# Power set of the Universal Set

In set theories with a universal set $U$, can we take the powerset of $U$? In particular, can we look at $\mathcal{P}(U)$ and compare with $U$, i.e. can we say something like $U \in \mathcal{P}(U)$, $U \notin \mathcal{P}(U)$, $|U| < |\mathcal{P}(U)|$, or $U \neq \mathcal{P}(U)$?

• In particular, which set theory? Commented Jan 7, 2018 at 3:34
• Any set theory with a universal set $U$ Commented Jan 7, 2018 at 3:35

Yes, you can do this at least some of the time. There are certain things which have to be true - e.g. we must have $\mathcal{P}(U)\in U$ if $\mathcal{P}(U)$ is a set, and we must have $U\in\mathcal{P}(U)$ by definition of $\subseteq$ - but on other points we have some flexibility.

The standard example I think is Quine's New Foundations with urelements, NFU (note that NF alone is not known to be consistent - that said, Holmes has a claimed consistency proof). Since the formula $$\psi(x)\equiv x=x$$ is stratified, we may apply stratified comprehension to it; this gives us the universal set $U$. Similarly, the formula $$\varphi(x; z)\equiv\forall y(y\in x\implies y\in z)$$ (here $z$ is a parameter) is always stratified, so applying stratified comprehension we have that powersets of arbitrary sets exist. In particular, taking $z=U$ we get the powerset of the universal set.

Now things get interesting. If there are no urelements (that is, in NF - then every set is a subset of $U$, hence $\mathcal{P}(U)=U$. However, since urelements are not sets of sets, we can have $\mathcal{P}(U)\subsetneq U$ in NFU!

Forster's book has more details on this sort of thing.

Some nonwell-founded set theorizing (§4.4) has mappings $$\wp(V) \rightarrow V$$ and vice versa. I'm not sure if this is a theorem of the system's axioms or an axiom itself (it appears to be a conjunction or alternation of axioms; the mappings correspond to an initial algebra and a final coalgebra, but I won't pretend to understand that difference!).

In paraconsistent set theory, things are even weirder. It appears to be the case here that a cardinal for ORD is equivalent to its own powerset while sort of being smaller than itself as such. Or at any rate, the situation is so exotic that it's hard to tell how the mapping between $$V$$ and $$\wp(V)$$ works except to say that it's not the same as usual!

In NBG set theory, there is a proper class of all sets, and $|U|=|\mathcal{P}(U)|.$

In ZFC + grothendieck universes, there is a set of all "small" sets $U$ with $|U|\lneq|\mathcal{P}(U)|.$ It is still a set, obeying Cantor's theorem.

So it can go either way.

• I guess I am more looking for set theories with an explicit universal set, not proper class of sets. I know New Foundations has a universal set... Commented Jan 7, 2018 at 4:21
• In a previous comment you said "any set theory with a universal set" Commented Jan 7, 2018 at 4:25
• Sorry, I should have been more specific. By Universal Set I mean a set $U$ containing all sets, including itself. Commented Jan 7, 2018 at 4:28
• If you have a specific question in mind, it would be good to write those specifics (and it's not too late to edit your question). It is also good not to "move the goalposts". Commented Jan 7, 2018 at 4:54
• @ziggurism I don't think this constitutes moving the goalposts - this is the standard meaning of the phrase "universal set." Commented Jan 7, 2018 at 8:38