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Let $G \subset \mathbb{C}$ be a region, $a \in G$ and $f : G \backslash\{a\} \to \mathbb{C}$ be injective and $\Omega = f(G \backslash\{a\})$ be bounded. Then $f$ has a removable singularity at $a$ and $f(a) \in \partial \Omega$.

This question is already answered here. I just have a silly question. Why $f(a) \in \partial \Omega$ is even possible? By open mapping theorem $\Omega$ is open, on the other hand $f$ can be extended to analytic function in $G$, but then $f(G) = \Omega \cup \{f(a)\}$ also has to be open, therefore $f(a)$ cannot be a boundary point for $\Omega$. Where is the mistake?

user285292
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1 Answers1

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The fundamental mistake is that the fact that a point $z\in \mathbb{C}$ is a boundary point for $\Omega \subseteq \mathbb{C}$ does not imply that $z$ is a boundary point for $\Omega \cup \{z\}$. Take the punctured unit disk $\mathbb{D}^* = \mathbb{D} \setminus \{0\}$, for example: $0$ is a boundary point for $\mathbb{D}^*$ since $0\in\overline{\mathbb{D}^*}$ and $0\in \overline{(\mathbb{D}^*)^c}$ (in fact, $0\in (\mathbb{D}^*)^c$). But $0\notin \overline{\mathbb{D}^c}$.

More generally, this will happen if and only if $z$ is an isolated point of $\mathbb{\Omega}^c$, so that $z$ is in the closure of $\mathbb{\Omega}$ but sufficiently small punctured neighborhoods of $z$ are contained in $\Omega$.

mheldman
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  • Sorry for bringing this up after 2 years but I have a question. How can any sufficiently small neighborhood of $z$ be contained in $\Omega$? Isn't $z$ an isolated point?

    Also, last line of your first paragraph, in the brackets. Do you mean "in fact, $0 \notin \mathbb{D}^*$?

     – Coward Aug 19 '20 at 20:31
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    Those are good questions. As for the (in fact.. ) parenthetical, I have no idea what I meant by that. Obviously $0\notin \mathbb{D}^*$; that was the whole point. I’ll just delete it.

    For the first question, what I said was correct: This will happen iff any sufficiently small punctured neighborhood of $z$ is contained in $\Omega$, key word being punctured. Meaning that, for any open set $U$ sufficiently small containing $z$, $U^* = U\setminus {z} \subseteq \Omega$.

     –  mheldman Aug 20 '20 at 21:07
  • @Coward Oh, I think I meant to say (in fact, $0\in (\mathbb{D}^*)^c$). It’s fixed now. Let me know if you have any more questions. –  mheldman Aug 20 '20 at 21:17