Asymptotic dimension and small subsets in locally compact topological groups

Abstract

We prove that for a coarse space $X$ the ideal $\mathcal{S }(X)$ of small subsets of $X$ coincides with the ideal $\mathcal{D }_<(X)=\{A\subset X:\mathrm{asdim}(A)<\mathrm{asdim}(X)\}$ provided that $X$ is coarsely equivalent to a Euclidean space $\mathbb{R }^n$. Also we prove that for a locally compact Abelian group $X$, the equality $\mathcal{S }(X)=\mathcal{D }_<(X)$ holds if and only if the group $X$ is compactly generated.

1 Introduction

In this paper we study the interplay between the ideal $\mathcal{S }(X)$ of small subsets of a coarse space $X$ and the ideal $\mathcal{D }_{<}(X)$ of subsets of asymptotic dimension less than $\mathrm{asdim}(X)$ in $X$. We show that these two ideals coincide in spaces that are coarsely equivalent to $\mathbb{R }^n$, in particular, they coincide in each compactly generated locally compact abelian group.

Let us recall that a coarse space is a pair $(X,\mathcal{E })$ consisting of a set $X$ and a coarse structure $\mathcal{E }$ on $X$, which is a family of subsets of $X\times X$ (called entourages) satisfying the following axioms:

• (A) each $\varepsilon \in \mathcal{E }$ contains the diagonal $\Delta _X=\{(x,y)\in X^2:x=y\}$ and is symmetric in the sense that ${\varepsilon }={\varepsilon }^{-1}$ where ${\varepsilon }^{-1}=\{(y,x):(x,y)\in {\varepsilon }\}$;

• (B) for any entourages ${\varepsilon },\delta \in \mathcal{E }$ there is an entourage $\eta \in \mathcal{E }$ that contains the composition$\delta \circ {\varepsilon }=\{(x,z)\in X^2:\exists y\in X$ with $(x,y)\in {\varepsilon }$ and $(y,z)\in \delta \}$;

• (C) a subset $\delta \subset X^2$ belongs to $\mathcal{E }$ if $\Delta _X\subset \delta =\delta ^{-1}\subset {\varepsilon }$ for some ${\varepsilon }\in \mathcal{E }$.

A subfamily $\mathcal{B }\subset \mathcal{E }$ is called a base of the coarse structure $\mathcal{E }$ if

$$\begin{aligned} \mathcal{E }=\{{\varepsilon }\subset X^2:\exists \delta \in \mathcal{B } \text{ with } \Delta _X\subset {\varepsilon }={\varepsilon }^{-1}\subset \delta \}. \end{aligned}$$

A family $\mathcal{B }$ of subsets of $X^2$ is a base of a (unique) coarse structure if and only if it satisfies the axioms (A), (B).

Each subset $A$ of a coarse space $(X,\mathcal{E })$ carries the induced coarse structure $\mathcal{E }_A=\{{\varepsilon }\cap A^2:{\varepsilon }\in \mathcal{E }\}$. Endowed with this structure, the space $(A,\mathcal{E }_A)$ is called a subspace of $(X,\mathcal{E })$.

For an entourage ${\varepsilon }\subset X^2$, a point $x\in X$, and a subset $A\subset X$ let $B(x,{\varepsilon })=\{y\in X:(x,y)\in {\varepsilon }\}$ be the ${\varepsilon }$ -ball centered at $x$, $B(A,{\varepsilon })=\bigcup _{a\in A}B(a,{\varepsilon })$ be the ${\varepsilon }$-neighborhood of $A$ in $X$, and $\mathrm{diam}(A)=A\times A$ be the diameter of $A$. For a family $\mathcal{U }$ of subsets of $X$ we put $\mathrm{mesh}(\mathcal{U })=\bigcup _{U\in \mathcal{U }}\mathrm{diam}(U)$.

Now we consider two basic examples of coarse spaces. The first of them is any metric space $(X,d)$ carrying the metric coarse structure whose base consists of the entourages $\{(x,y)\in X^2:d(x,y)<{\varepsilon }\}$ where $0\le {\varepsilon }<\infty$. A coarse space is metrizable if its coarse structure is generated by some metric.

The second basic example is a topological group $G$ endowed with the left coarse structure whose base consists of the entourages $\{(x,y)\in G^2:x\in yK\}$ where $K=K^{-1}$ runs over compact symmetric subsets of $G$ that contain the identity element $1_G$ of $G$. Let us observe that the left coarse structure on $G$ coincides with the metric coarse structure generated by any left-invariant continuous metric $d$ on $G$ which is proper in the sense that each closed ball $B(e,R)=\{x\in G:d(x,e)\le R\}$ is compact. In particular, the coarse structure on $\mathbb{R }^n$, generated by the Euclidean metric coincides with the left coarse structure of the Abelian topological group $\mathbb{R }^n$.

Now we recall the definitions of large and small sets in coarse spaces. Such sets were introduced in [4] and studied in [13, §11] and [2]. A subset $A$ of a coarse space $(X,\mathcal{E })$ is called

• large if $B(A,{\varepsilon })=X$ for some ${\varepsilon }\in \mathcal{E }$;

• small if for each large set $L\subset X$ the set $L\setminus A$ remains large in $X$.

It follows that the family $\mathcal{S }(X)$ of small subsets of a coarse space $(X,\mathcal{E })$ is an ideal. A subfamily $\mathcal{I }\subset \mathcal{P }(X)$ of the power-set of a set $X$ is called an ideal if $\mathcal{I }$ is additive (in the sense that $A\cup B\in \mathcal{I }$ for all $A,B\in \mathcal{I }$) and downwards closed (which means that $A\cap B\in \mathcal{I }$ for all $A\in \mathcal{I }$ and $B\subset X$).

Small sets can be considered as coarse counterparts of nowhere dense subsets in topological spaces, see [2]. It is well-known [8, 7.4.18] that the ideal of nowhere dense subsets in a Euclidean space $\mathbb{R }^n$ coincides with the ideal generated by closed subsets of topological dimension $<n$ in $\mathbb{R }^n$. The aim of this paper is to prove a coarse counterpart of this fundamental fact.

For this we need to recall [14, 9.4] the definition of the asymptotic dimension $\mathrm{asdim}(X)$ of a coarse space $X$.

Definition 1.1

The asymptotic dimension $\mathrm{asdim}(X)$ of a coarse space $(X,\mathcal{E })$ is the smallest number $n\in {\omega }$ such that for each entourage ${\varepsilon }\in \mathcal{E }$ there is a cover $\mathcal{U }$ of $X$ such that $\mathrm{mesh}(\mathcal{U })\subset \delta$ for some $\delta \in \mathcal{E }$ and each ${\varepsilon }$-ball $B(x,{\varepsilon })$, $x\in X$, meets at most $n+1$ sets $U\in \mathcal{U }$. If such a number $n\in {\omega }$ does not exist, then we put $\mathrm{asdim}(X)=\infty$.

In Theorem 2.7 we shall prove that

$$\begin{aligned} \mathrm{asdim}(A\cup B)\le \max \{\mathrm{asdim}(A),\mathrm{asdim}(B)\} \end{aligned}$$

for any subspaces $A,B$ of a coarse space $X$. This implies that for every number $n\in {\omega }\cup \{\infty \}$ the family $\{A\subset X:\mathrm{asdim}(A)<n\}$ is an ideal in $\mathcal{P }(X)$. In particular, the family

$$\begin{aligned} \mathcal{D }_<(X)=\{A\subset X:\mathrm{asdim}(A)<\mathrm{asdim}(X)\} \end{aligned}$$

is an ideal in $\mathcal{P }(X)$. According to [5, 9.8.4], $\mathrm{asdim}(\mathbb{R }^n)=n$ for every $n\in {\omega }$.

The main result of this paper is:

Theorem 1.2

For every $n\in \mathbb{N }$ the ideal $\mathcal{S }(X)$ of small subsets in the space $X=\mathbb{R }^n$ coincides with the ideal $\mathcal{D }_{<}(X)$.

Theorem 1.2 will be proved in Sect. 5 with help of some tools of Combinatorial Topology. In light of this theorem the following problem arises naturally:

Problem 1.3

Detect coarse spaces $X$ for which $\mathcal{S }(X)=\mathcal{D }_<(X)$.

It should be mentioned that the class of coarse spaces $X$ with $\mathcal{S }(X)=\mathcal{D }_<(X)$ is closed under coarse equivalences.

A function $f:X\rightarrow Y$ between two coarse spaces $(X,\mathcal{E }_X)$ and $(Y,\mathcal{E }_Y)$ is called

• coarse if for each $\delta _X\in \mathcal{E }_X$ there is ${\varepsilon }_Y\in \mathcal{E }_Y$ such that for any pair $(x,y)\in \delta _X$ we get $(f(x),f(y))\in {\varepsilon }_Y$;

• a coarse equivalence if $f$ is coarse and there is a coarse map $g:Y\rightarrow X$ such that $\{(x,g\circ f(x)):x\in X\}\subset {\varepsilon }_X$ and $\{(y,f\circ g(y)):y\in Y\}\subset {\varepsilon }_Y$ for some entourages ${\varepsilon }_X\in \mathcal{E }_X$ and ${\varepsilon }_Y\in \mathcal{E }_Y$.

