Sequences and filters of characters characterizing subgroups of compact abelian groups

M. Beiglböck^{∗}, C. Steineder^{†} and R. Winkler^{‡}

Institute of Discrete Mathematics and Geometry, TU Vienna, Wiedner Hauptstraße 8-10, 1040 Wien, Austria

Abstract

Let H be a countable subgroup of the metrizable compact abelian group G and f : H → 𝕋 = ℝ/ℤ a (not necessarily continuous) character of H. Then there exists a sequence (χ_{n})_{n=1}^{∞} of (continuous) characters of G such that lim_{n→∞} χ_{n}(α) = f(α) for all α ∈ H and (χ_{n}(α))_{n=1}^{∞} does not converge whenever α ∈ G \ H. If one drops the countability and metrizability requirement one can obtain similar results by using filters of characters instead of sequences. Furthermore the introduced methods allow to answer questions of Dikranjan et al.

MSC: 22C05; 43A40; 54A20 Key words: Characterizing sequences, Compact abelian groups, Duality theory, Filters

Acknowledgements: The authors are very grateful to András Biró and Vera T. Sós for substantial discussions concerning this article. Moreover they thank Dikran Dikranjan and Gabriel Maresch for valuable remarks and the Austrian Science Foundation FWF for the support through project no. S8312.

1 Introduction

1.1 Motivation

In [4] several techniques have been developed to prove the existence of sequences (k_{n})_{n=1}^{∞} of positive integers characterizing countable subgroups H of the circle group 𝕋 = ℝ/ℤ in the sense that for α ∈ 𝕋,

α ∈ H ⟺ lim_{n→∞} k_{n}α = 0.

∗mathias.beiglboeck@tuwien.ac.at †christian.steineder@tuwien.ac.at ‡reinhard.winkler@tuwien.ac.at

These methods were extended in [5] to show that if H is generated freely by finitely many elements, a characterization is possible in an even stronger sense: One can choose a characterizing sequence such that ∑_{n=1}^{∞} ∥k_{n}α∥ < ∞ for α ∈ H, while lim sup_{n→∞} ∥k_{n}α∥ ≥ 1/4 for α ∈ 𝕋 \ H. (For x = r + ℤ ∈ ℝ/ℤ, r ∈ ℝ, the norm ∥x∥ denotes the distance between r and the nearest integer.) In [15] arbitrary subgroups of 𝕋 were characterized by filters on its dual ℤ. This approach was used in [3] to extend the results from [5]. A different approach to the characterization of finitely generated dense subgroups of compact abelian groups by sums has recently been introduced in [6] and [7]. Dikranjan et. al. investigated related questions concerning the characterization of subgroups of more general topological abelian groups G (cf. [1], [2], [8], [9]). In this article we lift the techniques of [3] to this general setting and answer questions stated in [2] and [9]. Our Theorems 3.1 and 3.6 have contemporaneously and independently been proved by Dikranjan and Kunen, cf. [11]. They also treat questions related to descriptive set theory. More results in this directions were, for instance, also obtained by Eliaš in [12].

1.2 Content of the paper

In Section 2 we modify the filter method from [15] for our purposes. Theorem 2.1 essentially states that arbitrary subgroups of compact abelian groups G can be characterized by filters on the (discrete) Pontryagin dual Ĝ of G. (Such filters are intended to be the neighborhood filters of 0 w.r.t. precompact group topologies on Ĝ.) In Section 3 this characterization is used to prove Theorem 3.1: Among the compact abelian groups G exactly the metrizable ones have the property that every countable subgroup H is characterizable in the sense that there is a sequence of characters (χ_{n})_{n=1}^{∞} in Ĝ such that

α ∈ H ⟺ lim_{n→∞} χ_{n}(α) = 0.

This solves Problem 5.3 from [9]. Theorem 3.6 (which solves Problem 5.1 and Question 5.2 from [9]) states that in an arbitrary compact abelian group every countable subgroup is the intersection of subgroups characterizable in the above sense. Section 4 is motivated by Question 5.2 from [2]: Consider the case G = 𝕋 and Ĝ = ℤ. It was established in [4] that for any countable H ≤ 𝕋 there exists a sequence of integers k_{1} < k_{2} < … characterizing H. Is it possible to choose the k_{n} in such a way that the quotients \frac{k_{n+1}}{k_{n}} are bounded? We answer this question affirmatively by proving a stronger assertion, cf. Theorem 4.1. It states that, in a certain sense, characterizing sequences can be arbitrarily close to having positive density. This seems to be best possible insofar as (apart from trivial cases) characterizing sequences always have density 0. Theorem 4.2 is a counterpart to Theorem 4.1 and describes how sparse characterizing sequences of subgroups of 𝕋 can be. In Section 5 we introduce a refined characterization of subgroups of a compact metrizable group G by sequences: For a sequence (χ_{n})_{n=1}^{∞} in Ĝ we consider the set H of all α ∈ H for which (χ_{n}(α))_{n=1}^{∞} converges (not necessarily to 0 ∈ 𝕋). H is easily seen to be a subgroup of G and the pointwise limit is a (not necessarily continuous) homomorphism f : H → 𝕋. Theorem 5.1 gives a complete description of the situation: Given any subgroup H of a metrizable compact abelian group G and any homomorphism f : H → 𝕋 there is a sequence (χ_{n})_{n=1}^{∞} in Ĝ such that χ_{n} → f pointwise on H. If H is countable then one can even achieve that H is exactly the set of convergence. If G is a compact (not necessarily metrizable) group and H is an arbitrary (not necessarily countable) subgroup of G, this result is still valid when the convergence of sequences is replaced by the more general convergence of filters. By considering the trivial homomorphism f ≡ 0 we see that Theorem 5.1 nicely extends Theorem 3.1. Furthermore this result allows to construct counterexamples to Question 5.4 from [2]. (For the complete statement of this question cf. Section 5.)

1.3 Conventions and notation

If not stated otherwise, G is always an infinite locally compact abelian group. (For finite G most assertions turn out to be trivial.) Since we are only interested in abelian groups, we use additive notation. In particular we shall do so in the group 𝕋 = ℝ/ℤ where characters χ : G → 𝕋 take their values. Elements of G will be denoted by α, β, . . .. If H is a (not necessarily closed) subgroup of G, we write H ≤ G. If A ⊆ G is any subset, ⟨A⟩ denotes the subgroup generated by A. For finite A = {α_{1}, …, α_{n}} ⊆ G and M ∈ ℕ we put ⟨A⟩_{M} := {∑_{i=1}^{n} k_{i}α_{i} : k_{i} ∈ ℤ, |k_{i}| ≤ M}. Recall that a group compactification of (any topological group) G is defined to be a pair (ι, C) where C is a compact group and ι : G ↦ C is a continuous homomorphism with dense image. Relative topologies on G induced by group compactifications are called precompact. The so called Bohr compactification (ι_{bG}, bG) of G is the compactification of G which is maximal in the sense that for each compactification (ι, C) of G there is a continuous homomorphism ϕ : bG → C with ϕ ∘ ι_{bG} = ι_{C}. Ĝ = {χ : χ is a continuous homomorphism from G to 𝕋} denotes the dual group of G, equipped with the compact open topology. By Pontryagin’s Duality The-

orem (cf. for instance [10] or [13]) we know that for LCA groups G ≅ ˆ Ĝ in the algebraic as well as in the topological sense via the canonical mapping α ↦ x_{α}, x_{α} : χ ↦ χ(α). We take G_{d} to be G endowed with the discrete topology. Duality theory can be

applied to construct the Bohr compactification bG of G by setting bG := { \atop (Ĝ)_{d}} and ι_{B} : α ↦ x_{α}. Accordingly, the Bohr compactification of Ĝ is Ĝ_{d}. It is natural to call the precompact topology on G induced by bG the Bohr topology. On the dual group Ĝ the Bohr topology can be described by the so called Bohr sets which are defined by

B_{(α_{1},...,α_{t},ε)} := {χ ∈ Ĝ : ∥χ(α_{i})∥ ≤ ε for i ∈ {1, 2, …, t}} ,

where α_{1}, …, α_{t} ∈ G and ε > 0. These sets generate the neighborhood filter of 0 in Ĝ endowed with the Bohr topology. Further we put B_{(α_{1},...,α_{t},ε)}(E) := B_{(α_{1},...,α_{t},ε)} ∩ E for E ⊆ Ĝ. For α ∈ G and B ⊆ Ĝ we write ∥αB∥ := sup{∥χ(α)∥ : χ ∈ B}.

2 Characterizing filters

We will make use of so called filter limits: Let S be any set, let F be a filter on S, let y be a point in a topological space X and let f : S → X be function. Then

F − lim_{s} f(s) = y

iff for every neighboorhood U of y, {s ∈ S : f(s) ∈ U} ∈ F. We remark that filter limits are more general then limits along sequences: For a sequence (x_{n})_{n=1}^{∞} in X put

F_{(x_{n})_{n=1}^{∞}} = {A ⊆ X : ∃ m ∈ ℕ such that {x_{n} : n ≥ m} ⊆ A}

Then F_{(x_{n})_{n=1}^{∞}} − lim_{s} f(s) exists iff lim_{n→∞} f(x_{n}) exists and in this case they coincide.

Let H ≤ G be a subgroup of the compact abelian group G. Our task is to show that H can be characterized by a filter F_{H} on Ĝ in the sense that we have F_{H} − lim_{χ} χ(β) = 0 iff β ∈ H. It is clear that for all α ∈ G and all ε > 0 the set B_{(α,ε)} has to be an element of F_{H} to assure convergence for elements of H. By the filter properties of F_{H} the intersection of finitely many such sets will again be an element of F_{H}. Thus it would be natural to define F_{H} to be the filter generated by the sets B_{(α_{1},...,α_{t},ε)} where α_{1}, …, α_{t} ∈ H, ε > 0. This definition yields the minimal filter with the required property and corresponds to the precompact group topology on Ĝ induced by H. Later it will be important to us that we may neglect finite sets of characters. Therefore we will also take all cofinite sets to be elements of F_{H}. This leads to the following definition:

F_{H} := \{F ⊆ Ĝ : \begin{matrix}∃ α_{1}, …, α_{t} ∈ H, ε > 0, Γ ⊆ Ĝ, |Γ| < ∞ \\ such that B_{(α_{1},...,α_{t},ε)}(Ĝ \ Γ) ⊆ F \end{matrix} \} .

Theorem 2.1 Let G be an infinite compact abelian group, let H be a subgroup of G and let the filter F_{H} be defined as above. Then for all β ∈ G

F_{H} − lim_{χ} χ(β) = 0 ⟺ β ∈ H.

In the course of the proof we will employ the following lemma which will also be useful later on:

Lemma 2.2 Let G be a compact abelian group. Then Ĝ is dense in Ĝ_{d} w.r.t. pointwise convergence. Thus, for any countable subset H of G and any χ ∈ Ĝ_{d} there exists a sequence (χ_{n})_{n=1}^{∞} in Ĝ such that χ_{n}(α) → χ(α) (n → ∞) for all α ∈ H.

