On $T$-characterized subgroups of compact Abelian groups

We say that a subgroup $H$ of an infinite compact Abelian group $X$ is {\it $T$-characterized} if there is a $T$-sequence $\mathbf{u} =\{u_n \}$ in the dual group of $X$ such that $H=\{x\in X: \; (u_n, x)\to 1 \}$. We show that a closed subgroup $H$ of $X$ is $T$-characterized if and only if $H$ is a $G_\delta$-subgroup of $X$ and the annihilator of $H$ admits a Hausdorff minimally almost periodic group topology. All closed subgroups of an infinite compact Abelian group $X$ are $T$-characterized if and only if $X$ is metrizable and connected. We prove that every compact Abelian group $X$ of infinite exponent has a $T$-characterized subgroup which is not an $F_{\sigma}$-subgroup of $X$ that gives a negative answer to Problem 3.3 in [10].

that every countable subgroup of a compact metrizable Abelian group is characterized. It is natural to ask whether a closed subgroup of a compact Abelian group is characterized. The following easy criterion is given in [10]: Fact 1.3. [10] A closed subgroup H of a compact Abelian group X is characterized if and only if H is a G δ -subgroup. In particular, X/H is metrizable and the annihilator H ⊥ of H is countable.
The next fact follows easily from Definition 1.2: Fact 1.4. ( [9], see also [10]) Every characterized subgroup H of a compact Abelian group X is an F σδ -subgroup of X, and hence H is a Borel subset of X. Facts 1.3 and 1.4 inspired in [10] the study of the Borel hierarchy of characterized subgroups of compact Abelian groups. For a compact Abelian group X denote by Char(X) (respectively, SF σ (X), SF σδ (X) and SG δ (X)) the set of all characterized subgroups (respectively, F σ -subgroups, F σδ -subgroups and G δ -subgroups) of X. The next fact is Theorem E in [10]: Fact 1.5. [10] For every infinite compact Abelian group X, the following inclusions hold: SG δ (X) Char(X) SF σδ (X) and SF σ (X) ⊆ Char(X).
The inclusion (1.1) inspired the following question: Question 1.6. [10, Problem 3.3] Does there exist a compact Abelian group X of infinite exponent whose all characterized subgroups are F σ -subsets of X?
Main results. It is important to emphasize that there is no any restriction on a sequence u in Definition 1.2. If a characterized subgroup H of a compact Abelian group X is dense, then, by Fact 1.1, a characterizing sequence is also a T B-sequence. But if H is not dense, we can not expect in general that a characterizing sequence of H is a T -sequence. Thus it is natural to ask: Question 1.7. For which characterized subgroups of compact Abelian groups one can find characterizing sequences which are also T -sequences?
This question is of independent interest because every T -sequence u naturally defines the group topology τ u satisfying the following dual property: Fact 1.8. [20] Let H be a characterized subgroup of an infinite compact Abelian group X by a T -sequence u. Then ( X, τ u ) ∧ = H(= s u (X)) and n( X, τ u ) = H ⊥ algebraically.
This motivates us to introduce the following notion: Definition 1.9. Let H be a subgroup of a compact Abelian group X. We say that H is a T -characterized subgroup of X if there exists a T -sequence u = {u n } n∈ω in X such that H = s u (X).
Denote by Char T (X) the set of all T -characterized subgroups of a compact Abelian group X. Clearly, Char T (X) ⊆ Char(X). Hence, if a T -characterized subgroup H of X is closed it is a G δ -subgroup of X by Fact 1.3. Note also that X is T -characterized by the zero sequence.
The main goal of the article is to obtain a complete description of closed T -characterized subgroups (see Theorem 1.10) and to study the Borel hierarchy of T -characterized subgroups (see Theorem 1.18) of compact Abelian groups. In particular, we obtain a complete answer to Question 1.7 for closed characterized subgroups and give a negative answer to Question 1.6.
