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 $\Omega$.
Best Answer
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