Timeline for Does every ultrafilter has single limit imply Hausdorff separation
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2012 at 13:55 | vote | accept | Jialiang He | ||
| Apr 17, 2012 at 13:55 | vote | accept | Jialiang He | ||
| Apr 17, 2012 at 13:55 | |||||
| Apr 17, 2012 at 13:55 | vote | accept | Jialiang He | ||
| Apr 17, 2012 at 13:55 | |||||
| Apr 14, 2012 at 6:50 | comment | added | Martin Sleziak | I just thought it is good to mention that $p$ belongs to the closure of every element of $U$ is the definition of cluster point of $U$. For ultrafilters, the limit and cluster point are equivalent notins, for arbitrary filter they may differ. | |
| Apr 14, 2012 at 3:23 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |