proof of Casorati-Weierstrass theorem ProofOfCasoratiWeierstrassTheorem 2003-04-03 11:36:35 2003-04-03 11:36:35 Proof Casorati-Weierstrass theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Assume that $a$ is an essential singularity of $f$. Let $V\subset U$ be a punctured neighborhood of $a$, and let $\lambda\in\mathbb{C}$. We have to show that $\lambda$ is a limit point of $f(V)$. Suppose it is not, then there is an $\epsilon>0$ such that $|f(z)-\lambda|>\epsilon$ for all $z\in V$, and the function $$ g:V\to\mathbb{C}, z\mapsto\frac{1}{f(z)-\lambda} $$ is bounded, since $|g(z)|=\frac{1}{|f(z)-\lambda|}<\epsilon^{-1}$ for all $z\in V$. According to Riemann's removable singularity theorem, this implies that $a$ is a removable singularity of $g$, so that $g$ can be extended to a holomorphic function $\bar g:V\cup\{a\}\to\mathbb C$. Now $$ f(z)=\frac{1}{\bar g(z)}-\lambda $$ for $z\neq a$, and $a$ is either a removable singularity of $f$ (if $\bar g(z)\neq 0$) or a pole of order $n$ (if $\bar g$ has a zero of order $n$ at $a$). This contradicts our assumption that $a$ is an essential singularity, which means that $\lambda$ must be a limit point of $f(V)$. The argument holds for all $\lambda\in\mathbb{C}$, so $f(V)$ is dense in $\mathbb{C}$ for any punctured neighborhood $V$ of $a$. To prove the converse, assume that $f(V)$ is dense in $\mathbb{C}$ for any punctured neighborhood $V$ of $a$. If $a$ is a removable singularity, then $f$ is bounded near $a$, and if $a$ is a pole, $f(z)\to\infty$ as $z\to a$. Either of these possibilities contradicts the assumption that the image of any punctured neighborhood of $a$ under $f$ is dense in $\mathbb C$, so $a$ must be an essential singularity of $f$.