# Why does a meromorphic function in the (extended) complex plane have finitely many poles?

Let $f$ be meromorphic in $\mathbb{C} \cup \{\infty\}$. Why must $f$ have only finitely many poles?

Edit: Renamed question following the comments.

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The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $\mathbb C\cup\{\infty\}$ is usually called the extended complex plane or the Riemann sphere. –  Did Oct 5 '11 at 9:12
As an explicit counterexample to the title question, consider $\frac{1}{2 - e^z}$. –  Qiaochu Yuan Oct 5 '11 at 15:22

By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.

The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)

Since $\mathbb C \cup \{\infty\}$ is compact, any discrete and closed subset is discrete and compact, hence finite.

(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $\mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $\mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $\infty$. A typical example is given by $f(z) = 1/\sin z$.)

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As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets). –  AnonymousCoward Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere. –  AnonymousCoward Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf –  wqr Mar 29 '13 at 3:30