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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?

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  • $\begingroup$ 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. $\endgroup$ – Malice Vidrine Feb 5 at 5:49
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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, $\bigcup A$.

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