Martin Goldstern, Haim Judah and Saharon Shelah- Strong Measure Zero Sets Without Cohen Reals

8/3/2019 Martin Goldstern, Haim Judah and Saharon Shelah- Strong Measure Zero Sets Without Cohen Reals

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STRONG MEASURE ZERO SETS

WITHOUT COHEN REALS

January 1991

Martin Goldstern1

Bar Ilan University

Haim Judah1

Bar Ilan University and U.C. Chile

Saharon Shelah1,2

Hebrew University of Jerusalem

ABSTRACT. If ZFC is consistent, then each of the following

are consistent with ZFC + 20 = 2:1. X IR is of strong measure zero iff |X| 1 + there

is a generalized Sierpinski set.

2. The union of 1 many strong measure zero sets is astrong measure zero set + there is a strong measure

zero set of size 2.

1 The authors thank the Israel Foundation for Basic Research, Israel Academy of Science.2 Publication 438


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Goldstern, Judah, Shelah: Strong measure zero sets without Cohen reals

0. Introduction

In this paper we continue the study of the structure of strong measure zero sets. Strong

measure zero sets have been studied from the beginning of this century. They were dis-

covered by E. Borel, and Luzin, Sierpinski, Rothberger and others turned their attention

to the structure of these sets and proved very interesting mathematical theorems about

them. Most of the constructions of strong measure zero sets involve Luzin sets, which

have a strong connection with Cohen reals (see [6]). In this paper we will show that this

connection is only apparent; namely, we will build models where there are strong measure

zero sets of size c without adding Cohen reals over the ground model.

Throughout this work we will investigate questions about strong measure zero sets under

the assumption that c = 20 = 2. The reason is that CH makes many of the questions

we investigate trivial, and there is no good technology available to deal with most of our

problems when 20 > 2.

0.1 Definition: A set X IR of reals has strong measure zero if for every sequence

i : i < of positive real numbers there is a sequence xi : i < of real numbers such

that

X

i


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0.4 Fact: Assume H is a dominating family, i.e., for all f there is h H suchthat n f(n) < h(n). Then:

(1) If has index H, then X is a strong measure zero set.

(2) IfX is a strong measure zero set, then there is a sequence with index Hsuch that X X .

0.5 Definition: A set of reals X IR is a GLuzin (generalized Luzin) set if for every

meager set M IR, X M has cardinality less than c. X is a generalized Sierpinski setif set if for every set M IR of Lebesgue measure 0, X M has cardinality less than c.

0.6 Fact: (a) If c is regular, and X is GLuzin, then X has strong measure zero.

(b) A set of mutually independent Cohen reals over a model M is a GLuzin set.

(c) If c > 1 is regular, and X is a GLuzin set, then X contains Cohen reals over L.

Proof: See [6].

0.7 Theorem: [6] Con(ZF) implies Con(ZFC + there is a GLuzin set which is not strongmeasure zero).

0.8 Theorem: [6] Con(ZF) implies Con(ZFC + c > 1 + X [IR]c, X a strong measurezero set + there are no GLuzin sets).

In theorem 0.16 we will show a stronger form of 0.8.

0.9 Definition:

(1) Let Unif(S) be the following statement: Every set of reals of size lessthan c is a strong measure zero set.

(2) We say that the ideal of strong measure zero sets is c-additive, or Add(S),

if for every < c the union of many strong measure zero sets is a strongmeasure zero set. (So Add(S) Unif(S).)

0.10 Remark: Rothberger ([13] and [12]) proved that the following are equivalent:

(i) Unif(S)

(ii) for every h : , for every F [

n h(n)]


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0.11 Theorem: If Unif(S) and d = c, then there exists a strong measure zero set of size

c.

We start the proof by proving the following

0.12 Fact: If d = c, then there is a set {fi : i < c} of functions in

such that for everyg , the set

{i < c : fi g}

has cardinality less than c.

Proof of the fact: We build fi : i < c by transfinite induction. Let = {gj : j < c}.

We will ensure that for j < i, fi < gj . This will be sufficient.

But this is easy to achieve, as for any i, the family {gj : j < i} is not dominating, so thereexists a function fi such that for all j < i, for infinitely many n, fi(n) > gj(n).

This completes the proof of 0.12.

0.13 Proof of 0.11: Using d = c, let fi : i < c be a sequence as in 0.12. Let F :

[0, 1] Q be a homeomorphism. (Q is the set of rational numbers.) We will show that

X := {F(fi) : i < c} is a strong measure zero set.

Let n : n < be a sequence of positive numbers. Let {rn : n } = Q. ThenU1 :=

n(rn 2n, rn + 2n) is an open set. So K := [0, 1] U1 is closed, hence compact.

As K rng(F), also F1(K) is a compact set. So for all n the projection ofF1(K)

to the nth coordinate is a compact (hence bounded) subset of , say g(n). So

F1K {f : f g}

Let Y := X U1 K. Then Y F(F1(K)) {F(fi) : fi g} is (by assumption onfi : i < c) a set of size < c, hence has strong measure zero. So there exists a sequence ofreal numbers xn : n < such that Y U2, where

U2 :=

n

(xn 2n+1, xn + 2n+1)

and X U1 U2. So X is indeed a strong measure zero set.

In section 2 we will build models where Add(S) holds and the continuum is bigger than1 without adding Cohen reals. First we will show in 3.4:

0.14 Theorem: If ZFC is consistent, then

ZFC + c = 2 + S = [IR]1 + there are no Cohen reals over L

is consistent.

Note that c = 2 and S = [IR]1 implies

(1) Add(S). (Trivially)

(2) b = d = 1. (By 0.11)

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The same result was previously obtained by Corazza[3]. In his model the nonexistence of

strong measure zero sets of size c is shown by proving that every set of size c can be mapped

uniformly continuously onto the unit interval (which is impossible for a strong measure

zero set). Thus, the question arises whether is possible to get a model of S = [IR] b + Add(S) + no real is Cohen over Lis consistent.

(Note that by 0.11, d = c + Add(S) implies that there is a strong measure zero set of

size c.)

0.17 Notation: We use standard set-theoretical notation. We identify natural numbers

n with their set of predecessors, n = {0, . . . , n 1}. AB is the set of functions from A intoB, A


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We call a set H dominating, if g h H g < h.

M is the ideal of meager subsets of IR (or of 2). S is the ideal of strong measure zero sets.For any ideal J P(IR), Add(J) abbreviates the statement: The union of less than cmany sets in J is in J. Cov(J) means that the reals cannot be covered by less than cmany sets in J.

Iff is a function, dom(f) is the domain off, and rng(f) is the range off. For A dom(f),f|A is the restriction of f to A. For 2


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1. A few well known facts

We collect a few more or less well known facts about forcing, for later reference.

1.1 Definition: An ultrafilterU on is called a P-Point, if for any sequence An : n

of sets in U there is a set A in U that is almost contained in every An (i.e., n A An isfinite).

1.2 Definition: For any ultrafilter U on , we define the P-point game G(U) as follows:

There are two players, IN and NOTIN. The game consists

of many moves.

In the n-th move, player NOTIN picks a set An U, and playerIN picks a finite set an An.

Player IN wins if after many moves,

n an U.

We write a play (or run) of G(U) as

A0; a0 A1; a1 A2; . . . .

It is well known that an ultrafilter U is a P-point iff player NOTIN does not have a winning

strategy in the P-point game.

For the sake of completeness, we give a proof of the nontrivial implication (which isall we will need later):

Let U be a P-point, and let be a strategy for player NOTIN. We will construct a run ofthe game in which player NOTIN followed , but IN won.

Let A0 be the first move according to . For each n, let An be the set of all responses of

player notin according to in an initial segment of a play of length n in which playerIN has played only subsets of n:

An := {Ak : k n, A0; a0 A1; . . . ; ak1 Ak is aninitial segment of a play in which NOTIN

obeyed , and a0, . . . , ak1 n}

Note that A0 = {A0}, and for all n, An is a finite subset of U.

As U is a P-point, there is a set X U such that for all A

n An, X A is finite.

Let X A0 n0, and for k > 0 let nk satisfy

nk > nk1 and A Ank1 X A nk

Either

k[n2k, n2k+1) U, or

k[n2k+1, n2k+2) U.

Without loss of generality we assume

k[n2k, n2k+1) U.

Now define a play A0; a0 A1; a1 A2; . . . of the game G(U) by induction as follows:

A0 is given.

Given Aj , let aj := Aj [n2j , n2j+1) and let Aj+1 be s response to aj.

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Then as a0, . . . , aj1 n2j , we have X Aj n2j for all j. Therefore for all j wehave X [n2j , n2j+1) (Aj n2j) [n2j, n2j+1) = Aj [n2j , n2j+1) = aj . So

j aj

X

j[n2j , n2j+1) U.

Thus player IN wins the play A0; a0 A1; a1 A2; . . . in which player NOTIN obeyed.

1.3 Definition: We say that a forcing notion Q preserves P-points, if for every P-point

ultrafilter U on , | QU generates an ultrafilter, i.e. | Q x P() u U (u x or u x).

[9] defined the following forcing notion:

1.4 Definition: Rational perfect set forcing, RP is defined as the set of trees p , |

V = .

As a consequence, | 2If X , |X| 1, then there is < 2 such that X V.

We also recall the following facts about iteration of proper forcing notions:

1.8 Lemma: Assume CH, and let P, Q : < 2 be a countable support iterationsuch that for all < 2, | Q is a proper forcing notion of size c.

Then

(1) < 2: | c = 1. (see [14, III 4.1])

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(2) | 2c 2. (This follows from 1.7 and (1))(3) For all 2, P is proper [14, III 3.2] and satisfies the 2-cc. (See [14,

III 4.1])

(4) | 2V1 = 1. (See [14, III 1.6])

In [2, 4.1] the following is proved:

1.9 Lemma: Assume P, Q : < 2 is as in 1.8, and for all < 2:

| Q preserves P-points.

Then for all 2, P preserves P-points.

1.10 Definition: We say that a forcing notion Q is -bounding, if the set of old

functions is a dominating family in the generic extension by Q, or equivalently,

| Qf

g

V n f(n) < g(n)[14, V 4.3] proves:

1.11 Lemma: Assume P, Q : < 2 is as in 1.8, and for all < 2:

| Q is-bounding and -proper.

Then for all 2, P is -bounding.