Proof: As explained in the introduction, the compact group Ĝ_{d} is, with the set theoretic inclusion as dense embedding, the Bohr compactification of the discrete group Ĝ. This proves the first part. Thus for H = {α_{1}, α_{2}, …} ⊆ G and every n ∈ ℕ there is a χ_{n} ∈ Ĝ with ∥χ_{n}(α_{i})−χ(α_{i})∥ < \frac{1}{n} for all i ∈ \{1, 2, …, n\}. It follows that χ_{n} → χ pointwise on H. □

Proof of Theorem 2.1: The definition of F_{H} guarantees that F −lim_{χ} ∥χ(β)∥ = 0 for all β ∈ H. For the converse we prove that, given β ∉ H, for all α_{1}, …, α_{t} ∈ H and every ε > 0 there exist infintely many characters χ ∈ B_{(α_{1},...,α_{t},ε)} with ∥χ(β)∥ ≥ 1/4 which implies that {χ ∈ Ĝ : ∥χ(β)∥ < 1/4} ∉ F. First we see that there exists at least one such character: Consider the Bohr compactification Ĝ_{d} of Ĝ. Ĝ_{d} separates subgroups and points of G. Hence there exists some ϕ ∈ Ĝ_{d} such that

ϕ(α) = 0 for all α ∈ ⟨α_{1}, …, α_{t}⟩ and c = ϕ(β) ≠ 0,

w.l.o.g. ∥ϕ(β)∥ ≥ 1/3 (otherwise take an appropriate multiple 2ϕ, 3ϕ, …). By Lemma 2.2 ϕ can be approximated arbitrarily well on finitely many points by a character. Thus we find some χ ∈ Ĝ such that ∥χ(α_{i})∥ ≤ ε, 1 ≤ i ≤ t, ∥χ(β)∥ > 1/4. Next we prove that for ε > 0 each B_{(α_{1},...,α_{t},ε)} contains infinitely many χ with ∥χ(β)∥ ≥ 1/4. Let U := {(χ(α_{1}), …, χ(α_{t}), χ(β)) : χ ∈ Ĝ} ≤ 𝕋^{t+1}. We distinguish two cases:

1. U is finite, say U = {u_{1}, …, u_{k}}. There is some i, say i = 1, with u_{1} := (0, …, 0, c). Then the sets

Υ_{i} := {χ ∈ Ĝ : (χ(α_{1}), …, χ(α_{t}), χ(β)) = u_{i}} , i = 1, …, k,

and particularly Υ_{1} are infinite, or

2. U is an infinite subgroup of 𝕋^{t+1}. But then each point of U is an accumulation point. In both cases we find infinitely many χ with the required property. □

3 Characterizing countable subgroups

We solve Problem 5.3 from [9]: For which compact abelian G can every countable subgroup H be characterized by a sequence of characters?

For A ⊆ Ĝ we write lim_{χ∈A} χ(β) = 0 iff {χ ∈ A : χ(β) ≥ ε} is finite for all ε > 0. (I.e. instead of the characterizing sequence (χ_{n})_{n=1}^{∞} we consider the characterizing set A = {χ_{n} : n ∈ ℕ}.)

Theorem 3.1 Let G be an infinite compact abelian group and let H ≤ G be a countable subgroup. Then the following statements are equivalent:

(i) G is metrizable.

(ii) There exists a countable set A ⊆ Ĝ, such that

β ∈ H ⟺ lim_{χ∈A} χ(β) = 0.

Remark. The proof of Theorem 3.1 actually shows that (if G is metrizable) for every σ < 1/3 the characterizing set A can be chosen in such a way that β ∉ H implies lim sup_{χ∈A} ∥χ(β)∥ ≥ σ. Using a diagonalization argument it is not difficult to achieve lim sup_{χ∈A} ∥χ(β)∥ ≥ 1/3 and it is easy to see that this is best possible.

The proof of (i) ⟹ (ii) employs several lemmas which we formulate now and verify at the end of this section. According to our assumptions, in these lemmas G is an infinite compact abelian metrizable group.

Lemma 3.2 Let τ ∈ 𝕋 and n ∈ ℕ. Assume that ∥iτ∥ ≤ σ < 1/3 for all i ∈ {1, 2, …, n}. Then ∥τ∥ ≤ σ/n.

Lemma 3.3 Assume that γ_{1}, …, γ_{d} ∈ G freely generate H ≤ G. For arbitrary nonempty open sets I_{1}, …, I_{d} in 𝕋 there exists χ ∈ Ĝ such that χ(γ_{i}) ∈ I_{i} for all i ∈ {1, 2, …, d}.

Lemma 3.4 Let α_{1}, …, α_{t} ∈ G, ε > 0 and σ < 1/3.

1. For all finite Γ ⊆ Ĝ and all β ∈ G

∥βB_{(α_{1},...,α_{t},ε)}(Ĝ \ Γ)∥ ≤ σ ⟹ β ∈ ⟨α_{1}, …, α_{t}⟩.

2. Moreover there exists M ∈ ℕ such that for all finite Γ ⊆ Ĝ and all β ∈ G

∥βB_{(α_{1},...,α_{t},ε)}(Ĝ \ Γ)∥ ≤ σ ⟹ β ∈ ⟨α_{1}, …, α_{t}⟩_{M}.

3. If V ⊇ ⟨α_{1}, …, α_{t}⟩_{M} is an open subset of G then for all finite Γ ⊆ Ĝ there exists a finite set E ⊆ Ĝ \ Γ such that for β ∈ G

∥βB_{(α_{1},...,α_{t},ε)}(E)∥ ≤ σ ⟹ β ∈ V.

Lemma 3.5 Let R_{1} ⊆ R_{2} ⊆ … be finite subsets of G. There exists a sequence of open sets V_{n} ⊆ G, n ∈ ℕ such that 1. V_{n} ⊇ R_{n}.

2. lim inf_{n→∞} V_{n} = ⋃_{m=1}^{∞} ⋂_{n=m}^{∞} V_{n} = ⋃_{n=1}^{∞} R_{n}.

Proof of Theorem 3.1:

(i) ⟹ (ii): We will first construct the set A ⊆ Ĝ and then prove that β ∈ H iff lim_{χ∈A} χ(β) = 0. Let H =: {α_{t} : t ∈ ℕ} and pick ε = σ ∈ (0, 1/3). Using Lemma 3.4,2 we can choose a sequence (M_{t})_{t=1}^{∞} such that for every finite Γ ⊆ Ĝ and all β ∈ G

∥βB_{(α_{1},...,α_{t},ε)}(Ĝ \ Γ)∥ ≤ ε ⟹ β ∈ ⟨α_{1}, …, α_{t}⟩_{M_{t}}.

Next put for t ∈ ℕ, R_{t} := ⟨α_{1}, …, α_{t}⟩_{M_{t}} and define V_{t} ⊇ R_{t} according to Lemma 3.5 such that lim inf_{t→∞} V_{t} = H. Using Lemma 3.4,3 we choose a finite set E_{1} ⊆ Ĝ such that ∥βB_{(α_{1},ε)}(E_{1})∥ ≤ ε implies β ∈ V_{1}. By employing Lemma 3.4,3 again, we find E_{2} ⊆ Ĝ \ E_{1} such that ∥βB_{(α_{1},α_{2},ε)}(E_{2})∥ ≤ ε implies β ∈ V_{2}. Continuing in this fashion we arrive at a sequence (E_{t})_{t=1}^{∞} of disjoint subsets of Ĝ such that for each t ∈ ℕ

∥βB_{(α_{1},...,α_{t},ε)}(E_{t})∥ ≤ ε ⟹ β ∈ V_{t}.

Finally we put A := ⋃_{t=1}^{∞} B_{(α_{1},...,α_{t},ε)}(E_{t}). Assume that β ∈ H. To prove lim_{χ∈A} ∥χ(β)∥ = 0 note that, for arbitrary n ∈ ℕ, there exists T = T (n) ∈ ℕ such that iβ ∈ {α_{t} : t ≤ T } for all i ∈ {1, 2, …, n}. Thus whenever χ ∈ B_{(α_{1},...,α_{t},ε)}(E_{t}) for some t ≥ T we have ∥iχ(β)∥ ≤ ε for all 1 ≤ i ≤ n. By Lemma 3.2 this yields ∥χ(β)∥ ≤ ε/n. Since n was arbitrary we get lim_{χ∈A} ∥χ(β)∥ = 0. Conversely assume that lim sup_{χ∈A} ∥χ(β)∥ < ε for some β ∈ G. Then for all but finitely many t ∈ ℕ we have ∥βB_{(α_{1},...,α_{t},ε)}(E_{t})∥ ≤ ε. Thus there exists t_{0} ∈ ℕ such that β ∈ V_{t} for all t ≥ t_{0} which yields β ∈ H by the choice of the sequence (V_{t})_{t=1}^{∞}.

(ii) ⟹ (i): Let H ≤ G be an arbitrary countable subgroup characterized by the countable set A ⊆ Ĝ. Define Λ := ⟨A⟩ and

Λ^{0} := {g ∈ G : χ(g) = 0 for all χ ∈ Λ},

the annihilator of Λ. Clearly Λ^{0} ≤ H, thus |Λ^{0}| ≤ ℵ_{0}. Since Λ̂ ≅ G/Λ^{0} we have w(G/Λ^{0}) = w(Λ̂) = |Λ| = ℵ_{0}, where w denotes the topological weight, i.e. the least cardinal number of an open basis (cf. [13], 24.10 and 14.14). Hence G/Λ^{0} and Λ^{0} have at most countable weight and therefore also G ([13], 5.38), implying that G is metrizable. □

Let G be a compact abelian group. In [9] subgroups characterized by a sequence (χ_{n})_{n=1}^{∞} in Ĝ are denoted by

s_{(χ_{n})_{n=1}^{∞}}(G) := {α ∈ G : lim_{n→∞} χ_{n}(α) = 0} .

Furthermore such subgroups are called basic g-closed subgroups. According to Theorem 3.1 every countable subgroup of G is basic g-closed iff G is metrizable. A group H ≤ G is called g-closed if it is representable as the intersection of basic g-closed subgroups. The next theorem deals with g-closed subgroups and solves Problem 5.1 from [9].

Theorem 3.6 Every countable subgroup H of a compact abelian group G is g-closed.

Proof: For arbitrary β ∈ G \ H there is a χ ∈ Ĝ_{d} with χ(α) = 0 for all α ∈ H and ∥χ(β)∥ ≥ \frac{1}{3}. Thus Lemma 2.2 immediately yields a sequence of (χ_{n}^{β})_{n=1}^{∞} in Ĝ characterizing a subgroup

H_{β} := s_{(χ_{n}^{β})_{n=1}^{∞}}(G) = {α ∈ G : lim_{n→∞} χ_{n}^{β}(α) = 0} ≤ G

with H ≤ H_{β} and β ∉ H_{β}. Thus H = ⋂_{β∈G\H} H_{β}. □

Proofs of Lemmas 3.2 to 3.5 :

We assume the group G to be compact abelian and metrizable. Lemma 3.2 is elementary, so we skip the proof.

