Prove $\partial A = \partial \bar A$ if and only if int $(\bar A) =$ int$(A)$

Question

Consider a set $A \subseteq \mathbb{R}^n$. Let $\partial A$ denote the boundary of $A$ and $\bar A$ denote the closure of $A$. Some examples for the case when $\partial A \ne \partial \bar A$ can be found here: Example for $\Omega$ so that $\partial\Omega \neq \partial\bar{\Omega}$

Prove $\partial A = \partial \bar A$ if and only if int $(\bar A) =$ int$(A)$.

Answer 1

$A=(0,1]$. Then $A$ and $\overset {-} A$ both have the same boundary (namely $\{0,1\}$) but $A$ is not open so it is not the interior of $\overset {-} A$. Answer to the revised question: if $\overset {-} A $ and $A$ have the same interior then $\partial \overset {-} A =\overset {-} A \setminus int(\overset {-} A)=\overset {-} A \setminus int( A)=\partial A$. Conversely suppose $\partial \overset {-} A= \partial A$. Then $\overset {-} A \setminus int( A)=\overset {-} A \setminus int( \overset {-} A)$. Subtract both sides from $\overset {-} A $ to get $int( A)=int( \overset {-} A)$. [If $S \subset E$ and $T \subset E$ with $E\setminus S=E\setminus S$ then $S=T$ because $E\setminus (E\setminus S)=S$ and $E\setminus (E\setminus T)=T$].

Answer 2

$A = (0,1) \cup (1,2)$,
$\overline{A} = [1,3]$,
$\partial A = \{0,1,2\}$,
$\partial \overline{A} = \{0,2\}$.