Bounded holomorphic functions on $\mathbb{C} \smallsetminus K$ are constant.

Question

Suppose $K$ is a countable closed subset of the complex plane $\mathbb{C}$ and let $f$ be a bounded holomorphic function on $\mathbb{C}\smallsetminus K$. Why must $f$ be a constant?

Answer 1

If $K$ is finite, all its points are isolated, and since $f$ is bounded, removable singularities. Removing the singularities, we obtain a bounded entire function, which is constant by Liouville's theorem.

If $K$ is countably infinite, it contains isolated points (a closed set without isolated points must have the cardinality of $\mathbb{C}$). Let $I_0$ be the set of isolated points of $K$, and $K_1 = K \setminus I_0$. Again, the singularities in the points of $I_0$ are removable. After removing them, we have a bounded holomorphic function $f_1$ on $\mathbb{C}\setminus K_1$. The set $K_1$ is countable (finite or countably infinite), and closed (since it consists of the limit points of $K$).

If $K_n \neq \varnothing$, we can iterate the continuation process. Let $I_n$ be the set of isolated points of $K_n$, and $K_{n+1} = K_n \setminus I_n$. By removing the removable singularities in $I_n$, we obtain a bounded holomorphic function $f_{n+1}$ on $\mathbb{C}\setminus K_{n+1}$.

If we don't reach $K_m = \varnothing$ after finitely many steps, let $K_{\omega} = \bigcap\limits_{n\in\mathbb{N}} K_n$. $K_\omega$ is a closed countable set, and we have a bounded holomorphic function $f_\omega$ on $\mathbb{C}\setminus K_\omega$ ($f_\omega(z) = f_n(z)$ if $z \notin K_n$ is a well-defined holomorphic function there).

For every countable ordinal $\alpha$, continue the construction, if $\alpha$ is a successor ordinal, $\alpha = \beta+1$, let $I_\beta$ the set of isolated points of $K_\beta$, and $K_\alpha = K_\beta \setminus I_\beta$, remove the removable singularities in $I_\beta$ to obtain a bounded holomorphic function on $\mathbb{C}\setminus K_\alpha$. If $\alpha$ is a limit ordinal, let

$$K_\alpha = \bigcap_{\beta\in\alpha} K_\beta;\quad f_\alpha(z) = f_\beta(z) \text{ if } \beta\in\alpha \text{ and } z \notin K_\beta.$$

Let $\omega_1$ be the first uncountable ordinal. Then we can form $K_{\omega_1}$ in the manner above, and we must have $K_{\omega_1} = \varnothing$, since up to then, there have been uncountably many steps where if $K_\alpha$ was not yet empty, at least one point was removed.

Then $f_{\omega_1}$ is a bounded entire function, hence constant.

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If $K$ would have been supposed to be closed and discrete, we would not have needed transfinite induction, that would have given the result after the first step.