Timeline for Linear orders with only short cuts
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2011 at 7:11 | vote | accept | nick | ||
| Sep 30, 2011 at 23:10 | comment | added | nick | Yes, this is a convincing counterexample. On a second reading, I realized that I misinterpreted the question. The point is not that the only cuts have countable cofinality, but that all cuts with cofinality $(\lambda, \kappa)$ with $\lambda$ and $\kappa$ uncountable actually have cofinality &(\lambda, \lambda)& - that is, all cuts with uncountable cofinality have the same cofinality in the lower cut and in the upper cut with the reverse order. | |
| Sep 30, 2011 at 23:07 | vote | accept | nick | ||
| Oct 1, 2011 at 7:11 | |||||
| Sep 29, 2011 at 23:32 | comment | added | François G. Dorais | But this is not a "proper" cut if the $\omega_1$ sequence has a supremum in $I$. | |
| Sep 29, 2011 at 22:53 | history | edited | Ramiro de la Vega | CC BY-SA 3.0 |
deleted 9 characters in body
|
| Sep 29, 2011 at 22:47 | history | answered | Ramiro de la Vega | CC BY-SA 3.0 |