Does the existence of Grothendieck Universe contradict with axiom of choice?

Question

I am a beginner for set theory and I find something interesting:

Let $ U $ be a Grothendieck universe, then $ (U, \subseteq) $ is a partially ordered set. For every nonempty chain $ \mathcal{K} $ on it, $ \bigcup\mathcal{K} $ is an upper bound of $ \mathcal{ K } $. Thus by Hausdorff's maximal principle, which is equivalent to axiom of choice, the partially ordered set $ (U,\subseteq) $ has a maximal element $ M $. Put $ M^+=M\cup\{M\} $, since $ U $ is a Grothendieck universe, so $ M^+ $ is in $ U $. But by axiom of regularity, we have $ M \subsetneq M^+ $, and this contradicts with that $ M $ is a maximal element.

Does this means that the existence of Grothendieck universe contradicts with axiom of choice? Has this ever been considered before?

Answer 1

The problem with this proof is that you don't necessarily have that $\bigcup \mathcal{K}$ is in $U$ again. Universes aren't closed under all unions, just unions indexed by elements in $U$.

For example, if we consider $V_\omega$ (the collection of hereditarily finite sets) as a countable Grothendieck universe and let $0 = \emptyset$ and $n+1 = n \cup \{n\}$, then the collection $\{0, 1, 2, 3 \dots\}$ is a chain in $V_\omega$, but the union is an infinite set, hence not in $V_\omega$.

Answer 2

It is not true that the poset $(U,\subseteq)$ has a maximal element. This is because a chain can stretch all the way through $U$; take, for example, $\mathcal{K}=U\cap\mathsf{Ord}$ as the set of ordinals in $U$.

All "short" chains through $(U,\subseteq)$ - that is, all chains which themselves are elements of the universe $U$ - have upper bounds, but that's not enough to get Zorn to apply (we can't even fix this by "working inside $U$" since $(U,\subseteq)$ is itself not a set in $U$).

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It may help to replace $U$ with the set $\mathsf{HF}$ of hereditarily finite sets. This set (also called $V_\omega$ or $L_\omega$) is easily checked to satisfy all the properties of universes that are relevant for your idea, as well as the axiom of choice regardless of whether choice holds in $V$ itself. This is a much more transparent setting for these sorts of "overreach" ideas.