I’d like to introduce and discuss the *otherworldly* cardinals, a large cardinal notion that frequently arises in set-theoretic analysis, but which until now doesn’t seem yet to have been given its own special name. So let us do so here.

I was put on to the topic by Jason Chen, a PhD student at UC Irvine working with Toby Meadows, who brought up the topic recently on Twitter:

In response, I had suggested the *otherworldly* terminology, a play on the fact that the two cardinals will both be worldly, and so we have in essence two closely related worlds, looking alike. We discussed the best way to implement the terminology and its extensions. The main idea is the following:

**Main Definition.** An ordinal $\kappa$ is *otherworldly* if $V_\kappa\prec V_\lambda$ for some ordinal $\lambda>\kappa$. In this case, we say that $\kappa$ is *otherworldly to* $\lambda$.

It is an interesting exercise to see that every otherworldly cardinal $\kappa$ is in fact also *worldly*, which means $V_\kappa\models\text{ZFC}$, and from this it follows that $\kappa$ is a strong limit cardinal and indeed a $\beth$-fixed point and even a $\beth$-hyperfixed point and more.

**Theorem.** Every otherworldly cardinal is also worldly.

**Proof.** Suppose that $\kappa$ is otherworldly, so that $V_\kappa\prec V_\lambda$ for some ordinal $\lambda>\kappa$. It follows that $\kappa$ must in fact be a cardinal, since otherwise it would be the order type of a relation on a set in $V_\kappa$, which would be isomorphic to an ordinal in $V_\lambda$ but not in $V_\kappa$. And since $\omega$ is not otherworldly, we see that $\kappa$ must be an uncountable cardinal. Since $V_\kappa$ is transitive, we get now easily that $V_\kappa$ satisfies extensionality, regularity, union, pairing, power set, separation and infinity. The only axiom remaining is replacement. If $\varphi(a,b)$ obeys a functional relation in $V_\kappa$ for all $a\in A$, where $A\in V_\kappa$, then $V_\lambda$ agrees with that, and also sees that the range is contained in $V_\kappa$, which is a set in $V_\lambda$. So $V_\kappa$ agrees that the range is a set. So $V_\kappa$ fulfills the replacement axiom. $\Box$

**Corollary.** A cardinal is otherworldly if and only if it is fully correct in a worldly cardinal.

**Proof.** Once you know that otherworldly cardinals are worldly, this amounts to a restatement of the definition. If $V_\kappa\prec V_\lambda$, then $\lambda$ is worldly, and $V_\kappa$ is correct in $V_\lambda$. $\Box$

Let me prove next that whenever you have an otherworldly cardinal, then you will also have a lot of worldly cardinals, not just these two.

**Theorem.** Every otherworldly cardinal $\kappa$ is a limit of worldly cardinals. What is more, every otherworldly cardinal is a limit of worldly cardinals having exactly the same first-order theory as $V_\kappa$, and indeed, the same $\alpha$-order theory for any particular $\alpha<\kappa$.

**Proof.** If $V_\kappa\prec V_\lambda$, then $V_\lambda$ can see that $\kappa$ is worldly and has the theory $T$ that it does. So $V_\lambda$ thinks, about $T$, that there is a cardinal whose rank initial segment has theory $T$. Thus, $V_\kappa$ also thinks this. And we can find arbitrarily large $\delta$ up to $\kappa$ such that $V_\delta$ has this same theory. This argument works whether one uses the first-order theory, or the second-order theory or indeed the $\alpha$-order theory for any $\alpha<\kappa$. $\Box$

**Theorem.** If $\kappa$ is otherworldly, then for every ordinal $\alpha<\kappa$ and natural number $n$, there is a cardinal $\delta<\kappa$ with $V_\delta\prec_{\Sigma_n}V_\kappa$ and the $\alpha$-order theory of $V_\delta$ is the same as $V_\kappa$.

**Proof.** One can do the same as above, since $V_\lambda$ can see that $V_\kappa$ has the $\alpha$-order theory that it does, while also agreeing on $\Sigma_n$ truth with $V_\lambda$, so $V_\kappa$ will agree that there should be such a cardinal $\delta<\kappa$. $\Box$

**Definition.** We say that a cardinal is *totally otherworldly*, if it is otherworldly to arbitrarily large ordinals. It is *otherworldly beyond $\theta$*, if it is otherworldly to some ordinal larger than $\theta$. It is *otherworldly up to $\delta$*, if it is otherworldly to ordinals cofinal in $\delta$.