Two coarse spaces $X,Y$ are called coarsely equivalent if there is a coarse equivalence $f:X\rightarrow Y$.

Proposition 1.4

Assume that coarse spaces $X,Y$ are coarsely equivalent. Then

• (1) $\mathrm{asdim}(X)=\mathrm{asdim}(Y)$;

• (2) $\mathcal{D }_<(X)=\mathcal{S }(X)$ if and only if $\mathcal{D }_{<}(Y)=\mathcal{S }(Y)$.

This proposition will be proved in Sect. 3. Combined with Theorem 1.2 it implies:

Corollary 1.5

If a coarse space $X$ is coarsely equivalent to a Euclidean space $\mathbb{R }^n$, then $\mathcal{D }_{<}(X)=\mathcal{S }(X)$.

Problem 1.3 can be completely resolved for locally compact Abelian topological groups $G$, endowed with their left coarse structure. First we establish the following general fact, which will be proved in Sect. 4.

Theorem 1.6

For each topological group $X$ endowed with its left coarse structure we get $\mathcal{D }_<(X)\subset \mathcal{S }(X)$.

We recall that a topological group $G$ is compactly generated if $G$ is algebraically generated by some compact subset $K\subset G$.

Theorem 1.7

For an Abelian locally compact topological group $X$ the following conditions are equivalent:

• (1) $\mathcal{S }(X)=\mathcal{D }_<(X)$;

• (2) $X$ is compactly generated;

• (3) $X$ is coarsely equivalent to the Euclidean space $\mathbb{R }^n$ for some $n\in {\omega }$.

This theorem will be proved in Sect. 6.

Remark 1.8

The equivalence (1)$\Leftrightarrow$(2) in Theorem 1.7 does not hold beyond the class of Abelian groups. The simplest counterexample is the free group $F_2$ with two generators, endowed with the discrete topology. Any infinite cyclic subgroup $Z\subset F_2$ has infinite index in $F_2$ and hence is small, yet $\mathrm{asdim}(Z)=\mathrm{asdim}(F_2)=1$.

A less trivial example is the wreath product $A\wr \mathbb{Z }$ of a non-trivial finite abelian group $A$ and $\mathbb{Z }$. The group $A\wr \mathbb{Z }$ has asymptotic dimension 1 (see [9]) and the subgroup $\mathbb{Z }$ is small in $A\wr \mathbb{Z }$ and has $\mathrm{asdim}(\mathbb{Z })=1=\mathrm{asdim}(A\wr \mathbb{Z })$. Let us recall that the group $A\wr \mathbb{Z }$ consists of ordered pairs $((a_i)_{i\in \mathbb{Z }},n)\in \big (\oplus ^{\mathbb{Z }}A)\times \mathbb{Z }$ and the group operation on $A\wr \mathbb{Z }$ is defined by

$$\begin{aligned} ((a_i),n)*((b_i),m)=((a_{i+m}+b_i), n+m). \end{aligned}$$

The group $A\wr \mathbb{Z }$ is finitely-generated and meta-abelian but is not finitely presented, see [3]. Groups which are coarsely equivalent to abelian groups were studied in [1].

Problem 1.9

Is $\mathcal{S }(X)=\mathcal{D }_<(X)$ for each connected Lie group $X$? For each discrete polycyclic group $X$?

2 The asymptotic dimension of coarse spaces

In this section we present various characterizations of the asymptotic dimension of coarse spaces. First we fix some notation. Let $(X,\mathcal{E })$ be a coarse space, ${\varepsilon }\in \mathcal{E }$ and $A\subset X$. We shall say that $A$ has diameter less than ${\varepsilon }$ if $\mathrm{diam}(A)\subset {\varepsilon }$ where $\mathrm{diam}(A)=A\times A$. A sequence $x_0,\dots ,x_m\in X$ is called an ${\varepsilon }$ -chain if $(x_i,x_{i+1})\in {\varepsilon }$ for all $i<m$. In this case the finite set $C=\{x_0,\dots ,x_m\}$ also will be called an ${\varepsilon }$ -chain. A set $C\subset X$ is called ${\varepsilon }$-connected if any two points $x,y\in C$ can be linked by an ${\varepsilon }$-chain $x=x_0,\dots ,x_m=y$. The maximal ${\varepsilon }$-connected subset $C(x,{\varepsilon })\subset X$ containing a given point $x\in X$ is called the ${\varepsilon }$ -connected component of $x$. It consists of all points $y\in X$ that can be linked with $x$ by an ${\varepsilon }$-chain $x=x_0,\dots ,x_m=y$.

A family $\mathcal{U }$ of subsets of $X$ is called ${\varepsilon }$ -disjoint if $(U\times V)\cap {\varepsilon }=\emptyset$ for any distinct sets $U,V\in \mathcal{U }$. Each natural number $n$ is identified with the set $\{0,\dots ,n-1\}$.

We shall study the interplay between the asymptotic dimension introduced in Definition 1.1 and the following modification:

Definition 2.1

The colored asymptotic dimension $\mathrm{asdim}_{col}(X)$ of a coarse space $(X,\mathcal{E })$ is the smallest number $n\in {\omega }$ such that for every entourage $\varepsilon \in \mathcal{E }$ there is a cover $\mathcal{U }$ of $X$ such that $\mathrm{mesh}(\mathcal{U })\subset \delta$ for some $\delta \in \mathcal{E }$ and $\mathcal{U }$ can be written as the union $\mathcal{U }=\bigcup _{i\in n+1}\mathcal{U }_i$ of $n+1$ many ${\varepsilon }$-disjoint subfamilies $\mathcal{U }_i$. If such a number $n\in {\omega }$ does not exist, then we put $\mathrm{asdim}_{col}(X)=\infty$.

Without lost of generality we can assume that the cover $\mathcal{U }=\bigcup _{i\in n+1}\mathcal{U }_i$ in the above definition consists of pairwise disjoint sets. In this case we can consider the coloring $\chi : X\rightarrow n+1=\{0,\dots ,n\}$ such that $\chi ^{-1}(i)=\bigcup \mathcal{U }_i$ for every $i\in n+1$. For this coloring every $\chi$-monochrome ${\varepsilon }$-connected subset $C\subset X$ lies in some $U\in \mathcal{U }$ and hence has diameter $\mathrm{diam}(C)\subset \mathrm{diam}(U)\subset \mathrm{mesh}(\mathcal{U })\subset \delta$. A subset $A\subset X$ is $\chi$ -monochrome if $\chi (A)$ is a singleton. Thus we arrive to the following useful characterization of the colored asymptotic dimension.

Proposition 2.2

A coarse space $(X,\mathcal{E })$ has $\mathrm{asdim}_{col}(X)\le n$ for some number $n\in {\omega }$ if and only if for any $\varepsilon \in \mathcal{E }$ there are a coloring $\chi : X\rightarrow n+1$ and an entourage $\delta \in \mathcal{E }$ such that each $\chi$-monochrome $\varepsilon$-chain $C\subset X$ has $\mathrm{diam}(C)\subset \delta$.

Proof

The “only if” part follows from the above discussion. To prove the “if” part, for every ${\varepsilon }\in \mathcal{E }$ we need to construct a cover $\mathcal{U }=\bigcup _{i\in n+1}\mathcal{U }_i$ such that $\mathrm{mesh}(\mathcal{U })\in \mathcal{E }$ and each family $\mathcal{U }_i$ is ${\varepsilon }$-disjoint. By our assumption, there is a coloring $\chi :X\rightarrow n+1$ and an entourage $\delta \in \mathcal{E }$ such that each $\chi$-monochrome ${\varepsilon }$-chain $C\subset X$ has $\mathrm{diam}(C)\subset \delta$.

For each $x\in X$ let $C_\chi (x,{\varepsilon })$ be the set of all points $y\in X$ that can be linked with $x$ by a $\chi$-monochrome ${\varepsilon }$-chain $x=x_0,x_1,\dots ,x_m=y$. It follows that $\mathrm{diam}(C_\chi (x,{\varepsilon }))\subset \delta$. For every $i\in n+1$ consider the ${\varepsilon }$-disjoint family $\mathcal{U }_i=\{C_\chi (x,{\varepsilon }): x\in \chi ^{-1}(i)\}$. It is clear that $\mathcal{U }=\bigcup _{i\in n+1}\mathcal{U }_i$ is a cover with $\mathrm{mesh}(\mathcal{U })\subset \delta \in \mathcal{E }$, witnessing that $\mathrm{asdim}_{col}(X)\le n$. $\square$

Now we are ready to prove the equivalence of two definitions of asymptotic dimension. For metrizable coarse spaces this equivalence was proved in [5, 9.3.7].

Proposition 2.3

Each coarse space $(X,\mathcal{E })$ has $\mathrm{asdim}(X)=\mathrm{asdim}_{col}(X)$.

Proof

To prove that $\mathrm{asdim}(X)\le \mathrm{asdim}_{col}(X)$, put $n=\mathrm{asdim}_{col}(X)$ and take any entourage ${\varepsilon }\in \mathcal{E }$. By Definition 2.1, for the entourage ${\varepsilon }\circ {\varepsilon }\in \mathcal{E }$ we can find a cover $\mathcal{U }=\bigcup _{i\in n+1}\mathcal{U }_i$ with $\mathrm{mesh}(\mathcal{U })\in \mathcal{E }$ such that each family $\mathcal{U }_i$ is ${\varepsilon }\circ {\varepsilon }$-disjoint. We claim that each ${\varepsilon }$-ball $B(x,{\varepsilon })$, $x\in X$, meets at most one set of each family $\mathcal{U }_i$. Assuming that $B(x,{\varepsilon })$ meets two distinct sets $U,V\in \mathcal{U }_i$, we can find points $u\in U$ and $v\in V$ with $(x,u),(x,v)\in {\varepsilon }$ and conclude that $(u,v)\in {\varepsilon }\circ {\varepsilon }$, which is not possible as $\mathcal{U }_i$ is ${\varepsilon }\circ {\varepsilon }$-disjoint. Now we see that the ball $B(x,{\varepsilon })$ meets at most $n+1$ element of the cover $\mathcal{U }$ and hence $\mathrm{asdim}(X)\le n$.

The proof of the inequality $\mathrm{asdim}_{col}(X)\le \mathrm{asdim}(X)$ is a bit longer. If the dimension $n=\mathrm{asdim}(X)$ is infinite, then there is nothing to prove. So, we assume that $n\in {\omega }$. To prove that $\mathrm{asdim}_{col}(X)\le n$, fix an entourage ${\varepsilon }\in \mathcal{E }$. Let ${\varepsilon }^0=\Delta _X$ and ${\varepsilon }^{k+1}={\varepsilon }^k\circ {\varepsilon }$ for $k\in {\omega }$. Since $\mathrm{asdim}(X)\le n$, for the entourage ${\varepsilon }^{n+1}\in \mathcal{E }$ we can find a cover $\mathcal{U }$ of $X$ such that $\delta =\mathrm{mesh}(\mathcal{U })\in \mathcal{E }$ and each ${\varepsilon }^{n+1}$-ball $B(x,{\varepsilon }^{n+1})$ meets at most $n+1$ many sets $U\in \mathcal{U }$. For every $i\le n+1$ and $x\in X$ consider the subfamily $\mathcal{U }(x,{\varepsilon }^i)=\{U\in \mathcal{U }:B(x,{\varepsilon }^i)\cap U\ne \emptyset \}$ of $\mathcal{U }$. It follows that $1\le |\mathcal{U }(x,{\varepsilon }^i)|\le |\mathcal{U }(x,{\varepsilon }^{i+1})|\le n+1$ for every $0\le i\le n$. Consequently, $|\mathcal{U }(x,{\varepsilon }^i)|=i$ for some $i\le n+1$. Let $\chi (x)$ be the maximal number $k\le n$ such that $|\mathcal{U }(x,{\varepsilon }^{k+1})|=k+1$. In such a way we have defined a coloring $\chi :X\rightarrow n+1=\{0,\dots ,n\}$.

To finish the proof it suffices to show that any $\chi$-monochrome $\varepsilon$-chain $C=\{x_0,\dots ,x_m\}\subset X$ has $\mathrm{diam}(C)\subset \delta \circ \varepsilon ^{n+1}$. Let $k=\chi (x_0)$ be the color of the chain $C$. It follows that $|\mathcal{U }(x_i,{\varepsilon }^{k+1})|=k+1$ for all $x_i\in C$. We claim that $\mathcal{U }(x_i,{\varepsilon }^{k+1})=\mathcal{U }(x_{i+1},{\varepsilon }^{k+1})$ for all $i<m$. Assuming the converse, we would get that $|\mathcal{U }(x_i,{\varepsilon }^{k+1})\cup \mathcal{U }(x_{i+1},{\varepsilon }^{k+1})|\ge k+3$ and then the family $\mathcal{U }(x_i,{\varepsilon }^{k+2})\supset \mathcal{U }(x_i,{\varepsilon }^{k+1})\cup \mathcal{U }(x_{i+1},{\varepsilon }^{k+1})$ has cardinality $|\mathcal{U }(x_i,{\varepsilon }^{k+2})|\ge k+3$, which implies that $|\mathcal{U }(x_i,{\varepsilon }^i)|=i$ for some $k+3\le i\le n+1$. But this contradicts the definition of $k=\chi (x_i)$. Hence $\mathcal{U }(x_i,{\varepsilon }^{k+1})=\mathcal{U }(x_{0},{\varepsilon }^{k+1})$ for all $i\le m$ and then $C\subset B(U,{\varepsilon }^{k+1})$ for every $U\in \mathcal{U }(x_0,{\varepsilon }^{k+1})$. Now we see that $\mathrm{diam}(C)\subset \mathrm{diam}(U)\circ {\varepsilon }^{k+1}\subset \delta \circ {\varepsilon }^{n+1}$. $\square$

Propositions 2.2 and 2.3 imply:

Corollary 2.4

A coarse space $(X,\mathcal{E })$ has asymptotic dimension $\mathrm{asdim}(X)\le n$ for some $n\in {\omega }$ if and only if for any $\varepsilon \in \mathcal{E }$ there are $\delta \in \mathcal{E }$ and a coloring $\chi :X\rightarrow n+1$ such that any $\chi$-monochrome $\varepsilon$-chain $C\subset X$ has $\mathrm{diam}(C)\subset \delta$.

This corollary can be generalized as follows (cf. [6]).

Proposition 2.5

A coarse space $(X,\mathcal{E })$ has $\mathrm{asdim}(X)\le n$ for some $n\in {\omega }$ if and only if for any entourage $\varepsilon \in \mathcal{E }$ there is an entourage $\delta \in \mathcal{E }$ such that for any finite set $F\subset X$ there is a coloring $\chi :F\rightarrow n+1$ such that each $\chi$-monochrome $\varepsilon$-chain $C\subset F$ has $\mathrm{diam}(C)\subset \delta$.

Proof

This proposition will follow from Corollary 2.4 as soon as for any ${\varepsilon }\in \mathcal{E }$ we find $\delta \in \mathcal{E }$ and a coloring $\chi :X\rightarrow n+1$ such that each $\chi$-monochrome ${\varepsilon }$-chain in $X$ has diameter less that $\delta$.

By our assumption, there is an entourage $\delta \in \mathcal{E }$ such that for every finite subset $F\subset X$ there is a coloring $\chi _F:F\rightarrow n+1$ such that each $\chi _F$-monochrome ${\varepsilon }$-chain in $F$ has diameter less that $\delta$. Extend $\chi _F$ to a coloring $\tilde{\chi }_F:X\rightarrow n+1$.

Let $\mathcal{F }$ denote the family of all finite subsets of $X$, partially ordered by the inclusion relation $\subset$. The colorings $\tilde{\chi }_F$, $F\in \mathcal{F }$, can be considered as elements of the compact Hausdorff space $K=\{0,\dots ,n\}^X$ endowed with the Tychonoff product topology. The compactness of $K$ implies that the net $\{\tilde{\chi }_F\}_{F\in \mathcal{F }}$ has a cluster point $\chi \in K$, see [8, 3.1.23]. The latter means that for each finite set $F_0\in \mathcal{F }$ and a neighborhood $O(\chi )\subset K$ there is a finite set $F\in \mathcal{F }$ such that $F\supset F_0$ and $\tilde{\chi }_F\in O(\chi )$.

We claim that the coloring $\chi :X\!\rightarrow \! n+1$ has the required property: each $\chi$-monochrome ${\varepsilon }$-chain $C\subset X$ has $\mathrm{diam}(X)\subset \delta$. Observe that the finite set $C$ determines a neighborhood $O_C(\chi )\!=\!\{f\in K:f|C\!=\!\chi |C\}$, which contains a coloring $\tilde{\chi }_F$ for some finite set $F\supset C$. The choice of the coloring $\chi _F\!=\!\tilde{\chi }_F|F$ guarantees that the set $C\subset F$ has $\mathrm{diam}(C)\subset \delta$. $\square$

Proposition 2.5 admits the following self-generalization.

Theorem 2.6

A coarse space $(X,\mathcal{E })$ has $\mathrm{asdim}(X)\le n$ for some $n\in {\omega }$ if and only if for any entourage $\varepsilon \in \mathcal{E }$ there is an entourage $\delta \in \mathcal{E }$ such that for any finite ${\varepsilon }$-connected subset $F\subset X$ there is a coloring $\chi :F\rightarrow n+1$ such that each $\chi$-monochrome $\varepsilon$-chain $C\subset F$ has $\mathrm{diam}(C)\subset \delta$.

