It is one of my goals for 2019 to attend a Letter Writing Social. I've decided that if I can't find one near me or travel to one in a different city, then I will host my own somewhere in Minneapolis or St Paul, Minnesota.
What is a Letter Writing Social? I can only offer my best guess since I haven't actually been to one, but in my mind, it is a gathering place where the host provides supplies, tools and resources for those who want to gather and create some mail – a letter, a postcard, a typed note – something personal and creative. Ideally, the setting is conducive to letter writing and having a few typewriters sitting around for people to use is certainly aesthetically pleasing, while also serving as a functional tool.
Do you know of any other Letter Writing Socials? Have you been to one before? Would you like to attend one? What do you imagine they're like? I'd love to hear from you and get your input on this.


The promo was created for the 2009 CMT Music Awards.
After creating this and another promo Justin was able to attend the event with his wife. Here are a few pictures from while at the show.
While there Justin sat next to Pete Stacker who did the voice over for the promo. A very nice guy, Pete does voice overs for everything from movies to commercials. Check him out and have a listen to his work on his website.


text:
HomeIndustry ReleasesNetradyne and First Student Partner to Help Create Safer Roads and...
Netradyne and First Student Partner to Help Create Safer Roads and Communities
SAN DIEGO and CINCINNATI, – Netradyne, a leader in artificial intelligence (AI) and edge computing focusing on driver and fleet safety, and First Student, the leading school transportation provider in North America, today announced a partnership to help create safer roads and transportation for students. Through this relationship, First Student will initially deploy Driveri vision-based driver recognition devices at selected locations, with plans to evaluate future expansion.
Driveri will be implemented as part of First Student's DriverScore™ program. This initiative measures a driver's performance to help improve the quality and safety of transportation service for school districts and their families.
Driveri is the most advanced vision-based driver recognition and fleet safety solution, built to reinforce positive driving behavior and pinpoint coaching and improvement opportunities. In working with Netradyne, First Student will help champion its drivers to remain the safest operators on the road. In addition, there also is the ability for Netradyne's technology to interface with the Seon, First Student's onboard bus camera system.
Through Driveri, actionable information will be put into the hands of location leadership. Managers will have a complete understanding of a driver's performance, including key performance indicators and a "Green Zone" which indicates when the driver is following training and driving safely. First Student drivers will receive feedback about their performance, including positive recognition and rewards.
"At First Student, we believe our drivers are our best safety feature, and we invest heavily in their training and development," said First Student Chief Operating Officer Dean Suhre. "By partnering with Netradyne, we further demonstrate our commitment to providing the safest ride to and from school. We look forward to using this cutting-edge technology to elevate our DriverScore program and safety culture by helping our drivers sharpen their skills behind the wheel while also recognizing safe driving behavior."
"Our partnership with First Student aligns with our core mission of creating safer roads for all," said Adam Kahn, President, Fleet Business and Certified Transportation Professional (CTP), Netradyne. "While school buses are often regarded as the best option for transporting students to school, Netradyne's technology can play a massive role in making the ride even safer. Our advanced vision technology can help drivers mitigate risks and create a safety-driven environment–immediately upon deployment."
About Netradyne, Inc.
Netradyne harnesses the power of Computer Vision and Edge Computing to revolutionize the modern-day transportation ecosystem. Netradyne is an industry leader in fleet safety solutions, immediately improving driver behavior and fleet performance, setting the commercial vehicle driving standards. Netradyne collects and analyzes more data points and meaningful information than any other fleet safety organization so customers can improve retention, increase profitability, enhance safety, and enable end-to-end transparency. Organizations trust Netradyne to build a positive, safe, and driver-focused culture, so they can take their business to the next level. For more information visit https://www.netradyne.com/
About First Student:
As the leading school transportation solutions provider in North America, First Student strives to provide the best start and finish to every school day. First Student completes five million student journeys each day, moving more passengers than all U.S. airlines combined. With a team of highly-trained drivers and the industry's strongest safety record, First Student delivers reliable, quality services including full-service transportation and management, special-needs transportation, route optimization and scheduling, maintenance, and charter services with a fleet of about 40,000 buses. For more information, please visit firststudentinc.com
First Student
Buses with Detroit DD5 and DD8 Engines Have Fewer Regens, Improved Fuel Efficiency
(STN Podcast E87) It's All About People: Zeroed in on Student Safety & School Bus Driver Hiring
GreenPower Appoints K. Neal as a School Bus Dealer for the State of Maryland and Washington D.C.
Zūm CEO Ritu Narayan Receives 2023 Power of Women Award for Revolutionizing Student Transportation
Southern California Plays Host for First in-Person Meeting of the Year
Durham School Services Donates Bus to the West Jefferson Hills School District for an Innovative Mobile "Chill Room" for Mental Health Services
Blue Bird Welcomes New Dealership to Serve South Carolina
First Student Reaches 1,000,000 Electric School Bus Miles

redpajama_set_name: RedPajamaCommonCrawl


text:
ePistolarium
ePistolarium: A Web-based Humanities' Collaboratory on Correspondences
Circulation of Knowledge and Learned Practices in the 17th-century Dutch Republic (CKCC) investigates the circulation of knowledge in the 17th-century Dutch Republic. A multi-disciplinary project team consisting of historians, literature researchers, linguists and computer scientists works together in this project and created a web-based Humanities' Collaboratory on Correspondences. This project, is carried out thanks to a NWO Medium investment subsidy and with CLARIN subsidies to make the resources available withing the CLARIN domain. A consortium of Dutch universities and cultural heritage institutions is building a web-based collaboratory (an online space for asynchronous collaboration) around a corpus of 20.000 letters of scholars who lived in the 17th-century Dutch Republic to answer the research question: how did knowledge circulate in the 17th century? Hereto, it will be necessary to analyze this large amount of correspondence systematically. Based on this (extendable) corpus, we will implement a content processing workflow that consists of iterative cycles of conceptual analysis, enrichment with several layers of annotation and visualization. With advice from CLARIN-EU in the first stage of the project a demonstrator was developed which implements techniques of keyword extraction. The second stage consists of evaluating existing more complex tools en techniques that can tackle one or more aspects of the targeted grammatical, content-related, and network complexity analysis, annotation, and visualization. The phase shall identify a set of tools that can be readily utilized in CKCC, as well as tools that need to be adapted or extended to the needs of CKCC; in short, by the end of this phase resources, requirements and risks shall become clear (deadline: December 2010). In the third stage the collaboratory is further developed according to the description in the CKCC project goals, centering around the technique of concept extraction. These three stages constitute the Work Package Analysis Tools, the core of the CKCC project, which was supported by CLARIN-NL. Other Work Packages provide data and software tools needed to create a complete system: the digital corpus of letters (WP6), the editing collaboratory that will contain the letters (WP1), and the archiving environment for data and software (WP2).
CKCC
CLARIN National Project
CLARIN-NL
CLARIN centre
Huygens ING
Research domain
Tool task
corpus searching
Research Phase
Tool status
Input Language
Access Contact
/mailto:ckcc@huygens.knaw.nl/
Project Contact
prof. dr. Charles van den Heuvel
Creator Contact
CKCC at the CLARIN-NL Portal
http://portal.clarin.nl/node/1961
Ravenek, W, van den Heuvel, C and Gerritsen, G. 2017. The ePistolarium: Origins and Techniques. In: Odijk, J and van Hessen, A. (eds.) CLARIN in the Low Countries, Pp. 317–323. London: Ubiquity Press. DOI: https://doi.org/10.5334/bbi.26. License: CC-BY 4.0
CMDI File Link
Inventory Scope


text:
Mayor Charlie Hales announces support for labor activist who took refuge to avoid deportation
Updated Sep 23, 2014 ; Posted Sep 23, 2014
By Kasia Hall | The Oregonian/OregonLive
Francisco Aguirre seeks asylum at NE Portland church Francisco Aguirre, who is originally from El Salvador, is seeking asylum at a Northeast Portland church after he said federal immigration agents went to his home to detain him.
Portland Mayor Charlie Hales announced his support Tuesday for Francisco Aguirre, a local labor activist originally from El Salvador who took refuge Friday in a Northeast Portland church to avoid deportation.
The 35-year-old Fairview resident has been staying at Augustana Lutheran Church after he said Immigration and Customs Enforcement agents went to his home to detain him. ICE agents do not make arrests in sensitive locations such as schools and churches, according to agency officials.
Aguirre was involved in the Workers' Organizing Committee that went on to found Voz Workers' Rights Education Project, a nonprofit organization that mostly helps male Latino immigrants find work in Portland. He currently serves as the MLK Jr. Worker Center coordinator for the group.
On Tuesday Hales said Aguirre has been a community leader in Portland.
"Our community benefits from the work done by Francisco and by the Voz Workers Rights Education Project," the statement said. "And while immigration is a federal issue, I believe Francisco should remain in the United States, and in Portland, until his case can reach a humane conclusion."
Hales also said mayors across the nation are calling on Congress to address comprehensive immigration reform.
President Barack Obama announced that he wouldn't take any executive action on immigration legislation until after the November election. Since then churches elsewhere in the country have been offering sanctuary to illegal immigrants.
Aguirre was deported to El Salvador in 2000 after a conviction for drug trafficking offenses, ICE said in a statement. Aguirre came to the agency's attention again in August after he was arrested on suspicion of driving under the influence in Clackamas County, ICE said.
Aguirre's lawyer, Stephen Manning, said Francisco has applied for a U Visa, which provides legal status to victims of certain crimes who help authorities investigate crimes. He said the application and review process takes months.
"ICE should give him the opportunity to go through that process before actively trying to deport him," he said. "It doesn't make sense to deport him if he's going to get approved."
-- Kasia Hall


