On T-Characterized Subgroups of Compact Abelian Groups

Abstract

A sequence ${\left\{u_n\right\}}_{n\in \omega }$ in abstract additively-written Abelian group G is called a T-sequence if there is a Hausdorff group topology on G relative to which ${lim}_n{u}_n=0$. We say that a subgroup H of an infinite compact Abelian group X is T-characterized if there is a T-sequence $\mathbf{u}=\left\{u_n\right\}$ 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 Dikranjan and Gabriyelyan (Topol. Appl. 2013, 160, 2427–2442).

1. Introduction

Notation and preliminaries: Let X be an Abelian topological group. We denote by $\widehat{X}$ the group of all continuous characters on X, and $\widehat{X}$ endowed with the compact-open topology is denoted by $X^{\wedge }$. The homomorphism ${\alpha }_X:X\to X^{\wedge \wedge }$, $x\mapsto (\chi \mapsto (\chi ,x\left)\right)$, is called the canonical homomorphism. Denote by $\mathbf{n}\left(X\right)={\cap }_{\chi \in \widehat{X}}\ker \left(\chi \right)=ker\left({\alpha }_X\right)$ the von Neumann radical of X. The group X is called minimally almost periodic ($MinAP$) if $\mathbf{n}\left(X\right)=X$, and X is called maximally almost periodic ($MAP$) if $\mathbf{n}\left(X\right)=\left\{0\right\}$. Let H be a subgroup of X. The annihilator of H we denote by $H^{\perp }$, i.e., $H^{\perp }=\{\chi \in X^{\wedge }:(\chi ,h)=1$ for every $h\in H\}$.

Recall that an Abelian group G is of finite exponent or bounded if there exists a positive integer n, such that $ng=0$ for every $g\in G$. The minimal integer n with this property is called the exponent of G and is denoted by $\exp \left(G\right)$. When G is not bounded, we write $\exp \left(G\right)=\infty$ and say that G is of infinite exponent or unbounded. The direct sum of ω copies of an Abelian group G we denote by $G^{\left(\omega \right)}$.

Let $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ be a sequence in an Abelian group G. In general, no Hausdroff topology may exist in which $\mathbf{u}$ converges to zero. A very important question of whether there exists a Hausdorff group topology τ on G, such that $u_n\to 0$ in $(G,\tau )$, especially for the integers, has been studied by many authors; see Graev [1], Nienhuys [2], and others. Protasov and Zelenyuk [3] obtained a criterion that gives a complete answer to this question. Following [3], we say that a sequence $\mathbf{u}=\left\{u_n\right\}$ in an Abelian group G is a T-sequence if there is a Hausdorff group topology on G in which $u_n$ converges to zero. The finest group topology with this property we denote by ${\tau }_{\mathbf{u}}$.

The counterpart of the above question for precompact group topologies on $\mathbb{Z}$ is studied by Raczkowski [4]. Following [5,6] and motivated by [4], we say that a sequence $\mathbf{u}=\left\{u_n\right\}$ is a $TB$-sequence in an Abelian group G if there is a precompact Hausdorff group topology on G in which $u_n$ converges to zero. For a $TB$-sequence $\mathbf{u}$, we denote by ${\tau }_{b\mathbf{u}}$ the finest precompact group topology on G in which $\mathbf{u}$ converges to zero. Clearly, every $TB$-sequence is a T-sequence, but in general, the converse assertion does not hold.

While it is quite hard to check whether a given sequence is a T-sequence (see, for example, [3,7,8,9,10]), the case of $TB$-sequences is much simpler. Let X be an Abelian topological group and $\mathbf{u}=\left\{u_n\right\}$ be a sequence in its dual group $X^{\wedge }$. Following [11], set:

$$s_{\mathbf{u}}\left(X\right)=\{x\in X:(u_n,x)\to 1\}.$$

In [5], the following simple criterion to be a $TB$-sequence was obtained:

Fact 1 ([5]). A sequence $\mathbf{u}$ in a (discrete) Abelian group G is a $TB$-sequence if and only if the subgroup $s_{\mathbf{u}}\left(X\right)$ of the (compact) dual $X=G^{\wedge }$ is dense.

Motivated by Fact 1, Dikranjan et al. [11] introduced the following notion related to subgroups of the form $s_{\mathbf{u}}\left(X\right)$ of a compact Abelian group X:

Definition 2 ([11]). Let H be a subgroup of a compact Abelian group X and $\mathbf{u}=\left\{u_n\right\}$ be a sequence in $\widehat{X}$. If $H=s_{\mathbf{u}}\left(X\right)$, we say that $\mathbf{u}$ characterizes H and that H is characterized (by $\mathbf{u}$).

Note that for the torus $\mathbb{T}$, this notion was already defined in [12]. Characterized subgroups have been studied by many authors; see, for example, [11,12,13,14,15,16]. In particular, the main theorem of [15] (see also [14]) asserts 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 [13]:

Fact 3 ([13]). A closed subgroup H of a compact Abelian group X is characterized if and only if H is a $G_{\delta }$-subgroup. In particular, $X/H$ is metrizable, and the annihilator $H^{\perp }$ of H is countable.

The next fact follows easily from Definition 2:

Fact 4 ([17], see also [13]). Every characterized subgroup H of a compact Abelian group X is an $F_{\sigma \delta }$-subgroup of X, and hence, H is a Borel subset of X.

Facts 3 and 4 inspired in [13] the study of the Borel hierarchy of characterized subgroups of compact Abelian groups. For a compact Abelian group X, denote by $\mathrm{Char}\left(X\right)$ (respectively, ${\mathrm{SF}}_{\sigma }\left(X\right)$, ${\mathrm{SF}}_{\sigma \delta }\left(X\right)$ and ${\mathrm{SG}}_{\delta }\left(X\right)$) the set of all characterized subgroups (respectively, $F_{\sigma }$-subgroups, $F_{\sigma \delta }$-subgroups and $G_{\delta }$-subgroups) of X. The next fact is Theorem E in [13]:

Fact 5 ([13]). For every infinite compact Abelian group X, the following inclusions hold:

$${\mathrm{SG}}_{\delta }\left(X\right)⫋\mathrm{Char}\left(X\right)⫋{\mathrm{SF}}_{\sigma \delta }\left(X\right)and{\mathrm{SF}}_{\sigma }\left(X\right)\neg \subseteq \mathrm{Char}\left(X\right).$$

If in addition X has finite exponent, then:

$$\mathrm{Char}\left(X\right)⫋{\mathrm{SF}}_{\sigma }\left(X\right).$$

(1)

The inclusion Equation (1) inspired the following question:

Question 6 (Problem 3.3 in [13]). Does there exist a compact Abelian group X of infinite exponent all of whose characterized subgroups are $F_{\sigma }$-subsets of X?

Main results: It is important to emphasize that there is no restriction on the sequence $\mathbf{u}$ in Definition 2. If a characterized subgroup H of a compact Abelian group X is dense, then, by Fact 1, a characterizing sequence is also a $TB$-sequence. However, if H is not dense, we cannot expect in general that a characterizing sequence of H is a T-sequence. Thus, it is natural to ask:

Question 7. For which characterized subgroups of compact Abelian groups can one find characterizing sequences that are also T-sequences?

This question is of independent interest, because every T-sequence $\mathbf{u}$ naturally defines the group topology ${\tau }_{\mathbf{u}}$ satisfying the following dual property:

Fact 8 ([18]). Let H be a subgroup of an infinite compact Abelian group X characterized by a T-sequence $\mathbf{u}$. Then, ${(\widehat{X},{\tau }_{\mathbf{u}})}^{\wedge }=H(=s_{\mathbf{u}}\left(X\right))$ and $\mathbf{n}(\widehat{X},{\tau }_{\mathbf{u}})=H^{\perp }$ algebraically.

This motivates us to introduce the following notion:

Definition 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 $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ in $\widehat{X}$, such that $H=s_{\mathbf{u}}\left(X\right)$.

Denote by ${\mathrm{Char}}_T\left(X\right)$ the set of all T-characterized subgroups of a compact Abelian group X. Clearly, ${\mathrm{Char}}_T\left(X\right)\subseteq \mathrm{Char}\left(X\right)$. Hence, if a T-characterized subgroup H of X is closed, it is a $G_{\delta }$-subgroup of X by Fact 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 10) and to study the Borel hierarchy of T-characterized subgroups (see Theorem 18) of compact Abelian groups. In particular, we obtain a complete answer to Question 7 for closed characterized subgroups and give a negative answer to Question 6.

Note that, if a compact Abelian group X is finite, then every T-sequence $\mathbf{u}$ in $\widehat{X}$ is eventually equal to zero. Hence, $s_{\mathbf{u}}\left(X\right)=X$. Thus, X is the unique T-characterized subgroup of X. Therefore, in what follows, we shall consider only infinite compact groups.

The following theorem describes all closed subgroups of compact Abelian groups that are T-characterized.