(We may even replace -proper by proper, see [14], [4])

The following is trivial to check:

1.12 Fact: Assume Q is a forcing notion that preserves P-points or is -bounding. Then

| QThere are no Cohen reals over V

1.13 Definition: A forcing notion P is strongly -bounding, if there is a sequence

n : n of binary reflexive relations on P such that for all n :

(1) p n q p q.(2) p n+1 q p n q.

(3) Ifp0 0 p1 1 p2 3 , then there is a q such that n pn+1 n q.(4) Ifp | is an ordinal, and n , then there exists q n p and a finite

set A Ord such that Q| A.

1.14 Definition: (1) If P, Q : < is an iteration of strongly-bounding forcing

notions, F finite, n , p, q P, we say that p F,n q iff p q and F q|| p() n q().

(2) A sequence pn, Fn : n is called a fusion sequence if Fn : n is an increas-

ing family of finite subsets of , pn : n is an increasing family of conditions in P,

n pn n,Fn pn+1 and

n dom(pn)

n Fn.

Note that 1.13 is not a literally a strengthening of Baumgarters Axiom A (see [1]), as

we do not require that the relations n are transitive, and in (2) we only require pn+1 n qrather than pn+1 n+1 q. Nevertheless, the same proof as in [1] shows the following fact:

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1.15 Fact:

(1) If the sequence pn, Fn : n is a fusion sequence, then there exists a

condition q P such that for all n , pn+1 Fn,n q.

(2) If is a P-name of an ordinal, n , F P finite, then for all p thereexists a condition q n,F p and a finite set A of ordinals such that q| A.

(3) If X is a P-name of a countable set of ordinals, n , F P finite,then for all p there exists a condition q n,F p and a countable set A of

ordinals such that q| X A.

The next fact is also well known:

1.16 Fact: Let B be the random real forcing. Then B is strongly -bounding.

[Proof: Conditions in B are Borel subsets of [0, 1] of positive measure, p q iff p q.

We let p n q iff p q and (p q) 10n1(p), where is the Lebesgue measure.

Then if p0 0 p1 1 , letting q :=

npn we have for all n, all k n, (pk pk+1) 10k1(pk) 10k1(pn), so (pn q) 10n1 + 10n2 + 2 10n1(pn).

In particular, (q) 0.8 (p0), so q is a condition, and q n1 pn for all n > 0.

Given a name , an integer n and a condition p such that p| is an ordinal, let A be

the set of all ordinals such that [[ = ]] p has positive measure ([[]] is the booleanvalue of the statement , i.e. the union of all conditions forcing ). Since

A ([[ =

]] p) = (p) there is a finite subset F A such that letting q := p

A[[ = ]] wehave (q) (1 10n1)(p). So q n p and q| F.]

We will also need the following lemma from [17, 5, Theorem 9]:

1.17 Lemma: Every stationary S 2 can be written as a union of 2 many disjointstationary sets.

Finally, we will need the following easy fact (which is true for any forcing notion Q)

1.18 Fact: If f is a Q-name for a function from to , | Q f / V, and r0, r1 are any twoconditions in Q, then there are l , j0 = j1, r0 r0, r

1 r1 such that r

0| f

(l) = j0,

r1| f(l) = j1.

[Proof: There are a function f0 and a sequence r0 = r0 r1 of conditions in Q such

that for all n, rn| f

|n = f0|n. Since r1| f

/ V, r1| l f

(l) = f0(l). There is a condition

r1 r1 such that for some l and some j1 = f0(l), r1| f

(l) = j1. Let j0 := f0(l), and

let r0 := rl+1.]

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2 H-perfect treesIn this section we describe a forcing notion P TH that we will use in an iteration in the

next section. We will prove the following properties of P TH:

(a) P TH is proper and

-bounding.(b) P TH preserves P-points.

(c) P TH does not increase strong measure zero sets defined in the ground

model.

(d) P TH makes the reals of the ground model (and hence, by (c), the union of

all strong measure zero sets defined in the ground model) a strong measure

zero set.

2.1 Definition: For each function H with domain satisfying n 1 < |H(n)| < ,we define the forcing P TH, the set of H-perfect trees to be the set of all p satisfying

(A) p 1}

(2) Theheight

of a node p P TH is the number of splitting nodes strictlybelow :

htp() := |{ : split(p)}|

(Note that htp() ||.)(3) For p P TH, k , we let the kth splitting level of p be the set of splitting

nodes of height k.

splitk(p) := { split(p) : htp() = k}

(Note that split0(p) = {stem(p)}.)(4) For u , we let

splitu

(p) :=ku

splitk(p)

2.3 Remarks:

(i) Since H(n) is finite, (3) just means that either has a unique successor

i, or succp() = H(||).)(ii) Letting H(n) = |H(n)|, clearly P TH is isomorphic to P TH (and the

obvious isomorphism respects the functions htp(), p,k splitk(p),etc)

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2.4 Remark: If we let H(n) = for all n, then 2.1(A)(D) define RP, rational perfect

set forcing. The definitions in 2.2 make sense also for this forcing. Since we will not use

the fact that H(n) is finite before 2.12, 2.52.11 will be true also for RP.

2.5 Fact: Let p, q P TH, n , ,

htp() = n, which is impossible. Hence .

(5) follows easily from (1).

