Timeline for Does this property characterize straight lines in the plane?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2023 at 21:31 | history | edited | Alessandro Della Corte | CC BY-SA 4.0 |
Best described the difficulty
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| Jan 5, 2023 at 21:26 | history | edited | Alessandro Della Corte | CC BY-SA 4.0 |
Best described the difficulty
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| S Aug 19, 2021 at 6:40 | history | suggested | Alessandro Della Corte | CC BY-SA 4.0 |
Added a disclaimer because the argument of the answer is not complete.
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| Aug 18, 2021 at 21:19 | review | Suggested edits | |||
| S Aug 19, 2021 at 6:40 | |||||
| Dec 23, 2020 at 10:15 | comment | added | Alessandro Della Corte | @IlkkaTörmä I hope the proof can be fixed taking into account the objections, but if not maybe it would be better to add a disclaimer at the beginning of the answer, not to mislead someone interested in thinking that the problem is solved. | |
| Nov 27, 2020 at 15:09 | comment | added | user44143 | @A.DellaCorte and Ilkka Torma, will you remove the comments that are no longer relevant? And Ilkka Torma, will you provide an overview of the proof, or some indication of an insight that leads there? | |
| Nov 24, 2020 at 21:19 | comment | added | Alessandro Della Corte | In Lemma 3 you write: "every point of $S_x$ is within distance $2$ of $\ell_y$, which constrains the direction of $\ell_y$ to an interval whose length approaches $0$ as the length $\| \gamma(-x) - \gamma(x) \|$ of $S_x$ grows". It seems to me that you are taking for granted that the direction of $S_x$ remains the same, or at least has vanishingly small perturbations, when $x\to\infty$. But if you don't justify this in advance, I'm afraid you are only proving that $\ell_y$ tends to be parallel to $S_x$. | |
| Nov 24, 2020 at 16:15 | comment | added | Ilkka Törmä | @A.DellaCorte You're right, I meant to define a new point, and the line I defined didn't work. Hopefully this can be a final version. | |
| Nov 24, 2020 at 16:08 | history | edited | Ilkka Törmä | CC BY-SA 4.0 |
deleted 2 characters in body
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| Nov 24, 2020 at 15:48 | history | edited | Ilkka Törmä | CC BY-SA 4.0 |
added 113 characters in body
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| Nov 24, 2020 at 14:39 | comment | added | Alessandro Della Corte | Some remarks: 1. In fact you seem to assume that $\gamma$ escapes at $\infty$, but if not (provided the argument works as a whole), then there exists the "rightmost" line, say, orthogonal to $\ell^{\pm}$ and intersecting $\gamma(\mathbb{R})$, which you can exclude again by Lemma 2. | |
| Nov 24, 2020 at 13:30 | comment | added | Ilkka Törmä | @A.DellaCorte I found the time to edit the proof. Hopefully it is now more understandable. | |
| Nov 24, 2020 at 13:29 | history | edited | Ilkka Törmä | CC BY-SA 4.0 |
Clarified Lemma 3, fixed small issues in Lemmas 2 and 4.
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| Nov 23, 2020 at 14:59 | comment | added | Liviu Nicolaescu | If you allow for the curve to be semialgebraic then the existence of $\ell_\pm$ is guaranteed (semialgebraicity prohibits heavy oscillation). | |
| Nov 21, 2020 at 21:58 | comment | added | Ilkka Törmä | Here "has bounded distance from" means "is contained in a bounded-radius ball around" | |
| Nov 21, 2020 at 21:51 | comment | added | Ilkka Törmä | @A.DellaCorte If the direction oscillates by more than some fixed nonzero angle, then for large enough $x<y$ that witness the oscillation, the intersection of the convex hulls of $\ell_x^+ \cup \ell_x^-$ and $\ell_y^+ \cup \ell_y^-$ has bounded distance from $\gamma(0)$, a contradiction. | |
| Nov 21, 2020 at 20:06 | comment | added | Ilkka Törmä | @A.DellaCorte I'll try to elaborate. For each $x$ we have three parallel lines: $\ell = \ell(\gamma(-x),\gamma(x))$, $\ell^+$ and $\ell^-$, where $\ell$ and the set $C = \gamma([-x,x])$ are between the others. If the distance between $\ell^+$ and $\ell^-$ is greater than 2, then one of them, say $\ell^+$, has distance greater than 1 from $\ell$. Pick any $\gamma(b)$ from $C \cap \ell^+$, which is nonempty. Then $\gamma(-x), \gamma(x) \notin \bar D(\gamma(b))$ because the former are in $\ell$ and the latter is in $\ell^+$, and $C$ is on one side of $\ell^+$ by definition. | |
| Nov 21, 2020 at 18:10 | comment | added | Ilkka Törmä | @A.DellaCorte 1. Hmm, you're right that it's not clear that the lines converge. But it should be clear that their direction converges, and then we can take as $\ell^\pm$ the closest parallel pair of lines in that direction that enclose $\gamma$. They should have all the relevant properties. I'll edit the proof when I get the chance. 2. See my other comment. | |
| Nov 21, 2020 at 17:56 | comment | added | Ilkka Törmä | @GabeK If the distance is greater than 2, then one of the lines has distance greater than 1 from $\gamma(-x)$ and $\gamma(x)$, and so does any intersection point with $\gamma([-x,x])$. | |
| Nov 21, 2020 at 17:46 | comment | added | Gabe K | Can you explain further why the lines must be separated by a distance that is at most 2 in Lemma 3? I don't understand that step. Thanks! | |
| Nov 21, 2020 at 15:38 | history | answered | Ilkka Törmä | CC BY-SA 4.0 |