Marek Balcerzak, Andrzej Roslanowski and Saharon Shelah- Ideals without ccc

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    Ideals without ccc

    Marek Balcerzak

    Institute of Mathematics

    Lodz Technical University

    90-924 Lodz, Poland

    Andrzej Roslanowski

    Institute of Mathematics

    The Hebrew University of Jerusalem

    Jerusalem, Israel

    and

    Mathematical Institute of Wroclaw University

    50384 Wroclaw, Poland

    Saharon Shelah

    Institute of Mathematics

    The Hebrew University of Jerusalem

    Jerusalem, Israel

    and

    Department of Mathematics, Rutgers UniversityNew Brunswick, NJ 08854, USA

    October 6, 2003

    Research partially supported by KBN grant No PB 691/2/91 Research partially supported by KBN grant 1065/P3/93/04 Research supported by Basic Research Foundation of The Israel Academy of Sciences

    and Humanities. Publication number 512

    0

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    Abstract

    Let I be an ideal of subsets of a Polish space X, containing all singletonsand possessing a Borel basis. Assuming that I does not satisfy ccc, weconsider the following conditions (B), (M) and (D). Condition (B) statesthat there is a disjoint family F P(X) of size c, consisting of Borel setswhich are not in I. Condition (M) states that there is a Borel functionf : X X with f1[{x}] / I for each x X. Provided that X is a groupand I is invariant, condition (D) states that there exist a Borel set B / Iand a perfect set P X for which the family {B + x : x P} is disjoint.The aim of the paper is to study whether the reverse implications in thechain (D) (M) (B) not-ccc can hold. We build a -ideal onthe Cantor group witnessing (M)&(D) (Section 2). A modified versionof that -ideal contains the whole space (Section 3). Some consistencyresults on deriving (M) from (B) for nicely defined ideals are established(Sections 4 and 5). We show that both ccc and (M) can fail (Theorems1.3 and 5.6). Finally, some sharp versions of (M) for invariant ideals onPolish groups are investigated (Section 6).

    1 Introduction

    An ideal on a space X is a family I of subsets of X closed under finite unionsand subsets (i.e. A, B I A B I and A B, B I A I);-ideals are closed under countable unions. All ideals we consider are assumedto be non trivial, they do not contain the whole space X. Moreover we wantthem to contain all singletons {x} (x X). The ideal I on a Polish space Xis called Borel if it has a Borel basis (i.e. if for each set A I there is a Borelsubset B of X such that A B and B I).

    The most popular Borel -ideals (e.g. the ideal of meager sets or the ideal ofLebesgue null sets) satisfy the countable chain condition (ccc). This conditionsays that the quotient Boolean algebra of Borel subsets of the space modulo the-ideal is ccc (i.e. every family of disjoint Borel sets which do not belong tothe ideal is countable). In this paper we are interested in ideals which do notsatisfy this condition. The question that arises here is what can be the reasonsfor failing ccc. The properties (M) and (D) defined below imply that the idealdoes not satisfy ccc (and actually even more, see (B) below).

    Definition 1.1 LetI be an ideal on an uncountable Polish space X.

    1. We say thatI has property (M) if and only if

    there is a Borel measurable function f : X X with f1[{x}] / I foreach x X.

    2. Provided that X is a Polish Abelian group and I is invariant (i.e. A I

    and x X imply A + xdef= {a + x : a A} I),

    we say that I has property (D) if and only if

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    there are a Borel set B / I and a perfect (non-void) set P X such that(B + x) (B + y) = for any distinct x, y P.

    Properties (M) and (D) were introduced and investigated in [1]. It was observedthat (D)(M), if I is invariant in the group X. Of course, (M) implies thefollowing condition:

    (B) there is a family F P(X) of cardinality c (the size of the continuum)of pairwise disjoint Borel subsets of X that are not in I.

    In [1] Fremlins theorem stating the consistency of ((B)(M)) is shown. How-ever, it is unclear how his proof could be applied to the invariant case. Thefollowing questions arise (3 and 4 are posed in [1]):

    Problems we address 1.2

    1. Suppose I is a Borel (invariant) ideal on a Polish space (group) X, forwhich the ccc fails. Does I satisfy (B)?

    2. (Remains open) Is ((B)(M)) consistent, for some invariant ideal(-ideal)?

    3. (Remains open) Is ((B)(M)) provable in ZFC, for some ideal (-ideal)?

    4. Does (M)(D) hold for every invariant ideal (-ideal)?

    The present paper considers these questions. We mostly restrict ourselves to theCantor group 2 with the coordinatewise addition modulo 2 (denoted further

    by , or simply, by +).At first, let us show that question 1 of 1.2 can have the negative answer, ifwe do not require any additional properties of a -ideal I.

    Theorem 1.3 For each cardinal , < < c, there exists a Borel -idealI on 2 such that is the maximal cardinal for which one can find a disjointfamily of size of Borel sets in 2 that are not in I. Consequently, I does notsatisfy both ccc and (B), and it satisfies +-cc.

    Proof Pick pairwise disjoint nonempty perfect sets P 2, < . DefineI as follows:

    E 2 belongs to I if and only ifthere is a Borel set B 2 such that E B and for each < the

    intersection B P is meager in P.

    Obviously, I is a Borel -ideal on 2 and it does not satisfy ccc since each setP is not in I. Suppose that F is a family of pairwise disjoint I-positive Borelsets and |F| = +. Then there is an < such that

    |{E F : E P is non-meager in P}| = +.

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    This is impossible since the ideal of meager sets in P

    satisfies ccc.

    A similar idea but in a much more special form will be used to produce thenegative answer of a modified version of problem 1.2(1), where (B) is replacedby (M) and I is a -ideal on 2 with 12 definition (Conclusion 5.6). We stilldo not know what can happen if the invariance of I is assumed.

    Acknowledgment: We would like to thank the referee for very helpful com-ments on the presentation of the material.

    2 The minimal -ideal without property (D)

    In this section we are going to answer question 4 of 1.2 in negative. It is stated in

    [1] that Bukovsky has reformulated the question about (M)(D) by consideringthe -ideal I0 generated by the family

    F0 = {B 2 : B is Borel and there is a perfect set P 2 such that

    {B x : x P} is a disjoint family }.

