Timeline for Covering a compact subset by a tubular neighborhood of a smooth curve
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2020 at 23:34 | comment | added | Kosh | The question is precise. Your answer gives an intuitive idea of why should be true (in a very specific case), which is nothing but the intuitive idea that brought me to "conjecture" the question (was not taken from somewhere). I posted the question in the hope of some simple (maybe nonconstructive) proof. | |
| Jul 15, 2020 at 0:23 | comment | added | Rohil Prasad | I'm not going to belabor the point, but I don't really see why there is some deeper issue at work in this problem. You should edit your question to make it more clear what you want out of this, especially if there's a specific application in mind. | |
| Jul 14, 2020 at 8:18 | comment | added | Kosh | I am confident that someone will post a proper answer at some point. Thanks anyhow. | |
| Jul 14, 2020 at 1:38 | comment | added | Rohil Prasad | I suppose the "piecewise-linear" approach is only rigorous if the set is convex, but as mentioned above it is easy to adapt. If you would like an "algorithm", then you can just connect any pair of two points by curves that pairwise do not intersect (as long as $n \geq 3$). This is some complete graph, and then you just apply your favorite graph theory algorithm for extracting a path that visits all vertices exactly once. | |
| Jul 13, 2020 at 21:14 | comment | added | Kosh | The problem of the proof is not smoothness. It is to write down rigorously an "algorithm" which builds the curve (piecewise linear, even just continuous, is more than fine) given a linked sequence of balls (because one could be obliged to go through the same balls several times). I was hoping someone already faced such a question and could give a rigorous and simple answer :) The answer given by Rohil cover the simple part that anyone trying to prove the result would write down (I don't know how to express this without appearing rude, but it is not my intention) | |
| Jul 13, 2020 at 21:05 | comment | added | Pietro Majer | The third point can be simplified: one does not need smooth approximation. Instead, one can re-parametrize each affine segment of $\gamma$, in such a way that derivatives of all order vanish at any node, which makes the whole curve smooth, with the same image as $\gamma$. In other words, you can make any zig-zag curve smooth, if you stop whenever you need to make an angle. | |
| Jul 13, 2020 at 20:56 | comment | added | Jaap Eldering | Since $\Omega \subset \mathbb{R}^n$ is (path) connected, you can connect the $p_i$ inside it. That should help you construct $\gamma_K$ to satisfy all requirements. | |
| Jul 13, 2020 at 19:56 | comment | added | Kosh | Officially you cannot connect by a line $p_1$ and $p_2$ because this line can lie outside of $\Omega$. The idea is clear, the problem is the (maybe combinatorial) formalization of the existence of a path that visit all the centers of these balls without going out of $\Omega$. | |
| Jul 13, 2020 at 16:51 | history | answered | Rohil Prasad | CC BY-SA 4.0 |