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

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?

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