Compact subset of an open set in the complex plane I would just like to ask a question. Suppose $K$ is a compact subset of an open set $V$ in the complex plane, how would you prove that there exists an $r > 0$ such that the union $E$ of the closed discs $\bar{D}(z;r)$, $z\in K$ is a subset of $V$.
My proof is suppose the claim is false then there exists a sequence $\{z_n\}$ such that $\bar{D}(z_n;1/n)$ is not a subset of $V$ for all $n$. This sequence has a sub-sequence $\{z_{n_k}\}$ that converges to $z$ in $K$ (compact). $V$ open so, $\bar{D}(z;r) \subset V$ for some $r > 0$, but $\bar{D}(z_{n_k};1/{n_k}) \subset \bar{D}(z;r)$ for sufficiently large $n_k$ so we have a contradiction. 
Just wondering if this is correct and whether there is a more obvious proof.
Thank you.
 A: Your proof is good, here is another that avoids explicit sequences.
Let $\phi(x) = \inf_{w \in V^c} d(x,w)$. We see that $\phi(x) > 0$ iff $x \in V$ since $V$ is open. $\phi$ 'measures' the distance from $x$ to the complement of $V$.
Since $d(x,v) \le d(x,y)+d(y,v)$, we see that $\phi(x) \le d(x,y) + \phi(y)$,
and swapping $x,y$ shows that $|\phi(x)-\phi(y)| \le d(x,y)$, hence $\phi$ is
continuous.
Now let $\alpha = \min_{x \in K} \phi(x)$. The minimum exists because $K$ is
compact and $\phi$ is continuous. Also, because the minimum is attained for
some $x^* \in K \subset V$, we see that $\alpha = \phi(x^*) >0$.
Now suppose $0<r<\alpha$, $x \in K$ and $y$ is such that $d(x,y) \le r$.
Then $\phi(y) \ge \phi(x)-d(x,y)\ge \alpha-r > 0$ and so $y \in V$.
A: I will write this in a slightly different way, but the idea is similar to copper.hat's proof.
Question: Let $A$ be a closed set and $K$ be a compact set such that $A$ and $K$ are disjoint. Define the distance 
$$d(A,K) = \inf \{d(a,k): a\in A, k\in K\}$$
Prove there exists $r > 0$ such that $A$ and $\bigcup_{k\in K} \overline{D}(k;r)$ are disjoint.
Outline of proof.


*

*$d(A,K) = d(A,k)$, for some $k\in K$. (Because the distance function is continuous and achieves a minimum on the compact set $K$)

*$d(A,K) > 0$, because $A$ is closed and $A$, $K$ are disjoint.

*Let $r = \frac{1}{2}d(A,K)$.

*$A$ and $\bigcup_{k\in K} \overline{D}(k;r)$ are disjoint.  (Suppose they are not disjoint and $a\in \bigcup_{k\in K} \overline{D}(k;r)$.  Then, for some $k\in K$, $d(a,k)\leq r < d(A,K)$.  Contradiction.)