2.6 Definition: For p, q P TH, n , we let

(1) p q (q is stronger than p) iff q p.

(2) p n q iff p q and splitn(p) q. (So also splitk(p) q for all k < n.)

2.7 Fact: If p n q, n > 0, then stem(p) = stem(q).

2.8 Fact: Assume p, q P TH, n , p n q.

(0) For all q, htq() htp().

(1) For all k n, splitk(p) split(q).

(2) For all k < n, splitk(p) = splitk(q).

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(3) Ifp n q n r, then p n r.

Proof: (0) is clear.

(1): Let splitk(p) for some k < n, then by 2.5(4) there is a , splitn(p) q,

so q.(2): Let splitk(p), then split(q). Clearly htq() htp() = k. Using (1) induc-tively, we also get htq() k.

(3): Let splitn(p). So q, htq() htp() = n. By 2.5(4), there is splitn(q),

. As r, r.

2.9 Definition and Fact: If p0 1 p1 2 p2 3 are conditions in P TH, then we call

the sequence pn : n < a fusion sequence. If pn : n < is a fusion sequence, then

(1) p :=

n pn is in P TH(2) For all n: pn n+1 p.

2.10 Fact:

(1) If p P TH, then p[] P TH, and p p[]. (See 0.19.)

(2) Ifp q are conditions in P TH, q, then p[] q[].

2.11 Fact: If for all splitn(p), q p[] is a condition in P TH, then

(1) q :=

splitn(p)

q is in P TH,

(2) q n p

(3) for all splitn(p), q[] = q.

2.12 Fact: If n , p P TH, then splitn(p) is finite.Proof: This is the first time that we use the fact that each H(n) is a finite set: Assume

that the conclusion is not true, so for some n and p, splitn(p) is infinite. Then also

T := {|k : splitn(p), k ||} p

is infinite. As T is a finitely splitting tree, there has to be an infinite branch b T. By2.5(2), there is b T, htp() > n. This is a contradiction to 2.5(1).

2.13 Fact: P TH is strongly-bounding, i.e.:

If

is a P TH-name for an ordinal, p P TH, n , then there exists a finite set A of

ordinals and a condition q P TH, p n q, and q| A.

Proof: Let C := splitn(p). C is finite. For each node C, let q p[] be a condition

such that for some ordinal q| = . Now let

q :=

C

q and A := { : C}

Since any extension of q must be compatible with some q[] (for some C), q| A.

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2.14 Corollary: P TH is proper (and indeed satisfies axiom A, so is -proper for any

< 1) and-bounding. Moreover, ifn , p P TH, a name for a set of ordinals,

then there exists a condition q n p such that

(1) Ifp | is finite, then there is a finite set A such that q| A.(2) Ifp | is countable, then there is a countable set A such that q| A.

Proof: Use 2.13 and 2.9.

Similarly to 2.13 we can show:

2.15 Fact: Assume that is a RP-name for an ordinal, p RP, n .

Then there exists a countable set A of ordinals and a condition q P TH, p n q, andq| A.

Proof: Same as the proof of 2.13, except that now the set C and hence also the set A may

be countable.

2.16 Fact: RP is proper (and satifies axiom A). Proof: By 2.15 and 2.9.

2.17 Definition: Let G P TH be a V-generic filter. Then we let g

be the P TH-name

defined by

g

:=

pG

p

We may write g

H or g

PTH for this name g. IfP TH is the th iterand Q in an iteration,

we write g

for g

H.

2.18 Fact: PTH forces that

(0) g is a function with domain ,(1) n g

(n) H(n).

(2) For all f V, if n f(n) H(n) then n f(n) = g

(n).

Furthermore, for all p P TH,

(3) p| { g

|n : n } is a branch through p.

(4) p| kn g

|n splitk(p)

Proof: (0) and (2) are straightforward density arguments. (1) and (3) follow immedaitely

from the definition of g

. (4) follows from (3) and 2.5(2), applied in VPTH .

2.19 Remark: Since Unif(S) is equivalent tofor every H : , for every F [

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We will show a stronger result in 3.3: If P := P2 is the limit of a countable support

iteration P, Q : < 2, where many Q are of the form P TH for some H, then

some bookkeeping argument can ensure that VP |= Add(S).

Since P TH is

-bounding, it does not add Cohen reals. The same is true for a countablesupport iteration of forcings of the form P TH. However, in 3.9 we will have to consider a

forcing iteration in which some forcing notions are of the form P TH, but others do add an

unbounded real. To establish that even these iterations do not add Cohen reals, we will

need the fact that the forcing notion P TH preserves many ultrafilters.

2.20 Definition: Let Q be a forcing notion, x a Q-name, p Q, p| x . We say thatx is an interpretation of x (above p), if for all n there is a condition pn p suchthat pn| x n = x

n.

2.21 Fact: Assume Q, p, x are as in 2.20. Then

(1) There exists x such that x is an interpretation of x above p.(2) Ifp p and x is an interpretation of x above p

, then x is an interpretation

of x above p.

2.22 Lemma: P TH preserves P-points, i.e.: If U V is a P-point ultrafilter on , then

| PTHU generates an ultrafilter.

Proof: Assume that the conclusion is false. Then there is a P TH-name for a subset of and a condition p0 such that

p0| PTHx U : |x | = |( x) | = 0.

For each p P TH we choose a set (p) such that

(p) is an interpretation of above p.

If there is an interpretation of above p that is an element of U, then

(p) U.

Note that if (p) U and p p, then also (p) U, since (by 2.21(2)) we could have

chosen (p) := (p). Hence either for all p (p) U, or for some p1 p0, all p p1,(p) / U. In the second case we let be a name for the complement of , and let

(p) = (p). Then (p) U for all p p1. Also, (p) is an interpretation of abovep.

So wlog for all p p1, (p) U for some p1 P TH, p1 p0.

We will show that there is a condition q p1 and a set a U such that q| a .Recall that as U is a P-point, player NOTIN does not have a winning strategy in theP-point game for U (see 1.2).

We now define a strategy for player NOTIN. On the side, player NOTIN will construct a

fusion sequence pn : n < and a sequence mn : n < of natural numbers.

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p0 is given.

Given pn, we let

An =

splitn+1(pn)

(pn[])

This set is in U. Player IN responds with a finite set an An. Let mn := 1 + max(an).

For each splitn+1(pn) there is a condition q pn[] forcing mn = (pn

[]) mn,so in particular

q| an mn

Let pn+1 =

splitn+1(pn)

q.

Then

() pn+1 n+1 pn and pn+1| an

This is a well-defined strategy for player NOTIN. As it is not a winning strategy, there

is a play in which IN wins. During this play, we have constructed a fusion sequence

pn : n < . Letting a :=

n an, q :=

npn, we have that a U, p0 q P TH (by 2.9),and q| a (by ()), a contradiction to our assumption.

The following facts will be needed for the proof that if we iterate forcing notions P TH with

carefully chosen functions H, then we will get a model where the ideal of strong measure

zero sets is c-additive.

2.23 Fact and Definition: Assume p P TH, u is infinite, v = u. Thenwe can define a stronger condition q by trimming p at each node in splitv(p). (See

2.2(4).) Formally, let = i : splitv(p) be a sequence satisfying i H(||) for all

splitv(p).

Then

p := { p : n dom() : If |n splitv(p), then (n) = i|n}

is a condition in P THProof: Let q := p. q satisfies (A)(B) of the definition 2.1 of P TH. The definition of p

immediately implies:(1) If splitv(p) q, then succq() = {i}.

(2) If splitu(p) q, then succq() = succp() = H(||).

(3) If q split(p), then p split(p), so succq() = succp() is a singleton.

Note that split(p) = splitu(p) splitv(p), so (1)(3) cover all possible cases for q.

So q also satisfies 2.1(C).

From (1)(3) we can also conclude:

(4) For all q: succq() = .

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To show that q P TH, we still have to check condition 2.1(D). So let q. Since u isinfinite, there is k u, k > ||. By (4), there is an infinite branch b q containing . By

2.5(2) there is b, htp() = k. Then , and split(q).

2.24 Fact: p | g & splitv

(p) g(||) = i (where g is a name for thegeneric branch defined in 2.18).

Proof: p| g

p and succp() = {i}.

To simplify notation, we will now assume that for all n, H(n) . (If H(n) are just

arbitrary finite sets as in 2.1, then we could prove analogous statements, replacing 0 and

1 by any two elements 0n = 1n of H(n).)

2.25 Definition: Let f

be a P TH-name for a function from to . We say that f

splits

on p, k if for all splitk(p) there are l and j1 = j0 such that

p[0]| f

(l) = j0

p[1]| f

(l) = j1

2.26 Remark: If f

splits on p, k, and q k+1 p, then f

splits on q, k.

(Proof: splitk(p) = splitk(q), and for splitk(p), p[i] q[

i].)

2.27 Lemma: If p| f

/ V, k , then there is q k+1 p such that f

splits on q, k.

Proof: For splitk(p), i {0, 1} we let i be the unique element of splitk+1(p) satisfying

i i.By 1.18, for each splitk(p) we can find conditions q0 p

[0], q1 p[1] and integers

l, j,0 = j,1 such that q0 | f(l) = j0, q1 | f

(l) = j1. If splitk+1(p) is not of the

form 0 or 1 for any splitk(p), then let q := p[].

By 2.11, q :=

splitk+1(p)

q is a condition, q k+1 p, and q = q[] for all splitk+1(p).

We finish the proof of 2.27 by showing that f

splits on q, k: Let splitk(p) = splitk(q).Then q[

0] = q[0] = q0 , so q[0]| f

(l) = j,0. Similarly, q

[1]| f

(l) = j,1.

2.28 Lemma: If p| f

/ V, then there is q p, f

splits on q, k for all k.

Proof: By 2.27, 2.26 and 2.9 (using a fusion argument).

2.29 Lemma: Assume Q is a strongly -bounding forcing notion. Let f

be a Q-name

for a function, p a condition, n , p| f

/ V. Then there exists a natural numer k such

that

() for all k2 there is a condition q n p, q| f

/ [].

We will write kp,n or kf

,p,n for the least such k. Note that for any k kp,n, () will also

hold.

Proof: Assume that this is false. So for some f

, n0, p0,

() k k k2 : (q n0 p0 q| f

/ [k])

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Let

T := {k|l : l k, k }

T is a finitely branching tree ( 2) of infinite height, so it must have an infinite branch.

