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I wonder what is the relationship between ZF + $V = L$ and Tarski–Grothendieck set theory, because I haven't found any bibliographic references. If they are compatible, it is possible to introduce $V = L$ using Tarski–Grothendieck as basis instead of ZF.

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    $\begingroup$ Someone will correct me if I'm wrong, but I believe that if $V$ is a model of $\mathsf{TG}$ then $L^V$ is a model of $\mathsf{TG} + V=L$, so they're equiconsistent. $\endgroup$ – James Hanson Oct 29 '20 at 3:33
  • $\begingroup$ I think that "there are a proper class of inaccessible cardinals" implies TG. So it is consistent with constructibility. $\endgroup$ – 喻 良 Oct 29 '20 at 5:08
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    $\begingroup$ Tarski-Grothendieck doesn't get talked about much in set theory since it's equivalent to $\mathsf{ZFC}$ + "There is a proper class of (strong) inaccessibles." In general, while the language of universes is quite natural for (say) category theory, on the set theory side the language of large cardinals is more convenient. $\endgroup$ – Noah Schweber Oct 29 '20 at 5:55
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(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

  1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

  2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trevor Wilson's previous answer for the detailed proof.)

Since being inaccessible is downward absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. (Thank you for Noah Schweber to point it out. Being inaccessible need not be upward absolute.) It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

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    $\begingroup$ +1. For the OP re: absoluteness: inaccessibility is more generally downwards absolute since it is a $\Pi_1$ property. All the "small" large cardinal properties (inaccessibility, Mahlo, weakly compact, ...) are similar. Things change drastically once we get to measurable cardinals: while to the best of our knowledge ZFC + "There is a measurable cardinal" is consistent, ZFC + V=L proves that there is no measurable cardinal. So measurability is not downwards-absolute: even if kappa is measurable in V, kappa won't be measurable in L. $\endgroup$ – Noah Schweber Oct 29 '20 at 5:46
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    $\begingroup$ Again for the OP: to see that inaccessibility is not upwards absolute, consider forcings of the form $Col(\omega,\kappa)$ (say). $\endgroup$ – Noah Schweber Oct 29 '20 at 5:56
  • $\begingroup$ @Noah Thank you for your comment. I fixed that part of my answer. $\endgroup$ – Hanul Jeon Oct 29 '20 at 5:57

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