Tomek Bartoszynski and Saharon Shelah- Strongly Meager and Strong Measure Zero Sets

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    STRONGLY MEAGER AND STRONG MEASURE ZERO SETS

    TOMEK BARTOSZYNSKI AND SAHARON SHELAH

    Abstract. In this paper we present two consistency results concerning theexistence of large strong measure zero and strongly meager sets.

    1. Introduction

    Let M denote the collection of all meager subsets of 2 and let N be the col-

    lection of all subsets of 2 that have measure zero with respect to the standardproduct measure on 2.

    Definition 1.1. Suppose thatX 2 and let+ denote the componentwise additionmodulo 2. We say thatX is strongly meager if for every H N, X+ H = {x + h :x X, h H} = 2.

    We say that X is a strong measure zero set if for every F M, X+ F = 2. LetSM denote the collection of strongly meager sets and let SN denote the collectionof strong measure zero sets.

    For a family of sets J P(R) letcov(J) = min {|A| : A J and

    A = 2}.

    non(J) = min {|X| : X J }.

    Strong measure zero sets are usually defined as those subsets X of 2

    such thatfor every sequence of positive reals {n : n } there exists a sequence of basicopen sets {In : n } with diameter of In smaller than n and X

    n In. The

    Galvin-Mycielski-Solovay theorem ([4]) guarantees that both definitions are yieldthe same families of sets.

    Recall the following wellknown facts. Any of the following sentences is consis-tent with ZFC,

    (1) SN = [2]0 , (Laver [7])(2) SN = [2]1 , (Corazza [3], Goldstern-Judah-Shelah [5])(3) SM = [2]0 . (Carlson, [2])(4) non(SN) = d = 20 > 1, cov(M) = 1 and there exists a strong measure

    zero set of size 20 . (Goldstern-Judah-Shelah [5])

    The proofs of the above results as well as all other results quoted in this paper canbe also found in [1].

    In this paper we will show that the following statements are consistent with ZFC:

    for any regular > 0, SM = [2]

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    2 TOMEK BARTOSZYNSKI AND SAHARON SHELAH

    non(SN) = 20 > 1, d = 1 and there is a strong measure zero set of size20 .

    2. SM may have large additivity

    In this section we will show that SM can be an ideal with large additivity. Let

    m = min{ : MA fails}.

    We will show that SM = [2] 0,if A [2]K() then for all except finitely many k there exists C 2k suchthat

    (1) |C| 2k ,(2) (Ak) + C = 2k.

    Proof Proof of this lemma can be found in [8] or [1].

    Definition 2.2. For each n let {Cnm : n, m } be an enumeration of allclopen sets in 2 of measure 2n. For a real r and n define an openset

    Hrn = m>nCmr(m).

    It is clear that Hrn is an open set of measure not exceeding 2n. In particular,

    Hr =n H

    rn is a Borel measure zero set of type G.

    Theorem 2.3. Let > 0 be a regular cardinal. It is consistent with ZFC that

    MA 0. Let > be a regular cardinal suchthat

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    STRONGLY MEAGER AND STRONG MEASURE ZERO SETS 3

    added at the step = 0. We will show that VP |= X + Hc = 2, which will endthe proof.

    Suppose that the above assertion is false. Let p P and let z be a P-namefor a real such that

    p z X + Hc.

    Let X = {x : < } and for each find p p and n such that

    p z x + Hcn

    .

    Let Y be a set of size such that

    (1) n = n for Y,(2) {dom(p) : Y} form a -system with root ,(3) p = p, for Y,(4) p(0) =

    s, with |

    s| = >

    n, for Y.

    Fix a subset X = {xj : j < K(2)} Y and let m be such thatCm + X = 2.

    Define condition p as

    p() =

    pj if = & dom(pj ), j < K(2

    )s m if = for < .On one hand p C

    m H

    cn, so p

    X + Hcn = 2

    . On the other hand,p pj , j K(2

    ), so p z X + Hcn. Contradiction.To finish the proof we show that VP |= add(M) = . First note that MA

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    4 TOMEK BARTOSZYNSKI AND SAHARON SHELAH

    It is easy to see (see [1]) that EE is proper (satisfies axiom A), and strongly

    bounding, that is ifp and n then there is q n p and a finite set F

    such that q F.In [5] it is shown that a countable support iteration of EE and rational perfect

    set forcing produces a model where there is a strong measure zero set of size 20 .In particular, one can construct (consistently) a strong measure zero of size 20

    without Cohen reals. The remaining question is whether such a construction canbe carried out without unbounded reals.

