Suppose a region $S$ is simply connected and contains the circle $C =\{z:|z-\alpha|=r\}$. Show then that $S$ contains the entire disc $D=\{z:|z-\alpha|\leq r\}$.
HINT OF THE BOOK: Show that since $S$ is open (by definition) and $C$ is compact, $S$ contains the annulus $B = {z:r−δ ≤ |z−α|≤r+δ}$ for some $δ>0$.
MY SOLUTION: Let $C=\{z:|z-a|=r\}$ and consider $\delta_{z}=\max \{t:D(z;t)\subseteq S\}$.
$\delta_{z}$ is a continuous function of $z\in C$ and $\delta=\min_{z\in C} \delta_{z}$ exists. Hence, the annulus $B={z:r-\delta\leq|z-\alpha|\leq r+\delta}$ is contained in $S$.
It follows that any $z_{0}\in D(\alpha;r)$ must belong to $S$. For any path $\gamma$ connecting $z$ to $\infty$ must intersect $C$, and, at that point, $d(\gamma,\widetilde{S})\geq \delta$.
Any suggestions to improve the exercise?
A region D is simply connected if its complement is “connected within to ∞.” That is, if for any $z_{0}∈\widetilde{D}$ and $E> 0$, there is a continuous curve γ (t), $0≤t<∞$ such that
(a) $d(γ(t),\widetilde{D}))
(b) $γ_{0}=z_{0}$,
(c) $lim_{t→∞}$ $γ(t)=∞$.