@ Applied General Topology c© Universidad Politécnica de Valencia Volume 4, No. 2, 2003 pp. 317–325 Five different proofs of extraresolvability of countable totally bounded groups G. Artico, V. I. Malykhin and U. Marconi ∗ Dedicated to Professor S. Naimpally on the occasion of his 70th birthday. Abstract. We give different proofs of extraresolvability for count- ably infinite topological spaces and in particular for totally bounded groups. 2000 AMS Classification: 54H11, 54F99, 22A05. Keywords: Resolvable space, weak sequence, totally bounded group. 1. Introduction. For a subset A of a topological space X, the closure and the derived set of A are denoted by A and A′, respectively. As usual, A is said to be discrete provided that A ∩ A′ = ∅; A is said to be nowhere dense if the interior of A is empty. A family A refines a family B if every element of A is contained in some element of B. In this paper every space is assumed to be T1 and dense in itself. The following elementary fact will be used in the sequel. Proposition 1.1. In a dense in itself topological space every discrete subset is nowhere dense. In 1943 E. Hewitt called a space resolvable if it has two disjoint dense subsets [8]. Subsequently, discussing a result of Sierpiński [17], J. G. Ceder introduced maximally resolvable spaces: he proved that locally compact spaces and spaces with a linearly ordered base of neighborhoods are maximally resolvable [3]. A space is said to be κ-resolvable if there exists a collection (resolution) of κ- many disjoint dense subsets. The resolution may be chosen in such a way that every set intersects every non- empty open subset in at least κ-many points. A space X is said to be maximally resolvable if it is ∆(X)-resolvable, where ∗Work supported by the grant “Progetti di ricerca di Ateneo” of the University of Padova. 318 G. Artico, V. I. Malykhin and U. Marconi ∆(X) is the minimum cardinal of a non-empty open set. Interesting results about finite, countable and maximal resolvability may be found in [6, 15, 9, 10]. In [12], the authors prove that each totally bounded topological group is maximally resolvable. The following definition is due to V. I. Malykhin (e.g., see [7]). Definition 1.2. A space X is called extraresolvable if there exists a family D of dense subsets of X such that |D| > ∆(X) and A ∩ B is nowhere dense whenever A and B are distinct elements of D. In [11] V. I. Malykhin proved that every countably infinite totally bounded group is extraresolvable (through an almost disjoint family of cardinality c). Recall that a family D of infinite subsets of a set is said to be almost disjoint if A∩B is finite whenever A and B are distinct elements of D. Below we give a small survey with different proofs of extraresolvability for countably infinite spaces and, in particular, for totally bounded groups. Each proof presents different techniques and results which may be useful in studying this topic. Each section title is given according to the main tool of resolvability used in that section. The neutral element of a group will always be denoted by e and every group topology is assumed to be T0 (hence completely regular). Furthermore, we work with groups which admit a totally bounded Hausdorff topology (e.g., every abelian group). 2. Discrete subsets. A subset M of a topological space X is said to be strongly discrete if for every x ∈ M there exists an open neighborhood V (x) of x such that V (a)∩V (b) = ∅ whenever a and b are different points of M. Definition 2.1. A point z of X is called an lsd-point if there exists a strongly discrete subset M such that z ∈ M′. The proof of the following propositions is straightforward. Proposition 2.2. A point z is an lsd-point iff there exists a set M, points of which have a disjoint system of open neighborhoods Θ = {V (x) : x ∈ M} such that z 6∈ ∪Θ and z ∈ M. Proposition 2.3. Let M be a strongly discrete subset of a regular space X. If z 6∈ M then M ∪{z} is strongly discrete. Proposition 2.4. Let Y be a subset of an Hausdorff space X. If Y ′ 6= ∅, then Y contains an infinite strongly discrete subset. In [16] P.L. Sharma and S. Sharma proved that if every point of a T1 space