Timeline for Least cardinality of a set of points in the plane
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2012 at 18:26 | vote | accept | Holowitz | ||
| Mar 7, 2012 at 16:43 | comment | added | Tony Huynh | @Emil: Apparently, J. Leech has an elementary argument, although I could not track it down. For a proof of a stronger result (only assuming that one side is rational and the squares of the other sides are rational) see this paper matwbn.icm.edu.pl/ksiazki/aa/aa62/aa6246.pdf | |
| Mar 7, 2012 at 16:41 | answer | added | Tony Huynh | timeline score: 4 | |
| Mar 7, 2012 at 16:37 | comment | added | Emil Jeřábek | @Tony: I can’t say I see this. Is there a simple argument I am missing? | |
| Mar 7, 2012 at 16:10 | comment | added | Tony Huynh | @Emil: Yes. If each of the sides of a triangle has rational length, then the set of points at rational distance from all three of its vertices is dense in the plane. | |
| Mar 7, 2012 at 16:01 | comment | added | Holowitz | @Emil Right, for the second question we have $4\leq K$, atleast if they are not colinear. | |
| Mar 7, 2012 at 15:45 | comment | added | Emil Jeřábek | Is there any set of three points that does not have this property? | |
| Mar 7, 2012 at 15:34 | history | edited | Holowitz | CC BY-SA 3.0 |
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| Mar 7, 2012 at 14:09 | history | edited | Holowitz | CC BY-SA 3.0 |
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| Mar 7, 2012 at 14:07 | comment | added | Boris Bukh | Three points suffice. Take $S′=\{(0,0),(1,0)\}$. Then any point with algebraic distances to $S'$ has algebraic coordinates (it lies on the intersection of circles with algebraic coefficients). Let $S=S'\cup\{(x,y)\}$, where $1,x,y$ are algebraically independent over $\mathbb{Q}$. | |
| Mar 7, 2012 at 13:53 | history | edited | Holowitz | CC BY-SA 3.0 |
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| Mar 7, 2012 at 13:46 | history | asked | Holowitz | CC BY-SA 3.0 |