Let $A \subset \mathbb{R}^{n}$ be a closed set. Does there exist an open set $O$ containing $A$, and a smooth function $f : O \to \mathbb{R} $ such that
$f(x) = 0$ for all $x \in A$,
$f(x) > 0$ and $\nabla f(x) \ne 0$ for all $x \in O \setminus A$,
and $f(x) \to \infty$ as $x \to \partial O$ ?
Not always possible.
Let $n = 1$. Take $A \subsetneq [0,1]$ be the usual ternary Cantor set.
Every point in $A$ is a limit point.
Let $O$ be any open set containing $A$, then $O$ must contain some interval of the form $(a,b)$ with $a,b\in A$. Since $f(a) = f(b) = 0$ is needed, if we want $f(x) > 0$ on $O\setminus A$, we must have that $f(x)$ attains an interior maximum where it has a critical point.
Similar examples can be built in higher dimensions.