Let f 2 be such that {f|j : j } T.Since f V but p0| Q f

/ V, there exists a name m of a natural number such thatp0| f|m = f

|m. Let q n0 p0 be such that for some m , q| m < m

.

Claim: For some k, q| f

/ [k]. This will contradict ().

Proof of the claim: We have q| f

|m = f|m. Since f|m T, there is a k m suchthat f|m = k|m. Hence q| f

|m = f|m = k|m, so q| f

/ [k|m]. But then also

q| f

/ [k].

This finishes the proof of the claim and hence of the lemma.

2.30 Lemma: Assume that Q is a strongly -bounding forcing notion, H is a dominating

family in V, and =

h

: h H has index H. Then

| Q

hH

k

[h(k)] V

Proof: Assume that for some condition p and some Q-name f

,

p| f

/ V & f

hH

n

[h(n)].

We will define a tree of conditions such that along every branch we have a fusion sequence.

Specifically, we will define an infinite sequence ln : n of natural numbers, and foreach n a finite sequence

p(0, . . . , n1) : 0 02, . . . , n1

ln12

of conditions satisfying

(0) p() = p

(1) For all n: 0 02, . . . , n1 ln12 : ln kp(0,...,n1),n.

(2) For all n: 0 02, . . . , n1 ln12 n ln2

(a) p(0, . . . , n1) n p(0, . . . , n1, n).(b) p(0, . . . , n1, n)| f

/ [n].

Given p(0, . . . , n1) for all 0 02, . . . , n1

ln12, we can find ln satisfying condition

(1). The by the definition ofkp(0,...,n1),n we can (for all n ln2) find p(0, . . . , n1, n).

Now let h H be a function such that for all n, h(n) > ln. Define a sequence n : n by n :=

h(n)|ln, and let pn := p(0, . . . , n). Then p p0 0 p1 1 , so there

exists a condition q extending all pn. So for all n, q| f

/ [n]. But then also for all n,q| f

/ [h(n)], a contradiction.

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Lemma 2.30 will be needed later to show that if we iterate focings of the form P TH together

with random real forcing, after 2 many steps we obtain no strong measure zero sets of

size 2. The proof (in 3.4) would be much easier if we could omit strongly from the

hypothesis of 2.30, i.e., if we could answer the following question positively:

2.31 Open Problem: Assume H is a dominating family (or even wlog H = ),and has index H. Let Q be an -bounding forcing notion. Does this imply

| Q

hH

n

[h(n)] V ?

2.32 Fact: Assume h : {0}, H(n) = h(n)2. Let H be a dominating

family, and let have index H. Let g be the name of the generic function added by

P TH .

Then

| PTH h Hk

[h(k)] n

[ g(n)]

Proof: Assume not, then there is a condition p such that

() p| h H

k

[h(k)]

n

[ g(n)]

Let h H be a function such that k split2k+1(p) h(||) h(k). This function

h will be a witness contradicting ().For split2k+1(p) let i succp() = H

(||) = h(||)2 be defined by i :=

h(k)|h(||).

(Note that h

(k) h(k)

2 and h(k) h

(||).)Let := i : split2k+1(p), k and let q := p.Then q| nk ( g

|n split2k+1(p) g

(n) = i g|n

h(k)) by 2.24.

Since also q| kn g

|n split2k+1(p), we get q| kn [k(k)] [ g

(n)]. This contra-

dicts ().

3 Two models of Add(S).Recall that S12 := { < 2 : cf() = 1}.

3.1 Lemma: Let P, Q : < 2 be an iteration of proper forcing noitions as in 1.8,

p P2 , A a P2-name. If p| A is a strong measure zero set, then there is a closedunbounded set C 2 and a sequence : C S12 such that each is a P-name,

and

p| 2 has index V and A

hV

n

[h(n)]

Proof: Let c be a P2-name for a function from 2 to 2 such that for all < 2,

| 2h V

h Vc() : n

h(n) h(n)2 & A

n

[h(n)]

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(Why does c exist? Working in V[G2 ], note that there are only 1 many functions in V, and for each such h there is a h as required in


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Thus

V2 |= A X

n

[h0 (n)]

n

[g(n)]

and we are done.

3.3 Corollary: Assume P2 is as above. Then | P2Add(S).

Proof: Let Ai : i 1 be a family of strong measure zero sets in V2 . To each i we can

associate a closed unbounded set Ci as in 3.2. Let C :=

i1

Ci, then also C is closed

unbounded, and for all C S12 ,

i1

Ai

n

[g(n)]. Again by 3.2,

i1

Ai is a strong

measure zero set.

Our first goal is to show that Unif(S) does not guarantee the existence of a strong measure

zero set of size c. Clearly the model for this should satisfy d = 1 (if c = 2), so we willconstruct a countable support iteration of -bounding forcing notions.

3.4 Theorem: If ZFC is consistent, then

ZFC + c = 2 + S = [IR]1 + no real is Cohen over L

+ there is a generalized Sierpinski set

is consistent.

Proof: We will start with a ground model V0 satisfying V = L. Let H := ( {0}) L =

{h : < 1}, and let H(n) = h(n)2.

Let S : < 1 be a family of disjoint stationary sets { < 2 : cf() = 1}.

