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We say that a cardinal $\kappa$ is universal if and only if $V_\kappa\models ZFC$. I need to prove that the least universal cardinal has countable cofinality.

My attemp: We can list $ZFC:=\{\phi_n:n\in\omega\}$. Define $A_n:=\{\phi_i:0\leq i\leq n\}$. Working in $V_\kappa$, we can use the reflection principle to find $\kappa_n<\kappa$ such that $(V_{\kappa_n}\models A_n)^{V_\kappa}$. Let $\lambda:=\sup\kappa_n$ so, if $cf(\kappa)>\omega$, then $\lambda<\kappa$. But $(V_\lambda\models ZFC)^{V_\kappa}$ which implies that $\lambda$ is a universal cardinal in $V_\kappa$. Now, to conclude my problem I want to see that $V_\lambda\models ZFC$, because this is contradictory with the way in which we chose $\kappa$. In this point is where I have doubts, Can I conclude $V_\lambda\models ZFC$?

YCB
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2 Answers2

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I don't see why $V_\lambda$ need satisfy $ZFC$: in general, the set of levels of $V$ satisfying some sentence need not be closed. However, a variation on this idea does work. Specifically, show the following:

$(*)\quad$ If $V_\kappa\models ZFC$ and $cf(\kappa)>\omega$, then for each axiom $\varphi$ of $ZFC$ the set $\{\alpha<\kappa: V_\alpha\models \varphi\}$ contains a club in the sense of $\kappa$.

Since the intersection of countably many clubs is again a club, and in particular nonempty, this will tells that if $V_\kappa\models ZFC$ and $cf(\kappa)>\omega$ then there is some $\alpha<\kappa$ with $V_\alpha\models ZFC$.

It's a good exercise to show that $(*)$ isn't quite trivial: we can find a sentence $\varphi$ such that it's consistent with $ZFC$ that the set $$\{\alpha<\omega_1: V_\alpha\models\varphi\}$$ is both stationary and co-stationary in $\omega_1$.

I don't know whether this is in fact a consequence of $ZFC$, however!

Noah Schweber
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Here is a slightly different approach that does it "all at once" like you were trying to.

Theorem If $\kappa$ is universal and has uncountable cofinality, then $\{\alpha<\kappa: V_\alpha\preceq V_\kappa\}$ is unbounded in $\kappa.$

Essentially, we want to attempt to prove the reflection theorem for all formulas at once and show that we can avoid the pitfalls.

Recall that the standard proof is a closure argument. At any rank $\alpha$, for any formula $\exists x\phi(x,\vec y)$ with existential quantifier out front, we can define a "Skolem function" $F(\vec y)$ for $\vec y\in V_\alpha$ that produces the least rank of a witness if one exists. Then we take the $\sup$ over $\vec y\in V_\alpha$ and then take the sup over all the existential formulas to get an $\alpha'>\alpha$ so that all the witnesses are in $V_{\alpha'}$. Then we iterate this $\omega$ times for closure so we can invoke the Tarski-Vaught lemma to show we have reflection.

The reason we can't prove the reflection theorem for all formulas at once is that we cannot talk about all formulas at once, or more exactly, we can't define a uniform satisfaction relation for them. When we work external to a set $V_\kappa$ we do have a uniform satisfaction relation, however there is still an obstacle since we cannot define this in $V_\kappa.$ Thus while the individual "Skolem functions" $F(\vec y)$ may be defined inside $V_\kappa,$ the set of them cannot.

But we can still take the sup over $\vec y$ for each individually inside $V_\kappa$, and then we have a countable collection of ordinals $<\kappa$ that we can take the sup of from the outside. This is where the assumption of uncountable cofinality comes in. (And then it comes in again when we iterate countably many times.)

spaceisdarkgreen
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