Note that, if a compact Abelian group X is finite, then every T -sequence u in X is eventually equal to zero. Hence s u (X) = X. Thus X is the unique T -characterized subgroup of X. So in what follows we shall consider only infinite compact groups.
The following theorem describes all closed subgroups of compact Abelian groups which are T -characterized.
Theorem 1. 10. Let H be a proper closed subgroup of an infinite compact Abelian group X. Then the following assertions are equivalent: It is natural to ask for which compact Abelian groups all their closed G δ -subgroups are T -characterized. The next theorem gives a complete answer to this question. Theorem 1.14. Let X be an infinite compact Abelian group. The following assertions are equivalent: (1) All closed G δ -subgroups of X are T -characterized; (2) X is connected.
By Corollary 2.8 of [10], the trivial subgroup H = {0} of a compact Abelian group X is a G δ -subgroup if and only if X is metrizable. So we obtain: Corollary 1.15. All closed subgroups of an infinite compact Abelian group X are T -characterized if and only if X is metrizable and connected. Theorems 1.10 and 1.14 are proved in Section 2. In the next theorem we give a negative answer to Question 1.6: Theorem 1.16. Every compact Abelian group of infinite exponent has a dense T -characterized subgroup which is not an F σ -subgroup.
As a corollary of the inclusion (1.1) and Theorem 1.16 we obtain: Corollary 1.17. For an infinite compact Abelian group X the following assertions are equivalent: (i) X has finite exponent; (ii) every characterized subgroup of X is an F σ -subgroup; (iii) every T -characterized subgroup of X is an F σ -subgroup. Therefore, Char(X) ⊆ SF σ (X) if and only if X has finite exponent.
In the next theorem we summarize the obtained results about the Borel hierarchy of T -characterized subgroups of compact Abelian groups. Theorem 1.18. Let X be an infinite compact Abelian group X. Then: (1) Char T (X) SF σδ (X); (2) SG δ (X) Char T (X) Char T (X); (3) SG δ (X) ⊆ Char T (X) if and only if X is connected; (4) Char T (X) SF σ (X) SF σ (X); (5) Char T (X) ⊆ SF σ (X) if and only if X has finite exponent.
We prove Theorems 1.16 and 1.18 in Section 3. The notions of g-closed and g-dense subgroups of a compact Abelian group X were defined in [13]. In the last section of the paper, in analogy to these notions, we define g T -closed and g T -dense subgroups of X. In particular, we show that every g T -dense subgroup of a compact Abelian group X is dense if and only if X is connected (see Theorem 4.2).
2. The Proofs of Theorems 1.10 and 1.14 The subgroup of a group G generated by a subset A we denote by A . Recall that a subgroup H of an Abelian topological group X is called dually closed in X if for every x ∈ X \ H there exists a character χ ∈ H ⊥ such that (χ, x) = 1. H is called dually embedded in X if every character of H can be extended to a character of X. Every open subgroup of X is dually closed and dually embedded in X by Lemma 3 of [25].
The next notion generalizes the notion of the maximal extension in the class of all compact Abelian groups introduced in [11].
Definition 2.1. Let G be an arbitrary class of topological groups. Let (G, τ ) ∈ G and H be a subgroup of G. The group (G, τ ) is called a maximal extension of (H, τ | H ) in the class G if σ ≤ τ for every group topology on G such that σ| H = τ | H and (G, σ) ∈ G.
Clearly, the maximal extension is unique if it exists. Note that in Definition 2.1 we do not assume that (H, τ | H ) belongs to the class G.
If H is a subgroup of an Abelian group G and u is a T -sequence (respectively, a T B-sequence) in H, we denote by τ u (H) (respectively, τ bu (H)) the finest (respectively, precompact) group topology on H generated by u. We use the following easy corollary of the definition of T -sequences.
