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Elliot Glazer's user avatar
Elliot Glazer's user avatar
Elliot Glazer
  • Member for 7 years, 7 months
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Is Global Choice conservative over Zermelo with Choice?
Yes, first build a well-founded model of Zermelo with Urelements, where the $a$'s and $b$'s are the urelements, and then quotient out the intended relationship in the resulting model
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Does there always exist a categorical extension of $ZFC_2$ with no set models?
$\kappa$ has the property $`` \exists \alpha < \kappa$ such that $V_{\alpha} \equiv_{\text{SOL}} V_{\kappa}"$ because $M$ thinks $j(\kappa)$ has that property.
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Is Global Choice conservative over Zermelo with Choice?
You’re right about the typo. I’m not assuming ill-foundedness in $V,$ I’m defining $b_p$ in terms of how they’re intended to be interpreted in $M.$ A formal definition would be by a quotient structure, same as how Quine atoms are included in the model.
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Does there always exist a categorical extension of $ZFC_2$ with no set models?
Yes measurables work, because for $j: V \rightarrow M$ with critical point $\kappa,$ $V_{\kappa+1}^M$ and $V_{j(\kappa)+1}^M$ have the same theory.
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Does there always exist a categorical extension of $ZFC_2$ with no set models?
If there’s $\mathfrak{c}^+$ inaccessibles, some $V_{\kappa}$ has the same theory as a smaller one. Now treat that as your $V.$
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How much choice is needed to prove this statement?
This is very similar to your last question, so I'm not going to write out a full proof, but this is also equivalent to "there is a function choosing an enumeration of each countable ordinal." If you have a countable closed set of rank $\alpha,$ you can canonically enumerate $\alpha$ by sending a basic open set $U$ to $\beta$ if there is a unique point of maximal rank in $U,$ and that rank is $\beta.$
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How much choice is necessary to prove this statement?
By the way, you might be interested to know that existence of an $\omega_1$-sequence of reals is equivalent to the weaker assertion that that there is a choice of injections $(\varphi_{\alpha})_{\alpha<\omega_1}: \alpha \rightarrow \mathbb{R}.$
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Can a Vopenka cardinal be supercompact?
If $\kappa$ is almost huge with target $\lambda,$ then $V_{\lambda}$ thinks that $\kappa$ is a supercompact Vopenka cardinal. See neugierde.github.io/cantors-attic/Huge
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