# $\mathcal U$ Grothendieck universe. Is $\mathcal{P(U)}$ a model for NBG?

Suppose we are in ZFC, let $\mathcal U$ be an uncountable Grothendieck universe and consider the set of its parts $\mathcal{P(U)}$. (I will index axioms as $(\mathcal U.n)$)

Note that if $x \in \mathcal U$, then $x \in\mathcal{P(U)}$, directly from $(\mathcal U.1)$ (i.e., $\mathcal U$ is a transitive set).

Further, if $x \in \mathcal{P(U)}$ and $|x|<|\mathcal U|$, then $x \in \mathcal U$. (This is not so trivial and require the axiom of choice. Actually, in ZFC, Grothendieck universes are equivalent to the so called transitive Tarski class; a sketch of the proof, in italian, can be found here).

Thus, $\mathcal{P}_<(\mathcal U) := \{x \in \mathcal{P(U)}\,|\,|x|<|\mathcal U|\} = \mathcal U$.

Denoting $\mathcal P_=(\mathcal U) := \{C \in \mathcal{P(U)}\,|\,|C|=|\mathcal U|\}$, we have $\mathcal{P(U)} = \mathcal{P}_<(\mathcal U) \cup \mathcal P_=(\mathcal U) = \mathcal U \cup \mathcal P_=(\mathcal U)$, the union being disjoint.

Define the binary relation $\in_{\mathcal U} \subseteq \mathcal{P(U)}\times \mathcal{P(U)}$ such that ${\in_{\mathcal U}}_{|\mathcal U \times \mathcal{P(U)}} = {\in_{\text{ZFC}}}_{|\mathcal U \times \mathcal{P(U)}}$ and ${\in_{\mathcal U}}_{|\mathcal P_=(\mathcal U) \times \mathcal{P(U)}} = \emptyset$. That is, $\in_{\mathcal U}$ restricted to $\mathcal U \times \mathcal{P(U)}$ is just the usual set-relation $\in_{\text{ZFC}}$ of ZFC (i.e., just $\in$, if you prefer); and there is no other couple in $\in_{\mathcal U}$.

Does $(\mathcal{P(U)}, \in_{\mathcal U})$ serve as a model for NBG?

As two-sorted theory, this is equivalent to choose $\mathcal{P(U)}$ as classes, as sets the parts $x \in \mathcal{P(U)}$ such that $x \in \mathcal U$ (i.e., the parts $x \subset U$ such that $|x| < |\mathcal U|$) and as binary relation the modified membership $\in_{\mathcal U}$ defined above.

• What is $\in_{ZFC}$? Also NBG is a two-sort theory, so you need to tell us what are the sets and what are the classes of the model. Feb 20 '13 at 22:52
• Your edit did not answer my question. What are the sets and what are the classes of your model? (Also no need to write "Edited" in your post, there is an automatic stamp telling the readers that the post had been edited.) Feb 20 '13 at 22:57
• I really think NBG can be described as a single-sort theory. Just define a set to be a class $x$ for which there exists a class $C$ such that $x \in C$. Feb 20 '13 at 23:03
• Hmm. Yes. I think that you're right. To the actual question, if $M$ is a model of ZFC, then $\text{Def}(M)$, the collection of all internally-definable classes is a model of NBG. I don't know about the full power set, though. I suppose it's a matter of verifying some axioms. Also is a Grothendieck universe a model of ZFC to begin with? (I know that some models of ZFC are Grothendieck universes) Feb 20 '13 at 23:05
• Sorry. I still don't see where the axiom of choice enters the game. This axiom just says that (1) No element of $\cal U$ can be mapped onto $\cal U$; (2) The range of a map from an element of $\cal U$ into $\cal U$ is an element of $\cal U$ - that is, $\cal U$ is a model of second-order replacement. What I don't see is where the axiom of choice gets into necessity here. Feb 21 '13 at 2:36

## 1 Answer

Yes, the power set of a Grothendieck universe is a model of NBG, and even of the stronger Morse-Kelley set theory.

• I suspected this is the answer, and I was hoping that you'd turn up to give it! Feb 21 '13 at 3:14
• Many thanks, Professor Blass. I think it is an interesting result, that makes clear the incredible expressive power of TG. Feb 21 '13 at 14:16