December 22nd, 2011, 11:02 PM  #1 
Newbie Joined: Jan 2011 Posts: 20 Thanks: 0  A Full Proof of Zorn's Lemma
Statement A (Zorn's Lemma) > Let S be a nonempty inductively ordered set. Then S has a maximal element  Let A be a set of nonempty totally ordered subsets of S. Then A is not empty since any subset of S with 1 element belongs to, if , define to be . Then A is partially ordered and is strictly inductively ordered. Let be a totally ordered subset of A. Let , then Z is totally ordered. Let , then and for some . Since T is totally ordered, say . Then and since is totally ordered this implies or , thus Z is totally ordered and is the supremum of T in A. To conclude the proof, I prove something else:  Statement B > "Let A be a non empty strictly inductively ordered set. Then A has a maximal element" Suppose A doesn't have a maximum element. For each such that Let such that , This contradicts a well known theorem, "Let A be a non empty partially ordered and strictly inductively ordered set. Let be an increasing function. Then there exists an element such that And the proof for Statement B is complete, A has a maximum element g, this means g is a maximum totally ordered subset of S. Now, let k be an upper bound for g in S. Then k is the maximum element of , for if and , then is totally ordered. Thus and , hence x = k (maximum element)  Let me know your thoughts on whether this proof is correct. 
December 25th, 2011, 09:28 AM  #2 
Senior Member Joined: Jun 2011 Posts: 298 Thanks: 0  Re: A Full Proof of Zorn's Lemma
You introduced too many sets into your work. consists of a chain such that or . Take you pick. You can create a set of ordinal numbers, say , and is isomorphic to the set of ordinals. has a maximal, and so does . 

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