**Theorem.** Every inaccessible cardinal $\delta$ is a limit of otherworldly cardinals that are each otherworldly up to and to $\delta$.

**Proof.** If $\delta$ is inaccessible, then a simple Löwenheim-Skolem construction shows that $V_\kappa$ is the union of a continuous elementary chain $$V_{\kappa_0}\prec V_{\kappa_1}\prec\cdots\prec V_{\kappa_\alpha}\prec \cdots \prec V_\kappa$$ Each of the cardinals $\kappa_\alpha$ arising on this chain is otherworldly up to and to $\delta$. $\Box$

**Theorem.** Every totally otherworldly cardinal is $\Sigma_2$ correct, meaning $V_\kappa\prec_{\Sigma_2} V$. Consequently, every totally otherworldly cardinal is larger than the least measurable cardinal, if it exists, and larger than the least superstrong cardinal, if it exists, and larger than the least huge cardinal, if it exists.

**Proof.** Every $\Sigma_2$ assertion is locally verifiable in the $V_\alpha$ hierarchy, in that it is equivalent to an assertion of the form $\exists\eta V_\eta\models\psi$ (for more information, see my post about Local properties in set theory). Thus, every true $\Sigma_2$ assertion is revealed inside any sufficiently large $V_\lambda$, and so if $V_\kappa\prec V_\lambda$ for arbitrarily large $\lambda$, then $V_\kappa$ will agree on those truths. $\Box$

I was a little confused at first about how two totally otherwordly cardinals interact, but now everything is clear with this next result. (Thanks to Hanul Jeon for his helpful comment below.)

**Theorem.** If $\kappa<\delta$ are both totally otherworldly, then $\kappa$ is otherworldly up to $\delta$, and hence totally otherworldly in $V_\delta$.

**Proof.** Since $\delta$ is totally otherworldly, it is $\Sigma_2$ correct. Since for every $\alpha<\delta$ the cardinal $\kappa$ is otherworldly beyond $\alpha$, meaning $V_\kappa\prec V_\lambda$ for some $\lambda>\alpha$, then since this is a $\Sigma_2$ feature of $\kappa$, it must already be true inside $V_\delta$. So such a $\lambda$ can be found below $\delta$, and so $\kappa$ is otherworldly up to $\delta$. $\Box$

**Theorem.** If $\kappa$ is totally otherworldly, then $\kappa$ is a limit of otherworldly cardinals, and indeed, a limit of otherworldly cardinals having the same theory as $V_\kappa$.

**Proof.** Assume $\kappa$ is totally otherworldly, let $T$ be the theory of $V_\kappa$, and consider any $\alpha<\kappa$. Since there is an otherworldly cardinal above $\alpha$ with theory $T$, namely $\kappa$, and because this is a $\Sigma_2$ fact about $\alpha$ and $T$, it follows that there must be such a cardinal above $\alpha$ inside $V_\kappa$. So $\kappa$ is a limit of otherworldly cardinals with the same theory as $V_\kappa$. $\Box$

The results above show that the consistency strength of the hypotheses are ordered as follows, with strict increases in consistency strength as you go up (assuming consistency):

- ZFC + there is an inaccessible cardinal
- ZFC + there is a proper class of totally otherworldly cardinals
- ZFC + there is a totally otherworldly cardinal
- ZFC + there is a proper class of otherworldly cardinals
- ZFC + there is an otherworldly cardinal
- ZFC + there is a proper class of worldly cardinals
- ZFC + there is a worldly cardinal
- ZFC + there is a transitive model of ZFC
- ZFC + Con(ZFC)
- ZFC

We might consider the natural strengthenings of otherworldliness, where one wants $V_\kappa\prec V_\lambda$ where $\lambda$ is itself otherworldly. That is, $\kappa$ is the beginning of an elementary chain of three models, not just two. This is different from having merely that $V_\kappa\prec V_\lambda$ and $V_\kappa\prec V_\eta$ for some $\eta>\lambda$, because perhaps $V_\lambda$ is not elementary in $V_\eta$, even though $V_\kappa$ is. Extending successively is a more demanding requirement.

