- #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:
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.
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.
Last edited: