A metric space with its transfinite asymptotic dimension omega + 1

TL;DR

This paper constructs a specific metric space demonstrating that the transfinite asymptotic dimension can reach omega+1, providing a counterexample to the omega conjecture in geometric group theory.

Contribution

It presents the first explicit example of a metric space with transfinite asymptotic dimension omega+1, disproving the omega conjecture.

Findings

Transfinite asymptotic dimension can be omega+1.

Counterexample to the omega conjecture.

Both dimensions are omega+1 in the constructed space.

Abstract

We construct a metric space whose transfinite asymptotic dimension and complementary-finite asymptotic dimension are both omega+1, where omega is the smallest infinite ordinal number. Therefore, we prove that the omega conjecture is not true.

Equations

$$\text{diam}U=\text{sup}\{d(x,y)|x,y\in U\}\text{ and }d(U,V)=\text{inf}\{d(x,y)|x\in U,y\in V\}.$$

$$\text{diam}\mathcal{U}\stackrel{△}{=}\text{sup}\{\text{diam}U|U\in \mathcal{U}\}\le R.$$

$$d(U,V)\ge r\text{for every}U,V\in \mathcal{U}\text{ with }U\ne V.$$

$$M^{\sigma }=\{\tau \in Fin\mathbb{N}|\tau \cup \sigma \in M\text{ and }\tau \cap \sigma =\varnothing \}.$$

$$\text{Ord}M=0$$

$$\text{Ord}M\le \alpha$$

$$\text{Ord}M=\alpha$$

$$\text{Ord}M=\infty$$

$$\begin{array}{rl}A(X,d)=\{\sigma \in Fin\mathbb{N}|& \text{ there are no uniformly bounded families }{\mathcal{U}}_i\text{ for }i\in \sigma \\ & \text{ such that each }{\mathcal{U}}_i\text{ is }i\text{-disjoint and }\bigcup _{i\in \sigma }{\mathcal{U}}_i\text{ covers }X\}.\end{array}$$

$$d(a,b)=\text{max}\{d_{\text{max}}(a^{\prime },b^{\prime }),c\},$$

$$\widetilde{U_i}=N_{m+1}(\bigcup {\mathcal{U}}_i)\text{ and }\widetilde{U_0}=\overline{[0,B+n{]}^{m+1}\setminus \bigcup _{1\le i\le m}\widetilde{U_i}}$$

$${\mathcal{V}}_1^{(n)}=\{[in+2jk,in+(2j+1)k]|i\in \mathbb{Z},j=1,2,...,m-1\},$$

$${\mathcal{V}}_2^{(n)}=\{[in+(2j-1)k,in+2jk]|i\in \mathbb{Z},j=1,2,...,m-1\}\cup \{[in+2mk-k,in+2mk+2k]|i\in \mathbb{Z}\}.$$

$${\mathcal{V}}_1^{(n)}=\{[in+2jk,in+(2j+1)k]|i\in \mathbb{Z},j=1,2,...,m\},$$

$${\mathcal{V}}_2^{(n)}=\{[in+(2j-1)k,in+2jk]|i\in \mathbb{Z},j=1,2,...,m\}\cup \{[in+2mk+k,in+2mk+3k]|i\in \mathbb{Z}\}.$$

$${\mathcal{W}}_0^{(n)}=\{(\prod _{i=1}^n{V}_i)\cap X_n|V_i\in {\mathcal{V}}_0^{(n)}\}$$

$${\mathcal{W}}_1^{(n)}=\{\prod _{i=1}^{j-1}\{n{n}_i\}\times V_j\times \prod _{i=j+1}^n\{n{n}_i\}|n_i\in \mathbb{Z},V_j\in {\mathcal{V}}_1^{(n)},j=2,3,\cdots ,n\}\cup \{V_1\times \prod _{i=2}^n\{n{n}_i\}|n_i\in \mathbb{Z},V_1\in {\mathcal{V}}_1^{(n)}\}$$

