Consider a smooth function $f:\Bbb R^n\to \Bbb R$, and a noncompact regular level subset $S=f^{-1}(c)$ for some regular value $c\in \Bbb R$. Then $S$ is a smooth hypersurface in $\Bbb R^n$, so it is an $(n-1)$-dimensional manifold. I want to show that $S$ is the interior of a compact manifold with boundary. I expect that for a sufficiently large closed disk $D\subset \Bbb R^n$ centered at the origin, the intersection $D\cap S$ is such a manifold with boundary, but I'm not sure. Is this statement true? or is there another approach? Thanks in advance.
Edit: According to the comment, there is a counterexample even in the case $n=1$. Actually my siuation is the following: I want to show that for $S=\{z_1^{n_1}+z_2^{n_2} +z_3^{n_3}=c\}$ in $\Bbb C^3$ (where $n_1,n_2,n_3$ are positive integers and $c$ is a nonzero complex constant, $S\cap D$ is a manifold with boundary whose interior is homeomorphic to $S$, where $D$ is a sufficiently large closed disk $D^6$ in $\Bbb C^3$. I thought this result can be generalized to the situation described above, but I was wrong.