Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

Show that $0$ is not an essential singularity.

This is a homework question. I know that there is a theorem saying that if $f$ has an essential singularity $a$, then for all but two $w$ in $\mathbb{C}$, in any neighborhood of $a$, $f$ attains the value $w$.

Using this theorem the question is quite simple, but I am not sure if I am meant to quote this theorem, because it was only mentioned in my lectures without a proof.

I wonder if this can be shown without using the theorem I quoted.

share|cite|improve this question

marked as duplicate by mrf, Davide Giraudo, Hagen von Eitzen, Javier Álvarez, Brandon Carter Feb 10 '13 at 17:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It's not quite clear if the other thread answers the question, because we don't know if the Baire theorem is OK to use. // @Montez: please make the body of the text self-contained; a question that is split between the title and the body is harder to read. // My guess is that you are expected to use Picard's theorem; otherwise why would it be stated in lecture at all? – user53153 Feb 10 '13 at 16:34
up vote 0 down vote accepted

Assume that $f$ did have an essential singularity at $0$. Let $A_n=B_{\frac{1}{n}}(0)-\{0\}$ be the punctured disc of radius $\frac{1}{n}$ centered at $0$. Then, Casorati-Weierstrass and open mapping theorem implies that $f(A_n)$ is open and dense, and so by the Baire Category Theorem $\displaystyle \bigcap_{n=1}^{\infty}f(A_n)$ is dense, and so in in particular, not empty. Note though that by assumption if $w\in \mathbb{C}$ there exists finitely many $z\in\mathbb{D}-\{0\}$ such that $f(z)=w$ and thus we see that for large enough $n$ we have that $w\notin f(A_n)$. Thus, $\displaystyle w\notin \bigcap_{n=1}^{\infty}f(A_n)$ and so $\displaystyle \bigcap_{n=1}^{\infty}f(A_n)=\varnothing$. This is a contradiction.

NOTE: The above is actually a proof that if $f$ has an essential singularity at $0$ then it must take infinitely many values infinitely many times on any open neighborhood of $0$.

share|cite|improve this answer

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