$${\mathcal{W}}_2^{(n)}=\{\prod _{i=1}^{j-1}\{n{n}_i\}\times V_j\times \prod _{i=j+1}^n\{n{n}_i\}|V_j\in {\mathcal{V}}_2^{(n)},j=2,3,\cdots ,n\}\cup \{V_1\times \prod _{i=2}^n\{n{n}_i\}|n_i\in \mathbb{Z},V_1\in {\mathcal{V}}_2^{(n)}\}$$

$${\mathcal{U}}_0^{(n)}={\mathcal{W}}_0^{(n)}\bigcup {\mathcal{W}}_1^{(n)},{\mathcal{U}}_1^{(n)}={\mathcal{W}}_2^{(n)}.$$

$${\mathcal{U}}_0=\bigcup _{n\ge 6k}{\mathcal{U}}_0^{(n)},{\mathcal{U}}_1=\bigcup _{n\ge 6k}{\mathcal{U}}_1^{(n)}.$$

$$\text{coasdim}(X_{\omega +1}\setminus \bigcup ({\mathcal{U}}_0\cup {\mathcal{U}}_1))=\text{asdim}(X_{\omega +1}\setminus \underset{n\ge 6k}{⨆}{X}_n)=\text{asdim}(\underset{n=1}{\overset{6k-1}{⨆}}{X}_n)<\omega ,$$

Full text

A metric space with its transfinite asymptotic dimension $\omega+1$

Yan Wu∗  Jingming Zhu*∗∗* 111 College of Mathematics Physics and Information Engineering, Jiaxing University, Jiaxing , 314001, P.R.China. ∗ E-mail: yanwu@mail.zjxu.edu.cn ${}^{\ast}\ast$ E-mail: jingmingzhu@mail.zjxu.edu.cn

Abstract. We construct a metric space whose transfinite asymptotic dimension and complementary-finite asymptotic dimension are both $\omega+1$, where $\omega$ is the smallest infinite ordinal number. Therefore, we prove that the omega conjecture is not true.

**Keywords ** Asymptotic dimension, Transfinite asymptotic dimension, Complementary-finite asymptotic dimension;

222 This research was supported by the National Natural Science Foundation of China under Grant (No.11871342,11301224,11326104,11401256,11501249)

1 Introduction

In coarse geometry, asymptotic dimension of a metric space is an important concept which was defined by Gromov for studying asymptotic invariants of discrete groups [1]. This dimension can be considered as an asymptotic analogue of the Lebesgue covering dimension. T. Radul defined the trasfinite asymptotic dimension (trasdim) which can be viewed as a transfinite extension for asymptotic dimension and gave examples of metric spaces with trasdim$=\omega$ and with trasdim$=\infty$, where $\omega$ is the smallest infinite ordinal number (see [2]). But whether there is a metric space $X$ with $\omega<$trasdim$(X)<\infty$ is still unknown so far. M. Satkiewicz stated ”omega conjecture”(see [3]). That is, $\omega\leq$trasdim$(X)<\infty$ implies trasdim$(X)=\omega$. In this paper, we prove that the ”omega conjecture” is not true by constructing a metric space $X_{\omega+1}$ with trasdim$(X_{\omega+1})=\omega+1$.

In the paper [4], we introduced another approach to classify the metric spaces with infinite asymptotic dimension, which is called complementary-finite asymptotic dimension (coasdim), and gave metric spaces $X_{k}$ with coasdim$(X_{k})=\omega+k$ and trasdim$(X_{k})=\omega$, where $k\in\mathbb{N}$. The metric space $X_{\omega+1}$ constructed in this paper is the first metric space $Y$ with trasdim$(Y)>\omega$ and trasdim$(Y)=$coasdim$(Y)$.

The paper is organized as follows: In Section 2, we recall some definitions and properties of transfinite asymptotic dimension. In Section 3, we introduce a concrete metric space $X_{\omega+1}$ and prove that transfinite asymptotic dimension and complementary-finite asymptotic dimension of $X_{\omega+1}$ are both $\omega+1$.

