Let $k$ be the smallest ordinal such that $V_k$ is a model of ZFC. I know that $k$ need not be inaccessible cardinal, and $k$ has cofinality $\omega$.

Then how big is $k$? How can we write down $k$ in terms of $\aleph$ or $\beth$? Since its confinality is $\omega$, how can we find an $\omega$-sequence to reach $k$?


Indeed the least such $\kappa$, if it exists, has a countable cofinality. However $\kappa$ is a $\beth$-fixed point. This means that $\kappa=\beth_\kappa$. So in particular it is a bit hard to write a cofinal sequence in an explicit form.

And note that every $\beth$-fixed point is an $\aleph$-fixed point: $\beth_\alpha\geq\aleph_\alpha\geq\alpha$ is provable in $\sf ZFC$. If $\alpha=\beth_\alpha$ then we have equality all across the board.

All we can do is prove that this cardinal has a countable cofinality, and therefore such sequence exists.


These cardinals are now called wordly. As Asaf pointed out, any such $\kappa$ is a beth fixed point. Note that $V_\kappa$ cannot see any $\omega$-sequence cofinal in $\kappa$, since $\mathsf{ZFC}$ proves that no $\omega$-sequence of ordinals is cofinal in the class $\mathsf{ORD}$ of all ordinals. This means that any witnessing sequence cannot be "easily" described, or else $V_\kappa$ would be able to identify it.

The following argument may help you understand why the cofinality is $\omega$: Let $\mathsf{ZFC}_n$ be the subtheory of $\mathsf{ZFC}$ resulting from restricting the axiom schema of replacement to $\Sigma_n$ formulas. Let $C_n$ be the class of cardinals $\kappa$ such that $V_\kappa$ models $\mathsf{ZFC}_n$. By the reflection theorem, each $C_n$ is cofinal in $\mathsf{ORD}$. In fact, since $\Sigma_n$ satisfaction is definable, each $C_n$ contains a club: Consider those $\kappa$ such that $V_\kappa$ is a $\Sigma_n$-elementary substructure of the universe $V$.

Suppose $\kappa$ is wordly. [Note that, since satisfaction is not definable -- this is Tarski's theorem, -- although we know, from outside, that each $C_n^{V_\kappa}$ contains a club in $\kappa$, in $V_\kappa$ we do not have access to the sequence $(C_n^{V_\kappa}\mid n<\omega)$.] If $\kappa$ has cofinality larger than $\omega$, then $\bigcap_n C_n^{V_\kappa}$ again contains a club in $\kappa$. If $\rho$ is any element of this intersection, then $\rho$ is again wordly, since it satisfies $\Sigma_n$ replacement for all $n$. In fact, if we take $\rho$ in the intersection of the clubs contained in the $C_n^{V_\kappa}$, then we have the stronger conclusion that $V_\rho\prec V_\kappa$.

It follows that no such $\kappa$ can be the first wordly cardinal, and therefore the smallest one has cofinality $\omega$ (and this is witnessed by the fact that there is an $\omega$-sequence of clubs with empty intersection). Easy modifications of the argument show that actually a very long initial segment of the sequence of wordly cardinals consists solely of cardinals of cofinality $\omega$. The first wordly cardinal of uncountable cofinality has cofinality $\omega_1$, the next one again has cofinality $\omega$, and again all wordly cardinals past this one have cofinality $\omega$ for a long stretch, then we see again one of cofinality $\omega_1$, etc.

  • $\begingroup$ @AlessandroIraci asks (in an answer, since he cannot comment yet): "I'm interested in this topic. Is there any reference for the proof which Andrés Caicedo sketched above? Thank you!" $\endgroup$ – user228113 Dec 16 '15 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.