If two nested open sets have the same nonempty boundary, are they the same set? Let $(X,d)$ be a metric space. Let $B_\epsilon(x)$ be the open ball of radius $\epsilon$ centered at $x$. For $x\in X$ and $\epsilon>0$, suppose that $V$ is an open set in $X$ with $V\subseteq B_\epsilon(x)$. 
If $V$ and $B_\epsilon(x)$ share the same (nonempty) boundary, must they be the same set?
 A: Let $X$ be the middle-thirds Cantor set with the Euclidean metric, and let $x=\epsilon=\frac13$. Then
$$B_\epsilon(x)=\left(0,\frac23\right)\cap X=\left(0,\frac13\right]\cap X\;,$$
whose boundary in $X$ is $\{0\}$. Let 
$$V=\left(0,\frac29\right)\cap X=\left(0,\frac19\right]\cap X\;;$$
then $V$ is open in $X$, $V\subsetneqq B_\epsilon(x)$, and $\operatorname{bdry}V=\{0\}=\operatorname{bdry}B_\epsilon(x)$.
A: Let us assume that bd($B_\epsilon(x)$) and bd($V$) are non-empty. Given, bd($B_\epsilon(x)$) = bd($V$). We have, $V \subseteq B_\epsilon(x)$.
Assume, if possible, that $x \notin V$ then $d(x,V)>0 \implies $ there exist $a \in bd(V)$ such that $d(x,a)=min_{z \in \overline{V}}{d(x,z)}>0$. Now, there exist an open ball around $a$ which lies entirely in $B_\epsilon(x)$(why?). Hence, $ a\notin bd(B_\epsilon(x))$, which is a contradiction to fact that $bd(B_\epsilon(x))=bd(V)$.
$\implies x \in V$. Similarly, for any $y \in B_\epsilon(x)$, the argument can be repeated. Therefore, $B_\epsilon(x) \subseteq V$    
