# Is the interior of a closed subset of the closure of an open set included into the open set?

Hi everyone: Suppose $S$ is a closed set and $\omega$ an open set both in $\mathbb{R}^{n}$ $(n\geq2)$. If $S$ is included into $\overline{\omega}$ (the closure of $\omega$), can we conclude that $\overset{\circ}{S}$ (the interior of $S$) is included into $\omega$? Thanks for your help.

Suppose that $\omega$ is an open dense subset of $\mathbb{R}^n$ (which is not all of $\mathbb{R}^n$), and $S=\mathbb{R}^n$. Then $S=\overline{\omega}$ but $S^o=S\not\subset\omega$.

• What if $S$ and $\omega$ are bounded? – M. Rahmat Feb 26 '16 at 7:16

No, we cannot. Let $\omega = B \setminus \{x\}$, where $x$ is a point and $B$ is an open ball containing $x$. Let $S = \overline{\omega}$. Then $x \in S^\circ$, but $x \not \in \omega$, thus $S^\circ \not \subseteq \omega$.

$\omega \subseteq \overline{\omega}$, since $\overline{\omega}$ is the smallest closed set containing $\omega$.

So $S\subseteq \overline{\omega}$ since $S$ is closed.

Also $\overset{\circ}{S}\subseteq S$ since it's the largest open set contained in $S$.

So $\overset{\circ}{S}\subseteq \overline{\omega}$ and $\omega \subseteq \overline{\omega}$

The result only follows in general if we know $\omega = \overline{\omega}$

• $\omega$ is open here not closed. – R_D Feb 26 '16 at 6:39
• apologies, I misread :/ – Walt van Amstel Feb 26 '16 at 6:41
• Carmichael is right: the answer is no. Just set $\omega=(0,1)\cup(1,2)$ and $S=[0,2]$! Thanks every one. – M. Rahmat Feb 26 '16 at 7:27