is an lsd-point, then the space is ℵ0-resolvable. In the following theorem we use some ideas of their construction. Theorem 2.5. Let X be a countably infinite regular space. If every point of X is an lsd-point, then X is extraresolvable through a collection of cardinality c. Extraresolvability 319 Proof. Let {zn : n ∈ ω} be a one to one numeration of the space X. Since z0 is an lsd-point, then there exist M and Θ = {V (x) : x ∈ M} as in Proposition 2.2. Put M0 = M and Θ0 = Θ. By Proposition 2.3, it is not restrictive to assume that z1 ∈ M0. Now we are going to describe the next step of the inductive construction, which is very familiar with the general step. For each x ∈ M0, still choose Mx and Θx as in Proposition 2.2. We can assume that ⋃ Θx ⊆ V (x) for each x ∈ M0. Let M1 = ⋃ {Mx : x ∈ M0}. The set M1 is strongly discrete with disjoint system of open neighborhoods Θ1 = ⋃ {Θx : x ∈ M1}. Notice that Θ1 refines Θ0. Still by Proposition 2.3, it is not restrictive to assume that z2 ∈ M0 ∪M1. In the general case, by repeating the process for every x ∈ Mn, we get sequences {Mn} and {Θn} satisfying the following: (1) zn+1 ∈ M0 ∪ . . .∪Mn, (2) Mn is strongly discrete with disjoint systems of open neighborhoods Θn, (3) Mn ⊆ M′n+1, (4) Θn+1 refines Θn. By (2) and (3), we get that ( ⋃ Θn+1)∩Mn = ∅; consequently, by (4), the sets Mn are mutually disjoint. For every infinite subset A ⊆ ω put X(A) = ⋃ n∈A Mn. By (1) and (3), X(A) is dense in X. Since the sets Mn are mutually disjoint, then X(A) ∩X(B) = X(A ∩ B). Consequently X(A) ∩ X(B) is nowhere dense whenever A ∩ B is finite (Proposition 1.1). Thus if A is an almost disjoint family of cardinality c of infinite subsets of ω, then X(A) = {X(A) : A ∈ A} is a collection of cardinality c which ensures the extraresolvability of X. � Theorem 2.6. Every countably infinite totally bounded Hausdorff group is extraresolvable. Proof. I. V. Protasov [13] constructed a strongly discrete subset D such that e ∈ D′ in every totally bounded group topology. Consequently the identity (hence every element) is an lsd-point and the conclusion follows from Theorem 2.5. � 3. Weak sequences. A weak sequence on X is a countably infinite disjoint family F of finite subsets of X. We say that F converges to a point x if {F ∈ F : F ∩V = ∅} is finite for each neighborhood V of x. Proposition 3.1. Let X be a countably infinite space such that every point admits a weak sequence converging to it. Then there exist a weak sequence converging to every point of X. Proof. Let X = {zn : n ∈ ω} be a one to one numeration of X and for each n let Fn be a weak sequence converging to zn. We shall construct a new weak sequence {Km : m ∈ ω} which converges to each element of the space. By induction, suppose that K0, . . . ,Km−1 have been already defined. The set 320 G. Artico, V. I. Malykhin and U. Marconi T = ⋃ i ∆(X) and |A∩B| < nwd(X) whenever A and B are distinct elements of A. W. W. Comfort and S. Garcia-Ferreira proved that if d(X) = |G| ≥ ω, then G is strongly extraresolvable [5]. The previous Theorem 3.5 proves that a countably infinite totally bounded group is strongly extraresolvable through a family of cardinality c. 4. Talagrand’s theorem. In this section we shortly present the proof of Theorem 3.5 given in [11]. We identify a subset A ⊆ X with its characteristic function χA ∈ 2X, where 2X has the product topology. If F denotes a family of subsets of X, then the subspace B(F) = {χF : F ∈F}⊆ 2X is called the binary space of F. M. Talagrand [18] proved that for a free filter F on a set X the following conditions are equivalent: (a) B(F) is meager (as a subset of 2X). (b) There exists a sequence Kn of mutually disjoint finite subsets of X such that the set {n : Kn ∩F = ∅} is finite for each F ∈F. An accurate reading of the proof of Talagrand’s theorem shows that it suffices to assume that the family F of non-empty subsets of X satisfy the following condition: if A ∈F and B ⊇ A then B ∈F. A subset L of an infinite group G is called large if there exists a finite subset K ⊆ G such that KL = LK = G. In [1] the authors proved that the binary space B(L) of all large subsets of a countably infinite group G is meager. So, according to the general form of Talagrand’s theorem, there exists a sequence {Kn} of mutually disjoint finite subsets of G such that {n : Kn ∩ L = ∅} is finite for every large subset L of G. As each non-empty open subset of a totally bounded group is large, we deduce the following proposition. Proposition 4.1. Let G be a countably infinite group. There exists a weak sequence {Kn} which converges at each point of G with respect to any Hausdorff totally bounded group topology. 