Consider $\mathbb{R}^n$ with Euclidean metric $d$. Let $\overline{B}(x_0,r)=\{x\in{X}|d(x_0,x)\leq{r}\}$. Show that for any $x\in{\mathbb{R}^n}$ and $r>0$, $\overline{B}(x_0,r)$ is the closure of the open ball ($\overline{B}(x_0,r)=\overline{B(x_0,r)}$). Give an example showing that this is not necessarily true for any metric space.
My proof:
Since $B(x_0,r)$ is open, $\overline{B(x_0,r)}=B(x_0,r)+Bd(B(x_0,r))=\{x\in{X}|d(x_0,x)<r\}+\{x\in{X}|d(x_0,x)=r\}=\overline{B}(x_0,r)$.
My problem is that I didn't use the fact that it's the Euclidean metric, so I can find no example where it doesn't work. Can someone tell me where I went wrong?
Best Answer
If you are not familiar with the topological definition and properties of continuity then ignore all the previous and just prove that the set $\{x:d(x_0,x) \leq r\}$ is closed using the sequential characterization of closedness.
Now $B(x_0,r) \subseteq \{x:d(x,x_0) \leq r\}$ thus $\overline{B(x_0,r)} \subseteq\{x:d(x,x_0) \leq r\}$ because the closure is the intersection of all closed sets that contain $B(x_0,r)$ as a subset.
Now let $z=(z_1,z_2...z_n) \in \{x:d(x,x_0) \leq r\}$
$\bullet$ If $d(z,x_0)<r$ then $z \in B(x_0,r) \subseteq \overline{B(x_0,r)}$
$\bullet$ If $d(z,x_0)=r$ then consider the sequence $z_m=(\frac{m}{m+1})z+\frac{1}{m+1}x_0$.
We have that $$\sqrt{(\frac{m}{m+1}z_1+\frac{x_{0,1}}{m+1}-x_{0,1})^2+....+(\frac{m}{m+1}z_n+\frac{x_{0,n}}{m+1}-x_{0,n}})^2$$ $$=\sqrt{(\frac{m}{m+1})^2(z_1-x_{0,1})^2+...+(z_n-x_{0,n})^2)}$$ $$=\frac{m}{m+1}d(z,x_0)=\frac{m}{m+1}r<r$$ thus $z_m \in B(x_0,r), \forall m \in \mathbb{N}$
Also we have that $z_n \to z$ with respect to $d$
So for every $z \in \{x:d(x_0,x) \leq r\}$ such that $d(x_0,z)=r$ we found a sequece $z_m \in B(x_0,r)$ such that $z_m \to z$
Thus $z \in \overline{B(x,r)}$
From the two bullets we conclude that $\{x:d(x_0,z) \leq r\} \subseteq \overline{B(x_0,r)}$
Now as a counterexample in a general metric space: