Constructing balleans

TL;DR

This paper introduces new constructions of balleans, a type of coarse space, from families of balleans, and analyzes the compatible coarse structures on a set given a bornology.

Contribution

It presents three novel constructions of balleans—bornological products, bouquets, and combs—and studies the extremal coarse structures compatible with a given bornology.

Findings

Defined and analyzed bornological products, bouquets, and combs of balleans.

Characterized smallest and largest compatible coarse structures on a set.

Provided new methods for constructing and understanding balleans.

Abstract

A ballean is a set endowed with a coarse structure. We introduce and explore three constructions of balleans from a pregiven family of balleans: bornological products, bouquets and combs. We analyze the smallest and the largest coarse structures on a set $X$ compatible with a given bornology on $X$.

Equations

$$\big{\{}\big{(}(x_{\alpha})_{\alpha<\kappa},(y_{\alpha})_{\alpha<\kappa}\big{)}\in X\times X:\forall\alpha<\kappa\;\;(x_{\alpha},y_{\alpha})\in E_{\alpha}\big{\}},$$

$$E_1,\dots ,E_n\in \{F\cup F^{-1}\cup \{(x,y)\}\cup {△}_X:F\in \mathfrak{F},x,y\in X\}.$$

$$\mathrm{add}(\mathcal{B})=\min \{\mathcal{A}\subset \mathcal{B}:\bigcup \mathcal{A}\notin \mathcal{B}\},$$

$$\mathrm{cof}(\mathcal{B})=\min \{\mathcal{C}\subset \mathcal{B}:\forall B\in \mathcal{B}\exists C\in \mathcal{C}B\subset C\}\text{mbox}and$$

$$\mathrm{cov}(\mathcal{B})=\min \{|\mathcal{C}|:\mathcal{C}\subset \mathcal{B},\bigcup \mathcal{C}=X\}$$

$$\mathrm{add}({\mathcal{B}}_X)=\mathrm{cof}({\mathcal{B}}_X)=\mathrm{cof}({\mathcal{B}}_Y)=\mathrm{add}({\mathcal{B}}_Y).$$

$$X_{\mathcal{B}}=\big{\{}(x_{\alpha})_{\alpha\in A}\in\prod_{\alpha\in A}X_{\alpha}:\{\alpha\in A:x_{\alpha}\neq e_{\alpha}\}\in\mathcal{B}\big{\}},$$

$$\big{\{}\big{(}(x_{\alpha})_{\alpha\in A},(y_{\alpha})_{\alpha\in A}\big{)}\in X_{\mathcal{B}}\times X_{\mathcal{B}}:\forall\alpha\in B\;\;(x_{\alpha},y_{\alpha})\in E_{\alpha}\big{\}}$$

$$\bigvee_{\alpha\in A}X_{\alpha}:=\big{\{}(x_{\alpha})_{\alpha\in A}\in X_{\mathcal{B}}:|\{\alpha\in A:x_{\alpha}\neq e_{\alpha}\}|\leq 1\big{\}}$$

$$X\underset{\alpha \in A}{\perp \perp }{X}_{\alpha }:=(X\times \{e\})\cup \bigcup _{\alpha \in A}(\{\alpha \}\times X_{\alpha })$$

$${\mathcal{B}}_{X\times Y}=\{B\subset X\times Y:B\subset B_X\times B_Y\text{mbox}forsome{B}_X\in {\mathcal{B}}_X,B_Y\in {\mathcal{B}}_Y\}.$$

$$E[X\times \{y_0\}]\cap E[\{x_0\}\times Y]\subset E[B_{<\kappa }\times \{y_0\}]$$

$$D\times \{y\}\subset (X\times y\})\cap E[\{x_0\}\times Y]\subset (X\times Y_{\alpha })\cap V\subset (X\times Y_{\alpha })\setminus U\subset D_{\alpha }\times Y,$$

$$\mathrm{cof}({\mathcal{B}}_X)\le |\{D_{\alpha }{\}}_{\alpha \in \kappa }|\le \kappa =\mathrm{cov}({\mathcal{B}}_Y)<\mathrm{add}({\mathcal{B}}_X),$$

Full text

1. Introduction

Given a set $X$, a family $\mathcal{E}$ of subsets of $X\times X$ is called a coarse structure on $X$ if

• •

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

• •

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 $E^{\prime}\in\mathcal{E}$.

