Given a subset $B$ of a metric space $(M, d)$, show that $B$ is open iff $\partial B = \overline{B}\setminus B$.

I feel like that this question shouldn't be too hard to prove, but I'm not sure how to do it.

Attempt: I know how to prove in left to right direction. Suppose $B$ is an open set. That means $B=\operatorname{int}(B)$. We know $\partial B=\overline B\setminus\operatorname{int}(B)$. However, $B$ is just $\operatorname{int}(B)$ in this case. Thus $\partial B=\overline B\setminus B$.

However, I don't know how to prove this in the opposite direction.

(Weak attempt): Assume $\partial B=\overline B\setminus B$. We also know from a known fact that $\partial B=\overline B\setminus\operatorname{int}(B)$. If $\partial B=\overline B\setminus B$ and $\partial B=\overline B\setminus\operatorname{int}(B)$ this would imply $B=\operatorname{int}(B)$, which would imply $B$ is open.


If $B$ is not open, then take $x\in B$ such that $x\notin\mathring{B}$. Then every neighborhood of $x$ intersects $B^\complement$ and therefore, $x\in\overline{B^\complement}$. Since $x\in B$, then, in particular, $x\in\overline B$. So, $x\in\overline B\cap\overline{B^\complement}=\delta B$. But $x\notin\overline B\setminus B$, and this proves that $\delta B\neq\overline B\setminus B$.

  • $\begingroup$ Nice proof by contrapositive. $\endgroup$ – kemb Sep 28 '17 at 6:05
  • $\begingroup$ Does my proof work too, or it wrong? $\endgroup$ – kemb Sep 28 '17 at 6:05
  • $\begingroup$ Yes, it works too. $\endgroup$ – José Carlos Santos Sep 28 '17 at 6:11

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