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

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 $\bigcup_{i \in I} x_i \in \mathscr{U}$.

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.