Exercise 8.1.5

Prove that Zorn’s Lemma is equivalent to the statement: If A is a system of sets such that, for each B A which is linearly ordered by , B A , then A has an -maximal element.

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Proof. (→) Suppose Zorn’s Lemma and let A be a system of sets where B A for any B that is linearly ordered by . We know that is a partial order on A . So let B be any -chain of A . Then we know that B A , and we also claim that B is an upper bound of B . To see this, consider any X B and any x X , so that clearly x B . Hence X B since x was arbitrary. This shows that B is an upper bound of B since X was arbitrary. Since B was an arbitrary chain, this shows that ( A , ) is an ordered set where every chain has an upper bound. Thus by Zorn’s Lemma there is a -maximal element of A as desired.

(←) Now suppose that A has a -maximal element for any system of sets A such that B A for any B A where B is linearly ordered by . Consider any ordered set ( A , ) and let C be the set of all chains of A . Let B be any subset of C that is linearly ordered by , and consider any x and y in B . Then there are X and Y in B such that x X and y Y . Since B is linearly ordered by we have that either X Y or Y X . In the former case we have x X Y so that both x and y are in Y . Hence x and y are comparable in since Y B C so that Y is a -chain. A similar argument shows that x and y are comparable if Y X . Since x and y were arbitrary this shows that B is a -chain so that B C .

Thus C is a system of sets that meet the criteria of the initial supposition since B was arbitrary. Hence C has a -maximal element. Since ( A , ) was an arbitrary ordered set and we have shown that the set of all chains of ( A , ) has a -maximal element, Zorn’s Lemma follows from Exercise 8.1.4. □

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2024-07-15 11:42
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