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December 22nd, 2011, 11:02 PM   #1
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A Full Proof of Zorn's Lemma

Statement A (Zorn's Lemma) -> Let S be a non-empty inductively ordered set. Then S has a maximal element

Let A be a set of non-empty 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.
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December 25th, 2011, 09:28 AM   #2
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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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