Timeline for Which axioms of ZF are used for finite choice?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2014 at 9:01 | comment | added | Joel David Hamkins | Yes, I realized my set-theory superpowers after casting my vote. | |
| Aug 26, 2014 at 3:38 | comment | added | Asaf Karagila♦ | @Joel: Perks of having a gold badge in the tag, you get a supervote about closing and reopening duplicates. | |
| Aug 26, 2014 at 1:07 | answer | added | Joel David Hamkins | timeline score: 5 | |
| Aug 26, 2014 at 0:17 | comment | added | Joel David Hamkins | Oh, my vote to re-open was evidently successful... | |
| Aug 26, 2014 at 0:15 | history | reopened | Joel David Hamkins set-theory Users with the set-theory badge or a synonym can single-handedly close set-theory questions as duplicates and reopen them as needed. | ||
| Aug 26, 2014 at 0:15 | comment | added | Joel David Hamkins | The question was marked as a duplicate, but actually none of the answers over there answer the specific question here about which fragment of ZF suffices to prove AC for finite families. The answers at the other question are instead concerned mainly with category-theoretic or non-classical logic perspectives on AC for finite families, which is interesting, but seem less relevant to the specific question here. So I have voted to reopen. I'll post an answer if it is re-opened. | |
| Aug 25, 2014 at 23:31 | history | closed |
Stefan Kohl♦ S. Carnahan♦ |
Duplicate of Can we disallow finite choice? | |
| Aug 25, 2014 at 17:44 | comment | added | Joel David Hamkins | Oh yes, I agree completely. I made the point merely because I have a feeling that those who want a model where finite choice fails want it to fail for a family that is "really" finite, that is, finite in the meta-theory. | |
| Aug 25, 2014 at 17:41 | comment | added | Asaf Karagila♦ | @Joel: But that's a big "provided". Since one can easily manufacture a model where this fails using compactness. Of course the failure is at non-standard finiteness, but still it makes finite choice unprovable (internally). | |
| Aug 25, 2014 at 17:40 | comment | added | Joel David Hamkins | @AsafKaragila Right. But notice also that you don't even need induction in the theory to prove the instances of AC for finite families, provided that the size of the family is finite in the meta-theory. That is, even without induction we can prove AC for families of size 1, size 2, size 3, and so on, as a theorem scheme. | |
| Aug 25, 2014 at 17:32 | comment | added | მამუკა ჯიბლაძე | What about Vopěnka's principle in this context? | |
| Aug 25, 2014 at 17:29 | comment | added | Asaf Karagila♦ | Since we use induction to prove finite choice, we need to somehow disallow induction. I suppose disallowing sufficiently many replacement/separation axioms will do the trick. But when you remove these axioms you really just get left with a very weak theory, not something I would call "set theory" in any serious meaning of the term. | |
| Aug 25, 2014 at 17:20 | review | Close votes | |||
| Aug 25, 2014 at 23:31 | |||||
| Aug 25, 2014 at 17:15 | comment | added | Joel David Hamkins | By speaking of definable elements, the question seems to conflate the question of whether there is a choice function with whether there is a computable/definable choice function. Even when elements are not definable, one can prove the existence of choice functions for finite families by induction on the size of the family. This is trivial if the family has only one set. If one adds a nonempty set to a family of $n$ nonempty sets, then every choice function on the previous family extends to the new family, one extension for each element of the newly added set. This uses very little of ZF. | |
| Aug 25, 2014 at 16:22 | review | First posts | |||
| Aug 25, 2014 at 17:03 | |||||
| Aug 25, 2014 at 16:20 | history | asked | Doly Garcia | CC BY-SA 3.0 |