Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Question: is "Idea" below flawed?

Let $\gamma$ be a closed curve in the complex plane. It may intersect itself, and required only to be continuous (no differentiability assumptions).

The image of $\gamma$ is compact, $\mathbb{C} \setminus \gamma$ is partitioned into at most countably many open connected components $A_i$.

Show: $\mathbb{C} \setminus A_i$ is connected, for every component $A_i$.

Idea: $\gamma$ is path connected. So if you can connect each point in $\mathbb{C} \setminus A_i$ to a point on $\gamma$ then $\mathbb{C} \setminus A_i$ is path connected. Fix a point $w^*$ on $\gamma$. Let $z$ be any point in $A_j$, where $j \ne i$. Draw a segment from $z$ to $w*$. Show using continuity and compactness that the segment must hit $\gamma$ before it hits a point in any other component.

I worked out the details, and it seems OK, but I'm suspicious because it seems too easy.

FYI, this is related to a problem in Ahlfors' Complex Analysis which is to show that the complement of a path in $\mathbb{C}$ is made up of some bounded and simply connected regios and a single 2-connected unbounded region.

share|cite|improve this question
up vote 2 down vote accepted

Every connected component of $\mathbb{C}\backslash A_i$ must interest $\gamma,$ which is connected. Therefore $\mathbb{C}\backslash A_i$ is connected for every $i.$

share|cite|improve this answer
Is it obvious that every component of $\mathbb{C}\setminus A_i$ has to intersect $\gamma$? – bryanj Jan 8 '14 at 18:57
Yes, how could it fail to? – Igor Rivin Jan 8 '14 at 19:00
Good point :) I guess it should be obvious. Thanks! – bryanj Jan 8 '14 at 19:06

The claim is incorrect if you mean to take the complement of the closure of $A_i$. Consider the union of the following two circles in the plane: the circle $(x-1)^2+y^2=1$ and $(x-2)^2+y^2=4$. This can easily be parametrized as a single closed curve. Now choose the earring-shaped region between the two circles for your $A_i$. The complement of its closure is not connected.

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.