My question is regarding the notion of balls in metric spaces, and specifically about their diameters. If $(X,d)$ is a metric space and $A \subset X$, then the diameter of $A$ is defined by
$$
d(A) = \sup \{ d(a_1,a_2) : a_1 \text{ and } a_2 \in A \}.
$$
I wanted to get a "feel" for the definition; so, I tried to verify that the diameter of a ball of radius $r > 0$ in $\mathbb{R}^n$ is exactly $2r$ as per the above definition, and I managed to do this after some effort.
Then, I wondered whether this necessarily happens in every metric space.
Is it possible to give an example of metric space where the diameter of a ball is strictly smaller than the radius?
I am not sure how to go about finding a set $X$ with a metric $d$ such that this condition holds. I guess I am stuck mainly because I am gathering all my intuition from the case of $\mathbb{R}^n$ with the standard metric, and admittedly haven't got a feel for how abstract metric spaces behave. Any help is appreciated.
Best Answer
Consider the discrete metric $d$ on a set $X$:
$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 1,&\text{if }x\ne y\;. \end{cases}$$
Consider the ball of radius $r=1/2$ centered at $x$
Then $B(x,r)=\{x\}$
Now by definition, $\operatorname{diam} A = \sup\{ d(a,b) : a, b \in A \}$
Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$