Sequential coarse structures of topological groups

TL;DR

This paper introduces a new coarse structure on topological groups based on converging sequences and explores its properties, showing that in abelian groups with non-trivial converging sequences, the asymptotic dimension is infinite.

Contribution

It defines a novel coarse structure on topological groups using converging sequences and analyzes its asymptotic dimension in abelian groups.

Findings

If G is discrete, the coarse structure is finitary.

In abelian groups with non-trivial converging sequences, the asymptotic dimension is infinite.

Abstract

We endow a topological group $(G, \tau)$ with a coarse structure defined by the smallest group ideal $S_{\tau} $ on $G$ containing all converging sequences with their limits and denote the obtained coarse group by $(G, S_{\tau})$. If $G$ is discrete then $(G, S_{\tau})$ is a finitary coarse group studding in Geometric Group Theory. The main result: if a topological abelian group $(G, \tau)$ contains a non-trivial converging sequence then $asdim \ (G, S_{\tau})= \infty $.

Equations

$$FS(b_n{)}_{n\in \omega }=\left\{\sum _{i\in F}{b}_i:F\in [\omega {]}^{<\omega }\right\}$$

$$\sum _{i\in H}{b}_i=\sum _{i\in F}{b}_i⟹F=H.$$

$$\sum _{s=0}^k{t}_{i_s}{b}_{i_s}+t_{n+1}{b}_{n+1}\notin A_{k+1}\text{mbox}for{i}_0<\dots <i_k\le n\text{mbox}and$$

$$y=b_1+\dots +b_k\text{mbox}andz=c_1+\dots +c_k,$$

$$a_{m_1}+y\notin A_{m+1}\text{mbox}and{a}_{m_1}+z\notin A_{m+1},$$

Full text

Sequential coarse structures of topological groups

Igor Protasov

Abstract. We endow a topological group $(G,\tau)$ with a coarse structure defined by the smallest group ideal $S_{\tau}$ on $G$ containing all converging sequences with their limits and denote the obtained coarse group by $(G,S_{\tau})$. If $G$ is discrete then $(G,S_{\tau})$ is a finitary coarse group studding in Geometric Group Theory. The main result: if a topological abelian group $(G,\tau)$ contains a non-trivial converging sequence then asdim $(G,S_{\tau})=\infty$. We study metrizability, normality and functional boundedness of sequential coarse groups and put some open questions.

**MSC: ** 22A15, 54E35.

Keywords: Coarse structure, group ideal, asymptotic dimension, Hamming space.

1. Introduction

Let $X$ be a set. A family $\mathcal{E}$ of subsets of $X\times X$ is called a *coarse structure * if

• •

each $E\in\mathcal{E}$ contains the diagonal $\bigtriangleup_{X}$, $\bigtriangleup_{X}=\{(x,x):x\in X\}$;

• •

if $E$, $E^{\prime}\in\mathcal{E}$ then $E\circ E^{\prime}\in\mathcal{E}$ and $E^{-1}\in\mathcal{E}$, where $E\circ E^{\prime}=\{(x,y):\exists z((x,z)\in E,\ \ (z,y)\in E^{\prime})\}$, $E^{-1}=\{(y,x):(x,y)\in E\}$;

• •

if $E\in\mathcal{E}$ and $\bigtriangleup_{X}\subseteq E^{\prime}\subseteq E$ then $E^{\prime}\in\mathcal{E}$;

• •

for any $x,y\in X$, there exists $E\in\mathcal{E}$ such that $(x,y)\in E$.

A subset $\mathcal{E}^{\prime}\subseteq\mathcal{E}$ is called a base for $\mathcal{E}$ if, for every $E\in\mathcal{E}$, there exists $E^{\prime}\in\mathcal{E}^{\prime}$ such that $E\subseteq E^{\prime}$. For $x\in X$, $A\subseteq X$ and $E\in\mathcal{E}$, we denote $E[x]=\{y\in X:(x,y)\in E\}$, $E[A]=\cup_{a\in A}\ \ E[a]$ and say that $E[x]$ and $E[A]$ are balls of radius $E$ around $x$ and $A$.

