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Dec 10, 2022 at 11:15 comment added David Roberts @Asaf ah, I see :-)
Dec 10, 2022 at 9:32 comment added Asaf Karagila @David: I was joking of course, since it started with Benno's proof that WISC fails in the Gitik model, and then our papers improved this. So here we start with a super-Reinhardt, and surely we can engineer a model without using large cardinals too.
Dec 10, 2022 at 4:37 comment added David Roberts @AsafKaragila I'm not sure why the consistency strength should be high. My WISC paper only started with a well-pointed boolean topos with nno, which is more-or-less as strong as Bounded Zermelo (sans AC). Having an analogue of a collapse forcing to make WISC true while violating other similar axioms would be interesting, though...
Dec 10, 2022 at 4:36 comment added David Roberts @HanulJeon Maybe the best place currently is the nLab, eg ncatlab.org/nlab/show/WISC#relationships_to_other_axioms ncatlab.org/nlab/show/presentation+axiom#consequences ncatlab.org/nlab/show/axiom+of+multiple+choice ncatlab.org/nlab/show/axiom+of+choice#Variants etc
Dec 9, 2022 at 19:25 comment added Asaf Karagila @David might be a better candidate to answer that, to be honest. I have not thought about these since about a decade.
Dec 9, 2022 at 19:03 comment added Hanul Jeon @Asaf Sounds interesting. Is there any good source to check some list of weak choice principles used in algebraic set theory? (I know CoSHEP, but I am curious if there is any form of weak choices they considered.)
Dec 9, 2022 at 14:36 comment added Asaf Karagila @David: I'm sure we can engineer a symmetric extension (via class forcing) to get WISC without SVC. But it seems to me that all of the proofs in these algebraic set theory weak choice principles must begin with some enormous consistency strength... :-)
Dec 9, 2022 at 1:21 comment added Gabe Goldberg @DavidRoberts Yes, that's right.
Dec 8, 2022 at 20:14 comment added David Roberts @Gabe so, to be clear: a super Reinhardt cardinal implies WISC and also $\neg$SVC?
Dec 8, 2022 at 1:35 comment added Gabe Goldberg @DavidRoberts Yes, the Dorais version is (close to) the one I had in mind. The proof that WISC holds for wellorderable codomains under a Reinhardt cardinal is what I called the wellordered collection lemma in a paper named something like “Measurable cardinals and choiceless axioms.” Showing that full WISC follows from Lowenheim-Skolem cardinals is actually much easier. (It’s open whether Reinhardt is lower in consistency strength than Reinhardt plus a proper class of Lowenheim-Skolems.)
Dec 8, 2022 at 1:11 comment added David Roberts @GabeGoldberg I don't know! There are two ways to state WISC that I know of: "For every set X there is a set of surjections to it such that [etc]", or the version François Dorais figured out and gives in mathoverflow.net/a/99934/4177 It would be very cool if WISC and SVC are separated by large cardinal axioms like this.
Dec 8, 2022 at 0:01 comment added Gabe Goldberg @DavidRoberts I think the answer to your question is yes (starting with even stronger assumptions). For example, a super Reinhardt should outright imply WISC (if I’m not misunderstanding what WISC says). In fact, doesn’t WISC follow from a proper class of Lowenheim-Skolem cardinals? Also I think a Reinhardt implies WISC for wellordered codomains.
Dec 6, 2022 at 19:17 vote accept Hanul Jeon
Dec 6, 2022 at 19:17 comment added Hanul Jeon @Noah It is a name for a workshop we held this September.
Dec 6, 2022 at 18:20 comment added Noah Schweber @AsafKaragila I'm sorry, I have to know: what is this cheese you speak of? (Mathematical cheese has become sort of a meme for my students, so I need to tell them about this.)
Dec 6, 2022 at 16:09 answer added Gabe Goldberg timeline score: 9
Dec 6, 2022 at 16:02 comment added Hanul Jeon @Asaf Unfortunately I didn't recall what happened at the last day of CHEESE. I should check my handwritings.
Dec 6, 2022 at 7:12 comment added Asaf Karagila Wasn't this one of the questions we identified at the end of cheese?
Dec 6, 2022 at 6:27 comment added David Roberts If this doesn't work, it would be cool to know if adding WISC is relatively consistent over ZF+Reinhardt, since that's a pretty much minimal choice principle that is needed for various things in category theory.
Dec 6, 2022 at 4:05 history edited Hanul Jeon CC BY-SA 4.0
edited body
Dec 6, 2022 at 4:05 comment added Hanul Jeon (But using two different notations in a single post could be confusing, so I changed the notation.)
Dec 6, 2022 at 4:04 comment added Hanul Jeon @JoelDavidHamkins Both notations are used, and I found that papers by Gabriel and Schlutzenberg use the latter notation more.
Dec 6, 2022 at 3:54 comment added Joel David Hamkins Nice question. But do you mean $j^\omega(\kappa)$ in place of $\kappa^\omega(j)$?
Dec 6, 2022 at 3:46 history asked Hanul Jeon CC BY-SA 4.0