Finally, let us prove Addition Theorem for the asymptotic dimension. For metrizable spaces this theorem is well known; see [14, 9.13] or [5, 9.7.1].

Theorem 2.7

For any subspaces $A,B$ of a coarse space $(X,\mathcal{E })$ we get

$$\begin{aligned} \mathrm{asdim}(A\cup B)\le \max \{\mathrm{asdim}(A),\mathrm{asdim}(B)\}. \end{aligned}$$

Proof

Only the case of finite $n=\max \{\mathrm{asdim}(A),\mathrm{asdim}(B)\}$ requires the proof. Without loss of generality the sets $A$ and $B$ are disjoint. To show that $\mathrm{asdim}(A\cup B)\le n$ we shall apply Corollary 2.4. Fix any entourage ${\varepsilon }\in \mathcal{E }$. Since $\mathrm{asdim}(A)\le n$ there are an entourage $\delta _A\in \mathcal{E }$ and a coloring $\chi _A:A\rightarrow n+1$ such that each $\chi$-monochrome ${\varepsilon }$-chain in $A$ has diameter less that $\delta _A$. Since $\mathrm{asdim}(B)\le n$, for the entourage ${\varepsilon }_B={\varepsilon }\circ \delta _A\circ {\varepsilon }$ there are an entourage $\delta _B\in \mathcal{E }$ and a coloring $\chi _B:A\rightarrow \{0,\dots ,n\}$ such that each $\chi$-monochrome ${\varepsilon }_B$-chain in $B$ has diameter less that $\delta _B$.

The union of the colorings $\chi _A$ and $\chi _B$ yields the coloring $\chi :A\cup B\rightarrow \{0,\dots ,n\}$ such that $\chi |A=\chi _A$ and $\chi |B=\chi _B$. We claim that each $\chi$-monochrome ${\varepsilon }$-chain $C=\{x_0,\dots ,x_m\}\subset A\cup B$ has $\mathrm{diam}(C)\subset \delta$ where $\delta =\delta _A\circ {\varepsilon }\circ \delta _B\circ {\varepsilon }\circ \delta _A$. Without loss of generality, the points $x_0,\dots ,x_m$ of the chain $C$ are pairwise distinct.

If $C\subset A$, then $C$, being a $\chi _A$-monochrome ${\varepsilon }$-chain in $A$ has $\mathrm{diam}(C)\subset \delta _A\subset \delta$ and we are done. So, we assume that $C\not \subset A$. In this case $b=|C\cap B|\ge 1$ and we can choose a strictly increasing sequence $0\le k_1<k_2<\cdots <k_b\le m$ such that $\{x_{k_1},\dots ,x_{k_b}\}=C\cap B$. Then $\{x_0,\dots ,x_{k_1-1}\}$, being a $\chi _A$-monochrome ${\varepsilon }$-chain in $A$, has diameter less that $\delta _A$. Consequently, the ${\varepsilon }$-chain $\{x_0,\dots ,x_{k_1}\}$ has diameter less that $\delta _A\circ {\varepsilon }\subset {\varepsilon }_B$. By the same reason the ${\varepsilon }$-chain $\{x_{k_b},\dots ,x_m\}$ has diameter less that ${\varepsilon }\circ \delta _A\subset {\varepsilon }_B$ and for every $1\le i<b$ the ${\varepsilon }$-chain $\{x_{k_i},\dots ,x_{k_{i+1}}\}\subset \{x_{k_i}\}\cup A\cup \{x_{k_{i+1}}\}$ has diameter less than ${\varepsilon }\circ \delta _A\circ {\varepsilon }={\varepsilon }_B$. Then $\{x_{k_1},\dots ,x_{k_b}\}$, being a $\chi _B$-monochrome ${\varepsilon }_B$-chain in $B$, has diameter less that $\delta _B$. Now we see that the ${\varepsilon }$-chain $C=\{x_0,\dots ,x_m\}$ has $\mathrm{diam}(C)\subset \delta _A\circ {\varepsilon }\circ \delta _B\circ {\varepsilon }\circ \delta _A=\delta$. $\square$

The characterization Theorem 2.6 will be applied to prove the following theorem which was known [7, 2.1] in the context of countable groups.

Theorem 2.8

If $G$ is a topological group endowed with its left coarse structure, then

$$\begin{aligned} \mathrm{asdim}(G)=\sup \,\{\mathrm{asdim}(H):H \mathrm{{is\, a\, compactly\, generated\, subgroup\, of}} G\}. \end{aligned}$$

Proof

Let $n=\sup \,\{\mathrm{asdim}(H):H {\text{ is } \text{ a } \text{ compactly } \text{ generated } \text{ subgroup } \text{ of }} G\}.$ It is clear that $n\le \mathrm{asdim}(G)$. The reverse inequality $\mathrm{asdim}(G)\le n$ is trivial if $n=\infty$. So, we assume that $n<\infty$. To prove that $\mathrm{asdim}(G)\le n$, we shall apply Theorem 2.6. Let $\mathcal{E }$ be the left coarse structure of the topological group $G$. Given any entourage ${\varepsilon }\in \mathcal{E }$, we should find an entourage $\delta \in \mathcal{E }$ such that for each finite ${\varepsilon }$-connected subset $F\subset G$ there is a coloring $\chi :F\rightarrow n+1$ such that each $\chi$-monochrome ${\varepsilon }$-chain $C\subset F$ has $\mathrm{diam}(C)\subset \delta$.

By the definition of the coarse structure $\mathcal{E }$, for the entourage ${\varepsilon }\in \mathcal{E }$ there is a compact subset $K_{\varepsilon }=K_{\varepsilon }^{-1}\subset G$ such that ${\varepsilon }\subset \{(x,y)\in G^2:x\in yK_{\varepsilon }\}$. Let $H$ be the subgroup of $G$ generated by the compact set $K_{\varepsilon }$, $\mathcal{E }_H$ be the left coarse structure of $H$, and ${\varepsilon }_H=\{(x,y)\in H^2:x\in yK_{\varepsilon }\}\in \mathcal{E }_H$. Since $\mathrm{asdim}_{col}(H)=\mathrm{asdim}(H)\le n$, by Proposition 2.2, there is a coloring $\chi _H:H\rightarrow n+1$ and an entourage $\delta _H\in \mathcal{E }_H$ such that each $\chi _H$-monochrome ${\varepsilon }$-chain $C\subset H$ has diameter $\mathrm{diam}(C)\subset \delta _H$. By the definition of the coarse structure $\mathcal{E }_H$, there is a compact subset $K_\delta =K_\delta ^{-1}\ni 1_H$ of $H$ such that $\{(x,y)\in H\times H:x\in yK_\delta \}$.

We claim that the entourage $\delta =\{(x,y)\in G\times G:x\in yK_\delta \}$ satisfies our requirements. Let $F$ be a finite ${\varepsilon }$-connected subset of $G$. Then for each point $x_0\in F$ we get $F\in x_0 H$ and hence $x_0^{-1}F\subset H$. So, we can define a coloring $\chi :F\rightarrow n+1$ letting $\chi (x)=\chi _H(x_0^{-1}x)$ for $x\in F$. If $C\subset F$ is a $\chi$-monochrome ${\varepsilon }$-chain, then $x_0^{-1}C$ is a $\chi _H$-monochrome ${\varepsilon }_H$-chain in $H$ and hence $\mathrm{diam}(x_0^{-1}C)\subset \delta _H$. The latter means that for any points $c,c^{\prime }\in C$ we get $(x_0^{-1}c,x_0^{-1}c^{\prime })\in \delta _H\subset \{(x,y)\in H\times H:x\in yK_\delta \}$ and hence $x_0^{-1}c\in x_0^{-1}c^{\prime }K_\delta$ and $c\in c^{\prime }K_\delta$, which means that $(c,c^{\prime })\in \delta$ and hence $\mathrm{diam}(C)\subset \delta$. $\square$

3 Proof of Proposition 1.4

Let $f:X\rightarrow Y$ be a coarse equivalence between two coarse spaces $(X,\mathcal{E }_X)$ and $(Y,\mathcal{E }_Y)$. Then there is a coarse map $g:Y\rightarrow X$ such that $\{(x,g\circ f(x)):x\in X\}\subset \eta _X$ and $\{(y,f\circ g(y)):y\in Y\}\subset \eta _Y$ for some entourages $\eta _X\in \mathcal{E }_X$ and $\eta _Y\in \mathcal{E }_Y$. It follows that $B(f(X),\eta _X)=Y$ and $B(g(Y),\eta _X)=X$.

• 1. First we prove that $\mathrm{asdim}(X)=\mathrm{asdim}(Y)$. Actually, this fact is known [14, p.129] and we present a proof for the convenience of the reader. By the symmetry, it suffices to show that $\mathrm{asdim}(X)\le \mathrm{asdim}(Y)$. This inequality is trivial if $n=\mathrm{asdim}(Y)$ is infinite. So, assume that $n<\infty$. By Propositions 2.2 and 2.3, the inequality $\mathrm{asdim}(X)\le n$ will be proved as soon as for each ${\varepsilon }_X\in \mathcal{E }_X$ we find $\delta _X\in \mathcal{E }_X$ and a coloring $\chi _X:X\rightarrow n+1$ such that each $\chi _X$-monochrome ${\varepsilon }_X$-chain $C\subset X$ has diameter $\mathrm{diam}(C)\subset \delta _X$.