The hanging basket is filled with beautiful mixture of artificial silk flowers including pink geranium, daisies, yellow petunias, mini chrysanthemum and other greenery.
If you are not a keen gardener, lack of time or unable to tend your garden, artificial flower baskets are just for you. They are easy and convenient option to decorate your outdoor living space - gardens & patio areas, holiday homes as well as workplaces and industrial buildings. They bring instant colour to brighten the place for you to enjoy whole year round. You can't kill them, you don't need to worry about watering them, just brush dust from time to time. Baskets come complete with metal chain so all you need to just hang and enjoy .
The arrangement is hand made and measurements may slightly vary from one item to another. All measurements given below are approximate.
Should you want us to add a butterfly to the hanging basket please select the addition "Add a butterfly".


text:
\section{Introduction}\label{section Introduction}

The second author gave a series of seminar lectures on coarse structures
at University of Tennessee
based on \cite{Roe lectures} and \cite{Hof}. After that it became apparent there is a need
for another approach to coarse structures, an approach more suitable for geometers
and topologists. This paper is an attempt to do just that.

Recall that the {\it star} $\text{St}(B,{\mathcal U})$ of a subset $B$ of $X$ with respect
to a family ${\mathcal U}$ of subsets of $X$ is the union of those elements of ${\mathcal U}$
that intersect $B$.
Given two families ${\mathcal B}$ and ${\mathcal U}$ of subsets of $X$,
$\text{St}({\mathcal B},{\mathcal U})$ is the family $\{\text{St}(B,{\mathcal U})\}$, $B\in{\mathcal B}$,
of all stars of elements of ${\mathcal B}$ with respect to ${\mathcal U}$.

\begin{Def}\label{LSStructureDef}
A {\it large scale structure} ${\mathcal LSS}_X$ on a set $X$ is a non-empty set of families ${\mathcal B}$
of subsets of $X$ (called {\it uniformly bounded}
or {\it uniformly ${\mathcal LSS}_X$-bounded} once ${\mathcal LSS}_X$ is fixed)
satisfying the following conditions:
\begin{enumerate}
\item ${\mathcal B}_1\in{\mathcal LSS}_X$ implies ${\mathcal B}_2\in{\mathcal LSS}_X$ if each element of ${\mathcal B}_2$
consisting of more than one point
is contained in some element of ${\mathcal B}_1$.
\item ${\mathcal B}_1,{\mathcal B}_2\in{\mathcal LSS}_X$ implies $\text{St}({\mathcal B}_1,{\mathcal B}_2)\in{\mathcal LSS}_X$.
\end{enumerate}
\end{Def}

We think of (2) above as a generalization of the triangle inequality.

The {\it trivial extension} $e({\mathcal B})$ of a family ${\mathcal B}$ is defined as
${\mathcal B}\cup \{\{x\}\}_{x\in X}$. Recall that ${\mathcal B}$ is a {\it refinement} of ${\mathcal B}'$
if every element of ${\mathcal B}$ is contained in some element of ${\mathcal B}'$.
Thus, the meaning of (1) of \ref{LSStructureDef}
is that if ${\mathcal B}\in{\mathcal LSS}_X$, then all refinements of $e({\mathcal B})$ also belong to ${\mathcal LSS}_X$.

\begin{Prop}\label{PropsOfLSS}
Any large scale structure ${\mathcal LSS}_X$ on $X$ has the following
properties:
\begin{enumerate}
\item ${\mathcal B}\in{\mathcal LSS}_X$ if each element of ${\mathcal B}$ consists of at most one point.
\item ${\mathcal B}_1,{\mathcal B}_2\in{\mathcal LSS}_X$ implies ${\mathcal B}_1\cup{\mathcal B}_2\in{\mathcal LSS}_X$.
\end{enumerate}
\end{Prop}
{\bf Proof. } 1). Pick any ${\mathcal B}_1\in{\mathcal LSS}_X$ and notice ${\mathcal B}_2:={\mathcal B}$
satisfies (1) of \ref{LSStructureDef}.
\par 2). Let ${\mathcal B}_i':=e({\mathcal B}_i)$ for $i=1,2$.
Observe ${\mathcal B}_i'\in {\mathcal LSS}_X$. Therefore ${\mathcal B}_3=\text{St}({\mathcal B}_1',{\mathcal B}_2')\in{\mathcal LSS}_X$
and notice any element of ${\mathcal B}_1\cup{\mathcal B}_2$ is contained
in an element of ${\mathcal B}_3$.
\hfill $\blacksquare$

We have two basic examples of large scale structures induced by other structures
on $X$. The first one deals with metric spaces, so let us point out
there is no need to restrict ourselves to metrics assuming
only finite values. To emphasize that, let us call $d\colon X\times X\to R_+\cup\infty$
an {\it $\infty$-metric} if it satisfies all the regular axioms of a metric
(with the understanding that $x+\infty=\infty$).
Notice that $\infty$-metrics have the advantage over regular metrics in the fact
that one can easily define the {\it disjoint union} $\bigoplus\limits_{s\in S}(X_s,d_s)$
of any family of $\infty$-metric spaces $(X_s,d_s)$. Namely, put $d(x,y)=\infty$
if $x$ and $y$ belong to different spaces $X_s$ and $X_t$ (those are assumed
to be disjoint). Conversely, any $\infty$-metric space $(X,d)$ is the disjoint
union of its finite components $(C,d|C)$ (two elements belong to the same
finite component if $d(x,y) < \infty$).

\begin{Prop}\label{MetricLSS}
Any $\infty$-metric space $(X,d)$ has a natural large scale structure
${\mathcal LSS}(X,d)$ defined as follows:
\par  ${\mathcal B}\in {\mathcal LSS}(X,d)$ if and only if there is $M
> 0$ such that all elements of ${\mathcal B}$ are of diameter at most $M$.
\end{Prop}
{\bf Proof. } If ${\mathcal B}_1 \in {\mathcal LSS}(X, d)$ and for each $B_\beta \in {\mathcal B}_2$
consisting of more than one point there is a $B_\alpha \in {\mathcal B}_1$
containing $B_\beta$, then $\text{diam}(B_\beta) \leq \text{diam}(B_\alpha) \leq
M$ for each $B_\beta \in {\mathcal B}_2$, whence ${\mathcal B}_2 \in {\mathcal LSS}(X, d)$. If
${\mathcal B}_1, {\mathcal B}_2 \in {\mathcal LSS}(X, d)$ then there are $M_1, M_2 > 0$ such
that $\text{diam}(B_\alpha) \leq M_1$ and $\text{diam}(B_\beta) \leq M_2$ for
all $B_\alpha \in {\mathcal B}_1, B_\beta \in {\mathcal B}_2$. Thus for any $B_\alpha
\in {\mathcal B}_1$, $\text{diam}(\text{St}(B_\alpha, {\mathcal B}_2)) \leq 2M_2 + M_1$, whence
$\text{St}({\mathcal B}_1, {\mathcal B}_2) \in {\mathcal LSS}(X, d)$. It follows that ${\mathcal LSS}(X, d)$ is
a large scale structure. \hfill $\blacksquare$

One can generalize \ref{MetricLSS} as follows:
Given certain families ${\mathcal F}$ of positive functions from an $\infty$-metric space $X$ to reals
one can define $LSS(X,{\mathcal F})$ by declaring ${\mathcal B}\in {\mathcal LSS}(X,{\mathcal F})$
if and only if there is $f\in {\mathcal F}$ such that ${\mathcal B}$ refines the family of balls
$\{B(x,f(x))\}_{x\in X}$.

