I think the proposition is true. Here is what I thought. Let $\Omega$ be an open set in a metric space $X$.
$\partial\Omega=\overline{\Omega}\setminus\Omega^\circ=\overline{\Omega}\setminus\Omega$ since $\Omega$ is open. On the other hand, $\partial\overline{\Omega}=\overline{\overline{\Omega}}\setminus{\overline{\Omega}}^\circ=\overline{\Omega}\setminus{\overline{\Omega}}^\circ$ by the well-known theorem that the closure is closed. Now we prove ${\overline{\Omega}}^\circ=\Omega$. Since $\overline{\Omega}=\partial\Omega\cup\Omega$, our purpose is to prove no interior point of $\overline{\Omega}$ lies in $\partial\Omega$. If so, there is a point $x\in\overline{\Omega}$ and its neighborhood $U(x,\delta)\subset\partial\Omega$, which implies $U(x,\delta)\cap\Omega=\emptyset$. That contradicts the fact that $x$ is a limit point of $\Omega$. ${\overline{\Omega}}^\circ=\Omega$ is true.
In conclusion, $\partial\Omega=\overline{\Omega}\setminus\Omega=\overline{\Omega}\setminus{\overline{\Omega}}^\circ=\partial\overline{\Omega}$. The proof is complete.
Please help me to check whether the proof above is correct or not? Thank you all very much.
Best Answer
${\overline{\Omega}}^\circ=\Omega$ is false. In your argument you only have $U(x,\delta) \subset \overline {\Omega}$ not $U(x,\delta) \subset \partial {\Omega}$
$\mathbb R \setminus \{0\}$ is an open set whose boundary is $\{0\}$. But its closure is $\mathbb R$ whose boundary is empty.