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I'm trying to prove the following: Let $f$ be meromorphic on the open connected set $\Omega\subseteq\mathbb{\hat{C}}$, and let $A$ be the set of its poles in $\Omega$ then the accumulation points of $A$ are on the boundary of $\Omega$.

Given definition: $f:\Omega\to\mathbb{\hat{C}}$ is meromorphic on $\Omega$ if at each point of that set, $f$ is either holomorphic or has a pole; or, if $f\equiv\infty$. I'm not sure if this is valid

Attempt: Let $f\not\equiv\infty$ (otherwise there is nothing to be done). Suppose that there is an accumulation point $z$ of $A$ inside of $\Omega$. Then this implies that $1/f\left(z\right)\equiv 0$. Which implies that $f(z)\equiv\infty$. A contradiction. Thus $z$ must be on the boundary of $\Omega$.

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  • $\begingroup$ Looks fine to me, provided you fill in the details with isolation of zeros and so on. $\endgroup$ – Chappers Mar 30 '15 at 2:28
  • $\begingroup$ Something like, since $A$ is a set of poles, and each pole is isolated by definition, for each $x\in A$ there exists $r_x>0$ such that $B(x,r_x)$ contains no other points of $A$ besides $x$. Thus $A$ will form a set of isolated zeros of the function $1/f$. So that if $z\in\Omega$ is an accumulation point of $A$ then $1/f\equiv 0.$ etc... $\endgroup$ – mi986 Mar 30 '15 at 3:12
  • $\begingroup$ Possible duplicate of Limit point of poles is essential singularity? Am I speaking nonsense? $\endgroup$ – Alex M. Mar 18 '17 at 20:20
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Suppose there is an accumulation point (say $z_0)$. Then $f$ has an isolated singularity there or is analytic. Suppose it has an isolated singularity at $z_0$. Then There is a $r \gt 0$ such that $f$ is analytic in $B(z_0,r)-{z_0}$. But any such ball will intersect the sequence of poles and hence a contradiction. Similarly show that it can't be analytic there

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  • $\begingroup$ If $f$ is analytic at $z_0$, wouldn't that imply there exists $r>0$ such that $f$ is analytic on $B(z_0,r)$, which would give the same contradiction? (The ball will intersect the sequence of poles.) $\endgroup$ – mi986 Mar 30 '15 at 3:08
  • $\begingroup$ yes.. that's the argument. $\endgroup$ – tattwamasi amrutam Mar 30 '15 at 3:26

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