One family of interest is all $f$ such that $\lim\limits_{x\to\infty}\frac{f(x)}{d(x,x_0)}=0$,
where $x_0$ is a fixed point in a metric space $X$
(if $X$ is an $\infty$-metric space, one needs to look at each finite component separately). That leads to the {\it sublinear large scale
structure} on $X$.

\begin{Prop}\label{GroupLSS}
Any group $(X,\cdot)$ has a natural large scale structure
${\mathcal LSS}_l(X,\cdot)$ defined as follows:
\par ${\mathcal B}\in {\mathcal LSS}_l(X,\cdot)$ if and only if there
is a finite subset $F$ of $X$ such that ${\mathcal B}$ refines the shifts $\{x\cdot F\}_{x\in X}$ of $F$.
\end{Prop}
{\bf Proof. } Notice that if ${\mathcal B}\ne\emptyset$ refines $\{x\cdot F\}_{x\in X}$
for some finite subset $F$ of $X$, then $e({\mathcal B})$ also refines $\{x\cdot F\}_{x\in X}$.
\par
Suppose ${\mathcal B}_i$ refines $\{x\cdot F_i\}_{x\in X}$ for $i=1,2$,
where $F_1$ and $F_2$ are finite subsets of $X$. We may enlarge $F_2$
and assume it is symmetric ($y\in F_2$ implies $y^{-1}\in F_2$).
\par
Let $F$ be the set of all products $x\cdot y\cdot z$, where $x\in F_1$ and $y,z\in F_2$.
Given $B\in {\mathcal B}_1$ pick $a\in X$ such that $B\subset a\cdot F_1$.
If $B'\in {\mathcal B}_2$ and $u\in B\cap B'$, choose $y\in X$ so that $B'\subset y\cdot F_2$.
Thus $u=a\cdot f_1=y\cdot f_2$, where $f_1\in F_1$ and $f_2\in F_2$.
Therefore $y=a\cdot f_1\cdot f_2^{-1}$
and $B'\subset a\cdot F$ proving that $\text{St}(B,{\mathcal B}_2)\subset a\cdot F$.
\hfill $\blacksquare$

\begin{Rem}\label{LeftRightIssue}
Notice that any group $(X,\cdot)$ has another natural large scale structure
${\mathcal LSS}_r(X,\cdot)$ defined as follows:
\par ${\mathcal B}\in {\mathcal LSS}_r(X,\cdot)$ if and only if there
is a finite subset $F$ of $X$ such that ${\mathcal B}$ refines the shifts $\{F\cdot x\}_{x\in X}$ of $F$.
\par Clearly, the two structures coincide if $X$ is Abelian. However, they may differ even for
finitely presented virtually Abelian groups.
\par Consider $X=\langle a,t\mid t^2=1 \text{ and } tat=a^2\rangle$.
Notice every element of $X$ has unique representation as $t^ua^v$, where $u=0,1$.
If ${\mathcal LSS}_l(X,\cdot)={\mathcal LSS}_r(X,\cdot)$, then for $E=\{1,t\}$ there is a finite subset $F$ of $X$
such that for each $x\in X$ there is $y\in X$ satisfying $x\cdot E\subset F\cdot y$.
Pick $k\ge 1$ such that all elements of $F$ can be represented as $t^ua^v$ so that $u=0,1$ and $|v|\leq k$. Put $x=ta^{6k}$ and choose $y\in X$ satisfying $x\cdot E\subset F\cdot y$.
There is $c=0,1$ and $i$ so that $x=t^ca^iy$ and $|i|\leq k$.
Also, there is $d=0,1$ and $j$ so that $x\cdot t=t^da^jy$ and $|j|\leq k$.
\par Case 1: $c=1$. Now $y=a^{6k-i}$ and $d=0$, so $y=a^{-j}ta^{6k}t=a^{12k-j}$.
That means $6k-i=12k-j$ and $6k=j-i$ contradicting $|i|,|j|\leq k$.
\par Case 2: $c=0$. Now $y=a^{-i}ta^{6k}=ta^{6k-2i}$ and $d=1$, so
$y=a^{-j}ta^{6k}t=ta^{12k-2j}$. Thus $12k-2j=6k-2i$ and $6k=2j-2i$
contradicting $|i|,|j|\leq k$.
\end{Rem}

To create a large scale structure on a set $X$ all one needs is a family
${\mathcal LSS}_X'$ satisfying conditions resembling finite additivity and (2) of \ref{LSStructureDef}.

\begin{Prop}\label{GeneratingLSS}
If ${\mathcal LSS}_X'$ is a set of families in $X$ such that ${\mathcal B}_1,{\mathcal B}_2\in{\mathcal LSS}_X'$
implies existence of ${\mathcal B}_3\in{\mathcal LSS}_X'$ such that
 ${\mathcal B}_1\cup{\mathcal B}_2\cup\text{St}({\mathcal B}_1,{\mathcal B}_2)$ refines ${\mathcal B}_3$, then
the family ${\mathcal LSS}_X$ of all refinements of trivial extensions of elements of ${\mathcal LSS}_X'$ forms a large scale structure on $X$.
\end{Prop}
{\bf Proof. } It suffices to show that, given ${\mathcal B}_1,{\mathcal B}_2\in{\mathcal LSS}_X'$, $\text{St}(e({\mathcal B}_1),e({\mathcal B}_2))$
is a refinement of the trivial extension $e({\mathcal B}_3)$ for some ${\mathcal B}_3\in{\mathcal LSS}_X'$.
Choose ${\mathcal B}_3\in{\mathcal LSS}_X'$ so that $\text{St}({\mathcal B}_1\cup{\mathcal B}_2,{\mathcal B}_1\cup{\mathcal B}_2)$
refines it.

Given $B\in{\mathcal B}_1$ notice $\text{St}(B,e({\mathcal B}_2))=B\cup\text{St}(B,{\mathcal B}_2)$
is contained in $\text{St}(B,{\mathcal B}_1\cup{\mathcal B}_2)$.
Also, $\text{St}(x,e({\mathcal B}_2))$ is either a point or there is $B\in{\mathcal B}_2$
containing $x$ in which case $\text{St}(x,{\mathcal B}_2)$ is contained in $\text{St}(B,{\mathcal B}_1\cup{\mathcal B}_2)$.
\hfill $\blacksquare$
\begin{Rem}\label{RemOnGenLSS}
The family ${\mathcal LSS}_X$ in \ref{GeneratingLSS} is said to be {\it generated} by ${\mathcal LSS}_X'$.
A good example is the {\it discrete large scale structure} on any set $X$
generated by all ${\mathcal B}$ such that $\bigcup{\mathcal B}$ is finite.
\end{Rem}

Let us show the analog of Theorem 2.55 (p.34) in \cite{Roe lectures}.
Notice the simplicity of our proof.

\begin{Thm} \label{MetrizationThm}
Given a large scale structure ${\mathcal LSS}_X$ on a set $X$ the following
conditions are equivalent:
\begin{itemize}
\item[a.] There is an $\infty$-metric $d_X$ on $X$ such that
${\mathcal LSS}_X={\mathcal LSS}(X,d_X)$.
\item[b.] ${\mathcal LSS}_X$ is generated by a countable set.
\end{itemize}
\end{Thm}
{\bf Proof. } a)$\implies$b) is obvious as any ${\mathcal LSS}(X,d_X)$ is generated
by the family of $i$-balls, $i\ge 1$.
\par a)$\implies$b). Pick a sequence ${\mathcal B}_i\in{\mathcal LSS}_X$ generating ${\mathcal LSS}_X$.
Without loss of generality we may assume $\text{St}({\mathcal B}_i,{\mathcal B}_i)$
refines ${\mathcal B}_{i+1}$ for all $i\ge 1$.
Define the $\infty$-metric $d_X$ on $X$ by setting $d_X(x,y)$ (if $x\ne y$)
equal the smallest $i$ such that there is $B\in{\mathcal B}_i$ containing both $x$ and $y$.
If no such $i$ exists, put $d_X(x,y)=\infty$.
\par To show the triangle inequality notice that
$0 < d_X(x,y)\leq d_X(y,z)\leq i$ implies $d_X(x,z)\leq i+1$
as both $x$ and $z$ belong to $\text{St}(y,{\mathcal B}_i)$ which is contained
in some $B\in{\mathcal B}_{i+1}$.
\par Clearly ${\mathcal LSS}_X\subset {\mathcal LSS}(X,d_X)$ (each ${\mathcal B}_i$ refines
the family of $(i+1)$-balls in $(X,d_X)$).
Also, any family of $r$-balls in $(X,d_X)$ refines ${\mathcal B}_i$ for all $i > r$.
Thus ${\mathcal LSS}_X={\mathcal LSS}(X,d_X)$.
\hfill $\blacksquare$

\section{Coarse structures and their relation to large scale structures}\label{section coarse}

