# complex analysis topology problem

$$\underline{\textbf{Given:}}$$
(1) $$\;f\,$$ is a continuous function.
(2) $$\;f \,: \mathbb{C} \to \mathbb{C}.$$
(3) $$\;|f(z)| \to \infty\;$$ as $$\;|z| \to \infty.$$
(4) $$\;f(\mathbb{C})\;$$ is an open set.

$$\underline{\textbf{To Prove:}}$$
(5) $$\;f(\mathbb{C}) = \mathbb{C}.$$

$$\underline{\textbf{My Request:}}$$
I provide context and analysis below. I welcome any hint, guide, or proof of the problem. It will help me if you keep your responses basic (i.e. consistent with the context).

$$\underline{\textbf{Context:}}$$
I am recreationally self-studying "An Introduction to Complex Function Theory" : Bruce Palka : 1991. My query represents the very last exercise from chapter 2 of this book (i.e. exercise 5.37).

Chapter 1 of this book focuses on the complex number system and chapter 2 focuses on the rudiments of plane topology. Within chapter 2 are definitions, theorems and (end-of-chapter) exercises that focus on
A. disks, open/closed sets, boundary points, sequences, convergence, accumulation points
B. continuity, limits
C. connected/disconnected sets, domains, components of open sets
D. bounded sets and sequences, Cauchy sequences, compact sets, uniform continuity.

A set is connected $$\;\Leftrightarrow\;$$ the set is not disconnected.
A set $$\,S\,$$ is disconnected $$\;\Leftrightarrow\; \exists \;$$ disjoint open sets $$\;U, \,V \;\ni$$
$$(S \bigcap U) \neq \phi \neq (S \bigcap V)\;$$ and $$\;S \;\subseteq \;(U \bigcup V).$$

$$b\,$$ is a boundary point of a set $$\,S\; \equiv \;$$
$$\forall \;r > 0, \Delta(0,r)\;$$ contains at least one element from $$\,S\,$$
and one element that is not in $$\,S.$$

I interpret premise (3) above as:
$$\forall \;\epsilon > 0 \;\exists \;r > 0 \;\ni \;|z| \geq r \;\Rightarrow \;|f(z)| > \epsilon.$$
Please let me know if you disagree with my interpretation.

I completed all prior exercises in chapter 2. The problem (exercise 5.37) provides a hint:
Assume $$\;G = f(\mathbb{C}) \neq \mathbb{C}\;$$ and use exercise 5.25 from chapter 2 to get a contradiction.

Exercise 5.25 :
If $$\;G, \,D \;$$ are domains (i.e. open, connected sets) $$\;\ni$$
$$G \subseteq D\;$$ and $$\;G \neq D,\;$$ then $$\;\partial G \;\bigcap D \;\neq \;\phi$$
(i.e. $$\,D\,$$ contains at least one of the boundary points of $$\,G$$).

$$\underline{\textbf{My Analysis:}}$$

First of all, let $$\;h(z) = |f(z)|.$$
From the theorems in chapter 2, I know that $$\;h(z)$$ is a continuous function,
and that $$\;h(\mathbb{C})\;$$ is therefore an unbounded, open, and connected set.
This means that $$\;\forall \;r > 0, \;\exists \;z \,\in \,\mathbb{C} \;\ni \;|f(z)| = r.$$

Second of all, following the hint, and assuming that $$\;G \neq \mathbb{C},\;$$
$$G$$ has a boundary point $$\,b \;\ni \;b \,\not\in \,G.$$
Let $$\;r \equiv |b|,\;$$ and $$\;K(0,r) \equiv \{ \,z \,: \;|z| = r \,\}.$$
Let $$\;K_1 \equiv G \bigcap K(0,r)\;, K_2 \equiv [K(0,r) \sim K_1].$$

From previous exercises in chapter 2, I know that both $$\;K(0,r)\;$$ and $$\;\overline{G}\;$$ are connected sets.
However, I do not (as of yet) know whether any of $$\;\partial G, \;K_1, \;$$ or $$\;K_2 \;$$ are connected sets.
Further, even if I did know this, I don't see how it would help.

Since $$\,b\,$$ is a boundary point of $$\,G,\,$$ both $$\,G\,$$ and $$\,(\mathbb{C} \sim G)\,$$
contain elements arbitrarily close to $$\,b.$$
I don't see how this helps either.

$$\underline{\textbf{Addendum-1 Reaction to Qiyu Wen's answer:}}$$
I followed almost all of the answer, but I have some questions.

In response to your paragraph: "I just want to mention a gap in your interpretation: ... so 0 is what we try to capture."

Q1:
What does $$\;\mathbb{C^*}\;$$ refer to?

Q2:
By "gap in your interpretation," are you referring to my presumption that
$$\forall \;\epsilon > 0 \;\exists \;r > 0 \;\ni \;|z| \geq r \;\Rightarrow \;|f(z)| > \epsilon$$
or are you referring to my analysis that concluded that
$$\;\forall \;r > 0, \;\exists \;z \,\in \,\mathbb{C} \;\ni \;|f(z)| = r?$$

If the former, then how (else) does one interpret
$$\;|f(z)| \to \infty\;$$ as $$\;|z| \to \infty?$$
If the latter, I know that since $$\,f(z)$$ is continuous, so is $$\,h(z) = |f(z)|.$$
Further, I know that since $$\,\mathbb{C}\,$$ is a connected set, so is $$\,h(\mathbb{C}).$$
I also know, from premise (3), that $$\,h(\mathbb{C})\,$$ is unbounded.
However, I now realize that I do not immediately know that $$\,{0} \,\in \,h(\mathbb{C}).$$
Is this what you are referring to?

