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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.

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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 $i.$

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

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