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In a metric space $X$ for all $x \in X, r > 0$ following is true: $B(x,r) \subseteq \overline{B(x,r)} \subseteq \overline{B}(x,r)$. Here $\overline{B(x,r)}$ is the closure of the open ball of center $x$, radius $r$ and $\overline{B}(x,r)$ is the closed ball of center $x$, radius $r$. The discrete metric gives an example where $B(x,r) = \overline{B(x,r)} \subsetneq \overline{B}(x,r) \ $ by having $X$ be a set with at least two elements, $x \in X$ and $r = 1$. My question is, is there an example where you have proper inclusion between all three sets?

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up vote 8 down vote accepted

How about $[0,1]\cup\{2\}$, with the standard distance and $B(1,1)$?

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