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

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.

• Hint: Use convolutions of the characteristic function with suitable bump-functions. Commented Jul 8 at 15:58
• @MoisheKohan Can you elaborate? Commented Jul 8 at 17:42
• Use mollification as explained here and take generic sup-level sets of the mollifiers. These will have smooth boundary by Sard's theorem. Commented Jul 8 at 17:49

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 $$W_{k+1}\subset int W_k$$.
• Thanks for your answer! Just some minor comments/questions. As definition for the opens $U_k$ we can take $\mathrm{int}(W_{k})=\{x:f_k(x)>t\}$, right? And how do we get the domains $U_k$ nested (since taking $t$ smaller makes $W_k=W_k(t)$ larger)? Commented Jul 9 at 13:40