Completeness of closed and open balls in a metric space Let $B_{r}(x) = \{y \in X \mid p(x,y) &lt; r\}$ 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 :)
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.
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 $a&lt;b$.
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.