A set $\mathscr{U}$ is a universe if the following conditions are met:



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> * For any $x \in \mathscr{U}$ we have $x \subseteq \mathscr{U}$
> * For any $x,y \in \mathscr{U}$ we have $\{x,y\} \in \mathscr{U}$,
> * For any $x \in \mathscr{U}$ we have $\mathcal{P}(x) \in \mathscr{U}$,
> * For any family $(x\_i)\_{i \in I}$ of elements $x\_i \in \mathscr{U}$ indexed by an element $I \in \mathscr{U}$ we have $\bigcup\_{i \in I} x\_i \in \mathscr{U}$.
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Grothendieck introduced an addition axiom $\mathscr{U}$A which says that every set $x$ is contained in some universe $\mathscr{U}$.


I've seen some authors use the concept of the *successor universe* $\mathscr{U}^+$ of a given universe $\mathscr{U}$. It is the smallest universe which contains $\mathscr{U}$. However, I'm not sure how to prove that such a thing exists in $\mathsf{ZFC}$ (provided that $\mathscr{U}$ exists in the first place). If we knew that for any two universes $\mathscr{U}$ and $\mathscr{V}$ we have either $\mathscr{U} \in \mathscr{V}$ or $\mathscr{V} \in \mathscr{U}$, it would be easy. But I'm not sure if we can prove that latter without showing first that universes are equivalent to $V\_\kappa$ for inaccessible cardinals $\kappa$.


**Edit.** The question, as evident from the accepted answer is turned out to be quite trivial. I apologize for that.



How do we formally construct the successor universe $\mathscr{U}^+$ of a universe $\mathscr{U}$ in $\mathsf{ZFC}$?
The intersection of a nonempty set of universes is a universe. Now, let $U$ be a universe, and suppose that there exists a universe $V$ with $U\in V$. Then, the set of all universes $W$ with $U\in W\subseteq V$ exists and is nonempty. Its intersection is a universe, and this is readily checked to be the smallest universe containing $U$.