2 Preliminaries

Our terminology concerning the asymptotic dimension follows from [5] and for undefined terminology we refer to [2]. Let $(X,d)$ be a metric space and $U,V\subseteq X$, let

[TABLE]

Let $R>0$ and $\mathcal{U}$ be a family of subsets of $X$. $\mathcal{U}$ is said to be $R$-bounded if

[TABLE]

In this case, $\mathcal{U}$ is said to be uniformly bounded. Let $r>0$, a family $\mathcal{U}$ is said to be* $r$-disjoint* if

[TABLE]

In this paper, we denote $\bigcup\{U~{}|~{}U\in\mathcal{U}\}$ by $\bigcup\mathcal{U}$ and denote $\{U~{}|~{}U\in\mathcal{U}_{1}\text{~{}or~{}}~{}U\in\mathcal{U}_{2}\}$ by $\mathcal{U}_{1}\cup\mathcal{U}_{2}$.

**Definition 2.1. ** A metric space $X$ is said to have finite asymptotic dimension if there is an $n\in\mathbb{N}$, such that for every $r>0$, there exists a sequence of uniformly bounded families $\{\mathcal{U}_{i}\}_{i=0}^{n}$ of subsets of $X$ such that the family $\bigcup_{i=0}^{n}\mathcal{U}_{i}$ covers $X$ and each $\mathcal{U}_{i}$ is $r$-disjoint for $i=0,1,\cdots,n$. In this case, we say that the asymptotic dimension of $X$ less than or equal to $n$, which is denoted by asdim$X\leq n$. We say that asdim$X=n$ if asdim$X\leq n$ and asdim$X\leq n-1$ is not true.

T. Radul generalized asymptotic dimension of a metric space $X$ to transfinite asymptotic dimension which is denoted by trasdim$(X)$ (see [2]). Let $Fin\mathbb{N}$ denote the collection of all finite, nonempty subsets of $\mathbb{N}$ and let $M\subseteq Fin\mathbb{N}$. For $\sigma\in\{\varnothing\}\bigcup Fin\mathbb{N}$, let

[TABLE]

Let $M^{a}$ abbreviate $M^{\{a\}}$ for $a\in\mathbb{N}$. Define the ordinal number Ord$M$ inductively as follows:

[TABLE]

Given a metric space $(X,d)$, define the following collection:

[TABLE]

The transfinite asymptotic dimension of $X$ is defined as trasdim$X$=Ord$A(X,d)$.

3 Main result

Let $A$ be a subset of a metric space $X$ and $\epsilon>0$, we denote $\{x\in X|d(x,A)<\epsilon\}$ by $N_{\epsilon}(A)$. For the unit cube $I^{n}$, facets is the subset with the $i$-th coordinate $x_{i}=0$ or $x_{i}=1$ for some integer $i$ $(1\leq i\leq n)$.

To prove the main result, we will use another version of Lebesgue theorem:

Theorem 3.1**.**

(see [8], Theorem 4.3)

Let the unit cube $I^{n}$ be covered by a family of closed sets $\{X_{i}\}_{1\leq i\leq n}$. Then some connected component of $X_{i}$ intersects both the corresponding opposite facets $A_{i}$ and $B_{i}$.

Lemma 3.1**.**

(see [4], Proposition 2.1)

Given a metric space $X$, let $l\in\mathbb{N}\cup\{0\}$, then the following conditions are equivalent:

• •

(1) trasdim$(X)\leq\omega+l$;

• •

(2) For every $k\in\mathbb{N}$, there exists $m=m(k)\in\mathbb{N}$ such that for every $n\in\mathbb{N}$, there are uniformly bounded families $\mathcal{U}_{-l},\mathcal{U}_{-l+1},\cdots,\mathcal{U}_{m}$ satisfying $\mathcal{U}_{i}$ is $k$-disjoint for $i=-l,\cdots,0$, $\mathcal{U}_{j}$ is $n$-disjoint for $j=1,2,\cdots,m$ and $\bigcup_{i=-l}^{m}\mathcal{U}_{i}$ covers $X$.

