Proving the Open Mapping Theorem for Continuous Functions on Complex Numbers

[DeadOriginal]

Homework Statement

Let a continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ satisfy $|f(\mathbb{C})|\rightarrow\infty$ as $|z|\rightarrow\infty$ and let $f(\mathbb{C})$ be an open set. Then $f(\mathbb{C})=\mathbb{C}$.

The Attempt at a Solution

Suppose for contradiction that $G=f(\mathbb{C})\not=\mathbb{C}$. Then $G\subset\mathbb{C}$ so $\partial G\cap\mathbb{C}\not=\emptyset$. (This fact was proved in class.)

The professor gave a hint that I should be using Bolzano-Weierstrauss and the fact that if $f$ is a continuous function and $z_{n}\rightarrow z$ then $f(z_{n})\rightarrow f(z)$. I started by contradiction because that was the hint in the book but I can't see how to incorporate the professor's hint.

 

[Dick]

Homework Statement

Let a continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ satisfy $|f(\mathbb{C})|\rightarrow\infty$ as $|z|\rightarrow\infty$ and let $f(\mathbb{C})$ be an open set. Then $f(\mathbb{C})=\mathbb{C}$.

The Attempt at a Solution

Suppose for contradiction that $G=f(\mathbb{C})\not=\mathbb{C}$. Then $G\subset\mathbb{C}$ so $\partial G\cap\mathbb{C}\not=\emptyset$. (This fact was proved in class.)

The professor gave a hint that I should be using Bolzano-Weierstrauss and the fact that if $f$ is a continuous function and $z_{n}\rightarrow z$ then $f(z_{n})\rightarrow f(z)$. I started by contradiction because that was the hint in the book but I can't see how to incorporate the professor's hint.

Ok, suppose the boundary of $f(\mathbb{C})$ is not empty. Then pick a point $z_0$ in the boundary. Then there is a sequence $z_n$ such that $f(z_n) \rightarrow z_0$, right? Now can you see that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow\infty$ means $z_n$ must be a bounded sequence? Apply Bolzano-Weierstrass from there.

 

[DeadOriginal]

Ok, suppose the boundary of $f(\mathbb{C})$ is not empty. Then pick a point $z_0$ in the boundary. Then there is a sequence $z_n$ such that $f(z_n) \rightarrow z_0$, right? Now can you see that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow\infty$ means $z_n$ must be a bounded sequence? Apply Bolzano-Weierstrass from there.

By Bolzano-Weierstrauss, there exists a convergent subsequence of $z_{n}$,$z_{n_{k}}$ which converges to some $z\in\mathbb{C}$. Then since $f$ is continuous, $f(z_{n_{k}})\rightarrow f(z)=z_{0}$.

I am not sure how to proceed. I keep thinking that the goal is to derive a contradiction by showing that $\partial G\cup\mathbb{C}=\emptyset$. I think the problem is that I can't picture what I am trying to show in my head. How would you recommend going about picturing this?

 

[shortydeb]

Suppose x is a complex number, and not in f(C). Let M > |x|. Then there exists N such that if |z| > N, then |f(z)| > M. The set {y: |y| ≤ N} is compact.

Edit: x is either a limit point of f(C), or it isn't. If it's not, choose M so that the open ball with radius M centered at (0,0) contains the closure of the open ball with radius r centered at x, such that every open ball centered at x with radius greater than r contains a point in f(C), and the open ball with radius r centered at x contains no points in f(C).

 

[Dick]

By Bolzano-Weierstrauss, there exists a convergent subsequence of $z_{n}$,$z_{n_{k}}$ which converges to some $z\in\mathbb{C}$. Then since $f$ is continuous, $f(z_{n_{k}})\rightarrow f(z)=z_{0}$.

I am not sure how to proceed. I keep thinking that the goal is to derive a contradiction by showing that $\partial G\cup\mathbb{C}=\emptyset$. I think the problem is that I can't picture what I am trying to show in my head. How would you recommend going about picturing this?

$f(z)=z_0$ mean $z_0$ is in $f(\mathbb{C})$. Can an element of an open set also belong to the boundary? Think about definitions.

 

[DeadOriginal]

Ah! Then since $z_{0}$ is contained in $f(\mathbb{C})$, this is a contradiction because $G$ does not contain its boundary. If it did then $G$ would also have to be closed and the only sets which are both open and closed in $\mathbb{C}$ are $\emptyset$ and $\mathbb{C}$ but since $G$ contains $z_{0}$, $G$ must be the entirety of $\mathbb{C}$.

 

[Dick]

Ah! Then since $z_{0}$ is contained in $f(\mathbb{C})$, this is a contradiction because $G$ does not contain its boundary. If it did then $G$ would also have to be closed and the only sets which are both open and closed in $\mathbb{C}$ are $\emptyset$ and $\mathbb{C}$ but since $G$ contains $z_{0}$, $G$ must be the entirety of $\mathbb{C}$.

Yes, that's it. The real trick is the argument that the $z_n$ must be bounded. That's how you use the limit property.