The pair $(X,\mathcal{E})$ is called a coarse space [14] or a ballean [9], [12].

Each subset $Y\subseteq X$ defines the subballean $(Y,\mathcal{E}_{Y})$, where $\mathcal{E}_{Y}$ is the restriction of $\mathcal{E}$ to $Y\times Y$. A subset $Y$ is called bounded if $Y\subseteq E[x]$ for some $x\in X$ and $E\in\mathcal{E}$.

A family $\mathcal{F}$ of subsets of $X$ is called $E$-bounded ($E$-disjoint) if, for each $A\in\mathcal{F}$, there exists $x\in X$ such that $A\subseteq E[x]$ $(E[A]\cap E[B]=\emptyset$ for all distinct $A,B\in\mathcal{F})$.

By the definition [14, Chapter 9], asdim $(X,\mathcal{E})\geq n$ if, for each $E\in\mathcal{E}$, there exist $F\in\mathcal{E}$ and $F$-bounded covering $\mathcal{M}$ of $X$ which can be partitioned $\mathcal{M}=\mathcal{M}_{0}\cup\ldots\cup\mathcal{M}_{n}$ so that each family $M_{i}$ is $E$-disjoint. If there exists the minimal $n$ with this property then asdim $(X,\mathcal{E})=n$, otherwise *asdim * $(X,\mathcal{E})=\infty$.

Given two coarse spaces $(X,\mathcal{E})$, $(X^{\prime},\mathcal{E}^{\prime})$, a mapping $f:X\longrightarrow X^{\prime}$ is called macro-uniform if, for each $E\in\mathcal{E}$, there exists $E^{\prime}\in\mathcal{E}^{\prime}$ such that $f(E[x])\subseteq E^{\prime}[f(x)]$ for each $x\in X$. If $f$ is a bijection such that $f$ and $f^{-1}$ are macro-uniform then $f$ is called an asymorphism.

Now let $G$ be a group. A family $\mathcal{I}$ of subsets of $G$ is called a group ideal [10], [12] if $G$ contains the family $[G]^{<\omega}$ of all finite subsets of $G$ and $A,B\in\mathcal{I}$, $\ \ C\subseteq A$ imply $AB^{-1}\in\mathcal{I}$, $C\in\mathcal{I}$. Every group ideal $\mathcal{I}$ defines a coarse structure on $G$ with the base $\{\{(x,y):x\in Ay\}:A\in\mathcal{I}\}$. We denote $G$ endowed with this coarse structure by $(G,\mathcal{I})$.

If $G$ is discrete then the coarse space $(G,[G]^{<\omega})$ is the main subject of Geometric Group Theory, see [5]. For coarse structures on $G$ defined by the ideal $[G]^{<\kappa}$, where $\kappa$ is a cardinal, see [11].

Every topological group $G$ can be endowed with a coarse structure defined by the ideal of all totally bounded subsets of $G$. These coarse structures were introduced and studied in [6]. For asymptotic dimensions of locally compact abelian groups endowed with coarse structures defined by ideals of precompact subsets see [7]. We recall that a subset $Y$ of a topological space is precompact if the closure of $Y$ is compact.

For a topological group $(G,\tau)$, we denote by $\mathcal{C}_{\tau}$ the group ideal of precompact subsets of $G$, and by $\mathcal{S}_{\tau}$ the minimal group ideal containing all converging sequences with their limits. Clearly, $\mathcal{S}_{\tau}\subseteq\mathcal{C}_{\tau}$.

2. Asympotic dimension

We recall [13] that a sequence $(a_{n})_{n\in\omega}$ in an abelian group $G$ is a $T$-sequence if there exists a Hausdorff group topology on $G$ in which $(a_{n})_{n\in\omega}$ converges to [math]. For a $T$-sequences $(a_{n})_{n\in\omega}$ on $G$, we denote by $\tau_{(a_{n})}$ the strongest group topology on $G$ in which $(a_{n})_{n\in\omega}$ converges to [math]. We put $A=\{0,a_{n},-a_{n}:n\in\omega\}$ and denote by $A_{n}$ the sum on $n$ copies of $A$.