Since the map $f:X\rightarrow Y$ is coarse, for the entourage ${\varepsilon }_X$ there is an entourage ${\varepsilon }_Y$ such that $\{(f(x),f(x^{\prime })):(x,x^{\prime })\in {\varepsilon }_X\}\subset {\varepsilon }_Y$. Since $\mathrm{asdim}(Y)=n$, for the entourage ${\varepsilon }_Y$ there is an entourage $\delta _Y\in \mathcal{E }_Y$ and a coloring $\chi _Y:Y\rightarrow n+1$ such that each $\chi _Y$-monochrome ${\varepsilon }_Y$-chain $C_Y\subset Y$ has diameter $\mathrm{diam}(C_Y)\subset \delta _Y$.

Since the function $g:Y\rightarrow X$ is coarse, for the entourage $\delta _Y$ there is an entourage $\delta _X^{\prime }$ such that $\{(g(y),g(y^{\prime })):(y,y^{\prime })\in \delta _Y\}\subset \delta _Y^{\prime }$. Put $\delta _X=\eta _X\circ \delta _X^{\prime }\circ \eta _Y$ and consider the coloring $\chi _X=\chi _Y\circ f:X\rightarrow n+1$ of $X$. We claim that each $\chi _X$-monochrome ${\varepsilon }_X$-chain $C_X\subset X$ has diameter $\mathrm{diam}(C_X)\subset \delta _X$. Then choice of ${\varepsilon }_Y$ guarantees that the set $C_Y=f(C_X)$ is an ${\varepsilon }_Y$-chain. Being $\chi _X$-monochrome, it has diameter $\mathrm{diam}(C_Y)\subset \delta _Y$. Then the set $C_X^{\prime }=g(C_Y)$ has diameter $\mathrm{diam}(C_X^{\prime })\subset \delta _X^{\prime }$. Now take any two points $c,c^{\prime }\in C_X$ and observe that the pairs $(c,g\circ f(c))$ and $(c^{\prime },g\circ f(c^{\prime }))$ belong to the entourage $\eta _X$. Consequently,

$$\begin{aligned} \quad (c,c^{\prime })\in \{(c,g\circ f(c))\}\circ \{(g\circ f(c),g\circ f(c^{\prime })\}\circ \{(g\circ f(c^{\prime }),c^{\prime })\}\subset \eta _X\circ \delta ^{\prime }_X\circ \eta _X=\delta _X \qquad \end{aligned}$$

which means that the ${\varepsilon }_X$-chain $C_X$ has diameter $\mathrm{diam}(C_X)\subset \delta _X$. So, $\mathrm{asdim}(X)\le n$.

• 2. The second statement of Proposition 1.4, follows Claims 3.1 and 3.4 proved below.

Claim 3.1

A subset $A\subset X$ and its image $f(A)\subset Y$ have the same asymptotic dimension $\mathrm{asdim}(A)=\mathrm{asdim}(f(A))$.

Proof

This claim follows from Proposition 1.4(1) proved above, since $A$ and $f(A)$ are coarsely equivalent. $\square$

Claim 3.2

A subset $A\subset X$ is large in $X$ if and only if its image $f(A)$ is large in $Y$.

Proof

If $A$ is large in $X$, then $B(A,{\varepsilon }_X)=X$ for some ${\varepsilon }_X\in \mathcal{E }_X$. Since $f$ is coarse, there exists ${\varepsilon }_Y\in \mathcal{E }_Y$ such that for each $(x_0,x_1)\in {\varepsilon }_X$ we get $(f(x_0),f(x_1))\in {\varepsilon }_Y$. It follows that $B(f(A),{\varepsilon }_Y)\supset f(Y)$ and $B(f(A),{\varepsilon }_Y\circ \eta _Y)=B(B(f(A),{\varepsilon }_Y),\eta _Y)\supset B(f(X),\eta _Y)=Y$, which means that $f(A)$ is large.

Now assume conversely that the set $f(A)$ is large in $Y$. Then $g\circ f(A)$ is large in $X$. Since $g\circ f(A)\subset B(A,\eta _X)$, we conclude that $A$ in large in $X$. So, $A$ is large in $X$ if and only if $f(A)$ is large in $Y$. $\square$

Claim 3.3

A subset $A\subset X$ is small if and only if for each entourage ${\varepsilon }_X\in \mathcal{E }_X$ the set $B(A,{\varepsilon }_x)$ is small.

Proof

The “if” part is trivial. To prove the “only if” part, assume that the set $A$ is small in $X$. To show that $B(A,{\varepsilon }_X)$ is small in $X$, it is necessary to check that for each large subset $L\subset X$ the complement $L\setminus B(A,{\varepsilon }_X)$ is large in $X$. Consider the set $L^{\prime }=(L\setminus B(A,{\varepsilon }_X))\cup A$ and observe that $L\subset B(L^{\prime },{\varepsilon }_X)$ and hence $L^{\prime }$ is large in $X$. Since $A$ is small, the set $L^{\prime }\setminus A=L\setminus B(A,{\varepsilon }_X)$ is large in $X$. $\square$

Claim 3.4

A subset $A\subset X$ is small in $X$ if and only if its image $f(A)$ is small in $Y$.

Proof

Assume that $A$ is small in $X$. To prove that $f(A)$ is small in $Y$, we need to check that for any large subset $L\subset Y$ the complement $L\setminus f(A)$ is large in $Y$. Claim 3.2 implies that the set $g(L)$ is large in $X$. By Claim 3.3, the set $B(A,\eta _X)$ is small in $X$ and hence the complement $g(L)\setminus B(A,\eta _X)$ remains large in $X$. By Claim 3.2 $f(g(L)\setminus B(A,\eta _X))$ is large in $Y$. We claim that $f(g(L)\setminus B(A,\eta _X))\subset B(L\setminus f(A),\eta _Y)$. Indeed, given point $y\in f(g(L)\setminus B(A,\eta _X))$, find a point $x\in g(L)\setminus B(A,\eta _X)$ such that $y=f(x)$ and a point $z\in L$ such that $x=g(z)$. We claim that $z\notin f(A)$. Assuming conversely that $z\in f(A)$, we get $x=g(z)\in g\circ f(A)\subset B(A,\eta _X)$, which contradicts the choice of $x$. So, $z\in L\setminus f(A)$ and $y=f\circ g(z)\in B(z,\eta _Y)\subset B(L\setminus f(A),\eta _Y)$.

Taking into account that the set $f(g(L)\setminus B(A,\eta _X))\subset B(L\setminus f(A),\eta _Y)$ is large in $Y$, we conclude that the set $L\setminus f(A)$ is large in $Y$ and hence $f(A)$ is small in $Y$.

Now assume that the set $f(X)$ is small in $Y$. Then the set $g\circ f(A)$ is small in $X$ and so are the sets $B(g\circ f(A),\eta _X)\supset A$. $\square$

4 Proof of Theorem 1.6

Let $G$ be a topological group and $\mathcal{E }$ be its left coarse structure. The inclusion $\mathcal{D }_{<}(G)\subset \mathcal{S }(G)$ will follow as soon as we prove that each non-small subset $A\subset G$ has asymptotic dimension $\mathrm{asdim}(A)=\mathrm{asdim}(G)$. We divide the proof of this fact into 3 steps.

Claim 4.1

There is an entourage ${\varepsilon }_A\in \mathcal{E }$ such that the set $G{\setminus } B(A,{\varepsilon }_A)$ is not large in $G$.

Proof

Since $A$ is not small, there is a large set $L\subset X$ such that the complement $L{\setminus } A$ is not large. Since $L$ is large in $X$, there is an entourage ${\varepsilon }_A\in \mathcal{E }$ such that $B(L,{\varepsilon }_A)=G$. We claim that the set $G{\setminus } B(A,{\varepsilon }_A)$ is not large. Assuming the opposite, we can find an entourage $\delta \in \mathcal{E }$ such that $B(G{\setminus } B(A,{\varepsilon }_A),\delta )=G$. Then for each $x\in G$ the ball $B(x,\delta )$ meets $G\setminus B(A,{\varepsilon }_A)$ at some point $y$. By the choice of ${\varepsilon }_A$, the ball $B(y,{\varepsilon }_A)$ meets the large set $L$ at some point $z$. It follows from $y\notin B(A,{\varepsilon }_A)$ that $z\in L\cap B(y,{\varepsilon })\subset L\cap (X{\setminus } A)=L{\setminus } A$ and hence $x\in B(L{\setminus } A,{\varepsilon }_A\circ \delta )$, which means that $L\setminus A$ is large in $X$. This is a required contradiction. $\square$

Claim 4.2

$\mathrm{asdim}(B(A,{\varepsilon }_A))=\mathrm{asdim}(A)$.

