How does a “power set” work on proper classes?

I am trying to understand what the notion of "power set" means when we consider a class $$A$$ that is not a set (a proper class), where the definition of the "power set" of $$A$$ is:

$$\mathcal{P}(A) = \{x: x \subseteq A\}$$

I have a theorem that states:

If $$A$$ is a set then $$x \in \mathcal{P}(A) \iff x \subseteq A$$.

So, unless this theorem was intentionally written in a confusing way, this seems to indicate that one of these implications breaks for $$A$$ being a proper class. But that also seems to contradict the definition. So... what's going on here?

Another piece of information I have is that $$\mathcal{P(U)} = \mathcal{U}$$ where $$\mathcal{U}$$ is the universe in which we are working (and $$\mathcal{U}$$ is not a set).

So what exactly does it mean to take the power set of a proper class?

• It's important to note that if you're working in a theory like ZF, and $A$ is a proper class, $\exists x(x=A)$ is false. So you can't even apply an instance of the power set axiom to $A$ to start with. In ZF, proper classes are a figure of speech, and not actually things in the theory. – Malice Vidrine Feb 5 at 5:49

If you're working in a theory that doesn't allow proper classes to be elements (as I suspect you are), the best we can do is to construct the class of subsets of $$A$$. This class will be a proper class, so we should probably call it the power class of $$A$$. Your theorem will then apply for sets $$x$$.
Note also that $$\mathcal{P}(\mathcal{U})$$ is the class of subsets of $$\mathcal{U}$$, i.e. $$\mathcal{U}$$ itself as required. If this seems at odd with Cantor's theorem, note the latter's proof constructs something that would be a proper class and hence no element of $$\mathcal{U}$$, so the theorem doesn't work for proper classes' power classes.
You might want to modify these ideas to interpret the union of a proper class, $$\bigcup A$$.