Proof verification for Casorati-Weierstrass theorem

Question

I posted a question really simillar to this one few days/week(s) ago. But after some thoughts and some further elaboration, I wrote my proof "more" formally and thus I wanted to post it again to clarify some details.

So, here it goes:

Theorem. Let $a$ be a essential singularity of $f$, then we have that, for any $A \in \mathbb{C}$ there exists a sequence $\left(z_k\right)_{k=0}^{+\infty}$ such that this sequence converges to $a$ and such that $f(z_k) \rightarrow A$.

Proof so far. By contradition: Suposse that $\exists A \in \mathbb{C}$ such that the conclusion of the theorem doesn't hold, i.e., $\exists \epsilon >0$ such that $|z-A|<\epsilon$ doesn't agree with the image of $f$.

Thus, $\forall z \neq a$, we have that $|f(z)-A|\geq\epsilon$. Now, consider the particular case of $g(z) = \frac{1}{f(z)-A}$ in a deleted neighbourhood of $a$. It obviously comes that $|g(z)|\leq \frac{1}{\epsilon}$. Since $g(z)$ is analytic in a deleted neighbourhood of $a$, $a$ is also a singular point of $g$ and $a$ is also a removable singularity (using Riemann's Theorem), once it's limited near $a$. From this, it urges that $\lim_{z\rightarrow a}g(z)$ exists, call it $g(a)$, and we also have that $g$ extends to a analytic function in $a$.

Now, if $g(a)=0$, we have that, we have that $\lim_{z\rightarrow a}f(z) = \infty$ and so, $a$ would be a pole of $f$, which is a contradiction.

If $g(a) \neq 0$, we have that $\lim_{z\rightarrow a}f(z) = A + \frac{1}{g(a)}$ which means, again using the Rimeann's Theorem, that $f$ has a removable singularity, which is (once again) a contradiction.

NOTE. Proving that there exists a sequence $\left(z_k\right)_{k=0}^{+\infty}$ such that $f(z_k) \rightarrow A$ is equivalent to proving that the image of $f$ is dense in $\mathbb{C}$.

My final doubts and needs of clarification. I can see clearly where this proves that $f(z_k)\rightarrow A$. What I am having trouble with is the first part of the theorem, i.e, seeing where this proves that $z_k \rightarrow a$. If I am missing something and someone could clarify it to me I would be really thankfull. Equivalentelly, if this proof doesn't prove the first part of theorem I would also be thankfull if someone justified it! Thanks for all the help in advance.

Answer 1

If you are using a proof by contradiction. You would not prove that $z_k \to a$. When using a proof by contradiction in this case, you can say:

There exists $A \in \mathbb{C}$ such that any sequence $z_k \to a$ we have $f(z_k) \not \to A$

Note the emphasis on any sequence $z_k \to a$. You get to work with this assumption. This piece of information allows you to make the claim that there exists a deleted neighborhood around $a$ such that $|f(z) - A| \geq \epsilon$ for all $z$ in this deleted neighborhood.