# Casorati-Weierstrass theorem-related proof

Theorem. If $$f$$ has an essential singularity at $$p$$, then for any $$a \in \mathbb{C}$$ there exists a sequence $$\{z_k\}$$ such that: $$\begin{equation*} \lim_{n\rightarrow +\infty}z_n = p \quad \wedge \quad \lim_{n\rightarrow +\infty}f(z_n) = a \end{equation*}$$
Proof I have so far. I believe that, so far I was able to prove the second condition, i.e., $$\hspace{.2cm}\lim_{n\rightarrow +\infty}f(z_n) = a\hspace{.2cm}$$ by contradition taking $$\begin{equation*} g(z) = \frac{1}{f(z)-a} \end{equation*}$$
and then considering two cases $$\overline{g}(p)\neq 0$$ and $$\overline{g}(p)=0$$. I believe this part is somehow related to the Casorati-Weierstrass theorem.

The part where I am struggling is with the first condition, i.e., $$\begin{equation*} \lim_{n\rightarrow +\infty}z_n = p \end{equation*}$$

If someone could help me with this one, I would be really thankfull. (I understand this should be, somehow, related to the definition of convergence of a sequence).

Prove by contradiction. If the conclusion fails then there exists $$r>0$$ and $$\delta >0$$ such that $$|f(z)-a| \geq \delta$$ for $$0<|z-p| . Let $$g=\frac 1 {f-a}$$ Then $$g$$ is bounded near $$p$$ so it has a removable singularity at $$p$$. Obviously, $$g$$ is not identically $$0$$ in $$D(p,r)$$ so it has a zero of finite order (at most) at $$p$$. But then $$f(z)=a+\frac 1g$$ has a pole at $$p$$.

• I don't understand how this proves that \begin{equation*} \lim_{n\rightarrow +\infty}z_n = p \end{equation*} Isn't this a prove for what I have already shown? Thanks for your help
– xyz
Jun 8, 2021 at 23:20
• Try to write down the the negation of what you are asked to prove. That will give you existence of $r$ and $\delta$ as I have stated. @rpd Jun 8, 2021 at 23:23
• I understand that. But aren't you proving (by contradition) that \begin{equation*} \lim_{n\rightarrow +\infty} f(z_n) = a \hspace{.1cm} ? \end{equation*} I wanted to prove what I've typed on my first comment, since I've already done this part. If I am missing something I am sorry!
– xyz
Jun 8, 2021 at 23:25
• Suppose what I stated in my first sentence is false. Take $r=\delta=\frac 1 n$. Then we get $z_n$ such that $0<|z_n-p| <\frac 1 n$ and $|f(z_n)-a| <\frac 1 n$. Doesn't that give both $z_n \to p$ and $f(z_n) \to a$? @rpd Jun 8, 2021 at 23:29
• You are right indeed! Thanks for your help!
– xyz
Jun 8, 2021 at 23:29