# Example for $\Omega$ so that $\partial\Omega \neq \partial\bar{\Omega}$

Let $\Omega$ be an open subset of $\mathbb R^n$, with $\Omega \neq \emptyset$ and $\Omega \neq \mathbb R^n$.

Can you give an example where $\partial\Omega \neq \partial\bar{\Omega}$, and how can one exclude this situation?

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a) disk minus a radius; b) assume that the boundary is a smooth surface. – user31373 May 24 '12 at 14:53

Take any open subset $G \subsetneq \mathbb{R}^n$ nonempty. Pick $x\in G$. Let $\Omega = G\setminus \{x\}$ which is clearly open. Clearly $x\in \partial\Omega$ but $x\not\in\partial\bar{\Omega}$.

A necessary and sufficient condition to rule this out is to require that $\partial\Omega = \partial\left( \bar\Omega^c\right)$. In other words, it suffices that every neighborhood of any point $y\in \partial\Omega$ contains a point $z$ in the interior of the complement of $\Omega$.

If you just want sufficient conditions:

• As Leonid alluded to it suffices that $\partial\Omega$ is a smooth surface
• in fact it being a Lipschitz hypersurface would also do.
• Otherwise you may require $\Omega$ to be convex
• or that $\partial\Omega$ satisfies an exterior cone condition.

Do you have an application in mind?

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