TITLE: The 'class version' of almost disjoint sets: can it fail?
QUESTION [5 upvotes]: I have a question about 'class versions' of almost disjoint sets.   To even state what I'm after, I need to go beyond what one can state in theories like NBG or MK.      I'm wondering about the status of the following:
There is a class C of proper classes of sets with the following two properties
(1) C is almost disjoint in the sense that every two different members of C intersect in a set,
and yet
(2)  The size of C is larger than the size of the class V of sets in the sense that there is no injection of C into V.
I don't have a preferred theory in mind in which to discuss these, I'm interested in hearing about what the possible approaches could be.
It is clear that (1) + (2) is consistent from an inaccessible, but my question is whether these are provable in any natural system.  Even more, I'd be interested in whether there are models where there is a class of classes C where (1) + (2) fails.
The reason that I'm interested in these is that they are connected to results on functors on the category of classes that I'm thinking about.
Once the matter of almost disjoint classes is cleared up, I might be back with a follow up on the question that motivates this.

REPLY [7 votes]: Here is a way to formalize the almost-disjointness phenomenon in GBC. You don't need an
inaccessible cardinal.
Theorem. In any model of GBC, there is definable
transformation of classes $X\mapsto X^*$, such that if $X\neq Y$,
then $X^*$ and $Y^*$ are almost disjoint, and there is no
definable map from sets $a\mapsto X_a$ such that every $X^*$ is
some $X_a$.
(What I mean by a transformation of classes is that there is a
formula $\phi(X,Y)$ with only first-order quantifiers and class
variables $X$ and $Y$ such that GBC proves that every class $X$
has one and only one class $Y=X^*$ for which $\phi(X,X^*)$ holds.)
Proof. Let $X^*=\left\{X\cap V_\alpha
\mid\alpha\in\text{Ord}\right\}$. If $X\neq Y$, then $X\cap
V_\alpha\neq Y\cap V_\alpha$ for all sufficiently large ordinals
$\alpha$, and so $X^*\cap Y^*$ is a set. Thus, they are almost
disjoint.
But I claim that there is no definable map $a\mapsto X_a$ that is
surjective onto the classes $X^*$. For any map $a\mapsto X_a$, let
$A=\{\ a\ \mid a\notin X_a\ \}$, and it is easy to see that $A\neq
X_a$ for any $a$. Since $X=\bigcup X^*$ for any $X$, it follows
that $A^*\neq X_a^*$ for any set $a$. QED
In particular, there can be no "injection" of the classes $X^*$
into the sets, since by inverting that we would get such a
forbidden surjection $a\mapsto X_a^*$.

REPLY [7 votes]: There is an easy way to construct a bijection between the collection of all classes and a certain collection of almost-disjoint classes: send each class $X$ to the class $\{X \cap V_\alpha : \alpha \in \mathrm{Ord}\}.$ Theories with collections of classes should support this construction with relative ease since this argument makes no essential use of collections of classes. So the question becomes whether the collection of classes is larger than $V$. This is true, provided some relatively basic combinatorics.
By the usual diagonal argument, no map from $V$ into the collection of classes can be onto: given such a map $F$, consider the class $R = \{x \in V : x \notin F(x)\}$. This requires comprehension with higher-order parameters, so this could potentially fail in a very weak theory with collections of classes.
Since the second fact talks about surjections rather than injections, we also need the fact that if there is an injection from the collection of all classes into $V$ then there is a surjection from $V$ onto the collection of all classes. This is trivial, given such an injection, map each set to its preimage if there is one, or to $\varnothing$ (say) if there is none.
Without further knowledge about your theory of collections of classes, there is no way to know whether these three pieces work. However, I suspect whatever you have in mind satisfies all three pieces.