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

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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.

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  • $\begingroup$ 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

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