Tag Archives: worldly cardinals

Every worldly cardinal admits a Gödel-Bernays structure

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My Oxford student Emma Palmer and I have been thinking about worldly cardinals and Gödel-Bernays GBC set theory, and we recently came to a new realization.

Namely, what I realized is that every worldly cardinal $\kappa$ admits a Gödel-Bernays structure, including the axiom of global choice. That is, if $\kappa$ is worldly, then there is a family $X$ of sets so that $\langle V_\kappa,\in,X\rangle$ is a model of Gödel-Bernays set theory GBC including global choice.

For background, it may be helpful to recall Zermelo’s famous 1930 quasi-categoricity result, showing that the inaccessible cardinals are precisely the cardinals $\kappa$ for which $V_\kappa$ is a model of second-order set theory $\text{ZFC}_2$.

If one seeks only the first-order ZFC set theory in $V_\kappa$, however, then this is what it means to say that $\kappa$ is a worldly cardinal, a strictly weaker notion. That is, $\kappa$ is worldly if and only if $V_\kappa\models\text{ZFC}$. Every inaccessible cardinal is worldly, by Zermelo’s result. But more, every inaccessible is a limit of worldly cardinals, and so there are many worldly cardinals that are not inaccessible. The least worldly cardinal, for example, has cofinality $\omega$. Indeed, the next worldly cardinal above any ordinal has cofinality $\omega$.

Meanwhile, to improve slightly on Zermelo, we can observe that if $\kappa$ is inaccessible, then $V_\kappa$ is a model of Kelley-Morse set theory when equipped with the full second-order complement of classes. That is, $\langle V_\kappa,\in,V_{\kappa+1}\rangle$ is a model of KM.

This is definitely not true when $\kappa$ is merely worldly and not inaccessible, however, for in this case $\langle V_\kappa,\in,V_{\kappa+1}\rangle$ is never a model of KM nor even GBC when $\kappa$ is singular. The reason is that the singularity of $\kappa$ would be revealed by a short cofinal sequence, which would be available in the full power set $V_{\kappa)+1}=P(V_\kappa)$, and this would violate replacement.

So the question is:

Question. If $\kappa$ is worldly, then can we equip $V_\kappa$ with a suitable family $X$ of classes so that $\langle V_\kappa,\in,X\rangle$ is a model of GBC?

The answer is Yes!

What I claim is that for every worldly cardinal $\kappa$, there is a definably generic well order $\newcommand\slantleq{\mathrel{⩽}}\slantleq$ of $V_\kappa$, so that the subsets definable in $\langle V_\kappa,\in,\slantleq\rangle$ make a model of GBC.

To see this, consider the class forcing notion $\newcommand\P{\mathbb{P}}\P$ for adding a global well order $\slantleq$, as $V_\kappa$ sees it. Conditions are well orders of some $V_\alpha$ for some $\alpha<\kappa$, ordered by end-extension, so that lower rank sets always preceed higher rank sets in the resulting order.

I shall prove that there is a well-order $\slantleq$ that is generic with respect to dense sets definable in $\langle V,\in\rangle$.

For this, let us consider first the case where the worldly cardinal $\kappa$ has countable cofinality. In this case, we can find an increasing sequence $\kappa_n$ cofinal in $\kappa$, such that each $\kappa_n$ is $\Sigma_n$-correct in $V_\kappa$, meaning $V_{\kappa_n}\prec_{\Sigma_n}V_\kappa$.

In this case, we can build a definably generic filter $G$ for $\P$ in a sequence of stages. At stage $n$, we can find a well order up to $\kappa_n$ that meets all $\Sigma_n$ definable dense classes using parameters less than $V_{\kappa_n}$. The reason is that for any such definable dense set, we can meet it below $\kappa_n$ using the $\Sigma_n$-correctness of $\kappa_n$, and so by considering various parameters in turn, we can altogether handle all parameters below $V_{\kappa_n}$ using $\Sigma_n$ definitions. That is, the $n$th stage is itself an iteration of length $\kappa_n$, but it will meet all $\Sigma_n$ definable dense sets using parameters in $V_{\kappa_n}$.

Next, we observe that the ultimate well-order of $V_\kappa$ that arises from this construction after all stages is fully definably generic, since any definition with arbitrary parameters in $V_\kappa$ is a $\Sigma_n$ definition with parameters in $V_{\kappa_n}$ for some large enough $n$, and so we get a definably generic well order $\slantleq$. Therefore, the usual forcing argument shows that we get GBC in the resulting model $\langle V_\kappa,\in,\text{Def}(V_\kappa)\rangle$, as desired.

