# Does the reverse of Picard's great theorem hold?

The Picard's great theorem says:

Asume that $f$ is holomorph in a punctured neighborhood $\Omega$ of $c\in\mathbb C$. Furthermore assume that $f$ has a essential singularity at $c$, then for any neighborhood $U\subseteq\Omega$ of $c$ $\mathbb C\setminus f(U)$ has atmost one element.

Does it hold if we reverse the second part? That is:

Asume that $f$ is holomorph in a punctured neighborhood $\Omega$ of $c\in\mathbb C$. Furthermore assume that for every neighborhood $U\subseteq\Omega$ of $c$ $\mathbb C\setminus f(U)$ has at most one element, then $f$ has a essential singularity at $c$.

If yes: how do I prove it? If no: is there a counter example?

• As far as I know the picard theorem is one of the most difficult result in complex Analysis. For I would first say No but i dont have any justifaction Commented May 19, 2017 at 11:43
• I have a feeling this is true - essential singularities are characterised by the limit $\lim_{z \to c} f(z)$ not existing, and it looks like the assumption should show that directly. Commented May 19, 2017 at 11:47
• Yes, the converse holds (and is much easier to prove). If the singularity is not essential, then $\mathbb{C}\setminus f(U)$ contains a disk for every small enough $U$. Commented May 19, 2017 at 11:48

1. Suppose that $c$ is a removable singularity. Then, by Riemann, $f$ is bounded in a punctured neighborhood $U$ of $c$. Hence $\mathbb C\setminus f(U)$ has more than one element, a contradiction.
2. Suppose that $c$ is a pole. Then $|f(z)| \to \infty$ for $z \to c$. Hence there is a punctured neighborhood $U$ of $c$ such that $|f(z)| \ge 1$ for all $z \in U$. Hence $\{w \in \mathbb C: |w|<1\} \subset \mathbb C\setminus f(U)$ , a contradiction.