Lemma 2.2. For a sequence u in an Abelian group G the following assertions are equivalent: (1) u is a T -sequence in G; (2) u is a T -sequence in every subgroup of G containing u ; (3) u is a T -sequence in u . In this case, u is open in τ u (and hence u is dually closed and dually embedded in (G, τ u )), and (G, τ u ) is the maximal extension of ( u , τ u ( u ) in the class TAG of all Abelian topological groups.
Proof. Evidently, (1) implies (2) and (2) implies (3). Let u be a T -sequence in u . Let τ be the topology on G whose base is all translates of τ u ( u )-open sets. Clearly, u converges to zero in τ . Thus u is a T -sequence in G. So (3) implies (1).
Let us prove the last assertion. By the definition of τ u we have also τ ≤ τ u , and hence τ | u = τ u ( u ) ≤ τ u | u . Thus u is open in τ u , and hence it is dually closed and dually embedded in (G, τ u ) by [25,Lemma 3.3]. On the other hand, τ u | u ≤ τ u ( u ) = τ | u by the definition of τ u ( u ). So τ u is an extension of τ u ( u ). Now clearly, τ = τ u and (G, τ u ) is the maximal extension of ( u , τ u ( u ) in the class TAG.
For T B-sequences we have the following: Lemma 2.3. For a sequence u in an Abelian group G the following assertions are equivalent In this case, the subgroup u is dually closed and dually embedded in (G, τ bu ), and (G, τ bu ) is the maximal extension of ( u , τ bu ( u )) in the class of all precompact Abelian groups.
Proof. Evidently, (1) implies (2) and (2) implies (3). Let u be a T B-sequence in u . Then ( u , τ bu ( u )) ∧ separates the points of u . Let τ be the topology on G whose base is all translates of τ bu ( u )-open sets. Then ( u , τ bu ( u )) is an open subgroup of (G, τ ). It is easy to see that (G, τ ) ∧ separates the points of G. Since u converges to zero in τ , it is also converges to zero in τ + , where τ + is the Bohr topology of (G, τ ). Thus u is a T B-sequence in G. So (3) implies (1).
The last assertion follows from Proposition 1.8 and Lemma 3.6 in [11].
For a sequence u = {u n } n∈ω of characters of a compact Abelian group X set The following assertions is proved in [10]: The next two lemmas are natural analogues of Lemmas 2.2(ii) and 2.6 of [10].
Lemma 2.5. Let X be a compact Abelian groups and u = {u n } n∈ω be a T -sequence in X. Then Proof. Set H := s u (X) and K := K u . Let q : X → X/K be the quotient map. Then the adjoint homomorphism q ∧ is an isomorphism from (X/K) ∧ onto K ⊥ in X ∧ . For every n ∈ ω, define the character u n of X/K as follows: Clearly, u m is a T -sequence in G, τ u = τ um and s u (X) = s um (X) for every natural number m.
Lemma 2.6. Let K be a closed subgroup of a compact Abelian group X and q : X → X/K be the quotient map.
Note that the adjoint homomorphism q ∧ is an isomorphism from ( . This follows from the following chain of equivalences. By definition, The last equivalence is due to the inclusion K ⊆ H. Conversely, let H := q −1 ( H) be a T -characterized subgroup of X and a T -sequence u = {u n } n∈ω characterize H. Proposition 2.5 of [10] implies that we can find m ∈ N such that K ⊆ K um . So, taking into account that H = s u (X) = s um (X) for every natural number m, without loss of generality we can assume that K ⊆ K u . By Lemma 2.5, H/K u is a T -characterized subgroup of X/K u . Denote by q u the quotient homomorphism from is T -characterized in X/K by the previous paragraph of the proof. The next theorem is an analogue of Theorem B of [10], and it reduces the study of T -characterized subgroups of compact Abelian groups to the study of T -characterized ones of compact Abelian metrizable groups: Conversely, let H contain a closed G δ -subgroup K of X such that H/K is a T -characterized subgroup of the compact metrizable group X/K. Then H is a T -characterized subgroup of X by Lemma 2.6.