Proof of Lemma 3.3: Assume that

A := Ĝ⟨γ_{1}, …, γ_{d}⟩ = {χ(α) : χ ∈ Ĝ, α ∈ ⟨γ_{1}, …, γ_{d}⟩}

is not dense in 𝕋^{d}, i.e. H := \overline{A} ⪇ 𝕋^{d}. There is a nontrivial character of 𝕋^{d} vanishing on H, i.e. a nonzero vector h = (h_{1}, …, h_{d}) ∈ ℤ^{d} such that ∑_{i=1}^{d} h_{i}x_{i} = 0 holds for all x = (x_{1}, …, x_{d}) ∈ H. Fix an arbitrary χ ∈ Ĝ and put x_{i} = χ(γ_{i}). Then

0 = ∑_{i=1}^{d} h_{i}χ(γ_{i}) = χ (∑_{i=1}^{d} h_{i}γ_{i}) .

Since this holds for all χ ∈ Ĝ we have ∑_{i=1}^{d} h_{i}γ_{i} = 0, contradicting the independence of the free generators γ_{i}, 1 ≤ i ≤ d. □

Proof of Lemma 3.4: Let B_{0} := B_{(α_{1},...,α_{t},ε)}(Ĝ \ Γ).

1. Let F = F_{⟨α_{1},...,α_{t}⟩} be the filter of Theorem 2.1 characterizing ⟨α_{1}, …, α_{t}⟩ and let δ > 0 be arbitrary. Under the assumption ∥βB_{0}∥ ≤ σ < 1/3 we have to show that

F_{δ} := {χ ∈ Ĝ \ Γ : ∥χ(β)∥ ≤ δ} ∈ F.

Choose m ∈ ℕ such that δ ≥ σ/m and let B_{1} := B_{(α_{1},...,α_{t},ε/m)}(Ĝ \ Γ). By definition of F we have B_{1} ∈ F. For all χ ∈ B_{1}, i ∈ {1, 2, …, m}, we have iχ ∈ B_{0}. Thus ∥iχ(β)∥ ≤ σ for all i ∈ {1, 2, …, m} and Lemma 3.2 yields ∥χ(β)∥ ≤ σ/m < δ. Thus B_{1} ⊆ F_{δ} and hence F_{δ} ∈ F.

2. Assume that H := ⟨α_{1}, …, α_{t}⟩ is infinite (otherwise the assertion follows immediately). Since H is a finitely generated abelian group there exists a decomposition H = T ⊕ F where F is freely generated by γ_{1}, …, γ_{d} and T = ⟨ν_{1}, …, ν_{l}⟩ = ⊕_{i=1}^{l}⟨ν_{i}⟩ is the torsion subgroup of H. Hence ⟨ν_{i}⟩ ≅ ℤ_{e_{i}} for some e_{i} ∈ ℕ and

⟨α_{1}, …, α_{t}⟩ = ⟨γ_{1}, …, γ_{d}⟩ ⊕ ⟨ν_{1}, …, ν_{l}⟩ ≅ ℤ^{d} ⊕ ⨁_{i=1}^{l}ℤ_{e_{i}}.

Let δ > 0 be such that ∥χ(γ_{i})∥ ≤ δ for i ∈ {1, 2, …, d} and χ(ν_{j}) = 0 for j ∈ {1, 2, …, l} implies ∥χ(α_{k})∥ ≤ ε for k ∈ {1, 2, …, t}. Pick now any β ∈ G with ∥βB_{0}∥ ≤ σ < 1/3. By 1. above we have β ∈ H, thus β = ∑_{i=1}^{d} r_{i}γ_{i} +∑_{j=1}^{l} s_{j}ν_{j} for some r_{i} ∈ ℤ, s_{j} ∈ {0, 1, …, e_{j} − 1}, i ∈ {1, 2, …, d}, j ∈ {1, 2, …, l}. Let e := ∏_{j=1}^{l} e_{j}. By Lemma 3.3 there exist infinitely many χ ∈ eĜ := {eχ′ : χ′ ∈ Ĝ} such that

sign(r_{i})χ(γ_{i}) ∈ [\frac{1}{3 ∑_{j=1}^{d} |r_{j}|}, \frac{2}{3 ∑_{j=1}^{d} |r_{j}|}] + ℤ

holds for i ∈ {1, 2, …, d}. Therefore we have

r_{i}χ(γ_{i}) ∈ [\frac{|r_{i}|}{3 ∑_{j=1}^{d} |r_{j}|}, \frac{2|r_{i}|}{3 ∑_{j=1}^{d} |r_{j}|}] + ℤ

for all i ∈ {1, 2, …, d}. Summing up and using that χ(ν_{j}) = 0 for j ∈ {1, 2, …, l} this leads to

χ(β) = ∑_{i=1}^{d} r_{i}χ(γ_{i}) ∈ [\frac{1}{3}, \frac{2}{3}] + ℤ.

Thus χ ∉ B_{(α_{1},...,α_{t},ε)} and hence there is j ∈ {1, …, t} with ∥χ(α_{j})∥ > ε and therefore δ < \frac{2}{3 ∑_{j=1}^{d} |r_{j}|}. Equivalently ∑_{i=1}^{d} |r_{i}| < \frac{2}{3δ}. So there are only finitely many choices for β and we may put an universal bound M on the coefficients in the linear combination β = ∑_{i=1}^{r} k_{i}α_{i}.

3. Clearly, the set

I := {β ∈ G : ∥βB_{0}∥ ≤ σ} = ⋂_{χ∈B_{0}}{γ ∈ G : ∥χ(γ)∥ ≤ σ}

is closed and by 2. we have I ⊆ ⟨α_{1}, …, α_{t}⟩_{M} ⊆ V. Thus I ∩ V^{c} = ∅. By compactness of G there exists a finite set E ⊆ B_{0} such that

⋂_{χ∈E}{γ ∈ G : ∥χ(γ)∥ ≤ σ} ∩ V^{c} = ∅.

