Timeline for Requirement for connected sets
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 17, 2019 at 17:05 | comment | added | Robert Israel | @cherry $v$ is in $V$, which is open. Thus there is a neighbourhood of $v$ which is disjoint from $U$, so $v \notin \overline{U}$. | |
| Mar 16, 2019 at 6:37 | comment | added | cherry | Can you explain how v does not belongs to closure of U. I don't think anything is stopping from this. Of course then s would be zero. | |
| Mar 15, 2019 at 14:53 | history | edited | Todd Trimble | CC BY-SA 4.0 |
typo
|
| Mar 15, 2019 at 14:33 | history | answered | Robert Israel | CC BY-SA 4.0 |