To prove Theorem 1.10 we need the following: Since H is also characterized it is a G δ -subgroup of X by Fact 1.3. We have to show that H ⊥ admits a MinAP group topology.
Our idea of the proof is the following. Set G := X. By Fact 1.8, H ⊥ is the von Neumann radical of (G, τ u ). Now assume that we found another T -sequence v which characterizes H and such that v = H ⊥ (maybe v = u). By Fact 1.8, we have n(G, τ v ) is open, and hence it is dually closed and dually embedded in (G, τ v ). Hence n( v , τ v | v ) = n(G, τ v )(= v ) by Lemma 4 of [16]. So ( v , τ v | v ) is MinAP. Thus H ⊥ = v admits a MinAP group topology, as desired.
We find such a T -sequence v in 4 steps (in fact we show that v has the form u m for some m ∈ N).
Step 1. Let q : X → X/K u be the quotient map. For every n ∈ ω, define the character u n of X/K u by the equality u n = u n • q (this is possible since K u ⊆ ker(u n )). As it was shown in the proof of Lemma 2.5, the sequence u = { u n } n∈ω is a T -sequence which characterizes H/K u in X/K u . Set X := X/K u and H := H/K u . So that H = s u ( X). By [23, 5.34 and 24.11] and since K u ⊆ H, we have By Fact 1.3, X is metrizable. Hence H is also compact and metrizable, and G := X is a countable Abelian group by [23, 24.15]. Since H is a proper closed subgroup of X, (2.2) implies that G is non-zero. We claim that G is countably infinite. Indeed, suppose for a contradiction that G is finite. Then X/K u = X is also finite. Now Fact 2.4 implies that u is a finite subgroup of G. Since u is a T -sequence, u must be eventually equal to zero. Hence H = s u (X) = X is not a proper subgroup of X, a contradiction.
Step 2. We claim that there is a natural number m such that the group ( u m , τ u | um ) = ( u m , τ um | um ) is MinAP.
Indeed, since G is countably infinite, we can apply Fact 1.8. So H = ( G, τ u ) ∧ algebraically. Since H and ( G, τ u ) ∧ are Polish groups (see Fact 2.8), H and ( G, τ u ) ∧ are topologically isomorphic by the uniqueness of the Polish group topology. Hence ( G, τ u ) ∧∧ = H ∧ is discrete. As it was noticed before the proof, the natural homomorphism α : ( G, τ u ) → ( G, τ u ) ∧∧ is continuous. Since ( G, τ u ) ∧∧ is discrete we obtain that the von Neumann radical ker( α) of ( G, τ u ) is open in τ u . So there exists a natural number m such that u n ∈ ker( α) for every n ≥ m. Hence u m ⊆ ker( α). Lemma 2.2 implies that the subgroup u m is open in ( G, τ u ), and hence it is dually closed and dually embedded in ( G, τ u ). Now Lemma 4 of [16] yields u m = ker( α) and ( u m , τ u | um ) is MinAP.
Step 3. Set v = {v n } n∈ω , where v n = u n+m for every n ∈ ω. Clearly, v is a T -sequence in G characterizing H, τ u = τ v and K u ⊆ K v . Let t : X → X/K v and r : X/K u → X/K v be the quotient maps. Analogously to Step 1 and the proof of Lemma 2.5, the sequence where t ∧ , r ∧ and q ∧ are the adjoint homomorphisms to t, r and q respectively.
Since q ∧ and r ∧ are embeddings, we have is MinAP as well.
Step 4. By the second exact sequence in (2.1) applying to v, Fact 1.8 and since Thus H ⊥ = v , and hence H ⊥ admits a MinAP group topology generated by the T -sequence v.
(2) ⇒ (1): Since H is a G δ -subgroup of X, H is closed by [10, Proposition 2.4] and X/H is metrizable (due to the well known fact that a compact group of countable pseudocharacter is metrizable). Hence H ⊥ = (X/H) ∧ is countable. Since H ⊥ admits a MinAP group topology, H ⊥ must be countably infinite. By Theorem 3.8 of [21], H ⊥ admits a MinAP group topology generated by a T -sequence u = { u n } n∈ω . By Fact 1.8, this means that s u (X/H) = {0}. Let q : X → X/H be the quotient map. Set u n = u n • q = q ∧ ( u n ). Since q ∧ is injective, u is a T -sequence in X by Lemma 2.2. We have to show that H = s u (X). By definition, x ∈ s u (X) if and only if (2)⇔(3) follows from Theorem 3.8 of [21]. The theorem is proved.
Proof of Theorem 1.14. (1) ⇒ (2): Suppose for a contradiction that X is not connected. Then, by [23, 24.25], the dual group G = X ∧ has a non-zero element g of finite order. Then the subgroup H := g ⊥ of X has finite index. Hence H is an open subgroup of X. Thus H is not T -characterized by Corollary 1.13. This contradiction shows that X must be connected.
The next proposition is a simple corollary of Theorem B in [10].
Proposition 2.9. The closureH of a characterized (in particular, T -characterized) subgroup H of a compact Abelian group X is a characterized subgroup of X.
Proof. By Theorem B of [10], H contains a compact G δ -subgroup K of X. ThenH is also a G δ -subgroup of X.
ThusH is a characterized subgroup of X by Theorem B of [10].
In general we cannot assert that the closureH of a T -characterized subgroup H of a compact Abelian group X is also T -characterized as the next example shows.
Example 2.10. Let X = Z(2) × T and G = X = Z(2) × Z . It is known (see the end of (1) in [14]) that there is a T -sequence u in G such that the von Neumann radical n(G, τ u ) of (G, τ u ) is Z (2)  Recall that a Borel subgroup H of a Polish group X is called polishable if there exists a Polish group topology τ on H such that the inclusion map i : (H, τ ) → X is continuous. Let H be a T -characterized subgroup of a compact metrizable Abelian group X by a T -sequence u = {u n } n∈ω . Then, by [16,Theorem 1], H is polishable by the metric (3.1) ρ(x, y) = d(x, y) + sup{|(u n , x) − (u n , y)|, n ∈ ω}, where d is the initial metric on X. Clearly, the topology generated by the metric ρ on H is finer than the induced one from X.
To prove Theorem 1.16 we need the following three lemmas. For a real number x we write [x] for the integral part of x and x for the distance from x to the nearest integer. We also use the following inequality proved in [15] Lemma 3.1. Let {a n } n∈ω ⊂ N be such that a n → ∞ and a n ≥ 2, n ∈ ω. Set u n = k≤n a n for every n ∈ ω. Then u = {u n } n∈ω is a T -sequence in X = T, and the T -characterized subgroup H = s u (T) of T is a dense non-F σ -subset of T.
Proof. We consider the circle group T as R/Z and write it additively. So that d(0, x) = x for every x ∈ T. Recall (see, for example, the proof of Lemma 1 in [15]) that every x ∈ T has the unique representation in the form where 0 ≤ c n < a n and c n = a n − 1 for infinitely many indices n. It is known [1] (see also (12) in the proof of Lemma 1 of [15]) that x with representation (3.3) belongs to H if and only if (3.4) lim n→∞ c n a n (mod 1) = 0.
Hence H is a dense subgroup of T. Thus u is even a T B-sequence in Z by Fact 1.1.
We have to show that H is not an F σ -subset of T. Suppose for a contradiction that H is an F σ -subset of T. Then H = ∪ n∈N F n , where F n is a compact subset of T for every n ∈ N. Since H is a subgroup of T, without loss of generality we can assume that F n − F n ⊆ F n+1 . Since all F n are closed in (H, ρ) as well, the Baire theorem implies that there are 0 < ε < 0.1 and m ∈ N such that F m ⊇ {x : ρ(0, x) ≤ ε}.
Fix arbitrarily l > 0 such that 2 u l−1 < ε 20 . For every natural number k > l, set Then, for every k > l, we have This inequality and (3.2) imply that , for every k > l.
For every s ∈ ω and every natural number k > l, we estimate |1 − (u s , x k )| as follows. Case 1. Let s < k. Set q = max{s + 1, l}. By the definition of x k , we have 2π [(u s · x k ) (mod 1)] = 2π u s k n=l 1 u n · (a n − 1)ε 20 (mod 1) < 2π k n=q u s u n · (a n − 1)ε 20 This inequality and (3.2) imply Case 2. Let s ≥ k. By the definition of x k , we have In particular, (3.7) implies that x k ∈ H for every k > l. Now, for every k > l, (3.1) and (3.5)-(3.7) imply Thus x k ∈ F m for every natural number k > l. Clearly, Since F m is a compact subset of T, we have x ∈ F m . Hence x ∈ H. On the other hand, we have lim n→∞ 1 a n · (a n − 1)ε 20 (mod 1) = ε 20 = 0.
So (3.4) implies that x ∈ H. This contradiction shows that H = s u (T) is not an F σ -subset of T.
For a prime number p, the group Z(p ∞ ) is regarded as the collection of fractions m/p n ∈ [0, 1). Let ∆ p be the compact group of p-adic integers. It is well known that ∆ p = Z(p ∞ ).
Lemma 3.2. Let X = ∆ p . For an increasing sequence of natural numbers 0 < n 0 < n 1 < . . . such that n k+1 − n k → ∞, set Then the sequence u = {u k } k∈ω is a T -sequence in Z(p ∞ ), and the T -characterized subgroup H = s u (∆ p ) is a dense non-F σ -subset of ∆ p .
Following [17, 2.2], for every ω = (a n ) ∈ ∆ p and every natural number k > 1, set where j k = n k if 0 < a n k < p − 1, and otherwise j k = min{j : either a s = 0 for j < s ≤ n k , or a s = p − 1 for j < s ≤ n k }. In [17, 2.2] it is shown that (3.9) ω ∈ s u (∆ p ) if and only if n k − m k → ∞.
We have to show that H is not an F σ -subset of ∆ p . Suppose for a contradiction that H = ∪ n∈N F n is an F σ -subset of ∆ p , where F n is a compact subset of ∆ p for every n ∈ N. Since H is a subgroup of ∆ p , without loss of generality we can assume that F n − F n ⊆ F n+1 . Since all F n are closed in (H, ρ) as well, the Baire theorem implies that there are 0 < ε < 0.1 and m ∈ N such that F m ⊇ {x : ρ(0, x) ≤ ε}.
Since F m is a compact subset of ∆ p , we have ω ∈ F m . Hence ω ∈ H. On the other hand, it is clearly that m k ( ω) = n k − s for every k ≥ l + 1. Thus for every k ≥ l + 1, n k − m k ( ω) = s → ∞. Now (3.9) implies that ω ∈ H. This contradiction shows that H is not an F σ -subset of ∆ p .
Then u is a T -sequence in G, and the T -characterized subgroup H = s u (X) is a dense non-F σ -subset of X.
Proof. Set H := s u (X). In [17, 2.3] it is shown that (3.17) ω = (a n ) ∈ s u (X) if and only if a n b n → 0.
So n∈ω Z(b n ) ⊆ H. Thus H is dense in X. Now Fact 1.1 implies that u is a T -sequence in G.
We have to show that H is not an F σ -subset of X. Suppose for a contradiction that H = ∪ n∈N F n is an F σ -subset of X, where F n is a compact subset of X for every n ∈ N. Since H is a subgroup of X, without loss of generality we can assume that F n − F n ⊆ F n+1 . Since all F n are closed in (H, ρ) as well, the Baire theorem yields that there are 0 < ε < 0.1 and m ∈ N such that F m ⊇ {ω ∈ X : ρ(0, ω) ≤ ε}.