By [13, Theorem 2.3.11], $(G,\tau_{(a_{n})})$ is complete. Hence, a subset $S$ of $G$ is totally bounded in $(G,\tau_{(a_{n})})$ if and only if $S$ is precompact.

We use the following three theorems proved by the author in [3].

Theorem 1. For any $T$-sequences $(a_{n})_{n\in\omega}$ on $G$, the family $\{F+A_{n}:F\in[G]^{<\omega}\}$, $n\in\omega\}$ is a base for the ideal $\mathcal{C}_{\tau(a_{n})}$ and $\mathcal{C}_{\tau(a_{n})}=\mathcal{S}_{\tau(a_{n})}$. If $G$ is generated by the set $\{a_{n}:n\in\omega\}$ then $\{A_{n}:n\in\omega\}$ is a base for $\mathcal{C}_{\tau(a_{n})}$.

Proof. Apply Lemma 2.3.2 from [13]. $\Box$

Given an arbitrary subset $S$ of $G$, the Cayley graph Cay $(G,S)$ is a graph with the set of vertices $G$ and the set of edges $\{(x,y):x-y\in S\cup(-S)\}$.

Theorem 2. If a $T$-sequences $(a_{n})_{n\in\omega}$ generates $G$ then the coarse group $(G,\mathcal{C}_{\tau(a_{n})})$ is asymorphic to Cay $(G,\{a_{n}:n\in\omega\})$.

Proof. Apply and Theorem 1 and Theorem 5.1.1 from [12]. $\Box$

We recall that the Hamming space $\mathbb{H}$ is the set $[\omega]^{<\omega}$ endowed with the metric $h(F,H)=|F\bigtriangleup H|$. To see that asdim $\mathbb{H}=\infty$, it suffices to find an asymorphic copy of $\mathbb{N}$ in $\mathbb{H}$ and observe that $\mathbb{H}$ is asymorphic to $\mathbb{H}\times\mathbb{H}$.

Example 1. Let $G$ be the direct sum of groups $\{\langle a_{n}\rangle:n\in\omega\}$ of order 2. Clearly, $(a_{n})_{n\in\omega}$ is a $T$-sequence on $G$. By Theorem 1, the canonical bijection between $(G,\mathcal{C}_{\tau(a_{n})})$ and the Hamming space $\mathbb{H}$ of all finite subsets of $\omega$ is an asymorphism.

A $T$-sequences $(a_{n})_{n\in\omega}$ is called trivial if $a_{n}=0$ for all but finitely many $n\in\omega$.

Theorem 3. For any non-trivial $T$-sequences $(a_{n})_{n\in\omega}$ on $G$, the coarse group $(G,\mathcal{C}_{\tau(a_{n})})$ contains a subspace asymorphic to the Hamming space $\mathbb{H}$ so asdim $(G,\mathcal{C}_{\tau(a_{n})})=\infty$.

Proof. Without loss of generality, we suppose that $\{a_{n}:n\in\omega\}$ generates $G$ and $a_{n}\neq 0$ for each $n\in\omega$.

Given an arbitrary $T$-sequence $(b_{n})_{n\in\omega}$ in $G$, we denote

[TABLE]

and say that $(b_{n})_{n\in\omega}$ is FS-strict if, for any $H,F\in[\omega]^{<\omega}$,

[TABLE]

We note that $(b_{n})_{n\in\omega}$ is FS-strict if, for each $n\in\omega$,

[TABLE]

We assume that $(b_{n})_{n\in\omega}$ is FS-strict and

[TABLE]

Then the canonical bijection $f:\mathbb{H}\longrightarrow FS(b_{n}),\ \ f(H)=\sum_{i\in H}b_{i}$ is an asymorphism.