Theorem 10. Let H be a proper closed subgroup of an infinite compact Abelian group X. Then, the following assertions are equivalent:

(1)

H is a T-characterized subgroup of X; 

(2)

H is a $G_{\delta }$-subgroup of X, and the countable group $H^{\perp }$ admits a Hausdorff MinAPgroup topology; 

(3)

H is a $G_{\delta }$-subgroup of X and one of the following holds:

(a) 

$H^{\perp }$ has infinite exponent;

(b) 

$H^{\perp }$ has finite exponent and contains a subgroup that is isomorphic to $\mathbb{Z}{\left(exp\left(H^{\perp }\right)\right)}^{\left(\omega \right)}$.

Corollary 11. Let X be an infinite compact metrizable Abelian group. Then, the trivial subgroup $H=\left\{0\right\}$ is T-characterized if and only if $\widehat{X}$ admits a Hausdorff MinAP group topology.

As an immediate corollary of Fact 3 and Theorem 10, we obtain a complete answer to Question 7 for closed characterized subgroups.

Corollary 12. A proper closed characterized subgroup H of an infinite compact Abelian group X is T-characterized if and only if $H^{\perp }$ admits a Hausdorff MinAP group topology.

If H is an open proper subgroup of X, then $H^{\perp }$ is non-trivial and finite. Thus, every Hausdorff group topology on $H^{\perp }$ is discrete. Taking into account Fact 3, we obtain:

Corollary 13. Every open proper subgroup H of an infinite compact Abelian group X is a characterized non-T-characterized subgroup of X.

Nevertheless (see Example 1 below), there is a compact metrizable Abelian group X with a countable T-characterized subgroup H, such that its closure $\overline{H}$ is open. Thus, it may happen that the closure of a T-characterized subgroup is not T-characterized.

It is natural to ask for which compact Abelian groups all of their closed $G_{\delta }$-subgroups are T-characterized. The next theorem gives a complete answer to this question.

Theorem 14. Let X be an infinite compact Abelian group. The following assertions are equivalent:

(1)

All closed $G_{\delta }$-subgroups of X are T-characterized;

(2)

X is connected.

By Corollary 2.8 of [13], the trivial subgroup $H=\left\{0\right\}$ of a compact Abelian group X is a $G_{\delta }$-subgroup if and only if X is metrizable. Therefore, we obtain:

Corollary 15. All closed subgroups of an infinite compact Abelian group X are T-characterized if and only if X is metrizable and connected.

Theorems 10 and 14 are proven in Section 2.

In the next theorem, we give a negative answer to Question 6:

Theorem 16. Every compact Abelian group of infinite exponent has a dense T-characterized subgroup, which is not an $F_{\sigma }$-subgroup.

As a corollary of the inclusion Equation (1) and Theorem 16, we obtain:

Corollary 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 $F_{\sigma }$-subgroup;

(iii)

every T-characterized subgroup of X is an $F_{\sigma }$-subgroup.

Therefore, $\mathrm{Char}\left(X\right)\subseteq {\mathrm{SF}}_{\sigma }\left(X\right)$ 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 18. Let X be an infinite compact Abelian group X. Then:

(1)

${\mathrm{Char}}_T\left(X\right)⫋{\mathrm{SF}}_{\sigma \delta }\left(X\right)$;

(2)

${\mathrm{SG}}_{\delta }\left(X\right)\bigcap {\mathrm{Char}}_T\left(X\right)⫋{\mathrm{Char}}_T\left(X\right)$;

(3)

${\mathrm{SG}}_{\delta }\left(X\right)\subseteq {\mathrm{Char}}_T\left(X\right)$ if and only if X is connected;

(4)

${\mathrm{Char}}_T\left(X\right)\bigcap {\mathrm{SF}}_{\sigma }\left(X\right)⫋{\mathrm{SF}}_{\sigma }\left(X\right)$;

(5)

${\mathrm{Char}}_T\left(X\right)\subseteq {\mathrm{SF}}_{\sigma }\left(X\right)$ if and only if X has finite exponent.

We prove Theorems 16 and 18 in Section 3.

The notions of $\mathfrak{g}$-closed and $\mathfrak{g}$-dense subgroups of a compact Abelian group X were defined in [11]. In the last section of the paper, in analogy to these notions, we define ${\mathfrak{g}}_T$-closed and ${\mathfrak{g}}_T$-dense subgroups of X. In particular, we show that every ${\mathfrak{g}}_T$-dense subgroup of a compact Abelian group X is dense if and only if X is connected (see Theorem 37).

2. The Proofs of Theorems 10 and 14

The subgroup of a group G generated by a subset A we denote by $\langle A\rangle$.

Recall that a subgroup H of an Abelian topological group X is called dually closed in X if for every $x\in X\setminus H$, there exists a character $\chi \in H^{\perp }$, such that $(\chi ,x)\ne 1$. 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 [19].

The next notion generalizes the notion of the maximal extension in the class of all compact Abelian groups introduced in [20].

Definition 19. Let $\mathcal{G}$ be an arbitrary class of topological groups. Let $(G,\tau )\in \mathcal{G}$ and H be a subgroup of G. The group $(G,\tau )$ is called a maximal extension of $(H,\tau {|}_H)$ in the class $\mathcal{G}$ if $\sigma \le \tau$ for every group topology on G, such that ${\sigma |}_H{=\tau |}_H$ and $(G,\sigma )\in \mathcal{G}$.

Clearly, the maximal extension is unique if it exists. Note that in Definition 19, we do not assume that $(H,\tau {|}_H)$ belongs to the class $\mathcal{G}$.

If H is a subgroup of an Abelian group G and $\mathbf{u}$ is a T-sequence (respectively, a $TB$-sequence) in H, we denote by ${\tau }_{\mathbf{u}}\left(H\right)$ (respectively, ${\tau }_{b\mathbf{u}}\left(H\right)$) the finest (respectively, precompact) group topology on H generated by $\mathbf{u}$. We use the following easy corollary of the definition of T-sequences.

Lemma 20. For a sequence $\mathbf{u}$ in an Abelian group G, the following assertions are equivalent:

(1)

$\mathbf{u}$ is a T-sequence in G;

(2)

$\mathbf{u}$ is a T-sequence in every subgroup of G containing $\langle \mathbf{u}\rangle$;

(3)

$\mathbf{u}$ is a T-sequence in $\langle \mathbf{u}\rangle$.

In this case, $\langle \mathbf{u}\rangle$ is open in ${\tau }_{\mathbf{u}}$ (and hence, $\langle \mathbf{u}\rangle$ is dually closed and dually embedded in $(G,{\tau }_{\mathbf{u}})$), and $(G,{\tau }_{\mathbf{u}})$ is the maximal extension of $(\langle \mathbf{u}\rangle ,{\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$ in the class $\mathbf{TAG}$ of all Abelian topological groups.

Proof. Evidently, (1) implies (2) and (2) implies (3). Let $\mathbf{u}$ be a T-sequence in $\langle \mathbf{u}\rangle$. Let τ be the topology on G whose base is all translationsof ${\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$-open sets. Clearly, $\mathbf{u}$ converges to zero in τ. Thus, $\mathbf{u}$ is a T-sequence in G. Therefore, (3) implies (1).

Let us prove the last assertion. By the definition of ${\tau }_{\mathbf{u}}$, we have also $\tau \le {\tau }_{\mathbf{u}}$, and hence, ${\tau |}_{\langle \mathbf{u}\rangle }={\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)\le {\tau }_{\mathbf{u}}{|}_{\langle \mathbf{u}\rangle }$. Thus, $\langle \mathbf{u}\rangle$ is open in ${\tau }_{\mathbf{u}}$, and hence, it is dually closed and dually embedded in $(G,{\tau }_{\mathbf{u}})$ by [19] (Lemma 3.3). On the other hand, ${\tau }_{\mathbf{u}}{{|}_{\langle \mathbf{u}\rangle }\le {\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)=\tau |}_{\langle \mathbf{u}\rangle }$ by the definition of ${\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$. Therefore, ${\tau }_{\mathbf{u}}$ is an extension of ${\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$. Now, clearly, $\tau ={\tau }_{\mathbf{u}}$, and $(G,{\tau }_{\mathbf{u}})$ is the maximal extension of $(\langle \mathbf{u}\rangle ,{\tau }_{\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$ in the class $\mathbf{TAG}$.  ☐

For $TB$-sequences, we have the following:

Lemma 21. For a sequence $\mathbf{u}$ in an Abelian group G, the following assertions are equivalent:

(1)

$\mathbf{u}$ is a $TB$-sequence in G;

(2)

$\mathbf{u}$ is a $TB$-sequence in every subgroup of G containing $\langle \mathbf{u}\rangle$;

(3)

$\mathbf{u}$ is a $TB$-sequence in $\langle \mathbf{u}\rangle$.

In this case, the subgroup $\langle \mathbf{u}\rangle$ is dually closed and dually embedded in $(G,{\tau }_{b\mathbf{u}})$, and $(G,{\tau }_{b\mathbf{u}})$ is the maximal extension of $(\langle \mathbf{u}\rangle ,{\tau }_{b\mathbf{u}}\left(\langle \mathbf{u}\rangle \right))$ in the class of all precompact Abelian groups.

