Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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) $lim_{t→∞}$ $γ(t)=∞$.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.