Construct a countable support iteration P, Q : < 2 satisfying

(1) For all even < 2:

| PFor some h : {0}, letting H(n) =h(n)2, Q = P TH.

(2) If S, then | Q = P TH .(3) For all odd < 2:

| P Q = random real forcing.

By 1.11 (or as a consequence of 1.15), P2 is-bounding, so |

2H is a dominating

family. By 1.8(3) and 1.6 the assumptions of 3.3 are satisfied, so | 2Add(S). Also,

| 2c = 2 and there are no Cohen reals over L. Letting X be the set of random reals

added at odd stages, X is a generalized Sierpinski set: Any null set H V2 is coveredby some G null set H

that coded in some intermediate model. As coboundedly many

elements of X are random over this model, |H X| |H X| 1.

To conclude the proof of 3.4, we have to show

V2 |= If X IR is of strong measure zero, then |X| < c.

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Since H is a dominating family, by 0.4 it is enough to show that in V2 the following holds:

If has index H, then |X| 1.

We will show: If V has index H, then X V. (This is sufficient, by 1.7.)

Assume to the contrary that G2 is a generic filter, V, and in V[G2 ] there is > ,f V


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We will define sequences pn : n < ,

Fn : n < ,

n : n < ,

s

n : n pin : n , i {0, 1},

such that the following hold: For each n, pn, p0n, p

1n are conditions in P, n is an even

ordinal < , Fn is a finite subset of n, n is an integer, and sn is a Pn-name for anelement of 0 we

will have:

(1) pn1 Fn,n pn.

(2) Fn n, Fn1 Fn+1,

k dom(pk)

k Fk.

(3) n1 Fn.(4) pn|n| sn = stem(pn(n)) = stem(pn1(n))

(5) For i {0, 1}, pin = pn pn(n)

[sni]

.(6) pn|n| nl < n j0 = j1 i {0, 1} : p

in| n, f

(l) = ji.

Note that (5) implies:

(5) pn|n| pin P/Gn ,

and (6) implies

(6) For all n2: pn|n| i {0, 1} : pin| n, f|n =

[Proof of (6) (6): In Vn , let i {0, 1} be such that (l) = ji, where l is as in (6).]Finally, let q =

npn. Then q|n| stem(pn(n)) = stem(q(n)) = sn by (1), (3) and (4)

and 2.7. Let h H be a function such that for all n, n < h(n). So for all n, h

(n)|nis a well-defined member of n2.

For each n, let in be a Pn-name of an element of {0, 1} such that

(6) pn|n| pinn | f

|n = h

(n)|n

Now define a condition q as follows: For / {n : n }, q() = q(), and

q(n) = q()[snin

]

(This is a Pn-name.)

Claim: q q p (this is clear) and q| f

/

n

[h

(n)].

To prove this claim, let G P be a generic filter containing q, and assume f := f

[G] is

in [h

(n)]. Let in := in[Gn ]. Now q G implies pn G, so in particular pn|n Gn .

Note that stem(q(n)) = stem(pn(n)) = sn, so q G &pn G implies p

inn G. Also,

by (6) we have q| f

/

n[sf(n)], a contradiction.

This finishes the proof of 3.5 modulo the construction of the sequences pn, Fn, etc.

First we fix enumerations dom(r) = {mr : m } for all r P. We will write mn for

mpn .

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Assume pn1 is given. Let Fn := n (Fn1 {mk : k < n, m < n} {n1}). This willtake care of (2) and (3).

To define pn, first work in V[Gn ], where pn1|n Gn .

We let sn := stem(pn1(n)).

We let r0 := pn1 pn1(n)[sn0], and r1 := pn1 pn1(n)[

sn1].

By 1.18, we can find l, j0 = j1 and r0, r1 such that r

i ri, and r

i| f

(l) = ji.

We now define a condition r P/Gn as follows:

r|n = pn1|n.

r(n) = r0(n) r

1(n)

{pn1(n)

[sni]

: i succpn1(n)(sn) {0, 1}}.

(So stem(r(n)) = sn.)

If dom(pn1) dom(r0) dom(r1) and > n, we let r() be a P -name

such that

pn1|n | n

| n,

For i in {0, 1}: If sni gn , then r() = ri(),

and if gn extends neither sn0 nor sn

1,

then r() = pn1()

(We write gn for gQn , the branch added by the forcing Qn .)

This is a condition in P/Gn . Note that we have the following:

(i) stem(pn(n)) = sn = stem(pn1(n))

(ii) For i {0, 1}, r ri(n) ri.

(iii) r n pn1.Coming back to V, we can find names r, . . . , such that the above is forced by pn1|n.Now let r be a condition in Pn satisfying the following:

(a) r Fn,n pn1|n.

(b) For some countable set A , r| dom( r) A.

(c) For some n , r| l < n.

We can find a condition r satisfying (a)(c) by 1.15.

Finally, let pn := r r. So pn|n = r.

And let pin be defined by (5).

Why does this work?

First we check (1): pn1|n Fn,n pn|n by (a), and pn1 pn, because pn|n| pn =

r r r pn1 (by (iii)). So pn1 Fn,n pn.

(2) and (3) are clear.

Proof of (4): pn|n| stem(pn(n)) = stem((r r)(n)) = stem( r(n)) = sn.

