Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

By Riemann mapping Theorem, we have that there is a comformal mapping from a half plane to a unit disk.

That means, there is a homeomorphism from a half plane to a unit disk.

However, homeomorphism preserves the compactness.

Then, can we conclude from here that a half plane is compact? (Which is a contradiction since a half plane is not closed and bounded.)

There should be some error that I am making in this logic, but I can't find it..

Any comment would be grateful!

share|improve this question
add comment

2 Answers 2

The open half plane $\{x+iy\in\mathbb C\mid y>0\}$ maps to the open unit disc, which is not compact. You could map the close half plane $\{x+iy\in\mathbb C\mid y\ge 0\}$ to the closed disc minus a point, which is not compact. Or you could map the compactified closed half plane to the closed disc (which is compact). Only the last type of half-plane, i.e. $\{x+iy\in\mathbb C\mid y\ge0\}\cup\{\infty\}$ is compact.

share|improve this answer
    
Thank you! I was confused to think that an open disk is compact. You fixed this confusion! Thank you! –  Emily Oct 11 '12 at 18:06
    
I thought so that you wer just mixing something up. Glad to help. –  Hagen von Eitzen Oct 11 '12 at 18:42
add comment

This is a well-known theorem and has stood the test of time. If you think you have found a problem with the theorem then the chances are that you have actually found a problem with your understanding of the theorem. In this case, the open half plane (e.g. $\Re(z) > 0$) is mapped onto the open disk (e.g. $|z| < 1$).

An example of such a map is as follows:

$$z \mapsto \frac{1-z}{1+z} \, . $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.