Problem: The diameter of a metric space $(X,d)$ is defined to be:
\[
\operatorname{diam}(X,d) := \sup \{ d(x,y) : x,y \in X \}
\]
Compute the following diameters, justifying your answer:
$\operatorname{diam}(\mathbb{R}^n, d)$, where $d$ is the usual (Euclidean) distance;
$\operatorname{diam}(\mathbb{R}^n, \bar{d})$, where
\[
\bar{d}(x,y) = \frac{d(x,y)}{1 + d(x,y)}.
\]
Show that for any real number $a $, there exists $\mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n$ such that $d(\mathbf{x} ,\mathbf{y} ) > \alpha $. Note that $d((a,0,0,0,\dots, 0), (0,0,\dots, 0)) = a$.
Show that the set $\{ \bar{d} (\mathbf{x} ,\mathbf{y} ): \mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n \} $ is bounded above by $1$ and finally show that $1$ is the least upper bound to conclude that the diameter is $1$.
Solution:
We claim that the diameter of $\mathbb{R} ^n$ with respect to the Euclidean metric is $+\infty $. For that, we need to show that given any positive real number $\alpha $, there exists $\mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n$ such that $d(\mathbf{x} ,\mathbf{y} ) > \alpha $. Take
\[
\mathbf{x} = (\alpha + 1 , 0,0,\dots, 0) \text{ and } \mathbf{y} = \mathbf{0} .
\]
Then
\[
d(\mathbf{x} ,\mathbf{y} ) = \left\lVert \mathbf{x} - \mathbf{y} \right\rVert = \alpha + 1 > \alpha .
\]
Thus, $\operatorname{diam}\left( \mathbb{R} ^n, d \right) = +\infty $.
This also implies that $\mathbb{R} ^n$ is unbounded with respect to the Euclidean metric.
We claim that the diameter of $\mathbb{R} ^n$ with respect to $\bar{d} $ is $1$. Note that for any $\mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n$,
\[
\bar{d} \left( \mathbf{x} , \mathbf{y} \right) = \frac{d(\mathbf{x},\mathbf{y})}{1 + d(\mathbf{x} ,\mathbf{y} )} \leq 1.
\]
Therefore $1$ is an upper bound of the set $\{ \bar{d} (\mathbf{x} ,\mathbf{y} ): \mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n \} $. We now will show that it is the least upper bound. Suppose that $a \in \mathbb{R} $ be such that $a < 1$ and $\bar{d} (\mathbf{x} ,\mathbf{y} ) \leq a$ for any $\mathbf{x} , \mathbf{y} \in \mathbb{R} ^n$. So for any $\mathbf{x} ,\mathbf{y} \in \mathbb{R} ^n$, we have
\begin{align*}
\frac{d(\mathbf{x} ,\mathbf{y} )}{1 + d(\mathbf{x} ,\mathbf{y} )} \leq a & \implies d(\mathbf{x} ,\mathbf{y} ) \leq a + ad(\mathbf{x} ,\mathbf{y} ) \\
& \implies (1- a) d(\mathbf{x} , \mathbf{y} ) \leq a \\
& \implies d(\mathbf{x} ,\mathbf{y} ) \leq \frac{a}{1 - a} < +\infty .
\end{align*}
This shows that $\left( \mathbb{R} ^n, d \right) $ is bounded, contradicting the first part. Thus, no such $a < 1$ exists and hence $\sup \left\{ \bar{d} (\mathbf{x} ,\mathbf{y} ): \mathbf{x} , \mathbf{y} \in \mathbb{R} ^n \right\} = 1$.
This also implies that $\left( \mathbb{R}^n, \bar{d} \right)$ is bounded.