Timeline for Is a certain subset of the disc a convex set?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Dec 8, 2015 at 4:44 | comment | added | Christian Remling | @EricTowers: I've reorganized, maybe it's clearer now. | |
| Dec 8, 2015 at 4:44 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Dec 8, 2015 at 4:20 | comment | added | Christian Remling | @EricTowers: Yes, this is right of course: deleting a small interval slices off a small piece from my chord set (the set bounded by the deleted arc and the secant). The point is that everything else stays in the chord set (the argument is clumsily described in my answer, but a picture should clarify), so it remains convex. | |
| Dec 8, 2015 at 4:10 | comment | added | Eric Towers | I'm having a "slow" regarding your point (2). It would seem that points in the interior of the chord having the same endpoints as the deleted open interval are detached from the rest of the chords (except via the endpoints). I.e., there is no chord with an endpoint in the deleted arc, so there is no chord meeting the interior points of this chord. (... and/or I'm missing why the union of the triangular complements you mention don't leave an empty ellipse-like region near that chord.) | |
| Dec 8, 2015 at 1:26 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Dec 8, 2015 at 1:15 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Dec 8, 2015 at 0:20 | history | answered | Christian Remling | CC BY-SA 3.0 |