Proof

Observe that the identity embedding $i:A\rightarrow B(A,{\varepsilon }_A)$ is a coarse equivalence. The coarse inverse $j:B(A,{\varepsilon }_A)\rightarrow A$ to $i$ can be defined by choosing a point $j(x)\in B(x,{\varepsilon }_A)\cap A$ for each $x\in B(A,{\varepsilon }_A)$. Now we equality $\mathrm{asdim}(B(A,{\varepsilon }_A))=\mathrm{asdim}(A)$ follows from the invariance of the asymptotic dimension under coarse equivalences, see Proposition 1.4. $\square$

Claim 4.3

$\mathrm{asdim}(G)=\mathrm{asdim}(A)$.

Proof

The inequality $\mathrm{asdim}(A)\le \mathrm{asdim}(G)$ is trivial. So, it suffices to check that $\mathrm{asdim}(G)\le n$ where $n=\mathrm{asdim}(A)=\mathrm{asdim}(B(A,{\varepsilon }_A))$. If $n$ is infinite, then there is nothing to prove. So, we assume that $n\in {\omega }$.

For the proof of the inequality $\mathrm{asdim}(G)\le n$, we shall apply Theorem 2.6. Given any ${\varepsilon }\in \mathcal{E }$ we should find $\delta \in \mathcal{E }$ such that for each finite ${\varepsilon }$-connected subset $F\subset G$ there is a coloring $\chi :F\rightarrow n+1$ such that each $\chi$-monochrome ${\varepsilon }$-chain $C\subset F$ has $\mathrm{diam}(C)\subset \delta$. By the definition of the left coarse structure $\mathcal{E }$ we lose no generality assuming that ${\varepsilon }=\{(x,y)\in G\times G:x\in yK_{\varepsilon }\}$ for some compact subset $K_{\varepsilon }=K_{\varepsilon }^{-1}\subset G$ containing the neutral element $1_G$ of $G$. In this case the entourage ${\varepsilon }$ is left invariant in the sense that for each pair $(x,y)\in {\varepsilon }$ and each $z\in G$ the pair $(zx,zy)$ belongs to ${\varepsilon }$.

Since $\mathrm{asdim}_{col}(B(A,{\varepsilon }_A))=\mathrm{asdim}(B(A,{\varepsilon }_A))\le n$, for the entourage ${\varepsilon }\in \mathcal{E }$, there are an entourage $\delta \in \mathcal{E }$ and a coloring $\chi _A:B(A,{\varepsilon }_A)\rightarrow n+1$ such that each $\chi$-monochrome ${\varepsilon }$-chain $C\subset B(A,{\varepsilon }_A)$ has $\mathrm{diam}(C)\subset \delta$, see Proposition 2.2. By the definition of the left coarse structure $\mathcal{E }$, we lose no generality assuming that $\delta =\{(x,y)\in G\times G:x\in yK_\delta \}$ for some compact set $K_\delta =K_\delta ^{-1}\ni 1_G$ of $G$, which implies that the entourage $\delta$ is left invariant.

Now take any finite ${\varepsilon }$-connected subset $F\subset G$. Replacing $F$ by $F\cup F^{-1}\cup \{1_G\}$ we can assume that $F=F^{-1}\ni 1_G$. Since the set $G\setminus B(A,{\varepsilon }_A)$ is not large, there is a point $z\notin (G\setminus B(A,{\varepsilon }_A))F$. Then $zF^{-1}$ is disjoint with $G\setminus B(A,{\varepsilon }_A)$ and hence $zF=zF^{-1}\subset B(A,{\varepsilon }_A)$. So, it is legal to define a coloring $\chi :F\rightarrow n+1$ by the formula $\chi (x)=\chi _A(zx)$ for $x\in F$. Taking into account the left invariance of the entourages ${\varepsilon }$ and $\delta$, it is easy to see that each $\chi$-monochrome ${\varepsilon }$-chain $C\subset F$ has diameter $\mathrm{diam}(C)\subset \delta$. By Propositions 2.2 and 2.3, $\mathrm{asdim}(G)=\mathrm{asdim}_{col}(G)\le n=\mathrm{asdim}(A)$. $\square$

5 Proof of Theorem 1.2

We need to prove that a subset $A\subset \mathbb{R }^n$ is small if and only if it has asymptotic dimension $\mathrm{asdim}(A)<\mathrm{asdim}(\mathbb{R }^n)=n$. The “if” part of this characterization follows from the inclusion $\mathcal{D }_{<}(\mathbb{R }^n)\subset \mathcal{S }(\mathbb{R }^n)$ proved in Theorem 1.6. To prove the “only if” part, we need to recall some (standard) notions of Combinatorial Topology [10, 12].

On the Euclidean space $\mathbb{R }^n$ we shall consider the metric generated by the sup-norm $\Vert x\Vert =\max _{i\in n}|x(i)|$.

By the standard $n$ -dimensional simplex we understand the compact convex subset

$$\begin{aligned} \Delta =\left\{ (x_0,\dots ,x_n)\in [0,1]^{n+1}: \sum _{i=0}^nx_i=1\right\} \subset \mathbb{R }^{n+1} \end{aligned}$$

of the Euclidean space $\mathbb{R }^{n+1}$ endowed with the sup-norm. For each $i\le n$ by $v_i:n+1\rightarrow \{0,1\}\subset \mathbb{R }$ we denote the vertex of $\Delta$ defined by $v_i^{-1}(1)=\{i\}$. For each vertex $v_i$ of $\Delta$ consider its star

$$\begin{aligned} \mathcal{S }t_\Delta (v_i)=\{x\in \Delta :x(i)>0\} \end{aligned}$$

and its barycentric star

$$\begin{aligned} \mathcal{S }t^{\prime }_\Delta (v_i)=\left\{ x\in \Delta :x(i)=\max _{j\le n}x(j)\right\} \subset \mathcal{S }t_\Delta (v_i). \end{aligned}$$

It is clear that $\bigcup _{i=0}^n \mathcal{S }t^{\prime }_\Delta (v_i)=\Delta$ while $\bigcap _{i=0}^n\mathcal{S }t^{\prime }_\Delta (v_i)=\{b_\Delta \}$ is the singleton containing the barycenter

$$\begin{aligned} b_\Delta =\tfrac{1}{n+1}\sum _{i=0}^nv_i \end{aligned}$$

of the simplex $\Delta$.

Claim 5.1

$\bigcap _{i=0}^n B(St^{\prime }_\Delta (v_i),{\varepsilon })\subset B(b_\Delta ,n{\varepsilon })$ for each positive real number ${\varepsilon }$.

Proof

Given any vector $x\in \bigcap _{i=0}^n B(St^{\prime }_\Delta (v_i),{\varepsilon })$, for every $i\le n$ we can find a vector $y\in \mathcal{S }t^{\prime }_\Delta (v_i)$ with $\Vert x-y\Vert <{\varepsilon }$. Then $|x_i-y_i|\le \Vert x-y\Vert <{\varepsilon }$ and hence $x_i>y_i-{\varepsilon }=\max _{j\le n}y_j-{\varepsilon }\ge \frac{1}{n+1}-{\varepsilon }$. On the other hand,

$$\begin{aligned} x_i=1-\sum _{j\ne i}x_j<1-\sum _{j\ne i}\left( \frac{1}{n+1}-{\varepsilon }\right) =1-\frac{n}{n+1}+n{\varepsilon }=\frac{1}{n+1}+n{\varepsilon }. \end{aligned}$$

So, $\Vert x-b_\Delta \Vert <n{\varepsilon }$. $\square$

Now we are going to generalize Claim 5.1 to arbitrary simplexes. By an n-dimensional simplex in $\mathbb{R }^n$ we understand the convex hull $\sigma =\mathrm{conv}(\sigma ^{(0)})$ of an affinely independent subset $\sigma ^{(0)}\subset \mathbb{R }^n$ of cardinality $|\sigma ^{(0)}|=n+1$. Each point $v\in \sigma ^{(0)}$ is called a vertex of the simplex $\sigma$. The arithmetic mean

$$\begin{aligned} b_\sigma =\frac{1}{n+1}\sum _{v\in \sigma ^{(0)}}v \end{aligned}$$

of the vertices is called the barycenter of the simplex $\sigma$. By $\partial \sigma$ we denote the boundary of the simplex $\sigma$ in $\mathbb{R }^n$. Observe that the homothetic copy $\frac{1}{2}b_\sigma +\frac{1}{2}\sigma =\{\frac{1}{2}b_\sigma +\frac{1}{2}x:x\in \sigma \}$ of $\sigma$ is contained in the interior $\sigma \setminus \partial \sigma$ of $\sigma$. For each vertex $v\in \sigma ^{(0)}$ let

$$\begin{aligned} \mathcal{S }t_\sigma (v)=\sigma \setminus \mathrm{conv}(\sigma ^{(0)}{\setminus }\{v\}) \end{aligned}$$

be the star of $v$ in $\sigma$.