    Theorem 3.2 ([5]). Suppose that {P, Q : < 2} is a countable support itera-tion of proper, strongly -bounding forcing notions. Then

    VP2 |= SN [R]1 .

    The theorem above shows that using countable support iteration we cannotbuild a model with a strong measure zero set of size > d. Since countable support

    iteration seems to be the universal method for constructing models with 20 = 2the above result seems to indicate that a strong measure zero set of size > d cannotbe constructed at all. Strangely it is not the case.

    Theorem 3.3. It is consistent thatnon(SN) = 20 > d = 1 and there are strongmeasure zero sets of size 20 .

    Proof Suppose that V |= CH and = 0 > 1. Let P be a countablesupport product of copies of EE. The following facts are well-known (see [6])

    (1) P is proper,(2) P satisfies 2-cc,(3) P is -bounding,(4) for f V[G] there exists a countable set A , A V such that

    f V[GA].It follows from (3) that VP |= d = 1. Moreover, (1) and (2) imply that 20 = in VP .

    For a set X 2 VP let supp(X) be a set such that X V[Gsupp(X)].Note that supp(X) is not determined uniquely, but we can always choose it so that|supp(X)| = |X| +0.

    Lemma 3.4. Suppose that X 2 VP and supp(X) = . Then VP |= X SN

    Note that this lemma finishes the proof. Clearly the assumptions of the lemmaare met for all sets of size < and also for many sets of size .

    Proof We will use the following characterization (see [1]):

    Lemma 3.5. The following conditions are equivalent.

    (1) X 2 has strong measure zero.(2) For every f there exists g (2

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    STRONGLY MEAGER AND STRONG MEASURE ZERO SETS 5

    n . Let pf p be any condition such that \ {kn : n } dom(pf()). Wewill check that

    pf P x X n xf(n) = G()(kn)f(n),

    where G is the canonical name for the generic object. Take x X and r pf.Find n such that kn dom

    r()

    . Let r r and s be such that

    (1) supp(r) supp(X)(2) r rsupp(X),(3) r P xkn = s.

    Let

    r() =

    r() if =

    r() {(kn, s)} if = .

    It is easy to see that r xf(n) = G()(kn)f(n). Since f and x were arbitrarywe are done.

    References

    [1] Tomek Bartoszynski and Haim Judah. Set Theory: On the Structure of the Real Line. A.K.Peters, 1995.

    [2] Timothy J. Carlson. Strong measure zero and strongly meager sets. Proceedings of the Amer-ican Mathematical Society, 118(2):577586, 1993.

    [3] Paul Corazza. The generalized Borel conjecture and strongly proper orders. Transactions ofthe American Mathematical Society, 316(1):115140, 1989.

    [4] Galvin, Fred and Mycielski, Jan and Solovay, Robert. Strong measure zero sets. Notices ofAmerican Mathematical Society, 1973.

    [5] Martin Goldstern, Haim Judah, and Saharon Shelah. Strong measure zero sets without Cohenreals. The Journal of Symbolic Logic, 58(4):13231341, 1993.

    [6] Martin Goldstern and Saharon Shelah. Many simple cardinal invariants. Archive for Mathe-matical Logic, 32:203221, 1993.

    [7] Richard Laver. On the consistency of Borels conjecture. Acta Math., 137(3-4):151169, 1976.[8] G. Lorentz. On a problem of additive number theory. Proceedings of the American Mathemat-

    ical Society, 5:838841, 1954.

    Department of Mathematics and Computer Science, Boise State University, Boise,

    Idaho 83725 U.S.A.

    E-mail address: [email protected], http://math.idbsu.edu/~tomek

    Department of Mathematics, Hebrew University, Jerusalem, Israel

    E-mail address: [email protected], http://math.rutgers.edu/~shelah/