,i. W.l. ofg.

p0 {i < 1 | p,i G}

= 1 .

For every < + and i < 1 let dompi = {,i, | < i < } and pi(,i,) =

,i, >2. As 2


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cov(, 2, 2, 1) = , so there is a set b of cardinality 1 such that the following set

is unbounded in +:

A = { A0 | (1i)( < i)(,i, b) and ( < 0)(0, b)} .

We actually replace in cov by


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By basic probability some q,in|n


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Definition 2.1. P consists of all triples p = p0, p1, p2 so that

(1) p0

(2) p1 is a function with domain contained in p0

(3) p2 is a function defined over inaccessibles

(4) for every inaccessible |p0 | < , | domp1 | < and | domp2 | <

(5) for every domp1 p1()

(6) every element ofp0 is an ordinal of cofinality 0

(7) for every limit ordinal if cf > 0, then p0 is not stationary in and if cf = 0

then \(p ) is unbounded in

(8) for every domp2 p2() is a closed subset of disjoint with p0.

Definition 2.2. Let p, q P p = p0, p1, p2 and q = q0, q1, q2. Then p q iff

(1) p1 q1

(2) domp2 dom q2 and for every dom q2 p2() is an end extension of q2()

(3) p0 q0

(4) for every < , if is an inaccessible or a limit of inaccessibles and is the least

inaccessible above then p0 [, ) is an end extension of q0 [,

).

The forcing P is intended to add three objects. Thus, the first coordinates of P are

producing a subset S of which is stationary in V[S] and reflecting only in inaccessibles.

The second coordinate is responsible for a kind of diamond sequence over S and the last

coordinate adds clubs preventing reflection of S at inaccessibles and its stationarity.

The forcing P destroys the measurability of once used over V = L[]. It is bad for our

purpose. We are going to use a certain subforcing ofP which will preserve measurability

and contain the projection of P to the first two coordinates. But first let us study basic

properties of P.

Let P0 = {p0 | p1, p2 p0, p1, p2 P}, P01 = p0, p1|p2 p0, p1, p2 P}. Let

be an inaccessible. We denote by P the set

{p0 , p1 , p2 | p0, p1, p2 P}

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and by P\ the set

{p0\, p2 [, ), p2 [ + 1, ) | p0, p1, p2 P} ,

P0 , P01 and P0\, P01\ are defined similarly.

The following is standard.

Claim 2.3. Let be an inaccessible then the following holds

(1) P = P P\

(2) P0 = P0 P0\

(3) P01 = P01 P01\.

Let < be a limit ordinal and Q a forcing notion.

Consider the following game Game (Q, ):

I q1 q3 ...........................

..............

....................................... II q2 q q

where Players I, II are building an increasing sequence of elements of Q, I at even stages and

II at odds. If at some stage < II cannot continue i.e. there is no q above {q| < }

then I wins. Otherwise II wins.

Claim 2.4. The player II has a winning strategy in the game Game (P\, +) for every

inaccessible .

Proof: Let be an inaccessible. We define a winning strategy for Player II in the

Game (P\, +).

Let > be an inaccessible but not limit one. Denote by the supremum of

inaccessibles below .

Let p P\. We define p to be the condition obtained from p = p0, p1, p2 by adding

sup(p0 [, )) + sup(p2()) to p2() if p0 [, ) = or p0 [, ) = but p0 is

unbounded in , where runs over inaccessibles above which are not limit inaccessibles

and sup(p2()) = 0 whenever domp2.

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Now we define to be dependent only on the last move of I at successive stages of

the game. Set (p+1) = p+1. If + is limit and the game up to

p1 p3

p2 p

was played according to . Then set (p | < ) = the closure of


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Proof: Let D| < + be the list of dense open subsets of j(P)\ of N. Let N

be a winning strategy for Player II in Game (j(P)\, +). It exists by Claim 2.4 applied

in N to j(P). Play the game from V so that I plays at stage + 1 an element P+1 of D

which is above p , where < +

. We will finish with a desired N-generic set. Force with P over V. Let G be a generic subset. We denote

{p0 | p1, p2 p0, p1, p2

G} by S. For every S let A =

{p1() | p0, p1, p2 G and domp1} and

for inaccessible let C =

{p2() | p0, p1, p2 G and domp2}. Then S ,

and for every inaccessible C is a club of disjoint to S.

Claim 2.7. S is a stationary nonreflecting subset of in V[S, A | S, C |

inaccessible and < ].

Proof: C | < are witnessing the nonreflection. Suppose that S is nonstationary

in V[S, A | S, C | inaccessible and < ]. Return back to V and work

with names. Suppose for simplicity that the empty condition forces the nonstationarity

of S. Let C

be a name of witnessing club. Pick an elementary submodel N of V22 such

that P, C

N, |N| < and N is an ordinal of cofinality 0. Let = N and

n | n < be a cofinal in sequence. Now by induction we construct an increasing

sequence pi | i < of conditions of P (i.e. P without the information on a club of

disjoint to S) such that for every i <

(a) pi N

(b) pi decides the first element of C

above i

(c) sup(pi)0 i.

where pi = (pi)0, (pi)1, (pi)2

Now, in V, let

p = i


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V. Thus it is easy to extend j to the embedding of V[S, A | S]. This insures the

measurability of .

