For example, in the metric space $E=(0,1)\cup(1,2)\cup(2,3)$ with normal distance, $(1,2)=\{x:d(x,\frac{3}{2})<\frac{1}{2}\}$ is an open ball, and $(1,2)$ is also a closed set. However, we also have $(1,2)= \{x:d(x,\frac{3}{2})\leq \frac{1}{2}\}$, which is a closed ball, so $(1,2)$ cannot be served as a supportive example.
I would also like to know the answer of the mirror question: Can a closed ball be an open set and cannot be interpreted as an open ball at the same time?
Thanks!
Best Answer
Yes. Consider the set$$[2,\infty)\cup\left\{1,0,-2(=-1-1),-1-\frac12,-1-\frac13,-1-\frac14,\ldots\right\}.$$Then the closed ball with center $0$ and radius $1$ is an open set, but not an open ball.
Now, fro the question from the title, consider$$(-\infty,-2)\cup(-2,-1)\cup(-1,0]\cup\{2\}.$$Then the open ball with center $0$ and radius $2$ is a closed set which is not an closed ball.