Selection principles and the Minimal Tower problem

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1 Note i Matematica 22, n. 2, 2003, Selection principles an the Minimal Tower problem Boaz Tsaban i Department of Mathematics an Computer Science, Bar-Ilan University, Ramat-Gan 52900, Israel tsaban@macs.biu.ac.il, Receive: 13/01/2003; accepte: 28/07/2003. Abstract. We stuy iagonalizations of covers using various selection principles, where the covers are relate to linear quasiorerings (τ-covers. This inclues: equivalences an nonequivalences, combinatorial characterizations, critical carinalities an constructions of special sets of reals. This stuy leas to a solution of a topological problem which was suggeste to the author by Scheepers (an state in [15] an is relate to the Minimal Tower problem. We also introuce a variant of the notion of τ-cover, calle τ -cover, an settle some problems for this variant which are still open in the case of τ-covers. This new variant introuces new (an tighter topological an combinatorial lower bouns on the Minimal Tower problem. Keywors: Gerlits-Nagy property γ-sets, γ-cover, ω-cover, τ-cover, tower, selection principles, Borel covers, open covers. MSC 2000 classification: 03E05, 54D20, 54D80. 1 Introuction 1.1 Combinatorial spaces We consier zero-imensional sets of real numbers. For convenience, we may consier other spaces with more evient combinatorial structure, such as the Baire space N N of infinite sequences of natural numbers, an the Cantor space N {0, 1} of infinite sequences of bits (both equippe with the prouct topology. The Cantor space can be ientifie with P (N using characteristic functions. We will often work in the subspace P (N of P (N, consisting of the infinite sets of natural numbers. These spaces, as well as any separable, zeroimensional metric space, are homeomorphic to sets of reals, thus our results about sets of reals can be thought of as talking about this more general case. i This paper constitutes a part of the author s octoral issertation at Bar-Ilan University.

2 54 B. Tsaban 1.2 Selection principles Let U an V be collections of covers of a space X. The following selection hypotheses have a long history for the case when the collections U an V are topologically significant. S 1 (U, V: For each sequence {U n } n N of members of U, there is a sequence {V n } n N such that for each n V n U n, an {V n } n N V. S fin (U, V: For each sequence {U n } n N of members of U, there is a sequence {F n } n N such that each F n is a finite (possibly empty subset of U n, an n N F n V. U fin (U, V: For each sequence {U n } n N of members of U which o not contain a finite subcover, there exists a sequence {F n } n N such that for each n F n is a finite (possibly empty subset of U n, an { F n } n N V. We make the convention that The space X is infinite an all covers we consier are assume not to have X as an element. An ω-cover of X is a cover such that each finite subset of X is containe in some member of the cover. It is a γ-cover if it is infinite, an each element of X belongs to all but finitely many members of the cover. Following [7] an [13], we consier the following types of covers: O (respectively, B: The collection of countable open (respectively, Borel covers of X. Ω (respectively, B Ω : The collection of countable open (respectively, Borel ω-covers of X. Γ (respectively, B Γ : The collection of countable open (respectively, Borel γ-covers of X. The inclusions among these classes can be summarize as follows: B Γ B Ω B Γ Ω O These inclusions an the properties of the selection hypotheses lea to a complicate iagram epicting how the classes efine this way interrelate. However, only a few of these classes are really istinct. Figure 1 contains the istinct ones among these classes, together with their critical carinalities, which were erive in [7] an in [13]; see efinition in Section 3. The only unsettle implications in this iagram are marke with otte arrows.

3 Selection principles an the Minimal Tower problem 55 S 1 (B Γ, B Γ b S 1 (Γ, Γ b S 1 (B Γ, B Ω U fin (Γ, Γ b S 1 (Γ, Ω?? S fin (Γ, Ω U fin (Γ, Ω S 1 (B Γ, B S 1 (Γ, O U fin (Γ, O S 1 (B Ω, B Γ p S 1 (Ω, Γ p S fin (B Ω, B Ω S 1 (B Ω, B Ω cov(m S 1 (Ω, Ω cov(m S fin (Ω, Ω S 1 (B, B cov(m S 1 (O, O cov(m Figure 1. The surviving classes 1.3 τ-covers A cover of a space X is large if each element of X is covere by infinitely many members of the cover. Following [15], we consier the following type of cover. A large cover U of X is a τ-cover of X if for each x, y X we have either x U implies y U for all but finitely many members U of the cover U, or y U implies x U for all but finitely many U U. A quasiorering on a set X is a reflexive an transitive relation on X. It is linear if for all x, y X we have x y or y x. A τ-cover U of a space X inuces a linear quasiorering on X by: x y x U y U for all but finitely many U U. If a countable τ-cover is Borel, then the inuce = { x, y : x y} is a Borel subset of X X. We let T an B T enote the collections of countable open an Borel τ-covers of X, respectively. We have the following implications. B Γ B T B Ω B Γ T Ω O

4 56 B. Tsaban There is a simple hierarchy between the selection principles: For each U, V in {O, Ω, T, Γ} or in {B, B Ω, B T, B Γ }, we have that S 1 (U, V S fin (U, V U fin (U, V. The implication S fin (U, V U fin (U, V nees a little care when V is T or B T : It hols ue to the following lemma. 1 Lemma. Assume that U = n N F n, where each F n is finite, is a τ-cover of a space X. Then either F n = X for some n, or else V = { F n } n N is also a τ-cover of X. Proof. Assume that F n X for all n. Then, as U is an ω-cover of X, so is V. In particular, V is a large cover of X. Now fix any x, y X such that x U y U for all but finitely many U U, an let F = {n : ( U F n x U an y U}. Then F is finite an contains the set of n s such that x F n an y F n. 1.4 Equivalences The notion of τ-covers introuces seven new pairs namely, (T, O, (T, Ω, (T, T, (T, Γ, (O, T, (Ω, T, an (Γ, T to which any of the selection operators S 1, S fin, an U fin can be applie. This makes a total of 21 new selection hypotheses. Fortunately, some of them are easily eliminate, using the arguments of [11] an [7]. We will repeat the reasoning briefly for our case. The etails can be foun in the cite references. First, the properties S 1 (O, T an S fin (O, T imply S fin (O, Ω, an thus hol only in trivial cases (see Section 6 of [17]. Next, S fin (T, O is equivalent to U fin (T, O, since if the finite unions cover, then the original sets cover as well. Now, since finite unions can be use to turn any countable cover which oes not contain a finite subcover into a γ-cover [7], we have the following equivalences 1 : U fin (T, Γ = U fin (Γ, Γ, U fin (O, T = U fin (Ω, T = U fin (T, T = U fin (Γ, T, U fin (T, Ω = U fin (Γ, Ω; an U fin (T, O = U fin (Γ, O. In Corollary 6 we get that S 1 (T, Γ = S fin (T, Γ. We are thus left with eleven new properties, whose positions with respect to the other properties are escribe in Figure 2. In this Figure, as well as in the one to come, there still exist quite many unsettle possible implications. 1 We ientify each property with the collection of sets satisfying this property. Thus, for properties P an Q, we may write X P, P Q, etc.

5 Selection principles an the Minimal Tower problem 57 S 1 (Γ, Γ b S 1 (Γ, T? U fin (Γ, Γ b S fin (Γ, T? U fin (Γ, T x S 1 (Γ, Ω S fin (Γ, Ω U fin (Γ, Ω S 1 (Γ, O U fin (Γ, O S fin (T, T? S fin (T, Ω S 1 (T, Γ t S 1 (T, T? S 1 (T, Ω? S 1 (T, O? S fin (Ω, T p S fin (Ω, Ω S 1 (Ω, Γ p S 1 (Ω, T p S 1 (Ω, Ω cov(m S 1 (O, O cov(m Figure 2. The surviving classes for the open case 1.5 Equivalences for Borel covers For the Borel case we have the iagram corresponing to Figure 2, but in this case, more equivalences are known [13]: S 1 (B Γ, B Γ = U fin (B Γ, B Γ, S 1 (B Γ, B Ω = S fin (B Γ, B Ω = U fin (B Γ, B Ω, an S 1 (B Γ, B = U fin (B Γ, B. In aition, each selection principle for Borel covers implies the corresponing selection principle for open covers. This paper is ivie into two parts. Part 1 consists of Sections 2 4, an Part 2 consists of the remaining sections. In Section 2 we stuy subcover-type properties an their applications to the stuy of the new selection principles. In Section 3 we characterize some of the properties in terms of combinatorial properties of Borel images. In Section 4 we fin the critical carinalities of most of the new properties, an apply the results to solve a topological version of the minimal tower problem, which was suggeste to us by Scheepers an state in [15]. It seems that some new mathematical tools are require to solve some of the remaining open problems, as the special properties of τ-covers usually o not

6 58 B. Tsaban allow application of stanar methos evelope uring the stuy of classical selection principles. For this very reason, we believe that these are the important problems which must be aresse in the future. However, we suggest in the secon part of this paper two relaxations of the notion of τ-cover, which are easier to work with an may turn out useful in the stuy of the original problems. We emonstrate this by proving results which are still open for the case of usual τ-covers. Part 1: τ-covers 2 Subcovers with stronger properties 2 Definition. Let X be a set of reals, an U, V collections of covers of X. We say that X satisfies ( U V (rea: U choose V if for each cover U U there exists a subcover V U such that V V. Observe that for any pair U, V of collections of countable covers we have that the property S fin (U, V implies ( U V. Gerlits an Nagy [6] prove that for U = Ω an V = Γ, the converse also hols, in fact, S 1 (Ω, Γ = ( Ω Γ. But in general the property ( U V can be strictly weaker than Sfin (U, V. A useful property of this notion is the following. 3 Lemma (Cancellation Laws. For collections of covers U, V, W, (1 ( U V ( V W ( U W, (2 ( U V Sfin (V, W S fin (U, W, (3 S fin (U, V ( V W Sfin (U, W; an (4 ( U V S1 (V, W S 1 (U, W, (5 If W is close uner taking supersets, then S 1 (U, V ( V W S1 (U, W. Moreover, if U V W, then equality hols in (1 (5. Proof. (1 is immeiate. To prove (2, we can apply S fin (V, W to V- subcovers of the given covers. (4 is similar to (2. (3 Assume that U n U, n N, are given. Apply S fin (U, V to choose finite subsets F n U n, n N, such that V = n N F n V. By ( V W, there exists a subset W of V such that W W. Then for each n W F n is a finite (possibly empty subset of U n, an n N (W F n = W W. To prove (5, observe that the resulting cover V contains an element of W, an as W is close uner taking supersets, V W as well.

7 Selection principles an the Minimal Tower problem 59 It is clear that reverse inclusion (an therefore equality hol in (1 (5 when U V W. 4 Corollary. Assume that U V. Then the following equivalences hol: (1 S fin (U, V = ( U V Sfin (V, U. (2 If V is close uner taking supersets, then S 1 (U, V = ( U V S1 (V, U. Proof. We prove (1. Clearly S fin (U, V implies ( U V an Sfin (V, U. On the other han, by applying the Cancellation Laws (2 an then (3 we have that ( U S fin (V, U S fin (U, U V ( U S fin (U, V. V 2.1 When every τ-cover contains a γ-cover 5 Theorem. The following equivalences hol: (1 S 1 (T, Γ = ( T Γ Sfin (Γ, T, (2 S 1 (B T, B Γ = ( B T BΓ Sfin (B Γ, B T, Proof. (1 By the Cancellation Laws 3, ( T Γ Sfin (Γ, T S fin (Γ, Γ. In [7] it was prove that S fin (Γ, Γ = S 1 (Γ, Γ. Thus, ( T Γ Sfin (Γ, T ( T Γ S1 (Γ, Γ, which by the Cancellation Laws is a subset of S 1 (T, Γ. The other irection is immeiate. (2 is similar. 6 Corollary. The following equivalences hol: (1 S 1 (T, Γ = S fin (T, Γ; (2 S 1 (B T, B Γ = S fin (B T, B Γ. Using similar arguments, we have the following. 7 Theorem. The following equivalences hol: (1 S 1 (Ω, Γ = ( T Γ Sfin (Ω, T; (2 S 1 (B Ω, B Γ = ( B T BΓ Sfin (B Ω, B T.

8 60 B. Tsaban 2.2 When every ω-cover contains a τ-cover 8 Theorem. The following inclusions hol: (1 ( Ω T Sfin (Γ, T. (2 ( B Ω B T Sfin (B Γ, B T. Proof. We will prove (1 (the proof of (2 is ientical. Assume that X satisfies ( Ω T. If X is countable then it satisfies all of the properties mentione in this paper. Otherwise let x n, n N, be istinct elements in X. Assume that U n = {Um} n m N, n N, are open γ-covers of X. Define Ũn = {Um n \ {x n }} m N. Then U = n N Ũn is an open ω-cover of X, an thus contains a τ-cover V of X. Let be the inuce quasiorering. 9 Lemma. If X, has a least element, then V contains a γ-cover of X. Proof. Write V = {V n } n N. Let x 0 be a least element in X,. Consier the subsequence {V nk } k N consisting of the elements V n such that x 0 V n. Since τ-covers are large, this sequence is infinite. For all x X we have x 0 x, thus x 0 V n x V n for all but finitely many n. Since x 0 V nk for all k, we have that for all but finitely many k, x V nk. There are two cases to consier. Case 1. For some n x n is a least element in X,. Then V contains a γ- cover Ṽ of X. In this case, for all n x n belongs to all but finitely many members of Ṽ, thus Ṽ Ũn is finite for each n, an W = {U : ( nu \ {x n } Ṽ} is a γ-cover of X. Case 2. For each n there exists x x n with x x n. For each n, U n is a γ-cover of X, thus x belongs to all but finitely many members of V Ũn. Since x n oes not belong to any of the members in V Ũn, V Ũn must be finite. Thus, W = {U : ( nu \ {x n } Ṽ} is a τ-cover of X. If ( Ω T Sfin (T, Ω, then by Corollary 4 ( Ω T = Sfin (Ω, T. 10 Problem. Is ( Ω T = Sfin (Ω, T? 3 Combinatorics of Borel images In this section we characterize several properties in terms of Borel images in the spaces N N an P (N, using the combinatorial structure of these spaces.