To construct the desired sequence $(b_{n})_{n\in\omega}$ we rewrite $(2)$ in the following equivalent form

[TABLE]

We put $b_{0}=a_{0}$ and assume that $b_{0},\ldots,b_{n}$ have been chosen. We show how to choose $b_{n+1}$ to satisfy $(1)$ and

[TABLE]

[TABLE]

We assume that there exists a subsequence $(c_{m})_{m\in\omega}$ of $(a_{n})_{n\in\omega}$ such that $t_{i_{0}}b_{i_{0}}+\ldots+t_{i_{k}}b_{i_{k}}+tc_{m}\in A_{k+1}$ for $t\in\{1,-1\}$ and for each $m\in\omega$. Every infinite subset of $A_{k+1}$ has a limit point in $A_{k}$. Hence, $t_{i_{0}}b_{i_{0}}+\ldots+t_{i_{k}}b_{i_{k}}\in A_{k+1}$ contradicting the choice of $b_{0},\ldots,b_{n}$. Thus, $b_{n+1}$ can be taken from $\{a_{m_{0}},a_{m_{0}+1},\ldots\}$ for some $m_{0}\in\omega$. $\Box$

Theorem 4. Let $G$ be a group and let $\mathcal{I}$ be a group ideal on $G$. Assume that there exists a family $\mathcal{F}$ of group ideals on $G$ such that $\mathcal{J}\subseteq\mathcal{I}$ for each $\mathcal{J}\in\mathcal{F}$ and, for each $A\in\mathcal{I}$, there exists $\mathcal{J}\in\mathcal{F}$ such that $A\in\mathcal{J}$. If asdim $(G,\mathcal{J})\geq n$ for each $\mathcal{J}\in\mathcal{F}$ then asdim $(G,\mathcal{I})\geq n$.

Proof. Given any $A\in\mathcal{I}$ and $n\in\omega$, we choose $\mathcal{J}\in\mathcal{F}$ such that $A\in\mathcal{J}$ and a uniformly bounded covering $\mathcal{M}$ of $(G,\mathcal{J})$ and a partition $\mathcal{M}_{0},\ldots,\mathcal{M}_{n}$ of $\mathcal{M}$ witnessing asdim $(G,\mathcal{J})\geq n$. Then these $\mathcal{M},\mathcal{M}_{0},\ldots,\mathcal{M}_{n}$ say that asdim $(G,\mathcal{I})\geq n$. $\Box$

Theorem 5. If a topological abelian group $(G,\tau)$ contains a non-trivial converging sequence then asdim $(G,\mathcal{S}_{\tau})=\infty$.

Proof. We denote $\mathcal{F}=\{\mathcal{S}_{\tau(a_{n})}:(a_{n})$ is a non-trivial sequence in $(G,\tau)$ converging to [math] in $(G,\tau)\}$. Let $A\in\mathcal{S}_{\tau}$. By the definition of $\mathcal{S}_{\tau}$, there exists a subset $B\in\mathcal{S}_{\tau}$, $A\subseteq B$ and a finite number $(a_{1n})_{n\in\omega},\ldots,(a_{mn})_{n\in\omega}$ of sequences converging to [math] in $(G,\tau)$ such that $B$ can be obtained from the sets $\{a_{1n},-a_{1n}:n\in\omega\},\ldots,\{a_{mn},-a_{mn}:n\in\omega\}$ by the finite number of additions of these sets and join of finite subsets of $G$. We choose a sequence $(b_{n})_{n\in\omega}$ converging to [math] in $(G,\tau)$ and containing each $(a_{1n})_{n\in\omega},\ldots,(a_{mn})_{n\in\omega}$ as a subsequence. Then $A\in\mathcal{S}_{\tau(b_{n})}$, $\mathcal{S}_{\tau(b_{n})}\subseteq\mathcal{S}_{\tau}$. Apply Theorems 3 and 4. $\Box$

We note that Theorem 5 answers Question 2 from [3].

**Question 1. ** Let $(G,\tau)$ be a countable non-discrete metrizable abelian group. Does $(G,\mathcal{S}_{\tau})$ contain an asymorphic copy of $\mathbb{H}$?

In [4], the authors ask about asymptotic dimension of $(G,\mathcal{C}_{\tau})$, where $G$ is an infinite cyclic subgroup of the circle. We put this question in more genera form.

