Timeline for Is it still impossible to partition the plane into Jordan curves without choice?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2010 at 16:51 | vote | accept | Benoît Kloeckner | ||
| Apr 14, 2010 at 16:42 | comment | added | Benoît Kloeckner | @Nate Elredge: here I consider non constant curves only. It seems the most reasonnable choice: if you want the Jordan theorem to apply to all Jordan curves, then certainly you don't want to call a point a Jordan curve. | |
| Apr 14, 2010 at 15:52 | comment | added | Nate Eldredge | Somewhat off-topic question: a point is not a Jordan curve? Wikipedia's definitions seem to allow this (see articles on "Jordan Curve" and "Curve"). | |
| Apr 14, 2010 at 14:46 | history | edited | Reid Barton | CC BY-SA 2.5 |
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| Apr 14, 2010 at 14:40 | answer | added | Miguel | timeline score: 2 | |
| Apr 14, 2010 at 12:38 | comment | added | Joel David Hamkins | For the R^3 case, here is an interesting article: mscand.dk/article.php?id=77. They partition R^3 into unlinked congruent circles, and also consider other more arbitrary families of curves. Partitioning R^3 into circles is constructive, but for the more exotic partitions, the Axiom of Choice is used. | |
| Apr 14, 2010 at 12:13 | answer | added | t3suji | timeline score: 4 | |
| Apr 14, 2010 at 12:04 | comment | added | damiano | Instead of using the axiom of choice, choose a dense countable set of points in the plane. Associated to the first point in the set, there is a circle. Inside this circle, find the point in your dense countable set with least index and repeat. This might "guide" you through the circles, without using the axiom of choice. | |
| Apr 14, 2010 at 11:37 | history | asked | Benoît Kloeckner | CC BY-SA 2.5 |