9 Selection principles an the Minimal Tower problem The combinatorial structures A quasiorer is efine on the Baire space N N by eventual ominance: f g if f(n g(n for all but finitely many n. A subset Y of N N is calle unboune if it is unboune with respect to. Y is ominating if it is cofinal in N N with respect to, that is, for each f N N there exists g Y such that f g. b is the minimal size of an unboune subset of N N, an is the minimal size of a ominating subset of N N. Define a quasiorer on P (N by a b if a \ b is finite. An infinite set a N is a pseuo-intersection of a family Y P (N if for each b Y, a b. A family Y P (N is a tower if it is linearly quasiorere by, an it has no pseuointersection. t is the minimal size of a tower. A family Y P (N is centere if the intersection of each (nonempty finite subfamily of Y is infinite. Note that every tower in P (N is centere. A centere family Y P (N is a power if it oes not have a pseuointersection. p is the minimal size of a power. 3.2 The property ( B T B Γ For a set of reals X an a topological space Z, we say that Y is a Borel image of X in Z if there exists a Borel function f : X Z such that f[x] = Y. The following classes of sets were introuce in [8]: P: The set of X R such that no Borel image of X in P (N is a power, B: The set of X R such that every Borel image of X in N N is boune (with respect to eventual omination; D: The set of X R such that no Borel image of X in N N is ominating. For a collection J of separable metrizable spaces, let non(j enote the minimal carinality of a separable metrizable space which is not a member of J. We also call non(j the critical carinality of the class J. The critical carinalities of the above classes are p, b, an, respectively. These classes have the interesting property that they transfer the carinal inequalities p b to the inclusions P B D. 11 Definition. For each countable cover of X enumerate bijectively as U = {U n } n N we associate a function h U : X P (N, efine by h U (x = {n : x U n }.

10 62 B. Tsaban U is a large cover of X if, an only if, h U [X] P (N. As we assume that X is infinite an is not a member of any of our covers, we have that each ω-cover of X is a large cover of X. The following lemma is a key observation for the rest of this section. Note that h U is a Borel function whenever U is a Borel cover of X, an h U is continuous whenever all elements of U are clopen. 12 Lemma ([15]. Assume that U is a countable large cover of X. (1 U is an ω-cover of X if, an only if, h U [X] is centere. (2 U contains a γ-cover of X if, an only if, h U [X] has a pseuointersection. (3 U is a τ-cover of X if, an only if, h U [X] is linearly quasiorere by. Moreover, if f : X P (N is any function, an A = {O n } n N is the clopen cover of P (N such that x O n n x, then for U = {f 1 [O n ]} n N we have that f = h U. This Lemma implies that P = ( B Ω BΓ [13]. 13 Corollary. P = ( B Ω B T ( BT BΓ. It is natural to efine the following notion. T: The set of X R such that no Borel image of X in P (N is a tower. 14 Theorem. T = ( B T BΓ. Proof. See [15] for the clopen version of this theorem (a straightforwar usage of Lemma 12. The proof for the Borel case is similar. 15 Corollary. non( ( B T BΓ = non( ( T Γ = t. Proof. By Theorem 14, t non( ( B T BΓ. In [15] we efine T to be the collection of sets for which every countable clopen τ-cover contains a γ-cover, an showe that non(t = t. But ( B T ( BΓ T Γ T. Clearly, P T. The carinal inequality p t b suggests pushing this further by showing that T B; unfortunately this is false. Sets which are continuous images of Borel sets are calle analytic. 16 Theorem. Every analytic set satisfies T. In particular, N N T. Proof. Accoring to [15], no continuous image of an analytic set is a tower. In particular, towers are not analytic subsets of P (N. Since Borel images of analytic sets are again analytic sets, we have that every analytic set satisfies T. The following equivalences hol [13]: S 1 (B Γ, B Γ = B,

11 Selection principles an the Minimal Tower problem 63 S 1 (B Γ, B = D; S 1 (B Ω, B Γ = P. Theorem 16 rules out an ientification of T with any of the selection principles. However, we get the following characterization of S 1 (B T, B Γ in terms of Borel images. 17 Theorem. S 1 (B T, B Γ = T B. Proof. By the Cancellation Laws an Theorem 14, ( BT S 1 (B T, B Γ = S 1 (B Γ, B Γ = T B. B Γ 3.3 The property ( B Ω B T For a subset Y of P (N an a P (N, efine Y a = {y a : y Y }. If all sets in Y a are infinite, we say that Y a is a large restriction of Y. 18 Theorem. For a set X of real numbers, the following are equivalent: (1 X satisfies ( B Ω B T (2 For each Borel image Y of X in P (N, if Y is centere, then there exists a large restriction of Y which is linearly quasiorere by. Proof. 1 2: Assume that Ψ : X P (N is a Borel function, an let Y = Ψ[X]. Assume that Y is centere, an consier the collection A = {O n } n N where O n = {a : n a} Y for each n N. If the set a = {n : Y = O n } is infinite, then a is a pseuointersection of Y an we are one. Otherwise, by removing finitely many elements from A we get that A is an ω-cover of Y. Setting U n = Ψ 1 [O n ] for each n, we have that U = {U n } n N is a Borel ω-cover of X, which thus contains a τ-cover {U an } n N of X. Let a = {a n } n N, an efine a cover V = {V n } n N of X by { U n n a V n = otherwise Then V is a τ-cover of X, an by Lemma 12, Ψ[X] a = h U [X] a = h V [X] is linearly quasiorere by.

12 64 B. Tsaban 2 1: Assume that U = {U n } n N is an ω-cover of X. By Lemma 12, h U [X] is centere. Let a = {a n } n N be a large restriction of h U [X] which is linearly quasiorere by, an efine V as in 1 2. Then h U [X] a = h V [X]. Thus all elements in h V [X] are infinite (i.e., V is a large cover of X, an h V [X] is linearly quasiorere by (i.e., V is a τ-cover of X. Then V \ { } U is a τ-cover of X. 19 Remark. Replacing the Borel sets by clopen sets an Borel functions by continuous functions in the last proof we get that the following properties are equivalent for a set X of reals: (1 Every countable clopen ω-cover of X contains a τ-cover of X. (2 For each continuous image Y of X in P (N, if Y is centere, then there exists a large restriction of Y which is linearly quasiorere by. We o not know whether the open version of this result is true. 3.4 The property U fin (B Γ, B T 20 Definition. A family Y N N satisfies the exclue mile property if there exists g N N such that: (1 For all f Y, g f; (2 For each f, h Y, one of the situations f(n g(n < h(n or h(n g(n < f(n is possible only for finitely many n. 21 Theorem. For a set X of real numbers, the following are equivalent: (1 X satisfies U fin (B Γ, B T ; (2 Every Borel image of X in N N satisfies the exclue mile property. Proof. 1 2: For each n, the collection U n = {Um n : m N}, where Um n = {f N N : f(n m}, m N, is an open γ-cover of N N. Assume that Ψ is a Borel function from X to N N. By stanar arguments we may assume that Ψ 1 [Um] n X for all n an m. Then the collections U n = {Ψ 1 [Um] n : m N}, n N, are Borel γ-covers of X. For all n, the sequence {Um} n m N is monotonically increasing with respect to, therefore as large subcovers of τ-covers are also τ-covers we may use S 1 (B Γ, B T instea of U fin (B Γ, B T to get a τ-cover U = {Ψ 1 [Um n n ]} n N for X. Let g N N be such that g(n = m n for all n. For all x X, as U is a large cover of X, we have that Ψ(x Ug(n n (that is, Ψ(x(n g(n for infinitely many n. Let be the linear quasiorering of X inuce by the τ-cover U. Then

13 Selection principles an the Minimal Tower problem 65 for all x, y X, either x y or y x. In the first case we get that for all but finitely many n Ψ(x(n g(n Ψ(y(n g(n, an in the secon case we get the same assertion with x an y swappe. This shows that Ψ[X] satisfies the exclue mile property. 2 1: Assume that U n = {Um n : m N}, n N, are Borel covers of X which o not contain a finite subcover. Replacing each Um n with the Borel set k m U k n we may assume that the sets U m n are monotonically increasing with m. Define a function Ψ from X to N N so that for each x an n: Ψ(x(n = min{m : x U n m}. Then Ψ is a Borel map, an so Ψ[X] satisfies the exclue mile property. Let g N N be a witness for that. Then the sequence U = {Ug(n n } n N is a τ-cover of X: For each x X we have that g Ψ(x, thus U is a large cover of X. Moreover, for all x, y X, we have by the exclue mile property that at least one of the assertions Ψ(x(n g(n < Ψ(y(n or Ψ(y(n g(n < Ψ(y(n is possible only for finitely many n. Then the first assertion implies that x y, an the secon implies y x with respect to U. 22 Remark. The analogue clopen version of Theorem 21 also hols. We o not know whether there exist an analogue characterization of U fin (Γ, T (the open version in terms of continuous images. 4 Critical carinalities 23 Theorem. non(s fin (B T, B Ω = non(s fin (T, Ω =. Proof. S fin (B Ω, B Ω S fin (B T, B Ω S fin (T, Ω S fin (Γ, Ω, an accoring to [7] an [13], non(s fin (B Ω, B Ω = non(s fin (Γ, Ω =. 24 Theorem. non(s 1 (B T, B Γ = non(s 1 (T, Γ = t. Proof. By Theorem 5, S 1 (T, Γ = ( T Γ S1 (Γ, Γ, thus by Corollary 15, non(s 1 (T, Γ = min{non( ( T Γ, non(s1 (Γ, Γ} = min{t, b} = t. The proof for the Borel case is similar. 25 Definition. x is the minimal carinality of a family Y N N which oes not satisfy the exclue mile property. Therefore b x. A family Y P (N is splitting if for each infinite a N there exists s Y which splits a, that is, such that the sets a s an a \ s are infinite. s is the minimal size of a splitting family. In [14] it is prove that x = max{s, b}. 26 Theorem. non(u fin (B Γ, B T = non(u fin (Γ, T = x.

14 66 B. Tsaban Proof. By Theorem 21, non(u fin (B Γ, B T = x. Thus, our theorem will follow from the inclusion U fin (B Γ, B T U fin (Γ, T once we prove that non (U fin (Γ, T x. To this en, consier a family Y N N of size x which oes not satisfy the exclue mile property, an consier the monotone γ-covers U n, n N, of N N efine in the proof of Theorem 21. Then, as in that proof, we cannot extract from these covers a τ-cover of Y. Thus, Y oes not satisfy U fin (Γ, T. 27 Definition. Let κ ωτ be the minimal carinality of a centere set Y P (N such that for no a P (N, the restriction Y a is large an linearly quasiorere by. It is easy to see (either from the efinitions or by consulting the involve selection properties that κ ωτ an p = min{κ ωτ, t}. In [14] it is prove that in fact κ ωτ = p. 28 Lemma. non( ( B Ω ( B T = non( Ω T = p. Proof. Let P ωτ enote the property that every clopen ω-cover contains a γ-cover. Then ( B Ω ( B T Ω T Pωτ. By Theorem 18 an Remark 19, non( ( B Ω B T = non(p ωτ = κ ωτ = p. 29 Theorem. non(s fin (B Ω, B T = non(s fin (Ω, T = p. Proof. By Corollary 4 an Theorem 23, ( Ω non(s fin (Ω, T = min{non(, non(s fin (T, Ω} = T = min{κ ωτ, } = κ ωτ = p. (1 The proof for the Borel case is the same. 5 Topological variants of the Minimal Tower problem Let c enote the size of the continuum. The following inequalities are well known [4]: p t b c. For each pair except p an t, it is well known that a strict inequality is consistent. 30 Problem (Minimal Tower. Is it provable that p = t? This is one of the major an olest problems of infinitary combinatorics. Allusions to this problem can be foun in Rothberger s works (see, e.g., [10]. We know that S 1 (Ω, Γ S 1 (T, Γ, an that non(s 1 (Ω, Γ = p, an non(s 1 (T, Γ = t. Thus, if p < t is consistent, then it is consistent that