    Then I0 is the minimal invariant -ideal without property (D). Observe that I0is not trivial since F0 is contained in the ideal of measure zero sets (as well asin the ideal of meager sets).

    Theorem 2.1 There is a continuous function f : 2 2 such that

    (x 2)(f1[{x}] / I0).

    Consequently, the -ideal I0 has property (M) and does not have property (D).

    The rest of this section will be devoted to the proof of the above theorem. Wewill break it into several steps presented in consecutive subsections, some ofthese steps may be interesting per se. For simplicity, we shall write + for theaddition in 2; also + will be used for the addition of finite sequences of zerosand ones.

    2.1 A combinatorial lemma

    We start with defining the function f which existence is postulated in 2.1. Itsconstruction is very simple and based on the following (essentially elementary)

    observation.

    Lemma 2.2 For each n there are N and a subset C 2N suchthat every n translates of C have non-empty intersection, and likewise for C =

    2N \ C.

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    Proof Let n 1. Since log2

    (2x 1) < x for each x > 0, we can choose such that

    maxg[1,2n]

    log2(2g 1)

    g< < 1.

    Then 2g 1 < 2g for every g [1, 2n]. Next, take an integer N n such that

    nN + 1 < 2N(1 )

    (this is possible since 1 > 0). We claim that this N is good for our n. To

    show that there exists a suitable set C 2N we will estimate the number ofall bad sets. Fix for a moment a sequence s0, . . . , sn1 2

    N. We want to

    give an upper bound for the number of all subsets D of 2N such that

    eitherk

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    In the next steps we will show that the function f is as required in 2.1.Since, obviously, f is continuous, what we have to prove is that for every y 2

    its pre-image f1[{y}] is not in the ideal I0. As in the proof we will use theproperties of the sets Ci stated in (), () above only, it should be clear fromsymmetry considerations, that it suffices to show that the set

    Hdef= {x 2 : (i )(f(x)(i) = 1)}

    is not in I0. (Note that the set H consists of those sequences x which satisfyx[ni, ni+1) Ci for all i in .)

    2.2 A Baire topology on H

    At this step, for each sequence P = Pn : n of perfect subsets of 2

    , weintroduce a topology = (P) on H. Let the sequence P be fixed in this andthe next subsections.

    Let Pn = Pn + Pn = {x + y : x, y Pn} and Tn = {xm : x P

    n , m }

    for n . Note that Pn is a perfect set, Tn 2

    < is a perfect tree and itsbody (i.e. the set of all infinite branches through the tree) is [Tn ] = P

    n .

    In order to define the desired topology we need to consider tree order-ings, say , with domain a positive integer n, and compatible with the naturalordering of integers, together with an assignment of integers (k, ) to pairs

    {k, } [n]2, where k is the immediate predecessor of relative to (and thus,in particular, k < ). Note that the tree ordering can be determined fromthe mapping alone; is undefined when k is not the immediate predecessor

    of . Such a mapping will be called a tree mapping with domain n, and weshall reserve the letter , with subscripts and/or superscripts to denote treemappings.

    A sequence s 2< will be called acceptable if it belongs to the tree of Hand dom(s) is some ni. Thus if s is acceptable and dom(s) = ni then for eachj < i, the restriction s[nj , nj+1) is in the set Cj.

    Let S consist of all sequences = , s0, s1, . . . , sn1 where is a treemapping on n, the sj s are acceptable with the same domain ni n, andsk + s T

    (k,) for all (k, ) such that (k, ) is defined. We also set n = n()

    and i = i().

    Lemma 2.3 Suppose that = , s0, . . . , sn()1 S and j > i(). Thenthere are t0, . . . , tn()1 2

    nj such that s0 t0, . . . , sn()1 tn()1 and

    , t0, . . . , tn()1 S (where s t means that the sequence t is a properextension of s). If j is sufficiently large, t0, . . . , tn()1 can be chosen pairwisedistinct.

    Proof Let i = i(), n = n(). Let be the tree ordering on n determinedby the tree mapping (so is the immediate successor of k if and only if

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    (k, ) is defined). We will consider the case j = i + 1 since, for greater numbersj, simple induction works.What we have to do is to find sequences rk Ci (for k < n) such that

    if k < < n and is the immediate -successor of kthen (sk

    rk) + (sr) T(k,).

    For each k, < n such that k is the immediate predecessor of (so in particu-

    lar k < ) we choose sequences rk,l 2[ni, ni+1) such that (sk+s)

    rk, T(k,)

    (possible as sk + s T(k,) and T

    (k,) is a perfect tree). The sequences rk,

    are our candidates for the sums rk + r:

    () if we decide what is rk then we will put r = rk + rk,.

    As is a tree ordering it follows that if we keep the above rule then the choiceof the sequence r0 determines all the sequences r1, . . . , rn1. Why? Take k < n.Then there exists the unique sequence 0 = k0 < k1 < . . . < km = k such thatki+1 is the immediate successor of ki (for i < m) and therefore, by (),

    rk = r0 + rk0,k1 + rk1,k2 + . . . + rkm1,km .

    So choosing r0 we have to take care of the demand that rk Ci for all of thesequences rk (for k < n). Thus we have to find r0 Ci such that

    () if 0 = k0 < . . . < km < n and ki+1 is the immediate successor of ki(for i < m)

    then r0 Ci + rk0,k1 + rk1,k2 + . . . + rkm1,km .

    Why can we find such an r0? Each positive k < n appears as the largest elementin exactly one sequence k0, . . . , km as in (), so we get n 1 translations ofCiin (). Thus, considering one more trivial translation (the identity function)we have n ni of them. Applying condition () of the choice of Ci, ni+1 we

    may find a suitable r0 2[ni, ni+1).Now, as we stated before, the choice of r0 and () determine all sequences

    r0, r1, . . . , rn1. Moreover

    if 0 = k0 < .. . < km = k < n is a sequence as in ()then rk = r0 + rk0,k1 + . . . rkm1,km Ci

    (we use the fact that the addition and the subtraction in 2< coincide).Define tk = skrk for k < n. We immediately get , t0, . . . , tn1 S.Finally suppose that k, < n are such that k is the immediate predecessor

    of and 0 < . Take j > i and rk,, r

    k, 2

    [ni, nj) such that

    (sk + s)rk,, (sk + s)

    rk, T(k,) 2

    nj andrk,nj1 = r

    k,nj1 but

    rk,[nj1, nj) = rk,[nj1, nj)

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    (possible as T(k,)

    is perfect). If we now repeat the procedure described earlier

    choosing rk,[nm, nm+1) as rk, at the stages m < j 1 then, extending thesequences from nj1 to nj we may use either r

    k,[nj1, nj) or r

    k,[nj1, nj).

    Consequently we may make sure that the respective sequence r is distinct fromr0 . Repeating this for all pairs 0 < < n we may get that all the finalextensions t are distinct. The Lemma is proved.

    Note that if = , s0, . . . , sn1 S, n + 1 ni() and m < then

    , s0, . . . , sn1, s0 S, where is such that [n]2 = , (0, n) = m and

    (k, ) is undefined in all remaining cases. (Remember that finite sequencesconstantly equal to 0 are in Tm.) Thus we may extend each element of S(to an element of S) getting both longer sequences si and the number of these

    sequences (i.e. n()) larger. It should be remarked here that S is nonempty it is easy to give examples of elements of S with n() = 1.For S we define the basic set U() as

    {x0 H : (x1, . . . , xn()1 H)(s0 x0, s1 x1, . . . , sn()1 xn()1)and (j > i())(, x0nj, . . . , xn()1nj S)}

    where = , s0, . . . , sn()1. It follows from 2.3 that each U() is a non-emptysubset of H.

    Proposition 2.4 If 1, 2 S, x0 U(1) U(2) then, for some S, wehave x0 U() U(1) U(2).Consequently the family {U() : S} forms a (countable) basis of a topologyon H. [We will denote this topology by (P) or just if P is understood.]

    Proof For j = 0, 1, let j = j, sj0, . . . , sjnj1 and let x

    j1, . . . , x

    jnj1 H

    witness that x0 U(j). We shall define . Put n() = n0 + n1 1 and

    i = i() = max{i(0), i(1), n()} + 1. Define a partial mapping from [n()]2

    to as follows

    [n0]2 = 0,(n0 1 + k, n0 1 + ) = 1(k, ) if 0 < k < < n1, {k, } dom(1),(0, n0 1 + ) = 1(0, ) if 0 < < n1, {0, } dom(1),(k, ) is undefined in the remaining cases.

    It should be clear that is a tree mapping on n(). Finally, put

    = , x0ni, x01ni, . . . , x

    0n01ni, x

    11ni, . . . , x

    1n11ni.

    By the choice of i and we easily check that S and x0 U() is witnessedby x01, . . . , x

    0n01, x

    11, . . . x

    1n11 and that U() U(0) U(1).

    To conclude the proof of the proposition note that, since each x0 H isin some U() (take n() = 1 and s0 = x0n5), the family {U() : S} is a(countable) basis of a topology on H.

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    Proposition 2.5 The topology is stronger than the product topology of 2 re-stricted to H. Consequently, all (ordinary) Borel subsets of H are Borel relativeto the topology .

    Proof Since the sets [t] = {x 2 : t x} for t i>1

    2ni form a basis of

    the natural topology in 2, it is enough to show that s 2ni , i > 1 implies[s] H . But for this observe that if [s] H = then n() = 1, i() = i ands0 = s generate S such that U() = [s] H.

    Proposition 2.6 H, is a Baire space, actually each basic -open set U()is not -meager.

    Proof First note that each U() (for S) is a projection of a compactsubset of Hn() and hence it is a (non-empty) compact set (in the naturaltopology). Suppose to the contrary that for some S we have U() =k

    Nk, where each Nk is -nowhere dense. Then we may inductively choose

    0, 1, . . . S such that 0 = and

    U(k+1) Nk = , and U(k+1) U(k) for each k .

    It is possible as if S, N is -nowhere dense then there is a non-empty-open set U U() \ N (remember that U() = ). But the sets U()for S constitute the basis of the topology , so we find S withU() U U() \ N.Now, by the compactness of the sets U(k) we get

    =k

    U(k) U() \k

    Nk

    a contradiction finishing the proof of the proposition.

    2.3 non-meager subsets of H

    Proposition 2.7 IfB H is anon-meager set with the Baire property (withrespect to ) then for every m there are distinct x, y Pm such that theintersection (B + x) (B + y) is non-empty.

    Proof Let m and B H be a non-meager set with the Baireproperty. Then for some S the set U() \ B is -meager. Let U() \ B =k

    Nk where each Nk is -nowhere dense. Let = , s0, . . . sn()1. Put

    = , s0, . . . , sn()1, s0, . . . , sn()1

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    where is a tree mapping on 2n() given by

    [n()]2 = ,(n() + k, n() + ) = (k, ) if k < < n(), {k, } dom(),(0, n()) = m,(k, ) is undefined in the remaining cases.

    Easily, is indeed a tree mapping and S(remember that s0+s0 = 0 Tm).

    For a partial function 0 from [n]2 to , n > n() let (0)

    be defined by

    dom((0)) = {{(k), ()} : {k, } dom(0)} and

    (0)(k, ) = ((k), ()),

    where : n n is a permutation of n given by(0) = n(), (i + 1) = i for 0 i < n() and (i) = i for n() < i < n.

    Note that if 0 is a tree mapping on n > n() such that {0, n()} dom(0)then (0)

    is a tree mapping too. Hence, if 0 = 0, t0, . . . , tn1 S, n() < nand 0(0, n()) is defined then

    (0) def= (0)

    , tn(), t0, . . . , tn()1, tn()+1, . . . , tn1 is in S too.

    Claim 2.7.1 Suppose that 0 = 0, t0, . . . , tn1 S is such that n > n()and 0(0, n()) is defined. Let N H be a nowhere dense set. Then there is+ = +, s+0 , . . . , s

    +k1 S such that

    1. U(+) U(0) \ N, U((+)) U((0)) \ N, and

    2. i(+) > i(0), k = n(+) > n(0) = n, +[n(0)]2 = 0, and

    3. if < n then t s+ .

    [The operation () is as defined before.]

    Proof of the claim: Since N is nowhere dense we find 1 in S such thatU(1) U(0) \ N. Applying the procedure from the proof of Proposition 2.4(with an arbitrary x0 U(1)) we may find 2 = 2, s

    20, . . . , s

    2n(2)1

    S such

    that U(2) U(1) and i(2) > i(0), n(2) > n(0), 2[n(0)]2 = 0, t s

    2

    for < n. Now we look at (2) S and we choose 3 S such that

    U(3) U((2)) \ N U((0)

    ) \ N.

    Next, similarly as 2, we get 4 = 4, s40, . . . , s4n(4)1

    Swith the correspond-

    ing properties with respect to (2) and 3. So, in particular, 4[n(2)]

    2 =(2)

    , s2 s4+1 for < n(), s

    2n() s

    40 and s

    2 s

    4 for n() < < n(2).

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    Finally we apply the inverse operation to () and we get + S such that(+) = 4. (This is possible, as the only j (0, n()) for which the value of4(0, j) is defined, is j = 1.) It should be clear that the

    + is as required in theclaim.

    Now, by induction on k < , we choose nk, ski (for i < nk), k and k suchthat

    (i) n0 = 2n(), s0i = s0n()+i = si (for i < n()), 0 =

    ,

    [So = 0, s00, . . . , s0n01 = 0.]

    (ii) k = k, s0k, . . . , s0nk1 S, U(k) U(), U((k)

    ) U(),

    [Here, (k) is the element of S obtained from k by moving skn() to the

    first place, see the definition of (0), (0) above.]

    (iii) nk < nk+1, i(k) < i(k+1), ski s

    k+1i for i < nk, k = k+1[nk]

    2,

    (iv) U(k+1)Nk = U((k+1))Nk = and s

    k+1i (for i < nk+1) are pairwise

    distinct.

    The first step of the construction is fully described in the demand (i) above (notethat then (ii) is satisfied as U((0)

    ) U() and (iii), (iv) are not relevant).Suppose that we have defined k etc. Apply 2.7.1 to k, Nk standing for 0,N there to get k+1 (corresponding to

    + there). It should be clear that therequirements (ii)(iv) are satisfied except perhaps the last demand of(iv) thesequences sk+1i do not have to be pairwise distinct. But this is not a problem

    as by Lemma 2.3 we may take care of this extending them further.Let xi =

    k

    ski for i < . Then xi H (by (iii)+(ii)), x0 k

    U(k),

    xn() k

    U((k)) (by the definition of (k)

    ). Hence, by (ii)+(iv),

    x0, xn() U() \k

    Nk B

    and by the last part of(iv) they are distinct. Since k(0, n()) = m for each k < (by (iii)) we may apply the definition ofS to conclude that x0 + xn() [T

    m].

    So, x0 + xn() = x + y for some x, y Pm. As x0 = xn(), also x = y. We finishthe proof of the proposition noting that x0 + x = xn() + y and x0 + x B + x,

    xn() + y B + y, as required.

    2.4 Conclusion of the proof of Theorem 2.1

    As we stated at the end of the subsection 2.1, to conclude Theorem 2.1 it isenough to show that the set H defined there is not in the ideal I0. Suppose

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    to the contrary that H can be covered by the union n

    Bn

    of Borel (in the

    standard topology) subsets Bn of 2, each Bn from the family F0. Then, forevery n we may pick up a perfect set Pn 2

    witnessing Bn F0. LetP = Pn : n and let = (P) be the topology on H determined by Pin Proposition 2.4. We know that the space H is not meager in this topology(by Proposition 2.6) and therefore one of the sets Bn H is non-meager, sayBn0 H. But Bn0 H is Borel in the standard topology of H, so it has theBaire property. Applying Proposition 2.7 to it we conclude that there aredistinct x, y Pn0 such that the intersection (Bn0 + x) (Bn0 + y) is non-empty,a contradiction to the choice of Pn0 .

    3 Non-Borel caseNow, let us omit the assumption that the sets of the family F0 defined in theprevious section are Borel. We thus get

    F0 = {A 2 : there is a perfect set P 2 such that

    {A x : x P} is a disjoint family }.

    Let I0 be the -ideal generated by F0 . It turns out that I

    0 is not a proper

    ideal.

    Theorem 3.1 2 I0 .

    Proof Here + and will stand for the respective operations on ordinals,while the addition and subtraction in 2 are denoted by , , respectively. Letus define an increasing sequence : < + of ordinals as follows:

    0 = 0, n+1 = n + c for n , = sup

    nn, +n+1 = +n + c for n .

    Let P 2 be a perfect set independent in the Cantor group; cf. [12]. Pickpairwise disjoint perfect sets Pn (for n ) such that

    n

    Pn = P and fix

    enumerations Pn = {x : n < n+1} (for n ; so xs are distinct).Extend P to H, a maximal independent set called here a Hamel basis. We mayassume that |H \ P| = c. Let H \ P = {x : < +}, where xs aredistinct. Consequently, H = {x : < +}.

    Now, let us define y for < +, as follows:

    if < +, put y = x,

    if n < n+1, n , put

    y = x x+n+(n)+1 . . . x+n+(n)+n.

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    Then {y

    : < +

    } forms a Hamel basis. Each y 2 has the uniquerepresentation of the form

    uy

    y where uy + is finite. For m let

    Am = {y 2 : |uy| = m}.

    Obviously, 2 =

    mAm and

    (1) Am Am

    i2m

    Ai.

    The proof will be complete, if we show that

    (2) {Am x : x Pm+1} is a disjoint family.

    To get this, observe first that(3) if x, x Pn, x = x

    then x x A2n+2.

    Indeed, let x = x and x = x for some , such that n < n+1,

    n < n+1 and = . Since

    x = y x+n+(n)+1 . . . x+n+(n)+nx = y x+n+(n)+1 . . . x+n+(n)+n,

    we conclude that x x A2n+2 (remember that n = n) and (3) isproved. To show (2), take distinct x, x Pm+1 and suppose that the intersec-tion (Am x) (Am x

    ) is non-empty. Then a x = b x for some a, b Am.In other words, a b = x x. By (1), we have x x

    i2mAi and, by (3),

    we get x x A2m+4, a contradiction.

