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Message from discussion Need help with proof of Zorn Lemma
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Jim Heckman reply-to  
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 More options Nov 3 2009, 10:28 am
Newsgroups: sci.math
From: "Jim Heckman" <rot13(reply-to)@none.invalid>
Date: Tue, 3 Nov 2009 10:28:25 GMT
Local: Tues, Nov 3 2009 10:28 am
Subject: Re: Need help with proof of Zorn Lemma
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On  1-Nov-2009, agapito6...@aol.com
wrote in message
<269a34a3-3630-4094-aa9a-a1701107f...@t2g2000yqn.googlegroups.com>:

> It states:  If X is a partially ordered set such that every chain in X
> has an upper bound, then X contains a maximal element.  I'm trying to
> understand Halmos' proof.  To summarize, instead of dealing with X, he
> deals with Y, which is the set of all chains in X, and, after a long
> series of contortions,  proves the existence of a maximal element  in
> Y.   Can someone please help explain:

> 1.- How is this procedure equivalent to proving the existence of a
> maximal element in X?

> 2.- How or where is the original hypothesis invoked?

Forget Halmos' explanation of this.  In my opinion, he goes
completely off the rails here, hopelessly complicating what's
really quite simple:

A maximal element in Y is a maximal chain in X, that is, a chain
that's not a proper subset of any chain.  (Remember, Y is partially
ordered by inclusion.)  Let M be such a chain.  By the hypothesis of
Zorn's Lemma, M has an upper bound u in X, that is, m <= u for all
m in M.  Then u is a maximal element in X, for otherwise there would
be some v in X with u < v, so m <= u < v for all m in M, and M
would be a proper subset of the chain M U {v}, contradicting the
maximality of M.

> In the proof, the only basic principle invoked is the Axiom of
> Choice.  Any help appreciated.

Actually, I rather like the proof in general, despite all its
contortions.  It makes it very clear that you needn't invoke
well-ordering per se to prove Zorn's Lemma.  (Though of course
Halmos' function g:Y -> Y, combined with his requirement that the
union of any chain in a "tower" T be an element of T, is equivalent
to well-ordering the minimal tower T_0.)

--
Jim Heckman


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