Timeline for Efficiency of covers
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2022 at 10:53 | history | undeleted | Dominic van der Zypen | ||
| May 8, 2022 at 19:37 | history | deleted | Dominic van der Zypen | via Vote | |
| May 7, 2022 at 13:32 | comment | added | Dominic van der Zypen | Thanks @bof - I failed to see that line of argument. If you want, you can post this as an answer (and I'll accept and upvote it). If you don't want to, I'll delete the question after a few hours. | |
| May 7, 2022 at 9:54 | comment | added | bof | $|C\setminus D|+|D\setminus E|+|E\setminus C|=|D\setminus C|+|E\setminus D|+|C\setminus E|$ | |
| May 7, 2022 at 9:45 | comment | added | bof | Never mind covers, you can't have sets $C,D,E$ with $|C\setminus D|\lt|D\setminus C|$, $|D\setminus E|\lt|E\setminus D|$, and $|E\setminus C|\lt|C\setminus E|$. This follows from the fact that (at least if the axiom of choice holds) for cardinal numbers $w,x, y,z$, if $w\lt x$ and $y\lt z$ then $w+y\lt x+z$. | |
| May 7, 2022 at 8:42 | comment | added | Jukka Kohonen | Just checking: $|\cdot|$ denotes cardinality as usual? With finite covers the "more efficient than" relation would be equivalently $|C| < |D|$, so apparently you are looking for infinite covers. | |
| May 7, 2022 at 8:10 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |