simple question about test function ( Sobolev Spaces )

Let $\Omega \subset R^n$ a open set. Let ${\Omega}^{'} \subset \Omega$ a open subset of $R^n$. Supose that $\overline{{\Omega}^{'}}$ is a compact subset of ${\Omega}$. Let $\varphi \in C^{\infty}_{0} ({\Omega}^{'} )$.

If I define $\tilde{\varphi}$ by the equations $\tilde{\varphi} (x) = 0$ if $x \in \Omega - {\Omega}^{'}$ and $\tilde{\varphi} (x) = {\varphi} (x)$ if $x \in {\Omega}^{'}$ , is true that $\tilde{\varphi} \in C^{\infty}_{0} ({\Omega}) ?$ . I think the affirmation is true because the distance of the sets ${\Omega}$ and ${\Omega}^{'}$ is positive ( is positive because ${\Omega}$ is open and $\overline{{\Omega}^{'}}$ is compact ).

Someone can give me a hint ?

Thank you !

That the support of $\tilde{\phi}$ is an compact set contained in $\Omega$ it is trivial. Let's show that $\tilde{\phi}$ is $C^\infty$. If $x\in \Omega\setminus\overline{\Omega'}$, then you can find a neighbourhood $V$ of $x$ such that $\phi=0$ in $V$, hence $\phi$ is $C^\infty$ in $\Omega\setminus\overline{\Omega'}$. If $x\in \partial\Omega'$, then we rembember that by definition, there exist an neighbourhood $U$ (an neighbourhood in $\Omega'$, because $\phi\in C_0^\infty(\Omega'))$ of $\partial\Omega'$ such that $\phi(x)=0$ if $x\in U$, therefore, the same argument applies.