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My attempt: ($A \subseteq \mathbb{R}^d$) We know by definition that $\overline{A} = \partial A \cup A $. Hence $\overline{ \mathbb{R}^d \setminus A} = \partial [ \mathbb{R}^d \setminus A ]\cup \mathbb{R}^d \setminus A$. Obviously, have $\partial A = \partial [\mathbb{R}^d \setminus A]$. Hence:

$$ \overline{A} \cap \overline{ \mathbb{R}^d \setminus A } = [\partial A \cup A] \cap[\partial A \cup \mathbb{R}^d \setminus A ] = \partial A \cup( A \cap \mathbb{R}^d \setminus A) = \partial A \cup \varnothing = \partial A$$

Is this a correct approach? Any feedback would be extremely appreciated. thanks

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    $\begingroup$ It seems correct. $\endgroup$ – J.R. Feb 8 '14 at 21:20
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    $\begingroup$ And your definition of $\partial$ is...? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 8 '14 at 21:21
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It depends on your definition of $\partial A$ how obvious $\partial A=∂(\Bbb R^n\setminus A)$ is. If $\partial A$ is defined as the set of points $x$ such that any neighborhood of $x$ intersects both $A$ and $X\setminus A$, then yes, it is obvious.

So your proof is correct.

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