Are $\mathbb{C}\backslash[0,1]$ and $\mathbb{C}\backslash \left(\bigcup\limits_{n\in\mathbb{N}_{>0}}\{t \cdot \exp(2 \pi i /n):t\in[0,1/n]\} \right)$ biholomorphic?
A friend told me, they are biholomorphic, but he didn't know the map. Could you help me?
In my opinion they are not biholomorphic, because the stuff we are deleting at the second is the harmonic series, if I summ the length of the Intervals. Does it work, if we replace $1/n$ above with $1/n^2$?
The two domains are biholomorphically equivalent, but I can't give you an explicit biholomorphism between them, unfortunately.
To see that they are biholomorphically equivalent, note that by adjoining $\infty$ to both of them, we obtain two simply connected domains $U_1 = \hat{\mathbb{C}}\setminus [0,\,1]$ and $U_2 = \hat{\mathbb{C}} \setminus \left(\bigcup_{n > 0} \{ t\cdot\exp{2\pi i/n} : t \in [0,\,1/n]\}\right)$ in the Riemann sphere.
Since the complement of both, $U_1$ and $U_2$, contains more than one point, by the Riemann mapping theorem, there are biholomorphisms $\varphi_1 \colon \mathbb{D} \to U_1$ and $\varphi_2 \colon \mathbb{D} \to U_2$. Let $z_1 = \varphi_1^{-1}(\infty)$ and $z_2 = \varphi_2^{-1}(\infty)$. Let $\psi \in \operatorname{Aut}(\mathbb{D})$ with $\psi(z_1) = z_2$. Then $\varphi = \varphi_2 \circ \psi \circ \varphi_1^{-1}$ is a biholomorphism $U_1 \to U_2$ with $\varphi(\infty) = \infty$. Restricting $\varphi$ to $\mathbb{C}\setminus [0,\,1]$ gives a biholomorphism between the two domains in question.
