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?

  • $\begingroup$ 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

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, $\bigcup A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.