Mathematics Asked by Jesse Elliott on January 17, 2021
A cardinal $kappa$ is said to be worldly if $V_kappa$ is a model of ZFC. Let us (potentially) generalize this by saying an ordinal $alpha$ is worldly if $V_alpha$ is a model of ZFC. The existence of a worldly ordinal implies the existence of a transitive model of ZFC, hence a countable ordinal $alpha$ such that $L_alpha$ is a model of ZFC. But I’m not sure in that case if there must then be a worldly cardinal.
My questions are as follows.
Hopefully it’s my last set theory question for a while. 🙂
If $V_alpha$ is a model of ZFC, then $alpha$ must be a cardinal, and much more. In fact it must be a strong limit cardinal, a $beth$-fixed point, a fixed point in the enumeration of $beth$-fixed points, and have any other strong limit property of this sort.
To see this, observe that ZFC proves that the $beth$-hierarchy is unbounded, but also we can show that the $beth$ hierarchy (and the Von-Neumann hierarchy) is absolute for a "full" model of the form $V_alpha,$ since the model's power set operator is the same as the real one. So it follows that $alpha$ is a strong limit. And the stronger properties follow from similar considerations.
In a little more detail, if $beta <alpha,$ then $P(beta) in V_alpha$ since $P(beta)$ is just two ranks higher than $beta,$ and so since being a subset is absolute, $P(beta)^{V_alpha}=P(beta).$ ZFC proves $2^{|beta|}$ exists, which relativized to $V_alpha$ is the least ordinal in $V_alpha$ that has a bijection in $V_alpha$ with $P(beta).$ And being a bijection is absolute so this is a real bijection, thus there is an ordinal in $V_alpha$ that is in one-to-one correspondence with $P(beta),$ so $2^{|beta|}<alpha.$
Correct answer by spaceisdarkgreen on January 17, 2021
Yes, every worldly ordinal is a cardinal.
Suppose $alpha$ were a worldly ordinal that is not a cardinal, so that $kappa := |alpha| < alpha$. In particular, since $kappa$ is a set of rank $kappa < alpha$, we have $kappa in V_alpha$.
Let $f in V$ be a bijection from $alpha$ to $kappa$. By pushing forward the $in$ well-ordering on $alpha$, we get a relation $R$ on $kappa$ which is a well-ordering of type $alpha$. Since $R$ is a subset of $kappa times kappa$, it has rank $kappa+3$ or something like that. Now since $kappa < alpha$, and $alpha$, being worldly, is not a successor, we also have $kappa+3 < alpha$. So $R in V_alpha$. But ZFC proves there exists an ordinal isomorphic to $(kappa, R)$, so this ordinal must exist in $V_alpha$, and it must be $alpha$. This is a contradiction.
Answered by Nate Eldredge on January 17, 2021
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