Timeline for Result that follows from ZFC and not ZF but are strictly weaker than choice
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2012 at 12:23 | answer | added | Mike Shulman | timeline score: 3 | |
| Nov 26, 2012 at 7:23 | answer | added | AoC | timeline score: 11 | |
| Nov 25, 2012 at 21:42 | answer | added | none | timeline score: 2 | |
| Nov 25, 2012 at 15:27 | answer | added | Todd Trimble | timeline score: 5 | |
| Nov 25, 2012 at 15:04 | history | edited | Todd Trimble | CC BY-SA 3.0 |
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| Nov 25, 2012 at 15:01 | comment | added | Todd Trimble | @Asaf: that's an interesting discussion! | |
| Nov 25, 2012 at 13:56 | comment | added | Asaf Karagila♦ | @Todd, I actually posted a question about this: mathoverflow.net/questions/104016/… | |
| Nov 25, 2012 at 13:15 | answer | added | arsmath | timeline score: 3 | |
| Nov 25, 2012 at 13:08 | comment | added | Todd Trimble | @Asaf: I agree. The OP made an edit after I made the comment which reflects what you just said, and I'm happy now. Something that's interesting is that principles that seem to "require choice" in this shorthand sense often wind up being elevated to choice principles in their own right, and people wind up studying the mathematics of ZF + weakened choice principle, as you were perhaps delicately suggesting at the end. | |
| Nov 25, 2012 at 11:31 | comment | added | Asaf Karagila♦ | Todd, it's actually quite reasonable to say that "The countable union of countable sets is not provable without the axiom of choice" as a shorthand to "We cannot prove it from ZF, but we can prove it from ZFC". There are weak choice principles which really cannot be formulated in terms of partial choice or bounded choice, and the only way to "pinpoint" their strength is a tautological "$\varphi$ is not provable from ZF without assuming $\varphi$ holds." of some sort. | |
| Nov 25, 2012 at 11:29 | answer | added | Asaf Karagila♦ | timeline score: 4 | |
| Nov 25, 2012 at 8:37 | answer | added | Stefan Geschke | timeline score: 4 | |
| Nov 25, 2012 at 6:27 | answer | added | Andreas Blass | timeline score: 9 | |
| Nov 25, 2012 at 4:17 | history | edited | David Corwin | CC BY-SA 3.0 |
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| Nov 25, 2012 at 4:09 | comment | added | Todd Trimble | I'm not certain I understand "require choice to prove". For instance, one can prove countable choice from dependent choice; we don't require the full axiom of choice to prove it. We "require" AC if AC is necessary to prove some statement P, but I can't think of a meaning this might have that isn't the same as AC is a necessary consequence of P. | |
| Nov 25, 2012 at 3:58 | history | asked | David Corwin | CC BY-SA 3.0 |