# closed disk in open set

Let $A\subseteq\mathbb C$ be an open set, and let $\overline D$ be a closed disk contained in $A$. Why there is always another open disk $B$ such that $\overline D\subset B\subseteq A$ ?

In short: because $\overline{D}$ is compact (unless $D = A = \mathbb{C}$). If $A = \mathbb{C}$, there is nothing to show, so let's assume $A \neq \mathbb{C}$. Consider the function

$$\varphi(z) = \operatorname{dist}(z, \mathbb{C}\setminus A).$$

That is a continuous function, and vanishes nowhere on $\partial D$. Since $\partial D$ is compact, there is an $\varepsilon > 0$ such that $\varphi(z) > \varepsilon$ for all $z \in\partial D$. Then the triangle inequality says that

$$B_\varepsilon(D) = \{z : \operatorname{dist}(z,D) < \varepsilon\}$$

is contained in $A$.

• Is dist as distance function? :) – mrs Aug 6 '13 at 10:59
• Yes, $\operatorname{dist}(z,\,M) = \inf \{\lvert z - m\rvert : m \in M\}$. – Daniel Fischer Aug 6 '13 at 11:03
• I don't quite understand why the triangle inequality is needed. – Stefan Hamcke Aug 6 '13 at 17:08