Timeline for Can iterating countable unions give every set? (ZF)
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2011 at 20:46 | comment | added | Clinton Conley | Great! I don't have access to the paper tonight and was going to check tomorrow, but you saved me the trouble. Thanks again! | |
| Jul 12, 2011 at 20:42 | history | edited | François G. Dorais | CC BY-SA 3.0 |
belated addendum
|
| Jul 27, 2010 at 0:19 | vote | accept | CommunityBot | ||
| Jul 24, 2010 at 10:46 | comment | added | Joel David Hamkins | In my first comment above, I meant to ask about non-well-orderable cardinalities (since ZF has Foundation axiom, we have well-founded $\in$). But I guess you are right that even if every set is a countable union of strictly smaller sets, then for the reason you say one cannot seem to induct on rank to show every set is reachable. | |
| Jul 23, 2010 at 22:16 | comment | added | François G. Dorais | Some of the subsets might have the same rank as S. More generally, the argument works for the wellfounded part of the cardinal partial order, this might include all cardinals even if choice fails but I'm not sure about this particular model. | |
| Jul 23, 2010 at 21:49 | comment | added | Joel David Hamkins | If you did have that every set was a countable union of smaller sets, then you could get the desired answer to the question by inducing on rank (which is well-founded in ZF) rather than cardinality. | |
| Jul 23, 2010 at 11:57 | comment | added | Joel David Hamkins | What a great model! But is there any hope for us to have all cardinalities (including the non-well-founded ones) having cofinality $\omega$? That is, is it consistent with ZF that every set is a countable union of smaller sets? | |
| Jul 23, 2010 at 1:47 | history | edited | François G. Dorais | CC BY-SA 2.5 |
addendum
|
| Jul 23, 2010 at 0:54 | history | answered | François G. Dorais | CC BY-SA 2.5 |