Question: For a metric space $(X,d)$ is it always true that the closure of $B_r(a)$ is equal to $\{{
x \in X:d(x,a)\leq r}\}$?
My attempt at a counter example:
Take the discrete metric, $d_0(x,y)=\begin{cases}
1 & \text{if} \hspace{0.2cm}x\neq y\\
0 & \text{if} \hspace{0.2cm} x=y
\end{cases}$ for all $x,y \in X$. Now consider the open disk of radius $1$ around $a \in X$, $B_1(a)=\{{
x \in X:d_0(x,a)< 1}\}=\{a\}$. I believe that the closure of this singleton set in $X$ is just the set itself, $\{a\}$, but $\{{
x \in X:d_0(x,a)\leq 1}\}=X$. It is not necessarily true that the entire set $X$ is equal to the singleton set $\{a\}$, completing the counterexample.
I'm usually wrong for some reason or the other, so please correct me!
Best Answer
As I wrote in the comments, your example is just fine. I will provide a distinct one. Take $X=[0,1]\cup[2,3]$, endowed with its usual metric. Then $2\notin\overline{B_1(1)}$, but $2\in\{x\in X\,|\,d(x,1)\leqslant1\}.$