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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.

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

This question was marked as an exact duplicate of an existing 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
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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 $0$.

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