Elements $E\in\mathcal{E}$ of the coarse structure are called entourages on $X$.

For $x\in X$ and $E\in\mathcal{E}$ the set $E[x]:=\{y\in X:(x,y)\in\mathcal{E}\}$ is called the ball of radius $E$ centered at $x$. Since $E=\bigcup_{x\in X}\{x\}\times E[x]$, the entourage $E$ is uniquely determined by the family of balls $\{E[x]:x\in X\}$. A subfamily $\mathcal{B}\subset\mathcal{E}$ is called a base of the coarse structure $\mathcal{E}$ if each set $E\in\mathcal{E}$ is contained in some $B\in\mathcal{B}$.

The pair $(X,\mathcal{E})$ is called a coarse space [1] or a ballean [2], [3]. In [2] every base of a coarse structure, defined in terms of balls, is called a ball structure. We prefer the name balleans not only by the author’s rights but also because a coarse spaces sounds like some special type of topological spaces. In fact, balleans can be considered as non-topological antipodes of uniform topological spaces. Our compromise with [1] is in usage the name coarse structure in place of the ball structure.

In this paper, all balleans under consideration are supposed to be connected: for any $x,y\in X$, there is $E\in\mathcal{E}$ such $y\in E[x]$. A subset $Y\subseteq X$ is called bounded if $Y=E[x]$ for some $E\in\mathcal{E}$, and $x\in X$. The family $\mathcal{B}_{X}$ of all bounded subsets of $X$ is a bornology on $X$. We recall that a family $\mathcal{B}$ of subsets of a set $X$ is a bornology if $\mathcal{B}$ contains the family $[X]^{<\omega}$ of all finite subsets of $X$ and $\mathcal{B}$ is closed under finite unions and subsets. A bornology $\mathcal{B}$ on a set $X$ is called unbounded if $X\notin\mathcal{B}$.

Each subset $Y\subseteq X$ defines a subbalean $(Y,\mathcal{E}|_{Y})$ of $(X,\mathcal{E})$, where $\mathcal{E}|_{Y}=\{E\cap(Y\times Y):E\in\mathcal{E}\}$. A subbalean $(Y,\mathcal{E}|_{Y})$ is called large if there exists $E\in\mathcal{E}$ such that $X=E[Y]$, where $E[Y]=\bigcup_{y\in Y}E[y]$.

Let $(X,\mathcal{E})$, $(X^{\prime},\mathcal{E}^{\prime})$ be balleans. A mapping $f:X\to X^{\prime}$ is called *coarse (or macrouniform) * if, for every $E\in\mathcal{E}$, there exists $E^{\prime}\in\mathcal{E}$ 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 coarse then $f$ is called an asymorphism. If $(X,\mathcal{E})$ and $(X^{\prime},\mathcal{E}^{\prime})$ contains large asymorphic subballeans then they are called coarsely equivalent.

For coarse spaces $(X_{\alpha},\mathcal{E}_{\alpha})$, $\alpha<\kappa$, their product is the Cartesian product $X=\prod_{\alpha<\alpha}X_{\alpha}$ endowed with the coarse structure generated by the base consisting of the entourages

[TABLE]

where $(E_{\alpha})_{\alpha<\kappa}\in\prod_{\alpha<\kappa}\mathcal{E}_{\alpha}$.

A class $\mathfrak{M}$ of balleans is called a variety if $\mathfrak{M}$ is closed under formation of subballeans, coarse images and Cartesian products. For characterization of all varieties of balleans, see [4].

Given a family $\mathfrak{F}$ of subsets of $X\times X$, we denote by $\mathcal{E}$ the intersection of all coarse structures, containing each $F\cup\bigtriangleup_{X}$, $F\in\mathfrak{F}$, and say that $\mathcal{E}$ is generated by $\mathfrak{F}$. It is easy to see that $\mathcal{E}$ has a base of subsets of the form $E_{1}\circ E_{1}\circ\ldots\circ E_{n}$ , where

[TABLE]

By a pointed ballean we shall understand a ballean $(X,\mathcal{E})$ with a distinguished point $e_{*}\in X$.

2. Metrizability and normality

Every metric $d$ on a set $X$ defines the coarse structure $\mathcal{E}_{d}$ on $X$ with the base $\{\{(x,y):d(x,y)<n\}:n\in\mathbb{N}\}$. A ballean $(X,\mathcal{E})$ is called metrizable if there is a metric $d$ on such that $\mathcal{E}=\mathcal{E}_{d}$.

