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Timeline for Arguments against large cardinals

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Jan 18, 2017 at 12:00 comment added Joseph Van Name Nevertheless, I do not consider the fact that large cardinals imply that the Laver tables behave chaotically as strong evidence that large cardinals are inconsistent, but some people might.
Jan 18, 2017 at 11:55 comment added Joseph Van Name The freeness of $(1)_{n}\in\varprojlim_{n}A_{n}$ where $A_{n}$ is the $n$-th classical Laver table (a result requiring large cardinals) implies that the classical and generalized Laver tables behave chaotically. For example, computer calculations show that $\text{FM}_{n}^{-}(x,2^{y})$ is periodic for $n\leq 48$, but large cardinals prove that this periodicity must eventually end or change (see boolesrings.org/jvanname/… for information about $\text{FM}_{n}^{-}$).
Dec 14, 2016 at 23:10 comment added Joseph Van Name I think that one can define the $*$ and $\circ$ operations along with $\equiv^{\gamma}$ without referring to choice, but I am unsure about how well the theory of the algebras of elementary embeddings would work out without the Laver-Steel theorem. I therefore do not see an inconsistency arising from algebras of elementary embeddings from the ZF Reinhardt cardinals as being too much more likely than an inconsistency arising from the algebras of I1-I3 embeddings or from the n-huge cardinals.
Dec 14, 2016 at 23:03 comment added Joseph Van Name The Laver-Steel theorem can be reformulated algebraically to say that whenever $[j_{n}]\in\mathcal{E}_{\lambda}/\equiv^{\gamma}$ for all n, there is some $n$ with $[j_{0}]*...*[j_{n}]=Id_{V_{\lambda}}.$ In fact, the Laver-Steel theorem is needed to prove that $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ is always permutative. Without the Laver-Steel theorem, I am not able to prove very much structure about $\mathcal{E}_{\lambda}/\equiv^{\gamma}$.
Dec 14, 2016 at 23:00 comment added Joseph Van Name I am unaware of any work on the algebras of elementary embeddings that arise from Reinhardt cardinals (nor do I have much intuition about choiceless set theory). One is able to show that there is no elementary embedding $j:V\rightarrow V$ from the Laver-Steel theorem which may be considered as an algebraic result. The Laver-Steel theorem states that if $j_{n}:V_{\lambda}\rightarrow V_{\lambda}$ for all $n$ and $\lambda$ is a limit of inaccessibles, then $\sup_{n}j_{0}*...*j_{n}=\lambda$. In particular, we conclude that $\lambda$ must have countable cofinality (hence the Kunen inconsistency).
Dec 14, 2016 at 22:50 comment added Joseph Van Name Noah. An upvote in your case is warranted since you think that such an inconsistency arising from the algebras with $crit((x*x)*y)>crit(x*y)$ is plausible or possible.
Dec 12, 2016 at 22:08 comment added Noah Schweber I've upvoted, not because I'm convinced, but because of how excited I am about the possibility of this intuition leading to a discovery of inconsistency; not sure if that's in keeping with your instructions, and if you say so I'll un-upvote. Incidentally, something that would drastically increase my worries: can you find an even worse algebraic disconnect assuming ZFC + a Reinhardt? (That is, I'm wondering whether this line of argument could lead - admittedly in retrospect - to the intuition that Reinhardts are "even more" inconsistent than $I_0$s.)
Dec 7, 2016 at 21:54 comment added Joseph Van Name Noah. I have edited my answer to include more motivation for why one would expect the finite permutative LD-systems should always be embeddable into some $\mathcal{E}_{\lambda}/\equiv^{\gamma}.$
Dec 7, 2016 at 5:04 history edited Joseph Van Name CC BY-SA 3.0
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Dec 6, 2016 at 20:10 comment added Noah Schweber (That said, I find your post and these algebras extremely interesting, and would upvote if not for your paragraph explaining when to upvote.)
Dec 6, 2016 at 20:10 comment added Noah Schweber Thinking more about it I've retracted my upvote, since I don't share the intuition that every finite reduced permutative LD-system should be a subalgebra of some $\mathcal{E}_\lambda/\equiv^\gamma$, which seems to be key here; can you explain why this is a reasonable intuition to have?
Dec 6, 2016 at 17:36 comment added David Roberts @JosephVanName I can't make any informed judgement about the technicalities of your argument, but this would be incredibly cool if it worked. (Note that I haven't voted because I can't say either way)
Dec 6, 2016 at 17:20 comment added Joseph Van Name Noah Schweber. I have edited the answer to define $\equiv^{\gamma}$ which is essentially equivalence of elementary embeddings up to $\gamma.$
Dec 6, 2016 at 17:19 history edited Joseph Van Name CC BY-SA 3.0
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S Dec 6, 2016 at 16:03 history answered Joseph Van Name CC BY-SA 3.0
S Dec 6, 2016 at 16:03 history made wiki Post Made Community Wiki by Joseph Van Name