# Continous function defined by supremum of Radius of disc

Let $S\subset \mathbb{C}$ be a simply connected domain (i.e. every point in complement of $S$ can be connected to $\infty$). Let $C=\{z:|z-\alpha|=r\}$. Would you help me to show that for all $z\in S$, $\delta(z)=\sup \{u:D(z,u)\subset S\}$ is continuous function?

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Suppose $z, w \in S$ and $w \in D(z, u)\subseteq S$. It is not hard to show, using the triangle inequality, that $$D(w, u - |w-z|) \subseteq D(z,u).$$ Taking the supremum over $u$ yields $$\delta(w) \geq \delta(z) - |w-z|,$$ or rearranging terms $$\delta(z) - \delta(w) \leq |w-z|.$$ Everything we have done is symmetric in $z$ and $w$, so we actually have $$|\delta(z) - \delta(w)| \leq |w-z|,$$ and thus $\delta$ is continuous.
Wait, we can't claim that we can find u such that $w\in D(z,u) \subset S$. Consider the case when $\delta (z)<|z-w|$ – beginner Dec 7 '12 at 0:25
For proving continuity we need only consider points $z, w$ that are very close together. Assuming $\delta(z)$ is not zero then we can indeed find a $w \in S$ such that there is a disk around $w$ which contains $z$ and a disk around $z$ which contains $w$. Once you have done this, the argument is as I said. – Travis Dec 7 '12 at 19:00