*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.