Timeline for Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$?
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| when toggle format | what | by | license | comment | |
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| Apr 25, 2022 at 13:22 | comment | added | Alessandro Codenotti | @NikWeaver this is exactly what Pittman's proof does (see the first comment below bof's answer) | |
| Apr 25, 2022 at 13:20 | comment | added | Nik Weaver | What about: we can find a small circle centered at the vertex, such that each of the three prongs intersects it. Each prong then has an earliest point where it intersects the circle, so there are three non-intersecting paths from the vertex to the circle. Now use the Jordan curve theorem to get three disjoint open regions bounded by the prongs and circle. | |
| Apr 25, 2022 at 12:41 | comment | added | Piotr Hajlasz | The unclear part of your argument is "divide into three regions". The set is just homeomorphic to Y so it can be very complicated. Your argument is well known when the Y is say piecewise linear. | |
| Apr 25, 2022 at 11:50 | history | answered | Per Alexandersson | CC BY-SA 4.0 |