Notice, that is not measurable in V[S, A | S, C | < , inaccessible]

since S is a stationary nonreflecting subset of .Now, over a model V[S, A | S] we are going to force a Boolean algebra B such

that:

(a) is still measurable in V[S, A | S, B]

(b) j(B)

G(B) is isomorphic to the adding ofj()-Random reals, where j the elemen-

tary embedding ofV[S, A | S, B] into its ultrapower and G(B) a generic subset

of B.

First let us review some basics of product measure algebras. We refer to D. Fremlin

[Fr2] for detailed presentation.

Suppose that B is a -algebra, i.e. a Boolean algebra all of whose countable suprema

exist. A measure on B is a function : B [0, 1] so that: (a) (1B) = 1, and (b)

whenever {bn | n } B with bn bn = 0 for n = m, then

Vnbn

=n

(bn). If

in addition is positive (i.e. (b) = 0 iff b = 0), then we say that B, is a measure

algebra. A measure algebra is always a complete Boolean algebra.

Suppose now that I is a set, and Bi, i for i I are measure algebras. Call C iI

Bi

a cylinder iff C(i) is the unit element of Bi, except for a finite number of coordinates i.

Let B iI

Bi be the -algebra generated by the cylinders. It is known that there is a

unique measure on B so that (C) =iI

i(C(i)) for any cylinder C. may not be

positive, but there is a standard strategy: Let I = {b B | (b) = 0}. Then I is an ideal,

and B = B/I as usual is a -algebra consisting of equivalence classes [b] for b B (where

[b] = [c] iff the symmetric difference (b c) (c b)I). We can define a positive measure

on B by: ([b]) = (b). Thus, B, is a measure algebra, called the product measure

algebra of the Bi, is.

Let 2 be the basic measure algebra P(2), where is the measure: () = 0, ({0}) =

({1}) = 12 , and ({0, 1}) = 1. For any set I, let 2I denote the product measure algebra of

I copies of2. We can then force with 2I with the natural proviso: b is a stronger condition

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than c iff 0 < b c in 2I. This forcing obviously has the 1-c.c.

For I = , 2I is just the usual random real forcing and for I = 2I is the -random

real forcing. Let us denote them by Random and Random() respectively.

We consider the -algebras B i


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(n) | n < and an | n < so that for every n <

(a) n | n < is a cofinal in sequence

(b) an is a countable subset of [n, n+1)

(c) (n) is a measure over Bn+1 an respecting the ideal In+1 an, i.e. for every

X Bn+1 an (n)(X) = 0 iff X In+1

(d) n: Random Bn+1 an is a measure algebra isomorphism.

Denote Bn+1 an by B(n). Let us define a measure (n) over B(n). Thus for everyn < let us change the value (n)(n({0})) from 1/2 to 1

12n2

and those of n({1})

from 1/2 to 12n2 . Let

(n) be the measure obtained from (n) in such a fashion. Clearly,

such local changes have no effect on the set of measure zero. Namely, for every X B(n)

(n)(X) = 0 iff(n)(X) = 0.Define now the measure over B as the product measure of the measure algebras

B(n), (n) (n < ) together with all the rest, i.e.Bn+1 ([n, n+1)\an), n+1 ([n, n+1)\an) .

We claim that =

n


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every < I V2. So B V2. We will show the following claim which has a proof

similar to 2.7.

Claim 2.9. Random() does not embed into B in V1 and also in V2.

Proof: Notice that V2 and V1 have the same reals. So if is an embedding of Random()

into B in V2 then will be also such embedding in V1. Hence let us prove the claim for

V1.

Suppose otherwise. Let : Random() B witnessing embedding. Back in V let

us work with names. Let

be a name of and assume for simplicity that the empty

condition forces this.

Pick N and n | n < to be as in Claim 2.7 with

replacing C.We define sequences of conditions of P of {pn | n < } N of ordinals n | n 2 | the condition (1lRandom, ) is compatible with every element of G

(Q1)}.

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For , >2 let us write if the sequence extends the sequence . The following

two claims are obvious.

Claim 3.2.

is a Q1-name of a nonempty subset of>2 closed under initial segments

with no -maximal element and hence a tree.

Claim 3.3. Q the Cohen real is an -branch of .

Claim 3.4. There is no p Q1 and >2 such that p Q1 for every v >2 v

implies v

.

Proof: Suppose otherwise. Let p, be witnessing this. Then above p the forcing notion

Q1 is a complete subordering of Random. But it has to add a real. Hence it is isomorphic

to Random which is impossible by [Gi-Sh1].

Using -boundness of Q1, we can find p0 Q1 and a function h : such that

p0 Q1 for every n < and n2 there is v, v h(n)2 such that v /

. Let us

assume for simplicity that this p0 is just the weakest condition of Q1.

