- #1

Gear300

- 1,213

- 9

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:

1. For a set X, take the power set of x, P(X)

2. Consider the subset

3. Group

(∃x*: (A ε x*) ∧ (B ε x*) ) ⇔ ( A∪B ε

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*) = ∪

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

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

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.

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 X3. 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*}AM(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 in*C*. Thus, M(y*)∪M_{y}' is also a chain in*C*, 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 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.

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