Theorem 1** ([5]).**

A ballean $(X,\mathcal{E})$ is metrizable if and only if $\mathcal{E}$ has a countable base.

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

Two subset $Y,Z$ of $X$ are called asymptotically disjoint (separated) if, for every $E\in\mathcal{E}$, $E[Y]\cap E[Z]$ is bounded ($Y$ and $Z$ have disjoint asymptotic neighbourhoods).

A ballean $(X,\mathcal{E})$ is called normal [6] if any two asymptotically disjoint subsets of $X$ are asymptotically separated. Every ballean $(X,\mathcal{E})$ with linearly ordered base of $\mathcal{E}$ is normal. In particular, every metrizable ballean is normal, see [6].

A function $f:X\to\mathbb{R}$ is called slowly oscillating if for any $E\in\mathcal{E}$ and $\varepsilon>0$, there exists a bounded subset $B$ of $X$ such that $\mathrm{diam}\,f(E[x])<\varepsilon$ for each $x\in X\setminus B$.

Theorem 2** ([6]).**

A ballean $(X,\mathcal{E})$ is normal if and only if, for any two disjoint and asymptotically disjoint subsets $Y,Z$ of $X$, there exists a slowly oscillating function $f:X\to[0,1]$ such that $f(Y)\subset\{0\}$ and $f(Z)\subset\{1\}$.

For any unbounded bornology $\mathcal{B}$ on a set $X$ the cardinals

[TABLE]

are called the additivity, the cofinality, and the covering number of $\mathcal{B}$, respectively. It is well-known (and easy to see) that $\mathrm{add}(\mathcal{B})\leq\mathrm{cov}(\mathcal{B})\leq\mathrm{cof}(\mathcal{B})$.

The following theorem was proved in [3, 1.4].

Theorem 3**.**

If the product $X\times Y$ of balleans $X,Y$ is normal then

[TABLE]

Theorem 4**.**

Let $X$ be the Cartesian product of a family $\mathcal{F}$ of metrizable balleans. Then the following statements are equivalent:

1. (1)

$X$* is metrizable;* 2. (2)

$X$* is normal;* 3. (3)

all but finitely many balleans from $\mathcal{F}$ are bounded.

Proof.

We need only to show $(2)\Rightarrow(3)$. Assume the contrary. Then there exists a family $(Y_{n})_{n<\omega}$ of unbounded metrizable balleans such that the Cartesian product $Y=\prod_{n\in\omega}Y_{n}$ is normal. On the other hand, $\mathrm{add}(\mathcal{B}_{Y})\leq\mathrm{add}(\mathcal{B}_{Y_{0}})=\aleph_{0}$ and a standard diagonal argument shows that $\mathrm{cof}(\mathcal{B}_{Y})>\aleph_{0}$, contradicting Theorem 3. ∎

3. Bornological products

Let $\{(X_{\alpha},\mathcal{E}_{\alpha}):\alpha\in A\}$ be an indexed family of pointed balleans and let $\mathcal{B}$ be a bornology on the index set $A$. For each $\alpha\in A$ by $e_{\alpha}$ we denote the distinguished point of the ballean $X_{\alpha}$.

The $\mathcal{B}$-product of the family of pointed balleans $\{X_{\alpha}:\alpha\in A\}$ is the set

[TABLE]

endowed with the coarse structure $\mathcal{E}_{\mathcal{B}}$, generated by the base consisting of the entourages

[TABLE]

where $B\in\mathcal{B}$ and $(E_{\alpha})_{\alpha\in B}\in\prod_{\alpha\in B}\mathcal{E}_{\alpha}$.

For the bornology $\mathcal{B}=\mathcal{P}_{A}$ consisting of all subsets of the index set $A$, the $\mathcal{B}$-product $X_{\mathcal{B}}$ coincides with the Cartesian product $\prod_{\alpha\in A}(X_{a},\mathcal{E}_{a})$ of coarse spaces $(X_{\alpha},\mathcal{E}_{\alpha})$.

If each $X_{\alpha}$ is the doubleton $\{0,1\}$ with distnguished point $e_{\alpha}=0$, then the $\mathcal{B}$-product is called the $\mathcal{B}$*-macrocube * on $A$. If $|A|=\omega$ and $\mathcal{B}=[A]^{<\omega}$ then we get the well-known Cantor macrocube, whose coarse characterization was given by Banakh and Zarichnyi in [10].