Let T = {T | T >2 is a tree and for every n < , n2 there is v such that

v h(n)2 and v / T}. Consider also Tm = {T m2 | T T} for m < . T can be viewed

as a tree if we identify it with

m2. Then, clearly,

Q1

is an branch of T .

Claim 3.5. Suppose that n < , q0 Random, n2. Then there are m < ,

q, v0, v1, t0, t1 such that

(a) q Random and q q0.

(b) v0, v1 >2

(c) t0, t1 m2 and t0 = t1.

(d) (q, vi)

m2 = ti for i < 2.

Proof: Find first some q q0 and v0 deciding

m2. Let t0 be the decided

value, i.e. (q, v0)

m2 = t0. By the Claim 3.4 there will be m < and v ,

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v m2\t0. Find some (q, v1) (q, v) deciding

m2. Let t1 be the forced value, i.e.

(q, v1)

m2 = t1. Since (q, v1) v1

, we have (q, v1) v = v1 m

. But

this means to = t1.

Claim 3.6. Suppose that n, k < and q0 Random. Then there are q Random,

m < , v, | n2, < k and t, |

n2, < k such that

(a) q q0

(b) m n

(c) for every 1, 2 n2, 1, 2 < k t1,1 = t2,2 iff (1, 1) = (2, 2)

(d) for every n2, < k

v, >2 , t, T

m and (q, v,)

m2 = t, .

Proof: Just use the previous claim enough times. Thus, first, we generate a tree of

k (2n + 1) possibilities for one n2 and then we repeat the argument of Claim 3.5 on

all s.

Claim 3.7. For every n < , k < , q Random and E > 0 there are m < ,

q q, {q | < } Random pairwise disjoint, v,,j |

n2, < , j < k and

t,,j | n2, < , j < k such that

(a) Lb(q) 1 E (Lb denotes the Lebesgue measure)

(b) q =

2 v,,j and (q, v,,j)

m2 = t,,j .

(d) for every <

t1,j1 = t2,,j2 iff (1, j2) = (2, j2) .

Proof: We define by induction qs using Claim 3.6. Thus ifqi | i is defined then we

apply Claim 3.6 to 2\i

qi. The process stops after we reach s.t. Lb(


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Claim 3.8. For every n < and E > 0 there are m, n m < and a function

H : Tm 2 such that for every n2 and i 2 we can find qi,, i, < , qi, | <

i,

and vi, | < i, such that for every i < 2 and < i,

(a) vi,

>

2

(b) qi,, qi, Random and qi, =

2, v

such that (q, v)

m

2 = t. Let {q,t, | < ,t } be a maximal antichain sub-

set of I,t. Let q,t =


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in H. It is 1, as the value is always 1 and is 1 12k . In order to prove the last

inequality, let us use {q | < } of Claim 3.7. Thus

Lbq,t | t T

m and H(t) = i

=


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Claim 3.9. Q is a Cohen real over V.

Proof: It is enough to show the following:

for every E > 0 and >2 the following holds:

for every j < large enough and >2 of the length > j there are q, q | < and

v | < such that

(a) q, q | < are in Random

(b) q =

length and 2j < E2 . Let be given. We

choose by induction on k [j, length ) a set ak and q | ak such that

(a) ak nk2 is nonempty.

(b) aj is a singleton extending

(c) ak+1( nk ak) and ak ak+1 (

).

(d) for every ak

(q, )

k1=j

H

m2

= ()

i.e. (q, )

[j,k 1) = [j,k 1)

(e) for aj q = 2

(f) for ak q | ak+1 is an antichain of Random above q and

{Lb(q) |

ak+1}Lb(q) 1

12k

.

There is no problem in caring on this induction. This completes the proof of and

hence also the theorem.

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References

[Fr] D. Fremlin, Real-valued measurable cardinals in Set Theory of the Reals

H. Judah ed., Israel Math. Conf. Proceedings (1993), 151-305.

[Fr2] D. Fremlin, Measure Algebras, in J.D. Monk ed., Handbook of Boolean

Algebras, North-Holland (1989), 876-980.

[Gi-Sh1] M. Gitik and S. Shelah, On simple forcing notions and forcing with ideals,

Israel J. Math. 68(2) (1989) 129-160.

[Gi-Sh2] M. Gitik and S. Shelah, More on simple forcing notions and forcing with

ideals Ann. of Pure and Appl. Logic 59 (1993), 219-238.

[Pr] K. Prikry, Ideals and powers of cardinals, Bull. AMS 81 (1975), 907-909.

[So] R. Solovay, Real-valued measurable cardinals, in: D. Scott, ed. Axiomatic

Set Theory, Proc. Symp. Pure Math. 13(1), (1970), 397-428.

[Sh1] S. Shelah, Cardinal and Arithmetic, Oxford Logic Guides 29, Oxford Sci-

ence Publications 1994.

[Sh430] S. Shelah, Further Cardinal Arithmetic, [Sh430]

[Sh460] S. Shelah, The Generalized Continuum Hypothesis revisited, [Sh460], to

appear.

[Sh580] S. Shelah, On Strong Covering, [Sh580].

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moti gitik and saharon shelah- more on real-valued measurable cardinals and forcing with ideals

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