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Timeline for On a weak choice principle

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Aug 7, 2020 at 0:39 answer added David Roberts timeline score: 3
Aug 7, 2020 at 0:33 history edited David Roberts CC BY-SA 4.0
Fixed second broken link
Aug 7, 2020 at 0:10 history edited David Roberts CC BY-SA 4.0
Fixed broken link
Dec 14, 2017 at 6:22 vote accept David Roberts
Jun 18, 2012 at 23:35 comment added David Roberts Of course, what was I thinking? Not enough sleep...
Jun 18, 2012 at 23:10 comment added François G. Dorais @David: You mean that Con(ZFC + there is an almost huge cardinal) implies Con(ZF + ¬WISC). (In addition to the corrected conclusion, note the C in the hypothesis. In the absence of choice some large cardinals lose much of their might or even become completely meaningless...)
Jun 18, 2012 at 22:57 answer added Asaf Karagila timeline score: 3
Jun 18, 2012 at 22:44 comment added Andrés E. Caicedo Hi David. Yes, definitely. And the answer to your question to Asaf is yes as well (and much more).
Jun 18, 2012 at 22:43 comment added Asaf Karagila David, reviewing Monro's paper "INDEPENDENCE RESULTS CONCERNING DEDEKIND-FINITE SETS" it seems to me that the model constructed in section 4 has the property that a proper class of mutually incomparable sets can be mapped onto $\omega$. This breaks WISC badly, methinks.
Jun 18, 2012 at 22:40 comment added David Roberts @Andres - ah, that is nice. Can I confidently say $Con(ZF + \exists \textrm{almost huge cardinal}) \Rightarrow \neg WISC$?
Jun 18, 2012 at 22:35 comment added David Roberts @Asaf 2 - the set of surjections that the axiom guarantees does not necessarily form an antichain - it is itself only a poset.
Jun 18, 2012 at 22:33 comment added Asaf Karagila David, I think that if there is only a set of regular cardinals you could perhaps use similar proofs to show that all sharps exist. I haven't got the slightest clue, though. The proof itself uses the fact that we can contradict the covering lemma, so if we have set-many regulars, I would guess that we use similar tricks to show that the cover lemma fails for $L[A]$ for all $A$. This is a wild stab in the dark, though.
Jun 18, 2012 at 22:26 comment added David Roberts @Asaf - so can I assume a cardinal around the size of $0^\sharp$ a lower bound for showing the class of regular cardinals only forms a set?
Jun 18, 2012 at 22:22 comment added Asaf Karagila To add on the previous question, does your question asks if we can bound these chains without choice, that is find a weak choice principle which will guarantee that every set has a maximal anti-chain of size $\leq|A|$ in $\leq^\ast$ above it - or ask whether or not we can have WISC but the anti-chains have an unbounded size in cardinality...
Jun 18, 2012 at 22:17 comment added Asaf Karagila Robert, or anyone else knowledgeable on this, does WISC is really the assertion that in $\leq^\ast$ (order cardinals by surjections, instead of injections) above any $X$ there is a maximal anti-chain which is a set? I would believe that to violate that you would have to do some class-forcing. Perhaps Monro's model with a proper class which is D-finite could be a suitable model for this. I'll give it some thought...
Jun 18, 2012 at 21:35 comment added Asaf Karagila @Francois: Magidor proved that if both $\omega_1$ and $\omega_2$ are singular then $0^\#$ exists. So there is some need for large cardinals for sure. I am not sure about exact strength, but I think that the measure is about a Woodin or two for every two consecutive singulars... So for all [but a set] of them singular you probably need a whole lotta Woodin.
Jun 18, 2012 at 21:21 comment added Andrés E. Caicedo (Sorry for all the typos.)
Jun 18, 2012 at 21:19 comment added Andrés E. Caicedo (...) If one is only interested in "a lot of consecutive cardinals are singular" (for a reasonable interpretation of "a lot"), then $\omega$ Woodins (or, equivalently, $AD^{L({\mathbb R})}$) suffice. This was shown by Apter, and additional details and references can be found in the Busche-Schindler paper. (By the way, the reason I wrote that they obtain "a tiny bit more" is because they also get that every set in the inner model of the forcing extension has a sharp, which pushes the consistency strength past just $\omega$ Woodins.)
Jun 18, 2012 at 21:15 comment added Andrés E. Caicedo (...) available at Ralf's page. There, they prove (a tiny bit more than): If every uncountable (well-ordered) cardinal is singular, then there in a forcing extension there is an inner model where $AD^{L({\mathbb R})}$ holds. In large cardinal terms, what they obtained is a tiny bit beyond $\omega$ Woodin cardinals. We expect significant more strength is needed, but core model inductions in the absence of choice are very tricky, and their results are state of the art. (...)
Jun 18, 2012 at 21:09 comment added Andrés E. Caicedo Hi François. No, this is not known. The large cardinal assumption we currently need to force something like Gitik's result is an almost huge cardinal. This is much weaker than Gitik's original assumptions, but still in the stratosphere. We cannot currently prove any lower bounds that involve large cardinals at or beyond the level of superstrong, and almost huges are well past that. The best result to date is due to Daniel Busche and Ralf Schindler, in "The strength of choiceless patterns of singular and weakly compact cardinals, Ann. Pure Appl. Logic, 159 (2009), pp. 198-248" (...)
Jun 18, 2012 at 18:56 answer added François G. Dorais timeline score: 4
Jun 18, 2012 at 15:15 comment added David Roberts Whoops, in my last comment I meant "...WISC proves stronger results, so there exist models which violate WISC use less large cardinals". And @François - good point!
Jun 18, 2012 at 14:03 comment added François G. Dorais @Asaf: Is it known that Gitik's large cardinal hypotheses are necessary?
Jun 18, 2012 at 7:41 comment added David Roberts Yes, this possibility did occur to me after I posted the edit, but I couldn't be sure. There is always the possibility that WISC proves stronger results, and so uses less large cardinals, but that's not what I asked...
Jun 18, 2012 at 6:36 comment added Asaf Karagila David, if you only have a set of regular cardinals then you would need about the same amount of large cardinals as in Gitik's model.
Jun 18, 2012 at 1:05 history edited David Roberts CC BY-SA 3.0
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Jun 17, 2012 at 6:29 comment added godelian Oh no, just wanted to point it out in case you haven't heard about it, but it's better to update the question perhaps, since the links are updated. Also, if I understand correctly, there is a large cardinal assumption used to construct the model where WISC fails. One could then try to ask if this is really necessary.
Jun 17, 2012 at 0:22 comment added David Roberts Did you want to make an answer out of your comment?
Jun 17, 2012 at 0:21 comment added David Roberts Thanks, godelian. I was given a copy of this preprint when it appeared, and I had forgotten about this question of mine, else I would have updated it.
Jun 16, 2012 at 21:47 comment added godelian I happened to attend today a talk by Benno van den Berg at a workshop and was incidentally led to this article of his: staff.science.uu.nl/~berg0002/papers/WISC.pdf where he proves the independence of WISC from ZF by showing that WISC implies the existence of arbitrarily large regular cardinals. I checked again the nlab page and found out that it was already updated with this information, which answers the bonus question.
Aug 31, 2011 at 4:11 answer added Mike Shulman timeline score: 2
Aug 30, 2011 at 3:20 history edited David Roberts CC BY-SA 3.0
added 258 characters in body
Aug 29, 2011 at 14:49 comment added David Roberts Something funny happened to my comments; they've disappeared! I've edited the question as per suggestions, anyway.
Aug 29, 2011 at 4:01 history edited David Roberts CC BY-SA 3.0
Clarifies some points as per comments
Aug 29, 2011 at 2:17 comment added user5810 Since it is implied by AC, how can WISC "be interpreted as saying Choice is violated" at all?
Aug 29, 2011 at 1:38 comment added François G. Dorais How is WISC related to the other better known choice principles (COSHEP, SVC, AMC, etc.)? I can't think of any that are weaker, but some may be incomparable.
Aug 29, 2011 at 1:35 comment added François G. Dorais I think you need to set some limitations on 'weaker choice principle'. There is a very easy trick: restrict WISC to hold for a particular non-trivial set (often $\mathbb{R}$ but sometimes $\mathcal{P}(\mathbb{R})$ or higher up, if necessary). Assuming this consequence of WISC is nontrivial, it will surely be much weaker.
Aug 28, 2011 at 22:33 history asked David Roberts CC BY-SA 3.0