Timeline for First visit of intervals for an irrational rotation
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2022 at 15:26 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Dec 6, 2022 at 15:24 | comment | added | Saúl RM | True, I wrote $\varepsilon$ instead of $|c-a_n|$. Now it should work | |
| Dec 6, 2022 at 15:22 | comment | added | Alessandro Della Corte | "...we use that $d(z^{-a},z^{-b})>\varepsilon$". I don't get this. Looks like, setting $\eta=\min_{a,b<k}{d(z^a,z^b)}$, you just need $|c-a_n|<\eta$, which you assumed. If the intersection is nonempty then there is a point whose $a$- and $b$-images are less than $\eta$ apart, which is a contradiction if $a,b<k$. | |
| Nov 26, 2022 at 15:17 | comment | added | Saúl RM | @RonniePavlov Btw about the non disjoint case, my argument still works if the interior of $I_n$ is not contained in the closure of $J_n$: you can define $J_n'$ as the closure of $J_n$, $I_n'$ as the interior of $I_n\setminus J_n$, and then apply my argument to $I_n',J_n'$ | |
| Nov 25, 2022 at 20:02 | comment | added | Ronnie Pavlov | Got it! I am interested in the non disjoint case because there I think it's purely a measure theoretic argument. Will try to write details. | |
| Nov 25, 2022 at 19:50 | comment | added | Saúl RM | Oh no, I meant making $J_n$ the closure of $I_n$ and making $I_n$ the interior of $J_n$. So basically swapping $I_n$ and $J_n$. But yes, that was expressed in a confusing manner | |
| Nov 25, 2022 at 19:49 | comment | added | Ronnie Pavlov | You said you could even make $J_n$ closure of $I_n$ and I'm saying I don't think you can with your proof, that's all. | |
| Nov 25, 2022 at 19:47 | comment | added | Saúl RM | I don't understand. By definition, $I_n$ and $J_n$ are disjoint, so $R^{-a}I_n$ and $R^{-a}J_n$ are disjoint. Oh, or maybe you meant if we can extend the problem to $I_n,J_n$ not being disjoint. In that case I would have to think about which conditions you can impose on them, I guess. At first I also tried a measure theoretic argument, but in the end this is what worked | |
| Nov 25, 2022 at 19:45 | comment | added | Ronnie Pavlov | In fact I'm pretty sure that under the OP restrictions on length, the set of pts. visiting $J_n$ before $I_n$ ultimately has Lebesgue measure zero (not needing the assumption that $I_n$ and $J_n$ adjacent and disjoint, which is needed for Saul's very clever proof), but I haven't been able to get time to write details. | |
| Nov 25, 2022 at 19:44 | comment | added | Ronnie Pavlov | I suppose I could be confused, but I don't think you can work with $I_n$ and $J_n$ not disjoint. Clearly then for $a=b$, $R^{-a} I_n$ and $R^{-b}J_n$ are not disjoint, and I think you need them to be for small $a,b$ in your proof. | |
| Nov 25, 2022 at 19:43 | comment | added | Saúl RM | I added a bit more detail to make that clear | |
| Nov 25, 2022 at 19:42 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Nov 25, 2022 at 19:23 | comment | added | Saúl RM | Yes, that is the important ingredient, and also that they are disjoint and $I_n\cup J_n$ is contained in a small interval with length convergent to $0$ when $n\to\infty$ (this is to ensure that $R(J_n)$ does not contain $I_n$ for all $n$, or similar cases) | |
| Nov 25, 2022 at 19:16 | comment | added | Alessandro Della Corte | Ok, I finally had the time to properly read your answer. You are right! In my previous comments I confused $S^1\setminus X$ with the set of points that ultimately first visit $I_n$. In conclusion, a comeagre subset of $S^1$ is made of points that infinitely many times visit $I_n$ first and infinitely many times visit $J_n$ first, and the only really important ingredient is that they are open and nonempty. | |
| Nov 25, 2022 at 19:10 | vote | accept | Alessandro Della Corte | ||
| Nov 25, 2022 at 18:26 | comment | added | Saúl RM | I wouldn't describe it as "larger". I haven't even used that the $I_n$ are smaller than the $J_n$ when $n$ goes to infinity, in fact you can change $I_n$ to be the interior of $J_n$ and change $J_n$ be the closure of $I_n$ and the result still holds | |
| Nov 25, 2022 at 18:19 | comment | added | Alessandro Della Corte | Yeah but why the set of points that first visit the small one should be topologically larger than the set of those visiting first the large one? Anyway, I'll check it later, thanks. | |
| Nov 25, 2022 at 18:12 | comment | added | Saúl RM | You are right about the typo. I guess it's strange? Honestly I didn't think about it that much, since for example meager sets can have full measure | |
| Nov 25, 2022 at 18:10 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Nov 25, 2022 at 18:07 | comment | added | Alessandro Della Corte | Still didn't have time to look closely, but two things: 1. $B_n:=A_n$ must be just $B_n:=$... right? 2. the conclusion is a bit suspicious, as by your reasoning $\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty B_n$ is not only nonempty, but even comeager. That's strange, no? | |
| Nov 25, 2022 at 17:03 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Nov 25, 2022 at 16:55 | history | answered | Saúl RM | CC BY-SA 4.0 |