Let $f$ be a meromorphic function on the Riemann-sphere $\hat{\mathbb{C}}$. I want to show that $f$ can only have finitely many poles. I know the set of poles is discrete, and if I can prove that this set is closed, I am done since a closed discrete subset of a compact set is neccesarily finite. For closedness, I run into several issues.
I was thinking, the set of poles is the complement of $U$, the set of points where $f$ is holomorphic. If would be a function on just $\mathbb{C}$, I would be done since I would know that $U$ is open. Is the same true for the $\hat{\mathbb{C}}$?
Best Answer
A meromorphic function $f$ on a region $D \subset \hat{\mathbb{C}}$ is defined as a function such that each $z \in D$ has an open neighborhood $U \subset D$ such that either $f$ is holomorphic on $U$ or $f$ is holomorphic on $U \setminus \{ z \}$ and has a pole in $z$.
This implies that the set $P \subset D$ of poles of $f$ is discrete and closed in $D$ (note that $D \setminus P$ is open in $D$ by definition). The set $P$ may of course have accumulation points in $\hat{\mathbb{C}} \setminus D$ if the latter is nonempty.
In your question we have $D = \hat{\mathbb{C}}$ which is compact. Hence $P$ is a compact discrete set and therefore finite.
Note that this is an immediate consequence of the definition of a meromorphic function. "Holomorphic on $D$ with the exception of isolated poles" means in particular that the set of non-poles is open - otherwise it wouldn't make sense to speak about holomorphy in these points.