Proof. Evidently, (1) implies (2) and (2) implies (3). Let $\mathbf{u}$ be a $TB$-sequence in $\langle \mathbf{u}\rangle$. Then, ${(\langle \mathbf{u}\rangle ,{\tau }_{b\mathbf{u}}\left(\langle \mathbf{u}\rangle \right))}^{\wedge }$ separates the points of $\langle \mathbf{u}\rangle$. Let τ be the topology on G whose base is all translations of ${\tau }_{b\mathbf{u}}\left(\langle \mathbf{u}\rangle \right)$-open sets. Then, $(\langle \mathbf{u}\rangle ,{\tau }_{b\mathbf{u}}\left(\langle \mathbf{u}\rangle \right))$ is an open subgroup of $(G,\tau )$. It is easy to see that ${(G,\tau )}^{\wedge }$ separates the points of G. Since $\mathbf{u}$ converges to zero in τ, it also converges to zero in ${\tau }^{+}$, where ${\tau }^{+}$ is the Bohr topology of $(G,\tau )$. Thus, $\mathbf{u}$ is a $TB$-sequence in G. Therefore, (3) implies (1).

The last assertion follows from Proposition 1.8 and Lemma 3.6 in [20].  ☐

For a sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ of characters of a compact Abelian group X, set:

$$K_{\mathbf{u}}=\bigcap _{n\in \omega }ker\left(u_n\right).$$

The following assertions is proven in [13]:

Fact 22 (Lemma 2.2(i) of [13]). For every sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ of characters of a compact Abelian group X, the subgroup $K_{\mathbf{u}}$ is a closed $G_{\delta }$-subgroup of X and $K_{\mathbf{u}}={\langle \mathbf{u}\rangle }^{\perp }$.

The next two lemmas are natural analogues of Lemmas 2.2(ii) and 2.6 of [13].

Lemma 23. Let X be a compact Abelian group and $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ be a T-sequence in $\widehat{X}$. Then, $s_{\mathbf{u}}\left(X\right)/K_{\mathbf{u}}$ is a T-characterized subgroup of $X/K_{\mathbf{u}}$.

Proof. Set $H:=s_{\mathbf{u}}\left(X\right)$ and $K:=K_{\mathbf{u}}$. Let $q:X\to X/K$ be the quotient map. Then, the adjoint homomorphism $q^{\wedge }$ is an isomorphism from ${(X/K)}^{\wedge }$ onto $K^{\perp }$ in $X^{\wedge }$. For every $n\in \omega$, define the character ${\tilde{u}}_n$ of $X/K$ as follows: $({\tilde{u}}_n,q\left(x\right))=(u_n,x)$ (${\tilde{u}}_n$ is well-defined, since $K\subseteq ker\left(u_n\right)$). Then, $\tilde{\mathbf{u}}={\left\{{\tilde{u}}_n\right\}}_{n\in \omega }$ is a sequence of characters of $X/K$, such that $q^{\wedge }\left({\tilde{u}}_n\right)=u_n$. Since $\mathbf{u}\subset K^{\perp }$, $\mathbf{u}$ is a T-sequence in $K^{\perp }$ by Lemma 20. Hence, $\tilde{\mathbf{u}}$ is a T-sequence in ${(X/K)}^{\wedge }$ because $q^{\wedge }$ is an isomorphism.

We claim that $H/K=s_{\tilde{\mathbf{u}}}(X/K)$. Indeed, for every $h+K\in H/K$, by definition, we have $({\tilde{u}}_n,h+K)=(u_n,h)\to 1.$ Thus, $H/K\subseteq s_{\tilde{\mathbf{u}}}(X/K)$. If $x+K\in s_{\tilde{\mathbf{u}}}(X/K)$, then $({\tilde{u}}_n,x+K)=(u_n,x)\to 1.$ This yields $x\in H$. Thus, $x+K\in H/K$.  ☐

Let $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ be a T-sequence in an Abelian group G. For every natural number m, set ${\mathbf{u}}_m={\left\{u_n\right\}}_{n\ge m}$. Clearly, ${\mathbf{u}}_m$ is a T-sequence in G, ${\tau }_{\mathbf{u}}={\tau }_{{\mathbf{u}}_m}$ and $s_{\mathbf{u}}\left(X\right)=s_{{\mathbf{u}}_m}\left(X\right)$ for every natural number m.

Lemma 24. Let K be a closed subgroup of a compact Abelian group X and $q:X\to X/K$ be the quotient map. Then, $\tilde{H}$ is a T-characterized subgroup of $X/K$ if and only if $q^{-1}\left(\tilde{H}\right)$ is a T-characterized subgroup of X.

Proof. Let $\tilde{H}$ be a T-characterized subgroup of $X/K$, and let a T-sequence $\tilde{\mathbf{u}}={\left\{{\tilde{u}}_n\right\}}_{n\in \omega }$-characterized $\tilde{H}$. Set $H:=q^{-1}\left(\tilde{H}\right)$. We have to show that H is a T-characterized subgroup of X.

Note that the adjoint homomorphism $q^{\wedge }$ is an isomorphism from ${(X/K)}^{\wedge }$ onto $K^{\perp }$ in $X^{\wedge }$. Set $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$, where $u_n=q^{\wedge }\left({\tilde{u}}_n\right)$. Since $q^{\wedge }$ is injective, $\mathbf{u}$ is a T-sequence in $K^{\perp }$. By Lemma 20, $\mathbf{u}$ is a T-sequence in $\widehat{X}$. Therefore, it is enough to show that $H=s_{\mathbf{u}}\left(X\right)$. This follows from the following chain of equivalences. By definition, $x\in s_{\mathbf{u}}\left(X\right)$ if and only if:

$$(u_n,x)\to 1\iff ({\tilde{u}}_n,q\left(x\right))\to 1\iff q\left(x\right)\in \tilde{H}=H/K\iff x\in H.$$

The last equivalence is due to the inclusion $K\subseteq H$.

Conversely, let $H:=q^{-1}\left(\tilde{H}\right)$ be a T-characterized subgroup of X and a T-sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$-characterized H. Proposition 2.5 of [13] implies that we can find $m\in \mathbb{N}$, such that $K\subseteq K_{{\mathbf{u}}_m}$. Therefore, taking into account that $H=s_{\mathbf{u}}\left(X\right)=s_{{\mathbf{u}}_m}\left(X\right)$ for every natural number m, without loss of generality, we can assume that $K\subseteq K_{\mathbf{u}}$. By Lemma 23, $H/K_{\mathbf{u}}$ is a T-characterized subgroup of $X/K_{\mathbf{u}}$. Denote by $q_u$ the quotient homomorphism from $X/K$ onto $X/K_{\mathbf{u}}$. Then, $\tilde{H}=q_u^{-1}(H/K_{\mathbf{u}})$ is T-characterized in $X/K$ by the previous paragraph of the proof.  ☐

The next theorem is an analogue of Theorem B of [13], 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:

Theorem 25. A subgroup H of a compact Abelian group X is T-characterized if and only if H contains a closed $G_{\delta }$-subgroup K of X, such that $H/K$ is a T-characterized subgroup of the compact metrizable group $X/K$.

Proof. Let H be T-characterized in X by a T-sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ in $\widehat{X}$. Set $K:=K_{\mathbf{u}}$. Since K is a closed $G_{\delta }$-subgroup of X by Fact 22, $X/K$ is metrizable. By Lemma 23, $H/K$ is a T-characterized subgroup of $X/K$.

Conversely, let H contain a closed $G_{\delta }$-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 24.  ☐

As was noticed in [21] before Definition 2.33, for every T-sequence $\mathbf{u}$ in an infinite Abelian group G, the subgroup $\langle \mathbf{u}\rangle$ is open in $(G,{\tau }_{\mathbf{u}})$ (see also Lemma 20), and hence, by Lemmas 1.4 and 2.2 of [22], the following sequences are exact:

$$\begin{array}{cc}& 0\to (\langle \mathbf{u}\rangle ,{\tau }_{\mathbf{u}})\to (G,{\tau }_{\mathbf{u}})\to G/\langle \mathbf{u}\rangle \to 0,\hfill \\ & 0\to {\left(G/\langle \mathbf{u}\rangle \right)}^{\wedge }\to {(G,{\tau }_{\mathbf{u}})}^{\wedge }\to {(\langle \mathbf{u}\rangle ,{\tau }_{\mathbf{u}}{|}_{\langle \mathbf{u}\rangle })}^{\wedge }\to 0,\hfill \end{array}$$

(2)

where ${\left(G/\langle \mathbf{u}\rangle \right)}^{\wedge }\cong {\langle \mathbf{u}\rangle }^{\perp }$ is a compact subgroup of ${(G,{\tau }_{\mathbf{u}})}^{\wedge }$ and ${(\langle \mathbf{u}\rangle ,{\tau }_{\mathbf{u}})}^{\wedge }\cong {(G,{\tau }_{\mathbf{u}})}^{\wedge }/{\langle \mathbf{u}\rangle }^{\perp }$.