Recall that a {\it coarse structure} ${\mathcal C}$ on $X$ is a family of subsets $E$
(called {\it controlled sets})
of $X\times X$ satisfying the following properties:
\begin{enumerate}
\item The diagonal $\Delta=\{(x,x)\}_{x\in X}$ belongs to ${\mathcal C}$.
\item $E_1\in{\mathcal C}$ implies $E_2\in{\mathcal C}$ for every $E_2\subset E_1$.
\item $E\in{\mathcal C}$ implies $E^{-1}\in{\mathcal C}$, where $E^{-1}=\{(y,x)\}_{(x,y)\in E}$.
\item $E_1,E_2\in{\mathcal C}$ implies $E_1\cup E_2\in{\mathcal C}$.
\item $E, F\in{\mathcal C}$ implies $E\circ F\in{\mathcal C}$, where $E\circ F$
consists of $(x,y)$ such that there is $z\in X$ so that $(x,z)\in E$
and $(z,y)\in F$.
\end{enumerate}

\begin{Def}\label{BofEandEofB}
Given a family ${\mathcal B}$ of subsets of $X$ define $\Delta({\mathcal B})$
as $\bigcup\limits_{B\in{\mathcal B}}B\times B$.
Given $E\subset X\times X$ define ${\mathcal B}(E)$ as the family of
all $B\subset X$ such that $B\times B\subset E$.
\end{Def}

\begin{Lem} \label{BofEcircF}
Suppose ${\mathcal B}_1, {\mathcal B}_2$ are collections in $X$.
If $\Delta({\mathcal B}_i)\subset E_i$ for $i=1,2$,
then $\Delta(\text{St}({\mathcal B}_1,{\mathcal B}_2))\subset (E_2\circ E_1)\circ E_2$.
\end{Lem}
{\bf Proof. } Let $(x, y) \in \Delta(\text{St}({\mathcal B}_1, {\mathcal B}_2))$. Then for some $B
\in {\mathcal B}_1$ there are $B_x, B_y \in {\mathcal B}_2$, containing $x$ and $y$
respectively, such that there are $z_x \in B \cap B_x$ and $z_y \in
B \cap B_y$. Then
\begin{gather*}
(x, z_x) \in B_x \times B_x \subset \Delta({\mathcal B}_2) \subset E_2, \\
(z_y, y) \in B_y \times B_y \subset \Delta({\mathcal B}_2) \subset E_1, \\
(z_x, z_y) \in B \times B \subset \Delta({\mathcal B}_1) \subset E_1,
\end{gather*}
so there is a $z_x \in X$ such that $(x, z_x) \in E_2$ and $(z_x,
z_y) \in E_1$, whence $(x, z_y) \in E_2 \circ E_1$. But then there
is also a $z_y \in X$ such that $(z_y, y) \in E_2$, whence $(x, y)
\in (E_2 \circ E_1) \circ E_2$ as required. \hfill $\blacksquare$

\begin{Lem} \label{BofEcircF}
Suppose ${\mathcal B}_1, {\mathcal B}_2$ are collections in $X$.
If $E_i\subset \Delta({\mathcal B}_i)$ for $i=1,2$,
then $E_1\circ E_2\subset  \Delta(\text{St}({\mathcal B}_2,{\mathcal B}_1\cup{\mathcal B}_2))$.
\end{Lem}
{\bf Proof. } Suppose $(x,y)\in E_1\circ E_2$. There is $z$ such that
$(x,z)\in E_1$ and $(z,y)\in E_2$. Therefore
one has $B_1\in {\mathcal B}_1$ and $B_2\in {\mathcal B}_2$
so that $x,z\in B_1$ and $z,y\in B_2$.
Put $B_3=\text{St}(B_2,{\mathcal B}_1\cup{\mathcal B}_2)$ and notice $B_1\cup B_2\subset B_3$. Thus $x,y\in B_3$.
\hfill $\blacksquare$

\begin{Prop} \label{LSSInducesCS}
Every large scale structure ${\mathcal LSS}_X$ on $X$ induces a coarse structure ${\mathcal C}$ on $X$
as follows: A subset $E$ of $X\times X$ is declared controlled if and only if
there is ${\mathcal B}\in{\mathcal LSS}_X$ such that $E\subset \bigcup\limits_{B\in{\mathcal B}}B\times B$.
\end{Prop}
{\bf Proof. } By the remarks after Definition 1.1, all refinements of
$e({\mathcal B})$, for ${\mathcal B} \in {\mathcal LSS}_X$, themselves belong to ${\mathcal LSS}_X$, meaning
that $\{\{x\}\}_{x \in X}$ is a member of ${\mathcal LSS}_X$. Thus
\[\Delta \subset \bigcup_{B \in \{\{x\}\}}B \times B = \bigcup_{x \in X}\{x\}
\times \{x\}\] so $\Delta \in {\mathcal C}$. Let $E_1 \in {\mathcal C}$, so there is a
${\mathcal B} \in {\mathcal LSS}_X$ such that $E_1 \subset \Delta({\mathcal B})$. $E_2 \subset
E_1$ then $E_2 \subset \Delta({\mathcal B})$ also, whence $E_2 \in {\mathcal C}$. It
is clear that if $E \subset \Delta({\mathcal B})$ then $E^{-1} \subset
\Delta({\mathcal B})$, so $E^{-1} \in {\mathcal C}$. If $E_1, E_2 \in {\mathcal C}$ then there
are families ${\mathcal B}_1, {\mathcal B}_2 \in {\mathcal LSS}_X$ such that $E_1 \subset
\Delta({\mathcal B}_1)$ and $E_2 \subset \Delta({\mathcal B}_2)$. But
\begin{align*}
E_1 \cup E_2 \subset \Delta({\mathcal B}_1) \cup \Delta({\mathcal B}_2) & =
\left(\bigcup_{B \in {\mathcal B}_1}B \times B\right) \cup \left(\bigcup_{B \in {\mathcal B}_2}B \times B\right) \\
& = \bigcup_{B \in {\mathcal B}_1 \cup {\mathcal B}_2}B \times B \\
& = \Delta({\mathcal B}_1 \cup {\mathcal B}_2)
\end{align*}
and since, by Proposition 1.2 ${\mathcal B}_1 \cup {\mathcal B}_2 \in {\mathcal LSS}_X$, it
follows that $E_1 \cup E_2 \in {\mathcal C}$. Finally, let $E_1, E_2 \in {\mathcal C}$
and ${\mathcal B}_1, {\mathcal B}_2 \in {\mathcal LSS}_X$ again be as above. Then $E_1 \circ E_2
\subset \Delta(\text{St}({\mathcal B}_2,{\mathcal B}_1\cup {\mathcal B}_2))$, which, since both
${\mathcal B}_1\cup {\mathcal B}_2$ and ${\mathcal B}_2$ are members of ${\mathcal LSS}_X$, is itself a
member of ${\mathcal LSS}_X$, completing the proof. \hfill $\blacksquare$

\begin{Prop} \label{CSInducesLSS}
Every coarse structure ${\mathcal C}$ on $X$ induces a large scale structure ${\mathcal LSS}_X$
on $X$ as follows:
${\mathcal B}$ is declared uniformly bounded if and only if there is a controlled set $E$
such that $ \bigcup\limits_{B\in{\mathcal B}}B\times B\subset E$.
\end{Prop}
{\bf Proof. } Let $B_1 \in {\mathcal LSS}_X$; then there is a controlled set $E \in
{\mathcal C}$ such that $\Delta({\mathcal B}_1) \subset E$. Suppose ${\mathcal B}_2$ is a
family of subsets of $X$ such that for each $B_\beta \in {\mathcal B}_2$
consisting of more than one point there is a $B_\alpha \in {\mathcal B}_1$
containing $B_\beta$. Then $\Delta({\mathcal B}_2) \subset \Delta({\mathcal B}_1)\cup\Delta
\subset E\cup\Delta\in{\mathcal C}$, whence ${\mathcal B}_2 \in {\mathcal LSS}_X$. Now suppose that ${\mathcal B}_1, {\mathcal B}_2
\in {\mathcal LSS}_X$, so there are $E_1, E_2 \in {\mathcal C}$ such that $\Delta({\mathcal B}_1)
\subset E_1$ and $\Delta({\mathcal B}_2) \subset E_2$. But $(E_2 \circ E_1)
\circ E_1$ is controlled, and $\Delta(\text{St}({\mathcal B}_1, {\mathcal B}_2)) \subset
(E_2 \circ E_1) \circ E_2$, whence $\text{St}({\mathcal B}_1, {\mathcal B}_2) \in {\mathcal LSS}_X$. It
follows that ${\mathcal LSS}_X$ is indeed a large scale structure. \hfill
$\blacksquare$

\section{Higson functions and Higson compactification}\label{section Higson}

In this section we discuss relation of large scale structures on a topological space $X$ to compactifications of $X$. Our approach is quite different from that of
\cite{Roe lectures} (pp.26--31) for coarse structures and seems simpler.

