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.

1. $U_1,U_2,\dots$ are universes, and:
2. $U_1\in U_2 \in \dots$

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.

1. If $x\in A\in U$, then $x\in U$
2. If $I\in U$ and $(A_i)_{i\in I}$ is a family in $U$, then $\bigcup_{i\in I} A_i \in U$
3. If $A\in U$, then so is $A$'s powerset
4. If $A,B\in U$, then $\{A,B\}\in U$
5. $\mathbb{N}\in U$

(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.

• What do you call a universe? – nombre Oct 10 '15 at 9:37
• @nombre I amended my question. – Stefan Perko Oct 10 '15 at 9:47
• Axioms $1$, $4$ and $5$ imply $U$ is the class of sets: let $x$ be a set, $\{x\}$ is a set and $\{(n,x) \ | \ n \in \mathbb{N}\}$ is a surjection $\mathbb{N} \in U \rightarrow \{x\}$ so $\{x\}$ in $U$ so $x \in U$. What you probably want for $4$ is a restriction to definable $f$. Then one can prove that $U$ contains every ordinal so $U$ can't be a set. – nombre Oct 10 '15 at 13:10
• @nombre I'm a little bit confused about the axioms, as different sources specify different axioms (I got my original ones from "Handbook of categorical algebra") and I'm not sure, whether they are equivalent. I'm gonna change the 4th axiom to a weaker looking axiom to get the axioms, that Zhen Lin uses in this paper: arxiv.org/pdf/1304.5227v2.pdf – Stefan Perko Oct 10 '15 at 14:09
• If you have the axiom of replacement (and enough induction) then you can construct an increasing sequence of universes. – Zhen Lin Oct 10 '15 at 14:26

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.)
• Prove by induction on $n$ that for each $n$, there is a unique function $\varphi_n$ with domain $\{m\in\mathbb{N}:m<n\}$ which satisfies $\varphi_n(0)=\emptyset$ and $\varphi_n(m+1)=U(\varphi_n(m))$ for all $m<n-1$. Now define $\mathcal{U}$ to be the set of sets which are of the form $\varphi_n(m)$ for some $n\in\mathbb{N}$ and $m<n$ (this set exists by Replacement), and define $V$ to be the union of $\mathcal{U}$. The entire point of the Axiom of Replacement is that you can do this without already knowing the codomain of the function is a set. – Eric Wofsey Oct 10 '15 at 19:58