Timeline for Components of the complement of a compact set
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2022 at 18:15 | comment | added | Saúl RM | This is not a topological problem because $A$ can have infinitely many components with $K$ being the image of an injective curve $\gamma:[0,3]\to\mathbb{R}$: you can take $\gamma(\frac{1}{n})=(\frac{1}{n},0)$, $\gamma(3-\frac{1}{n})=(\frac{1}{n},1)$ for natural $n\geq1$, $\gamma(0)=(0,0)$, $\gamma(3)=(0,1)$ and define $\gamma$ in the rest of the domain so that it stays outside of the rectangle $[0,1]\times[-2,2]$ (and $r=1/2,R=\infty$). $\gamma$ can probably be chosen to be $C^\infty$ as well. | |
| S Feb 19, 2022 at 10:18 | history | suggested | CommunityBot | CC BY-SA 4.0 |
I changed a question with yes/no answer to a more general one.
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| Feb 18, 2022 at 23:42 | comment | added | M. Rahmat | @RyanBudney: Thanks. If the $\rho-$boundaries of $K$, i.e., the set of points have a distance $\rho$ to $K$, are all connected for sufficiently small $\rho$, it works. But maybe this condition is too demanding, and also, it is not a condition on $K$. | |
| Feb 18, 2022 at 19:30 | comment | added | M. Rahmat | @IosifPinelis: I was just curious. | |
| Feb 18, 2022 at 19:08 | comment | added | Iosif Pinelis | @M.Rahmat : Only indirectly. I was just trying to understand Ryan Budney's comments. | |
| Feb 18, 2022 at 19:07 | comment | added | Iosif Pinelis | @RyanBudney : Thank you for the reference. | |
| Feb 18, 2022 at 19:03 | comment | added | Ryan Budney | I don't know anything cute and minimal. But my first inclination would be to have a "nice" geometric regular neighbourhood. By "nice" I mean something like smooth or PL. And by "geometric" I'm talking about using the distance function you use in your question to define the regular neighbourhood. So maybe you could get away with an NDR pair $(\Bbb R^n, K)$ or something like that? | |
| Feb 18, 2022 at 18:49 | comment | added | M. Rahmat | @RyanBudney : Thanks. For 2) I meant if there are some necessary and sufficient conditions or at least some sufficient conditions. | |
| Feb 18, 2022 at 18:48 | comment | added | M. Rahmat | @IosifPinelis: Sorry, I am a little lost: is your discussion related to my question? | |
| Feb 18, 2022 at 18:24 | review | Suggested edits | |||
| S Feb 19, 2022 at 10:18 | |||||
| Feb 18, 2022 at 17:41 | comment | added | Ryan Budney | Check out Theorem 1.38 in Chapter 1 of Hatcher's algebraic topology notes. It gives a general prescription for construction of covering spaces. | |
| Feb 18, 2022 at 17:35 | comment | added | Iosif Pinelis | @RyanBudney : Thank you for your response. I have briefly looked at Ch.2 of Spanier's Algebraic Topology, but did not see answers to my questions. (Sorry, I have no knowledge of algebraic topology, just curious.) | |
| Feb 18, 2022 at 17:22 | comment | added | Ryan Budney | @IosifPinelis: Most introductory algebraic topology textbooks have the construction. The method I am thinking of is as a quotient space of a path-space construction. Once you see it, it has a very intuitive nature to it. I came across a homeless guy in New York (no formal mathematical training) who had near a hundred hand sketches of embeddings of finite-sheeted and universal covers of graphs into the plane. I imagine this could be turned into a fun semi-complicated game, like latin squares / sudoku. | |
| Feb 18, 2022 at 14:35 | comment | added | Iosif Pinelis | @RyanBudney : Where can one learn about necessary conditions and sufficient conditions for the existence of a universal cover, and about how to construct a universal cover when it is known to exist? | |
| Feb 18, 2022 at 5:37 | comment | added | Ryan Budney | 1) Yes, think about embedding the universal cover of $S^1 \vee S^1$ in the plane, i.e. $m=2$. Choose any embedding where the ends get close to touching. 2) Sure. Isn't your item (2) one such condition? | |
| Feb 18, 2022 at 1:42 | history | asked | M. Rahmat | CC BY-SA 4.0 |