Let $X_{1}=\mathbb{R}$, $X_{2}=(\mathbb{R}\times 2\mathbb{Z})\bigcup(2\mathbb{Z}\times\mathbb{R})$, $X_{3}=(\mathbb{R}\times 3\mathbb{Z}\times 3\mathbb{Z})\bigcup(3\mathbb{Z}\times\mathbb{R}\times 3\mathbb{Z})\bigcup(3\mathbb{Z}\times 3\mathbb{Z}\times\mathbb{R})$,… For $a=(a_{1},a_{2},...,a_{l})\in X_{l}$ and $b=(b_{1},b_{2},...,b_{k})\in X_{k}$ with $l\leq k$, let $a^{\prime}=(a_{1},...,a_{l},0,...,0)$,$b^{\prime}=(b_{1},...,b_{k},0,...,0)\in\bigoplus\mathbb{R}$ and put $c=0$ if $l=k$ and $c=l+(l+1)+...+(k-1)$ if $l<k$. Define a metric $d$ on $\bigsqcup X_{n}$ by

[TABLE]

where $d_{\text{max}}$ is the maximum metric in $\bigoplus\mathbb{R}.$ Let $X_{\omega+1}=(\bigsqcup X_{n},d)$.

Proposition 3.1**.**

trasdim$(X_{\omega+1})\leq\omega$ is not true.

Proof.

Suppose that trasdim$(X_{\omega+1})\leq\omega$. Then by Lemma 3.1, for every $k\in\mathbb{N}$, there exists $m=m(k)\in\mathbb{N}$ such that for every $n\in\mathbb{N}$, there exist $B=B(k,n)>0$ and $B$-bounded families $\mathcal{V}_{0},\mathcal{V}_{1},\cdots,\mathcal{V}_{m}$ such that $\mathcal{V}_{0}$ is $k$-disjoint, $\mathcal{V}_{j}$ is $(n+4m+4)$-disjoint for $j=1,2,\cdots,m$ and $\bigcup_{i=0}^{m}\mathcal{V}_{i}$ covers $X$. Let $\mathcal{U}_{i}=\{U\cap X_{m+1}\cap[0,B+n]^{m+1}|U\in\mathcal{V}_{i}\}$ for $i=0,1,2,\cdots,m$.

Without loss of generality, we assume that $n\in\mathbb{N}$, $n>4m+4$ and $\frac{B+n}{m+1}=p\in\mathbb{N}$. Then $[0,B+n]^{m+1}$ contains $(m+1)$-dimensional closed cubes $V_{1},V_{2},...,V_{p^{m+1}}$, where $V_{i}=\prod_{j=1}^{m+1}[(m+1)n_{j}^{(i)},(m+1)(n_{j}^{(i)}+1)]$, whose edges are contained in $X_{m+1}$ for $i=1,2,\cdots,p^{m+1}$.

Let $\mathcal{V}=\{V_{j}\}_{1\leq j\leq p^{m+1}}$ and $A_{i}$, $B_{i}$ be the pair of two opposite facets of $[0,B+n]^{m+1}$ for $i=1,2,\cdots,m+1$.

Let $\widetilde{\mathcal{U}_{i}}=\{V\in\mathcal{V}|V\cap U\neq\emptyset\text{ for some }U\in\mathcal{U}_{i}\}$ for $1\leq i\leq m$.

Then, by the definition, $\widetilde{\mathcal{U}_{i}}$ is $(n+4m+4)$-disjoint and $(B+2m+2)$-bounded subset family in $[0,B+n]^{m+1}$ for $1\leq i\leq m$.

