(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).

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."

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

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?

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

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.
I need to be clear about your reasoning in this paragraph.
Are you saying that since $\;h(\mathbb{C})\;$ is an open set,
$\exists \;\delta > 0 \;\ni \;\Delta(|w_0|, \delta) \,\subseteq\,h(\mathbb{C})?$

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?

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.

  • $\begingroup$ 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 $D$, $|w_0|\leq M$. $\endgroup$ – Qiyu Wen May 13 at 22:24
  • $\begingroup$ 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. $\endgroup$ – Qiyu Wen May 13 at 22:40
  • $\begingroup$ "open interval", not "over interval". $\endgroup$ – 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.

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

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