This E is as required. □

Proof of Lemma 3.5: Let ρ be a metric on G compatible with its topology. Since the sets R_{1} ⊆ R_{2} ⊆ … ⊆ G are finite there is a sequence (d_{n})_{n=1}^{∞} of positive reals decreasing to 0 such that

2d_{n} < min{ρ(α, α′) : α, α′ ∈ R_{n}, α ≠ α′}
d_{n} + d_{n+1} < min{ρ(α, α′) : α ∈ R_{n}, α′ ∈ R_{n+1} \ R_{n}}.

Define

V_{n} := {β ∈ G : ∃ α ∈ R_{n} with ρ(β, α) < d_{n}}.

By monotonicity of the sets R_{n}, β ∈ ⋃_{n=1}^{∞} R_{n} implies β ∈ ⋃_{m=1}^{∞} ⋂_{n=m}^{∞} V_{n}. Conversely, assume β ∈ ⋃_{m=1}^{∞} ⋂_{n=m}^{∞} V_{n} or, equivalently, that there exists an m with β ∈ V_{n} for all n ≥ m. According to the definition of the sets V_{n} there exists a unique α_{n} ∈ R_{n} such that ρ(α_{n}, β) < d_{n} for n ≥ m. Moreover the choice of the d_{n} guarantees that α_{m} = α_{m+1} = … and so ρ(β, α_{m}) = ρ(β, α_{n}) ≤ d_{n} → 0. Hence β = α_{m} ∈ R_{m} ⊆ ⋃_{n=1}^{∞} R_{n}. □

4 Thick and thin characterizing sequences

Question 5.2 from [2] asks: Does every countable subgroup H of 𝕋 admit a characterizing sequence (k_{n})_{n=1}^{∞} with bounded quotients, i.e. q_{n} = \frac{k_{n+1}}{k_{n}} ≤ C for all n ∈ ℕ and some C ∈ ℝ? We answer this question affirmatively by proving a stronger result. Which type of statement can be expected? Assume that α ∈ H is irrational. Then, by uniform distribution of the sequence (nα)_{n=1}^{∞}, the set of all k ∈ ℕ with ∥kα∥ < ε has density 2ε. Thus (with the exception of trivial cases) characterizing sequences have zero density. Furthermore the length of their gaps is tends to infinity. In particular the thickest characterizing sequences we can expect might have a density which converges to zero very slowly in some sense. This is the content of the following result.

Theorem 4.1 Let H ≤ 𝕋 be a countable subgroup and let (ε_{j})_{j=1}^{∞} be a sequence with 0 ≤ ε_{j} ≤ 1 that converges to 0 (arbitrarily slowly). Let ℕ be partitioned into nonempty intervals I_{j} = {i_{j}, i_{j} + 1, …, i_{j+1} − 1} with i_{0} = 0 and lim_{j→∞}(i_{j+1} − i_{j}) = ∞. Then there exists a sequence (k_{n})_{n=1}^{∞} of nonnegative integers characterizing H such that

\frac{|\{n : k_{n} ∈ I_{j}\}|}{|I_{j}|} ≥ ε_{j} for all j.

Proof: Let, according to Theorem 3.1 (or to [4]), c_{1} < c_{2} < … ∈ ℕ be any sequence characterizing H. We are going to construct a sequence d_{1} < d_{2} < … ∈ ℕ containing at least ε_{j}|I_{j}| elements in each I_{j} such that ∥k_{n}α∥ → 0 for all α ∈ H. Then A = {k_{1} < k_{2} < …} = {d_{n} : n ∈ ℕ} ∪ {c_{n} : n ∈ ℕ} clearly has the desired properties. Let H =: {α_{t} : t ∈ ℕ}, put I_{j}^{+}(0) := I_{j} and

I_{j}^{+}(t) := {k ∈ I_{j} : ∥kα_{i}∥ < 1/t for all i ∈ {1, 2, …, t}}

for t ≥ 1. For each j ∈ ℕ let t_{j} be the maximal t ∈ {0, 1, …, j} such that |I_{j}^{+}(t)| ≥ ε_{j}|I_{j}| and put

{d_{1} < d_{2} < …} = ⋃_{j=0}^{∞} I_{j}^{+}(t_{j}).

It suffices to show that t_{j} → ∞ for j → ∞ or, equivalently, that for each t_{0} ∈ ℕ there exists j_{0} such that for all j ≥ j_{0}

| {d ∈ I_{j} : ∥dα_{i}∥ < 1/t_{0} for all i = 1, …, t_{0}} | ≥ ε_{j}I_{j}.

Since ε_{j} → 0 for j → ∞ this is an immediate consequence of the well distribution (cf. [14]) of the sequence (ng)_{n=1}^{∞} in the closed subgroup G ≤ 𝕋^{t_{0}} generated by g = (α_{1}, …, α_{t_{0}}) ∈ 𝕋^{t_{0}}: The open subset O ⊆ G of all (β_{1}, …, β_{t_{0}}) with ∥β_{i}∥ < 1/t_{0} has positive Haar measure μ(O) and the set of all k ∈ ℤ with kg ∈ O has uniform density μ(O) > 0. □

Theorem 4.1 indeed answers the question about quotients: Take, for instance, i_{j} = j^{2} and choose the ε_{j} in such a way that at least one k_{n} lies in each I_{j}. Then the quotients q_{n} = \frac{k_{n+1}}{k_{n}} tend to 1. This example can be modified in many ways. It has been proved in [2] that q_{n} → ∞ implies that the corresponding characterized group H is uncountable. Thus, for a given countable H, characterizing sequences cannot be arbitrarily sparse in this sense. Nevertheless we have:

Theorem 4.2 Let H be a countable subgroup of 𝕋 and let m_{1} < m_{2} < … be an (arbitrarily fast) increasing sequence of positive integers. Then there is a characterizing sequence k_{1} < k_{2} < … for H with m_{n} < k_{n} for all n ∈ ℕ.