Let $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ be a T-sequence in an Abelian group G. It is known [10] that ${\tau }_{\mathbf{u}}$ is sequential, and hence, $(G,{\tau }_{\mathbf{u}})$ is a k-space. Therefore, the natural homomorphism $\alpha :={\alpha }_{(G,{\tau }_{\mathbf{u}})}:(G,{\tau }_{\mathbf{u}})\to {(G,{\tau }_{\mathbf{u}})}^{\wedge \wedge }$ is continuous by [23] (5.12). Let us recall that $(G,{\tau }_{\mathbf{u}})$ is MinAP if and only if $(G,{\tau }_{\mathbf{u}})=ker\left(\alpha \right)$.

To prove Theorem 10, we need the following:

Fact 26 ([16]). For each T-sequence $\mathbf{u}$ in a countably infinite Abelian group G, the group ${(G,{\tau }_{\mathbf{u}})}^{\wedge }$ is Polish.

Now, we are in a position to prove Theorem 10.

Proof of Theorem 10. $\left(1\right)\Rightarrow \left(2\right)$ Let H be a proper closed T-characterized subgroup of X and a T-sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$-characterizedH. Since H is also characterized, it is a $G_{\delta }$-subgroup of X by Fact 3. We have to show that $H^{\perp }$ admits a MinAP group topology.

Our idea of the proof is the following. Set $G:=\widehat{X}$. By Fact 8, $H^{\perp }$ is the von Neumann radical of $(G,{\tau }_{\mathbf{u}})$. Now, assume that we found another T-sequence $\mathbf{v}$ that characterizes H and such that $\langle \mathbf{v}\rangle =H^{\perp }$ (maybe $\mathbf{v}=\mathbf{u}$). By Fact 8, we have $\mathbf{n}(G,{\tau }_{\mathbf{v}})=H^{\perp }=\langle \mathbf{v}\rangle$. Lemma 20 implies that the subgroup $(\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle })$ of $(G,{\tau }_{\mathbf{v}})$ is open, and hence, it is dually closed and dually embedded in $(G,{\tau }_{\mathbf{v}})$. Hence, $\mathbf{n}(\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle })=\mathbf{n}(G,{\tau }_{\mathbf{v}})(=\langle \mathbf{v}\rangle )$ by Lemma 4 of [16]. Therefore, $(\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle })$ is MinAP. Thus, $H^{\perp }=\langle \mathbf{v}\rangle$ admits a MinAP group topology, as desired.

We find such a T-sequence $\mathbf{v}$ in four steps (in fact, we show that $\mathbf{v}$ has the form ${\mathbf{u}}_m$ for some $m\in \mathbb{N}$).

Step 1. Let $q:X\to X/K_{\mathbf{u}}$ be the quotient map. For every $n\in \omega$, define the character ${\tilde{u}}_n$ of $X/K_{\mathbf{u}}$ by the equality $u_n={\tilde{u}}_n\circ q$ (this is possible since $K_{\mathbf{u}}\subseteq ker\left(u_n\right)$). As was shown in the proof of Lemma 23, the sequence $\tilde{\mathbf{u}}={\left\{{\tilde{u}}_n\right\}}_{n\in \omega }$ is a T-sequence, which characterizes $H/K_{\mathbf{u}}$ in $X/K_{\mathbf{u}}$. Set $\tilde{X}:=X/K_{\mathbf{u}}$ and $\tilde{H}:=H/K_{\mathbf{u}}$. Therefore, $\tilde{H}=s_{\tilde{\mathbf{u}}}\left(\tilde{X}\right)$. By [24] (5.34 and 24.11) and since $K_{\mathbf{u}}\subseteq H$, we have:

$$H^{\perp }\cong {(X/H)}^{\wedge }\cong {\left(\tilde{X}/\tilde{H}\right)}^{\wedge }\cong {\tilde{H}}^{\perp }.$$

(3)

By Fact 3, $\tilde{X}$ is metrizable. Hence, $\tilde{H}$ is also compact and metrizable, and $\tilde{G}:=\widehat{\tilde{X}}$ is a countable Abelian group by [24] (24.15). Since H is a proper closed subgroup of X, Equation (3) implies that $\tilde{G}$ is non-zero.

We claim that $\tilde{G}$ is countably infinite. Indeed, suppose for a contradiction that $\tilde{G}$ is finite. Then, $X/K_{\mathbf{u}}=\tilde{X}$ is also finite. Now, Fact 22 implies that $\langle \mathbf{u}\rangle$ is a finite subgroup of G. Since $\mathbf{u}$ is a T-sequence, $\mathbf{u}$ must be eventually equal to zero. Hence, $H=s_{\mathbf{u}}\left(X\right)=X$ is not a proper subgroup of X, a contradiction.

Step 2. We claim that there is a natural number m, such that the group $(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{\tilde{\mathbf{u}}}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })=(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{{\tilde{\mathbf{u}}}_m}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })$ is MinAP.

Indeed, since $\tilde{G}$ is countably infinite, we can apply Fact 8. Therefore, $\tilde{H}={(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge }$ algebraically. Since $\tilde{H}$ and ${(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge }$ are Polish groups (see Fact 26), $\tilde{H}$ and ${(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge }$ are topologically isomorphic by the uniqueness of the Polish group topology. Hence ${(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge \wedge }={\tilde{H}}^{\wedge }$ is discrete. As was noticed before the proof, the natural homomorphism $\tilde{\alpha }:(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})\to {(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge \wedge }$ is continuous. Since ${(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})}^{\wedge \wedge }$ is discrete, we obtain that the von Neumann radical $ker\left(\tilde{\alpha }\right)$ of $(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})$ is open in ${\tau }_{\tilde{\mathbf{u}}}$. Therefore, there exists a natural number m, such that ${\tilde{u}}_n\in ker\left(\tilde{\alpha }\right)$ for every $n\ge m$. Hence, $\langle {\tilde{\mathbf{u}}}_m\rangle \subseteq ker\left(\tilde{\alpha }\right)$. Lemma 20 implies that the subgroup $\langle {\tilde{\mathbf{u}}}_m\rangle$ is open in $(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})$, and hence, it is dually closed and dually embedded in $(\tilde{G},{\tau }_{\tilde{\mathbf{u}}})$. Now, Lemma 4 of [16] yields $\langle {\tilde{\mathbf{u}}}_m\rangle =ker\left(\tilde{\alpha }\right)$, and $(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{\tilde{\mathbf{u}}}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })$ is MinAP.

Step 3. Set $\mathbf{v}={\left\{v_n\right\}}_{n\in \omega }$, where $v_n=u_{n+m}$ for every $n\in \omega$. Clearly, $\mathbf{v}$ is a T-sequence in G characterizing H, ${\tau }_{\mathbf{u}}={\tau }_{\mathbf{v}}$ and $K_{\mathbf{u}}\subseteq K_{\mathbf{v}}$. Let $t:X\to X/K_{\mathbf{v}}$ and $r:X/K_{\mathbf{u}}\to X/K_{\mathbf{v}}$ be the quotient maps. Analogously to Step 1 and the proof of Lemma 23, the sequence $\tilde{\mathbf{v}}={\left\{{\tilde{v}}_n\right\}}_{n\in \omega }$ is a T-sequence in $\widehat{X/K_{\mathbf{v}}}$, which characterizes $H/K_{\mathbf{v}}$ in $X/K_{\mathbf{v}}$, where $v_n={\tilde{v}}_n\circ t$. Since $t=r\circ q$, we have:

$$v_n={\tilde{v}}_n\circ t=t^{\wedge }\left({\tilde{v}}_n\right)=q^{\wedge }\left(r^{\wedge }\left({\tilde{v}}_n\right)\right),$$

where $t^{\wedge }$, $r^{\wedge }$ and $q^{\wedge }$ are the adjoint homomorphisms to t, r and q, respectively.

Since $q^{\wedge }$ and $r^{\wedge }$ are embeddings, we have $r^{\wedge }\left({\tilde{v}}_n\right)={\tilde{u}}_{n+m}$. In particular, $\langle \mathbf{v}\rangle \cong \langle \tilde{\mathbf{v}}\rangle \cong \langle {\tilde{\mathbf{u}}}_m\rangle and$:

$$(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{\tilde{\mathbf{u}}}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })=(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{{\tilde{\mathbf{u}}}_m}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })\cong (\langle \tilde{\mathbf{v}}\rangle ,{\tau }_{\tilde{\mathbf{v}}}{|}_{\langle \tilde{\mathbf{v}}\rangle })\cong (\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle }).$$

By Step 2, $(\langle {\tilde{\mathbf{u}}}_m\rangle ,{\tau }_{{\tilde{\mathbf{u}}}_m}{|}_{\langle {\tilde{\mathbf{u}}}_m\rangle })$ is MinAP. Hence, $(\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle })$ is MinAP, as well.

Step 4. By the second exact sequence in Equation (2) applying to $\mathbf{v}$, Fact 8, and since $(\langle \mathbf{v}\rangle ,{\tau }_{\mathbf{v}}{|}_{\langle \mathbf{v}\rangle })$ is MinAP (by Step 3), we have $H=s_{\mathbf{v}}\left(X\right)={(G,{\tau }_{\mathbf{v}})}^{\wedge }={\left(G/\langle \mathbf{v}\rangle \right)}^{\wedge }={\langle \mathbf{v}\rangle }^{\perp }$ algebraically. Thus, $H^{\perp }=\langle \mathbf{v}\rangle$, and hence, $H^{\perp }$ admits a MinAP group topology generated by the T-sequence $\mathbf{v}$.

$\left(2\right)\Rightarrow \left(1\right)$: Since H is a $G_{\delta }$-subgroup of X, H is closed by [13] (Proposition 2.4) and $X/H$ is metrizable (due to the well-known fact that a compact group of countable pseudo-character is metrizable). Hence, $H^{\perp }={(X/H)}^{\wedge }$ is countable. Since $H^{\perp }$ admits a MinAP group topology, $H^{\perp }$ must be countably infinite. By Theorem 3.8 of [9], $H^{\perp }$ admits a MinAP group topology generated by a T-sequence $\tilde{\mathbf{u}}={\left\{{\tilde{u}}_n\right\}}_{n\in \omega }$. By Fact 8, this means that $s_{\tilde{\mathbf{u}}}(X/H)=\left\{0\right\}$. Let $q:X\to X/H$ be the quotient map. Set $u_n={\tilde{u}}_n\circ q=q^{\wedge }\left({\tilde{u}}_n\right)$. Since $q^{\wedge }$ is injective, $\mathbf{u}$ is a T-sequence in $\widehat{X}$ by Lemma 20. We have to show that $H=s_{\mathbf{u}}\left(X\right)$. By definition, $x\in s_{\mathbf{u}}\left(X\right)$ if and only if:

$$(u_n,x)=({\tilde{u}}_n,q\left(x\right))\to 1\iff q\left(x\right)\in s_{\tilde{\mathbf{u}}}(X/H)\iff q\left(x\right)=0\iff x\in H.$$

(2)⇔(3) follows from Theorem 3.8 of [9]. The theorem is proven.  ☐

Proof of Theorem 14. $\left(1\right)\Rightarrow \left(2\right)$: Suppose for a contradiction that X is not connected. Then, by [24] (24.25), the dual group $G=X^{\wedge }$ has a non-zero element g of finite order. Then, the subgroup $H:={\langle g\rangle }^{\perp }$ of X has finite index. Hence, H is an open subgroup of X. Thus, H is not T-characterized by Corollary 13. This contradiction shows that X must be connected.

$\left(2\right)\Rightarrow \left(1\right)$: Let H be a proper $G_{\delta }$-subgroup of X. Then, H is closed by [13] (Proposition 2.4), and $X/H$ is connected and non-zero. Hence, $H^{\perp }\cong {(X/H)}^{\wedge }$ is countably infinite and torsion free by [24] (24.25). Thus, $H^{\perp }$ has infinite exponent. Therefore, by Theorem 10, H is T-characterized.  ☐

The next proposition is a simple corollary of Theorem B in [13].

Proposition 27. The closure $\overline{H}$ 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 [13], H contains a compact $G_{\delta }$-subgroup K of X. Then, $\overline{H}$ is also a $G_{\delta }$-subgroup of X. Thus, $\overline{H}$ is a characterized subgroup of X by Theorem B of [13].  ☐

In general, we cannot assert that the closure $\overline{H}$ of a T-characterized subgroup H of a compact Abelian group X is also T-characterized, as the next example shows.

Example 1. Let $X=\mathbb{Z}\left(2\right)\times \mathbb{T}$ and $G=\widehat{X}=\mathbb{Z}\left(2\right)\times \mathbb{Z}$. It is known (see the end of $\left(\mathbf{1}\right)$ in [7]) that there is a T-sequence $\mathbf{u}$ in G, such that the von Neumann radical $\mathbf{n}(G,{\tau }_{\mathbf{u}})$ of $(G,{\tau }_{\mathbf{u}})$ is $\mathbb{Z}\left(2\right)\times \left\{0\right\}$, the subgroup $H:=s_{\mathbf{u}}\left(X\right)$ is countable and $\overline{H}=\left\{0\right\}\times \mathbb{T}$. Therefore, the closure $\overline{H}$ of the countable T-characterized subgroup H of X is open. Thus, $\overline{H}$ is not T-characterized by Corollary 13.

We do not know the answers to the following questions:

Problem 28. Let H be a characterized subgroup of a compact Abelian group X, such that its closure $\overline{H}$ is T-characterized. Is H a T-characterized subgroup of X?

Problem 29. Does there exists a metrizable Abelian compact group that has a countable non-T-characterized subgroup?

3. The Proofs of Theorems 16 and 18

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 $i:(H,\tau )\to X$ is continuous. Let H be a T-characterized subgroup of a compact metrizable Abelian group X by a T-sequence $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$. Then, by [16] (Theorem 1), H is polishable by the metric:

$$\rho (x,y)=d(x,y)+sup\left\{\right|(u_n,x)-(u_n,y)|,n\in \omega \},$$

(4)

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 16 we need the following three lemmas.

For a real number x, we write $\left[x\right]$ for the integral part of x and $\parallel x\parallel$ for the distance from x to the nearest integer. We also use the following inequality proven in [25]:

$$\pi \left|\phi \right|\le |1-e^{2\pi i\phi }|\le 2\pi |\phi |,\phi \in \left[-\frac{1}{2},\frac{1}{2}\right).$$

(5)

Lemma 30. Let ${\left\{a_n\right\}}_{n\in \omega }\subset \mathbb{N}$ be such that $a_n\to \infty$ and $a_n\ge 2,n\in \omega$. Set $u_n={\prod }_{k\le n}{a}_n$ for every $n\in \omega$. Then, $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$ is a T-sequence in $X=\mathbb{T}$, and the T-characterized subgroup $H=s_{\mathbf{u}}\left(\mathbb{T}\right)$ of $\mathbb{T}$ is a dense non-$F_{\sigma }$-subset of $\mathbb{T}$.

Proof. We consider the circle group $\mathbb{T}$ as $\mathbb{R}/\mathbb{Z}$ and write it additively. Therefore, $d(0,x)=\parallel x\parallel$ for every $x\in \mathbb{T}$. Recall that every $x\in \mathbb{T}$ has the unique representation in the form:

$$x=\sum _{n=0}^{\infty }\frac{c_n}{u_n},$$

(6)

where $0\le c_n<a_n$ and $c_n\ne a_n-1$ for infinitely many indices n.

It is known [26] (see also (12) in the proof of Lemma 1 of [25]) that x with representation Equation (6) belongs to H if and only if:

$$\underset{n\to \infty }{lim}\frac{c_n}{a_n}\left(\mathrm{mod}1\right)=0.$$

(7)

Hence, H is a dense subgroup of $\mathbb{T}$. Thus, $\mathbf{u}$ is even a $TB$-sequence in $\mathbb{Z}$ by Fact 1.

We have to show that H is not an $F_{\sigma }$-subset of $\mathbb{T}$. Suppose for a contradiction that H is an $F_{\sigma }$-subset of $\mathbb{T}$. Then, $H={\cup }_{n\in \mathbb{N}}{F}_n$, where $F_n$ is a compact subset of $\mathbb{T}$ for every $n\in \mathbb{N}$. Since H is a subgroup of $\mathbb{T}$, without loss of generality, we can assume that $F_n-F_n\subseteq F_{n+1}$. Since all $F_n$ are closed in $(H,\rho )$, as well, the Baire theorem implies that there are $0<\epsilon <0.1$ and $m\in \mathbb{N}$, such that $F_m\supseteq \{x:\rho (0,x)\le \epsilon \}$.

Fix arbitrarily $l>0$, such that $\frac{2}{u_{l-1}}<\frac{\epsilon }{20}$. For every natural number $k>l$, set:

$$x_k:=\sum _{n=l}^k\frac{1}{u_n}·\left[\frac{(a_n-1)\epsilon }{20}\right].$$

Then, for every $k>l$, we have:

$$x_k=\sum _{n=l}^k\frac{1}{u_n}·\left[\frac{(a_n-1)\epsilon }{20}\right]<\sum _{n=l}^k\frac{1}{u_{n-1}}·\frac{\epsilon }{20}<\frac{1}{u_{l-1}}\sum _{n=0}^{k-l}\frac{1}{2^n}<\frac{2}{u_{l-1}}<\frac{\epsilon }{20}<\frac{1}{2}.$$

This inequality and Equation (5) imply that:

$$d(0,x_k)=\parallel x_k\parallel =x_k<\frac{\epsilon }{20},\mathrm{ for every }k>l.$$

(8)

For every $s\in \omega$ and every natural number $k>l$, we estimate $|1-(u_s,x_k\left)\right|$ as follows.

Case 1. Let $s<k$. Set $q=max\{s+1,l\}$. By the definition of $x_k$, we have:

$$\begin{array}{cc}\hfill 2\pi \left[(u_s·x_k)\right.\left.\left(\mathrm{mod}1\right)\right]& =2\pi \left[u_s\sum _{n=l}^k\frac{1}{u_n}·\left[\frac{(a_n-1)\epsilon }{20}\right]\left(\mathrm{mod}1\right)\right]<2\pi \sum _{n=q}^k\frac{u_s}{u_n}·\frac{(a_n-1)\epsilon }{20}\hfill \\ & <\frac{\pi \epsilon }{10}\left(1+\frac{1}{a_{s+1}}+\frac{1}{a_{s+1}{a}_{s+2}}+\frac{1}{a_{s+1}{a}_{s+2}{a}_{s+3}}+\cdots \right)\hfill \\ & <\frac{\pi \epsilon }{10}\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \right)=\frac{\pi \epsilon }{10}·2<\frac{2\epsilon }{3}<\frac{1}{2}.\hfill \end{array}$$

This inequality and Equation (5) imply:

$$|1-(u_s,x_k)|=\left|1-exp\left\{2\pi i·\left[(u_s·x_k)\left(\mathrm{mod}1\right)\right]\right\}\right|<\frac{2\epsilon }{3}.$$

(9)

Case 2. Let $s\ge k$. By the definition of $x_k$, we have:

$$|1-(u_s,x_k\left)\right|=0.$$

(10)

In particular, Equation (10) implies that $x_k\in H$ for every $k>l$.

Now, for every $k>l$, Equations (4) and (8)–(10) imply:

$$\rho (0,x_k)<\frac{\epsilon }{20}+\frac{2\epsilon }{3}<\epsilon .$$

Thus, $x_k\in F_m$ for every natural number $k>l$. Clearly,

$$x_k\to x:=\sum _{n=l}^{\infty }\frac{1}{u_n}·\left[\frac{(a_n-1)\epsilon }{20}\right]\mathrm{in}\mathbb{T}.$$

Since $F_m$ is a compact subset of $\mathbb{T}$, we have $x\in F_m$. Hence, $x\in H$. On the other hand, we have:

$$\underset{n\to \infty }{lim}\frac{1}{a_n}·\left[\frac{(a_n-1)\epsilon }{20}\right]\left(\mathrm{mod}1\right)=\frac{\epsilon }{20}\ne 0.$$

Therefore, Equation (7) implies that $x\notin H$. This contradiction shows that $H=s_{\mathbf{u}}\left(\mathbb{T}\right)$ is not an $F_{\sigma }$-subset of $\mathbb{T}$.  ☐

For a prime number p, the group $\mathbb{Z}\left(p^{\infty }\right)$ is regarded as the collection of fractions $m/p^n\in [0,1)$. Let ${\mathsf{\Delta }}_p$ be the compact group of p-adic integers. It is well known that $\widehat{{\mathsf{\Delta }}_p}=\mathbb{Z}\left(p^{\infty }\right)$.

Lemma 31. Let $X={\mathsf{\Delta }}_p$. For an increasing sequence of natural numbers $0<n_0<n_1<\cdots$, such that $n_{k+1}-n_k\to \infty$, set:

$$u_k=\frac{1}{p^{n_k+1}}\in \mathbb{Z}\left(p^{\infty }\right).$$

Then, the sequence $\mathbf{u}={\left\{u_k\right\}}_{k\in \omega }$ is a T-sequence in $\mathbb{Z}\left(p^{\infty }\right)$, and the T-characterized subgroup $H=s_{\mathbf{u}}\left({\mathsf{\Delta }}_p\right)$ is a dense non-$F_{\sigma }$-subset of ${\mathsf{\Delta }}_p$.

Proof. Let $\omega ={\left(a_n\right)}_{n\in \omega }\in {\mathsf{\Delta }}_p$, where $0\le a_n<p$ for every $n\in \omega$. Recall that, for every $k\in \omega$, [24] (25.2) implies:

$$(u_k,\omega )=exp\left\{\frac{2\pi i}{p^{n_k+1}}\left(a_0+p{a}_1+\cdots +p^{n_k}{a}_{n_k}\right)\right\}.$$

(11)

Further, by [24] (10.4), if $\omega \ne 0$, then $d(0,\omega )=2^{-n}$, where n is the minimal index, such that $a_n\ne 0$.

Following [27] (2.2), for every $\omega =\left(a_n\right)\in {\mathsf{\Delta }}_p$ and every natural number $k>1$, set:

$$m_k=m_k\left(\omega \right)=max\{j_k,n_{k-1}\},$$

where:

$$j_k=n_k\mathrm{ if }0<a_{n_k}<p-1,$$

and otherwise:

$$j_k=min\{j:\mathrm{ either }{a}_s=0\mathrm{ for }j<s\le n_k,\mathrm{ or }{a}_s=p-1\mathrm{ for }j<s\le n_k\}.$$

In [27] (2.2), it is shown that:

$$\omega \in s_{\mathbf{u}}\left({\mathsf{\Delta }}_p\right)\mathrm{ if and only if }{n}_k-m_k\to \infty .$$

(12)

Therefore, $H:=s_{\mathbf{u}}\left({\mathsf{\Delta }}_p\right)$ contains the identity $\mathbf{1}=(1,0,0,\cdots )$ of ${\mathsf{\Delta }}_p$. By [24] (Remark 10.6), $\langle \mathbf{1}\rangle$ is dense in ${\mathsf{\Delta }}_p$. Hence, H is dense in ${\mathsf{\Delta }}_p$, as well. Now, Fact 1 implies that $\mathbf{u}$ is a T-sequence in $\mathbb{Z}\left(p^{\infty }\right)$.

We have to show that H is not an $F_{\sigma }$-subset of ${\mathsf{\Delta }}_p$. Suppose for a contradiction that $H={\cup }_{n\in \mathbb{N}}{F}_n$ is an $F_{\sigma }$-subset of ${\mathsf{\Delta }}_p$, where $F_n$ is a compact subset of ${\mathsf{\Delta }}_p$ for every $n\in \mathbb{N}$. Since H is a subgroup of ${\mathsf{\Delta }}_p$, without loss of generality, we can assume that $F_n-F_n\subseteq F_{n+1}$. Since all $F_n$ are closed in $(H,\rho )$, as well, the Baire theorem implies that there are $0<\epsilon <0.1$ and $m\in \mathbb{N}$, such that $F_m\supseteq \{x:\rho (0,x)\le \epsilon \}$.

Fix a natural number s, such that $\frac{1}{2^s}<\frac{\epsilon }{20}$. Choose a natural number $l>s$, such that, for every natural number $w\ge l$, we have:

$$n_{w+1}-n_w>s.$$

(13)

For every $r\in \mathbb{N}$, set:

$${\omega }_r:=\left(a_n^r\right),\mathrm{ where }{a}_n^r=\left\{\begin{array}{cc}\hfill 1,& \mathrm{if }n=n_{l+i}-s\mathrm{ for some }1\le i\le r,\hfill \\ \hfill 0,& \mathrm{otherwise}.\hfill \end{array}\right.$$

Then, for every $r\in \mathbb{N}$, Equation (13) implies that ${\omega }_r$ is well defined and:

$$d(0,{\omega }_r)=\frac{1}{2^{n_{l+1}-s}}<\frac{1}{2^{n_l}}\le \frac{1}{2^l}<\frac{1}{2^s}<\frac{\epsilon }{20}.$$

(14)

Note that:

$$1+p+\cdots +p^k=\frac{p^{k+1}-1}{p-1}<p^{k+1}.$$

(15)

For every $k\in \omega$ and every $r\in \mathbb{N}$, we estimate $|1-(u_k,{\omega }_r\left)\right|$ as follows.

Case 1. Let $k\le l$. By Equations (11) and (13) and the definition of ${\omega }_r$, we have:

$$|1-(u_k,{\omega }_r\left)\right|=0.$$

(16)

Case 2. Let $l<k\le l+r$. Then, Equation (15) yields:

$$\frac{2\pi }{p^{n_k+1}}\left|p^{n_{l+1}-s}+\cdots +p^{n_k-s}\right|<\frac{2\pi }{p^{n_k+1}}·p^{n_k-s+1}=\frac{2\pi }{p^s}\le \frac{2\pi }{2^s}<\frac{\epsilon }{2}<\frac{1}{2}.$$

This inequality and the inequality Equations (5) and (11) imply:

$$|1-(u_k,{\omega }_r)|=\left|1-exp\left\{\frac{2\pi i}{p^{n_k+1}}\left(p^{n_{l+1}-s}+\cdots +p^{n_k-s}\right)\right\}\right|<\frac{\epsilon }{2}.$$

(17)

Case 3. Let $l+r<k$. By Equation (15), we have:

$$\begin{array}{cc}\hfill \frac{2\pi }{p^{n_k+1}}\left|p^{n_{l+1}-s}+\cdots +p^{n_{l+r}-s}\right|& <\frac{2\pi }{p^{n_k+1}}·p^{n_{l+r}-s+1}\hfill \\ & <\frac{2\pi }{p^{n_k+1}}·p^{n_k-s+1}=\frac{2\pi }{p^s}\le \frac{2\pi }{2^s}<\frac{\epsilon }{2}.\hfill \end{array}$$

These inequalities, Equations (5) and (11) immediately yield:

$$|1-(u_k,{\omega }_r)|=\left|1-exp\left\{\frac{2\pi i}{p^{n_k+1}}\left(p^{n_{l+1}-s}+\cdots +p^{n_{l+r}-s}\right)\right\}\right|<\frac{\epsilon }{2},$$

(18)

and:

$$|1-(u_k,{\omega }_r)|<\frac{2\pi }{p^{n_k+1}}·p^{n_{l+r}-s+1}\to 0,\mathrm{ as }k\to \infty .$$

(19)

Therefore, Equation (19) implies that ${\omega }_r\in H$ for every $r\in \mathbb{N}$.

For every $r\in \mathbb{N}$, by Equations (4), (14) and (16)–(18), we have:

$$\rho (0,{\omega }_r)=d(0,{\omega }_r)+sup\left\{\left|1-(u_k,{\omega }_r)\right|,k\in \omega \right\}<\frac{\epsilon }{20}+\frac{\epsilon }{2}<\epsilon .$$

Thus, ${\omega }_r\in F_m$ for every $r\in \mathbb{N}$. Evidently,

$${\omega }_r\to \tilde{\omega }=\left({\tilde{a}}_n\right)\mathrm{ in }{\mathsf{\Delta }}_p,\mathrm{ where }{\tilde{a}}_n=\left\{\begin{array}{cc}\hfill 1,& \mathrm{ if }n=n_{l+i}-s\mathrm{ for some }i\in \mathbb{N},\hfill \\ \hfill 0,& \mathrm{otherwise}.\hfill \end{array}\right.$$

Since $F_m$ is a compact subset of ${\mathsf{\Delta }}_p$, we have $\tilde{\omega }\in F_m$. Hence, $\tilde{\omega }\in H$. On the other hand, it is clear that $m_k\left(\tilde{\omega }\right)=n_k-s$ for every $k\ge l+1$. Thus, for every $k\ge l+1$, $n_k-m_k\left(\tilde{\omega }\right)=s\neg \to \infty$. Now, Equation (12) implies that $\tilde{\omega }\notin H$. This contradiction shows that H is not an $F_{\sigma }$-subset of ${\mathsf{\Delta }}_p$.  ☐

Lemma 32. Let $X={\prod }_{n\in \omega }\mathbb{Z}\left(b_n\right)$, where $1<b_0<b_1<\cdots$ and $G:=\widehat{X}={⨁}_{n\in \omega }\mathbb{Z}\left(b_n\right)$. Set $\mathbf{u}={\left\{u_n\right\}}_{n\in \omega }$, where $u_n=1\in \mathbb{Z}{\left(b_n\right)}^{\wedge }\subset G$ for every $n\in \omega$. Then, $\mathbf{u}$ is a T-sequence in G, and the T-characterized subgroup $H=s_{\mathbf{u}}\left(X\right)$ is a dense non-$F_{\sigma }$-subset of X.

Proof. Set $H:=s_{\mathbf{u}}\left(X\right)$. In [27] (2.3), it is shown that:

$$\omega =\left(a_n\right)\in s_{\mathbf{u}}\left(X\right)\mathrm{ if and only if }∥\frac{a_n}{b_n}∥\to 0.$$

(20)

Therefore, ${⨁}_{n\in \omega }\mathbb{Z}\left(b_n\right)\subseteq H$. Thus, H is dense in X. Now, Fact 1 implies that $\mathbf{u}$ is a T-sequence in G.

We have to show that H is not an $F_{\sigma }$-subset of X. Suppose for a contradiction that $H={\cup }_{n\in \mathbb{N}}{F}_n$ is an $F_{\sigma }$-subset of X, where $F_n$ is a compact subset of X for every $n\in \mathbb{N}$. Since H is a subgroup of X, without loss of generality, we can assume that $F_n-F_n\subseteq F_{n+1}$. Since all $F_n$ are closed in $(H,\rho )$, as well, the Baire theorem yields that there are $0<\epsilon <0.1$ and $m\in \mathbb{N}$, such that $F_m\supseteq \{\omega \in X:\rho (0,\omega )\le \epsilon \}$.

Note that $d(0,\omega )=2^{-l}$, where $0\ne \omega ={\left(a_n\right)}_{n\in \omega }\in X$ and l is the minimal index, such that $a_l\ne 0$. Choose l, such that $2^{-l}<\epsilon /3$. For every natural number $k>l$, set:

$${\omega }_k:=\left(a_n^k\right),\mathrm{ where }{a}_n^k=\left\{\begin{array}{cc}\hfill \left[\frac{\epsilon b_n}{20}\right],& \mathrm{ for every }n\mathrm{ such that }l\le n\le k,\hfill \\ \hfill 0,& \mathrm{ if either }1\le n<l\mathrm{ or }k<n.\hfill \end{array}\right.$$

Since $(u_n,{\omega }_k)=1$ for every $n>k$, we obtain that ${\omega }_k\in H$ for every $k>l$. For every $n\in \omega$, we have:

$$2\pi ·\frac{1}{b_n}\left[\frac{\epsilon b_n}{20}\right]<\frac{2\pi \epsilon }{20}<\epsilon <\frac{1}{2}.$$

This inequality and the inequality Equations (4) and (5) imply:

$$\begin{array}{cc}\hfill \rho (0,{\omega }_k)& =d(0,{\omega }_k)+sup\left\{\left|1-(u_n,{\omega }_k)\right|,n\in \omega \right\}\hfill \\ & \le \frac{1}{2^l}+max\left\{\left|1-exp\left\{2\pi i\frac{1}{b_n}\left[\frac{\epsilon b_n}{20}\right]\right\}\right|,l\le n\le k\right\}\hfill \\ & \le \frac{\epsilon }{3}+2\pi ·max\left\{\frac{1}{b_n}\left[\frac{\epsilon b_n}{20}\right],l\le n\le k\right\}<\frac{\epsilon }{3}+\frac{2\pi \epsilon }{20}<\epsilon .\hfill \end{array}$$

Thus, ${\omega }_k\in F_m$ for every natural number $k>l$. Evidently,

$${\omega }_k\to \tilde{\omega }={\left({\tilde{a}}_n\right)}_{n\in \omega }\mathrm{ in }X,\mathrm{ where }{\tilde{a}}_n=\left\{\begin{array}{cc}\hfill 0,& \mathrm{ if }0\le n<l,\hfill \\ \hfill \left[\frac{\epsilon b_n}{20}\right],& \mathrm{ if }l\le n.\hfill \end{array}\right.$$

Since $F_m$ is a compact subset of X, we have $\tilde{\omega }\in F_m$. Hence, $\tilde{\omega }\in H$. On the other hand, since $b_n\to \infty$, we have:

$$\underset{n\to \infty }{lim}∥\frac{{\tilde{a}}_n}{b_n}∥=\underset{n\to \infty }{lim}\frac{1}{b_n}\left[\frac{\epsilon b_n}{20}\right]=\frac{\epsilon }{20}\ne 0.$$

Thus, $\tilde{\omega }\notin H$ by Equation (20). This contradiction shows that H is not an $F_{\sigma }$-subset of X.  ☐

Now, we are in a position to prove Theorems 16 and 18.

Proof of Theorem 16. Let X be a compact Abelian group of infinite exponent. Then, $G:=\widehat{X}$ also has infinite exponent. It is well-known that G contains a countably-infinite subgroup S of one of the following form:

(a)

$S\cong \mathbb{Z}$;

(b)

$S\cong \mathbb{Z}\left(p^{\infty }\right)$;

(c)

$S\cong {⨁}_{n\in \omega }\mathbb{Z}\left(b_n\right)$, where $1<b_0<b_1<\cdots$.

Fix such a subgroup S. Set $K=S^{\perp }$ and $Y=X/K\cong S_d^{\wedge }$, where $S_d$ denotes the group S endowed with the discrete topology. Since S is countable, Y is metrizable. Hence, $\left\{0\right\}$ is a $G_{\delta }$-subgroup of Y. Thus, K is a $G_{\delta }$-subgroup of X. Let $q:X\to Y$ be the quotient map. By Lemmas 30–32, the compact group Y has a dense T-characterized subgroup $\tilde{H}$, which is not an $F_{\sigma }$-subset of Y. Lemma 24 implies that $H:=q^{-1}\left(\tilde{H}\right)$ is a dense T-characterized subgroup of X. Since the continuous image of an $F_{\sigma }$-subset of a compact group is an $F_{\sigma }$-subset, as well, we obtain that H is not an $F_{\sigma }$-subset of X. Thus, the subgroup H of X is T-characterized, but it is not an $F_{\sigma }$-subset of X. The theorem is proven.  ☐

Proof of Theorem 18. (1) Follows from Fact 5.

