$K$ compact set $\subset A$ , open set of $\mathbb{C}$, implies that there exists $D$(smooth boundary) s.t. $K \subset D\subset \bar{D} \subset A$ I consider this situation: I have an open set $A\subset \mathbb{C}$ and a compact set $K\subset A$ and  I want to prove that there exists an open set $D$ such that $K \subset D\subset \bar{D} \subset A$ $\textbf{with piecewise smooth boundary}$.
In order to prove this fact should I use the theorem of decomposition of an open set ( there exists a sequence of open sets $\{U_i\}$ such that ${\overline U _i}$ is compact and ${\overline U _i} \subset {U_{i + 1}}$ and $U = \bigcup\limits_i {{U_i}}$.)?
How can I use it?
In this way I can prove that there is always a set $D$ s.t. $K \subset D\subset \bar{D} \subset A$ and $\partial D$ is piecewise linear, since I can construct every set $U_i$ with a piecewise linear boundary, am I right?
Many thanks for the help!
 A: You can do it from scratch too. Let $(B_z)_{z\in K}$, $B_z=B(z,r_z)$ be a collection of open balls for each point of $K$ that fit in $A$. Then, let $(S_z)_{z\in K}$ be a collection of open squares centered at $z$ and side length $r_z$ $$S_z=(\Re z - \frac12{r_z},\Re z + \frac12r_z)\times(\Im z -\frac12r_z, \Im z + \frac12r_z).$$
Note that $z \in S_z \subset \overline{S_z} \subset B_z \subset A$. Now, because $K$ is compact and $(S_z)_{z\in K}$ is its cover, we can pick a finite subcover $(S_i)_{i=0,\ldots,n}$ and set $D=\bigcup_{i=0}^nS_i$.
I hope this helps $\ddot\smile$
A: Let's try something simpler. 
$A$ is open, thus $F=\mathbb C\smallsetminus A$, closed, and as $K$ is compact, with $K\cap F=\varnothing$, then
$$
d=\mathrm{dist}(K,F)=\inf\big\{\lvert z-w\rvert : z\in F\,\&\,w\in K\big\}>0.
$$
Define
$$
D=\{z\in\mathbb C: \mathrm{dist}(K,z)<d/2\}.
$$
Then $D$ is open and
$$
K\subset D \subset\overline{D}\subset A.
$$
A: Since $A$ is an open cover of $K$, for each point $z \in K$ there is an open ball of radius $2r_z$ which is entirely contained in $A$. Hence for each point $z$ of $K$ the open ball  of radius $r_z$ is also contained entirely in $A$. We denote by $\mathscr{F}$ the family of open balls $\{B(z, r_z)\}_{z\in K}$. By the Heine-Borel property, a finite subfamily of $\mathscr{F}$ which we denote
\begin{equation}
B(z_1,r_1), \, B(z_2,r_2), \, \ldots, \,B(z_i,r_i)
\end{equation}
covers $K$. Let $\rho=\min \{r_1, r_2, \ldots, r_i\}$. Notice that if $w \in K$, then $w$ is in one of the balls $B(z_j,r_j)$. Hence we have
\begin{equation}
B(z,\rho) \subset B(z_j,r_j+\rho) \subset B(z_j,2r_j) \, .
\end{equation}
The ball $B(z_j, 2r_j)$ was chosen so that it is completely contained in $A$. Since $w$ is an arbitrary point of $A$, this holds for all points of $K$. Since the set $\underset{z \in K}{\bigcup} B(z, \rho)$ is open, the result follows.
What is being used here is the Lebesgue Number Lemma.