    Remark: For the Cantor group, the operations and are identical. However,we have distinguished them since the same proof works for any Abelian Polishgroup admitting a perfect independent set. Note that a number of groupsdifferent from 2 are good: by [11], each connected Abelian Polish group whichhas an element of infinite order admits a perfect independent set.

    4 Getting (M) from not ccc

    In this section we try to conclude (consistently) the property (M) from theproperty (B) for nicely defined Borel ideals. The results here are complementary,in a sense, to Fremlins theorem mentioned in Introduction. (But note thathere we deal with ideals with simple definitions, while the ideal constructed byFremlin is very complicated.)

    Theorem 4.1 Let n 2. The following statement is consistent with ZFC+c = n:

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    if B 2 2 is a 13

    set such that, for some set A 2 of size2, the sections Bx, x A are nonempty pairwise disjoint (whereBx = {y 2

    : x, y B})then for some perfect set P 2, the sections Bx, x P arenonempty pairwise disjoint.

    Proof Start with the universe V satisfying CH.Let Cn be the (finite support) product ofn copies of the Cohen forcing notionC and let c : < n be the sequence of Cohen reals, Cn-generic over V.Work in V[c : < n].

    Clearly c = n. Suppose that B 2 2 is a 13 set and x : < 2 isa sequence of reals such that 1 < 2 < 2 implies that the sections Bx1 , Bx2are nonempty disjoint. Let U V [n] be such that the parameters of the

    13 formula (x, y) defining B are in V[c : U]. Next choose a sequenceU : < 2 V of countable subsets of n \ U such that each real x is inV[c : U U ]. Moreover, we demand that

    V[c : U U ] |= (z)(x , z)

    (remember that is 13; use Shoenfields absoluteness). By CH in V and the-lemma we find A [2]

    2 and a countable set U U such that {U \ U :

    A} is a disjoint family. We may assume that all sets U \ U are infinite (by

    adding more members). Let V = V[c : U].

    Each sequence c : U \ U is essentially one Cohen real; denote this

    real by d . Note that if0, 1, 2 A are distinct then d0 , d1 , d2 is CCC-generic over V. A real x 2 in the one Cohen real extension is the value

    of a Borel function f : 2 2 from the ground model at this Cohen real.Consequently, we find a sequence f : A V

    of Borel functions from 2

    into 2 such thatV[c : < n] |= x = f(d).

    By CH in V we find A [A]2 V and a Borel function f : 2 2, f V

    such that f = f for A. By Shoenfields absoluteness we have that fordistinct 0, 1, 2 A

    :

    V[d0 , d1 , d2 ] |= (z)[(f(d0), z) & (f(d1), z)] & (z)(f(d0), z) & (z)(f(d1), z)

    (remember the choice of U and U : the witness for (z)((x , z)) is in V[c :

    U U ] already).As the d s are Cohen reals over V

    and their supports are disjoint, by densityargument we get that in V:

    () CCC

    (z)[(f(c0), z) & (f(c1), z)] & (z)(f(c0), z) & (z)(f(c1), z) ,

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    where c0

    , c1

    , c2

    is the canonical CCC-name for the generic triple of Cohenreals. In V[c : < n] take a perfect set P 2 such that P P \ O forevery open dense subset O of 2 2 coded in V ( stands for the diagonal{x, x : x 2}). Then x, y is C C-generic over V for each distinctx, y P. (Such a perfect set is added by one Cohen real; the property that itis a perfect set of mutually Cohen reals over V is preserved in passing to anextension.) We claim that for distinct x, y P V[c : < n]:

    V[c : < n] |= (t)(f(x), t) & (t)(f(y), t) & (t)[(f(x), t) & (f(y), t)] .

    Why? As x, y is C C-generic, by upward absoluteness for 13 formulas and() we get

    V[c : < n] |= (t)(f(x), t) & (t)(f(y), t).Assume that

    V[c : < n] |= (t)((f(x), t) & (f(y), t)).

    The formula here is 13, so we have a real z 2 V[c : < 1] such that (by

    12 absoluteness)

    () V[x,y,z] |= (t)[(f(x), t) & (f(y), t)].

    This real is added by one Cohen real over V[x, y]. So we may choose z to bea Cohen real over V[x, y] (13-upward absoluteness again). Then x,y,z isCCC-generic over V and () contradicts ().

    Now, working in V[c : < n], we see that for distinct x, y P the sectionsBf(x), Bf(y) are disjoint and nonempty. Hence the function f is one-to-one onthe perfect P and we can easily get a required perfect P.

    Definition 4.2 We say that a Borel ideal I on 2 has a 1n definition if thereis a 1n-formula (x) such that

    (a) a is a Borel code and the set #a coded by a belongs to I.

    We say that I has a projective definition if it has a 1n definition for some n.

    Corollary 4.3 Letn 2. It is consistent with ZFC +c = n that:

    for each Borel idealI on 2 with a 13 definition, if there exists asequence B : < 2 of disjoint Borel sets not belonging to I thenI satisfies (M).

    In particular the statement

    (B) (M) for Borel ideals which are 13

    is consistent.

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    Proof Work in the model of 4.1 (so after adding n

    Cohen reals to a modelof CH). Suppose that I is a Borel ideal with 13 definition and let (x) be the13 formula witnessing it. Let

    (x, y) x is a Borel code & (x) & y #x.This is a 13 formula defining a

    13 subset B of 2

    2. If there are 2 pairwisedisjoint I-positive Borel sets then they determine 2 pairwise disjoint nonemptysections of the set B. By 4.1 we can find a perfect set P 2 such that{Bx : x P} is a family of disjoint nonempty sets. Define a function f : 2

    Pby:

    for y 2, if there is x P such that (x, y) B then f(y) is this(unique) x, otherwise f(y) = x0 where x0 is a fixed element of P.

