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> Let $\overline{B\_d (x, r)}$ be the closure of a ball in some metric space ($X, d$), and let $\hat{B}\_d (x, r)$ be the closed ball in some metric space ($X, d$). Prove, or disprove, that $\overline{B\_d (x, r)}$ = $\hat{B}\_d (x, r)$ in any metric space.
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I have a sense the answer is false when we use the discrete metric (professor told us if we ever want to "break" open/closed relationship, we should start with the discrete metric), but can't really understand how it would be false.

I was thinking maybe considering some radius $r > 1$. Then, $\overline{B\_d (x, r)}$ would be just $x$ itself (i.e. {$x$}), since we can't have a radius greater than $1$ with the discrete metric, right? But then wouldn't $\hat{B}\_d (x, r)$ be the same thing since we can't have distances greater than $1$..? I don't think I'm really understanding the discrete metric well. How would this statement be false wrt. the discrete metric? Any explanation/clarity would be very helpful. Thank you.