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 $\Omega$.

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.