The bounded coarse structure on a metric space is the topological coarse structure associated to its Higson compactification.

This post is rehearsal about Proposition 2.47 in Lectures in Coarse Geometry by Roe. Any comment is appreciated as well as usage of language.

PROPOSITION 2.47. Let $(X, d)$ be a proper metric space. Then the bounded coarse structure on $X$ is the topological coarse structure associated to its Higson compactification.

Poof.$\newcommand{\d}{\operatorname{d}}$
We only need to verify that the bounded coarse structure on $X$ is coarser than the topological coarse structure associated to its Higson compactification.

Suppose $E\subset X\times X$ is continuously controlled by $\operatorname{h} X$ but not boundedly controlled, then for each $n$ there is a pair $(x_n,y_n)\in E$ with $d(x_n,y_n)>2n$. Without loss of generality, suppose $x_n$ tends to infinity of $X$. Then $y_n$ tends to infinity of $Y$ since   $E$ is continuously controlled.

Fix a point $O$ in $X$. Because $X$ is a proper metric space (that is, any  bounded set is relatively compact), we can choose a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_1}=x_1$ and  $x_{n_{k+1}}$ is out of the closed ball $B(O, r_k)$ where $r_k=\max\{d(O, x_{n_k}), d(O,y_{n_k})\}+2k+2$. It is easy to see that  closed balls $B(x_{n_k}, k)$ are disjoint pairwise.

Define $f(x)=\begin{cases}\frac{k-d(x,x_{n_k})}{k} & \mbox{if}~ d(x,x_{n_k})\sup_{(x,y)\in F}d(x,y)$ and  $(x,y)$ is an arbitrary point in $F$.

– $d(x, x_{n_k})<k$ and $d(y, x_{n_k})<k$ for some $k$.  $|\d f(x,y)|=\frac{|d(x,x_{n_k)}-d(y,x_{n_k})|}{k}\leq \frac{d_0}{k}.$
– $d(x, x_{n_k})<k$ for some $k$ while $d(y, x_{n_k})\geq k$. $0\leq f(x)=\frac{k-d(x,x_{n_k})}{k}\leq \frac{d(y,x_{n_k})-d(x,x_{n_k})}{k}\leq \frac{d_0}{k}.$ If $d(y, x_{n_j})<j$ for some $j\neq k$, then $|\d f(x,y)|\leq \frac{2d_0}{\min \{j,k\}}.$ If $d(y, x_{n_j})\geq j$ for all $j$, then $|\d f(x,y)|\leq \frac{d_0}{k}.$
– $d(x, x_{n_k})\geq k$ and $d(y, x_{n_k})\geq k$ for all $k$.  $f(x)=f(y)=0$, so $\d f(x,y)=0$.

Let $B$ be the closed ball $B(O, r_k)(\supset \cup_{i=1}^k B(x_{n_i},i))$, then $K:=(B\times \overline{F^{-1}[B]})\cup (\overline{F[B]}\times B)$ is a compact set of $X\times X$. If $(x,y)\in F\backslash K$, then $x\notin B$ and $y\notin B$, thus $|\d f(x,y)|\leq \frac{2d_0}{k+1}.$ Therefore, $f\in C_h(X)$.

Hence $\d f(x_n,y_n)$ tends to zero as $(x_n,y_n)$ tends to infinity of $X\times X$. However, $\d f(x_{n_k}, y_{n_k})=-1$ by the definition of $f$, a contradiction.