In fact, $n$-dimensional simplexes can be alternatively defined as images of the standard $n$-dimensional simplex $\Delta$ under injective affine maps $f:\Delta \rightarrow \mathbb{R }^n$.

A map $f:\Delta \rightarrow \mathbb{R }^n$ is called affine if $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$ for any points $x,y\in \Delta$ and a real number $t\in [0,1]$. It is well-known that each affine function $f:\Delta \rightarrow \mathbb{R }^n$ is uniquely defined by its restriction $f|\Delta ^{(0)}$ to the set $\Delta ^{(0)}=\{v_i\}_{i\le n}$ of vertices of $\Delta$.

A map $f:\Delta \rightarrow \mathbb{R }^n$ will be called $b_\Delta$ -affine if for every $i\le n$ the restriction $f|\mathrm{conv}(\{b_\Delta \}{\cup }\Delta ^{(0)}{\setminus }\{v_i\})$ is affine. A $b_\Delta$-affine function $f:\Delta \rightarrow \mathbb{R }^n$ is uniquely determined by its restriction $f|\Delta ^{(0)}\cup \{b_\Delta \}$.

A function $f:X\rightarrow Y$ between metric spaces $(X,d_X)$ and $(Y,d_Y)$ is called Lipschitz if it its Lipschitz constant

$$\begin{aligned} \mathrm{Lip}(f)=\sup \left\{ \frac{d_Y(f(x),f(x^{\prime }))}{d_X(x,x^{\prime })}:x,x^{\prime }\in X,\;\;x\ne x^{\prime }\right\} \end{aligned}$$

is finite. A bijective function $f:X\rightarrow Y$ is bi-Lipschitz if $f$ and $f^{-1}$ are Lipschitz.

Claim 5.2

For any $n$-dimensional simplex $\sigma$ in $\mathbb{R }^n$ there is a real constant $L$ such that each $b_\Delta$-affine function $f:\Delta \rightarrow \sigma$ with $f(\Delta ^{(0)})=\sigma ^{(0)}$ and $f(b)\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$ is bijective, bi-Lipschitz and has $\mathrm{Lip}(f)\cdot \mathrm{Lip}(f^{-1})\le L$.

This claim can be easily derived from the fact that each $b_\Delta$-affine function $f:\Delta \rightarrow \sigma$ with $f(\Delta ^{(0)})=\sigma ^{(0)}$ is Lipschitz and its Lipschitz constant $\mathrm{Lip}(f)$ depends continuously on $f(b_\Delta )$.

Given an $n$-dimensional simplex $\sigma \subset \mathbb{R }^n$ and a point $b^{\prime }\in \sigma \setminus \partial \sigma$ in its interior, fix a $b_\Delta$-affine function $f:\Delta \rightarrow \sigma$ such that $f(\Delta ^{(0)})=\sigma ^{(0)}$ and $f(b_\Delta )=b^{\prime }$. For each vertex $v\in \sigma ^{(0)}$ consider its $b^{\prime }$ -barycentric star

$$\begin{aligned} \mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v)=f\big (\mathcal{S }t^{\prime }_\Delta (f^{-1}(v))\big )\subset \mathcal{S }t_\sigma (v). \end{aligned}$$

It is easy to see that the set $\mathcal{S }t_{\sigma ,b^{\prime }}(v)$ does not depend on the choice of the $b_\Delta$-affine function $f$.

Claim 5.3

For any $n$-dimensional simplex $\sigma$ in $\mathbb{R }^n$ there is a real constant $L$ such that for each point $b^{\prime }\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$ and each ${\varepsilon }>0$ we get $\sigma \cap \bigcap _{v\in \sigma ^{(0)}}B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),{\varepsilon })\subset B(b^{\prime },L{\varepsilon })$.

Proof

By Claim 5.2, there is a real constant $C$ such that each bijective $b_\Delta$-affine function $f:\Delta \rightarrow \sigma$ with $f(\Delta ^{(0)})=\sigma ^{(0)}$ and $f(b_\Delta )\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$ has $\mathrm{Lip}(f)\cdot \mathrm{Lip}(f^{-1})\le C$. Put $L=nC$. Given any point $b^{\prime }\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$, choose a bijective $b_\Delta$-affine function $f:\Delta \rightarrow \sigma$ such that $f(\Delta ^{(0)})=\sigma ^{(0)}$ and $f(b_\sigma )=b^{\prime }$. The choice of $C$ guarantees that $\mathrm{Lip}(f)\cdot \mathrm{Lip}(f^{-1})\le C$. Now observe that

$$\begin{aligned} \sigma \cap \bigcap _{v\in \sigma ^{(0)}}B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),{\varepsilon })&= \bigcap _{v\in \sigma ^{(0)}}f\circ f^{-1}\left( B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),{\varepsilon })\right) \\&\subset \bigcap _{v\in \sigma ^{(0)}}f\left( B(f^{-1}(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v)),\mathrm{Lip}(f^{-1}){\varepsilon })\right) \\&= \bigcap _{v\in \sigma ^{(0)}}f\left( B(\mathcal{S }t^{\prime }_{\Delta }(f^{-1}(v)),\mathrm{Lip}(f^{-1}){\varepsilon })\right) \\&= f\left( \bigcap _{v\in \Delta ^{(0)}}B\left( \mathcal{S }t^{\prime }_{\Delta }(v),\mathrm{Lip}(f^{-1}){\varepsilon }\right) \right) \\&\subset f\left( B(b_\Delta ,n\mathrm{Lip}(f^{-1}){\varepsilon })\right) \subset B\left( f(b_\Delta ),\mathrm{Lip}(f)\mathrm{Lip}(f^{-1})n{\varepsilon }\right) \\&= B(b^{\prime },Cn{\varepsilon })=B(b^{\prime },L{\varepsilon }). \end{aligned}$$

$\square$

Now consider the binary unit cube $K=\{0,1\}^n\subset \mathbb{R }^n$ endowed with the partial ordering $\le$ defined by $x\le y$ iff $x(i)\le y(i)$ for all $i<n$. Given two vectors $x,y\in \{0,1\}^n$, we write $x<y$ if $x\le y$ and $x\ne y$.

For every increasing chain $v_0<v_1<\ldots <v_n$ of points of the binary cube $K=\{0,1\}^n$, consider the simplex $\mathrm{conv}\{v_0,\ldots ,v_n\}$ and let $\mathcal{T }_K$ be the (finite) set of these simplexes. Next, consider the family $\mathcal{T }=\{\sigma +z:\sigma \in \mathcal{T }_K,\;z\in \mathbb{Z }^n\}$ of translations of the simplexes from the family $\mathcal{T }_K$, and observe that $\bigcup \mathcal{T }=\mathbb{R }^n$. For each point $v\in \mathbb{Z }^n$ let

$$\begin{aligned} \mathcal{S }t_{\mathcal{T }}(v)=\bigcup \{\mathcal{S }t_\sigma (v):v\in \sigma \in \mathcal{T }\} \end{aligned}$$

be the $\mathcal{T }$ -star of $v$ in the triangulation $\mathcal{T }$ of the space $\mathbb{R }^n$.

Now we are able to prove the “only if” part of Theorem 1.2. Assume that a subset $A\subset \mathbb{R }^n$ is small. Then there is a function $\varphi :(0,\infty )\rightarrow (0,\infty )$ such that for each $\delta \in (0,\infty )$ and a point $x\in \mathbb{R }^n$ there is a point $y\in \mathbb{R }^n$ with $B(y,\delta )\subset B(x,\varphi (\delta ))\setminus A$. The inequality $\mathrm{asdim}(A)<n$ will follow as soon as given any $\delta <\infty$ we construct a cover $\mathcal{U }$ of $A$ with finite $\mathrm{mesh}(\mathcal{U })=\sup _{U\in \mathcal{U }}\mathrm{diam}(U)$ such that each $\delta$-ball $B(a,\delta )$, $a\in A$, meets at most $n$ elements of the cover $\mathcal{U }$.

By Claim 5.3, there is a constant $L$ such that for each simplex $\sigma \in \mathcal{T }$, each point $b^{\prime }\in \frac{1}{2} b_\sigma +\frac{1}{2}\sigma$ and each ${\varepsilon }>0$ we get $\sigma \cap \bigcap _{v\in \sigma ^{(0)}}B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),{\varepsilon })\subset B(b^{\prime },L{\varepsilon })$.

Given any $\delta <\infty$, choose ${\varepsilon }>0$ so small that for any simplex $\sigma \in \mathcal{T }$ the following conditions hold:

• (1) $B\big (b_\sigma ,{\varepsilon }\varphi (L\delta )\big )\subset \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$;

• (2) for any $b^{\prime }\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$ and any vertex $v\in \sigma ^{(0)}$ the $2{\varepsilon }\delta$-neighborhood $B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),2{\varepsilon }\delta )$ lies in the $\mathcal{T }$-star $\mathcal{S }t_\mathcal{T }(v)$ of $v$.

