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Let $D$ be a bounded domain in $\mathbb{R}^{N}$ ($N\geq2$) and $E$ a closed subset of $D$ with empty interior. Show that the boundary of $D\setminus E$ is the union of $E$ and the boundary of $D$: $$\partial(D\setminus E)=\partial D\cup E.$$

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  • $\begingroup$ Have you tried anything so far? $\endgroup$ – Joppy Oct 8 '17 at 0:14
  • $\begingroup$ Yes, I found one. Please see below. $\endgroup$ – M. Rahmat Oct 8 '17 at 1:27
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Let $A:=D\setminus E$. Then \begin{align*} \partial A&=\overline{A}\setminus A\\ &=\overline{D}\setminus (D\setminus E)\\ &=\overline{D}\cap D^{\complement}\cup (\overline{D}\cap E)\\ &=\partial D\cup S, \end{align*} where the bar over the set indicates its closure and the top "C" is for the complement.

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