A cardinal $\kappa$ is a Vopěnka cardinal if Vopěnka's principle holds in $V_\kappa$.

Suppose that $\kappa$ and $\lambda$ are both Vopěnka cardinals with $\lambda > \kappa$. Must it be the case that $V_\lambda \vDash$ '$\kappa$ is a Vopěnka cardinal'?

(Understand that the predicate "… is a Vopěnka cardinal" after the turnstile as standing for a schema, since the informal claim is not directly expressible in $\mathsf{ZFC}$.)


First, let me clear up a few issues. There is a distinction between the Vopěnka principle and Vopěnka scheme, where the latter is expressed in ZFC as a scheme of assertions concerning the definable classes, and the former is expressed in GBC as a single statement, but quantifying over all classes. (See my paper J. D. Hamkins, The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme, arxiv:1606.03778, for extensive discussion of this.)

Unfortunately, the terminology in the literature on this point is a mess, with many authors using the Vopěnka scheme, but calling it the Vopěnka principle, and others using the full Vopěnka principle, and calling it the Vopěnka principle. Alas, some authors seem not to distinguish between the two principles.

Although Vopěnka himself had apparently meant the full class version, nevertheless it is common to find set theorists considering what I call the Vopěnka scheme.

The difference is important in the concept of Vopěnka cardinal, because the usual way is to define this is with the analogue of the full Vopěnka principle, not the Vopěnka scheme, as you have done. That is, $\kappa$ is a Vopěnka cardinal, if $V_\kappa$ satisfies the full-blown class version of Vopěnka's principle, using all subsets of $V_\kappa$ and not merely the definable ones. The corresponding cardinals for the Vopěnka scheme are called Vopěnka scheme cardinals, and they are strictly weaker in consistency strength.

But finally, it doesn't matter whether you have in mind Vopěnka cardinals or Vopěnka scheme cardinals, since both of these are expressible in ZFC as a single assertion (not a scheme). We use that satisfaction in a set structure is definable in ZFC for the scheme version. And furthermore, since these notions are completely determined by $V_\kappa$ and its subsets, any $V_\lambda$ with $\lambda>\kappa$ can tell whether or not $\kappa$ is a Vopěnka cardinal or a Vopěnka scheme cardinal.

So the answer is yes. And furthermore it is yes in a strong way, because there is no need for the larger cardinal $\lambda$ to be a Vopěnka cardinal itself.


I believe the answer is "yes" - $V_\lambda$ can compute $\varphi^{V_\kappa}$ for every formula $\varphi$ (since the full truth predicate for $V_\kappa$ lives already at around $V_{\kappa+2}$ or so), and any embedding between structures in $V_\kappa$ also lives in $V_\lambda$. Or am I missing a subtlety?

(Indeed, note that this doesn't depend on any properties of $\lambda$ other than $\lambda$ being large enough to define truth in $V_\kappa$.)


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.