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Let $(X,d)$ be a metric space such that $d$ is not the discrete metric. Let $x_0 \in X$, let $r>0$, and let $$B(x_0;r) \colon= \{ x \in X \colon d(x,x_0) < r \}$$ be the open ball with center $x_0$ and radius $r$, and let $$\tilde{B}(x_0;r) \colon= \{ x \in X \colon d(x,x_0) \leq r \}$$ be the corresponding closed ball.

Then what is (are) the necessary and sufficient condition(s) on $X$ and / or $d$ such that the closure $\overline{B(x_0;r)}$ of $B(x_0;r)$ is different from $\tilde{B}(x_0;r)$?

Rustyn
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Saaqib Mahmood
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  • Closed balls are the closures of open balls by definition, I don't understand how they could ever be different. Maybe I'm misinterpreting your question – Rustyn Sep 20 '14 at 03:07
  • @Rustyn: The closure of an open ball is closed, but not necessarily the same as the closed ball. The usual metric on $\mathbb{Z}$ shows this. The closure of $B(0,1)$ is ${0}$, but the closed unit ball is ${-1,0,1}$. (I make this slip every couple of years.) – copper.hat Sep 20 '14 at 03:38
  • @copper.hat Hah! Yes, you're right. In $\mathbb{R}^2$ they are the same no? I believe this has to do with isolated points, connectedness, and what separation axioms the space satisfies. I will put some more thought into it. – Rustyn Sep 20 '14 at 18:22
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    @Rustyn: With the usual metric they are. That presumably is where our intuition comes from. But you could choose $d(x,y) = \max(|x-y|,1)$ and then any closed unit ball is the entire space :-). – copper.hat Sep 20 '14 at 18:28
  • I am now upvoting this question because I see now that it's a good question. – Rustyn Sep 20 '14 at 18:31

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