Working in MK *(or some other not-too-strong class theory if you prefer)*, say that an **up-class** is a class of structures $\mathfrak{X}$ which is definable in $V$ (allowing parameters from $V$) and such that whenever $\mathcal{A}\in\mathfrak{X}$ and $i:\mathcal{A}\rightarrow\mathcal{B}$ is an embedding then $\mathcal{B}\in\mathfrak{X}$.


Consider the following principle of **i**nfinitary **a**xiomatization of **u**p-classes:



> 
> **(IAU)** Suppose $\mathfrak{X}$ is an up-class. Then there is some $\mathcal{L}\_{\infty,\infty}$-sentence $\varphi$ such that $\mathfrak{X}=\{\mathcal{A}:\mathcal{A}\models\varphi\}$.
> 
> 
> 


By a result of [Makowsky](https://www.jstor.org/stable/2273786), we get that Vopenka's principle implies IAU. Specifically, suppose $\mathfrak{X}$ is an up-class. Per Makowsky there is some $\kappa$ such that whenever $T$ is a finitary first-order all of whose models are in $\mathfrak{X}$ then there is some $T\_0\subseteq T$ of size $<\kappa$ all of whose models are in $\mathfrak{X}$. Meanwhile, note that whenever $\mathcal{A}\in\mathfrak{X}$ we have - since $\mathfrak{X}$ is an up-class - that the every model of the atomic diagram of $\mathcal{A}$ is also in $\mathfrak{X}$. Putting these together, letting $\sigma$ be the disjunction of all $\mathcal{L}\_{\kappa,\kappa}$-sentences which only have models in $\mathfrak{X}$ we get $Mod(\sigma)=\mathfrak{X}$.


However, the actual strength of IAU itself is unclear to me:


* How strong is IAU?


In particular, I'm curious whether IAU is compatible with V=L. I strongly suspect the answer is negative, but I don't see how to prove it at the moment.


Note that one point of flexibility here is that IAU says nothing about the syntactic complexity of the defining sentence, whereas the argument via Vopenka gives a very simple form: the $\sigma$ it builds is a disjunction of size $\le 2^\kappa$ of sentences, each of which consists of an existential quantifier over a $<\kappa$-length tuple followed by a $<\kappa$-length conjunction of literals - basically "$\Sigma\_2$ in the sense of $\mathcal{L}\_{\infty,\infty}$."



How strong is "all up-classes are infinitarily definable"?
I think IAU is equivalent to Vopěnka's principle. For the other direction, assume
Vopěnka's principle fails. Then there is a proper class of structures (WLOG graphs), none of which embeds into any other. Because this is a proper class, there is an injection of $\mathcal{L}\_{\infty,\infty}$ into it. In other words, for each sentence $\sigma \in \mathcal{L}\_{\infty,\infty}$ we may choose a structure $\mathcal{M}\_\sigma$ in such a way that for any two distinct sentences $\sigma,\tau \in \mathcal{L}\_{\infty,\infty}$, neither $\mathcal{M}\_\sigma$ nor $\mathcal{M}\_\tau$ embeds into the other.


Now we define the "diagonal" class of structures $D = \{\mathcal{M}\_\sigma \mid \sigma \in \mathcal{L}\_{\infty,\infty} \wedge \mathcal{M}\_\sigma \not\models \sigma\}$. I claim that its upward closure $D\mathord{\uparrow}$ does not have the form $\operatorname{Mod}(\tau)$ for any sentence $\tau \in \mathcal{L}\_{\infty,\infty}$. This is because $\mathcal{M}\_\tau \in D\mathord{\uparrow} \iff \mathcal{M}\_\tau \in D \iff \mathcal{M}\_\tau \not\models \tau \iff \mathcal{M}\_\tau \notin \operatorname{Mod}(\tau)$.


The key point is that $\mathcal{M}\_\tau \in D\mathord{\uparrow}$ implies $\mathcal{M}\_\tau \in D$ because no other structure $\mathcal{M}\_\sigma \in D$ can embed into $\mathcal{M}\_\tau$.


**Edit:** The argument can be modified to avoid relying on global choice, as follows.


Assume that we have a counterexample to Vopěnka's principle, meaning a proper class $\mathcal{C}$ of structures (WLOG graphs), none of which embeds into any other. For every sentence $\sigma \in \mathcal{L}\_{\infty,\infty}$, define the structure $\mathcal{M}\_\sigma = \langle V\_\lambda; \in, \{\sigma\}, \mathcal{C} \cap V\_\lambda\rangle$ for the least limit ordinal $\lambda$ that is greater than the rank of $\sigma$ (meaning just its Von Neumann rank as a set) and has the additional property that $A \cap \lambda$ has order type $\lambda$ where $A$ is the proper class of ordinals $\{\text{rank}(\mathcal{M}) \mid \mathcal{M}\in C\}$.


Note that for any two distinct sentences $\sigma, \tau \in \mathcal{L}\_{\infty,\infty}$ there is no *elementary* embedding $j : \mathcal{M}\_\sigma\to \mathcal{M}\_\tau$. If there were, then letting $\kappa$ be the critical point of $j$ (which exists because $j$ maps $\sigma$ to $\tau$) and letting $\mathcal{M} \in \mathcal{C}$ be a structure of rank equal to the $\kappa$th element of $A$, the restriction $j \restriction \mathcal{M}$ is an embedding of $\mathcal{M}$ into some other structure $j(\mathcal{M}) \in \mathcal{C}$ whose rank equals the $j(\kappa)$th element of $A$ and which is therefore different from $\mathcal{M}$. This contradicts the assumption that $\mathcal{C}$ was a counterexample to Vopěnka's principle.


To remove the word "elementary" above and ensure that for any two distinct sentences $\sigma, \tau \in \mathcal{L}\_{\infty,\infty}$ there is no embedding $j : \mathcal{M}\_\sigma\to \mathcal{M}\_\tau$, we just need to modify the structures $\mathcal{M}\_\sigma$ in the usual way by adding a relation for every first-order formula. (I didn't want to complicate the notation with this when defining the structures initially.)


Then we can define the diagonal class of structures $D = \{\mathcal{M}\_\sigma \mid \sigma \in \mathcal{L}\_{\infty,\infty} \wedge \mathcal{M}\_\sigma \not\models \sigma\}$ and show that its upward closure $D\mathord{\uparrow}$ does not have the form $\operatorname{Mod}(\tau)$ for any sentence $\tau \in \mathcal{L}\_{\infty,\infty}$ as before.


