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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 subsetCof the power set consisting of all chains in X

3. GroupCinto classes x*, where for any chains A,B εC,

(∃x*: (A ε x*) ∧ (B ε x*) ) ⇔ ( A∪B εC) ) 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 M_{y}', such that M(y*) ⊂ M_{y}'. However, by definition of (3), M(y*) ε y*. Also, M_{y}' is a chain inC. Thus, M(y*)∪M_{y}' is also a chain inC, and M_{y}' is also in y*. But this contradicts the maximality of M_{y}', so it must be the case that M(y*) = M_{y}' 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 εC) )

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.

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# I Zorn's Lemma

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