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?
Best Answer
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 $U$).
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.