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Let $A,B$ be open and connected subsets of a simply connected domain $Ω\subset \mathbb{C}$, such that

$$\partial Ω\cap \partial A\neq\emptyset,\partial Ω\cap \partial B\neq\emptyset, \partial A\cap \partial B\neq\emptyset$$ and the set

$$(\partial Ω\cap \partial A)\cup (\partial Ω\cap \partial B)$$ is a connected subset of $\partial Ω$. Does this imply that

$$A\cap B\neq \emptyset?$$

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Consider the case

\begin{align*} \Omega &= \mathbb{D}\\ A &= \{z \in \mathbb{D} \mid \operatorname{Im}(z) > 0\}\\ B &= \{z \in \mathbb{D} \mid \operatorname{Im}(z) < 0\}. \end{align*}

Note that we have

\begin{align*} \partial\Omega &= S^1\\ \partial A &= \{e^{i\theta} \mid 0 \leq \theta \leq \pi\}\cup[-1, 1]\\ \partial B &= \{e^{i\theta} \mid \pi \leq \theta \leq 2\pi\}\cup [-1, 1] \end{align*}

and therefore

\begin{align*} \partial\Omega\cap\partial A &= \{e^{i\theta} \mid 0 \leq \theta \leq \pi\} \neq \emptyset\\ \partial\Omega\cap\partial B &= \{e^{i\theta} \mid \pi \leq \theta \leq 2\pi\} \neq \emptyset\\ \partial A\cap \partial B &= [-1, 1] \neq \emptyset. \end{align*}

Furthermore, $(\partial\Omega\cap\partial A)\cup(\partial\Omega\cap\partial B) = S^1$ which is a connected subset of $\partial\Omega$, so all of your conditions are met. However, $A$ and $B$ are disjoint.


In the above example, $\partial\Omega\cap\partial A\cap\partial B = \{-1, 1\}$ is disconnected. If you want an example with $\partial\Omega\cap\partial A\cap\partial B$ connected, consider $\Omega = \mathbb{D}$, $A = \{z \in \mathbb{D} \mid |z - \frac{1}{2}| < \frac{1}{2}\}$, and $B = \Omega\setminus\bar{A}$.

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  • $\begingroup$ Thanks for the answer. The boundaries on the unit circle of these sets intersect on just two points. What if also the respective boundaries of the open without their endpoints intersect? $\endgroup$ Mar 14, 2016 at 23:19
  • $\begingroup$ I have provided an alternative example above which satisfies the additional condition $\partial\Omega\cap\partial A\cap\partial B$ connected. $\endgroup$ Mar 14, 2016 at 23:41

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