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Recall that $\kappa$ is a worldly cardinal if $V_\kappa$ is a model of $\sf ZFC$. While every worldly cardinal is a strong limit cardinal, it is not necessarily regular. The point being that the short cofinal sequence is not first-order definable, so Replacement is not violated.

In particular, the first worldly cardinal has countable cofinality, and in fact the first worldly cardinal which is a limit of worldly cardinal has countable cofinality (as do the worldly cardinals below it).

Consider the following statement "For every $x$ there is a transitive model $M$ such that $x\in M$". Clearly this statement follows from "There is a proper class of worldly cardinals". Does it also imply it, or at least is it equiconsistent with it?

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  • $\begingroup$ If your transitive model $M$ is a transitive $ZFC$ model, then you may find your answer in the "Transitive $ZFC$ Model" entry of Cantor's Attic (it's in the first paragraph of that entry). Does the argument contained therein answer your question? $\endgroup$ – Thomas Benjamin Apr 28 '17 at 8:00
  • $\begingroup$ Yes. Transitive model means a model of ZFC. The Cantor's Attic entry does have a section about the "Transitive model universe axiom". Nothing about its exact consistency strength, though. $\endgroup$ – Asaf Karagila Apr 28 '17 at 8:41
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The answer is no, because I claim that if $\kappa$ is worldly, then $V_\kappa$ thinks that every set is a member of a transitive model of ZFC.

To see this, note first that every worldly cardinal $\kappa$ is a beth-fixed point $\beth_\kappa=\kappa$ and furthermore $V_\kappa=H_\kappa$, the set of sets whose transitive closures have size less than $\kappa$. Now consider any $x\in V_\kappa$. By the Löwenheim-Skolem theorem, we can find an elementary substructure $X\prec V_\kappa$ with $x\subseteq X$ and $x\in X$, with $|X|=|\text{TC}(x)|<\kappa$. The transitive collapse $M$ of $X$ will be a model of ZFC containing $x$ as an element. And even though $\kappa$ is singular, and so perhaps $X$ is unbounded in $V_\kappa$, nevertheless we will have $M\in V_\kappa$ since it is small enough. For example, $M$ will have fewer than $\kappa$ many ordinals, and so $M\subseteq V_\beta$ for some $\beta<\kappa$ and hence $M\in V_{\beta+1}\subset V_\kappa$.

Incidentally, this axiom, that every set is an element of a transitive model of ZFC, has sometimes been put forth as an alternative to the Grothendieck universe axiom, asserting that there are unboundedly many inaccessible cardinals, since it captures much of the power of that axiom in its applications, by providing a robust small universe concept for any given set, while having considerably weaker large cardinal strength. But to use this axiom, one needs to make the move from Grothendieck universes to transitive models of ZFC, which do not necessarily compute the power set correctly and whose height may be singular, although this is not visible internally to the model.

The axiom has also been proposed as a natural arena in which to investigate the corresponding multiverse theory, a version of hyperversism, but with uncoutnable models. For this reason, I like the axiom very much. In some recent upcoming joint work with Øysten Linnebo, we analyze the modal logic of this collection of models, viewed as a Kripke model.

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  • $\begingroup$ Yes. I was waiting for you, Joel. I figured that same answer (and I wanted to edit and delete my question, hence my question on meta.MO), but I realized that I have to edit the undeleted version, and for some reason I decided it'd be best to leave you the task of writing a nice answer instead. $\endgroup$ – Asaf Karagila Apr 28 '17 at 13:19
  • $\begingroup$ Ha! Actually, for a brief moment I had deleted my answer, thinking I had made a mistake, and I tried to edit it, but couldn't while it was deleted, but then realized I hadn't made a mistake anyway, so I undeleted it. $\endgroup$ – Joel David Hamkins Apr 28 '17 at 13:20
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    $\begingroup$ Hi Joel, I ran into this post now, and I was curious whether or not this axiom is equiconsistent with "Ord is worldly", or at least "If $V_\kappa$ satisfies the axiom, then $\kappa$ is worldly", or something like that. I imagine that it is a bit weaker, kind of like how "Ord is such and such" tend to be slightly weaker than "there is a such and such cardinal" in a significant sense. But maybe you have an answer off hand? $\endgroup$ – Asaf Karagila Oct 31 '19 at 17:32
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    $\begingroup$ Let me take "Ord is worldly" to be the theory of all $\sigma$ for which ZFC+"$\kappa$ is worldly" proves $V_\kappa\models\sigma$. Then one of those $\sigma$'s will be the assertion that every set is in a countable transitive model of ZFC. So the axiom is a consequence of Ord is worldly. But it is not equiconsistent, since $\text{Con}(\text{ZFC}+\sigma)$ is also such a sentence. $\endgroup$ – Joel David Hamkins Nov 1 '19 at 10:27
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    $\begingroup$ Ah. I see. Since "every set is an element of a transitive model of ZFC" is a first-order statement, and it is true in $V_\kappa$ for a worldly $\kappa$, it is also true in some element of $V_\kappa$ (take a countable elementary submodel and collapse it, for example). Therefore this is weaker than however we want to understand "Ord is worldly". $\endgroup$ – Asaf Karagila Nov 1 '19 at 10:44

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