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Let $f$ be a mermorphic function on a domain $G$. I want to show that the amount of poles and roots have no accumulation points in $G$.

Someohow this is related to Can it be proved that a Meromorphic function only has a countable number of poles? Now I only need to do it for the zeros. I think because of the definition that the poles have to be isolated, I think it's not really provable.

Leviathan
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    Doesn't that post give you the answer? Poles are isolated singularities and zeros of a non-zero function are isolated. Any uncountable subset of the Complex plane has a limit point. So one can have at most countably-many . – gary Jul 13 '17 at 16:30
  • It answers my question of course, but how do I write this down? – Leviathan Jul 13 '17 at 16:31
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    I would use something to the effect of " Any uncountable subset of the plane must contain a limit point ( e.g., by Weirstrass' thm) and poles are isolated and zeros of a non-zero analytic function are isolated, therefore the amount of poles and zeros of a non-zero analytic function are at most countable. – gary Jul 13 '17 at 16:35
  • What is your definition of $f$ being meromorphic on $|z-a| < r$ ? If it is with the Laurent series then it is clear $a$ is not an accumulation point of poles of $f$. If it is $f = \frac{g}{h}$ where $g,h$ are analytic, then the Taylor series of $g,h$ show $a$ is not an accumulation point of their zeros. @gary – reuns Jul 13 '17 at 18:12
  • @gary which weierstrass theorem you mean? – Leviathan Jul 14 '17 at 12:02
  • @Leviathan: Every bounded infinite subset of Euclidean space has a limit point ( within the subset). Consider then the collection of balls $B(0,n); n=1,2,....$ , all finite. If S is uncountable, one of those balls will contain infinitely-many points, and therefore, by Weirstrass, a limit point. – gary Jul 14 '17 at 17:17

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