# $f$ meromorphic on $\mathbb{\hat{C}}$ $\implies$ $f$ has a finite number of poles

Setting: Let $f: \mathbb{\hat{C}}\rightarrow \mathbb{\hat{C}}$ be a meromorphic function. Let $\{p_k\}$ denote the set of poles of $f$ inside $\mathbb{\hat{C}}$.

Question: Why must $\{p_k\}$ be a finite set?

Attempt:

I know only how to reason that $\{p_k\}$ is a countable set. To see this, consider that the poles $p_k$ of $f$ are isolated by definition. Hence for each $p_k$ there is associated a neighborhood $N_{p_k}$ about $p_k$ s.t. there are no poles inside $N_{p_k}$. Then we have that

$$j \ne k \implies N_{p_k} \cap N_{p_j} = \emptyset$$

so that if $\{p_k\}$ were uncountable, we would have that $\bigcup N_{p_k} \subsetneq \mathbb{\hat{C}}$ which is absurd (an uncountable number of neighborhoods of $\mathbb{\hat{C}}$ must cover $\mathbb{\hat{C}}$).

So how do I show that $\{p_k\}$ is finite?

• The set of poles is a closed discrete subset of the sphere. The sphere is compact. Commented Apr 28, 2014 at 15:16
• Daniel, I used your comment below to construct a proof. Does it look right? Commented Apr 28, 2014 at 16:17
• See this question. Commented Apr 28, 2014 at 16:27
• Also, more directly (this is just the topology on $\hat{\mathbb{C}}$): we say $f$ has a pole at $\infty$ if $f(\frac 1z)$ has a pole at $0$. For this singularity to be isolated, say there is some neighborhood of radius $\frac 1\delta$ on which there are no other poles. Then $f(z)$ has no poles for $|z|>\delta$, and since the ball of radius $\delta$ is compact, there are finitely many poles in there.
– user59193
Commented Nov 26, 2014 at 9:42

1. Let $U = \mathbb{\hat{C}} - \{p_k\}$. Since $U$ is the complement of a set of discrete points in $\mathbb{\hat{C}}$, it is open.
2. Then $\{N_{p_k}\} \cup U$ is an open covering of the compact space $\mathbb{\hat{C}}$.
3. Then there exists a finite subcovering from $\{N_{p_k}\} \cup U$ which covers $\mathbb{\hat{C}}$ entirely.
4. Since the $\{N_{p_k}\}$ are a collection of disjoint sets (by construction), and since $U$ does not cover any of the $p_k$ (by construction), we have that none of the $N_{p_k}$ can be eliminated from the original covering. Hence $\{N_{p_k}\}$ must be a finite set already.
5. Hence $\{p_k\}$ is a finite set, as desired.
• There's a problem in point 1. Maybe. When you say "a set of discrete points in $\hat{\mathbb{C}}$", do you mean that the set in the subspace topology is discrete? Then that isn't sufficient to conclude that $U$ is open (e.g. $\mathbb{Z}$ is discrete in the subspace topology, but $\hat{\mathbb{C}}\setminus\mathbb{Z}$ is not open). You need that every point in $\hat{\mathbb{C}}$ has a neighbourhood that contains only finitely many $p_k$ to conclude that $U$ is open. Commented Apr 28, 2014 at 17:29
• Then I could say that $U$ is open since it is the domain of an analytic function $f$, and analytic functions have open domains by definition. Commented Apr 28, 2014 at 17:39