The remaining case occurs when kappa has uncountable cofinality. In this case, there is a club set $C\subseteq\kappa$ of ordinals $\gamma\in C$ with $V_\gamma\prec V_\kappa$. (We can just intersect the clubs $C_n$ of the $\Sigma_n$-correct cardinals.) Now, we build a well-order of $V_\kappa$ that is definably generic for every $V_\gamma$ for $\gamma\in C$. At limits, this is free, since every definable dense set in V_lambda with parameters below is also definable in some earlier $V_\gamma$. So it just reduces to the successor case, which we can get by the arguments above (or by induction). The next correct cardinal $\gamma$ above any ordinal has countable cofinality, since if one considers the next $\Sigma_1$-correct cardinal, the next $\Sigma_2$-correct cardinal, and so on, the limit will be fully correct and cofinality $\omega$.

The conclusion is that every worldly cardinal $\kappa$ admits a definably generic global well-order on $V_\kappa$ and therefore also admits a Gödel-Bernays GBC set theory structure $\langle V_\kappa,\in,X\rangle$, including the axiom of global choice.

The argument relativizes to any particular amenable class $A\subseteq V_\kappa$. Namely, if $\langle V_\kappa,\in,A\rangle$ is a model of $\text{ZFC}(A)$, then there is a definably generic well order $\slantleq$ of $V_\kappa$ such that $\langle V_\kappa,\in,A,\slantleq\rangle$ is a model of $\text{ZFC}(A,\slantleq)$, and so by taking the classes definable from $A$ and $\slantleq$, we get a GBC structure $X$ including both $A$ and $\slantleq$.

This latter observation will be put to good use in connection with Emma’s work on the Tarski’s revenge axiom, in regard to finding the optimal consistency strength for one of the principles.

The otherwordly cardinals

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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.

Worldly cardinals are not always downwards absolute

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UniversumI recently came to realize that worldly cardinals are not necessarily downward absolute to transitive inner models. That is, it can happen that a cardinal $\kappa$ is worldly in the full set-theoretic universe $V$, but not in some transitive inner model $W$, even when $W$ is itself a model of ZFC. The observation came out of some conversations I had with Alexander Block from Hamburg during his recent research visit to New York. Let me explain the argument.

A cardinal $\kappa$ is inaccessible, if it is an uncountable regular strong limit cardinal. The structure $V_\kappa$, consisting of the rank-initial segment of the set-theoretic universe up to $\kappa$, which can be generated from the empty set by applying the power set operation $\kappa$ many times, has many nice features. In particular, it is transitive model of $\newcommand\ZFC{\text{ZFC}}\ZFC$. The models $V_\kappa$ for $\kappa$ inaccessible are precisely the uncountable Grothendieck universes used in category theory.

Although the inaccessible cardinals are often viewed as the entryway to the large cardinal hierarchy, there is a useful large cardinal concept weaker than inaccessibility. Namely, a cardinal $\kappa$ is worldly, if $V_\kappa$ is a model of $\ZFC$. Every inaccessible cardinal is worldly, and in fact a limit of worldly cardinals, because if $\kappa$ is inaccessible, then there is an elementary chain of cardinals $\lambda<\kappa$ with $V_\lambda\prec V_\kappa$, and all such $\lambda$ are worldly. The regular worldly cardinals are precisely the inaccessible cardinals, but the least worldly cardinal is always singular of cofinality $\omega$.

The worldly cardinals can be seen as a kind of poor-man’s inaccessible cardinal, in that worldliness often suffices in place of inaccessibility in many arguments, and this sometimes allows one to weaken a large cardinal hypothesis. But meanwhile, they do have some significant strengths. For example, if $\kappa$ is worldly, then $V_\kappa$ satisfies the principle that every set is an element of a transitive model of $\ZFC$.

It is easy to see that inaccessibility is downward absolute, in the sense that if $\kappa$ is inaccessible in the full set-theoretic universe $V$ and $W\newcommand\of{\subseteq}\of V$ is a transitive inner model of $\ZFC$, then $\kappa$ is also inaccessible in $W$. The reason is that $\kappa$ cannot be singular in $W$, since any short cofinal sequence in $W$ would still exist in $V$; and it cannot fail to be a strong limit there, since if some $\delta<\kappa$ had $\kappa$-many distinct subsets in $W$, then this injection would still exist in $V$. So inaccessibility is downward absolute.

