# Proving the Maximum Modulus Principle using the Open Mapping Theorem

I was reading on Wikipedia that

"The maximum modulus principle can be viewed as a special case of the open mapping theorem, which states that a nonconstant holomorphic function maps open sets to open sets. If $|f|$ attains a local maximum at $a$, then the image of a sufficiently small open neighborhood of $a$ cannot be open. Therefore, $f$ is constant".

Could someone expand upon that? I don't follow why the image of a open neighborhood of a would not be open?

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Because it can't contain an open neighborhood of $|f(a)|$ by maximality. (Here you need to observe that the absolute value function is also open away from the origin, but this is not hard to see.) – Qiaochu Yuan Aug 10 '11 at 23:53

If $\frak{U}$ is an open set in $\mathbb{C}$, then $|\frak{U}|$ can have no greatest element. Thus, $|f(\frak{U})|$ is cannot have a greatest element since $f(\frak{U})$ is open.
Suppose that $f$ attains its maximum in $\frak{U}$; this means that for some $z_0\in\frak{U}$, $|f(z)|\le|f(z_0)|$ for all $z\in\frak{U}$. Thus, $|f(z_0)|$ is the greatest element of $|f(\frak{U})|$. This means that $f(\frak{U})$ is not open. Therefore, $f$ is constant.
As Qiaochu Yuan mentioned, this is not true if $\frak{U}$ contains the origin. – lhf Aug 11 '11 at 10:47
Plus the conclusion is that $f$ is constant, not a contradiction. – lhf Aug 11 '11 at 10:49
@lhf: Thanks! I forgot about the origin. I had assumed $f$ was non-constant, but never wrote it down. It does read more like the maximum modulus principle to conclude that $f$ is constant, so rather than introduce my hidden assumption, I have concluded that $f$ is constant. – robjohn Aug 11 '11 at 11:19
You have now phrased your answer differently but the original version had ${\frak{U}} \mbox{ open in } \mathbb C \Rightarrow |{\frak{U}}| \mbox{ open in } \mathbb R$ and this is not true if $\frak{U}$ contains the origin. – lhf Aug 11 '11 at 17:42