Now consider the closed cover

$$\begin{aligned} \widetilde{\mathcal{T }}=\left\{ \tfrac{1}{{\varepsilon }}\sigma :\sigma \in \mathcal{T }\right\} \end{aligned}$$

of the space $\mathbb{R }^n$ and observe that for each simplex $\sigma \in \widetilde{\mathcal{T }}$ we get

• $(1_{\varepsilon })$ $B\big (b_\sigma ,\varphi (L\delta )\big )\subset \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$;

• $(2_{\varepsilon })$ for any $b^{\prime }\in \frac{1}{2}b_\sigma +\frac{1}{2}\sigma$ and any vertex $v\in \sigma ^{(0)}$ the $2\delta$-neighborhood $B(\mathcal{S }t^{\prime }_{\sigma ,b^{\prime }}(v),2\delta )$ lies in the $\widetilde{\mathcal{T }}$-star $\mathcal{S }t_{\widetilde{\mathcal{T }}}(v)$ of $v$.

By the choice of the function $\varphi$, for each simplex $\sigma \in \widetilde{\mathcal{T }}$, there is a point $b_{\sigma }^{\prime }\in \mathbb{R }^n$ such that $B(b_{\sigma }^{\prime },L\delta )\subset B(b_\sigma ,\varphi (L\delta ))\setminus A$. The condition $(1_{\varepsilon })$ guarantees that

$$\begin{aligned} b_{\sigma }^{\prime }\in B(b_\sigma ,\varphi (L\delta )) \subset \tfrac{1}{2}b_\sigma +\tfrac{1}{2}\sigma . \end{aligned}$$

For every point $v\in \frac{1}{{\varepsilon }}\mathbb{Z }^n$ consider the set

$$\begin{aligned} \mathcal{S }t^{\prime }(v)=\bigcup \left\{ \mathcal{S }t_{\sigma ,b^{\prime }_\sigma }(v):\sigma \in \widetilde{\mathcal{T }},\;v\in \sigma ^{(0)}\right\} \subset \mathcal{S }t_{\widetilde{\mathcal{T }}}(v) \end{aligned}$$

and observe that $\mathcal{U }=\{\mathcal{S }t^{\prime }(v):v\in {\frac{1}{{\varepsilon }}}\mathbb{Z }^n\}$ is a cover of the Euclidean space $\mathbb{R }^n$. It follows that

$$\begin{aligned} \mathrm{mesh}(\mathcal{U })=\sup _{v\in {\varepsilon }^{-1}\mathbb{Z }^n}\mathrm{diam}(\mathcal{S }t^{\prime }(v))\le 2\sup _{v\in {\varepsilon }^{-1}\mathbb{Z }^n}\mathrm{diam}(\sigma )\le 2{\varepsilon }^{-1}\mathrm{diam}([0,1]^n)<\infty . \end{aligned}$$

It remains to check that each ball $B(a,\delta )$, $a\in A$, meets at most $n$ sets $U\in \mathcal{U }$.

Assume conversely that there are a point $a\in A$ and a set $V\subset {\varepsilon }^{-1}\mathbb{Z }^n$ of cardinality $|V|=n+1$ such that $B(a,\delta )\cap \mathcal{S }t^{\prime }(v)\ne \emptyset$ for each $v\in V$. Then $a\in \bigcap _{v\in V}B(\mathcal{S }t^{\prime }(v),\delta )$. It follows from $a\in \bigcap _{v\in V}B(\mathcal{S }t^{\prime }(v),\delta )\subset \bigcap _{v\in V}\mathcal{S }t_{\widetilde{\mathcal{T }}}(v)$ that $V$ coincides with the set $\sigma ^{(0)}$ of vertices of some simplex $\sigma \in \widetilde{\mathcal{T }}$ and $a$ lies in the interior of the simplex $\sigma$.

Next, we show that $a\in B(\mathcal{S }t^{\prime }_{\sigma ,b_\sigma ^{\prime }}(v),\delta )$ for each $v\in V$. In the opposite case, $a\in B(\mathcal{S }t^{\prime }_{\tau ,b_\tau ^{\prime }}(v),\delta )\subset B(\tau ,\delta )$ for some simplex $\tau \in \widetilde{\mathcal{T }}\setminus \{\sigma \}$ such that $v\in \tau ^{(0)}\setminus \sigma ^{(0)}$. Choose a vertex $u\in \sigma ^{(0)}\setminus \tau ^{(0)}$ and observe that the condition $(2_{\varepsilon })$ implies that $a\in B(\mathcal{S }t^{\prime }(u),\delta )\cap B(\tau ,\delta )=\emptyset$, which is a contradiction.

Finally, the choice of $L$ and $b_\sigma ^{\prime }$ yields the desired contradiction

$$\begin{aligned} a\in \sigma \cap \bigcap _{v\in \sigma ^{(0)}}B(\mathcal{S }t^{\prime }_{\sigma ,b_\sigma ^{\prime }}(v),\delta )\subset B(b_\sigma ^{\prime },L\delta )\subset \mathbb{R }^n\setminus A, \end{aligned}$$

completing the proof of the theorem.

6 Proof of Theorem 1.7

Given an Abelian locally compact topological group $G$ endowed with its left coarse structure, we need to prove the equivalence of the following statements:

• (1) $\mathcal{S }(G)=\mathcal{D }_<(G)$;

• (2) $G$ is compactly generated;

• (3) $G$ is coarsely equivalent to a Euclidean space $\mathbb{R }^n$ for some $n\in {\omega }$.

We shall prove the implications $(1){\Rightarrow }(2){\Rightarrow }(3){\Rightarrow }(1)$. The implication $(3){\Rightarrow }(1)$ follows from Corollary 1.5.

To prove that $(2){\Rightarrow }(3)$, assume that the group $G$ is compactly generated. By Theorem 24 [11, p.85], $G$ is topologically isomorphic to the direct sum $\mathbb{R }^n\times \mathbb{Z }^m\times K$ for some $n,m\in {\omega }$ and a compact subgroup $K\subset G$. Since the projection $\mathbb{R }^n\times \mathbb{Z }^m\times K\rightarrow \mathbb{R }^n\times \mathbb{Z }^m$ and the embedding $\mathbb{Z }^n\times \mathbb{Z }^m\rightarrow \mathbb{R }^n\times \mathbb{Z }^m$ are coarse equivalences, we conclude that $G$ is coarsely equivalent to $\mathbb{Z }^{n+m}$ and to $\mathbb{R }^{n+m}$.

To prove that $(1){\Rightarrow }(2)$, assume that $\mathcal{S }(G)=\mathcal{D }_{<}(G)$. First we prove that $G$ has finite asymptotic dimension. By the Principal Structure Theorem 25 [11, p.26], $G$ contains an open subgroup $G_0$ that is topologically isomorphic to $\mathbb{R }^n\times K$ for some $n\in {\omega }$ and some compact subgroup $K$ of $G_0$. The subgroup $G_0$ has asymptotic dimension $\mathrm{asdim}(G_0)=\mathrm{asdim}(\mathbb{R }^n)=n<\infty$. If $\mathrm{asdim}(G)=\infty$, then the quotient group $G/G_0$ has infinite asymptotic dimension and hence has infinite free rank. Then the group $G/G_0$ contains a subgroup isomorphic to the free abelian group $\oplus ^{\omega }\mathbb{Z }$ with countably many generators. It follows that $G$ also contains a discrete subgroup $H$ isomorphic to $\oplus ^{\omega }\mathbb{Z }$. Replacing $H$ by a smaller subgroup, if necessary, we can assume that $H$ has infinite index in $G$ and hence is small in $G$. Since $\mathrm{asdim}(H)=\infty =\mathrm{asdim}(G)$, we conclude that $\mathcal{S }(G)\ne \mathcal{D }_<(G)$, which is a desired contradiction showing that $\mathrm{asdim}(G)<\infty$.

By Theorem 2.8, there is a compactly generated subgroup $H\subset G$ with $\mathrm{asdim}(H)=\mathrm{asdim}(G)$. Since $H\notin \mathcal{D }_{<}(G)=\mathcal{S }(G)$, the subset $H$ is not small in $G$. Repeating the proof of Claim 4.1, we can show that the set $G\setminus B(H,{\varepsilon })$ is not large for some entourage ${\varepsilon }\in \mathcal{E }$. By the definition of the left coarse structure $\mathcal{E }$, there is a compact subset $K\subset G$ such that $B(H,{\varepsilon })\subset HK$. We claim that $K^{-1}HK=G$. Assuming the opposite, we can find a point $x\in G\setminus K^{-1}HK$ and consider the finite set $F=\{x,x^{-1},xx^{-1}\}=F^{-1}$. Since the set $G\setminus HK$ is not large, there is a point $z\in (G\setminus HK)F$. For this point $z$ we get $zF\cap (G\setminus HK)=\emptyset$ and hence $z\in zF\subset HK$. Then $x\in z^{-1}zF\subset z^{-1}HK\subset K^{-1}HHK=K^{-1}HK$, which is a contradiction. Now the compact generacy of the subgroup $H$ implies the compact generacy of the group $G=K^{-1}HK$.