Is there a sequence of universes in TG? Assume we are working in ZFC + Tarski's axiom (Every set is an element of some universe). I wonder, if there is a universe $U$ with a sequence $(U_n)_{n\in \mathbb{N}}$ in $U$, s.t.


*

*$U_1,U_2,\dots$ are universes, and:

*$U_1\in U_2 \in \dots$


If not or if thats unknown: Is "ZFC + Existence of the above sequence" (obviously) inconsistent or likely inconsistent for some reason?

A universe $U$ is a set, s.t.


*

*If $x\in A\in U$, then  $x\in U$

*If $I\in U$ and $(A_i)_{i\in I}$ is a family in $U$, then $\bigcup_{i\in I} A_i \in U$

*If $A\in U$, then so is $A$'s powerset

*If $A,B\in U$, then $\{A,B\}\in U$

*$\mathbb{N}\in U$


(I hope I got the axioms right, so that any universe provides a model for ZFC)

The motivation for having such a sequence is simply, that in category theory I would not need to define bigger and bigger universes, when the need arises. It is like defining the factorial once and for all, instead of defining $0!,1!,2!,\dots$ and so on and so forth.
 A: Yes, this is true.  Note that it is clear that any intersection of universes is a universe, so for any set $X$, there is a smallest universe $U(X)$ containing $X$ (namely, the intersection of all universes containing $X$).  Now define a sequence of sets $U_n$ by induction: $U_0=\emptyset$, $U_{n+1}=U(U_n)$.  Let $V=\bigcup_{n\in\mathbb{N}} U_n$ (this set exists by Replacement and Union), and let $U_\omega=U(V)$.  Then clearly $U_\omega$ satisfies your requirements.
In fact, by transfinite induction, this sequence can be continued through all the ordinals.  That is, there exist universes $U_\alpha$ for all ordinals $\alpha$ such that $U_\alpha\in U_\beta$ for $\alpha<\beta$.
(In fact, it is well-known that a set $U$ is a universe iff $U=V_\kappa$ for some inaccessible cardinal $\kappa$, so $U(X)$ is just $V_\kappa$ where $\kappa$ is the least inaccessible cardinal greater than the rank of $X$.  So Tarski's axiom is equivalent to the existence of arbitrarily large inaccessible cardinals.  In the construction above, $U_\alpha=V_{\kappa_\alpha}$, where $\kappa_\alpha$ is the $\alpha$th inaccessible cardinal.)