For relations between macrocubes and hyperballeans, see [8], [9].

Theorem 5**.**

Let $\mathcal{B}$ be a bornology on a set and let $X_{\mathcal{B}}$ be the $\mathcal{B}$-product of a family of unbounded metrizable pointed balleans. Then the following statements are equivalent:

1. (1)

$X_{\mathcal{B}}$* is metrizable;* 2. (2)

$X_{\mathcal{B}}$* is normal;* 3. (3)

$|A|=\omega$* and $\mathcal{B}=[A]^{<\omega}$.*

Proof.

To see that $(2)\Rightarrow(3)$, repeat the proof of Theorem 4. ∎

Theorem 6**.**

Let $\mathcal{B}$ be a bornology on a set $A$ and let $X_{\mathcal{B}}$ be the $\mathcal{B}$-product of a family $\{X_{\alpha}:\alpha\in A\}$ of bounded pointed balleans which are not singletons. The coarse space $X_{\mathcal{B}}$ is metrizable if and only if the bornology $\mathcal{B}$ has a countable base.

Proof.

Apply Theorem 1. ∎

Let $X$ be a macrocube on a set $A$ and $Y$ be a macrocube on a set $B$, $A\cap B=\emptyset$. Then $X\times Y$ is a macrocube on $X\cup Y$ and, by Theorem 3, $X\times Y$ needs not to be normal.

Question 1**.**

How can one detect whether a given macrocube is normal? Is a $\mathcal{B}$-macrocube on an infinite set $A$ normal provided that $\mathcal{B}\neq\mathcal{P}_{A}$ is a maximal unbounded bornology on $A$?

Let $\{X_{n}:n<\omega\}$ be a family of finite balleans, $\mathcal{B}=[\omega]^{<\omega}$. By [10], the $\mathcal{B}$-product of the family $\{X_{n}:n<\omega\}$ is coarsely equivalent to the Cantor macrocube.

Question 2**.**

Let $\{X_{\alpha}:\alpha\in A\}$ be a family of finite (bounded) pointed balleans and let $\mathcal{B}$ be a bornology on $A$. How can one detect whether a $\mathcal{B}$-product of $\{X_{\alpha}:\alpha\in A\}$ is coarsely equivalent to some macrocube?

4. Bouquets

Let $\mathcal{B}$ be a bornology on a set $A$ and let $\{(X_{\alpha},\mathcal{E}_{\alpha}):\alpha\in A\}$ be a family of pointed balleans. The subballean

[TABLE]

of the $\mathcal{B}$-product $X_{\mathcal{B}}$ is called the $\mathcal{B}$-bouquet of the family $\{(X_{\alpha},\mathcal{E}_{\alpha}):\alpha\in A\}$. The point $e=(e_{\alpha})_{\alpha\in A}$ is the distinguished point of the ballean $\bigvee_{\alpha\in A}X_{\alpha}$.

For every $\alpha\in A$ we identify the ballean $X_{\alpha}$ with the subballean $\{(x_{\beta})_{\beta\in A}\in X_{\mathcal{B}}:\forall\beta\in A\setminus\{\alpha\}\;\;x_{\beta}=e_{\beta}\}$ of $\bigvee_{\alpha\in A}X_{\alpha}$. Under such identification $\bigvee_{\alpha\in A}X_{\alpha}=\bigcup_{\alpha\in A}X_{\beta}$ and $X_{\alpha}\cap X_{\beta}=\{e\}=\{e_{\alpha}\}=\{e_{\beta}\}$ for any distinct indices $\alpha,\beta\in A$.

Applying Theorem 1, we get the following two theorems.

Theorem 7**.**

Let $\mathcal{B}$ be a bornology on a set $A$ and let $\{X_{\alpha}:\alpha\in A\}$ be a family of unbounded pointed metrizable balleans. The $\mathcal{B}$-bouquet $\bigvee_{\alpha\in A}X_{\alpha}$ is metrizable if and only if $|A|=\omega$ and $\mathcal{B}=|A|^{<\omega}$.

Theorem 8**.**

Let $\mathcal{B}$ be a bornology on a set $A$ and let $\{X_{\alpha}:\alpha\in A\}$ be a family of bounded pointed balleans, which are not singletons. The $\mathcal{B}$-bouquet $\bigvee_{\alpha\in A}X_{\alpha}$ is metrizable if and only if the bornology $\mathcal{B}$ has a countable base.

