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There are several ways to see this fact, which is a simple instance of Alexander duality. Here is the simplest I know.

Let $H$ be a compact smooth hypersurface in $\mathbb{R}^n$, whithout boundary, and $x\in \mathbb{R}^n\setminus H$. Then the radial projection $p_x : H \to \mathbb{S}^{n-1}$, $y\mapsto (y-x)/\|y-x\|$ has a degree mod $2$, say $d_x$, which may be defined as the number of elements mod $2$ of $p_x^{-1}(u)$ for almost all $u\in\mathbb{S}^{n-1}$ (this is well-defined by transversality theory). Then the subset of $x\in \mathbb{R}^n$ such that $d_x=1$ is your $\Omega$, a relatively compact open set with boundary $H$. EDIT : here one must assume $n>1$, as in Mohan's answer, otherwise the relative compactness might not hold.

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There are several ways to see this fact, which is a simple instance of Alexander duality. Here is the simplest I know.

Let $H$ be a compact smooth hypersurface in $\mathbb{R}^n$, whithout boundary, and $x\in \mathbb{R}^n\setminus H$. Then the radial projection $p_x : H \to \mathbb{S}^{n-1}$, $y\mapsto (y-x)/\|y-x\|$ has a degree mod $2$, say $d_x$, which may be defined as the number of elements mod $2$ of $p_x^{-1}(u)$ for almost all $u\in\mathbb{S}^{n-1}$ (this is well-defined by transversality theory). Then the subset of $x\in \mathbb{R}^n$ such that $d_x=1$ is your $\Omega$, a relatively compact open set with boundary $H$.