Q3:
In your statement "Since $$\,f(C)\,$$ is open, there exists an open disk cantered at $$\,w$$",
do you intend ...cantered at $$\,w_0$$?

Q4:
In the paragraph that begins: "Suppose, for contradiction, that $$\,w_0 \neq 0$$ ..."
I agree that both $$\;f(\mathbb{C})\;$$ and $$\;h(\mathbb{C})\;$$ are open sets,
since both $$\,f\,$$ and $$\,h = |f|\,$$ are continuous.
Are you saying that since $$\;h(\mathbb{C})\;$$ is an open set,
$$\exists \;\delta > 0 \;\ni \;\Delta(|w_0|, \delta) \,\subseteq\,h(\mathbb{C})?$$

Q5:
At the end of your first paragraph, you say "...Note that $$\,w_0 < M.$$"
Why? I agree that $$\,|f|\,$$ has a minimum at some point $$\,z_0\,$$
on the closed disk $$\,D,\,$$ but I don't understand why $$\,|f(z_0)|\,$$ must be $$\,< M.$$
If I'm right that this is unclear, I think that the analysis is immediately remedied
by simply taking any random $$\,z_1 \,\in \,\mathbb{C}\,$$ and then choosing $$\,M > |f(z_1)|.$$
What do you think?

Q6:
If I'm not mistaken, you ignored the hint given by the book.
Is there a relation between your approach and the book's hint?

$$\underline{\textbf{Addendum-2 Reaction to Qiyu Wen's response to my questions:}}$$
Way cool!

"...I'm working with $$\,f(C),\,$$ which I know is open, not $$\,h(C).$$"
"You pick a non-zero complex number, then in any ball about it"
"you can find another complex number closer to 0..."

It's a critical point.
When I first saw your approach, I was blind to the above idea.
When you emphasized it, it sunk in. Thanks

"...I didn't realize there's a hint. See my edited answer."
Again, thanks. The hint-approach, which is probably the intended answer, is something that I might eventually have stumbled into. Re your initial approach, there's no way that I would have thought of that on my own.

• Q1: $\mathbb{C}^*=\mathbb{C}-\{0\}$. Q2: Pretty much so. Even if you know $h$ to be open, you still only know its range is an over interval $(a,+\infty)$, not necessarily $(0,+\infty)$. Q3: Yes. Q4: I'm working with $f(\mathbb{C}$, which I know is open, not $h(\mathbb{C}$. You pick a non-zero complex number, then in any ball about it you can find another complex number closer to $0$. Q5: I said that $w_0\leq M$, not strictly less than. Since $M\in h(\mathbb{C})$ and $h(z)\neq M$ for all $z\notin D$, it must be that $h(z)=M$ for some $z\in D$. Since $|w_0|$ is the minimum on $D$, $|w_0|\leq M$. – Qiyu Wen May 13 at 22:24
• Anyway, keep in mind that $M$ could happen to be the minimum, so you don't want to say things like $|f(z)|<M$. Q6: I didn't realize there's a hint. See my edited answer. – Qiyu Wen May 13 at 22:40
• "open interval", not "over interval". – Qiyu Wen May 13 at 22:55

We first prove that $$f(z)=0$$ for some $$z\in\mathbb{C}$$. Pick $$M$$ in the range of $$|f|$$. Since $$|f(z)|\to+\infty$$ as $$|z|\to+\infty$$, there is a closed disk $$D$$ such that $$|f(z)|>M$$ for all $$z\notin D$$. By continuity of $$f$$ and compactness of $$D$$, $$|f|$$ has a minimum on $$D$$ at some $$z_0\in D$$. Put $$w_0=f(z_0)$$. Note that $$|w_0|\leq M$$. Hence $$|f(z_0)|$$ is a minimum of $$|f|$$ on $$\mathbb{C}$$.

Suppose, for contradiction, that $$w_0\neq 0$$. Since $$f(\mathbb{C})$$ is open, there exists an open disk cantered at $$w_0$$ and contained in the range of $$f$$, from which we can pick $$w$$ such that $$|w|<|w_0|$$, a contradiction.

Now let $$w$$ be any complex number and put $$g(z)=f(z)-w$$. Then $$g$$ also has the properties listed, so $$g(z)=0$$ for some complex number $$z$$. Hence $$f$$ is surjective.

I just want to mention a gap in your interpretation: you do not immediately know that $$|f(\mathbb{C})|$$ is open in $$\mathbb{R}$$, because $$z\mapsto |z|$$ is not an open map from $$\mathbb{C}$$ to $$\mathbb{R}$$. However, it is open on $$\mathbb{C}^*$$, so $$0$$ is what we try to capture.

This is how you can use the hint: Suppose that $$f(\mathbb{C})\neq\mathbb{C}$$. Let $$w_0$$ be a boundary point of $$f(\mathbb{C})$$. Let $$(w_k)$$ be a sequence in $$f(\mathbb{C})$$ that converges to $$w_0$$. Then you obtain a sequence $$(z_k)$$ such that $$f(z_k)=w_k$$. Use $$\lim|f(z)|\to+\infty$$ to show that $$(z_k)$$ is a sequence in a compact set, hence admits a convergent subsequence with limit $$z_0$$. Then by continuity of $$f$$ we have $$f(z_0)=w_0$$, contradicting $$w_0\notin f(\mathbb{C})$$.

These approaches are not that different. One way or another you want to show that $$f(\mathbb{C})$$ cannot have a boundary point.

• Fascinating approach, which I mostly understood. I do have questions. Please see my Addendum-1, which I have just added to my original query. – user2661923 May 13 at 19:00
• Thanks for the follow-up. Please see my Addendum-2, which I have just added to my query. – user2661923 May 13 at 23:34