Timeline for Is there any research on set theory without extensionality axiom?
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22 events
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| Nov 20 at 19:55 | comment | added | user76284 | @AsafKaragila I think the point is that collection seems more "orthogonal" to the other axioms than replacement does. For example, it doesn't need the notion of equality (which replacement does, because of the uniqueness quantifier). | |
| Aug 15, 2018 at 10:42 | comment | added | Joel David Hamkins | @MikeShulman No, I don't quite know the full extent of this phenomenon, and it is something that I have been wanting to understand much more fully. | |
| Aug 15, 2018 at 4:52 | comment | added | Mike Shulman | Do you know to what extent this fact is true in weaker theories than ZF? In particular, what about intuitionistic set theory with function-sets, but without powersets, collection, or unbounded separation (corresponding to the internal language of a $\Pi$-pretopos)? | |
| May 27, 2014 at 19:35 | comment | added | Joel David Hamkins | My perspective is that the fact that replacement ZF minus E is surprisingly weak, while collection ZF minus E is not, shows the value and perhaps more fundamental nature of collection over replacement. (And why are you fixed on the idea that ZFC must use replacement instead of collection? Certainly both axiomatizations are quite commonly found.) | |
| May 27, 2014 at 19:28 | comment | added | Asaf Karagila♦ | Joel, I think you may have misunderstood my previous comment. I meant to say that the fact that removing axioms weakens the consistency strength is actually a good thing. | |
| May 27, 2014 at 16:05 | comment | added | Joel David Hamkins | Oh, I completely agree with that. To my way of thinking, the fact that we can recover a companion ZFC model hiding beneath any model of ZFC-E simply affirms that we needn't have bothered omitting it in the first place. | |
| May 27, 2014 at 15:57 | comment | added | Asaf Karagila♦ | By the way, Joel, I'm not so sure that philosophically speaking, we should be in a rush to change $\sf ZFC$ to something that will end up equiconsistent when removing power set/foundation/extensionality. The fact that from $\sf ZFC$ we can always move to strictly weaker theories by removing axioms, and ensure the existence of models over which we can work what we need to work, seems like a very positive quality of $\sf ZFC$. | |
| May 27, 2014 at 13:56 | comment | added | Joel David Hamkins | I have edited to clarify the issue about replacement, which I am glad to learn about. Thanks, Asaf! | |
| May 27, 2014 at 13:56 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| May 27, 2014 at 13:39 | comment | added | Joel David Hamkins | I am sorry if there was a misunderstanding. Personally, I axiomatize ZF with collection+separation, and we seem to have more and more evidence that this is the right way to do it, because of the problems arising with various weakenings of the theory if one has only replacement. For example, the same issue arises when dropping foundation, and looking at the anti-foundational theories. See jdh.hamkins.org/…. | |
| May 27, 2014 at 13:34 | comment | added | Asaf Karagila♦ | Yes, but you don't make it clear in your answer. That's my point, and just to prove my point, see my first comment being baffled on this fact, that it didn't cross my mind to take $\sf ZF$ with collections rather than replacement, and I think I know at least a bit, even if not much more than a bit, about set theory. | |
| May 27, 2014 at 13:33 | comment | added | Joel David Hamkins | I disagree with that. For example, my paper on ZFC-powerset shows quite clearly that if you say "ZFC minus powerset" you definitely should NOT mean the version with replacement. I agree that people should not have to guess, but that isn't what is going on here, since both Scott and Friedman make clear what their theory is. And as it turns out, Friedman has the more robust theory. Scott's theory is flawed in the same way that ZFC-P with replacement is flawed, and Scott himself essentially shows this. | |
| May 27, 2014 at 13:31 | comment | added | Asaf Karagila♦ | Joel, whether or not that is true, the canonical formulation of $\sf ZFC$ is not with the collection schema, but with the replacement schema. Expecting people to immediately guess that in one situation or another we should replace one schema by another is unrealistic. In particular when the OP is not an expert in set theory, and may or may not know about the collection schema and when to use it instead of replacement. | |
| May 27, 2014 at 13:30 | comment | added | Joel David Hamkins | Oh, I see after posting my comment that Emil has also made the same point.... | |
| May 27, 2014 at 13:29 | comment | added | Joel David Hamkins | The explanation would seem to be that they have slightly different versions of ZF, with Scott using replacement rather than collection, and this seems to be key to the argument. That is, he gets replacement just because there are too many automorphisms. (And this is a very similar issue to that arising in my work on ZFC-power set jdh.hamkins.org/what-is-the-theory-zfc-without-power-set) The reviewer Robin Gandy briefly criticizes this in that paragraph, and I would think that the argument itself shows that Scott's version isn't what we should mean by ZF minus extensionality. | |
| May 27, 2014 at 13:22 | comment | added | Asaf Karagila♦ | @Emil: Yes, that is also pointed out in the review. But $\sf ZF$ is not usually formed with collection, but with replacement. You can't just say something like "ZF without extensionality" and mean "with collections", that's not the usual interpretation. | |
| May 27, 2014 at 13:20 | comment | added | Emil Jeřábek | @Asaf: ZFC - Ext with replacement is quite weak. ZFC - Ext with collection is equiconsistent with ZFC. | |
| May 27, 2014 at 13:18 | comment | added | Asaf Karagila♦ | ams.org/mathscinet-getitem?mr=163838 <-- See the last paragraph of the review. | |
| May 27, 2014 at 13:18 | comment | added | Joel David Hamkins | @AsafKaragila can you post a reference for that? It would seem to contradict this old result of Friedman's. | |
| May 27, 2014 at 13:14 | comment | added | Asaf Karagila♦ | If I recall correctly, the consistency of $\sf ZFC-Ext$ is strictly weaker than that of $\sf ZFC$. So your answer seems a bit strange. | |
| May 27, 2014 at 12:39 | comment | added | Joel David Hamkins | The article is on JSTOR: jstor.org/discover/10.2307/…. | |
| May 27, 2014 at 12:21 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |