Timeline for Belief in consistency of extremely large cardinals
Current License: CC BY-SA 3.0
16 events
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| Nov 3, 2020 at 10:08 | comment | added | Monroe Eskew | In my opinion, it is equally plausible to say that the large number of forcing results using large cardinals is equally as convincing as inner model theory. As I understand, the argument is essentially, “If they were inconsistent, the inconsistency would turn up in inner model theory.” Likewise, why not, “ If they were inconsistent, then we would be able to force some clear inconsistency about small cardinals”? | |
| Dec 2, 2016 at 11:14 | answer | added | Thomas Benjamin | timeline score: 7 | |
| Nov 29, 2016 at 21:52 | comment | added | მამუკა ჯიბლაძე | @Stefan Great, thanks! That is why I wanted to know - so then whichever way we go it seems like in some sense existence and nonexistence of Woodin cardinals are in some sense "equiconsistent" with each other, and I think this means that it is reasonable to "believe/disbelieve them with equal amount of faith"... | |
| Nov 29, 2016 at 18:45 | comment | added | Stefan Mesken | @მამუკა ჯიბლაძე Yes, this is possible. Let $\phi \equiv M_1^{\sharp} \text{ exists } + \text{ there is no Woodin cardinal}$. (If $\delta$ is the least Woodin cardinal in our universe, then $V_{\delta}$ satisfies this theory, so it is not outright inconsistent.) Now there is no Woodin cardinal in our universe, but we can linearly iterate the top extender of $M_1^\sharp$ out of the universe, thereby obtaining the inner model $M_1$ which has a Woodin cardinal (from its point of view - the Woodin cardinal of $M_1$ is countable in $V$). | |
| Nov 29, 2016 at 18:17 | comment | added | მამუკა ჯიბლაძე | @Stefan OK thank you, and sorry for too many questions but what I would like to ask more precisely was whether in your last statement one could moreover have $\phi$ which proves that certain large cardinal does not exist and at the same time if there is a model of $ZFC+\phi$ then there also provably is a model of $ZFC+\phi$ which possesses an inner model of $ZFC$ with such cardinal inside it. | |
| Nov 29, 2016 at 15:45 | comment | added | Stefan Mesken | @მამუკაჯიბლაძე What I mean is that starting from $\operatorname{ZFC} + \phi$ for some sufficiently strong condition $\phi$ (say $\phi \equiv \text{Axiom of Determinacy}$) we often can't prove that certain large cardinals must exists (in fact, we can prove that we can't prove that), but - at the same time - we can produce a certain inner model $\mathcal M$ such that inside $\mathcal M$ there provably are large cardinals. | |
| Nov 29, 2016 at 15:36 | comment | added | მამუკა ჯიბლაძე | @Stefan Thanks! Fascinating and confusing at the same time :D When you say "we - in general - don't get large cardinals in our universe", do you mean that this universe represents a model for some axioms rigorously forbidding certain large cardinals, or you mean rather that the latter are absent in the background model "by accident"? Or this distinction does not matter here? | |
| Nov 29, 2016 at 13:58 | comment | added | Stefan Mesken | [...] A famous result along those lines (due to Woodin) is the equiconsitency of $\operatorname{ZF} + \operatorname{AD}$ and $\operatorname{ZFC} + \text{ there are infinitely many Woodin cardinals}$. It's important to note that from a certain assumption $\dagger$ we - in general - don't get large cardinals in our universe, but that $\dagger$ allows us to prove the existence of certain large cardinals in a (canonical) inner model. Typically these large cardinals are in fact countable in our background universe - and hence far from being 'large' in that sense. | |
| Nov 29, 2016 at 13:52 | comment | added | Stefan Mesken | @მამუკაჯიბლაძე In Gödel's constructible universe $L$, only so called 'small large cardinals' can exist. Therefore we always have a canonical inner model with relatively few large cardinals and moreover, any inner model $M$ of $V$ has an inner model with the same large cardinals, since $L^{M}$ (i.e. $L$ constructed inside $M$) is just $L$. On the other hand, not only do large cardinals imply models with other large cardinals - all kinds of axiom imply large cardinals in inner models. [...] | |
| Nov 29, 2016 at 10:52 | comment | added | მამუკა ჯიბლაძე | Thanks, @Asaf and Cameron, your comments are helpful. One further question though, if I may. Is it known for which kind of large cardinal, having a model containing it one can build another model without it, and vice versa (from a model without it one can build another model with it)? | |
| Nov 29, 2016 at 10:43 | comment | added | Cameron Zwarich | @მამუკაჯიბლაძე Kunen's inconsistency theorem established the inconsistency of several otherwise plausible large cardinals relative to ZFC. I believe there is no other argument for establishing inconsistency of an otherwise plausible large cardinal axiom. As Asaf mentions, this is not known to hold relative to ZF alone, but conjectures of Woodin would establish some of the consequence's of Kunen's theorem relative to ZF in a much deeper way than Kunen's original result for ZFC. | |
| Nov 29, 2016 at 10:09 | comment | added | Asaf Karagila♦ | @მამუკაჯიბლაძე: We don't know any proof of inconsistency of "post-choice cardinals" (e.g. Berkeley cardinals and the likes of them). If one assumes the axiom of choice, then just taking a Reinhardt cardinal is already inconsistent. If one assumes the HOD Conjecture, then one can extrapolate some inconsistencies from ZFC to ZF again. | |
| Nov 29, 2016 at 9:23 | comment | added | მამუკა ჯიბლაძე | It would aid (for me) understanding your question if you could illustrate it with the most appropriate, in your opinion, nontrivial example of a seemingly feasible kind of cardinal whose existence is known to be inconsistent. | |
| Nov 29, 2016 at 9:16 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 29, 2016 at 3:51 | history | edited | Cameron Zwarich |
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| Nov 29, 2016 at 3:36 | history | asked | Cameron Zwarich | CC BY-SA 3.0 |