Timeline for Connectedness of the set having a fixed distance from a closed set 2
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2020 at 4:15 | comment | added | Anthony Quas | No. Everything on the sphere of radius 2 is at most 0.1 away from $F$. The finite part of $A$ consists of points of radius approximately 1: points of radius 1 are exactly 1 unit from the sphere of radius 2 and approximately 1 away from $F$. | |
| Dec 12, 2020 at 18:31 | comment | added | M. Rahmat | @ Anthony Quas. It seems to me that the component of $A$ that is inside the sphere of radius 2 is not a complete sphere, but a smaller copy of $F$ on the sphere of radius 2; i.e. it is a 0.1-dense continuous curve around the sphere of radius 1. Could you tell me what you think? | |
| Dec 12, 2020 at 0:49 | vote | accept | M. Rahmat | ||
| Dec 11, 2020 at 21:51 | history | edited | Anthony Quas | CC BY-SA 4.0 |
added 32 characters in body
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| Dec 11, 2020 at 21:50 | comment | added | Anthony Quas | 0.1-dense in the sphere means that for each point in the sphere, there is a point on the curve at most 0.1 away. I think the complement of $A$ has three components: two components where $d(x,F)>1$: one of these is approximately the unit ball; and the other one an infinite component consisting of everything outside (in the everyday sense) of a fattened version of $F$. The third component is things where the distance to $F$ is strictly between 0 and 1. | |
| Dec 11, 2020 at 21:44 | comment | added | M. Rahmat | @ Anthony Quas I see. Thanks. Your construction is very good, but could you please tell me what you mean by 0.1-dense continuous curve? I have also another question (it was not in the question that I asked but to better understand): Is the complement of $A $ in your construction still connected, or disconnected because of the component of $A$ that is inside the ball? Thanks for your explanation. | |
| Dec 11, 2020 at 20:19 | comment | added | D.S. Lipham | @M.Rahmat $F$ is the red line, and its complement is connected | |
| Dec 11, 2020 at 18:36 | comment | added | M. Rahmat | The complement of $F$ must be connected. It seems to me that in your counter example this not the case. Could you please clarify? | |
| Dec 11, 2020 at 7:11 | history | answered | Anthony Quas | CC BY-SA 4.0 |