I'm not sure if there is a written reference for that.

Let me give a quick summary of what I can tell. We know that classes are actually formulas with one free variable. We can therefore manipulate classes the same way we manipulate formulas, with the added fact that we can encode all sort of structure into the sets.

For example, if $A=\{x\mid A(x)\}$ and $B=\{x\mid B(x)\}$ then the union would be disjunction; the intersection would be conjuction. So far this is really just the same as formulas being manipulated.

If we wish to take a Cartesian product we can use the fact that we can define ordered pairs, and so $$A\times B=\{z\mid z=\langle x,y\rangle\land A(x)\land B(y)\}$$

This can become intensely complex using all sort of crazy formulas. What you *can't* do is talk about classes as actual objects. You cannot say that "there exists a subclass", or "the collection of all subclasses" - although you can talk about collection of all subsets: $$\mathcal P(A)=\{x\mid\forall y(y\in x\rightarrow A(y))\}=\{x\mid x\subseteq A\}$$

You can talk about the existence of subclasses *externally*, and if you're lucky enough you can prove they are definable internally as well. For example, you can talk about the class of ordinals or sets of ordinals externally, and since these are simple enough definitions this is also internally definable.

I suppose that if you wish to give some recent example which bothered you I could try and help to figure it out.

In the comments David asks about equivalence relations and quotient of equivalence relations. The problem here is that often the quotient space is a collection of equivalence classes, and if the equivalence classes are not sets then this object is not even definable in ZFC.

However we can use Scott's trick to overcome this issue. Scott's trick makes heavy use of the axiom of regularity, and its two common uses are defining cardinals in ZF and proving that one can talk about ultrapowers internally.

Suppose that $\varphi(x,y)$ is a formula defining an equivalence relation. We define $$A_\varphi=\{x\mid \varphi(x,A)\land\mathrm{rank}(x)\text{ is minimal for this property}\}$$
Namely if $\varphi(x,A)$ is true and $x\in A_\varphi$ then there is no $y$ whose rank is smaller than that of $x$ and $\varphi(y,A)$.

By the fact that all the sets in $A_\varphi$ have the same rank we can now show that this i a set, since it is bounded by some $V_\alpha$. It is possible that there is a canonical choice of representative, or it is possible that there is none. But we don't care.

Now we can talk about $C/\varphi = \{A_\varphi\mid A\in C\}$, as a collection of equivalence classes, and $A_\varphi=B_\varphi$ if and only if $\varphi(A,B)$ is true as wanted.

If $C/\varphi$ is a set (for example if you can prove there can only be set-many of equivalence classes) then using the axiom of choice - if it exists - it is possible to just pick a system of representatives, but since the equivalence classes themselves need not be sets, you would still have to use Scott's trick to define the collection from which you are choosing.

The subsequent is ugly, but it is a one way to avoid, if you want, some problems with proper classes in definition. We can define the category as some class $\mathcal{C}$ such that there exists class $\mathcal{A}$ of "arrows", for which $\mathcal{C}\subseteq\mathcal{A}\times\mathcal{A}\times\mathcal{A}$, and $(\beta,\alpha,\gamma)\in\mathcal{C}$ has a sense that arrows $\beta,\alpha$ can be composed, and $\beta\alpha=\gamma$. The corresponding axioms are obvious. Then $Ob(\mathcal{C})$ and hom-sets can be proper classes. Yet the set theory is narrow for category theory.

## Best Answer

The interpretation that proper classes are simply too "large" does not only entail that they do not have a well-defined cardinality, but also that they cannot squeeze inside another class/element. If $A$ is a class (proper or not) and $B \in A$, then $B$ must be a set. In other words, a proper class is not an element of any class.

When discussing classes, we usually do it under NBG Set Theory, which is just a little bit different from the usual ZFC Set Theory that you probably hear about all the time. Regardless, the theories $\mathsf{NBG}$ (with choice) and $\mathsf{ZFC}$ are so similar that in many cases we do not have to distinguish them apart - see this post.

Thus, to answer your questions (under $\mathsf{NBG}$):

Yes. This is a consequence of Axiom of Comprehension.

No, as by definition we must have $A \in \mathcal{P}(A)$, which implies that the proper class $A$ is an element of some class.

Yes. If $A$ is a proper class and $\leq$ is a partial order on $A$, then $\leq \; \subseteq A \times A$ (all elements of $\leq$ are sets).