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A worldly cardinal is a cardinal $\kappa$ such that $V_\kappa$ is a model of $\mathsf{ZFC}$. Please forgive me if this is very silly, but if $\mathsf{ZFC}$ is consistent (so there exists a model of $\mathsf{ZFC}$), must there exists a model of the form $V_\kappa$? Please note that I'm not asking about the existence of an inaccessible cardinal, which I know is not granted according to this question. Any help appreciated.

Jianing Song
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    No. There need not even be a transitive model of ZFC. (This is probably a duplicate, or close enough to one.) But even if there are transitive models of ZFC there won’t necessarily be a worldly cardinal: $V_\kappa$ for the least worldly $\kappa$ contains transitive models (certainly the minimal model) but no worldlies. – spaceisdarkgreen Oct 24 '23 at 21:36
  • @spaceisdarkgreen Thanks! – Jianing Song Oct 24 '23 at 22:07
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    In fact, the existence of a wordly cardinal is much stronger than the consistency of ZFC. The existence of a wordly cardinal implies not only Con(ZFC), but Con(ZFC+Con(ZFC)), or its iterates, and much more stronger principles like, "Every $\Pi^1_1$ theorem of $\mathsf{ZFC}$ is true." Also, what I stated follows from the existence of a transitive model of $\mathsf{ZFC}$, which is still strictly weaker than a wordly cardinal. – Hanul Jeon Oct 24 '23 at 22:13
  • (You can see that if $\kappa$ is a wordly cardinal, then for every $a\in V_\kappa$ there is a transitive model $M$ of $\mathsf{ZFC}$ such that $a\in M\in V_\kappa$.) – Hanul Jeon Oct 24 '23 at 22:13
  • I hadn’t looked at the post you linked till now: note that the exact same argument sketched there goes through without any modification but replacing “inaccessible” with “worldly”. (And it shows the stronger result I mentioned above that it’s independent of transitive models. And can be made even stronger by Hanul’s observation that you can easily extend to show every set is contained in a transitive model.) – spaceisdarkgreen Oct 24 '23 at 22:14
  • @HanulJeon Thanks for your explanation! – Jianing Song Oct 24 '23 at 22:41
  • @spaceisdarkgreen "$V_\kappa$ for the least worldly $\kappa$ contains ... no worldlies" Is this a sufficient argument? That particular $V_\kappa$ does not contain worldlies but by upward skolem there should be bigger models than $V_\kappa$ which could (should?) contain worldlies. – Damian Oct 25 '23 at 05:28
  • @Damian In order to show consistency of "there are no worldlies" with ZFC + (lots of) transitive models, it suffices to produce a model of ZFC with (lots of) transitive models and no worldlies. (On a side-note, the upward Lowenheim Skolem theorem produces models that might be structurally different but are elementarily equivalent, i.e. doesn't change the truth value of any first order statement, such as "there are no worldlies". So no, those particular models can't and shouldn't. But that's tangential to the main point.) – spaceisdarkgreen Oct 25 '23 at 06:45

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