Timeline for Subcovers without a choice set
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2017 at 15:02 | comment | added | Ramiro de la Vega | I meant to say "largest index" instead of "least index". | |
| Nov 10, 2017 at 14:55 | comment | added | Ramiro de la Vega | @BjørnKjos-Hanssen: No, it is not the same construction. To see that $|S\cap D|=1$ for $S \in \mathcal S_0$, note that if $d_\xi \in S$ then for each $\alpha>\xi$, $d_\alpha$ was chosen in $X \setminus S$. So $\xi$ is the least index (and hence the only index) for which $d_\xi \in S$. | |
| Nov 10, 2017 at 12:40 | comment | added | Dominic van der Zypen | Thank you Ramiro for putting this so clearly & concisely! | |
| Nov 10, 2017 at 12:39 | vote | accept | Dominic van der Zypen | ||
| Nov 10, 2017 at 10:48 | history | answered | Ramiro de la Vega | CC BY-SA 3.0 |