**Question 2. ** *Let $(G,\tau)$ be a countable non-discrete metrizable group. Is asdim $(G,\mathcal{C}_{\tau})=\infty$? *

**Example 2. ** Let $G$ be the direct sum of $\omega$ copies of $\mathbb{Z}_{2}$ endowed with the topology induced by the Tikhonov topology of $\mathbb{Z}_{2}^{\omega}$. Since asdim $(G,\mathcal{C}_{\tau})>0$ and $(G,\mathcal{C}_{\tau})$ is asymorphic to $(G,\mathcal{C}_{\tau})\times(G,\mathcal{C}_{\tau})$, we see that asdim $(G,\mathcal{C}_{\tau})=\infty$.

Let $(X,\mathcal{E})$ be a coarse space. A function $f:X\longrightarrow\{0,1\}$ is called slowly oscillating if, for every $E\in\mathcal{E}$ , there exists a bounded subset $B$ of $X$ such that $f|_{E[x]}=const$ for each $x\in X\setminus B$. We endow $X$ with the discrete topology, identify the Stone-$\check{C}$ech compactification $\beta X$ of $X$ with the set of ultrafilters on $X$ and denote $X^{\sharp}=\{p\in\beta X:$ each $P\in p$ is unbounded $\}$. We define an equivalence $\sim$ on $X^{\sharp}$ by the rule: $p\sim q$ if and only if $f^{\beta}(p)=f^{\beta}(q)$ for every slowly oscillating function $f:X\longrightarrow\{0,1\}$. The quotient $X^{\sharp}/\sim$ is called a space of *ends * or binary corona of $(X,\mathcal{E})$, see [12, Chapter 8].

Theorem 6. If a non-trivial $T$-sequences $(a_{n})_{n\in\omega}$ generates $G$ then the space of ends of $(G,\mathcal{C}_{\tau(a_{n})})$ is a singleton.

Proof. First we show that for every slowly oscillating function $f:G\longrightarrow\{0,1\}$ there exists an $m$ such that

[TABLE]

Indeed, by the definition of slow oscillation and Theorem 1, there exists $m\in\omega$ such that $f|_{x+A}=const$ for each $x\in G\setminus A_{m}$. We show nw that (5) holds true for this $m$.

We take arbitrary $y,z\in G\setminus A_{m}$. Since $A$ generates $G$ and contains 0, there exists an index $k$ such that $y,z\in A_{k}$, i.e.,

[TABLE]

for appropriate $\ \ b_{1},\ldots,b_{k}$, $\ \ c_{1},\ldots,c_{k}\in A$. By a property of $T$-sequences established at the end of the proof of Theorem 3, there exists a member $a_{m_{1}}$ of $(a_{n})$ such that

[TABLE]

since $(a_{n})$ is a $T$-sequence. Then $a_{m_{1}}+b_{2}+\ldots+b_{k}\notin A_{m}$, $\ \ a_{m_{1}}+c_{2}+\ldots+c_{k}\notin A_{m}$. Therefore,

[TABLE]

Repeating this trick $k$ times, we can replace $b_{2},\ldots,b_{k}$ and $c_{2},\ldots,c_{k}$, by appropriate members $a_{m_{2}},\ldots,a_{m_{k}}$ of $(a_{n})$, as before. Hence, we can replace $b_{2}+\ldots+b_{k}$ and $c_{2}+\ldots+c_{k}$, by $a_{m_{2}}+\ldots+a_{m_{k}}$ in (6). This obviously gives $f(y)=f(z)$ and proves (5).