Given a large scale structure ${\mathcal LSS}_X$ on $X$, a subset $B$ of $X$ is
{\it bounded} if $\{B\}\in{\mathcal LSS}_X$.

A bounded continuous function $f\colon X\to R$ is called {\it Higson}
if for every ${\mathcal B}\in{\mathcal LSS}_X$ and for every $\epsilon > 0$
there is a bounded subset $U$ of $X$ such that
$|f(x)-f(y)| < \epsilon$ for all $x,y\in B\setminus U$, $B\in{\mathcal B}$.

If $X$ is a topological space, then using Higson maps one can construct
a compact space $h(X,{\mathcal LSS}_X)$ and a natural map $i\colon X\to h(X,{\mathcal LSS}_X)$.
Namely, first we construct $i\colon X\to \prod\limits_{f}[\inf(f),\sup(f)]$
by sending $x$ to $\{f(x)\}_{f}$, and then we declare $h(X,{\mathcal LSS}_X)$ to be
the closure of $i(X)$ in $\prod\limits_{f}[\inf(f),\sup(f)]$.

It is of interest to investigate cases where $h(X,{\mathcal LSS}_X)$ is a compactification
of $X$ (called {\it Higson compactification} of $(X,{\mathcal LSS}_X)$), i.e. $i\colon X\to i(X)$
is a homeomorphism. Here is the simplest sufficient condition for $h(X,{\mathcal LSS}_X)$
to be a compactification.

\begin{Prop}\label{HigsonCExists}
Suppose $X$ is a Tychonoff space.
If ${\mathcal LSS}_X$ is a large scale structure such that the family
of all open and bounded subsets of $X$ forms a basis of $X$,
then $h(X,{\mathcal LSS}_X)$ is a compactification of $X$.
\end{Prop}
{\bf Proof. } It suffices to show that the family of Higson maps $f\colon X\to [0,1]$
separates points from closed sets. Indeed, given $x_0\in X\setminus A$,
where $A$ is closed, we find $U$ open and bounded such that
$x_0\in U\subset X\setminus A$. Any map $f\colon X\to [0,1]$ such that
$f(x_0)=1$ and $f|(X\setminus U)\equiv 0$ is a Higson map.
\hfill $\blacksquare$

In case of locally compact Tychonoff spaces $X$ we are interested
in the {\it Higson corona} $\nu(X,{\mathcal LSS}_X):=h(X,{\mathcal LSS}_X)\setminus X$ of $X$.

\begin{Cor}\label{HigsonCoronaExists}
Suppose $X$ is a locally compact Tychonoff space.
If ${\mathcal LSS}_X$ is a large scale structure such that all compact subsets of $X$ are bounded,
then $h(X,{\mathcal LSS}_X)$ is a compactification of $X$.
\end{Cor}
{\bf Proof. } Notice all open and relatively compact sets in $X$ are bounded
and form a basis of $X$.
\hfill $\blacksquare$

Given a compactification $c(X)$ of a locally compact Tychonoff space $X$
we are interested in constructing a large scale structure ${\mathcal LSS}(c(X),X)$ on $X$
satisfying the following two conditions:

\begin{itemize}
\item[a.] The bounded subsets of $X$ are precisely relatively compact
subsets of $X$.
\item[b.] The Higson maps of ${\mathcal LSS}(c(X),X)$ include restrictions $f|X$ of all continuous
maps $f:c(X)\to R$.
\end{itemize}

Notice $\text{St}(K,{\mathcal B})$ is bounded for every relatively compact $K$
and every ${\mathcal B}\in{\mathcal LSS}(c(X),X)$. That leads to the following definition.

\begin{Def}\label{DefOfProperFamily}
A family ${\mathcal B}$ is {\it proper} if $\text{St}(K,{\mathcal B})$ is relatively compact
for all relatively compact $K\subset X$. Notice that every $B\in{\mathcal B}$
is relatively compact in that case (consider $K$ consisting of a point in $B$).
\end{Def}

Recall $E\subset X\times X$ is {\it proper} provided both $E[K]$ and $E^{-1}[K]$
are relatively compact for all relatively compact $K\subset X$ (see Definition 2.1 on p.21
in \cite{Roe lectures}).

\begin{Lem} \label{EOfKForDeltaB}
If ${\mathcal B}$ is a family of subsets of $X$, then $\Delta({\mathcal B})[K]=\text{St}(K,{\mathcal B})$.
\end{Lem}
{\bf Proof. } Recall that $E[K]$ is the set of all $x'$ such that there is $x\in K$
satisfying $(x',x)\in E$. If $E=\Delta({\mathcal B})$ that means precisely
there is $B\in{\mathcal B}$ such that $x',x\in B$ and $x\in K$, i.e. $x'\in\text{St}(K,{\mathcal B})$.
\hfill $\blacksquare$

\begin{Cor} \label{ProperFamImpliesProperE}
${\mathcal B}$ is proper if and only if $\Delta({\mathcal B})$ is proper.
\end{Cor}

\begin{Prop}\label{StarsOfProperFamilies}
If ${\mathcal B}_1$ and ${\mathcal B}_2$ are two proper families,
then $\text{St}({\mathcal B}_1,{\mathcal B}_2)$ is a proper family.
\end{Prop}
{\bf Proof. } Notice $\text{St}(K,\text{St}({\mathcal B}_1,{\mathcal B}_2))\subset
\text{St}(\text{St}(K,{\mathcal B}_1),{\mathcal B}_2)\cup \text{St}(\text{St}(K,{\mathcal B}_2),{\mathcal B}_1)$
for every $K\subset X$. If $K$ is relatively compact,
so is $\text{St}(\text{St}(K,{\mathcal B}_1),{\mathcal B}_2)\cup \text{St}(\text{St}(K,{\mathcal B}_2),{\mathcal B}_1)$.
\hfill $\blacksquare$

A {\it Higson family} relative to compactification $c(X)$
is a proper family ${\mathcal B}$ satisfying the following property:
For any map $f:c(X)\to R$ and for any $\epsilon > 0$ there is
a relatively compact set $K$ in $X$ such that $|f(x)-f(y)| < \epsilon$
for all $x,y\in B\setminus K$, $B\in{\mathcal B}$.

\begin{Prop}\label{StarsOfHigsonFamilies}
If ${\mathcal B}_1$ and ${\mathcal B}_2$ are two Higson families,
then $\text{St}({\mathcal B}_1,{\mathcal B}_2)$ is a Higson family.
\end{Prop}
{\bf Proof. } Suppose $f:c(X)\to R$ is continuous
and $\epsilon > 0$.
Find a relatively compact set $K$ such that
$|f(x)-f(y)| < \epsilon/4$
for all $x,y\in B\setminus K$, $B\in{\mathcal B}_1$ or $B\in{\mathcal B}_2$.
Put $L=\text{St}(\text{St}(K,{\mathcal B}_1),{\mathcal B}_2))\cup K$.
Suppose $x,y\in \text{St}(B,{\mathcal B}_2)\setminus L$ for some $B\in{\mathcal B}_1$
and $|f(x)-f(y)| >\epsilon$. Clearly, both $x$ and $y$ cannot belong to $B$.
We will discuss the case of $x,y\in X\setminus B$, the other cases are similar.
Thus $x\in B_x\in{\mathcal B}_2$ and $y\in B_y\in{\mathcal B}_2$
so that there exist $a\in B\cap B_x$ and $b\in B\cap B_y$.
Notice $a,b\in X\setminus K$ (otherwise $x,y\in L$).
Therefore $|f(a)-f(b)| < \epsilon/4$,
$|f(a)-f(x)| < \epsilon/4$, and $|f(y)-f(b)| < \epsilon/4$
resulting in $|f(x)-f(y)| < 3\cdot\epsilon/4$, a contradiction.
\hfill $\blacksquare$

Define ${\mathcal LSS}(c(X),X)$ as consisting of all
Higson families ${\mathcal B}$.
It is a large scale structure as the trivial extension of a Higson family
is a Higson family and refinements of Higson families are Higson as well.
Notice every continuous $f\colon c(X)\to R$
restricts to  a Higson map $f|X$ of ${\mathcal LSS}(c(X),X)$.
\par
Using Example 2.34 on p.28 in \cite{Roe lectures} consider
the compactification $c(Z)$ of integers ($Z$ is equipped with the discrete
topology) obtained by identifying two different points $u$ and $v$
in the \v Cech-Stone corona $\beta(Z)\setminus Z$. Notice
${\mathcal LSS}(c(Z),Z)$
is the discrete large scale structure on $Z$ (generated by families
${\mathcal B}$ such that $\bigcup{\mathcal B}$ is finite) whose Higson functions
are all bounded functions $f:Z\to R$, a set larger than restrictions
$f|Z$ of all continuous maps $f\colon c(Z)\to R$.
Thus, the Higson compactification
of ${\mathcal LSS}(c(Z),Z)$ may be larger than $c(X)$.