(6): Let Gn be a generic filter containing pn|n. Work in V[Gn ]. We write r for r[Gn ],etc.

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We want to show p0n| n f(l) = j0. (p

1n| n f

(l) = j1 is similar.) As r0| n f

(l) = j0, it

is enough to see p0n r0.

First we note that p0n pn pn1. Also p0n = pn pn(n)

[sn0] pn = r r r.

Finally, p0n(n) = r(n)

[sn0]

= r

0(n).So p0n = p

0n r

0(n) r r

0(n) r

0, and we are done.

Our next model will satisfy

() Unif(S) + d = c = 2.

This in itself is very easy, as it is achieved by adding 2 Cohen reals to L. (Also Miller

[10] showed that Unif(S) + c = 2 + b = 1 is consistent.)Our result says that we can obtain a model for () (and indeed, satisfying Add(S)) without

adding Cohen reals. In particular, () does not imply Cov(M).3.9 Theorem: Con(ZFC) implies

Con(ZFC + c = d = 2 > b + Add(S) + no real is Cohen over L)Proof (sketch): We will build our model by a countable support iteration of length 2 where

at each stage we either use a forcing of the form P TH, or rational perfect set forcing. A

standard bookkeeping argument ensures that the hypothesis of 3.3 is satisfied, so we get

| 2Add(S). Using rational perfect set forcing on a cofinal set yields | 2d = c = 2.

Since all P-point ultrafilters from V0 are preserved, no Cohen reals are added.

Proof (detailed version): Let { < 2 : cf() = 1}

.First we claim that there is a countable support iteration P, Q : < 2 and a sequenceof names H

: < 2 : < 1 such that

(1) For all < 2, all < 1, H is a P-name.

(2) For all < 2, | {H : < 1} =

( {0, 1}).(3) For all < 2: If /


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which in turn is a consequence of 1.7.

So by 3.3, V2 |= Add(S).

Let G2 P2 be a generic filter, V2 = V[G2 ].

Again by 1.7, every H V2 of size 1 is a subset of some V, < 2, so Hcannot be a dominating family, as rational perfect set forcing Q+1 will introduce a real

not bounded by any function in H V V+1. Hence d = c = 2.

Finally, any P-point ultrafilter from V is generates an ultrafilter in V2 , so there are no

Cohen reals over V.

This ends the proof of 3.9.

REFERENCES.

[1] J. Baumgartner, Iterated forcing, in: Surveys in set theory (A. R. D. Mathias,editor), London Mathematical Society Lecture Note Series, No. 8, Cambridge

University Press, Cambridge, 1983.

[2] A. Blass and S. Shelah, There may be simple P1- and P2-points, and the Rudin-

Keisler order may be downward directed, Annals of pure and applied logic, 33

(1987), pp.213243.

[3] P. Corazza, The generalized Borel Conjecture and strongly proper orders, Transac-

tions of the American Mathematical Society, 316 (1989), pp.115140. .

[4] M. Goldstern, H. Judah, On Shelahs preservation theorem, preprint.

[5] H. Judah, S Shelah, H. Woodin, The Borel Conjecture, Annals of pure and applied

logic, to appear.

[6] H. Judah, Strong measure zero sets and rapid filters, Journal of Symbolic logic, 53

(1988), pp.393402.

[7] K. Kunen, Set Theory: An Introduction to Independence Proofs,

[8] A. Miller, Mapping a set of reals onto the reals, Journal of Symbolic logic, 48 (1983),

pp.575584.[9] A. Miller, Rational Perfect Set Forcing, in: Axiomatic Set Theory, Boulder, Co

1983, 143159, Contemporary Math 31, AMS, Providence RI 1984.

[10] A. Miller, Some properties of measure and category, Transactions of the American

Mathematical Society, 266,1 (1981), pp.93114.

[11] J. Pawlikowski, Power of transitive bases of measure and category, Proceedings of

the American Mathematical Society, 93 (1985), pp.719729.

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[12] Rothberger, Sur des families indenombrables de suites de nombres naturels et

les problemes concernant la proprietee C, Proc. Cambr. Philos. Soc., 37 (1941),

pp.109126.

[13] Rothberger, Eine Verscharfung der Eigenschaft C, Fundamenta Mathematicae, 30(1938), pp.5055.

[14] S. Shelah, Proper Forcing, Lecture Notes in Mathematics Vol. 942, Springer Verlag.

[15] S. Shelah, Proper and Improper Forcing, to appear in Lecture Notes in Mathematics,

Springer Verlag.

[16] R. Solovay and S. Tennenbaum, Iterated Cohen extensions and Souslins problem,

Annals of Mathematics, 94 (1971), pp.201245.

[17] R. Solovay, Real valued measurable cardinals, in: Axiomatic Set Theory, Proc.

Symp. Pure Math. 13 I (D. Scott, ed.), pp.397428, AMS, Providence RI, 1971.

Martin GOLDSTERN [email protected]

Haim JUDAH [email protected]

Department of Mathematics

Bar Ilan University

52900 Ramat Gan, Israel

Saharon SHELAH [email protected]

Department of Mathematics

Givat Ram

Hebrew University of Jerusalem

438 26 January 1991

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Martin Goldstern, Haim Judah and Saharon Shelah- Strong Measure Zero Sets Without Cohen Reals

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