Timeline for If a polyhedron in $\mathbb{R}^3$ has local intersections, does it also have more global intersections?
Current License: CC BY-SA 4.0
4 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Mar 12, 2023 at 18:48 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
deleted 83 characters in body
|
Mar 12, 2023 at 18:46 | comment | added | Anton Petrunin | Oh --- there was a mistake, I will fix it in a minute. | |
Mar 12, 2023 at 1:32 | comment | added | Omega Tree | Sorry, I don't completely follow your argument (though it's quite possible I'm misinterpreting your notation). Suppose $\alpha$ and $\beta$ are coplanar triangles, neither containing the other, and sharing an edge $\sigma$, as in this image: i.sstatic.net/vz3z7.png. Then following your construction I can pick $x$ and $y$ such that $\alpha^\prime = \alpha$ and $\beta^\prime$ is an edge sharing a vertex with $\alpha^\prime$. But in this case $\alpha^\prime$ and $\beta^\prime$ are not disjoint, hence there is no immediate contradiction. | |
Mar 11, 2023 at 16:16 | history | answered | Anton Petrunin | CC BY-SA 4.0 |