Timeline for Does locally nilpotent imply nilpotent for continuous self-maps of intervals?
Current License: CC BY-SA 4.0
13 events
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Sep 15, 2022 at 13:39 | comment | added | Christian Remling | @QiaochuYuan: I guess your concerns have been addressed already in the other comments, but here it is one more time anyway: $I_0=(a,b)$ is by definition a connected component of $\{ f>0\}$, so $f(a)=f(b)=0$. And then $I_1\subseteq [0,b_0]$ because I want to show that $f=0$ on some $[0,d]$, so I assume, to obtain a contradiction, that there are (arbitrarily small) $I_n$ arbitrarily close to zero. | |
Sep 15, 2022 at 8:43 | comment | added | Jochen Glueck | @YCor: Ah, I see. Thanks. | |
Sep 15, 2022 at 8:03 | comment | added | YCor | @JochenGlueck Otherwise for some $a>0$ you get the restriction of $f$ as $g:[0,a]\to [0,a]$ locally nilpotent with $g^{-1}(\{0\})=\{0\}$, and this is clearly absurd. | |
Sep 15, 2022 at 8:02 | history | edited | YCor | CC BY-SA 4.0 |
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Sep 15, 2022 at 6:58 | vote | accept | Dattier | ||
Sep 15, 2022 at 6:47 | comment | added | Jochen Glueck | Hmm, why does the invariance of $[0,a]$ for each $a$ imply that the zeros of $f$ accumulate at $0$? | |
Sep 15, 2022 at 6:39 | comment | added | Jochen Glueck | @QiaochuYuan: Re first paragraph, actually we don't need to replace $f$ with $f^k$, either: by the intermediate value theorem $f$ has a fixed point, and clearly $f$ cannot have any fixed point different from $0$. | |
Sep 15, 2022 at 4:40 | comment | added | Ville Salo | $f(\overline{I_0}) \ni 0$ is because the endpoints of $\overline{I_0}$ map to $0$ or the connected component $I_0$ would be larger. | |
Sep 15, 2022 at 4:38 | comment | added | Ville Salo | @QiaochuYuan: I think $\exists I_1: I_1 \subseteq [0, b_0]$ is because we are assuming (for a contradiction) that the $I_n$ do accumulate at $0$. | |
Sep 15, 2022 at 3:09 | comment | added | Qiaochu Yuan | I am having trouble understanding this argument (and perhaps others are too given the lack of upvotes). Why must $f(\bar{I_0})$ contain $0$, and why must there be some $I_1 \subseteq [0, b_0]$? I think I at least understand the last paragraph, and in the first paragraph we can avoid appealing to Sharkovski's theorem using Neil Strickland's idea in the comments to replace $f$ by $f^k$ as necessary. | |
Sep 15, 2022 at 0:26 | history | edited | Christian Remling | CC BY-SA 4.0 |
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Sep 14, 2022 at 22:37 | history | edited | Christian Remling | CC BY-SA 4.0 |
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Sep 14, 2022 at 22:28 | history | answered | Christian Remling | CC BY-SA 4.0 |