Finally, to prove the assertion of the theorem, pick $p,q\in X^{\sharp}$. In order to check that $p\sim q$ fix an arbitrary slowly oscillating function $f:X\longrightarrow\{0,1\}$. We have prove that $f^{\beta}(p)=f^{\beta}(q)$. Pick an $m$ with (5). Since $G\setminus A_{m}\in p\cap q$, for every $P\in p$ and for every $Q\in q$ we have $P_{1}:=P\setminus A_{m}\in p$ and $Q_{1}:=Q\setminus A_{m}\in q$ and $f|_{P_{1}}=f|_{Q_{1}}$ is constant in view of (5). This proves that $f^{\beta}(p)=f^{\beta}(q)$. $\Box$

**Question 3. ** Let $(G,\tau)$ be a countable non-discrete metrizable abelian group. Is the space of ends of $(G,\mathcal{S}_{\tau})$ a singleton? The same question for $(G,\mathcal{C}_{\tau})$.

3. Metrizability and normality

By [12, Theorem 2.1.1], a coarse space $(X,\mathcal{E})$ is metrizable if and only if $\mathcal{E}$ has a countable base.

Theorem 7. For an infinite abelian group $G$ and a $T$-sequence $(a_{n})_{n\in\omega}$ in $G$, $(G,\mathcal{S}_{\tau(a_{n})})$ is metrizable if and only if $G$ is countable.

Proof. Apply Theorem 1. $\Box$

Theorem 8. Let $G,$ be a non-discrete metrizable group. Then the coarse structure $\mathcal{E}_{\tau}$ of $(G,S)$ does not have a linearly ordered base. In particular, is not metrizable.

Proof. We assume that $\mathcal{E}_{\tau}$ has a linear base $\mathcal{E}$ and choose a sequence $(A_{n})_{n\in\omega}$ in $\mathcal{E}$ such that $A_{n}\subset A_{n+1}$ and the closure of $\bigcup_{n\in\omega}A_{n}$ is not compact. If $B\in\mathcal{E}$ then $B\subseteq A_{m}$ for some $m$ so $\{A_{n}:n\in\omega\}$ is a base for $\mathcal{E}_{\tau}$.

Now let $\{U_{n}:n\in\omega\}$ is a base of neighbourhoods of the identity of $(G,\tau)$. We choose an injective sequence $(b_{n})_{n\in\omega}$ in $G$ such that $b_{n}\in U_{n}$, $b_{n}\notin A_{n}$. Then $\{b_{n}:n\in\omega\}\in\mathcal{S}_{\tau}$ but $\{b_{n}:n\in\omega\}\setminus A_{n}\neq\emptyset$. Hence, $\{A_{n}:n\in\omega\}$ is not a base for $\mathcal{E}_{\tau}$ and we get a contradiction. $\Box$

Let $(X,\mathcal{E})$ be a ballean. A subset $U$ of $X$ is called an asymptotic neighbourhood of a subset $Y\subseteq X$ if, for every $E\in\mathcal{E}$, $E[Y]\setminus U$ is bounded.

Two subsets $Y,Z$ of $X$ are called

• •

asymptotically disjoint if, for every $E\in\mathcal{E}$, $E[Y]\cap E[Z]$ is bounded;

• •

asymptotically separated if $Y,Z$ have disjoint asymptotic neighbourhoods.

A ballean $(X,\mathcal{E})$ is called normal [8] if any two asymptotically disjoint subsets of $X$ are asymptotically separated. Every ballean with linearly ordered base is normal [8, Proposition 1.1].

We suppose that a non-discrete metrizable group $(G,\tau)$ is topologically isomorphic to the product $G_{1}\times G_{2}$ of infinite groups. Applying Theorem 8 and Theorem 1.4 from [1], we conclude that $(G,\mathcal{S}_{\tau})$ is not normal.

**Question 4. ** Let $(G,\tau)$ be a non-discrete metrizable group. Is the ballean $(G,\mathcal{S}_{\tau})$ non-normal?

4. Functional boundedness

Let $(X,\mathcal{E})$ be a coarse space. Following [2], we say that a function $f:(X,\mathcal{E})\longrightarrow\mathbb{R}$ is

• •

bornologous if $f(B)$ is bounded in $\mathbb{R}$ for each bounded subset $B$ of $X$;

• •

macro-uniform if, for every $E\in\mathcal{E}$, the supremum $\sup_{x\in X}\ diam\ f(E[x])$ is finite;

• •

eventually macro-uniform if, for every $E\in\mathcal{E}$, there exists a bounded subset $B$ of $X$ such that $\sup_{x\in X\setminus B}\ diamf(E[x])$ is finite;

• •

slowly oscillating if, for every $E\in\mathcal{E}$ and $\varepsilon>0$, there exists a bounded subset $B$ of $X$ such that

$diamf(E[x])<\varepsilon$ for each $x\in X\setminus B$.

We say that a coarse space $(X,\mathcal{E})$ is

• •

$b$*-bounded * if each bornologous function on $X$ is bounded;

• •

mu-bounded if each macro-uniform function on $X$ is bounded;

• •

emu- bounded if, for every macro-uniform function $f$ on $X$, there exists a bounded subset $B$ of $X$ such that $f$ is bounded on $X\setminus B$;

• •

so-bounded if, for every slowly oscillating function $f$ on $X$, there exists a bounded subset $B$ of $X$ such that $f$ is bounded on $X\setminus B$.

If $(G,\tau)$ is discrete then every function $f:(G,\mathcal{S}_{\tau})\longrightarrow\mathbb{R}$ is bornologous. If $(G,\tau)$ has a non-trivial converging sequence then there is a non-bornologous function $f:(G,\mathcal{S}_{\tau})\longrightarrow\mathbb{R}$.

Theorem 9. If $(G,\tau)$ is metrizable and compact then $(G,\mathcal{S}_{\tau})$ is $b$-bounded. If $(G,\tau)$ is countable and metrizable then $(G,\mathcal{S}_{\tau})$ is not $b$-bounded.

Proof. In the first case, we assume that there exists an unbounded bornologous function $f:(G,\mathcal{S}_{\tau})\longrightarrow\mathbb{R}$.

We choose a sequence $(a_{n})_{n\in\omega}$ in $G$ such that $|f(a_{n})|>n$. Passing to a subsequence, we may suppose that $(a_{n})_{n\in\omega}$ converges to some point $a$. Then $f$ is not bounded on the set $\{a,a_{n}:n\in\omega\}\in\mathcal{S}_{\tau}$ so $f$ is not bornologous.

In the second case, we denote by $H$ the completion of $G$ and choose $an$ injective sequence $(b_{n})_{n\in\omega}$ in $G$ converging to some point $h\in H\setminus G$. Then we define a function $f:(G,\tau)\longrightarrow\mathbb{R}$ by $f(b_{n})=n$ and $f(x)=0$ for each $x\in G\setminus\{b_{n}:n\in\omega\}$. Since $K\cap\{b_{n}:n\in\omega\}$ is finite for each compact subset $K$ of $G$, we see that $f$ is bornologous. Hence, $(G,\mathcal{S}_{\tau})$ is not $b$-bounded. $\Box$

If $(G,\tau)$ is discrete and $(G,\mathcal{S}_{\tau})$ is $mu$-bounded then $G$ is a Bergman group [2].

Theorem 10. If $(G,\tau)$ is metrizable and totally bounded then $(G,\mathcal{S}_{\tau})$ is $mu$-bounded.

Proof. We assume the contrary. Then there exists a macro-uniform function $f:(G,\mathcal{S}_{\tau})\longrightarrow\mathbb{R}^{+}$ such that $f(a_{n+1})-f(a_{n})>n$ for some sequence $(a_{n})_{n\in\omega}$ in $(G,\tau)$. Passing to a subsequence, we may suppose that $(a_{n})_{n\in\omega}$ converges to some point $h\in H$, where $H$ is the completion of $G$. We denote $b_{n}=a_{n}a_{n+1}^{-1}$ and observe that the sequence $(b_{n})_{n\in\omega}$ converges to the identity $e$ of $G$. We put $B=\{e,b_{n}:n\in\omega\}$ and note that $a_{n}\in Ba_{n+1}$. Since $B\in\mathcal{S}_{\tau}$, we see that $f$ is not macro-uniform.

**Question 5. ** Let be a compact metrizable abelian group. Is $(G,\mathcal{S}_{\tau})$ $\ so$-bounded? $\ mu$-bounded?