\section{Asymptotic dimension}\label{section asdim}

Large scale structures offer a very simple definition of
{\it asymptotic dimension}. Namely,
$\text{asdim}(X,{\mathcal LSS}_X)\leq n$ if ${\mathcal LSS}_X$ is generated by families ${\mathcal B}$ such that
the {\it multiplicity} of ${\mathcal B}$ is at most $n+1$ (that means each point $x\in X$
is contained in at most $n+1$ elements of ${\mathcal B}$).

It is well-known that for metric spaces
the condition $\text{asdim}(X)\leq n$ can be expressed by
one of the following equivalent conditions:

\begin{itemize}
\item[a.] for every uniformly bounded family ${\mathcal B}$ in $X$ there is
a uniformly bounded family ${\mathcal B}'$ on $X$ of which ${\mathcal B}$ is a refinement
such that the multiplicity of ${\mathcal B}'$ is at most $n+1$.
\item[b.] for every $r > 0$ there is
a decomposition of $X$ as $X_0\cup\ldots\cup X_n$
such that the family of $r$-components of each $X_i$
is uniformly bounded.
\end{itemize}

Our first observation is that one can generalize it
to $\infty$-metric spaces without changing the proof.

\begin{Prop} \label{EquivOfTwoDefinitionsOfAsdimMetric}
Suppose $(X,d)$ is an $\infty$-metric space.
If $n\ge 0$, then the following conditions are equivalent:
\begin{itemize}
\item[a.] for every uniformly bounded family ${\mathcal B}$ in $X$ there is
a uniformly bounded family ${\mathcal B}'$ on $X$ of which ${\mathcal B}$ is a refinement
such that the multiplicity of ${\mathcal B}'$ is at most $n+1$.
\item[b.] for every $r > 0$ there is
a decomposition of $X$ as $X_0\cup\ldots\cup X_n$
such that the family of $r$-components of each $X_i$
is uniformly bounded.
\end{itemize}
\end{Prop}

Actually, the concept of $\text{asdim}(X,d)\leq n$ for $\infty$-metric spaces has the benefit
that one can express what Bell-Dranishnikov \cite{BellDranish A Hurewicz Type}
call $\text{asdim}(X_s,d_s)\leq n$ {\it uniformly} for all $s\in S$
simply by stating $\text{asdim}(\bigoplus\limits_{s\in S}X_s)\leq n$.

We would like to generalize \ref{EquivOfTwoDefinitionsOfAsdimMetric}
to arbitrary large scale structures.
For that we need the notion of {\it ${\mathcal B}$-components}.
Those are equivalence classes of the relation $x\sim_{\mathcal B} y$
meaning that there is a finite chain of points $x_0=x,\ldots,x_k=y$
such that for every $i\ge 0$ (and $i\leq k-1$) there is
$B_i\in{\mathcal B}$ satisfying $x_i,x_{i+1}\in B_i$.

Our generalization of \ref{EquivOfTwoDefinitionsOfAsdimMetric}
has the advantage that its proof is by reduction to \ref{EquivOfTwoDefinitionsOfAsdimMetric}
which shows that the asymptotic dimension of arbitrary
large scale structures can be reduced to asymptotic dimension
of $\infty$-metric spaces.

\begin{Cor} \label{EquivOfTwoDefinitionsOfAsdim}
Suppose ${\mathcal LSS}_X$ is a large scale structure on a set $X$.
If $n\ge 0$, then the following conditions are equivalent:
\begin{itemize}
\item[a.] for every uniformly bounded family ${\mathcal B}$ in $X$ there is
a uniformly bounded family ${\mathcal B}'$ on $X$ of which ${\mathcal B}$ is a refinement
such that the multiplicity of ${\mathcal B}'$ is at most $n+1$.
\item[b.] for every uniformly bounded family ${\mathcal B}$ in $X$ there is
a decomposition of $X$ as $X_0\cup\ldots\cup X_n$
such that the family of ${\mathcal B}$-components of each $X_i$
is uniformly bounded.
\end{itemize}
\end{Cor}
{\bf Proof. } a)$\implies$b).
Given ${\mathcal B}\in{\mathcal LSS}_X$ construct inductively
a sequence of elements ${\mathcal B}_i\in{\mathcal LSS}_X$ satisfying the following conditions:
\begin{enumerate}
\item ${\mathcal B}_1={\mathcal B}$,
\item $\text{St}({\mathcal B}_i,{\mathcal B}_\i)$ is a refinement of ${\mathcal B}_{i+1}$ for each $i\ge 1$,
\item the multiplicity of ${\mathcal B}_i$ is at most $n+1$ for $i > 1$.
\end{enumerate}

Given two points $x,y\in X$ we define $d(x,y)$
as the smallest integer $i$ such that $x,y\in B\in {\mathcal B}_i$ for some $i$.
If such integer does not exist, we put $d(x,y)=\infty$.

Notice $\text{asdim}(X,d)\leq n$. Therefore one can decompose
$(X,d)$ as $X_0\cup\ldots\cup X_n$
such that the family of $2$-components of each $X_i$
is uniformly bounded by a fixed integer $M$.
That can be translated into ${\mathcal B}$-components of each $X_i$
being contained in an element of ${\mathcal B}_{M+1}$.

\par
b)$\implies$a). Given ${\mathcal B}_1$ put ${\mathcal B}_2=\text{St}(e({\mathcal B}_1),e({\mathcal B}_1))$
and find a decomposition of $X$ as $X_0\cup\ldots\cup X_n$
such that the family of ${\mathcal B}_2$-components of each $X_i$
is uniformly bounded. Consider ${\mathcal B}_3$ consisting of stars $\text{St}(C,{\mathcal B}_1)$,
where $C$ is a ${\mathcal B}_2$-component of some $X_i$.
Clearly, ${\mathcal B}_1$ refines ${\mathcal B}_3$, so it remains to show
that the multiplicity of ${\mathcal B}_3$ is at most $n+1$.
That follows from the observation that $\text{St}(C,{\mathcal B}_1)\cap \text{St}(C',{\mathcal B}_1)=
\emptyset$ for every two different ${\mathcal B}_2$-components $C$ and $C'$ of the same $X_i$
(otherwise $\text{St}(x,{\mathcal B}_1)$ would interesect both $C$
and $C'$ for any $x\in \text{St}(C,{\mathcal B}_1)\cap \text{St}(C',{\mathcal B}_1)$, a contradiction).
\hfill $\blacksquare$

Our final task is to generalize the Hurewicz Theorem for asymptotic dimension of \cite{BellDranish A Hurewicz Type} and \cite{BDLM}.

\par First let us point out that {\it large scale uniform} functions (or {\it bornologous functions}
in terminology of \cite{Roe lectures}) between metric spaces have a very simple
extension to large scale structures: $f\colon (X,{\mathcal LSS}_X)\to (Y,{\mathcal LSS}_Y)$
is large scale uniform if $f({\mathcal B})\in {\mathcal LSS}_Y$ for all ${\mathcal B}\in{\mathcal LSS}_X$.

\par Given a function $f\colon (X,{\mathcal LSS}_X)\to (Y,{\mathcal LSS}_Y)$ we need to define
the concept of $\text{asdim}(f)\leq n$. Since that has to do with $f^{-1}({\mathcal B})$
for ${\mathcal B}\in{\mathcal LSS}_Y$, let us define a natural large scale structure
on the disjoint union $\bigoplus\limits_{s\in S}A_s$ for any family $\{A_s\}_{s\in S}$
of subsets of $X$. Since we want the natural projection
$\bigoplus\limits_{s\in S}A_s\to X$ to be large scale uniform,
the natural choice is to call ${\mathcal B}$ uniformly bounded in $\bigoplus\limits_{s\in S}A_s$
if and only if there is ${\mathcal C}\in{\mathcal LSS}_X$ such that ${\mathcal B}|A_s$ refines ${\mathcal C}$
for all $s\in S$.

