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?
[Math] closed ball in euclidean space
general-topologymetric-spacesreal-analysis
Related Question
- [Math] A metric space is complete if for some $\epsilon \gt 0$, every $\epsilon$-ball in $X$ has compact closure.
- [Math] Open Ball in a Metric Space vs. Open Set in a Topological Space
- [Math] Prove that if the closure of each open ball in compact metric space is the closed ball with the same radius, then any ball in this space is connected
- [Math] Closure of Ball = Closed Ball in Normed Space
Best Answer
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.