Timeline for Is the limit set of a group action always closed?
Current License: CC BY-SA 3.0
7 events
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| Nov 27, 2013 at 10:00 | history | edited | Hao Chen | CC BY-SA 3.0 |
a wrong example
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| Nov 27, 2013 at 8:46 | comment | added | YCor | Thinking twice, considering an irrational rotation of the disc $X$ gives rise to an even simpler example: here $L$ is the complement of $\{0\}$ and thus is not closed, and the action is, this time, isometric. | |
| Nov 26, 2013 at 23:19 | comment | added | Hao Chen | @YvesCornulier Nice one. So the answer is no for general cases. Thanks! | |
| Nov 26, 2013 at 23:11 | comment | added | YCor | Here's an easy example: define, on the closed unit disc $X$ of the complex plane, $f(z)=u(|z|)z$ where $u(r)=e^{i\pi(1-r)}$. Clearly $f$ is a self-homeomorphism, defining an action of $\mathbf{Z}$ on $X$. Then $L$ is exactly the set of points in $X$ whose radius is irrational, and thus is not closed. | |
| Nov 26, 2013 at 23:07 | comment | added | Hao Chen | @YvesCornulier Thanks. Then, let's say that $X$ is a compact metric space. (the terms limit point / accumulation point / cluster point messed up ... ) | |
| Nov 26, 2013 at 22:59 | history | edited | Hao Chen | CC BY-SA 3.0 |
edited title
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| Nov 26, 2013 at 22:53 | history | asked | Hao Chen | CC BY-SA 3.0 |