15 Selection principles an the Minimal Tower problem 67 S 1 (Ω, Γ S 1 (T, Γ. Thus the following problem, which was suggeste to us by Scheepers, is a logical lower boun on the ifficulty of the Minimal Tower problem. 31 Problem ([15]. Is it consistent that S 1 (Ω, Γ S 1 (T, Γ? We also have a Borel variant of this problem. 32 Problem. Is it consistent that S 1 (B Ω, B Γ S 1 (B T, B Γ? We will solve both of these problems. For a class J of sets of real numbers with J J, the aitivity number of J is the minimal carinality of a collection F J such that F J. The aitivity number of J is enote a(j. 33 Lemma ([15]. Assume that Y P (N is linearly orere by, an for some κ < t, Y = α<κ Y α where each Y α has a pseuointersection. Then Y has a pseuointersection. 34 Theorem. a( ( B T BΓ = a( ( T Γ = t. Proof. By Theorem 14 an Lemma 33, we have that a( ( B T BΓ = a(t = t. The proof that a( ( T Γ = t is not as elegant an requires a back-an-forth usage ( of Lemma 12. Assume that κ < t, an let X α, α < κ, be sets satisfying T Γ. Let U be a countable open τ-cover of X = α<κ X α. Then h U [X] = α<κ h U[X α ] is linearly quasiorere by. Since each X α satisfies ( T Γ, for each α U contains a γ-cover of X α, that is, h U [X α ] has a pseuo-intersection. By Lemma 33, h U [X] has a pseuo-intersection, that is, U contains a γ-cover of X. 35 Theorem. a(s 1 (B T, B Γ = t. Proof. By Theorem 5, S 1 (B T, B Γ = ( B T BΓ S1 (B Γ, B Γ, an accoring to [1], a(s 1 (B Γ, B Γ = b. By Theorem 34, we get that a(s 1 (B T, B Γ min{t, b} = t. On the other han, by Theorem 24 we have a(s 1 (B T, B Γ non(s 1 (B T, B Γ = t. In [12] Scheepers proves that S 1 (Γ, Γ is close uner taking unions of size less than the istributivity number h. Consequently, we get that a(s 1 (T, Γ = t [1]. As it is consistent that S 1 (Ω, Γ is not close uner taking finite unions [5], we get a positive solution to Problem 31. We will now prove something stronger: Consistently, no class between S 1 (B Ω, B Γ an ( Ω T (inclusive is close uner taking finite unions. This solves Problem 31 as well as Problem 32.

16 68 B. Tsaban 36 Theorem (CH. There exist sets of reals A an B satisfying S 1 (B Ω, B Γ, such that A B oes not satisfy ( Ω T. In particular, S1 (T, Γ S 1 (Ω, Γ, an S 1 (B T, B Γ S 1 (B Ω, B Γ. Proof. By a theorem of Brenle [3], assuming CH there exists a set of reals X of size continuum such that all subsets of X satisfy P. (Recall that P = S 1 (B Ω, B Γ. As P is close uner taking Borel (continuous is enough images, we may assume that X [0, 1]. For Y [0, 1], write Y + 1 = {y + 1 : y Y } for the translation of Y by 1. As X = c an only c many out of the 2 c many subsets of X are Borel, there exists a subset Y of X which is not F σ neither G δ. By a theorem of Galvin an Miller [5], for such a subset Y the set (X \ Y (Y + 1 oes not satisfy S 1 (Ω, Γ. Set A = X \ Y an B = Y + 1. Then A an B satisfy S 1 (B Ω, B Γ, an A B oes not satisfy S 1 (Ω, Γ = ( ( Ω T T Γ. By Theorem 34, A B satisfies ( ( T Γ an therefore it oes not satisfies Ω T. But by Theorem 35, the set A B satisfies S 1 (B T, B Γ. 6 Special elements 6.1 The Cantor set C Let C R be the canonic mile-thir Cantor set. 37 Proposition. Cantor s set C oes not satisfy S fin (Γ, T. Proof. Ha it satisfie this property, we woul have by Theorem 16 that C ( T Γ Sfin (Γ, T = S 1 (Γ, Γ, contraicting [7]. Thus C satisfies S fin (T, Ω an U fin (Γ, T, an none of the other new properties. 6.2 A special Lusin set In Subsection 3.3 of [1] we construct, using cov(m = c, special Lusin sets of size c which satisfy S 1 (B Ω, B Ω. The meta-structure of the proof is as follows. At each stage of this construction we efine a set Yα which is a union of less that cov(m many meager sets, an choose an element x α G α \ Yα where G α is a basic open subset of N Z. 38 Theorem. If cov(m = c, then there exists a Lusin set satisfying S 1 (B Ω, B Ω but not U fin (Γ, T. Proof. We moify the aforementione construction so to make sure that the resulting Lusin set L oes not satisfy the exclue mile property. As we o not nee to use any group structure, we will work in N N rather than N Z.

17 Selection principles an the Minimal Tower problem Lemma. Assume that A is an infinite set of natural numbers, an f N N. Then the sets are meager subsets of N N. Proof. For each k, the sets M f,a = {g N N : [g f] A is finite} M f,a = {g N N : [f < g] A is finite} N k = {g N N : ( n > k n A f(n < g(n} Ñ k = {g N N : ( n > k n A g(n f(n} are nowhere ense in N N. Now, M f,a = k N N k, an M f,a = k N Ñk. Consier an enumeration f 2α : α < c of N N which uses only even orinals. At stage α for α even, let Y α be the set efine in [1], an let Ỹ α be the union of Y α an the two meager sets M fα,n = {g N N : [g f α ] is finite} M fα,n = {g N N : [f α < g] is finite}. Then Ỹα is a union of less than cov(m many meager sets. Choose x α G α \Ỹ α. In step α + 1 of the construction let Yα+1 be efine as in [1], an let Ỹ α+1 be the union of Yα+1 with the meager sets M fα,[f α<x α] = {g N N : [g f α < x α ] is finite} M fα,[x α f α] = {g N N : [x α f α < g] is finite}. Now choose x α+1 G α+1 \ Ỹ α+1. Then x α an x α+1 witness that f α oes not avoi miles in the resulting set L = {x α : α < c}. Consequently, L oes not satisfy U fin (Γ, T. The proof that L satisfies S 1 (B Ω, B Ω is as in [1]. 6.3 Sierpinski sets If a Sierpinski set satisfies ( T Γ, then it satisfies S1 (T, Γ but not S 1 (O, O. Such a result woul give another solution to Problem 31. However, as towers are null in the usual measure on N {0, 1}, it is not straightforwar to construct a Sierpinski set which is a member of ( T Γ. 40 Problem. Does there exist a Sierpinski set satisfying ( T Γ?

18 70 B. Tsaban 6.4 Unsettle implications The paper [13] rule out the possibility that any selection property for the open case implies any selection property for the Borel case. Some implications are rule out by constructions of [7] an [13]. Several other implications are eliminate ue to critical carinality consierations. 41 Problem. Which implications can be ae to the iagram in Figure 2 an to the corresponing Borel iagram? A summary of all unsettle implications appears in [18]. As a first step towars solving Problem 41, one may try to answer the following. 42 Problem. What are the critical carinalities of the remaining classes? Part 2: Variations on the theme of τ-covers 7 τ -covers The notion of a τ -cover is a more flexible variant of the notion of a τ-cover. 43 Definition. A family Y P (N is linearly refinable if for each y Y there exists an infinite subset ŷ y such that the family Ŷ = {ŷ : y Y } is linearly quasiorere by. A cover U of X is a τ -cover of X if it is large, an h U [X] (where h U is the function efine before Lemma 12 is linearly refinable. For x X, we will write x U for h U (x, an ˆx U for the infinite subset of x U such that the sets ˆx U are linearly quasiorere by. If U is a countable τ-cover, then h U [X] is linearly quasiorere by an in particular it is linearly refinable. Thus every countable τ-cover is a τ -cover. The converse is not necessarily true. Let T (B T enote the collection of all countable open (Borel τ -covers of X. Then T T Ω an B T B T Ω. Often problems which are ifficult in the case of usual τ-covers become solvable when shifting to τ -covers. We will give several examples. 7.1 Refinements One of the major tools in the analysis of selection principles is to use refinements an e-refinements of covers. In general, the e-refinement of a τ-cover is not necessarily a τ-cover. 44 Lemma. Assume that U T refines a countable open cover V (that is, for each U U there exists V V such that U V. Then V T. The analogous assertion for countable Borel covers also hols.

19 Selection principles an the Minimal Tower problem 71 Proof. Fix a bijective enumeration U = {U n } n N. Let ˆx U, x X, be as in the efinition of τ -covers. For each n let V n V be such that U n V n. We claim that W = {V n : n N} T. As W is an ω-cover of X, it is infinite; fix a bijective enumeration {W n } n N of W. For each n efine S n = {k : U k W n }, an S n = S n \ m<n S m. For each x X efine ˆx W by: n ˆx W S n ˆx U. Then each ˆx W is a subset of x W. Each ˆx W is infinite: For each W n1,..., W nk choose x i W ni, i = 1,..., k. Then {x, x 1,..., x k } W ni for all i = 1,..., k. As U is an τ -cover of X, there exists m ˆx U such that {x, x 1,..., x k } U m. Consier the (unique n such that m S n. Then U m W n ; therefore W n {W n1,..., W nk }, an in particular n {n 1,..., n k }. As m S n ˆx U, we have that n ˆx W. The sets ˆx W are linearly quasiorere by : Assume that a, b X. We may assume that â U ˆbU. As lim n min S n, we have that S n â U S n ˆb U for all but finitely many n. This shows that W is a τ -cover of X. Now, V is an extension of W by at most countably many elements. It is easy to see that an extension of a τ -cover by countably many open sets is again a τ -cover, see [17]. The first consequence of this important Lemma is that S fin (U, T (S fin (U, B T implies U fin (U, T (U fin (U, B T that is, the analogue of Lemma 1 hols. 45 Corollary. Assume that U = n N F n, where each F n is finite, is a τ - cover of a space X. Then either F n = X for some n, or else V = { F n } n N is also a τ -cover of X. Proof. U refines V. 7.2 Equivalences All equivalences mentione in Subsection 1.4 hol for τ -covers as well. In particular, the analogue of Theorem 5 hols (with a similar proof. 46 Corollary. The following equivalences hol: (1 S 1 (T, Γ = S fin (T, Γ; (2 S 1 (B T, B Γ = S fin (B T, B Γ. In fact, in the Borel case we get more equivalences in the case of τ -covers than in the case of τ-covers see Subsection 7.4.

20 72 B. Tsaban 7.3 Continuous images We now solve the problems mentione in Remarks 19 an 22 in the case of τ -covers. 47 Theorem. The following properties are equivalent for a set X of reals: (1 X satisfies ( Ω T ; (2 For each continuous image Y of X in P (N, if Y is centere, then Y is linearly refinable. Proof. 1 2: The proof for this is similar to the proof of 1 2 in Theorem : Assume that U is an ω-cover of X. Replacing each member of U with all finite unions of Basic clopen subsets of it, we may assume that all members of U are clopen (to unravel this assumption we will use the fact that T is close uner e-refinements. Thus, h U is continuous an Y = h U [X] is centere. Consequently, Y is linearly refinable, that is, U is a τ -cover of X. 48 Remark. The analogue assertion (to Theorem 47 for the Borel case, where open covers are replace by Borel covers an continuous image is replace by Borel image, also hols an can be prove similarly. As in [14], we will use the notation [f h] := {n : f(n g(n}. Then a subset Y N N satisfies the exclue mile property if, an only if, there exists a function h N N such that the collection {[f h] : f Y } is a subset of P (N an is linearly quasiorere by. 49 Definition. We will say that a subset Y N N satisfies the weak exclue mile property if there exists a function h N N such that the collection {[f h] : f Y } is linearly refinable. Recall that U fin (Γ, T = U fin (O, T. 50 Theorem. For a zero-imensional set X of real numbers, the following are equivalent: (1 X satisfies U fin (O, T ; (2 Every continuous image of X in N N satisfies the weak exclue mile property.

21 Selection principles an the Minimal Tower problem 73 Proof. We make the neee changes in the proof of Theorem 2.1 in [16]. 2 1: Assume that U n, n N, are open covers of X which o not contain finite subcovers. For each n, replacing each member of U n with all of its basic clopen subsets we may assume that all elements of U n are clopen, an thus we may assume further that they are isjoint. For each n enumerate U n = {U n m} m N. As we assume that the elements U n m, m N, are isjoint, we can efine a function Ψ from X to N N by Ψ(x(n = m x U n m. Then Ψ is continuous. Therefore, Y = Ψ[X] satisfies the weak exclue mile property. Let h N N, an for each f Y, A f [f h] be such that {A f : f Y } is linearly quasiorere by. For each n set F n = {U n k : k h(n}. We claim that U = { F n } n N is a τ -cover of X. We will use the following property. ( For each finite subset F of X an each n x F A Ψ(x, F F n. Let { F kn } n N be a bijective enumeration of U, an let f N N be such that for each n, F n = F kf(n. For each x X set ˆx U = f[a Ψ(x ]. We have the following. ˆx U is a subset of x U : Assume that f(n ˆx U, where n A Ψ(x. Then x F n = F kf(n, therefore f(n x U. ˆx U is infinite: Assume that f[a Ψ(x ] = {f(n 1,..., f(n k } where n 1,..., n k A Ψ(x. For each i k choose x i F, an set F = {x, x kf(ni 1,..., x k }. Then for all i k F F. Choose n kf(ni a F A Ψ(x. By property (, F F n = F kf(n, therefore f(n {f(n 1,..., f(n k }. But n A Ψ(x, thus f(n ˆx U, a contraiction. As the sets A Ψ(x are linearly quasiorere by, so are the sets ˆx U = f[a Ψ(x ]. 1 2: Since Ψ is continuous, Y = Ψ[X] also satisfies U fin (O, T. Consier the basic open covers U n = {Um} n m N efine by Um n = {f Y : f(n = m}. Then there exist finite F n U n, n N, such that either Y = F n for some n, or else V = { F n : n N} is a τ -cover of Y. The first case can be split into two sub-cases: If there exists an infinite set A N such that Y = F n, then for each n A the set {f(n : f Y } is finite, an we can efine { max{f(n : f Y } n A h(n = 0 otherwise

