A note on free vector balleans

TL;DR

This paper introduces the concept of free vector balleans, establishing their unique existence for any given ballean and linking their normality to the metrizability of the underlying space.

Contribution

It defines free vector balleans, proves their uniqueness, and characterizes their normality in terms of metrizability.

Findings

Existence and uniqueness of free vector balleans for any ballean.

Normality of free vector balleans is equivalent to metrizability.

Described the coarse structure of free vector balleans.

Abstract

A vector balleans is a vector space over $\mathbb{R}$ endowed with a coarse structure in such a way that the vector operations are coarse mappings. We prove that, for every ballean $(X, \mathcal{E})$, there exists the unique free vector ballean $\mathbb{V}(X, \mathcal{E})$ and describe the coarse structure of $\mathbb{V}(X, \mathcal{E})$. It is shown that normality of $\mathbb{V}(X, \mathcal{E})$ is equivalent to metrizability of $(X, \mathcal{E})$.

Equations

$$\mathcal{V}\times \mathcal{V}⟶\mathcal{V},(x,y)⟼x+yand\mathbb{R}\times \mathcal{V},(\lambda ,x)⟼\lambda x$$

$$\{{\mathcal{S}}_{n,\epsilon }+[-n,n]z:\epsilon \in \mathcal{E},n\in \mathbb{N}\}$$

Full text

Abstract. A vector balleans is a vector space over $\mathbb{R}$ endowed with a coarse structure in such a way that the vector operations are coarse mappings. We prove that, for every ballean $(X,\mathcal{E})$, there exists the unique free vector ballean $\mathbb{V}(X,\mathcal{E})$ and describe the coarse structure of $\mathbb{V}(X,\mathcal{E})$. It is shown that normality of $\mathbb{V}(X,\mathcal{E})$ is equivalent to metrizability of $(X,\mathcal{E})$.

**MSC: ** 46A17, 54E35.

Keywords: coarse structure, ballean, vector ballean, free vector ballean.

1. Introduction

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

• •

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

• •

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

• •

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

• •

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

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

The pair $(X,\mathcal{E})$ is called a coarse space [11] or a ballean [8], [10].

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 $Y\subseteq\varepsilon[x]$ for some $x\in X$ and $\varepsilon\in\mathcal{E}$.

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

Every metric $d$ on a set $X$ defines the *metric ballean * $(X,\mathcal{E}_{d})$, where $\mathcal{E}_{d}$ has the base $\{\{(x,y):d(x,y)<r\}:r\in\mathbb{R}^{+}\}$. We say that a ballean $(X,\mathcal{E})$ is metrizable if there exists a metric $d$ on $X$ such that $\mathcal{E}=\mathcal{E}_{d}$. In what follows, we consider $\mathbb{R}$ as a ballean defined by the metric $d(x,y)=|x-y|$.

Given two balleans $(X_{1},\mathcal{E}_{1})$, $(X_{2},\mathcal{E}_{2})$, we defines the product $(X_{1}\times X_{2},\mathcal{E})$, where $\mathcal{E}$ has the base $\mathcal{E}_{1}\times\mathcal{E}_{2}$.

Let $\mathcal{V}$ be a vector space over $\mathbb{R}$ and let $\mathcal{E}$ be a coarse structure on $\mathcal{V}$. Following [5], we say that $(\mathcal{V},\mathcal{E})$ is a vector ballean if the operation

[TABLE]

are coarse.

A family $\mathcal{I}$ of subsets of $\mathcal{V}$ is called a vector ideal if

(1)\ \ if $U,U^{\prime}\in\mathcal{I}$ and $W\subseteq U$ then $U\cup U^{\prime}\in\mathcal{I}$ and $W\in\mathcal{I}$;

(2)\ \ for every $U\in\mathcal{I}$, $U+U\in\mathcal{I}$;

(3)\ \ for any $U\in\mathcal{I}$ and $\lambda\in\mathbb{R}^{+}$, $[-\lambda,\lambda]\ U\in\mathcal{I}$, where $[-\lambda,\lambda]\ U=\cup\ \{\lambda^{\prime}\ U:\lambda^{\prime}\in[-\lambda,\lambda]\}$;

(4)\ \ $\cup\ \mathcal{I}=\mathcal{V}$.

A family $\mathcal{I}^{\prime}\subseteq\mathcal{I}$ is called a base for $\mathcal{I}$ if, for each $U\in\mathcal{I}$, there is $U^{\prime}\in\mathcal{I}^{\prime}$ such that $U\subseteq U^{\prime}$.

If $(\mathcal{V},\mathcal{E})$ is a vector ballean then the family $\mathcal{I}$ of all bounded subsets of $\mathcal{V}$ is a vector ideal. On the other hand, every vector ideal $\mathcal{I}$ on $\mathcal{V}$ defines the vector ballean $(\mathcal{V},\mathcal{E})$, where $\mathcal{E}$ is a coarse structure with the base $\{\{(x,y):x-y\in U\}:U\in\mathcal{I}\}$. Thus, we have got a bijective correspondence between vector balleans on $\mathcal{V}$ and vector ideas. Following this correspondence, we write $(\mathcal{V},\mathcal{I})$ in place of $(\mathcal{V},\mathcal{E})$.

Let $(\mathcal{V},\mathcal{I})$, $(\mathcal{V}^{\prime},\mathcal{I}^{\prime})$ be vector balleans. We note that a linear mapping $f:\mathcal{V}\longrightarrow\mathcal{V}^{\prime}$ is coarse if and only if $f(U)\in\mathcal{I}^{\prime}$ for each $U\in\mathcal{I}$.

In Section 2, we show that, for every baleen $(X,\mathcal{E})$, there exists the unique vector ideal $\mathcal{I}_{(X,\mathcal{E})}$ on the vector space $\mathbb{V}(X)$ with the basis $X$ such that

• •

$(X,\mathcal{E})$ is a subballean of $(\mathbb{V}(X),\mathcal{I}_{(X,\mathcal{E})})$;

• •

for every vector ballean $(\mathcal{V},\mathcal{I})$, every coarse mapping $(X,\mathcal{E})\longrightarrow(\mathcal{V},\mathcal{I})$ gives rise to the unique coarse linear mapping $(\mathbb{V}(X),\ \mathcal{I}_{(X,\mathcal{E})})\longrightarrow(\mathcal{V},\mathcal{I})$ .

We denote $\mathbb{V}(X,\mathcal{E})=(\mathbb{V}(X),\mathcal{I}_{(X,\mathcal{E})})$ and say that $\mathcal{I}_{(X,\mathcal{E})}$ and $\mathbb{V}(X,\mathcal{E})$ are free vector ideal and *free vector ballean * over $(X,\mathcal{E})$.

Free vector balleans can be considered as counterparts of free vector spaces studied in many papers, for examples, [1], [3], [4]. It should be mentioned that the free activity in topological algebra was initiated by the famous paper of Markov on free topological groups [6]. For free coarse groups see [9].

2. Construction

Given a ballean $(X,\mathcal{E})$, we consider $X$ as the basis of the vector space $\mathbb{V}(X)$. For each $\varepsilon\in\mathcal{E}$ and $n\in\mathbb{N}$, we set $\mathcal{D}_{\varepsilon}=\{x-y:(x,y)\in\varepsilon\}$ and denote by $\mathcal{S}_{n,\varepsilon}$ the sum of $n$ copies of $[-n,n]\mathcal{D}_{\varepsilon}$.

**Theorem 1. ** Let $(X,\mathcal{E})$ be a ballean and let $z\in X$. Then the family

[TABLE]

is a base of the free vector ideal $\mathcal{I}_{(X,\mathcal{E})}$.

Proof. We denote by $\mathcal{I}$ the family of all subsets $U$ of $\mathbb{V}(X)$ such that $U$ is contained in some $\mathcal{S}_{n,\varepsilon}+[-n,n]z$. Clearly, $\mathcal{I}$ satisfies (1), (2), (3) from the definition of a vector ideal. To see that $\cup\mathcal{I}=\mathbb{V}(X)$, we take an arbitrary $y\in X$ and choose $\varepsilon\in\mathcal{E}$ so that $(y,z)\in\varepsilon$. Then $y=(y-z)+z\in\mathcal{D}_{\mathcal{E}}+z$. In view of (2), (3), we conclude that $\cup\mathcal{I}=\mathbb{V}(X)$.

To show that $(X,\mathcal{E})$ is a subballean of $(\mathbb{V}(X),\ \mathcal{I})$, we denote by $\mathcal{E}^{\prime}$ the coarse structure of the ballean $(\mathbb{V}(X),\ \mathcal{I})$. Since $\mathcal{D}_{\mathcal{E}}\in\mathcal{I}$ for each $\varepsilon\in\mathcal{E}$, $\mathcal{E}=\mathcal{E}^{-1}$ we have $\mathcal{E}\subseteq\mathcal{E}^{\prime}|_{X}$. To verify the inclusion $\mathcal{E}^{\prime}\mid_{X}\subseteq\mathcal{E}$ we take $x,y\in X$, assume that $x-y\in\mathcal{S}_{n,\mathcal{E}}+[-n,n]z$ and show that $(x,y)\in\varepsilon^{n}$.

