Is every compact set equal to an intersection of nested bounded sets with smooth boundary?

Question

Let $K$ be a compact set of (say) the plane $\mathbb{R}^{2}$. Do there exist bounded open sets $(U_{n})_{n\in\mathbb{N}}$ with $C^\infty$-boundary $\partial U_{n}$ such that $$\ldots\subseteq U_{n+1}\subseteq U_{n}\subseteq\ldots\subseteq U_{1}$$ and $$K=\bigcap_{n\in\mathbb{N}}\overline{U}_{n}?$$

Hand-wavy attempt: Proceed inductively over $n\in\mathbb{N}$. By compactness we can find finitely many $x_{j}\in K$ such that $C_{n}:=\bigcup_{j}B(x_{j},1/n)\supseteq K$. Define $\widetilde{C}_{n}:=C_{n}\cap U_{n-1}$. There are finitely many points where $\partial\widetilde{C}_{n}$ is non-smooth. Since $K$ is compact, we can cut out a sufficiently small part around these non-smooth points (away from $K$) to make it smooth. Define $U_{n}$ to be the resulting set.

Answer 1

Here is the standard procedure for constructing such open subsets in $\mathbb R^n$. Let $C\subset \mathbb R^n$ be any closed subset. For $r>0$ define its closed $r$-neighborhood $$ \bar{N}_r(C):=\{x: d(x, C)\le r\}$$ and open $r$-neighborhood $$ N_r(C):=\{x: d(x, C)< r\}$$ in $\mathbb R^n$, where $d$ is the Euclidean metric. For each $\epsilon>0$ define the standard bump-function $\varphi_\epsilon$ on $\mathbb R^n$, whose support is the closed ball $\bar{B}(0, \epsilon)$, see e.g. here.

Let $\chi_k$ denote the characteristic function of $N_{1/k}(C)$ and take the convolution $$ f_k=\chi_k \star \varphi_{\epsilon}. $$ Then $f_k$ is smooth and the support of $f_k$ is exactly $\bar{N}_{\epsilon+1/k}(C)$. For concreteness, I will take $\epsilon=\frac{1}{2k}$. Then $f_k$ is identically equal to $1$ on $N_{1/(2k)}(C)$ and its support is $\bar{N}_{3/(2k)}(C)$. We also have inclusions $$ \bar{N}_{3/(2(k+1))}(C)\subset N_{3/(2k)}(C) \subset int \bar{N}_{3/(2k)}(C). $$ By Sard's theorem, for almost every $t\in (0,1)$, the subset $$ W_k:=\{x: f_k(x)\ge t\}\subset N_{3/(2k)}(C) $$ will be a closed subset with smooth boundary, $C\subset W_k$. If $C$ is compact, then $N_k(C)$ will be bounded, hence, $W_k$ will be compact. Clearly, $$ \bigcap_{k\ge 1} W_k= \bigcap_{k\ge 1} N_k(C)=C. $$ In order to get nested domains, one has to make a more careful choice of the parameter $t=t_k$, I will make it sufficiently close to $0$. Then, since $$ \bar{N}_{3/(2(k+1))}(C)\subset int \bar{N}_{3/(2k)}(C), $$ we will also get $W_{k+1}\subset int W_k$.