Note that d(0, ω) = 2 −l , where 0 = ω = (a n ) n∈ω ∈ X and l is the minimal index such that a l = 0. Choose l such that 2 −l < ε/3. For every natural number k > l, set , for every n such that l ≤ n ≤ k, 0, if either 1 ≤ n < l or k < n.
Since (u n , ω k ) = 1 for every n > k, we obtain that ω k ∈ H for every k > l. For every n ∈ ω we have This inequality and the inequalities (3.1) and (3.2) imply Thus ω k ∈ F m for every natural number k > l. Evidently, Since F m is a compact subset of X, we have ω ∈ F m . Hence ω ∈ H. On the other hand, since b n → ∞ we have lim n→∞ a n b n = lim Thus ω ∈ H by (3.17). This contradiction shows that H is not an F σ -subset of X.
Now we are in position to prove Theorems 1.16 and 1.18.
Proof of Theorem 1. 16. Let X be a compact Abelian group of infinite exponent. Then G := X also has infinite exponent. It is well-known that G contains a countably infinite subgroup S of one of the following form: Fix such a subgroup S. Set K = S ⊥ and Y = X/K ∼ = S ∧ d , where S d denotes the group S endowed with the discrete topology. Since S is countable, Y is metrizable. Hence {0} is a G δ -subgroup of Y . Thus K is a G δ -subgroup of X. Let q : X → Y be the quotient map. By Lemmas 3.1-3.3, the compact group Y has a dense T -characterized subgroups H which is not an F σ -subset of Y . Lemma 2.6 implies that H := q −1 ( H) is a dense T -characterized subgroups of X. Since the continuous image of an F σ -subset of a compact group is an F σ -subset as well, we obtain that H is not an F σ -subset of X. Thus the subgroup H of X is T -characterized but it is not an F σ -subset of X. The theorem is proved.
(2): By Lemma 3.6 in [10], every infinite compact Abelian group X contains a dense characterized subgroup H. By Fact 1.1, H is T -characterized. Since every G δ -subgroup of X is closed in X by Proposition 2.4 of [10], H is not a G δ -subgroup of X.
(4) follows from Fact 1.5. (5) follows from Corollary 1.17. It is trivial that Char T (X) ⊆ Char(X) for every compact Abelian group X. For the circle group T we have. Proof. We have to show only that Char(T) ⊆ Char T (T). Let H = s u (T) ∈ Char(T) for some sequence u in Z.
If H is infinite, then H is dense in T. So u is a T -sequence in Z by Fact 1.1. Thus H ∈ Char T (T). If H is finite, then H is closed in T. Clearly, H ⊥ has infinite exponent. Thus H ∈ Char T (T) by Theorem 1.10.
Note that, if a compact Abelian group X satisfies the equality Char T (X) = Char(X), then X is connected by Fact 1.3 and Theorem 1.14. This fact and Proposition 3.4 justify the next problem: Problem 3.5. Does there exists a connected compact Abelian group X such that Char T (X) = Char(X)? Is it true that Char T (X) = Char(X) if and only if X is connected?
For a compact Abelian group X, the set of all subgroups of X which are both F σδ -and G δσ -subsets of X we denote by S∆ 0 3 (X). To complete the study of the Borel hierarchy of (T -)characterized subgroups of X we have to answer to the next question.
Problem 3.6. Describe compact Abelian groups X of infinite exponent for which Char(X) ⊆ S∆ 0 3 (X). For which compact Abelian groups X of infinite exponent there exists a T -characterized subgroup H that does not belong to S∆ 0 3 (X)?
4. g T -closed and g T -dense subgroups of compact Abelian groups The following closure operator g of the category of Abelian topological groups is defined in [13]. Let X be an Abelian topological group and H its arbitrary subgroup. The closure operator g = g X is defined as follows The set of all T -sequences in the dual group X of a compact Abelian group X we denote by T s ( X). Clearly, T s ( X) X N . Let H be a subgroup of X. In analogy to the closure operator g, g-closure and g-density, the operator g T is defined as follows g T (H) :=