Suppose a region S is simply connected and contains the circle

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) $d(γ(t),\widetilde{D}))<E$ for all $t≥0$,

(b) $γ_{0}=z_{0}$,

(c) $lim_{t→∞}$ $γ(t)=∞$.

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Function names such as "min" get interpreted as a juxtaposition of variables names if you just write them out like that; that causes them to be italicized and to get the wrong spacing. To get proper formatting, you can use the predefined commands such as \min. If you need a name for which there isn't a predefined command, you can use e.g. \operatorname{Var} to produce $\operatorname{Var}$. –  joriki May 1 '12 at 6:17
A first problem is that the maximum in the definition of $\delta_z$ doesn't exist, since $S$ is open. I think you'll want to use something like half the supremum. In the last paragraph, is $z$ supposed to be $z_0$? Is $\widetilde S$ the complement of $S$? What's $d(\gamma,\widetilde S)$? I can't think of any interpretation that would let me follow that argument. (By the way, there were a lot more formatting problems; I suggest to take a look at my edits (by clicking on the "edited ..." time stamp) so you can avoid them next time.) –  joriki May 1 '12 at 6:30

1 Answer

You should tell us what you have been told about "simply connected" so that we know the rules of the game.

As a starter here is a proof of the HINT, even though I couldn't figure out the intention of this hint:

Each point $z\in C$ has a neighborhood $U_{2\delta(z)}(z)\subset S$, and as $C$ is compact there are finitely many $z_k\in C\$ $(1\leq k\leq N)$, such that $C\subset\bigcup_{1\leq k\leq N} U_{\delta(z_k)}(z_k)$. Put $\delta:=\min_{1\leq k\leq N}\delta(z_k)>0$. Then, given any $z\in B$, let $z'$ be its projection onto $C$. There is a $k\in[N]$ such that $$|z-z_k|\leq|z-z'|+|z'-z_k|<\delta+\delta(z_k)\leq 2\delta(z_k)\ ,$$ and it follows that $z\in U_{2\delta(z_k)}\subset S$.

Edit. Given your definition of "simply connected", after the proof of the HINT you can proceed as you have proposed: If $\epsilon<\delta$ then there is no curve from $z_0$ to $\infty$ of the required kind.

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