Theorem 9**.**

A bornological bouquet of any family of pointed normal balleans is normal.

Proof.

Let $\mathcal{B}$ be a bornology on a non-empty set $A$ and $X$ be the $\mathcal{B}$-bouquet of pointed normal balleans $X_{\alpha}$, $\alpha\in A$. Given two disjoint asymptotically disjoint sets $Y,Z\subset X$, we shall construct a slowly oscillating function $f:X\to[0,1]$ such that $f(Y)\subset\{0\}$ and $f(Z)\subset\{1\}$. The definition of the coarse structure on the $\mathcal{B}$-bouquet ensures that for every $\alpha\in A$ the subsets $Y\cap X_{\alpha}$ and $Z\cap X_{\alpha}$ are asymptotically disjoint in the coarse space $X_{\alpha}$, which is identified with the subspace $\{(x_{\beta})\in X:\forall\beta\in A\setminus\{\alpha\}\;\;x_{\beta}=e_{\beta}\}$ of the $\mathcal{B}$-bouquet $X$. By the normality of $X_{\alpha}$, there exists a slowly oscillating function $f_{\alpha}:X_{\alpha}\to[0,1]$ such that $f_{\alpha}(Y\cap X_{\alpha})\subset\{0\}$ and $f_{\alpha}(Z\cap X_{\alpha})\subset[0,1]$. Changing the value of $f_{\alpha}$ in the distinguished point $e_{\alpha}$ of $X_{\alpha}$, we can assume that $f_{\alpha}(e_{\alpha})=f_{\beta}(e_{\beta})$ for any $\alpha,\beta\in A$. Then the function $f:X\to[0,1]$, defined by $f{\restriction}X_{\alpha}=f_{\alpha}$ for $\alpha\in A$ is slowly ascillating and has the desired property: $f(Y)\subset\{0\}$ and $f(Z)\subset\{1\}$. By Theorem 2, the ballean $X$ is normal. ∎

5. Combs

Let $(X,\mathcal{E})$ be a ballean and $A$ be a subset of $X$. Let $\{(X_{\alpha},\mathcal{E}_{\alpha}):\alpha\in A\}$ be a family of pointed balleans with the marked points $e_{\alpha}\in X_{\alpha}$ for $\alpha\in A$.

The bornology $\mathcal{B}_{X}$ of the ballean $(X,\mathcal{E})$ induces a bornology $\mathcal{B}:=\{B\in\mathcal{B}_{X}:B\subset A\}$ on the set $A$. Let $\bigvee_{\alpha\in A}X_{\alpha}$ be the $\mathcal{B}$-bouquet of the family of pointed balleans $\{(X_{\alpha},\mathcal{E}_{\alpha}):\alpha\in A\}$, and let $e$ we denote the distinguished point of the bouquet $\bigvee_{\alpha\in A}X_{\alpha}$.

For for every $\alpha\in A$ we identify the ballean $X_{\alpha}$ with the subballean $\{(x_{\beta})_{\beta\in A}\in\bigvee_{\alpha\in A}X_{\alpha}:\forall\beta\in A\setminus\{\alpha\}\;\;x_{\beta}=e_{\beta}\}$ of $\bigvee_{\alpha\in A}X_{\alpha}$. Then $\bigvee_{\alpha\in A}X_{\alpha}=\bigcup_{\alpha\in A}X_{\alpha}$ and $X_{\alpha}\cap X_{\beta}=\{e\}=\{e_{\alpha}\}=\{e_{\beta}\}$ for any distinct indices $\alpha,\beta\in A$.

The suballean

[TABLE]

of the ballean $X\times\textstyle{\bigvee\limits_{\alpha\in A}}X_{\alpha}$ is called the comb with handle $X$ and spines $X_{\alpha}$, $\alpha\in A\subset X$. We shall identify the handle $X$ and the spines $X_{\alpha}$ with the subsets $X\times\{e\}$ and $\{\alpha\}\times X_{\alpha}$ in the comb $X\operatornamewithlimits{\perp\!\perp}\limits_{\alpha\in A}X_{\alpha}$.