Proof: Let (c_{n})_{n=1}^{∞} be any characterizing sequence of H. Put k_{2n} := c_{j_{n}} and k_{2n+1} := c_{j_{n}} + c_{n} where j_{n} is large enough in the sense that k_{2n} > m_{2n} and k_{2n+1} > m_{2n+1}. Clearly α ∈ H implies k_{n}α → 0. On the other hand, if β ∈ 𝕋 and k_{n}β → 0 then also (k_{2n+1} − k_{2n})β = c_{n}β → 0. (c_{n})_{n=1}^{∞} characterizes H, therefore β ∈ H. □

Theorem 4.2 implies that for any countable H ≤ 𝕋 there are sequences (k_{n})_{n=1}^{∞} characterizing H with lim sup_{n→∞} \frac{k_{n+1}}{k_{n}} = ∞: In Theorem 4.2 put m_{n} = n^{n} and let k_{1} < k_{2} < … be a characterizing sequence of H such that m_{n} ≤ k_{n} for all n ∈ ℕ. Then

sup_{n∈ℕ} \frac{k_{n+1}}{k_{n}} ≥ sup_{n∈ℕ} ^{n} \sqrt{∏_{i=1}^{n} \frac{k_{i+1}}{k_{i}}} ≥ sup_{n∈ℕ} ^{n} \sqrt{\frac{k_{n}}{k_{1}}} ≥ sup_{n∈ℕ} ^{n} \sqrt{\frac{n^{n}}{k_{1}}} = ∞.

For more sophisticated methods to generate sparse characterizing sequences we refer to [3] and [5]: E.g. for a countable subgroup H ≤ 𝕋 one can construct a characterizing sequence (k_{n})_{n=1}^{∞} such that for all r > 0 and α ∈ H, ∑_{n=1}^{∞} ∥k_{n}α∥^{r} < ∞.

The idea of the proof of Theorem 4.2 has further remarkable extensions. We will analyze them more detailed in the next section.

5 Groups as sets of convergence

The following theorem presents a generalized approach to the characterization of subgroups.

Theorem 5.1 Let G be a compact abelian group.

1. Let F be a filter on Ĝ. Then the set H of all α ∈ G for which F −lim_{χ} χ(α) exists is a subgroup of G. The mapping f : H ↦ 𝕋, α ↦ F − lim_{χ} χ(α) is a group homomorphism.

In particular if (χ_{n})_{n=1}^{∞} is a sequence in Ĝ, the set H of all α ∈ G for which lim_{n→∞} χ_{n}(α) exists is a subgroup and the mapping f : H ↦ 𝕋, α ↦ lim_{n→∞} χ_{n}(α) is a group homomorphism.

2. Let conversely H be a subgroup of G and let f : H ↦ 𝕋 be a homomorphism. Then there exists a filter F on Ĝ such that F − lim_{χ} χ(α) = f(α) for all α ∈ H and F − lim_{χ} χ(β) does not exist whenever β ∉ H.

3. If furthermore H ≤ G is countable then there exists a sequence (χ_{n})_{n=1}^{∞}

in Ĝ such that

χ_{n}(α) → f(α) for all α ∈ H.

4. If G is metrizable and H is countable then there exists a sequence (χ′_{n})_{n=1}^{∞}

in Ĝ such that

χ′_{n}(α) → f(α) for all α ∈ H

and (χ′_{n}(β))_{n=1}^{∞} does not converge if β ∉ H.

Proof: 1: Clear.

2. For α_{1}, …, α_{t} ∈ H and ε > 0 put

F (α_{1}, …, α_{t}, ε) := {χ ∈ Ĝ : ∥χ(α_{i}) − f(α_{i})∥ ≤ ε for i = {1, 2, …, t}}

and

F = F(H, f) := \{F ⊆ Ĝ : \begin{matrix}∃ α_{1}, …, α_{t} ∈ H, ∃ ε > 0 \\ such that F (α_{1}, …, α_{t}, ε) ⊆ F \end{matrix} \} .

We have to show that (a) F is a filter.

(b) For all α ∈ H : F − lim_{χ} χ(α) = f(α).

(c) For all β ∉ H : F − lim_{χ} χ(β) does not exist.

ad (a): Since the set F (α_{1}, …, α_{t}, β_{1}, …, β_{t}, min(ε_{1}, ε_{2})) ∈ F is contained in F (α_{1}, …, α_{t}, ε_{1})∩F (β_{1}, …, β_{t}, ε_{2}) it suffices to show that each F (α_{1}, …, α_{t}, ε) is not empty. There exists an extension of f : H ↦ 𝕋 to χ : G ↦ 𝕋 such that χ ∈ Ĝ_{d}. By Lemma 2.2 there is a χ′ ∈ Ĝ such that ∥χ′(α_{i}) − χ(α_{i})∥ ≤ ε for i = 1, …, t. Hence χ′ ∈ F (α_{1}, …, α_{t}, ε) ≠ ∅. ad (b): Let α ∈ H and U ∈ U(f(α)). There exists an ε > 0 such that ∥ξ − f(α)∥ < ε implies ξ ∈ U. χ ∈ F (α, ε) ∈ F implies ∥χ(α) − f(α)∥ < ε proving F − lim_{χ} χ(α) = f(α). ad (c): Let β ∉ H and F ∈ F be arbitrary. We will show that there exist χ_{1}, χ_{2} ∈ F such that ∥χ_{1}(β) − χ_{2}(β)∥ ≥ 1/4. F ∈ F implies that there exist α_{1}, …, α_{t} ∈ H and ε > 0 such that F (α_{1}, …, α_{t}, ε) ⊆ F. Note that there is a χ′ ∈ Ĝ_{d} with χ′(h) = 0 for all h ∈ H and χ′(β) ≥ 1/3. By Lemma 2.2 there exists a χ ∈ Ĝ such that ∥χ(α_{i})∥ < ε/2 for i = 1, …, t and χ(β) > 1/4. Pick χ_{1} ∈ F (α_{1}, …, α_{t}, ε/2) ⊆ F arbitrary and let χ_{2} = χ + χ_{1}. Then χ_{2} is also in F and ∥χ_{2}(β) − χ_{1}(β)∥ = ∥χ(β)∥ > 1/4.

3. Let H = {α_{t}, t ∈ ℕ}. The proof of 2. shows that for each n ∈ ℕ there is a χ_{n} ∈ Ĝ such that ∥χ_{n}(α_{i}) − f(α_{i})∥ < 1/n for i = 1, …, n. The sequence (χ_{n})_{n=1}^{∞} has the desired properties.

4. If G is metrizable and H is countable we know by Theorem 3.1 that there exists a sequence (χ̃_{n})_{n=1}^{∞} in Ĝ such that

∥χ̃_{n}(α)∥ → 0 iff α ∈ H.

Let furthermore (χ_{n})_{n=1}^{∞} be as in 3. and define χ′_{2n} := χ_{n} and χ′_{2n+1} := χ_{n} +χ̃_{n}. Then χ′_{n}(α) → f(α) for all α ∈ H. Conversely, for β ∉ H the sequence (χ′_{n}(β))_{n=1}^{∞} cannot converge: If χ′_{n}(β) → c for some c ∈ 𝕋, then χ̃_{n}(β) = χ′_{2n+1}(β) − χ′_{2n}(β) → 0. Hence β ∈ H, contradiction. □

We want to apply Theorem 5.1 to Question 5.4 in [2] which, in our notation, reads as follows. Let (c_{n})_{n=1}^{∞} be a sequence in ℤ. Are the subsequent conditions (i) and (ii) equivalent?

(i) There exists a precompact abelian group G ⊇ ℤ such that c_{n} → h in G and ⟨h⟩ ∩ ℤ = {0}.

(ii) There exists an infinite subgroup A ≤ 𝕋 such that c_{n}α → 0 holds for all α ∈ A.

Conditions (i) is obviously equivalent to (i’) below:

(i’) There exists a group compactification (ι, G) of ℤ such that ι is 1-1, ι(c_{n}) → h in G and ⟨h⟩ ∩ ι(ℤ) = {0}.

We remark first that (ii) implies (i’): Let ι : ℤ → 𝕋^{A}, n ↦ (nα)_{α∈A} and put G = \overline{ι(ℤ)}. Obviously (ι, G) is a compactification of ℤ and since A is infinite, ι is 1-1.

To see that the converse does not hold, pick α, β ∈ 𝕋, such that α and β are linearly independent over the rationals. Define a homomorphism f : ⟨α⟩ → ⟨β⟩, nα ↦ nβ. By Theorem 5.1 choose a sequence (c_{n})_{n=1}^{∞} in ℤ such that c_{n}α → f(α) = β and (c_{n}γ)_{n=1}^{∞} does not converge for γ ∈ 𝕋 \ ⟨α⟩. Then ι : ℤ → 𝕋, n ↦ nα gives rise to a group compactification of ℤ. Put h := β such that ι(c_{n}) = c_{n}α → β. Since α and β were chosen to be linearly independent we have ⟨h⟩ ∩ ι(ℤ) = ⟨β⟩ ∩ ⟨α⟩ = {0}. Thus (i′) is valid. On the other hand (ii) fails since c_{n}γ → 0 only for γ = 0.

For a different type of counterexample fix a prime p and consider the p-adic integers ℤ_{p}. Choose an arbitrary sequence (k_{n})_{n=1}^{∞} in {0, 1, …, p − 1} which contains infinitely many non zero elements and satisfies k_{1} = 1. Using this, put for each n ∈ ℕ, h_{2n} = h_{2n+1} = ∑_{i=1}^{n} k_{i}p^{i} and let c_{n} = p^{n} + h_{n}. Then

lim_{n→∞} c_{n} = lim_{n→∞} h_{n} = ∑_{i=1}^{∞} k_{i}p^{i} =: h ∈ ℤ_{p} \ ℤ.

Hence kh ∈ ℤ_{p} \ ℤ for all k ∈ ℤ \ {0}, so (i) holds. Next pick α ∈ 𝕋 such that lim_{n→∞} c_{n}α = 0. It follows that also lim_{n→∞}(c_{n+1} − c_{n})α = 0. Since h_{2n} = h_{2n+1} this yields p^{n}α → 0 so α = a/p^{l} + ℤ for some l ∈ ℕ and a ∈ {0, 1, …, p^{l} − 1}. But then, for all n ≥ l

∥αc_{2n}∥ = ∥(a/p^{l}) (p^{2n} + ∑_{i=1}^{n} k_{i}p^{i})∥ = ∥(a/p^{l}) ∑_{i=1}^{l} k_{i}p^{i}∥ .

The last term tends to 0 only if a = 0. Hence α = 0 and (ii) fails.

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