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If $X$ is a discrete space (metric).

Then the closure of a open ball $B_1(x)=\{x\}$ is $B_1(x)=\{x\}$, and the closed ball is $X$, therefore do not coincide.

You know another example such that:

(Closure) $\overline{B_\epsilon(x)=\{y\in{X}: d(x,y)<\epsilon\}}\neq{closed \ ball} $

Thank you all.

Calvin Lin
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user126033
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    One very common place you find this property is in non-archimedean fields, if you know what those are. For instance, in the $p$-adic numbers $\mathbb{Q}_p$ for $p$ a prime number. – froggie Mar 06 '14 at 06:04
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    Trivial extension of your example: Take $(-1,0] \cup [1,2)$ with the usual metric. Then $\bar{B}(0,1) = (-1,0]$, but the closed unit ball centred at $0$ is $(-1,0] \cup {1}$. – copper.hat Mar 06 '14 at 07:04
  • Take $X$ as the unit circle and its center. The open unit ball around the center is just the center itself as a singleton. The closed unit ball around the center is the whole space $X$. – k.stm Mar 06 '14 at 10:15

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