Timeline for Meeting a set of lines in $\mathbb{R}^n$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 7, 2017 at 10:51 | vote | accept | Dominic van der Zypen | ||
| S Aug 7, 2017 at 10:51 | history | suggested | Ali Taghavi |
I add a tag
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| Aug 7, 2017 at 10:11 | review | Suggested edits | |||
| S Aug 7, 2017 at 10:51 | |||||
| Aug 7, 2017 at 10:11 | answer | added | Ali Taghavi | timeline score: 1 | |
| Aug 7, 2017 at 8:11 | answer | added | Martin Sleziak | timeline score: 4 | |
| Aug 7, 2017 at 7:47 | comment | added | Gerhard Paseman | And if you don't like projection, well order the lines and proceed as follows: for the next unselected line, add to M all intersection points of that line with other unselected lines in the set of lines, and then throw all those lines out of the set and continue. Gerhard "Ensure They're All Lined Up" Paseman, 2017.08.07. | |
| Aug 7, 2017 at 7:40 | comment | added | Gerhard Paseman | Indeed, more generally for n=2, pick a line l, and let M be a subset of that line containing all intersections with non parallel lines to l, union a set of representatives of points, one for each line parallel to and different from l. This should work, and you may be able to project down to two dimensions in general. Gerhard "Has Met A Solution Now" Paseman, 2017.08.07. | |
| Aug 7, 2017 at 7:24 | comment | added | Dominic van der Zypen | @GerhardPaseman looks good to me - thanks...! | |
| Aug 7, 2017 at 7:23 | comment | added | Dominic van der Zypen | @MartinSleziak we almost wrote the same thing :) | |
| Aug 7, 2017 at 7:23 | comment | added | Gerhard Paseman | Actually, when n is 2 take for M the set of x and y axes. This works when the set of lines is rich enough, and for thinner sets of lines a subset of this M might be constructed. Gerhard "Is Working On Getting Thinner" Paseman, 2017.08.07. | |
| Aug 7, 2017 at 7:22 | comment | added | Martin Sleziak | Ok, so I missed that $\mathcal L$ is not necessarily finite. So the above works if $\mathcal L$ is countable (and it seems that in fact $|\mathcal L|<\mathfrak c$ suffice). | |
| Aug 7, 2017 at 7:22 | comment | added | Dominic van der Zypen | @Martin I think this works if for every line there is a point that is only covered by that line. But ${\cal L}$ may be such that this is not the case. | |
| Aug 7, 2017 at 7:19 | comment | added | Gerhard Paseman | This idea may not work if a line is covered by other lines in the collection. For n=2 and finitely many lines, one can take all intersections (or representatives in the case of no intersections), but for infinitely many lines (e.g. all of them) it is unclear to me, even with a well ordering of the lines. Gerhard "Isn't Meeting With A Solution" Paseman, 2017.08.07. | |
| Aug 7, 2017 at 7:09 | comment | added | Martin Sleziak | I might have misunderstood this, but it seems that it suffices to choose one point from each line in a such way that none of the chosen points is intersection of this line with some other line in $\mathcal L$. (In the other words, for each $l\in\mathcal L$ we choose a point from $l \setminus \bigcup_{k\in \mathcal L} k$.) | |
| Aug 7, 2017 at 7:04 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |