# Does Riemann Mapping theorem guarentee a half plane to be compact?

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!

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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.

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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

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:

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

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