Salutations, friends from afar. The question I have is mundane, but I felt I should be sure. It is basically to spot the insufficiency in this proof for Zorn's Lemma: If every chain in a partially ordered set M has an upper bound, then M contains a maximal element. Proof: 1. For a set X, take the power set of x, P(X) 2. Consider the subset C of the power set consisting of all chains in X 3. Group C into classes x*, where for any chains A,B ε C, (∃x*: (A ε x*) ∧ (B ε x*) ) ⇔ ( A∪B ε C ) ) is true. This grouping is reflexive and symmetric but not transitive. However, each class is a lattice, in which for any two sets A and B in a class x*, A∪B and A∩B are also in x*. 4. For any class x*, take the union of all elements (chains) in x*, M(x*) = ∪Aεx*A M(x*) is a maximal chain. To show this, suppose the contrary and assume a class y* where M(y*) is not maximal. Then there exists some chain My', such that M(y*) ⊂ My'. However, by definition of (3), M(y*) ε y*. Also, My' is a chain in C. Thus, M(y*)∪My' is also a chain in C, and My' is also in y*. But this contradicts the maximality of My', so it must be the case that M(y*) = My' and M(y*) must be maximal. By hypothesis, since M(y*) is a chain, it should have an upper bound m. Since it is a maximal chain, m ε M(y*) must be true. And so, we have our maximal element m. If the proof is at least intuitively right, then I suspect the insufficiency is at 3. To outline things: 1. Axiom of Power Set 2. Axiom of Schema 3. Perhaps a frivolous assumption of the Axiom of Schema 4. 4 is 4. So I think 3 must be a frivolous claim depending on whether or not (∃x*: (A ε x*) ∧ (B ε x*) ) ⇔ ( A∪B ε C ) ) can be resolved to a statement of the Axiom of Schema paired with the Axiom of Choice. The proof does not explicitly make use of the Axiom of Choice. I am familiar with proofs that make use of the Axiom of Choice, but I just wanted to be sure of this. Thanks.