Would it be possible to present a set in a discrete metric space as a ball? More concretely, suppose our metric space is $X = (1,2] \cap \Bbb{Q}$ equipped with the discrete metric, where the discrete metric is defined as:
$$d(x,y)=\begin{cases} 0, &\text{if}\;x=y;\\\\ 1,&\text{otherwise}.\end{cases}$$
Would it be possible to find a ball $B$ in $X$ such that $B = X$, for $X = (1,2] \cap \Bbb{Q}$ as above?
Other examples or general statements would also be welcome.
On
In general, if $(X,d)$ is a bounded metric space, then it can be expressed as an open ball. A bounded metric space is one for which there exists an $r > 0$ such that $X = B_r(x)$ for some $x \in X$.
If $d$ is the discrete metric on $X$, then $(X,d)$ is a bounded metric space, because for any $r \geq 1$, $X = B_r(x)$ for any $x \in X$.
If the set is a singleton or the entire space, yes.
These are the only options for a ball in the discrete space, so in all other options the answer is no.