    Note that the set {(x, y) B : x P} is Borel and has the property that itsprojection onto ys axes is one-to-one. Consequently the set

    {y 2 : (x P)((x, y) B)}

    is Borel and hence the function f is Borel. Thus f witnesses (M) for the idealI. (Note that no harm is done that the function is onto P instead of 2.)

    Remark: 1) If we start with a model for CH and add simultaneously nrandom reals over V (by the measure algebra on 2n) then in the resulting modelwe will have a corresponding property for 12 subsets of the plane 2

    2 andthe ideals with 12 definitions. The proof is essentially the same as in 4.1. Theonly difference is that the perfect set P is a perfect set of sufficiently randomreals. For a countable elementary submodel N V of H()V

    we choose by

    the Mycielski theorem (cf. [13]) a perfect set P V[r : < 2] such that eachtwo distinct members of P are mutually random over N.2) Note that the demand that the ideal I in 4.3 has to admit 2 disjoint BorelIpositive sets is not an accident. By 5.6, the failure of ccc is not enough.3) In the presence of large cardinals we may get more, see 4.4 below.

    Theorem 4.4 Assume that is a weakly compact cardinal and + is acardinal such that

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    ifB 2 2 is a projective set which has 2

    disjoint non-emptysections then B has a perfect set of disjoint non-empty sections.

    The point is that projective formulas are absolute between intermediate modelscontaining VP0 (see e.g. the explanations to [8, 6.5]). We finish as in 4.3.

    Theorem 4.5 Assume that I is a Borel ideal with 12 definition. Suppose thatthere exists a Borel function f : 2 2 such that f1[{x}] / I for somedistinct points x, < 2. Then there is a perfect set P 2

    such that

    (x P)(f1[{x}] / I).

    Consequently, the ideal I has property (M).

    Proof Let (x) be the 12 definition of the ideal I. Consider the relationE on 2:

    xEy (x = y or ((f1[{x}]) & (f1[{y}]))).

    This is an equivalence 12 relation with at least 2 classes. As E remains anequivalence relation after adding a Cohen real we may apply [6] (or [15]) to geta perfect set of pairwise nonequivalent elements. Removing at most one pointfrom this set we find the desired one.

    Remark: In the above theorem, if the ideal I has a 11 definition then it isenough to assume that there are 1 respective points x for f.

    5 An ideal on trees

    Here we present a Borel ideal I with 12 definition which (in ZFC) satisfiesthe 2cc but does not have the ccc. Thus if CH fails then this ideal cannothave the property (M). But we can conclude this even under CH, if we haveenough Cohen reals (see 5.6).

    Definition 5.1 LetTr be the set of all subtrees T of < and let

    Trw = {T Tr : T is well founded }.

    For a tree T Trw let hT : T 1 be the canonical rank function. For < 1

    putA = {T Trw : hT() = }.

    Clearly Tr is a closed subset of the product space 2(<

    ), so it is a Polish spacewith the respective topology. The set Trw is a 11-subset of it and the sets Aare Borel subsets of Tr.

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    The ideal I will live on the space Tr. To define it we introduce topologies on the sets A (for < 1). Fix < 1 for a moment. The basis of thetopology consists of all sets of the form

    U(n, T) = {T A : T n n = T n n & hTn

    n = hTn n}

    for T A, n . It should be clear that this is a (countable) basis of atopology.

    Lemma 5.2 1. A, is a Baire space.

    2. If > 0 then there is no isolated point in .

    3. Each Borel (in the standard topology) subset of A is -Borel.

    Proof 1) Suppose that Ok A are -open dense (for k < ) and thatT A, n . For each limit fix an increasing sequence (m) .Next define inductively sequences nk : k , Tk : k such that

    () nk < nk+1 < , Tk A, n0 = n, T0 = T,

    () U(nk+1, Tk+1) U(nk, Tk) Ok,

    () for each Tk n nkk ,

    if = hTk() is a limit ordinal then for each k there is Tk+1

    n nk+1k+1 such that () hTk+1(

    ) and , and

    if = hTk() is a successor ordinal then there is Tk+1 n

    nk+1

    k+1extending and such that hTk+1() = 1.

    Put T =k

    Tk n nkk , h

    =k

    hTkn nkk . Then h

    : T +1 is a rank

    function (so T is well founded). Moreover, the condition () guarantees thatit is the canonical rank function on T and hence T

    k

    Ok U(n, T).

    The assertions 2) and 3) should be clear.

    Definition 5.3 The ideal I consists of subsets of Borel sets B Tr such that

    (0 < < 1)(B A is -meager).

    Proposition 5.4 I is a non-trivial Borel ideal on Tr which does not satisfythe ccc but satisfies the 2cc.

    Proof As {A : < 1} is a partition of Trw and each A is not -meager (by 5.2), we get that I is a non-trivial -ideal on Tr. Since each setA is not in I

    , the ccc fails for I. If {B : 2} is a family of I-almost

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    disjoint Ipositive Borel sets then for some < 1

    and a set Z [2

    ]2 ,{B A : Z} is a family of -Borel non-meager (-meager)-almostdisjoint sets. This contradicts the fact that the topology has a countablebasis (and each basic set is not -meager).

    Now, we want to estimate the complexity of I.

    Proposition 5.5 The ideal I has a 12 definition.

    Proof Fix < 1. A basic open subset of A (in the topology ) isdetermined by a pair F, h, where F < is a finite tree and h : F + 1is a rank function such that h() = . The basic sets U(F, h) are Borel subsetsof A (in the standard topology). Let

    nwd(a, ) a is a Borel code &(F, h)(F, h)[F F & h h & U(F, h) #a = ].

    It should be clear that nwd(, ) is a 11-formula. Consequently, the formula

    a is a Borel code and the set #a A is -nowhere dense is 11. Let

    meager(a, ) a is a Borel code &(cn : n )[(n)nwd(cn, ) & #a A

    n

    #cn].

    This formula says that the intersection #a A is -meager; it is 12. On theother hand, its negation is equivalent to

    meager(a, ) a is not a Borel code or(F, h)(cn : n )[(n)nwd(cn, ) & U(F, h) #a n

    #cn]

    which is 12 too. Consequently,

    a is a Borel code such that #a A is -meager

    is absolute between all transitive models of (a large enough part of) ZFC con-taining a , , . . .. Now, we can give a 12-definition of the ideal I

    :

    (a) a is a Borel code &(E )[E is not well founded or , E |= ora is not encoded in , E or , E |= V = L[a] or, E |= ( 1)(meager(a, ))].

