# Proof verification for Casorati-Weierstrass theorem

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.

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.

• Do you happen to know how to prove the first part of the theorem then? i.e., to prove that $z_k \rightarrow a$ ?
– xyz
Jun 26, 2021 at 0:39
• Again, unless you are doing a direct proof you would not need to prove this.
– Mike
Jun 26, 2021 at 0:39
• Since you are using a proof by contradiction, you get to assume "there exists $A \in \mathbb{C}$ such that for any sequence $z_k \to a$ we have $f(z_k) \not \to f(a)$. You get to work with this statement throughout the proof. The contradiction provided in your post necessarily implies that such a sequence exists.
– Mike
Jun 26, 2021 at 0:50
• So I am also "proving" it using this contradiction? I am sorry to be stubborn...
– xyz
Jun 26, 2021 at 8:06
• In essence: yes
– Mike
Jun 27, 2021 at 13:03