Timeline for When does a homeomorphism split into essentially minimal homeomorphisms?
Current License: CC BY-SA 3.0
10 events
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| Nov 26, 2012 at 13:28 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Nov 26, 2012 at 13:17 | comment | added | Tom Goodwillie | I see. I misunderstood the question because you used "or" to mean exclusive "or". I have taken the liberty of editing the end of your "My Question" for clarity. | |
| Nov 26, 2012 at 13:12 | history | undeleted | Tom Goodwillie | ||
| Nov 26, 2012 at 13:02 | history | deleted | Tom Goodwillie | ||
| Nov 26, 2012 at 12:38 | comment | added | Gabor Szabo | I don't see how these maps satisfy the given property. Without loss of gerenality, we can work with $X=[0,1]$. If the endpoints are the only fixed points, either $\varphi(x)>x$ for all $x\neq 0,1$ or $\varphi(x)< x$ for all $x\neq 0,1$. Assume the latter. So for a given $x_0\in (0,1)$, the sequence $x_n=\varphi^n(x_0)$ is decreasing and bounded, hence it converges. Since the limit will be a fixed point, the limit is 0. Similarly, the sequence $y_n=\varphi^{-n}(x_0)$ will converge to 1. In particular, almost all orbit closures contain both minimal sets, which is ruled out from the beginning. | |
| Nov 26, 2012 at 4:08 | comment | added | Tom Goodwillie | Yes, sorry. In correcting a typo I also messed it up. Fixed now. | |
| Nov 26, 2012 at 4:07 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Nov 26, 2012 at 1:33 | comment | added | Joel David Hamkins | Don't you need still to have only the two fixed points (the endpoints), as in your original answer? Otherwise, this won't satisfy the hypothesis of having exactly two minimal invariant sets. | |
| Nov 26, 2012 at 1:13 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Nov 26, 2012 at 0:50 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |