TITLE: Why does a meromorphic function in the (extended) complex plane have finitely many poles?
QUESTION [19 upvotes]: Let $f$ be meromorphic in $\mathbb{C} \cup \{\infty\}$. Why must $f$ have only finitely many poles?
Edit: Renamed question following the comments.

REPLY [4 votes]: Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $\mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K=\{|z|\leq R\}$. Suppose there are countably infinite of them, then they form a bounded sequence in $\mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.