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Dec 22, 2020 at 23:37 history edited Noah Schweber
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Dec 18, 2020 at 3:09 history edited Noah Schweber CC BY-SA 4.0
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Dec 15, 2020 at 20:56 vote accept Noah Schweber
Dec 15, 2020 at 17:51 history edited Noah Schweber CC BY-SA 4.0
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Dec 15, 2020 at 13:14 comment added Asaf Karagila @Noah: Maybe it's time you start using real large cardinals like super-Reinhardt cardinals.
Dec 15, 2020 at 0:40 answer added Gabe Goldberg timeline score: 17
Dec 14, 2020 at 22:16 comment added Noah Schweber That said, there might be some clever way of "projecting" $I0$ down somehow, I don't know - this is far-out-there for me too. (The strongest large cardinal axiom I've ever used is Vopenka's principle, and even then only recently.)
Dec 14, 2020 at 22:16 comment added Noah Schweber @WillBrian $L(V_{\kappa+1})$ is enough to express that, since all we have to do is "bundle up" information already present in $V_{\kappa+1}$. However, I think $I0$ doesn't work here: if memory serves, the $\lambda$ for an $I0$ embedding has to have cofinality $\omega$. So we have to be in a situation where the e.e. of $Ord$ (or $\kappa$ rather) really doesn't extend to an e.e. for $V$ (or $V_\kappa$ rather). Which, per Kunen, isn't surprising.
Dec 14, 2020 at 21:38 comment added Will Brian Instead of $V_{\kappa+2}$, can $L(V_{\kappa+1})$ express "is a second order elementary embedding"? If so, does that mean that the $I0$ axiom gives you what you want? (This could well be a useless idea for easy reasons -- I'm in pretty unfamiliar territory here.)
Dec 14, 2020 at 21:18 comment added Asaf Karagila Always a pleasure to see someone giving a good answer to a nice question on MSE, and then making a nice question on MO as a result. That's quite the ideal way that the system should work, in my opinion.
Dec 14, 2020 at 21:11 comment added Noah Schweber @AlecRhea Really though the big obstacle (per my second comment) is that while we are allowed to use second-order formulas to define structure on $Ord$, we're not allowed to use second-order constructions from $Ord$. This is really the same obstacle that's preventing me from using Kunen here.
Dec 14, 2020 at 20:53 comment added Alec Rhea Ah, thank you for explaining. If we repeat the construction anywhere above $\omega$ only Cauchy completion works to produce a field, and the field will be non-Archimidean and thusly have many nontrivial automorphisms. It looks like from your second comment that we would probably have even more trouble coding these non-Archimedean fields into the ordinals, so this approach is likely fruitless. Interesting question!
Dec 14, 2020 at 20:53 comment added Noah Schweber @AlecRhea and actually it doesn't even do that: building the reals in that way does not amount to adding structure to $Ord$ directly. Rather, it gives an interpretation of $\mathbb{R}$ in $V_{\omega+1}$, and we still would need to find a way to code that directly into the ordinals - and it's not obvious how to do that.
Dec 14, 2020 at 20:49 comment added Noah Schweber @AlecRhea No, that would only block an elementary embedding from moving "small" ordinals. Nothing about that prevents for example an elementary embedding which fixes all the ordinals below the first inaccessible, for example.
Dec 14, 2020 at 20:44 comment added Alec Rhea You can build $\mathbb{R}$ as the Dedekind (or Cauchy) completion of the field of fractions of the Grothendieck ring of $\omega$; does that count as a reasonably simple expansion? (the final step is second order) I believe that since the only automorphism of $\mathbb{R}$ is the identity, this would mean that there are no nontrivial second order elementary embeddings since they would preserve the final step in that construction and thusly yield a nontrivial automorphism of $\mathbb{R}$, correct?
Dec 14, 2020 at 20:31 history asked Noah Schweber CC BY-SA 4.0