Notations:
\begin{align}
&B_x(r)\mathrm{~is~an~open~ball~centred~at~}x\mathrm{~with~radius~}r,\\\\
&\overline{B}_x(r)\mathrm{~is~a~closed~ball~centred~at~}x\mathrm{~with~radius~}r,\\\\
&{B_x}'(r)\mathrm{~is~the~set~of~limit~points~of~}B_x(r),\\\\
&\overline{B_x(r)}\mathrm{~is~the~closure~of~}B_x(r)\mathrm{~i.e.,~}\overline{B_x(r)}=B_x(r)\cup {B_x}'(r).
\end{align}
Show that closure of $B_x(r)$ is the corresponding closed ball, $\overline{B}_x(r)$ in $(\mathbb{R}^n,d)$ where $d$ is Euclidean distance.
Proof:
Part 1: $\Big{[}$To show: $\overline{B_x(r)}\subseteq \overline{B}_x(r)\Big{]}$
Let $y\in\overline{B_x(r)}$. If $y\in B_x(r)\subset \overline{B}_x(r)$, we have $y\in\overline{B}_x(r)$. Otherwise, $y\in{B_x}'(r)$ i.e., $y$ is a limit point of $B_x(r)$. Then,
$$\forall~\epsilon>0~\exists~z\ne y~\backepsilon~z\in B(x,r)\cap B(y,\epsilon)\Rightarrow\forall~\epsilon>0~\exists~z\ne y~\backepsilon~z\in \overline{B}(x,r)\cap B(y,\epsilon)$$
since, $B_x(r)\subset\overline{B}_x(r)$. Therefore $y$ is a limit point of $\overline{B}_x(r)$. Now, $\overline{B}_x(r)$ is a closed set and hence, contains all its limit points. We get $y\in{B_x}'(r)\Rightarrow y\in\overline{B}_x(r)$.
$$\therefore\overline{B_x(r)}\subseteq \overline{B}_x(r)$$
Part 2: $\Big{[}$To show: $\overline{B}_x(r)\subseteq \overline{B_x(r)}\Big{]}$
Let $y\in\overline{B}_x(r)$. Then, $\|y-x\|\le r\Rightarrow \|y-x\|<r \mathrm{~or~}\|y-x\|=r$. If $\|y-x\|\le r$ then, $y\in B_x(r)\subseteq\overline{B_x(r)}$$\Rightarrow y\in\overline{B_x(r)}$. Otherwise $\|y-x\|=r$. Then, suppose $y\notin\overline{B_x(r)}\supset B(x,r)$.
I was hoping to arrive at some contradiction which would imply our assumption is wrong and $y\in\overline{B_x(r)}$ but I can't understand what to do next.
I found quite lot of proofs on the site for counter-examples and proofs for the property holding or not holding for general metric spaces. My question is specifically regarding $\mathbb{R}^n$ with Euclidean norm/distance. I want to know if my approach is correct and how to complete it; I was also trying to avoid using the fact that the closure of a set is the smallest set containing it. If there are any alternate proofs using this property or without it, I would like to see them too.
Best Answer
If $\|y-x\| \leq r$ then $\|y_i-x\| <r$ for each $i$ where $y_i=x+(1-\frac 1 i)(y-x)$. Also $y_i \to x+(y-x)=y$. So $y$ is a limit point of $B(x,r)$.