Let

[TABLE]

Then $\{\widetilde{U}_{i}\}_{0\leq i\leq m}$ is a closed cover of $[0,B+n]^{m+1}$ with covering multiplicity at most $m+1$. By the Theorem 3.1, $\exists$ at least one $\widetilde{U}_{i}$ which intersects both the corresponding opposite facets $A_{i}$ and $B_{i}$. But every connected component in $\widetilde{U}_{i}$ for $1\leq i\leq m$ is $(B+2m+2)$-bounded, so $\widetilde{U_{0}}\neq\emptyset$ and there is a path $\alpha:I\to\widetilde{U}_{0}$ such that $\alpha(0)$ and $\alpha(1)$ belong to opposite facets $A_{i}$ and $B_{i}$ respectively.

Let $\mathcal{W}=\{V\in\mathcal{V}|V\cap\alpha(I)\neq\emptyset\}$. Then $\bigcup\mathcal{W}$ is a connected subset in $[0,B+n]^{m+1}$ which intersects both the corresponding opposite facets $A_{i}$ and $B_{i}$. But $\alpha(I)\subset\widetilde{U}_{0}$ implies that any $V\in\mathcal{V}$ can not intersect both $\alpha(I)$ and $\bigcup{\mathcal{U}_{i}}$ for any $1\leq i\leq m$. Then $(\bigcup\mathcal{W})\cap(\bigcup\mathcal{U}_{i})=\emptyset$ for $1\leq i\leq m$. So $\exists$ a path $\beta:I\to[0,B+n]^{m+1}$ such that $\beta(I)$ is a subset of the 1-dimensional skeleton of the $V$’s in $\mathcal{W}$(and hence a subset of $X_{m+1}$) and intersects both the corresponding opposite facets $A_{i}$ and $B_{i}$. Since $\bigcup\mathcal{W}\cap(\bigcup\mathcal{U}_{i})=\emptyset$ for $1\leq i\leq m$ and $\beta(I)\subset\bigcup\mathcal{W}$, $\beta(I)\cap(\bigcup\mathcal{U}_{i})=\emptyset$ for $1\leq i\leq m$. So $\beta(I)\subset(\bigcup\mathcal{U}_{0})\cap X_{m+1}(=\bigcup\mathcal{V}_{0})$ which is a contradiction with the fact that $\mathcal{V}_{0}$ is $B$-bounded and $k$-disjoint.

∎

**Definition 3.2. ** Every ordinal number $\gamma$ can be represented as $\gamma=\lambda(\gamma)+n(\gamma)$, where $\lambda(\gamma)$ is the limit ordinal or [math] and $n(\gamma)\in\mathbb{N}\cup\{0\}$. Let $X$ be a metric space, we define complementary-finite asymptotic dimension coasdim$(X)$ inductively as follows:

• •

coasdim$(X)=-1$ iff $X=\emptyset$,

• •

coasdim$(X)\leq\gamma$ iff for every $r>0$ there exist $r$-disjoint uniformly bounded families $\mathcal{U}_{0},...,\mathcal{U}_{n(\gamma)}$ of subsets of $X$ such that coasdim$(X\setminus\bigcup(\bigcup_{i=0}^{n(\gamma)}\mathcal{U}_{i}))<\lambda(\gamma)$,

• •

coasdim$(X)=\gamma$ iff coasdim$(X)\leq\gamma$ and for every $\beta<\gamma$, coasdim$(X)\leq\beta$ is not true.

• •

coasdim$(X)=\infty$ iff for every ordinal number $\gamma$, coasdim$(X)\leq\gamma$ is not true.

Remark 3.1.

• •

It is easy to see that for every $n\in\mathbb{N}\cup\{0\}$, coasdim$(X)\leq n$ if and only if asdim$(X)\leq n$.

• •

To simplify, we will abuse the notation a little bit. In the following, asdim$(X)<\omega$ means that asdim$(X)\leq n$ for some $n\in\mathbb{N}$.

Lemma 3.2**.**

(see [4], Theorem 3.2)

Let $X$ be a metric space, if coasdim$(X)\leq\omega+k$ for some $k\in\mathbb{N}$, then trasdim$(X)\leq\omega+k$. Especially, coasdim$(X)=\omega$ implies trasdim$(X)=\omega$.

Proposition 3.2**.**

trasdim$(X_{\omega+1})\leq\omega+1$.