Theorem 10**.**

The comb $X\operatornamewithlimits{\perp\!\perp}\limits_{\alpha\in A}X_{\alpha}$ is metrizable if the balleans $X$ and $X_{\alpha}$, $\alpha\in A$, are metrziable, and for each bounded set $B\subset X$ the intersection $A\cap B$ is finite.

Proof.

Applying Theorem 7, we conclude that the bouquet $\bigvee_{\alpha\in A}X_{\alpha}$ is metrizable. Then the comb $X\operatornamewithlimits{\perp\!\perp}\limits_{\alpha\in A}X_{\alpha}$ is metrizable being a subspace of the metrizable ballean $X\times\bigvee_{\alpha\in A}X_{\alpha}$. ∎

By analogy with Theorem 9 we can prove

Theorem 11**.**

The comb $X\operatornamewithlimits{\perp\!\perp}\limits_{\alpha\in A}X_{\alpha}$ is normal if the balleans $X$ and $X_{\alpha}$, $\alpha\in A$, are normal.

6. Coarse structures, determined by bornologies

Let $\mathcal{B}$ be a bornology on a set $X$. We say that a coarse structure $\mathcal{E}$ on $X$ is compatible with $\mathcal{B}$ if $\mathcal{B}$ coincides with the bornology $\mathcal{B}_{X}$ of all bounded subsets of $(X,\mathcal{E})$.

The family of all coarse structures, compatible with a given bornology $\mathcal{B}$ has the smallest and largest elements ${\Downarrow}\mathcal{B}$ and ${\Uparrow}\mathcal{B}$.

The smallest coarse structure ${\Downarrow}\mathcal{B}$ is generated by the base consisting of the entourages $(B\times B)\cup\bigtriangleup_{X}$, where $B\in\mathcal{B}$.

The largest coarse structure ${\Uparrow}\mathcal{B}$ consists of all entourages $E\subseteq X\times X$ such that $E=E^{-1}$ and $E[B]\in\mathcal{B}$ for each $B\in\mathcal{B}$.

An unbounded ballean $(X,\mathcal{E})$ is called

• •

discrete if $\mathcal{E}={\Downarrow}\mathcal{B}_{X}$,

• •

antidiscrete if $\mathcal{E}={\Uparrow}\mathcal{B}_{X}$,

where $\mathcal{B}_{X}$ is the bornology on $X$, generated by the coarse structure $\mathcal{E}$.

It can be shown that an unbounded ballean $(X,\mathcal{E})$ is discrete if and only if for every $E\in\mathcal{E}$ there exists a bounded set $B\subset X$ such that $E[x]=\{x\}$ for each $x\in X\backslash B$. In [3, Chapter 3] discrete balleans are called pseudodiscrete.

An unbounded ballean $(X,\mathcal{E})$ is called

• •

maximal if its coarse structure is maximal by inclusion in the family of all unbounded coarse structures on $X$;

• •

ultradiscrete if $X$ is discrete and its bornology $\mathcal{B}_{X}$ is maximal by inclusion in the family of all unbounded bornologies on $X$.

It is clear that each maximal ballean is antidiscrete. For maximal balleans, see [3, Chapter 10]. For any regular cardinal $\kappa$ the ballean $(\kappa,{\Uparrow}[\kappa]^{<\kappa})$ is maximal.

It can be shown that each maximal ballean is antidscrete and each ultradiscrete ballean is both discrete and antidiscrete.

Question 3**.**

Is there a ballean which is discrete and antidiscrete but not ultradiscrete?

A ballean $(X,\mathcal{E})$ is called ultranormal if $X$ contains no two unbounded asymptotically disjoint subsets. By [3, Theorem 10.2.1], every unbounded subset of a maximal ballean is large, which implies that each maximal ballean is ultranormal.

Example 1**.**

For every infinite set $X$, there exists a bornology $\mathcal{B}$ on $X$ such that the ballean $(X,{\Uparrow}\mathcal{B})$ is normal but not ultranormal.

Proof.

We take two free ultrafilters $p,q$ on $X$ such that for every mapping $f:X\to X$, there exists $P\in p$ such that $f(P)\notin q$. We put $\mathcal{B}=\{B\subseteq X:B\notin p,B\notin q\}$ and note that $\mathcal{B}$ is a bornology.