22 74 B. Tsaban so that A [f h] for each f Y, an we are one. Otherwise Y = F n for only finitely many n, therefore we may replace each F n satisfying Y = F n with F n =, so we are in the secon case. The secon case is the interesting one. V = { F n : n N} is a τ -cover of Y fix a bijective enumeration { F kn } n N of V an witnesses ˆf V, f Y, for that. Define h(n = max{m : Um n F n } for each n. Then the subsets {k n : n ˆf V } of [f h], f Y, are infinite an linearly quasiorere by. This shows that Y is linearly refinable. 7.4 Borel images Define the following notion. T : The set of X R such that for each linearly refinable Borel image Y of X in P (N, Y has a pseuointersection. By the usual metho we get the following. 51 Lemma. T = ( B T B Γ. Clearly T implies T. 52 Lemma. non(t = t. Proof. It is easy to see that non(t is the minimal size of a linearly refinable family Y P (N which has no pseuointersection. We will show that t non(t. Assume that Y P (N is a linearly refinable family of size less than t, an let Ŷ be a linear refinement of Y. As Ŷ Y < t, Ŷ has a pseuointersection, which is in particular a pseuointersection of Y. An application of Lemma 51 an the Cancellation Laws implies the following. 53 Theorem. S 1 (B T, B Γ = T B. We o not know whether T = T. In particular, we have the following (recall Theorem Problem. Is it true that every analytic set of reals satisfies T? Does N N T? We o not know whether S 1 (B Γ, B T = U fin (B Γ, B T or not. This can be contraste with the following result. 55 Theorem. For a set X of real numbers, the following are equivalent: (1 X satisfies S 1 (B Γ, B T, (2 X satisfies S fin (B Γ, B T, (3 X satisfies U fin (B Γ, B T ;

23 Selection principles an the Minimal Tower problem 75 (4 Every Borel image of X in N N satisfies the weak exclue mile property. Proof. Clearly : This can be prove like 1 2 in Theorem : Assume that U n = {Uk n : k N}, n N, are Borel γ-covers of X. We may assume that these covers are pairwise isjoint. Define a function Ψ : X N N so that for each x an n: Ψ(x(n = min{k : ( m k x U n m}. Then Ψ is a Borel map, an so Y = Ψ[X] satisfies the weak exclue mile property. Let h N N an A f [f h], f Y, be witnesses for that. Set U = {Uh(n n } n N. For each x X set ˆx U = A Ψ(x. Then ˆx U is infinite an ˆx U [Ψ(x h] x U for each x X, an the sets ˆx U are linearly quasiorere by. 7.5 Critical carinalities The argument of Theorem 23 implies that non(s fin (B T, B Ω = non(s fin (T, Ω =. 56 Theorem. non(s 1 (B T, B Γ = non(s 1 (T, Γ = t. Proof. By Theorem 53, non(s 1 (B T, B Γ = min{non(t, non(b} = min{t, b} = t. On the other han, S 1 (T, Γ implies S 1 (T, Γ, whose critical carinality is t. Define the following properties. X: The set of X R such that each Borel image of X in N N satisfies the exclue mile property. wx: The set of X R such that each Borel image of X in N N satisfies the weak exclue mile property. Recall that by Theorem 21, U fin (B Γ, B T = X. In Theorem 55 we prove that S 1 (B Γ, B T = wx. We o not know whether wx = X. 57 Problem. Is non(wx = x?

24 76 B. Tsaban 7.6 Finite powers In [7] it is observe that if U is an ω-cover of X, then for each k U k = {U k : U U} is an ω-cover of X k. Similarly, it is observe in [15] that if U is a τ-cover of X, then for each k U k is a τ-cover of X k. We will nee the same assertion for τ -covers. 58 Lemma. Assume that U is a τ -cover of X. Then for each k, U k is a τ -cover of X k. Proof. Fix k. Let U = {U n } n N be an enumeration of U, an let ˆx U x U, x X, witness that U is a τ -cover of X. For each x = (x 0,..., x k 1 X k efine (xˆ i U. ˆ x U k = i<k As the sets ˆx U are infinite an linearly quasiorere by, the sets ˆ x U k are also infinite an linearly quasiorere by. Moreover, for each n ˆ x U k an each i < k, n (x ˆ i U, an therefore x i U n for each i < k; thus x Un, k as require. In [7] it is prove that the classes S 1 (Ω, Γ, S 1 (Ω, Ω, an S fin (Ω, Ω are close uner taking finite powers, an that none of the remaining classes they consiere has this property. Actually, their argument for the last assertion shows that assuming CH, there exist a Lusin set L an a Sierpinski set S such that L L an S S can be mappe continuously onto the Baire space N N. Consequently, we have that none of the classes S 1 (Γ, T, S fin (Γ, T, U fin (Γ, T, S 1 (T, O, an their corresponing Borel versions, is close uner taking finite powers. We o not know whether the remaining 7 classes which involve τ-covers are close uner taking finite powers. 59 Theorem. S 1 (Ω, T an S fin (Ω, T are close uner taking finite powers. Proof. We will prove the assertion for S 1 (Ω, T ; the proof for the remaining assertion is similar. Fix k. In [7] it is prove that for each open ω-cover U of X k there exists an open ω-cover V of X such that the ω-cover V k of X k refines U. Assume that {U n } n N is a sequence of open ω-covers of X k. For each n choose an open ω-cover V n of X such that Vn k refines U n. Apply S 1 (Ω, T to extract elements V n V n, n N, such that W = {V n } n N T. By Lemma 58, W k is a τ -cover of X k. For each n choose U n U n such that Vn k U n. Then by Lemma 44, {U n } n N is a τ -cover of X.

