0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\delta B\neq\overline B\setminus B$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.