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Is it true that given any unbounded domain $D$ strictly in $\mathbb C$ and a boundary point $z_0$, and two real numbers $R > r > 0$, I can choose some $r' < r$ and $R' > R$ such that $\{ z \in D : r' < |z - z_0| < R'\}$ is a domain?

Here an open subset of $\mathbb C$ is called a domain if any two points in the subset can be joined by gluing straight broken line segments in the subset.

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No. Let $$D = \mathbb C \setminus \{ z = x + iy \mid x \in \{0\} \cup \{1/n \mid n \in \mathbb N \}, y \ge 0 \} .$$ Then $z_0 = 0$ is a boundary point of $D$. Now let $A$ be any annulus around $0$. Then $A \cap D$ has infinitely many connected components.

qualcuno
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Paul Frost
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  • I am not sure if $D$ is a domain here. – Trash Failure Apr 30 '21 at 10:57
  • @TrashFailure In fact each open and connected subset is a domain in your sense. See my answer to https://math.stackexchange.com/q/3960180. In the present case it is even simpler: Each point of $D$ can be connected by a vertical line segment with the line $L = {x+iy \mid y = -1}$. – Paul Frost Apr 30 '21 at 11:01