322 G. Artico, V. I. Malykhin and U. Marconi The third proof of Theorem 2.6 follows as in Theorem 3.2 by considering the family of sets G(A) = ⋃ n∈A Kn, where A ranges over an almost disjoint family of cardinality c. 5. Quadrosequences. The proof given here uses some ideas which are already present in Section 3. One interesting point is the result provided in Proposition 5.1. We denote by Xm = {x1, . . . ,xm} a one to one numeration of a set with m elements of a group G and by D(Xm) the set {xix−1k : i < k, k ≤ m}. Proposition 5.1. Let G = {gn : n ∈ ω} be a one to one numeration of a countably infinite group G. There exist infinite subsets Xn, n ∈ ω, such that the subsets Tn = {g1, . . . ,gn} ·D(Xn) are mutually disjoint. Proof. Let us consider a general step of a pyramidal inductive construction. Let us assume that some initial parts Xmkk = {xk,1, . . . ,xk,mk} have already been defined for each k ≤ n in such a way that the sets T mk k = {g1, . . . ,gk} ·D(X mk k ) are pairwise disjoint. The n + 1th step consists in adding a new point to every X mk k and starting with the first two points of Xn+1. • Adding a new point to some Xmkk , for k ≤ n. In this case, an added point x = xmk+1 must satisfy to the following conditions: giyx −1 6∈ Tmrr , i ≤ k, r 6= k, r ≤ n, y ∈ X mk k . The sets Tmrr vary during the process of adding these points. Such a point x does exist since the number of excluded conditions is finite. • Forming a new set X2n+1 = {a,b}. In this case the elements a = xn+1,1 and b = xn+1,2 must satisfy to the following conditions: giab −1 6∈ Tmkk , i ≤ n + 1, k ≤ n. This construction ends the proof. � Arguing as in Theorem 3.4, item (2), one obtains the following: Lemma 5.2. Let F be a finite subset of G and let X be a countably infinite subset of G. There exists an infinite subset Y of X such that the sets F ·Dm(Y ) are mutually disjoint. Now we deduce Theorem 3.5 from Proposition 5.1. Proof. By Theorem 3.4, item (1), g ∈ g · ⋃ {Dm(Xn) : m ∈ J} for every infinite subset J of ω. Consequently, if A is a subset of ω, the set T (A) = ⋃ n∈A ( ⋃ m∈A {g1, . . . ,gn} ·Dm(Xn) ) Extraresolvability 323 is dense whenever A is infinite. By Lemma 5.2, in Proposition 5.1 it is not restrictive to assume that, for each n ∈ ω, the family {{g1, . . . ,gn} ·Dm(Xn) : m ∈ ω} is disjoint. Since the sets Tn are mutually disjoint too, the set T (A) ∩T (B) = T (A∩B) = ⋃ n∈A∩B ( ⋃ m∈A∩B {g1, . . . ,gn} ·Dm(Xn) ) is finite whenever A∩B is finite. The required collection of sets is obtained by considering the almost disjoint family {T (A)}, where A ranges over an almost disjoint family of ω of cardinality c. � 6. Protasov method. In this section we use a method due to Protasov, by applying his argument to the countable case [14, 12]. As in Section 4, a subset L of a group G is said to be large if there exists a finite subset K of G such that KL = LK = G (e.g., see [1]). Sets of the form gK and Kg are called right and left circles of radius K and center g, respectively. The following criterium of [1] will be useful in the sequel. Proposition 6.1. The set G\S fails to be large if and only if S contains (right or left) circles of any finite radius. Proposition 6.2. Let G be a countably infinite group. There exists a weak sequence F = {Fn} such that, whenever A ⊆ ω is infinite and ω\A is infinite, both sets F(A) = ⋃ n∈A Fn and G\F(A) fail to be large. Proof. Let G = {gn} be a one to one numeration of G and let Gn = {gk : k < n}, for each n ∈ ω. Arguing by induction, we shall construct a weak sequence {Fn} in such a way that Fn contains a circle of radius Gn for each n. Let us assume that F0, . . . ,Fn−1 have been already defined. Since T = ⋃ i