To show that the ballean $(X,{\Uparrow}\mathcal{B})$ is normal, it suffices to prove that any disjoint unbounded sets $Y,Z\subset X$ are asymptotically separated. This will follow as soon as we show that $Y$ is an asymptotic neighborhood of $Y$ and $Z$ is an asymptotic neighborhood of $Z$ in the ballean $(X,{\Uparrow}\mathcal{B})$.

The definition of the bornology $\mathcal{B}\ni Y,Z$ ensures that one of the sets (say $Y$) belongs to the ultrafilter $p$ and the other (in this case $Z$) belongs to the ultrafilter $q$.

First we show that $Y$ is an asymptotic neighborhood of $Y$ in $(X,{\Uparrow}\mathcal{B})$. Given any entourage $E\in{\Uparrow}\mathcal{B}$, we should check that the set $E[Y]\setminus Y$ is bounded. Let $Y_{0}=\{y\in Y:E[y]\subset Y\}$ and $Y_{1}:=Y\setminus Y_{0}$. Since $Y=Y_{0}\cup Y_{1}\in p$, either $Y_{0}\in p$ or $Y_{1}\in p$.

If $Y_{0}\in p$, then $Y_{1}\in\mathcal{B}$ and $E[Y]\setminus Y\subset E[Y_{1}]$ is bounded in $(X,{\Uparrow}\mathcal{B})$.

If $Y_{1}\in p$, then for each $y\in Y_{1}$ we can choose an element $f(y)\in E[y]\setminus Y$. By the choice of the ultrafilters $p,q$, there exists a set $P\subseteq Y_{1}$, $P\in p$ such that $f(P)\notin q$. Hence, $f(P)\in\mathcal{B}$ and the set $P\subset E^{-1}[f(P)]$ is bounded, which contradicts $P\in p$. This contradiction shows that the case $Y_{1}\in p$ is not possible and $E[Y]\setminus Y$ is bounded.

By analogy we can prove that $E[Z]\setminus Z$ is bounded. Therefore, the disjoint sets $Y,Z$ have disjoint asymptotically separated neighborhoods and the ballean $X$ is normal.

Since $X$ can be written as the union $X=Y\cup Z$ of two disjoint (unbounded) sets $Y\in p$, $Z\in q$, the ballean $(X,{\Uparrow}\mathcal{B})$ is not ultranormal. ∎

By a bornological space we understand a pair $(X,\mathcal{B}_{X})$ consisting of a set $X$ and a bornology $\mathcal{B}_{X}$ on $X$. A bornological space $(X,\mathcal{B}_{X})$ is unbounded if $X\notin\mathcal{B}_{X}$. For two bornological spaces $(X,\mathcal{B}_{X})$ and $(Y,\mathcal{B}_{Y})$ their product is the bornological space $(X\times Y,\mathcal{B})$ endowed with the bornology

[TABLE]

The following theorem allows us to construct many examples of bornological spaces $(X,\mathcal{B})$ for which the coarse space $(X,{\Uparrow}\mathcal{B})$ is not normal.

Theorem 12**.**

Let $(X\times Y,\mathcal{B})$ be the product of two unbounded bornological spaces $(X,\mathcal{B}_{X})$ and $(Y,\mathcal{B}_{Y})$. If $\mathrm{cov}(\mathcal{B}_{Y})<\mathrm{add}(\mathcal{B}_{X})$, then the coarse space $(X\times Y,{\Uparrow}\mathcal{B})$ is not normal.

Proof.

Fix any point $(x_{0},y_{0})\in X\times Y$. Assuming that $\mathrm{cov}(\mathcal{B}_{Y})<\mathrm{add}(\mathcal{B}_{X})$, we shall prove that for a coarse structure $\mathcal{E}$ on $X\times Y$ is not normal if $\mathcal{E}$ has the following three properties:

1. (1)

$\mathcal{E}$ is compatible with the bornology $\mathcal{B}$; 2. (2)

for any $B_{Y}\in\mathcal{B}_{Y}$ there exists $E\in\mathcal{E}$ such that $X\times B_{Y}\subset E[X\times\{y_{0}\}]$; 3. (3)

for any $B_{X}\in\mathcal{B}$ there exists $E\in\mathcal{E}$ such that $B_{X}\times Y\subset E[\{x_{0}\}\times Y]$.

It is easy to see that the coarse structure ${\Uparrow}\mathcal{B}$ has these three properties.

