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Let $B_{r}(x) = \{y \in X \mid p(x,y) < r\}$ be an open ball and $\bar{B_{r}}(x) = \ [y \in X \mid p(x,y) \leq r\}$ be a closed ball.

Why is it that a closed ball is a complete metric space while an open ball may be an incomplete metric space?

I know a closed interval is complete and it makes sense why. Can we use that fact to show this for the closed ball? Can we also apply the fact that an open interval can be an incomplete metric space as well?

They seem so straightforward that I am having a hard time showing they are true :)

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  • $\begingroup$ Closed ball contains all its limit points, an open ball does not. $\endgroup$ – Matt A Pelto Sep 21 '14 at 17:10
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    $\begingroup$ A closed ball need not be complete. (Think of the rational numbers with the usual metric) $\endgroup$ – Thomas Sep 21 '14 at 17:10
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    $\begingroup$ Well, a closed ball is a complete metric space, in general, only if $X$ was a complete metric space. $\endgroup$ – Thomas Andrews Sep 21 '14 at 17:11
  • $\begingroup$ If $X$ is a complete metric space, then all you need to do is show that $\overline{B_r}(x)$ is closed to show it is complete. If $X$ is not complete, you can find a $\overline{B_r}(x)$ which is not complete. $\endgroup$ – Thomas Andrews Sep 21 '14 at 17:16
  • $\begingroup$ It's possible for an open ball to be closed, if the space is disconnected. For example, if $X=[0,1]\cup[3,4]$ the $B_{1}(\frac12)=[0,1]$ and thus is complete. $\endgroup$ – Thomas Andrews Sep 21 '14 at 17:19
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If $(X,d)$ is a complete metric space, and $A \subset X$ is closed, then $(A,d)$ is also a complete metric space: if $(x_n)$ is Cauchy and all points are from $A$, then in particular this is a Cauchy sequence in $X$ (as we use the same metric, and being Cauchy only depends on the distances between members of the sequence) and so has a limit $x$ in $X$. But as $A$ is closed, it must contain $x$ as well (closed sets are closed under limits). SO $d$-Cauchy sequences from $A$ converge in $(A,d)$, so it is a complete metric space.

Now show that a closed ball $\overline{B(x,r)}$ is always closed in any metric space (consider the complement and show it is open).

To see that an open ball need not be a complete metric space (in the same metric), it suffices to give a single example of this. So take $X = \mathbb{R}$, where $d$ is the usual metric $d(x,y) = |x-y|$, and show that $B(0,1)$ is not complete in this metric.

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  1. In $\mathbb Q$ with the standard metric the closed balls are of the form $[a,b]\cap\mathbb Q$ and they are NOT COMPLETE is $a<b$.

  2. If $X\ne\varnothing$ and $d$ is the discrete metric (i.e., $d(x,y)=1$, iff $x\ne 0$), then every open ball in $(X,d)$ is complete.

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  • $\begingroup$ Sure, but 1. is not a counterexample as the rationals as a whole are not complete. As to 2, of course open balls can be complete, but the question was to show that they need not be. $\endgroup$ – Henno Brandsma Sep 21 '14 at 18:08
  • $\begingroup$ @HennoBrandsma: Did you ever mention in your formulation of the problem that $X$ is complete? Try to be more consistent with what you write. $\endgroup$ – Yiorgos S. Smyrlis Sep 21 '14 at 18:21
  • $\begingroup$ I did, in my answer. $\endgroup$ – Henno Brandsma Sep 21 '14 at 18:59

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