closed ball in euclidean space In general metric spaces the closed ball is not the closure of an open ball. 
However, I read that in the Euclidean space with usual metric, closed ball is the closure of an open ball. I'm having trouble rigorously proving this? How can I show this?
 A: Hint: 
Suppose $B_x(r)$ and $B_x[r]$ are the respectively open and closed balls with centre $x$ and radius $r \gt 0$. If $y \in B_x[r]$ then $||y - x|| \le r \implies ||y - x|| \lt  r$ or $||y - x|| = r$. For the first case since $y \in B_x(r)$ it is easy to prove that $y$ is also in the closure since $ A \subseteq A^{\circ} $. 
For the second case, assume $y$ is not in some closed set $C$ which contains $B_x(r)$. Since $C$ is closed it contains its interior and boundary points and hence $ y $ is an exterior point for $C$ and hence for $ B_x(r) $. Now consider any neighbourhood of $y$.
Hence try to prove that $y$ is in every closed set which contains $ B_x(r) $ and you would be done.  
A: The only reason a closed ball might not be the closure of an open ball is when there is a "hole" between a point on the sphere and the centre.
Euclidean spaces have no "holes": if you take a point $x$ on a sphere of radius $r$ centred at a point $x_0$, then there is an interval connecting $x_0$ to $x$ and except $x$ it lies within an open ball of radius $r$.
The same argument works for normed spaces and general geodesic spaces, and with slight modification, for arbitrary length metric spaces.