Proof.

By the Lemma 3.2, it suffices to show that coasdim$(X_{\omega+1})\leq\omega+1$. i.e., for every $k>0$ there exist $k$-disjoint uniformly bounded families $\mathcal{U}_{0},\mathcal{U}_{1}$ of subsets of $X_{\omega+1}$ such that coasdim$(X_{\omega+1}\setminus\bigcup(\mathcal{U}_{0}\cup\mathcal{U}_{1}))<\omega$. Let $n\in\mathbb{N}$ and $n\geq 6k$, let $\mathcal{V}_{0}^{(n)}=\{[in-k,in+k]~{}|~{}i\in\mathbb{Z}\}.$

• •

If $[\frac{n-2k}{k}]=2m$, then let

[TABLE]

[TABLE]

• •

If $[\frac{n-2k}{k}]=2m+1$, then let

[TABLE]

[TABLE]

Note that $\mathcal{V}_{0}^{(n)}\cup\mathcal{V}_{1}^{(n)}$,$\mathcal{V}_{2}^{(n)}$ are $k$-disjoint, $3k$-bounded and $\mathcal{V}_{0}^{(n)}\cup\mathcal{V}_{1}^{(n)}\cup\mathcal{V}_{2}^{(n)}$ covers $\mathbb{R}$. Let

[TABLE]

[TABLE]

[TABLE]

It is easy to see that $\mathcal{W}_{i}^{(n)}$ are $k$-disjoint and $3k$-bounded for $i=0,1,2$. Let

[TABLE]

Since $\mathcal{V}_{0}^{(n)}\cup\mathcal{V}_{1}^{(n)}$ is $k$-disjoint and $3k$-bounded, $\mathcal{U}_{0}^{(n)}$ is $k$-disjoint and $3k$-bounded. For every $x\in X_{n}$, without loss of generality, we assume that $x=(x_{1},...,x_{n})\in\mathbb{R}\times n\mathbb{Z}\times...\times n\mathbb{Z}$. Then $x$ is in one of the following cases.

• •

$x_{i}\in n\mathbb{Z}\text{~{}for~{}}i=2,3,\cdots$ and $x_{1}\in[ni-k,ni+k]$ for some $i\in\mathbb{Z}$, it is easy to see that $x\in\bigcup\mathcal{W}_{0}^{(n)}$.

• •

$x_{i}\in n\mathbb{Z}\text{~{}for~{}}i=2,3,\cdots$ and $x_{1}\in V$ for some $V\in\mathcal{V}_{1}^{(n)}$, it is easy to see that $x\in\bigcup\mathcal{W}_{1}^{(n)}$.

• •

$x_{i}\in n\mathbb{Z}\text{~{}for~{}}i=2,3,\cdots$ and $x_{1}\in V$ for some $V\in\mathcal{V}_{2}^{(n)}$, it is easy to see that $x\in\bigcup\mathcal{W}_{2}^{(n)}$.

So $\mathcal{U}_{0}^{(n)}\cup\mathcal{U}_{1}^{(n)}$ covers $X_{n}$. Let

[TABLE]

Since $d(X_{i},X_{j})\geq n$ when $i,j>n$ and $i\neq j$, then $\mathcal{U}_{0},\mathcal{U}_{1}$ are $k$-disjoint , $3k$-bounded and $\mathcal{U}_{0}\cup\mathcal{U}_{1}$ covers $\bigsqcup_{n\geq 6k}X_{n}$. It follows that

[TABLE]

which implies coasdim$(X_{\omega+1})\leq\omega+1$. $\square$

∎

Remark 3.2.

• •

By the Proposition 3.1 and the Proposition 3.2, we can obtain that trasdim$(X_{\omega+1})=\omega+1$.

• •

By the Proposition 3.1 and the Lemma 3.2, we can obtain that coasdim$(X_{\omega+1})\leq\omega$ is not true. Moreover, we obtain that coasdim$(X_{\omega+1})=\omega+1$ by the Proposition 3.2.