The various degrees of hyper-inaccessibility are also downwards absolute to inner models, so that if $\kappa$ is an inaccessible limit of inaccessible limits of inaccessible cardinals, for example, then this is also true in any inner model. This downward absoluteness extends all the way through the hyperinaccessibility hierarchy and up to the Mahlo cardinals and beyond. A cardinal $\kappa$ is Mahlo, if it is a strong limit and the regular cardinals below $\kappa$ form a stationary set. We have observed that being regular is downward absolute, and it is easy to see that every stationary set $S$ is stationary in every inner model, since otherwise there would be a club set $C$ disjoint from $S$ in the inner model, and this club would still be a club in $V$. Similarly, the various levels of hyper-Mahloness are also downward absolute.

So these smallish large cardinals are generally downward absolute. How about the worldly cardinals? Well, we can prove first off that worldliness is downward absolute to the constructible universe $L$.

Observation. If $\kappa$ is worldly, then it is worldly in $L$.

Proof. If $\kappa$ is worldly, then $V_\kappa\models\ZFC$. This implies that $\kappa$ is a beth-fixed point. The $L$ of $V_\kappa$, which is a model of $\ZFC$, is precisely $L_\kappa$, which is also the $V_\kappa$ of $L$, since $\kappa$ must also be a beth-fixed point in $L$. So $\kappa$ is worldly in $L$. QED

But meanwhile, in the general case, worldliness is not downward absolute.

Theorem. Worldliness is not necessarily downward absolute to all inner models. It is relatively consistent with $\ZFC$ that there is a worldly cardinal $\kappa$ and an inner model $W\of V$, such that $\kappa$ is not worldly in $W$.

Proof. Suppose that $\kappa$ is a singular worldly cardinal in $V$. And by forcing if necessary, let us assume the GCH holds in $V$. Let $V[G]$ be the forcing extension where we perform the Easton product forcing $\newcommand\P{\mathbb{P}}\P$, so as to force a violation of the GCH at every regular cardinal $\gamma$. So the stage $\gamma$ forcing is $\newcommand\Q{\mathbb{Q}}\Q_\gamma=\text{Add}(\gamma,\gamma^{++})$.

First, I shall prove that $\kappa$ is worldly in the forcing extension $V[G]$. Since every set of rank less than $\kappa$ is added by some stage less than $\kappa$, it follows that $V_\kappa^{V[G]}$ is precisely $\bigcup_{\gamma<\kappa} V_\kappa[G_\gamma]$. Most of the $\ZFC$ axioms hold easily in $V_\kappa^{V[G]}$; the only difficult case is the collection axiom. And for this, by considering the ranks of witnesses, it suffices to show for every $\gamma<\kappa$ that every function $f:\gamma\to\kappa$ that is definable from parameters in $V_\kappa^{V[G]}$ is bounded. Suppose we have such a function, defined by $f(\alpha)=\beta$ just in case $\varphi(\alpha,\beta,p)$ holds in $V_\kappa^{V[G]}$. Let $\delta<\kappa$ be larger than the rank of $p$. Now consider $V_\kappa[G_\delta]$, which is a set-forcing extension of $V_\kappa$ and therefore a model of $\ZFC$. The fail forcing, from stage $\delta$ up to $\kappa$, is homogeneous in this model. And therefore we know that $f(\alpha)=\beta$ just in case $1$ forces $\varphi(\check\alpha,\check\beta,\check p)$, since these arguments are all in the ground model $V_\kappa[G_\delta]$. So the function is already definable in $V_\kappa[G_\delta]$. Because this is a model of $\ZFC$, the function $f$ is bounded below $\kappa$. So we get the collection axiom in $V_\kappa^{V[G]}$ and hence all of $\ZFC$ there, and so $\kappa$ is worldly in $V[G]$.

For any $A\of\kappa$, let $\P_A$ be the restriction of the Easton product forcing to include only the stages in $A$, and let $G_A$ be the corresponding generic filter. The full forcing $\P$ factors as $\P_A\times\P_{\kappa\setminus A}$, and so $V[G_A]\of V[G]$ is a transitive inner model of $\ZFC$.

But if we pick $A\of\kappa$ to be a short cofinal set in $\kappa$, which is possible because $\kappa$ is singular, then $\kappa$ will not be worldly in the inner model $V[G_A]$, since in $V_\kappa[G_A]$ we will be able to identify that sequence as the places where the GCH fails. So $\kappa$ is not worldly in $V[G_A]$.

In summary, $\kappa$ was worldly in $V[G]$, but not in the transitive inner model $W=V[G_A]$, and so worldliness is not downward absolute. QED