Timeline for A question about Transfinite Induction
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2013 at 19:52 | history | wiki removed | François G. Dorais | ||
| Nov 25, 2010 at 17:03 | comment | added | Kaveh | your questions is similar to the following: are the set of cardinals well-ordered? but I think in practice it would be difficult to use this in place of ordinal induction because we often use induction to prove a statement about objects defined by recursion. | |
| Nov 19, 2010 at 4:15 | answer | added | Andrés E. Caicedo | timeline score: 8 | |
| Nov 19, 2010 at 3:39 | answer | added | Pete L. Clark | timeline score: 4 | |
| Nov 19, 2010 at 3:24 | comment | added | Joel David Hamkins | Andres: what can one say about well-founded cardinalities that are not well-ordereable? For example, suppose $X$ is a set and all smaller cardinalities are well-orderable---must $X$ be well-orderable? | |
| Nov 19, 2010 at 3:16 | answer | added | Joel David Hamkins | timeline score: 10 | |
| Nov 19, 2010 at 2:55 | comment | added | Andrés E. Caicedo | @Nate: AC has nothing to do with this, in the context of this question, where "cardinal" means "initial ordinal". The result is an easy consequence of well-orderability. But, now that you mention this, I suppose one could ask the ZF question, where P is a property of (not necessarily well-ordered) "cardinalities". Then the answer is no, in general. I think an example would be to assume that there is an infinite Dedekind-finite set, and let P be the statement "the cardinalities are well-founded below me". | |
| Nov 19, 2010 at 2:51 | comment | added | Nate Eldredge | @Andres: Even without AC? | |
| Nov 19, 2010 at 2:32 | comment | added | Andrés E. Caicedo | Sure. Any cardinal is an ordinal. The usual version easily implies this one. | |
| Nov 19, 2010 at 2:30 | history | asked | Dong xiaowei | CC BY-SA 2.5 |