Is there a smooth compact set between any two compact sets? today I saw the following statement and of course believe that this is true but I don't know how to prove it rigorously (and neither do my colleagues).
Let $\Omega \subset \mathbb{R}^n$ be open and bounded and $K\subset \Omega$ compact. Then there is a set $A$ such that $K\subset A\subset \Omega$ with $\partial A$ smooth (smooth means at least $C^1$).
However, my first ansatz was: Mollify the characteristic function of $K$ such that its support is in $\Omega$ and denote this functions by $g$. Now my intention was to use the regular value theorem on $g$. Like $g^{-1}(1/2)$ should be a smooth manifold. But I can't show whether this point is a regular point nor do I know that $\{x :g(x)\geq 1/2\} \supset K$. Does anyone have a different or additional idea on this?
 A: For each $p\in K$, let $B(\delta_p,p)\subset\Omega$ where $B(\delta_p,p)$ is an ball of ratio $\delta_p$ and ceter $p$. Because $K$ is compact, you can find a finite number of balls $B(\delta_{p_i},p)$ such that $$K\subset\bigcup_{i=1}^mB(\delta_{p_i},p)\subset\Omega$$
To finish in we have to consider the points where any two balls intersect. Denote by $A=\cup_{i=1}^mB(\delta_{p_i},p)$.
Without loss of generality consider that $n=2$. The procedure is the same for $n>2$. Take two balls and consider their boundarys that are the circles $\partial B_1$ and $\partial B_2$. We can suppose that the points in the intersections of the circle lie whitin the axis $y$ in $\mathbb{R}^2$. Take some $\epsilon>0$ small and erase the part of the circles contained in $(-\epsilon,\epsilon)\times\mathbb{R}$. We now have two arcs of circles. We join these two arcs by two lines (if we were in dimension $n>2$, we could use a cylinder). 
Now we have something like this. By using a argument of partition of unity, you can regularize all the part that is the intersection between the lines and the circles. If you take $\epsilon>0$ small you still have that $A\subset \Omega$. 
A: First let $\Delta\Omega$ be the boundary of $\Omega$ as $\Omega$ is open and bounded $\Delta\Omega$ is compact and disjoint from $\Omega$ (hence does not intersect $K$).
Now as $K$ and $\Delta\Omega$ are non intersecting compact sets there exists some $\varepsilon>0$ such that no pair of points $x\in K, y\in\Delta\Omega$ have $\|x-y\|<2\varepsilon$ Furthermore we may choose a finite number of points $x_1,\dots, x_n\in K$ and $y_1,\dots y_m\in\Delta\Omega$  such that the open $\varepsilon$ balls $\mathcal B(\varepsilon, x_i)$ and $\mathcal B(\varepsilon, y_i)$ cover $K$ and $\Delta\Omega$ respectively. 
Now let $A'\subset\Omega$ be the set of points $p$ such that there exists some $i$ for which $\|p-x_i\|<\|p-y_j\|$ for every $j\leq m$.
Then for every point $p\in\Delta A'$ there must exist some pair $i,j$ such that $\|p=x_i\| = \|p=y_j\|>\varepsilon$.
Therefore $\Delta A'$ is a simplex and for every $p\in\delta A$ the ball $\mathcal B(p,\frac \varepsilon 2)$ is contained in $\Omega\setminus K$.
I can't remember a theorem off the top of my head, but I'm pretty sure there's a smooth modification of $\Delta A'$ that stays within $\frac\varepsilon 2$ of $A$.
