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Update. My new article grows out of and extends my 2010 answer to this question.

Statement 1 is proved by class forcing to add a club class $C$ avoiding the regular cardinals. This destroys the assertion "Ord is Mahlo" and therefore destroys the Vopenka principle, while preserving the Vopenka scheme because it does not add sets.

Statement 2 is proved by class forcing of the global axiom of choice. The difficult part is to show that the Vopenka principle holds true with respect the new classes definable from the generic filter. The proof involves the concept of a stretchable set $g\subset\kappa$ for an $A$-extendible cardinal, one which has the property that for every cardinal $\lambda<\kappa$ and every extension $h\subset\lambda$ with $h\cap\kappa=g$, there is an elementary embedding $j:\langle V_\lambda,\in,A\cap V_\lambda\rangle\to\langle V_\theta,\in,A\cap V_\theta\rangle$ such that $j(g)\cap\lambda=h$. Thus, the set $g$ can be stretched by an $A$-extendibility embedding so as to agree with any given $h$. This stretchability property is the $A$-extendibility analogue of the master condition technique in other large cardinal contexts.

Update. My new article grows out of and extends my 2010 answer to this question. The new part is the conservativity result, showing that the Vopěnka principle has the same first-order consequences as the strictly weaker Vopěnka scheme.

Statement 1 is proved by class forcing to add a club class $C$ avoiding the regular cardinals. This destroys the assertion "Ord is Mahlo" and therefore destroys the Vopěnka principle, while preserving the Vopěnka scheme because it does not add sets.

Statement 2 is proved by class forcing of the global axiom of choice. The difficult part is to show that the Vopěnka principle holds true with respect the new classes definable from the generic filter. The proof involves the concept of a stretchable set $g\subset\kappa$ for an $A$-extendible cardinal, one which has the property that for every cardinal $\lambda<\kappa$ and every extension $h\subset\lambda$ with $h\cap\kappa=g$, there is an elementary embedding $j:\langle V_\lambda,\in,A\cap V_\lambda\rangle\to\langle V_\theta,\in,A\cap V_\theta\rangle$ such that $j(g)\cap\lambda=h$. Thus, the set $g$ can be stretched by an $A$-extendibility embedding so as to agree with any given $h$. This stretchability property is the $A$-extendibility analogue of the master condition technique in other large cardinal contexts.

This is a very interesting question, and I finally have an answer. (I just had a fruitful conversation about Vopenka in my office with Andrew Brooke-Taylor, who will give a talk on Vopenka for our seminar tomorrow.)

The answer to your question is no, provided that VP is consistent, there can be no such definable class.

The reason has to do with the formalization of VP in set theory, whether one interprets it as a second order axiom or as a first-order scheme. Let me consider the set-theoretic versions of the Vopenka Principle, the assertion that every proper class ordinal-indexed sequence $\langle M_\alpha\mid\alpha\in\text{Ord}\rangle$ of mathematical structures in the same language has some $\alpha\lt\beta$ with an elementary embedding of $M_\alpha$ into $M_\beta$. The principle is equivalent if one restricts to the case where each $M_\alpha$ is a graph, although set-theorists often prefer to restrict to the case where each $M_\alpha$ is a rank-initial segment $V_{\gamma_\alpha}$ of the universe, sometimes with some additional predicates added on the side.

This way of stating the axiom is ambiguous until we specify how one is to interpret the quantification over classes ("for every class sequence..."). The standard formulation of VP insists that this is a second-order quantifier, ranging over all classes, and this can be expressed in GBC by a single statement about all classes. In ZFC, we can express a similar-but-actually-weaker principle as an axiom scheme, ranging over all possible definitions $\varphi$, as the statement "if for every ordinal $\alpha$ there is a unique structure $M_\alpha$ satisfying $\varphi(\alpha,M_\alpha)$, then...".

The same difference arises when defining the Vopenka cardinals. Namely, a cardinal $\kappa$ is a Vopenka cardinal if $\kappa$ is inaccessible and every $\kappa$-sequence of structures $\langle M_\alpha\mid\alpha\lt\kappa\rangle$ with $M_\alpha\in V_\kappa$ has some $\alpha\lt\beta$ with $M_\alpha$ embedding elementarily into $M_\beta$. Alternatively, we could define that $\kappa$ is an almost Vopenka cardinal if $V_\kappa$ satisfies the Vopenka scheme.

The difference in each case is the difference between the boldface concept and a lightface one. Vopenka cardinals are defined by reference essentially to $V_{\kappa+1}$, using all $\kappa$-sequences of structures whether the sequence is definable or not, whereas the almost Vopenka cardinals only consider the sequences definable (with parameters) over $V_\kappa$. And similarly for the full class versions: the full VP uses all classes, but the Vopenka scheme only uses definable classes.

My point is that the weaker definable versions are strictly weaker.

Theorem.

• If the Vopenka Principle holds, then there is a forcing extension $V[G]$ in which the Vopenka Principle fails, but the Vopenka scheme continues to hold.
• Every Vopenka cardinal $\kappa$ is a limit of $\kappa$ many almost-Vopenka cardinals. In particular, the smallest almost-Vopenka cardinal is not a Vopenka cardinal.

Proof. If VP holds, then we may force to add a closed unbounded class C of cardinals avoiding the regular cardinals. This kills the Mahlo-ness of the ordinals, without adding any new sets. Thus, V and V[C] have the same sets, and so V[C] continues to satisfy the Vopenka scheme. But V[C] does not satisfy the full VP, since that implies that Ord is Mahlo.

For the second statement, suppose $\kappa$ is a Vopenka cardinal. It follows that $\kappa$ is Mahlo, in fact it is $\kappa$-Mahlo. Thus, there are numerous inaccessible $\gamma\lt\kappa$ with $V_\gamma$ elementary in $V_\kappa$. Every such $\gamma$ is almost-Vopenka, since the scheme transfers down to $V_\gamma$ by elementarity. QED

This theorem implies that if the Vopenka Principle is consistent at all, then there are models where VP fails, but not for any definable counterexample. So there can be no such definable counterexample as you describe.

I anticipate that you may choose to fall back, and say that you didn't really want the full second-order Vopenka principle, but only the definable version anyway. In this case, a version of my argument will still apply. Namely, we can stratify the definable levels of Vopenka by the $\Sigma_n$ complexity of the definitions involved. The Vopenka scheme is asserting $\Sigma_n$-Vopenka for every natural number $n$ in the meta-theory. If this scheme holds, we have that ORD is definably Mahlo. By the Reflection theorem, we can find an inaccessible cardinal $\gamma$ such that $V_\gamma$ is $\Sigma_n$ elementary in $V$, and so it will continue to satisfy $\Sigma_n$-Vopenka, but not the full Vopenka scheme. If your definition is $\Sigma_n$, then it will therefore not produce a counterexample, even though the VP scheme fails in this model. This is a strong sense in which there can be no definable counterexample as you describe.

Update. My new article grows out of and extends my 2010 answer to this question.

• Joel David Hamkins, The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme, manuscript under review. (arxiv)

Abstract. The Vopěnka principle, which asserts that every proper class of first-order structures in a common language admits an elementary embedding between two of its members, is not equivalent over GBC to the first-order Vopěnka scheme, which makes the Vopěnka assertion only for the first-order definable classes of structures. Nevertheless, the two Vopěnka axioms are equiconsistent and they have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for first-order assertions in the language of set theory.

The Vopěnka principle is the assertion that for every proper class $\mathcal{M}$ of first-order $\mathcal{L}$-structures, for a set-sized language $\mathcal{L}$, there are distinct members of the class $M,N\in\mathcal{M}$ with an elementary embedding $j:M\to N$ between them. In quantifying over classes, this principle is a single assertion in the language of second-order set theory, and it makes sense to consider the Vopěnka principle in the context of a second-order set theory, such as Godel-Bernays set theory GBC, whose language allows one to quantify over classes. In this article, GBC includes the global axiom of choice.

In contrast, the first-order Vopěnka scheme makes the Vopěnka assertion only for the first-order definable classes $\mathcal{M}$ (allowing parameters). This theory can be expressed as a scheme of first-order statements, one for each possible definition of a class, and it makes sense to consider the Vopěnka scheme in Zermelo-Frankael ZFC set theory with the axiom of choice.

Because the Vopěnka principle is a second-order assertion, it does not make sense to refer to it in the context of ZFC set theory, whose first-order language does not allow quantification over classes; one typically retreats to the Vopěnka scheme in that context. The theme of my article is to investigate the precise meta-mathematical interactions between these two treatments of Vopěnka's idea.

Main Theorems.

1. If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka scheme holds, but the Vopěnka principle fails.
2. If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka principle holds.

Statement 1 is proved by class forcing to add a club class $C$ avoiding the regular cardinals. This destroys the assertion "Ord is Mahlo" and therefore destroys the Vopenka principle, while preserving the Vopenka scheme because it does not add sets.

Statement 2 is proved by class forcing of the global axiom of choice. The difficult part is to show that the Vopenka principle holds true with respect the new classes definable from the generic filter. The proof involves the concept of a stretchable set $g\subset\kappa$ for an $A$-extendible cardinal, one which has the property that for every cardinal $\lambda<\kappa$ and every extension $h\subset\lambda$ with $h\cap\kappa=g$, there is an elementary embedding $j:\langle V_\lambda,\in,A\cap V_\lambda\rangle\to\langle V_\theta,\in,A\cap V_\theta\rangle$ such that $j(g)\cap\lambda=h$. Thus, the set $g$ can be stretched by an $A$-extendibility embedding so as to agree with any given $h$. This stretchability property is the $A$-extendibility analogue of the master condition technique in other large cardinal contexts.

Corollaries.

1. Over GBC, the Vopěnka principle and the Vopěnka scheme, if consistent, are not equivalent.
2. Nevertheless, the two Vopěnka axioms are equiconsistent over GBC.
3. Indeed, the two Vopěnka axioms have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for assertions in the first-order language of set theory. $$\text{GBC}+\text{VP}\vdash\phi\qquad\text{if and only if}\qquad\text{ZFC}+\text{VS}\vdash\phi$$

See the edit history for my 2010 answer.