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I was reading through a proof of the fact that all meromorphic functions on $\hat {\mathbb{C}}$ are rational functions found here

http://math.haifa.ac.il/hinich/RSlec/lec1.pdf

and I didn't understand the justification behind the statement

"The set of poles of a meromorphic function is discrete, therefore, finite since $\hat{\mathbb{C}}$ is compact"

I understand that $\hat{\mathbb{C}}$ is compact, but I don't see how that gives us that the number of poles is finite.

for example, the analytic continuation of the gamma function has an infinite number of poles, and I thought that was meromorphic.

Thanks in advance for the help!

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  • $\begingroup$ To address your final point, the gamma function is meromorphic on $\mathbb{C}$, but not on $\hat{\mathbb{C}}$! $\endgroup$ – Eric Wofsey May 10 '18 at 4:19
  • $\begingroup$ And that's because on $\hat{\mathbb{C}}$ all of the poles form a sort of "cluster point" at infinity so the singularity there isn't isolated? I guess I'm just not sure why a singularity at infinity wouldn't be isolated since it's a single point. $\endgroup$ – JonHales May 10 '18 at 4:35
  • $\begingroup$ They don't form a "sort of" cluster point: they literally have a cluster point at $\infty$. It may be helpful to consider the change of coordinates $w=1/z$, which sends $z=\infty$ to $w=0$. In the $w$ coordinates, there are poles which accumulate at $0$, so there is a non-isolated singularity at $0$. $\endgroup$ – Eric Wofsey May 10 '18 at 4:38
  • $\begingroup$ Okay, I think I get it. Thank you for your patience and for explaining this multiple times! $\endgroup$ – JonHales May 10 '18 at 4:41
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If there are infinitely many poles, they have a limit point in $\widehat{\mathbb C}$. In your example, that limit point is $\infty$. The limit point is a singularity of the function, but not an isolated singularity, so the function is not meromorphic on $\widehat{\mathbb C}$.

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  • $\begingroup$ Thank you! I guess I'm still a little confused as to why I can't have a function with a pole at every integer. Aren't the integers isolated? Or am I missing something about the extended complex plane? $\endgroup$ – JonHales May 10 '18 at 4:19
  • $\begingroup$ @JonHales: The integers are isolated, but not closed: they accumulate at $\infty$. So a function with a pole at every integer also has a singularity at $\infty$. $\endgroup$ – Eric Wofsey May 10 '18 at 4:20

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