One then naturally wants longer and longer chains, and ultimately we find ourselves considering various notions of rank in the *rank elementary forest*, which is the relation $\kappa\preceq\lambda\iff V_\kappa\prec V_\lambda$. The otherworldly cardinals are simply the non-maximal nodes in this order, while it will be interesting to consider the nodes that can be extended to longer elementary chains.

I think the existence of a $\Sigma_2$-correct cardinal above a totally otherworldly cardinal implies the consistency of the existence of a totally otherworldly.

Let $\kappa$ be a totally otherworldly cardinal and $\lambda>\kappa$ be a $\Sigma_2$-correct cardinal. Take $\alpha<\lambda$ and $\beta>\alpha$ such that $V_\kappa\prec V_\beta$.

Since $V_{\beta+1}\models (\beta>\alpha\land V_\kappa\prec V_\beta)$, we have $V_{\beta+1}\models \exists\xi (\xi>\alpha\land V_\kappa\prec V_\xi)$.

By $\Sigma_2$-correctness of $\lambda$, there is $\gamma<\lambda$ with $\gamma>\alpha\land V_\kappa\prec V_\gamma)$. Hence $V_\lambda\models (\exists\xi (\xi>\alpha\land V_\kappa\prec V_\xi)$. Since $\alpha$ is arbitrary, $V_\lambda$ thinks $\kappa$ is totally otherworldly.

Since every totally otherworldly cardinal is $\Sigma_2$-correct, the existence of two totally otherworldly cardinals implies the consistency of the existence of a totally otherworldly cardinal.

Your comment was garbled a little, and I tried to edit it, but now I have become confused about your argument; perhaps I have made a mistake with editing—I apologize. We don’t know that $V_\lambda$ is worldly, so how does your conclusion work?

Since ZFC proves that the $\Sigma_2$-correct cardinals are unbounded, we cannot expect to prove in ZFC that if there is an totally otherworldly cardinal with a $\Sigma_2$-correct cardinal above, then Con(totally otherworldly), since this would violate the incompleteness theorem.

Oh, I see now. The Sigma_2 correct cardinal you intend to use is the other totally otherworldly cardinal. This seems to work perfectly and it resolves my issue. Thanks! I’ll update the post tomorrow.

Sorry for checking your reply lately. My initial statement is somewhat misleading: it should be as “if $\kappa$ is a totally otherworldly and $\lambda$ is a $\Sigma_2$-correct cardinal larger than $\kappa$, then $V_\lambda$ sees $\kappa$ is totally otherworldly.” As you pointed out, this statement itself implies no results about consistency strength.

If $\lambda$ is also worldly, then $V_\lambda$ is a model of ZFC with a totally otherworldly cardinal. You proved that a totally otherworldly cardinal is $\Sigma_2$-correct worldly cardinal, so we have the desired result.

Yes, that is how I understood your argument in the end, and this is how I have presented it in the revised blog post above (with thanks to you).

I guess the name “elementary cardinal” was taken…

Two questions/observations:

1. Are otherworldly cardinals downward absolute (we know that worldlies, in general, are not)?

2. We always knew that strength and size didn’t go hand in hand (e.g. strong vs. superstrong cardinals). But this is really crazy how big the least totally otherworldly is in the presence of other large cardinals. I guess then we then have “there exists a measurable/huge/whatever and a totally otherworldly” is stronger than “there exists a measurable/huge/whatever and an inaccessible” (since in the latter, the inaccessible is redundant).

For question 1, I guess you mean downward absolute to transitive inner models? If so, the answer is negative, and I think it is a little easier even than in the worldly case. Suppose $\kappa$ is otherworldly to $\lambda$. This is preserved by the forcing of the GCH, so we may assume GCH. Now force violations to the GCH with Easton forcing at every successor. This also preserves otherworldliness. But if we stop this latter forcing at $\kappa$, then this would be an inner model where $\kappa$ is no longer otherworldly, since the GCH holds above $\kappa$ but fails unboundedly often below.

(I made a few edits to this comment.)

For question 2, I agree.