# Finite number of poles meromorphic function on Riemann-sphere

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 $$\hat{\mathbb{C}}$$?

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.

• You say that $D\setminus P$ is open $\textit{by definition}$, but I don't see how this works. If $f$ is holomorphic at $z_0\in \mathbb{C}$, it is also holomorphic at an open neighbourhood of $z_0$, but how does the same work for $\infty$? – Johnduck Jun 3 at 16:36
• Holomorphic at $z_0$ means that $f$ is holomorphic on some an open neighborhood $U$ of $z_0$. In particular. it must be defined on the whole set $U$. Is your question what it means that $f$ is holomorphic at $\infty$? – Paul Frost Jun 3 at 17:28
• In that case I suggest you ask another question, perhaps "What is the definition of a holomorphic map $f : U \to \mathbb C$ when $U$ is an open neighborhood of $\infty$ in $\hat{\mathbb C}$?" – Paul Frost Jun 3 at 17:43
• Note that a pole at $\infty$ is perhaps easier to understand: It simply means that $f$ is holomorphic in $U \setminus \{\infty\}$ and $\lvert f(z) \rvert \to \infty$ as $\lvert z \rvert \to \infty$. – Paul Frost Jun 3 at 17:49
• Final remark: Let us accept that there is a reasonable interpretion of "$f$ holomorphic at $\infty$". No matter how it is defined, the essence of my answer is this. For a meromorphic $f$ there are two types of points $z \in D$. Type 1: $f$ is holomorphic in an open $U$ neigborhood of $z$. Type 2:$f$ is holomorphic in $U \setminus \{ z \}$, $U$ an open neigborhood of $z$, and has a pole at $z$. – Paul Frost Jun 3 at 18:47