0
$\begingroup$

The problem, as stated in the title, asks to prove that

Given a closed countable set $K\subset\mathbb{C}$, there's no biholomorphic map $f$ from $\mathbb{C}-K$to a subset of $\mathbb{D}$

Now, if the set is discrete, the result is easy: by the Riemann extension theorem, $f$ admits an extension to $\mathbb{C}$, which must be constant by Liouville's theorem.

If the set is not discrete, however, the result is not as trivial. My idea was: by Riemann, $f$ admits an extension to $\mathbb{C}-K'$, where $K'$ is the derived set of $K$. Since $K$ is countable, it is not perfect, so $K'\subsetneq K$, and we have extended $f$. Let $U_1$ be its domain and $K_1=K'$. We define, for every ordinal: $$ K_{\lambda+1}:=K_\lambda';\ U_{\lambda+1}:=\mathbb{C}-(K_{\lambda+1})$$ And for a limit ordinal $\lambda$: $$ K_{\lambda}:=\cap_{\alpha<\lambda} K_\lambda; U_\lambda:=\mathbb{C}-K_\lambda=\cup_{\alpha<\lambda} U_\alpha$$ It is easy to see that $f$ admits an extension to $U_\lambda$ for every ordinal $\lambda$. Since $K_{\lambda+1}\subsetneq K_\lambda$ and $K_0$ is countable, we have that $K_\lambda$ must reach $\emptyset$ before $\omega_1$, and thus $f$ admits an extension to $\mathbb{C}$, which is constant by Liouville.

Is my proof correct?

$\endgroup$
1
  • $\begingroup$ I am aware that this is not the most efficient proof, I'm just interested in verifying whether its correct. $\endgroup$ – Pelota Feb 22 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.