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| ``Elegant proof''
by ratboy on 2005-02-23 23:45:24 |
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| Thomas Forster, in a note preceding his fiddly proof of Zorn from AC, remarks "deducing Zorn from AC was a well-known fiddly task and in some ways remains so." Believing him, I didn't trouble to look beyond the obvious errors in the previous versions of this proof, so I was pleasantly surprised when a short, easy proof finally emerged.
There are some stylistic changes that I would make, but these are merely a matter of taste. A handy set-theoretic notation is f"Y ($f``Y$) for the image {f(y)|y in Y} of Y under the function f. There are actually two choice functions in play. First, there is your f on p"X. Define U:P(X)->P(X) by U(x)={y|y is an upper bound for x}; if C is the set of chains in X, then there is a choice function h on U"C. Let x be any member of X and define g:Ord->X by transfinite recursion:
g(0) = x,
g(a+1) = f(p(g(a))),
and
g(a) = h(U(g"a)) if a is a limit ordinal.
Then g is strictly increasing, etc.
If this proof originated with you, congratulations.
Forster's remark is on page 61 of _Logic, Induction and Sets_, Cambridge University Press, 2003. |
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