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.

  • $\begingroup$ 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 $\infty$? $\endgroup$ – Johnduck Jun 3 at 16:36
  • $\begingroup$ 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 $\infty$? $\endgroup$ – Paul Frost Jun 3 at 17:28
  • $\begingroup$ 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 $\hat{\mathbb C}$?" $\endgroup$ – Paul Frost Jun 3 at 17:43
  • $\begingroup$ 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 $\lvert z \rvert \to \infty$. $\endgroup$ – Paul Frost Jun 3 at 17:49
  • $\begingroup$ 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 $z$. $\endgroup$ – Paul Frost Jun 3 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.