We write $x-y=\lambda_{1}(x_{1}-y_{1})+\ldots+\lambda_{n}(x_{n}-y_{n})+\lambda_{n+1}z,(x_{i},y_{i})\in\varepsilon$, $\lambda_{i}\in[-n,n]$. Since $(\lambda_{1}-\lambda_{1})+\ldots+(\lambda_{n}-\lambda_{n})+\lambda_{n+1}=0$, we have $\lambda_{n+1}=0$ so $x-y=\lambda_{1}(x_{1}-y_{1})+\ldots+\lambda_{n}(x_{n}-y_{n})$. If $(x_{1},y_{1}),\ldots,x_{n},y_{n})\in\{x,y\}$ then the statement is evident because $(x_{i},y_{i})\in\varepsilon$. Assume that there exists $a\in\{x_{1},y_{1},\ldots,x_{n},y_{n}\}$ such that $a\notin\{x,y\}$. We take all items $\lambda_{i}(x_{i}-y_{i})$, $i\in I$ such that $a\in\{x_{i},y_{i}\}$, and denote by $s$ the sum of all these items. The coefficient before $a$ in the canonical decomposition of $s$ by the basis $X$ must be [math]. We take $\lambda_{k}(x_{k}-y_{k})$, $k\in I$. If $x_{k}=a$ or $y_{k}=a$ then we replace each $a$ in $\lambda_{i}(x_{i}-y_{i})$, $i\in I$, to $y_{k}$ or $x_{k}$ respectively. Then we get $x-y=\lambda_{1}(x_{1}^{\prime}-y_{1}^{\prime})+\ldots+\lambda_{n}(x_{n}^{\prime}-y_{n}^{\prime})$, $a\notin\{x_{1}^{\prime},y_{1}^{\prime},\ldots,x_{n}^{\prime}-y_{n}^{\prime}\}$ and $(x_{i}^{\prime},y_{i}^{\prime})\in\varepsilon^{2}$. Repeating this trick, we run into the case $x_{1},y_{1},\ldots,x_{n},y_{n}\in\{x,y\}$ and $(x_{i},y_{i})\in\varepsilon^{n}$.

To conclude the proof, we observe that $\mathcal{I}$ is the minimal vector ideal on $\mathbb{V}(X)$ such that $(X,\mathcal{E})$ is a subballean of $(\mathbb{V}(X),\mathcal{I})$. If $(\mathcal{V},\mathcal{I}^{\prime})$ is a ballean and $f:(X,\mathcal{E})\longrightarrow(\mathcal{V},\mathcal{I}^{\prime})$ is a coarse mapping then the linear extension $h:(\mathbb{V}(X),\mathcal{I})\longrightarrow$ $(\mathcal{V},\mathcal{I}^{\prime})$ of $f$ is coarse because $h^{-1}(\mathcal{I}^{\prime})$ is a vector ideal on $\mathbb{V}(X)$ and $\mathcal{I}\subseteq h^{-1}(\mathcal{I}^{\prime})$. $\ \ \ \Box$

3. Metrizability and normality

**Theorem 2. ** A ballean $\mathbb{V}(X,\mathcal{E})$ is metrizable if and only if $(X,\mathcal{E})$ is metrizable.

Proof. By [10, Theorem 2.1.1], $(X,\mathcal{E})$ is metrizable if and only if $\mathcal{E}$ has a countable base. Apply Theorem 1. $\ \ \ \Box$

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

Two subset $A,B$ of $X$ are called asymptotically disjoint (asymptotically separated) if, for every $\varepsilon\in\mathcal{E}$, the intersection $\varepsilon[A]\cap\varepsilon[B]$ is bounded ($A,B$ have disjoint asymptotic neighbourhoods).

A ballean $(X,\mathcal{E})$ is called normal [7] if any two asymptotically disjoint subsets of $X$ are asymptotically separated. Every metrizable ballean is normal.

Given an arbitrary ballean $(X,\mathcal{E})$, the family $\mathcal{B}_{X}$ of all bounded subsets of $X$ is called a bornology of $(X,\mathcal{E})$. A subfamily $\mathcal{B}^{\prime}\subseteq\mathcal{B}_{X}$ is called a base of $\mathcal{B}_{X}$ if, for every $B\in\mathcal{B}_{X}$, there exists $B^{\prime}\in\mathcal{B}_{X}$ such that $B\subseteq B^{\prime}$. The minimal cardinality of bases of $\mathcal{B}_{X}$ is denoted by $cof\ \mathcal{B}_{X}$.

**Theorem 3. ** For every ballean $(X,\mathcal{E})$, the free vector ballean $\mathbb{V}(X,\mathcal{E})$ is normal if and only if $\mathbb{V}(X,\mathcal{E})$ is metrizable.

Proof. For $|X|=1$, the statement is evident. Let $|X|>1$, $a\in X$, $L=\mathbb{R}a$, $Y=X\setminus\{a\}$. Applying Theorem 1, we conclude that the canonical isomorphism between $\mathbb{V}(X,\mathcal{E})$ and $L\times\mathbb{V}(Y,\mathcal{E}_{Y})$ is an asymorphism. If $\mathbb{V}(X,\mathcal{E})$ is normal then, by Theorem 1.4 from [2], $cof\ \mathcal{B}_{L}=cof\ \mathcal{B}_{Z}$, where $Z=\mathbb{V}(Y,\mathcal{E}_{Y})$. Since $cof\ \mathcal{B}_{L}=\aleph_{0}$, $\mathcal{B}_{Z}$ has a countable base. To conclude the proof, it suffices to note that $\mathcal{B}_{Z}$ is the vector ideal such that $\mathbb{V}(Y,\mathcal{E}_{Y})=(\mathbb{V}(Y),\mathcal{I})$. Hence $\mathcal{I}$ has a countable base and $\mathbb{V}(Y,\mathcal{E}_{Y})$ is metrizable by Theorem 2.1.1. from [10]. $\ \ \ \Box$