(2) By Lemma 3.6 in [13], every infinite compact Abelian group X contains a dense characterized subgroup H. By Fact 1, H is T-characterized. Since every $G_{\delta }$-subgroup of X is closed in X by Proposition 2.4 of [13], H is not a $G_{\delta }$-subgroup of X.

(3) Follows from Theorem 14 and the aforementioned Proposition 2.4 of [13].

(4) Follows from Fact 5.

(5) Follows from Corollary 17.  ☐

It is trivial that ${\mathrm{Char}}_T\left(X\right)\subseteq \mathrm{Char}\left(X\right)$ for every compact Abelian group X. For the circle group $\mathbb{T}$, we have:

Proposition 33. ${\mathrm{Char}}_T\left(\mathbb{T}\right)=\mathrm{Char}\left(\mathbb{T}\right)$.

Proof. We have to show only that $\mathrm{Char}\left(\mathbb{T}\right)\subseteq {\mathrm{Char}}_T\left(\mathbb{T}\right)$. Let $H=s_{\mathbf{u}}\left(\mathbb{T}\right)\in \mathrm{Char}\left(\mathbb{T}\right)$ for some sequence $\mathbf{u}$ in $\mathbb{Z}$.

If H is infinite, then H is dense in $\mathbb{T}$. Therefore, $\mathbf{u}$ is a T-sequence in $\mathbb{Z}$ by Fact 1. Thus, $H\in {\mathrm{Char}}_T\left(\mathbb{T}\right)$.

If H is finite, then H is closed in $\mathbb{T}$. Clearly, $H^{\perp }$ has infinite exponent. Thus, $H\in {\mathrm{Char}}_T\left(\mathbb{T}\right)$ by Theorem 10.  ☐

Note that, if a compact Abelian group X satisfies the equality ${\mathrm{Char}}_T\left(X\right)=\mathrm{Char}\left(X\right)$, then X is connected by Fact 3 and Theorem 14. This fact and Proposition 33 justify the next problem:

Problem 34. Does there exists a connected compact Abelian group X, such that ${\mathrm{Char}}_T\left(X\right)\ne \mathrm{Char}\left(X\right)$? Is it true that ${\mathrm{Char}}_T\left(X\right)=\mathrm{Char}\left(X\right)$ if and only if X is connected?

For a compact Abelian group X, the set of all subgroups of X that are both $F_{\sigma \delta }$- and $G_{\delta \sigma }$-subsets of X we denote by $\mathrm{S}{\mathsf{\Delta }}_3^0\left(X\right)$. To complete the study of the Borel hierarchy of (T-)characterized subgroups of X, we have to answer the next question.

Problem 35. Describe compact Abelian groups X of infinite exponent for which $\mathrm{Char}\left(X\right)\subseteq \mathrm{S}{\mathsf{\Delta }}_3^0\left(X\right)$. For which compact Abelian groups X of infinite exponent there exists a T-characterized subgroup H that does not belong to $\mathrm{S}{\mathsf{\Delta }}_3^0\left(X\right)$?

4. ${\mathfrak{g}}_T$-Closed and ${\mathfrak{g}}_T$-Dense Subgroups of Compact Abelian Groups

The following closure operator $\mathfrak{g}$ of the category of Abelian topological groups is defined in [11]. Let X be an Abelian topological group and H its arbitrary subgroup. The closure operator $\mathfrak{g}={\mathfrak{g}}_X$ is defined as follows:

$${\mathfrak{g}}_X\left(H\right):=\bigcap _{\mathbf{u}\in {\widehat{X}}^{\mathbb{N}}}\left\{s_{\mathbf{u}}\left(X\right):H\le s_{\mathbf{u}}\left(X\right)\right\},$$

and we say that H is $\mathfrak{g}$-closed if $H=\mathfrak{g}\left(H\right)$, and H is $\mathfrak{g}$-dense if $\mathfrak{g}\left(H\right)=X$.

The set of all T-sequences in the dual group $\widehat{X}$ of a compact Abelian group X we denote by ${\mathcal{T}}_s\left(\widehat{X}\right)$. Clearly, ${\mathcal{T}}_s\left(\widehat{X}\right)⫋{\widehat{X}}^{\mathbb{N}}$. Let H be a subgroup of X. In analogy to the closure operator $\mathfrak{g}$, $\mathfrak{g}$-closure and $\mathfrak{g}$-density, the operator ${\mathfrak{g}}_T$ is defined as follows:

$${\mathfrak{g}}_T\left(H\right):=\bigcap _{\mathbf{u}\in {\mathcal{T}}_s\left(\widehat{X}\right)}\left\{s_{\mathbf{u}}\left(X\right):H\le s_{\mathbf{u}}\left(X\right)\right\},$$

and we say that H is ${\mathfrak{g}}_T$-closed if $H={\mathfrak{g}}_T\left(H\right)$, and H is ${\mathfrak{g}}_T$-dense if ${\mathfrak{g}}_T\left(H\right)=X$.

In this section, we study some properties of ${\mathfrak{g}}_T$-closed and ${\mathfrak{g}}_T$-dense subgroups of a compact Abelian group X. Note that every $\mathfrak{g}$-dense subgroup of X is dense by Lemma 2.12 of [11], but for ${\mathfrak{g}}_T$-dense subgroups, the situation changes:

Proposition 36. Let X be a compact Abelian group.

(1)

If H is a ${\mathfrak{g}}_T$-dense subgroup of X, then the closure $\overline{H}$ of H is an open subgroup of X.

(2)

Every open subgroup of a compact Abelian group X is ${\mathfrak{g}}_T$-dense.

Proof. (1) Suppose for a contradiction that $\overline{H}$ is not open in X. Then, $X/\overline{H}$ is an infinite compact group. By Lemma 3.6 of [13], $X/\overline{H}$ has a proper dense characterized subgroup S. Fact 1 implies that S is a T-characterized subgroup of $X/\overline{H}$. Let $q:X\to X/\overline{H}$ be the quotient map. Then, Lemma 24 yields that $q^{-1}\left(S\right)$ is a T-characterized dense subgroup of X containing H. Since $q^{-1}\left(S\right)\ne X$, we obtain that H is not ${\mathfrak{g}}_T$-dense in X, a contradiction.

(2) Let H be an open subgroup of X. If $H=X$, the assertion is trivial. Assume that H is a proper subgroup (so X is disconnected). Let $\mathbf{u}$ be an arbitrary T-sequence, such that $H\subseteq s_{\mathbf{u}}\left(X\right)$. Since H is open, $s_{\mathbf{u}}\left(X\right)$ is open, as well. Now, Corollary 13 implies that $s_{\mathbf{u}}\left(X\right)=X$. Thus, H is ${\mathfrak{g}}_T$-dense in X.  ☐

Proposition 36(2) shows that ${\mathfrak{g}}_T$-density may essentially differ from the usual $\mathfrak{g}$-density. In the next theorem, we characterize all compact Abelian groups for which all ${\mathfrak{g}}_T$-dense subgroups are also dense.

Theorem 37. All ${\mathfrak{g}}_T$-dense subgroups of a compact Abelian group X are dense if and only if X is connected.

Proof. Assume that all ${\mathfrak{g}}_T$-dense subgroup of X are dense. Proposition 36(2) implies that X has no open proper subgroups. Thus, X is connected by [24] (7.9).

Conversely, let X be connected and H be a ${\mathfrak{g}}_T$-dense subgroup of X. Proposition 36(1) implies that the closure $\overline{H}$ of H is an open subgroup of X. Since X is connected, we obtain that $\overline{H}=X$. Thus, H is dense in X.  ☐

For ${\mathfrak{g}}_T$-closed subgroups, we have:

Proposition 38. Let X be a compact Abelian group.

(1)

Every proper open subgroup H of X is a $\mathfrak{g}$-closed non-${\mathfrak{g}}_T$-closed subgroup.

(2)

If every $\mathfrak{g}$-closed subgroup of X is ${\mathfrak{g}}_T$-closed, then X is connected.

Proof. (1) The subgroup H is ${\mathfrak{g}}_T$-dense in X by Proposition 36. Therefore, H is not ${\mathfrak{g}}_T$-closed. On the other hand, H is $\mathfrak{g}$-closed in X by Theorem A of [13].

(2) Item (1) implies that X has no open subgroups. Thus, X is connected by [24] (7.9).  ☐

We do not know whether the converse in Proposition 38(2) holds true:

Problem 39. Let a compact Abelian group X be connected. Is it true that every $\mathfrak{g}$-closed subgroup of X is also ${\mathfrak{g}}_T$-closed?