25 Selection principles an the Minimal Tower problem Strong properties Assume that {U n } n N is a sequence of collections of covers of a space X, an that V is a collection of covers of X. The following selection principle is efine in [17]. S 1 ({U n } n N, V: For each sequence {U n } n N where for each n U n U n, there is a sequence {U n } n N such that for each n U n U n, an {U n } n N V. The notion of strong γ-set, which is ue to Galvin an Miller [5], is a particular instance of the new selection principle, where V = Γ an for each n U n = O n, the collection of open n-covers of X (we use here the simple characterization given in [17]. It is well known that the γ-property S 1 (Ω, Γ oes not imply the strong γ-property S 1 ({O n } n N, Γ. It is an open problem whether S 1 (Ω, T implies S 1 ({O n } n N, T. The following notions are efine in [17]. A collection U of open covers of a space X is finitely thick if: (1 If U U an for each U U F U is a finite family of open sets such that for each V F U, U V X, then U U F U U. (2 If U U an V = U F where F is finite an X F, then V U. A collection U of open covers of a space X is countably thick if for each U U an each countable family V of open subsets of X such that X V, U V U. Whereas T is in general not finitely thick nor countably thick, T is both finitely an countably thick [17]. In [17] it is prove that if V is countably thick, then S 1 (Ω, V = S 1 ({O n } n N, V. Consequently, S 1 (Ω, T = S 1 ({O n } n N, T. 7.8 Closing on the Minimal Tower problem Clearly S 1 (T, Γ implies S 1 (T, Γ, an S 1 (B T, B Γ implies S 1 (B T, B Γ. So we now have new topological lower bouns on the Minimal Tower problem. 60 Problem. (1 Is S 1 (Ω, Γ = S 1 (T, Γ? (2 Is S 1 (B Ω, B Γ = S 1 (B T, B Γ? We also have a new combinatorial boun. 61 Definition. p is the minimal size of a centere family in P (N which is not linearly refineable. p. 62 Theorem. The critical carinalities of the properties ( B Ω B T an ( Ω T is Proof. This follows from Theorem 47 an Remark 47.

26 78 B. Tsaban The following can ( be prove either irectly from the efinitions or from the = Ω ( T T. equivalence ( Ω Γ Γ 63 Corollary. p = min{p, t}. Thus, if p < t is consistent, then p < t is consistent as well. We therefore have the following problem. 64 Problem. Is p = p? 7.9 The remaining Borel classes We are left with Figure 3 for the Borel case. S 1 (B Γ, B Γ S 1 (B Γ, B T S 1 (B Γ, B Ω S 1 (B Γ, B S fin (B T, B T S fin (B T, B Ω S fin (B Ω, B T S fin (B Ω, B Ω S 1 (B T, B Γ S 1 (B T, B T S 1 (B T, B Ω S 1 (B T, B S 1 (B Ω, B Γ S 1 (B Ω, B T S 1 (B Ω, B Ω S 1 (B, B Figure 3. 8 Sequences of compatible τ-covers When consiering sequences of τ-covers, it may be convenient to have that the linear quasiorerings they efine on X agree, in the sense that there exists a Borel linear quasiorering on X which is containe in all of the inuce quasiorerings. In this case, we say that the τ-covers are compatible. We thus have the following new selection principle: S 1 (T, V: For each sequence {U n} n N of countable open compatible τ-covers of X there is a sequence {U n } n N such that for each n U n U n, an {U n } n N V. The selection principle S fin (T, V is efine similarly. Replacing open by Borel gives the selection principles S 1 (B T, V an S fin (B T, V. The following

27 Selection principles an the Minimal Tower problem 79 implications hol: S 1 (T, V S fin (T, V ( T V S 1 (T, V S fin (T, V an similarly for the Borel case. For V = Γ the new notions coincie with the ol ones. 65 Proposition. The following equivalences hol: (1 S 1 (T, Γ = S 1 (T, Γ = S fin(t, Γ = S fin (T, Γ; (2 S 1 (B T, B Γ = S 1 (B T, B Γ = S fin (B T, B Γ = S fin (B T, B Γ. Proof. We will prove (1; (2 is similar. By Theorem 5, we have the following implications S 1 (T, Γ S fin (T, Γ ( T Γ Sfin (Γ, Γ = S 1 (T, Γ S 1 (T, Γ S fin (T, Γ 8.1 The class S fin (B T, B T 66 Definition. A τ-cover of X, is a τ-cover of X such that the inuce quasiorering contains. 67 Lemma. Let X, be a linearly quasiorere set of reals, an assume that every Borel image of in N N is boune (with respect to. Assume that U n = {Uk n : k N} are Borel τ-covers of X,. Then there exist finite subsets F n of U n, n N, such that n N F n is a τ-cover of X,. Proof. Fix a linear quasiorering of X, an assume that U n = {U n k : k N} are Borel τ-covers of X,. Define a Borel function Ψ from to N N by: Ψ(x, y(n = min{k : ( m k x U n m y U n m}. Ψ[] is boune, say by g. Now efine a Borel function Φ from X to N N by: Φ(x(n = min{k : g(n k an x U n k } Note that Φ[X] is a Borel image of in N N, thus it is boune, say by f. It follows that the sequence {Ug(n n,..., U f(n n : g(n f(n} n N is large, an is a τ-cover of X.

28 80 B. Tsaban Accoring to [13], the property that every Borel image is boune is equivalent to S 1 (B Γ, B Γ. 68 Lemma. Let P be a collection of spaces which is close uner taking Borel subsets, continuous images (or isometries, an finite unions. Then for each set X of real numbers, the following are equivalent: (1 Each Borel linear quasiorering of X satisfies P, (2 X 2 satisfies P. (3 There exists a Borel linear quasiorering of X satisfying P, Proof : The set = X 2 is a linear quasiorering of X. 3 2: If satisfies P, then so oes its continuous image = {(y, x : x y}. Thus, X 2 = satisfies P. 2 1: P is close uner taking Borel subsets. Thus, lemma 67 can be restate as follows. 69 Theorem. If X 2 satisfies S 1 (B Γ, B Γ, then X satisfies S fin (B T, B T. Proof. The property S 1 (B Γ, B Γ satisfies the assumptions of Lemma Problem. Assume that X satisfies S fin (B T, B T. Is it true that X 2 satisfies S 1 (B Γ, B Γ? Since b = cov(n = cof(n (in particular, the Continuum Hypothesis implies that S 1 (B Γ, B Γ is not close uner taking squares [13], a positive answer to Problem 70 woul imply that the property S 1 (B Γ, B Γ oes not imply S fin (B T, B T. Acknowlegements. I wish to thank Martin Golstern for reaing this paper an making several important comments. I also thank Saharon Shelah for the opportunity to introuce this work at his research seminar, an for the fruitful cooperation which followe [14]. Finally, I woul like to thank the organizers of the Lecce Workshop on Coverings, Selections an Games in Topology (June 2002 Cosimo Guio, Ljubisa D. R. Kočinac, an Marion Scheepers for inviting me to give a lecture on this paper in their workshop an for the kin hospitality uring the workshop. Ae in proof. Problem 42 was almost completely solve in: H. Milenberger, S. Shelah, an B. Tsaban, The combinatorics of τ-covers (see arxiv.org/abs/math.gn/ There remains exactly one unsettle critical carinality in the iagram. It follows from the results of that paper that the answer to the question before Theorem 55 is negative.