An open subset with smooth boundary I have a question about construction of nice covering.
Let $X \subset \mathbb{R}^d$ be a $C^{1}$-domain. That is, $X$ is a connected open subset of $\mathbb{R}^d$ which satisfies the following condition:
For any $p \in \partial X$, there exist a $r>0$ and a $C^{1}$ mapping $\gamma:\mathbb{R}^{d-1} \to \mathbb{R}$ such that (upon rotating and relabeling the coordinate axes if necessary) we have 
\begin{align*}
X \cap B(p,r)=\{y \mid \gamma(y_{1},\ldots,y_{d-1})<y_{d}\} \cap B(p,r),
\end{align*}
where $B(p,r)$ denotes the open ball centered at $p$ with radius $r>0$.
(Although there are several ways to define the $C^1$ domain, in this post we adopt this definition.)
My question
Let $D \subset \mathbb{R}^d$ be an unbounded $C^1$-domain. Let $K$ be a compact subset of $\bar{D}$. Then, can we construct an open subset $U \subset \mathbb{R}^d$ which satisfies the following?:


*

*$U \cap D$ is a bounded $C^{1}$-domain and $K \subset U$.


I think the existence of this open subset is not trivial.
Thanks for any information.
 A: More of a comment: given a compact $K \subset D$, there exists $V$ a union of finitely many closed cubes so that $K \subset \overset{\circ}{V}$ and $V\subset U$. Again, there exists $W$ a finite union of closed cubes so that $V \subset \overset{\circ}{W}$ and $W \subset U$. So you may substitute $K$ with $V$ and $U$ with $\overset{\circ}{W}$, and look for $C^{1}$ domain $U$,  $V \subset U \subset W$. This looks a little easier now. 
A: This is also not a complete answer, but I've thought about the problem for a while and maybe can give you some new ideas. 
1) On the manifold $M=\partial D$, define $f(p) = \vert p \vert^2$, the usual squared distance to the origin. Assuming a little more regularity ($M \in C^{d-2}$) one can apply Sard's theorem to $f$. In particular there is some $R>0$ such that $K$ lies in the ball $B(R)$ of radius $R$ and $R$ is a regular values of $f$, meaning that $M\cap \partial B(R) = f^{-1}(R)$ is a (closed) manifold. Let's decompose it into its connected components: $M\cap \partial B(R) = \Gamma_1 \coprod ... \coprod \Gamma_m$.
2) Over each $\Gamma$ I picture the situation as follows: (I am confident that this is quite accurate for $d=3$, but of course needs some more thought in higher dimensions.)

Now the obvious idea is to try and cap $D$ off to obtain a $C^1-$domain $U$ with $ D \cap B(R) \subset U \subset D \cap B(R+1)$
3) Let's consider a toy problem in dimension $d=3$. Assume $\Gamma$ is a closed curve in the $xy-$plane of $\mathbb{R}^3$ (and not on the sphere) which bounds a domain $\Lambda$. Now consider a function $f: \bar \Lambda \rightarrow [0,\infty)$ with $f(\Gamma) = 0$ and its graph $M \subset \mathbb{R}^3$ over $\bar \Lambda$. The question from earlier translates to: Does there exists a function $f_1: \bar \Lambda \rightarrow [0,\infty)$ with $f_1 \le 1$ such that $f = f_1$ in a neighbourhood of $\Gamma$? 
Coming up with this is not too hard: Let $\rho: [0,\infty) \rightarrow [0,\infty) $ be a smooth function with $\rho(t) = 1$ for $0 \le t < 1/3$ and $\rho(t) = 0 $ for $t> 2/3$. Then you can put $f_1(x) = \rho(f(x))f(x) +(1-\rho(f(x))$.
I hope this helps a bit.
