Weierstrass-Casorati Theorem

Theorem
Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.

Let $f$ have an essential singularity at $a$.

Then:
 * $\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$.

Proof
, suppose $a = 0$ and $r = 1$.

$\exists s < 1$ such that:
 * $f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.

Then, by definition of dense subset:
 * $\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$

Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:
 * $\map \varphi z = \dfrac 1 {\map f z - z_0}$

is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because:


 * $\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$

Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).

Let $\map \varphi 0 \ne 0$.

Then:
 * $\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$

and the singular point of $f$ was removable.

Otherwise, let the power series of $\varphi$ be written:
 * $\ds \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$

Then as $\varphi \ne 0$:
 * $E = \set {k \in \N: a_k \ne 0} \ne \O$

Let $p = \min E$.

Then $0$ is a pole of order $p$ of $f$.

In each case, the assumption that:
 * $\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.

Hence the result, by Proof by Contradiction.

Also known as
It is also known as the Casorati-Weierstrass Theorem.