By the definition of the cardinal $\kappa=\mathrm{cov}(\mathcal{B}_{Y})$, there there is a family $\{Y_{\alpha}\}_{\alpha\in\kappa}\subset\mathcal{B}_{Y}$ such that $\bigcup_{\alpha\in\kappa}Y_{\alpha}=Y$.

Assume that $\mathcal{E}$ is a coarse structure on $X\times Y$ satisfying the conditions (1)–(3). First we check that the sets $X\times\{y_{0}\}$ and $\{x_{0}\}\times Y$ are asymptotically disjoint in $(X\times Y,\mathcal{E})$. Given any entourage $E\in\mathcal{E}$, we should prove that the intersection $E[X\times\{y_{0}\}]\cap E[\{x_{0}\}\times Y]$ is bounded. By the condition (1), for every $\alpha\in\kappa$ the bounded set $E^{-1}[E[\{x_{0}\}\times Y_{\alpha}]]$ is contained in the product $B_{\alpha}\times Y$ for some bounded set $B_{\alpha}\in\mathcal{B}_{X}$. Since $\kappa<\mathrm{add}(\mathcal{B}_{X})$, the union $B_{<\kappa}:=\bigcup_{\alpha\in\kappa}B_{\alpha}$ belongs to the bornology $\mathcal{B}_{X}$. Given any point $(u,v)\in E[X\times\{y_{0}\}]\cap E[\{x_{0}\}\times Y]$, find $x\in X$ and $y\in Y$ such that $(u,v)\in E[(x,y_{0})]\cap E[(x_{0},y)]$. Since $Y=\bigcup_{\alpha\in\kappa}Y_{\alpha}$, there exists $\alpha\in\kappa$ such that $y\in Y_{\alpha}$. Then $(x,y_{0})\in E^{-1}[E[(x_{0},y)]]\subset E^{-1}[E[\{x_{0}\}\times Y_{\alpha}]]\subset B_{\alpha}\times Y\subset B_{<\kappa}\times Y$ and hence $(u,v)\in E[(x,y_{0})]\subset E[B_{<\kappa}\times\{y_{0}\}]$, which implies that the intersection

[TABLE]

is bounded in $(X\times Y,\mathcal{E})$.

Assuming that the coarse space $(X\times Y,\mathcal{E})$ is normal, we can find disjoint asymptotical neighborhoods $U$ and $V$ of the asymptotically disjoint sets $X\times\{y_{0}\}$ and $Y\times\{x_{0}\}$. By the condition (2), for every $\alpha\in\kappa$ there exists an entourage $E_{\alpha}\in\mathcal{E}$ such that $X\times Y_{\alpha}\subset E_{\alpha}[X\times\{y_{0}\}]$. Since $U$ is an asymptotic neighborhood of the set $X\times\{y_{0}\}$ in $(X\times Y,\mathcal{E})$, the set $(X\times Y_{\alpha})\setminus U\subset E_{\alpha}[X\times\{y_{0}\}]\setminus U$ is bounded in $(X\times Y,\mathcal{E})$. Now the condition (1) implies that $(X\times Y_{\alpha})\setminus U\subset D_{\alpha}\times Y$ for some bounded set $D_{\alpha}\in\mathcal{B}_{X}$.

We claim that the family $\{D_{\alpha}\}_{\alpha\in\kappa}$ is cofinal in $\mathcal{B}_{X}$. Indeed, given any bounded set $D\in\mathcal{B}_{X}$, use the condition (3) and find a entourage $E\in\mathcal{E}$ such that $D\times Y\subset E[\{x_{0}\}\times Y]$. Since $V$ is an asymptotic neighborhood of the set $\{x_{0}\}\times Y$, the set $E[\{x_{0}\}\times Y]\setminus V$ is bounded in $(X\times Y,\mathcal{E})$ and the condition (1) ensures that it has bounded projection onto $Y$. Since $Y\notin\mathcal{B}_{Y}$, we can find a point $y\in Y$ such that $X\times\{y\}$ is disjont with $E[\{x_{0}\}\times Y]\setminus V$. Find $\alpha\in\kappa$ with $y\in Y_{\alpha}$. Then $(X\times\{y\})\cap E[\{x_{0}\}\times Y]\subset V$ and hence

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which yields the desired inclusion $D\subset D_{\alpha}$. Therefore,

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which contradicts the known inequality $\mathrm{add}(\mathcal{B}_{X})\leq\mathrm{cof}(\mathcal{B}_{X})$. ∎