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Exercise 8.1.5
Prove that Zorn’s Lemma is equivalent to the statement: If is a system of sets such that, for each which is linearly ordered by , , then has an -maximal element.
Answers
Proof. Suppose Zorn’s Lemma and let be a system of sets where for any that is linearly ordered by . We know that is a partial order on . So let be any -chain of . Then we know that , and we also claim that is an upper bound of . To see this, consider any and any , so that clearly . Hence since was arbitrary. This shows that is an upper bound of since was arbitrary. Since was an arbitrary chain, this shows that is an ordered set where every chain has an upper bound. Thus by Zorn’s Lemma there is a -maximal element of as desired.
Now suppose that has a -maximal element for any system of sets such that for any where is linearly ordered by . Consider any ordered set and let be the set of all chains of . Let be any subset of that is linearly ordered by , and consider any and in . Then there are and in such that and . Since is linearly ordered by we have that either or . In the former case we have so that both and are in . Hence and are comparable in since so that is a -chain. A similar argument shows that and are comparable if . Since and were arbitrary this shows that is a -chain so that .
Thus is a system of sets that meet the criteria of the initial supposition since was arbitrary. Hence has a -maximal element. Since was an arbitrary ordered set and we have shown that the set of all chains of has a -maximal element, Zorn’s Lemma follows from Exercise 8.1.4. □