    The above is a sentence carrying a large part of information on ZFC, inparticular for each transitive set M, M |= should imply that M is A-adequatefor all A M and the absoluteness of 11,

    11 formulas holds for M (see [7, Ch. 2,

    15]). Thus, ifE is a well founded relation on , a is a real encoded in E (which

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    means that a belongs to the transitive collapse ofE) and , E |= +V = L[a]then the transitive collapse of E is L[a] for some < 1. Now, if (a) holdsthen for all < 1 such that L[a] is a-adequate we have

    L[a] |= ( < 1)meager(a, ).

    The absoluteness implies that if L[a] |= meager(a, ) then meager(a, ). Con-sequently, (a) implies ( < 1)meager(a, ). Suppose now that a is a Borelcode for which (a) fails. Then for some < 1 (suitably closed) we have

    L[a] |= ()(meager(a, ))

    and hence for some < we get

    L[a] |= meager(a, ).

    Once again the absoluteness implies that meager(a, ) holds. Consequently,(a) implies that there is < 1 such that meager(a, ) and finally,

    (a) a is a Borel code & #a I.

    It should be clear that (x) is a 12 formula.

    Conclusion 5.6 Assume that either CH fails or every 12 set of reals has theBaire property. Then the ideal I does not have the property (M) [and: itsatisfies the 2-cc, it does not satisfy the ccc and has a

    12 definition].

    Proof If we are in the situation of CH then (M) cannot hold because of2-cc. So assume that all

    12 sets of reals have Baire property.

    Suppose that f : 2

    2

    is a Borel function (of course, it is coded by a real).Let c : < 2 be a C2-generic sequence of Cohen reals. In V[c : < 2]we have that for sufficiently large , f1[{c}] I

    . Since the definition of I

    is absolute (and it involves no parameters),

    V[c : < 2] |= f1[{c}] I

    implies L[f][c ] |= f1[{c}] I

    .

    By density arguments and the ccc ofC2 we conclude that

    L[f] |= C f1[{c}] I

    where c is the canonical C-name for the generic Cohen real. Our assumptionthat 12 sets of reals have the Baire property is equivalent to the statement

    for each real r there exists a Cohen real over L[r]

    (see e.g. [9]). Let c V be a Cohen real over L[f]. Then

    L[f][c] |= f1[{c}] I.

    By Shoenfield absoluteness we conclude V |= f1[{c}] I. Consequently, fcannot witness the property (M) for I.

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    6 Other variants of (M)The following natural question concerning the hereditary behavior of our con-ditions (B), (M) and (D) arises:

    If an ideal I on X satisfies one of those conditions and E / I isa Borel subset of X, is it true that I P(E) satisfies the samecondition (on E; at this moment we consider the ideal on the spaceE and modify respectively the sense of (B), (M) and (D)))?

    When we do not assume a group structure on X and the invariance of I, onecan easily construct I which makes the answer negative. If the invariance of Iis supposed, the problem becomes less trivial.

    Example: Let I be the ideal of null sets with respect to 1-dimensional Haus-dorff measure (cf. [5]) on R2 treated as an additive group. Clearly, I is in-variant. It satisfies (M) since the fibers of the continuous function given byf(x, y) = x are the lines {x} R, for x R, which are not in I. However,1-dimensional Hausdorff measure on R coincides with the linear Lebesgue mea-sure, so I P({x} R) does not satisfy (B) since ccc works there.

    Further, we shall concentrate on some hereditary versions of (M) connectedwith open sets.

    Definition 6.1 Assume that I is an ideal on an uncountable Polish space Xand that each open nonempty subset of X is not in I.

    1. We say thatI has property (M) relatively to a set E X (in short, I has(M) rel E ) if and only ifthere is a Borel function f : E X whose all fibers f1[{x}], x X, arenot in I.

    2. We say thatI has property (M) if and only ifit has property (M) rel U for each nonempty open set U X.

    3. Letx X. We say that I has property (Mx) if and only ifit has property (M) rel U for each open neighborhood U of x.

    4. We say thatI has property (M) if and only ifthere is a Borel function f : X X such that f1[{x}] U / I for anyx X and open non-void U X.

    Remarks:

    1. Plainly, (M) in the sense of Definition 1.1 is (M) rel X. Moreover, if Eis Borel, (M) rel E implies (M) rel X since one can consider any Borelextension of the respective Borel function f : E X to the whole X.

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    2. Obviously, I has (M) iff it has (Mx

    ) for each x from a fixed dense set inX. In some cases (Mx) satisfied for one point x implies (M

    ). This holdsif x is taken from a fixed dense set whose any two points s, t have home-omorphic open neighborhoods Us, Ut and the homeomorphism preservesI. In particular, if X is a group, x X is fixed, Q is a dense set in Xand I is Q-invariant (i.e. E I and t Q imply E + t I) then (Mx)guarantees (M).

    3. Studies of (M) were initiated in [2] where some examples are given. Con-dition (M) for the ideal of nowhere dense sets means the existence of aBorel function f from X onto X with dense fibers. For X = (0, 1) suchfunctions are known as being strongly Darboux (cf. [4]). From Mauldinsproof in [10] it follows that the -ideal generated by closed Lebesgue null

    sets satisfies (M

    ) (for details, see [2]).4. Evidently, (M) (M) (M). It is interesting to know whether those

    implications can be reversed. A simple method producing ideals (-ideals)with property (M) and without (M) follows from the example given in[2]. If one part of an ideal I, defined on a Borel set B X has (M) relB, and the remaining part, defined on X\ B with int(X\ B) = has not(M) rel X \ B (for instance, ccc holds there) then the ideal is as desired.If we want I to be a -ideal invariant in the group X then the situationis different.

    Theorem 6.2 Suppose that I is an invariant -ideal on R. If the ideal I hasthe property (M) then it has the property (M).

    Proof We start with two claims of a general character.

    Claim 6.2.1 Suppose that f : R R is a Borel function. Then there existsa perfect set P R such that f1[P] is both meager and Lebesgue null.

    Proof of the claim: It follows directly from the fact that each perfect set can bedivided into continuum disjoint perfect sets and both the ideal of meager setsand the ideal of null sets satisfy ccc.

    Claim 6.2.2 Suppose that U R is an open set and A R is a meager nullset. Then there exist disjoint sets Dk U and reals xk (for k ) such thatk

    Dk is nowhere dense and A k

    Dk + xk.

    Proof of the claim: First we prove the claim under the assumption that A isnowhere dense (and null). For this choose an interval J U. Since A is null wecan find (open) intervals Jk : k such that

    A k

    Jk andk

    |Jk| < |J|

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    (here |J| stands for the length of J). Choose disjoint open intervals Jk

    J (fork ) such that |Jk | = |Jk| (for k ). Let reals xk be such that J

    k + xk = Jk.

    For k we putDk = (A Jk) xk.

    Clearly A k

    Dk + xk. Since the sets Dk are nowhere dense and contained

    in disjoint open intervals we get that their unionk

    Dk is nowhere dense.

    If now A is just a meager null set then we represent it as a union A =n

    An,

    where each set An is nowhere dense and null. Take disjoint open sets Un U(for n ) and apply the previous procedure to each pair An, Un gettingsuitable Dnk , x

    nk . Then D

    nk , x

    nk : n, k is as required for A and U (proving

    the claim).

    Suppose now that I is an invariant -ideal on R and that f : R R isa Borel function witnessing the property (M) for I. By Claim 6.2.1 we find aperfect set P R such that f1[P] is both meager and null.Let {Un : n } be an enumeration of all rational open intervals in R. As finiteunion of nowhere dense sets is nowhere dense, we may apply Claim 6.2.2 andchoose inductively (by induction on n ) closed sets Dn,k and reals xn,k suchthat

    (a) Dn,k (for n, k ) are pairwise disjoint

    and for each n :

    (b) k

    Dn,k is a nowhere dense subset of Un,

    (c) f1[P] k

    Dn,k + xn,k.

    Now we define a function f : R P by

    f(x) =

    f(x + xn,k) if x Dn,k, n, k , and f(x + xn,k) P,y0 otherwise,

    where y0 is a fixed element of P. Clearly the function f is Borel. We claim

    that it witnesses the property (M) for I. Why? Suppose that U R is anopen nonempty set and y P. Take n such that Un U. We know thatf1[P]

    k

    Dn,k + xn,k and f1[{y}] / I. Since I is -complete, we find

    k such thatf1[{y}] (Dn,k + xn,k) / I.

    As I is translation invariant we get

    (f1[{y}] xn,k) Dn,k / I.

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    Clearly (f1[{y}]xn,k

    )Dn,k

    (f)1[{y}]U so the last set does not belongto I. The theorem is proved.

    In the above proof the use of Lebesgue measure is important. In Theorem 5.2we can replace R by e.g. 2 but we do not know if we can have a correspondingresult for all Polish groups. Moreover we do not know if we can omit theassumption of-completeness ofI (i.e. prove the theorem for (finitely additive)ideals on R which do not contain nonempty open sets).For the question of dependences between (M), (M) and (M) the followingsimple theorem seems useful.

    Theorem 6.3 Let I be an ideal on X and {Un : n } - a base of open setsin X. For f : X X define Hn(f) = {y X : Un f

    1[{y}] / I}.

    1. Letx X. Condition (Mx) holds iff there is a Borel function f : X Xsuch that, for each n with x Un, the set Hn(f) contains a perfectset.

    2. Condition (M) holds iff there is a Borel function f : X X such thateach set Hn(f), n , contains a perfect set.

    3. Condition (M) holds iff there is a Borel function f : X X such thatn

    Hn(f) contains a perfect set.

    4. Suppose that I is a -ideal, f : X X is a Borel function such thatf1[{x}] / I for all x X, and each set Hn(f), n , either is countable

    or contains a perfect set. Then (M

    x) holds for some x X.

    Proof 1. Necessity. Fix n such that x Un. Then (M) rel Un holds. Itsuffices to extend the respective Borel function g : Un X to a Borel functiondefined on the whole X.Sufficiency. Fix n such that x Un. Let P Hn(f) be a perfect set andlet B = f1[P]. Extend fB to a Borel function g : Un P. Consider a Borelfunction h from P onto X. Then h g witnesses (M) rel Un.The proofs of 2 and 3 are analogous.

    4. Suppose it is not the case. Thus for each x X choose nx such thatI has not (M) rel Unx . For T = {nx : x X} we get X =

    nT

    Un. Since I is

    a ideal, therefore, by the properties of f, we get X =nT

    Hn(f). Hence, by

    the assumption, there is n T such that Hn(f) contains a perfect set. Now, asin the proof of 1, we infer that I has (M) rel Un, a contradiction.

    Example: Let I2 (I) be a Mycielski ideal (defined in [14]) on 2 ( ,

    respectively) generated by the respective system {Kt : t 2

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    (M). However, it is not clear if I

    does. Let us modify the proof of (M) for Ito get (M) rel [s] where [s], for s

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    [8] Judah H., Roslanowski A., Martins Axiom and the continuum, The Journalof Symbolic Logic, 60(1995), 374-392.

    [9] Judah H., Shelah S., 12 sets of reals, Annals of Pure and Applied Logic42(1989), 207-223.

    [10] Mauldin R.D., The Baire order of the functions continuous almost every-where, Proc. Amer. Math. Soc. 41(1973), 535-540.

    [11] Mauldin R.D., On the Borel subspaces of algebraic structures, IndianaUniv. Math. J. 29(1980), 261-265.

    [12] Mycielski J., Independent sets in topological algebras, Fund. Math.55(1964), 139-147.

    [13] Mycielski J., Algebraic independence and measure, Fund. Math. 61(1967),165-169.

    [14] Mycielski J., Some new ideals of sets on the real line, Colloq. Math.20(1969), 71-76.

    [15] Shelah S., On co--Souslin relations, Israel Journal of Mathematics47(1984), 139-153.