[example: Balls in a Discrete Metric Space]
Let $(X,d)$ be a set with the discrete metric, fix $x\in X$, and let $r>0$. We compute the open ball
\begin{align*}
B(x,r)=\{y\in X:d(x,y)<r\}.
\end{align*}
First suppose $0<r\le 1$. Since $d(x,x)=0$ and $0<r$, we have
\begin{align*}
d(x,x)=0<r,
\end{align*}
so $x\in B(x,r)$. If $y\ne x$, then the definition of the discrete metric gives $d(x,y)=1$. Because $r\le 1$, the inequality $1<r$ is false, so $d(x,y)<r$ is false. Hence no point $y\ne x$ belongs to $B(x,r)$, and therefore
\begin{align*}
B(x,r)=\{x\}.
\end{align*}
Now suppose $r>1$. For $y\in X$, either $y=x$ or $y\ne x$. If $y=x$, then
\begin{align*}
d(x,y)=d(x,x)=0<r.
\end{align*}
If $y\ne x$, then
\begin{align*}
d(x,y)=1<r.
\end{align*}
In both cases $y\in B(x,r)$, so every point of $X$ lies in the ball. Thus
\begin{align*}
B(x,r)=X.
\end{align*}
The radius $1$ is the threshold: balls of radius at most $1$ isolate the centre, while balls of radius greater than $1$ contain the whole space.
[/example]