Timeline for Monotonic bijections of rational numbers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 22, 2013 at 22:52 | comment | added | Andreas Blass | Concerning "not such a 'smooth' structure as the reals," note that any monotonic bijection of $\mathbb Q$ to itself extends (uniquely) to a monotonic homeomorphism of $\mathbb R$ to itself. So a certain amount of "smoothness" of $\mathbb R$ reaches $\mathbb Q$. | |
| Jul 22, 2013 at 21:31 | history | edited | user9072 |
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| Feb 2, 2013 at 19:52 | comment | added | Pietro Majer | You can biject two given countable dense subsets of $\mathbb{R}$ by an entire diffeomorphism: mathoverflow.net/questions/42460/… | |
| Jan 30, 2013 at 19:20 | answer | added | Robert Israel | timeline score: 5 | |
| Jan 30, 2013 at 17:56 | comment | added | Joerg Zimmermann | Enlightening constructions, Emil, thank you very much. So even if the rationals have not such a "smooth" structure as the reals, there is still a large degree of flexibility in reparameterizing stochastic models having a rational parameter. Have to think about the implications... | |
| Jan 30, 2013 at 17:31 | vote | accept | Joerg Zimmermann | ||
| Jan 30, 2013 at 17:12 | comment | added | Emil Jeřábek | In fact, if $F$ is a countable set of functions, you can arrange that the the value of the bijection in the $n$-th chosen point disagrees with the values given by the first $n$ functions from $F$. You will get a monotone bijection which differs from any $f\in F$ in all but finitely many points. For example, $F$ can be the set of all rational functions with rational coefficients. | |
| Jan 30, 2013 at 17:09 | history | edited | Joerg Zimmermann | CC BY-SA 3.0 |
Changing the requirement from continuous to monotonic, because that fits better what I'm interested in.
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| Jan 30, 2013 at 17:03 | comment | added | Emil Jeřábek | Consider the zig-zag construction showing that any two countable dense linear orders are isomorphic, and apply it with $\mathbb Q$ on both sides. You have a lot of freedom in the construction: in particular, you can do it in such a way that the image or preimage being chosen is slightly off from the position it would have if the bijection were linear between the nearest points where the function has already been fixed (I hope this description is clear). You will end up with a monotone bijection which is not linear on any nonempty interval. | |
| Jan 30, 2013 at 16:58 | answer | added | Jeremy Rickard | timeline score: 11 | |
| Jan 30, 2013 at 16:48 | comment | added | Jeremy Rickard | For example, swap the intervals $(\sqrt{2},\sqrt{2}+1)$ and $(\sqrt{2}+1,\sqrt{2}+2)$, leaving everything outside these intervals fixed. | |
| Jan 30, 2013 at 16:39 | comment | added | Joerg Zimmermann | You're right, I'm not interested in these (at least not at the moment), so I should add monotonic as a requirement. Can you please provide or point to a non-monotonic continuous bijection from $\mathbb{Q}$ to itself? Thanks a lot in advance! | |
| Jan 30, 2013 at 16:30 | comment | added | Jeremy Rickard | There are loads of non-monotonic continuous bijections from $\mathbb{Q}$ to itself that don't extend to continuous functions on $\mathbb{R}$. But the way you phrase the question suggests that you may not be interested in these? | |
| Jan 30, 2013 at 16:01 | history | asked | Joerg Zimmermann | CC BY-SA 3.0 |