Let us adopt the notation of $\bigoplus{\mathcal B}$ for the disjoint union of any family
${\mathcal B}$. Now, $\text{asdim}(f)\leq n$ means that $\text{asdim}(\bigoplus f^{-1}({\mathcal B}))\leq n$
for all ${\mathcal B}\in{\mathcal LSS}_Y$.

\begin{Thm} \label{HurThm}
If $f\colon (X,{\mathcal LSS}_X)\to (Y,{\mathcal LSS}_Y)$ is a large scale uniform
function, then
$$\text{asdim}(X,{\mathcal LSS}_X)\leq \text{asdim}(f)+\text{asdim}(Y,{\mathcal LSS}_Y).$$
\end{Thm}
{\bf Proof. } Let $\text{asdim}(f)=n$ and $\text{asdim}(Y,{\mathcal LSS}_Y)=k$.

Suppose ${\mathcal B}_1\in{\mathcal LSS}_X$ is a cover. Let us construct by induction
a sequence of covers ${\mathcal B}_i\in{\mathcal LSS}_X$ and a sequence of covers ${\mathcal C}_i\in{\mathcal LSS}_Y$
satisfying the following conditions:
\begin{enumerate}
\item $\text{St}({\mathcal B}_i,{\mathcal B}_i)$ refines ${\mathcal B}_{i+1}$ for all $i\ge 1$.
\item $f({\mathcal B}_i)$ refines ${\mathcal C}_i$.
\item The multiplicity of ${\mathcal C}_i$ is at most $k+1$.
\item The cover of $\bigoplus f^{-1}({\mathcal C}_i)$ induced by ${\mathcal B}_i$
refines a cover of multiplicity at most $n+1$ that is
a refinement of the cover of $\bigoplus f^{-1}({\mathcal C}_i)$ induced by ${\mathcal B}_{i+1}$.
\item $\text{St}({\mathcal C}_i,{\mathcal C}_i)$ refines ${\mathcal C}_{i+1}$ for all $i\ge 1$.
\end{enumerate}

Define the $\infty$-metric $d_X$ on $X$ by setting $d_X(x,y)$
equal the smallest $i$ such that there is $B\in{\mathcal B}_i$ containing both $x$ and $y$.
If no such $i$ exists, put $d_X(x,y)=\infty$.
Create a $\infty$-metric $d_Y$ on $Y$ the same way using the sequence ${\mathcal C}_i$.
Notice the following properties of $f\colon (X,d_X)\to (Y,d_Y)$:
\begin{itemize}
\item[a.] $\text{asdim}(Y,d_Y)\leq n$.
\item[b.] $\text{asdim}(f)\leq n$.
\item[c.] $f\colon (X,d_X)\to (Y,d_Y)$ is large scale uniform.
\end{itemize}
Indeed, ${\mathcal LSS}(X,d_X)$ is generated by ${\mathcal B}_i$'s and ${\mathcal LSS}(Y,d_Y)$
is generated by ${\mathcal C}_i$'s (see the proof of \ref{MetrizationThm}), so a) and c) follow.
Similarly, b) holds.

Since the proof of Hurewicz Theorem in \cite{BDLM} is valid for $\infty$-metric
spaces, one concludes $\text{asdim}(X,d_X)\leq n+k$.
In particular there is a uniformly bounded family ${\mathcal U}$ in $(X,d_X)$
such that ${\mathcal B}_1$ refines ${\mathcal U}$ and the multiplicity of ${\mathcal U}$ is at most
$k+n+1$. Notice ${\mathcal U}$ refines ${\mathcal B}_M$ for some large $M$.
Thus, ${\mathcal U}\in{\mathcal LSS}_X$ which completes the proof.
\hfill $\blacksquare$

\section{\v Svarc-Milnor Lemma}

\cite{BDM} gives a simple proof of \v Svarc-Milnor Lemma.
Let us use the approach of this paper to offer an explanation of assumptions
in the \v Svarc-Milnor Lemma.

\par Given a function $f\colon X\to Y$ and given a large scale structure
${\mathcal LSS}_Y$ on $Y$ let us define {\it the induced large scale structure $f^\ast({\mathcal LSS}_Y)$}
on $X$ as that generated by $f^{-1}({\mathcal B})$, ${\mathcal B}\in {\mathcal LSS}_Y$.

\begin{Lem} \label{LeftActionsByIso}
If a group $(G,\cdot)$ acts on the left by isometries on a metric space $(X,d)$,
then ${\mathcal LSS}_l(G,\cdot)\subset f^\ast({\mathcal LSS}(X,d))$ for any $x_0\in X$
and $f(g):=g\cdot x_0$ for $g\in G$.
\end{Lem}
{\bf Proof. } Suppose $F\subset G$ is finite. Put $r=\max\{d(x_0,h\cdot x_0)\mid h\in F\}$.
Given $g\in G$ let $U=B(g\cdot x_0,r)$. It suffices to show
$f(g\cdot F)\subset U$. That is obvious as $d(g\cdot h\cdot x_0,g\cdot x_0)=
d(h\cdot x_0,x_0) < r$ as $h\in F$.
\hfill $\blacksquare$

\begin{Lem} \label{LeftActionsByCocompact}
Suppose a group $(G,\cdot)$ acts on the left by isometries on a metric space $(X,d)$, $x_0\in X$
and $f(g):=g\cdot x_0$ for $g\in G$.
$${\mathcal LSS}_l(G,\cdot)=f^\ast({\mathcal LSS}(X,d))$$
 if and only if for any bounded
subset $U$ of $G\cdot x_0$ containing $x_0$
the set $\{g\in G\mid (g\cdot U)\cap U\ne\emptyset\}$
is finite.
\end{Lem}
{\bf Proof. } In view of \ref{LeftActionsByIso}, we need to analyze $f^\ast({\mathcal LSS}(X,d))\subset {\mathcal LSS}_l(G,\cdot)$. It holds if and only if, for any $r > 0$,
there is a finite subset $F_r$ of $G$ such that for any $x\in G\cdot x_0$
there is $g_x\in X$ so that $f^{-1}(B(x,r))\subset g_x\cdot F_r$.
\par Put $U=B(x_0,r)$ and assume $F_r=\{g\in G\mid (g\cdot U)\cap U\ne\emptyset\}$
is finite.
If $x=g_x\cdot x_0$, then $f^{-1}(B(x,r))=\{g\in G\mid d(g\cdot x_0,g_x\cdot x_0) < r\}
=\{g\in G\mid g_x^{-1}g\cdot x_0\in B(x_0,r)\}\subset g_x\cdot F_r$.
\par Assume that, for any $r > 0$,
there is a finite subset $F_r$ of $G$ and $g_0\in X$ so that $f^{-1}(B(x_0,r))\subset g_0\cdot F_r$. Consider $U=B(x_0,r)$ (any bounded subset of $G\cdot x_0$
is contained in such ball). If $h\cdot x_0\in (g\cdot U)\cap U$,
then $h\in f^{-1}(B(x_0,r))$, so $h\in g_0\cdot F_r$.
Also, $g^{-1}\cdot h\in g_0\cdot F_r$
which means the set $\{g\in G\mid (g\cdot U)\cap U\ne\emptyset\}$
is finite.
\hfill $\blacksquare$

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The miracle of flying amidst a soft blue cushion white, in a sunny heavenly domain came to a sudden end on February 11 1992 when an aircraft crashed in the city of Hengelo close by. This aircraft, busy with a routine training flight took off from neighbour city airport 8km away at Enschede. There are thousands of flights every year that pass without incident in the busy skies above us. Very occasionally something goes wrong; this was my recollection of Enschede when I first came to live in this area.


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Bleached's First Album Since Getting Sober Is Their Best Yet — And One of 2019's Strongest Rock Records
The L.A. duo's hook-filled 'Don't You Think You've Had Enough?' pulls influence from disco, post-punk and Paramore.
The day before Bleached played Panorama Festival in 2017, the stage they were supposed to play on collapsed.
That didn't stop them from putting on a big rock 'n' roll show at the New York City fest. Guitarists/core members Jennifer and Jessie Clavin were loud, punchy, and on-point — their chemistry as strong as you'd expect from two sisters who had been in the same bands for over a decade. There were some rock-star stances; their transitions from song to song were snappy; they even worked in a cover of a Nevermind deep cut amongst their own peppy garage rock. I caught up with the Clavins for a brief interview after the show, and they were in especially good spirits, chatting about their new EP and zine as the Panorama dance tent thumped nearby.
Two years later, leading up to Don't You Think You've Had Enough? (out today on Dead Oceans), the Clavins admit the new LP is the first they've written since both deciding to go sober, as alcohol was threatening to derail their lives. Seeing them at the fest, I had no idea they'd recently made this leap of faith, but now it's clear how it's improved… everything.
"[Panorama] was a perfect example where I was super nervous before we played," says Jennifer, the older of the Clavin sisters and the band's lead vocalist and lyricist. There was once a time when they'd perform close to a blackout level of drunk, partially for the punk cool, moreso to cover for nagging insecurities that they have since faced.
"Drinking and playing music was so romantic — that's just what I thought you did," says Jessie, the band's guitarist and chief instrumentalist. "A lot of times as a drinker I just wanted to numb feelings," Jennifer remembers. "Once we got onstage [at Panorama], all my nerves went away and I was having a great time."
Don't You Think You've Had Enough? is one of the year's best rock albums, and far more than a survival tale. It's Bleached's third full-length, and if you count Mika Miko — the cult-famous punk quintet both Clavins played in throughout the 2000s — their fifth LP, alongside several handfuls of EPs and singles. They've been trafficking in Black Flag and Blondie-indebted dive bar rawk for well over a decade, and on that note, the new album turns a major corner. On 2016's Welcome the Worms, Bleached battled through drinking, drugs and abusive relationships with ferocious, open-road riffage; today they rock out when necessary, but with space for rubbery disco bass lines, whistles, cowbell, and arty minimalism.
Recent single "Hard To Kill" sounds like LCD Soundsyetem getting down to business and writing a three-and-a-half-minute pop song. "Kiss You Goodbye" and its call-and-response breakdown discinctly sounds like a band that's become friends, collaborators and touring partners with After Laughter-era Paramore. It all ends with "Shitty Ballet," a bare-bones rocker they had the guts to leave sounding like a demo.
Calling in from their L.A. home base last week, the Clavins explain to Billboard just how they wound up at this point in their career.
You were just on tour with the Hives and Refused. Are you honorary Swedes now?
Jennifer: Our mom's side of the family is Swedish. We wanted to make a button that says "Our Mom Is Swedish," but we forgot.
Jessie: When we were on tour, it hit us what showmen they are. I was like, we definitely need to channel these live performances.
Jennifer: The Hives' showmanship is next-level good. Everyone plays a role — it's not one guy carrying the whole show. Refused were really cool because they kept talking about back in the day when stuff was really punk. Seeing where they're at now, it was really inspiring.
You toured Europe with Paramore in 2017. What did you take away from that?
Jennifer: Once again we were on tour with a band that really kills it live… We're touring with these bands that have a huge crew and budget. Then it's just the four of us with maybe a tour manager. Sometimes it's a little intimidating because I feel we're just left to fend for ourselves.
Jessie: Because Paramore has three guitarists onstage, that tour opened me up to a lot of guitar tones. We [sometimes] have Jen just singing so she can run around and interact with the audience more and I'm just left on guitar. I was like, I really want to explore with different tones, make things more peaceful and simple.
Jennifer: Watching how hard they give it, I really connect with that, because when I play I also tap into this place where I'm putting on a show. I think it's important to channel your best performance, just lose yourself in the live show. That's also been a bit of a struggle because when I was a teenager I could really lose myself because I didn't care if I was messing up or singing poorly because that's the point, I'm putting on a show. Now there's a side of me that wants to put on a good, professional show as well as the fun side. Trying to find that balance is something I'm working on.
From starting in the punk world during the Mika Miko days, it sounds like you're learning where it makes sense to take Bleached today.
Jessie: Yeah, even going back to our roots… Some post-punk I never realized was pretty. Some has really cool, dance-y guitars. It's not this crazy, heavy punk.
"Hard To Kill" on the new album is so dance-y, with the cowbell and everything! I can see how touring with Paramore could have rubbed off there.
Jennifer: It was fun to watch the audience get really into it, too. If we're going to be playing so many live shows, I want to have fun doing this. It inspired me to want to write dancier stuff. Plus Jessie and I are super into disco and post-punk. It made a lot of sense.
You worked with a handful of collaborators and co-writers on your new album. Who are some you want to highlight?
Jennifer: "Shitty Ballet" we wrote with this girl Madi [Diaz], and it was really cool. I'd never met her before — our manager [Christian Stavros] was like, "Oh she's really awesome, you guys should try to write." [Diaz and I] were talking about what we both were going for and we were both dealing with a similar situation so the lyrics came really easily. Everyone was so stoked on the demo — just vocals and one guitar — that we were like, "This needs to be what the song sounds like." We also went to Nashville and ended up playing with some of our friends we met in Paramore: Joey [Howard] the bassist, and Joey [Mullen] the percussionist. We had so many demos.
Jessie: They jammed out some songs with us, which is so necessary. With Jen and I writing, we don't have a live band.
Sobriety played a crucial role for both of you in making this record. Were there defining moments that made you say, enough is enough, time for a change?
Jennifer: For me, there was a defining moment I don't feel comfortable sharing, but yes, there was that moment: What am I doing? Why am I living my life like this? I could potentially die. This needs to end. It was really scary, but also very hopeful. I've been sober for three-and-a-half years. I have a good support system. Noticing what sobriety has done to my life, how much it's improved my life, this is one of my favorite records we've ever written… No one ever gets sober and they're like, "Oh, my life sucks" [Laughs]. You have to take that leap of faith that it's gonna be okay. It's hard. For me, alcohol was my love. I had to let it go.
Jessie: Drinking and playing music was so romantic. That's just what I thought you did. I know people who didn't live past 27. Like, it's cool people get sober but [I thought,] "I'm not like that, I'm very different from everyone, I'm supposed to be partying like this!"
But I started seeing a lot of the damage I was making in my party days and at a very young age I was already doing a lot of crazy stuff. It just took me a little bit longer to see it. It wasn't an immediate thing. I guess needed more chaos in my life to finally be like, "Oh yeah I should try and get sober."
How is the writing and creative process different with a sober mind and body?
Jennifer: Honestly, it's not difficult. I thought it was gonna be difficult, but I kinda just let things come to me. I had to rely more on the power within myself to bring whatever needed to come out: a song, melody, guitar part. I used to think I had to get really fucked up to write the best music. I remember we would [be] up all night writing and then finally, pass out, wake up in the morning like, "Wait, what did we even write last night?" We'd listen back and be like, "Oh, that sounds like shit," or "Oh, that's cool."
When I record, I get this anxiety that everything's not gonna be the way I want it to be. I feel like it's because it's my art, I want to hold it and mold it and shape it exactly how I see it. But I know that's not possible. With alcohol and drugs you can kind of numb yourself. Numbing yourself allows you to let go of this control you wanted to have over your art. Being sober, I had to learn to let go and let this art take shape on its own. I can't control it, because it's something bigger than me.
What about performing? I know that alcohol can help take the edge off when you're in front of a crowd.
Jennifer: When I first started performing sober, I was coming offstage with this insane high. Like whoa, this is crazier than anything that would happen when I was performing [drunk], blacked out even. The worst would be when I'd play and then be like, "Oh my God, what did I do? I don't even know how I performed, if I was acting crazy or what." I hadn't thought about that in so long because now it's been three and a half years — now I'm used to performing sober.
I think when you drink and you perform, you force yourself out of your head, and when you're performing sober you can get stuck in your head a little bit. I think it's easy to get out of that. Once I'm onstage, I'm just looking at the members in my band and realizing we're all here together and there's nothing I could really do to fuck everything up. Maybe I'll fuck up one little thing, but no one's gonna notice… I don't have that much power to fuck up a whole song!
Along with your 2017 EP Can You Deal?, you compiled a zine featuring writing from other female musicians — Lizzo, Liz Phair, Alice Glass — about their experiences in the industry. As you made this album, did you find any of their contributions particularly influential?
Jennifer: I think the whole thing really resonated with me. It was super empowering to have all these women share that we've all been feeling a similar way. And I think just getting it off our chest and having a place to accept ourselves and have people actually hear what we were saying was so empowering. From that, I was able to move on to the next chapter of whatever I'm doing, and not keep dwelling on the fact of, yeah, people are gonna ask you what it's like to be a girl in a band. It's funny because now we don't get asked that question at all.
Was that what inspired the zine — always getting bombarded with that question?
Jennifer: Yeah. Even when we were younger in [Mika Miko], it was always, "So what is it like being girls in a band?" We were like, what do they mean by that? That can go in so many directions. I thought maybe it's because we're an all-girl punk band. Maybe they were just trying to talk about how the punk scene is very male-dominant, which I totally understand, but it just kept happening to Bleached. We were three girls and one guy. I was like, why are we still getting asked this? A lot of writers appreciated it was brought to our attention, which I thought was cool.
Don't You Think You've Had Enough? is your first album in three years. What are you hoping to achieve with this album that you hadn't before?
Jennifer: There's a part of me that's like, yeah, you should set your goals but there's [another] part of me that just wants to go with the flow. It feels a little overwhelming to be back in record cycle world. This is the longest amount of time I've ever had off — about a year — but I didn't know how to relax during that year because I was so used to touring. So I kept living in Bleached world. All of a sudden, I'm really in it again.